question
stringlengths 200
50k
| answer
stringclasses 1
value | source
stringclasses 2
values |
|---|---|---|
# Exam Reliability – Cronbach’s Alpha
## What is Cronbach’s Alpha?
Cronbach’s Alpha provides a measure of the internal consistency of a survey, test or exam.
Within an exam context it is used to evaluate how reliably the exam is measuring candidates’ ability, assuming that the content of the exam paper is consistently testing knowledge about a single subject matter/theme.
## How is Cronbach’s Alpha expressed?
Cronbach’s Alpha is expressed as a number up to 1 – generally this is between 0 and 1, but in extreme cases can be negative.
The closer the number is to 1, the more reliably the exam is measuring the candidates’ ability in that subject field. As a general guide a Cronbach’s Alpha:
• > 0.8 indicates strong reliability
• 0.6 – 0.8 indicates good reliability
• < 0.6 indicates candidates are not consistently scoring according to their ability.
## How is it calculated? (The nerdy bit)
At its heart, Cronbach’s Alpha is (1 – Si/St) where Si is the sum of item mark variances and St is the total mark variance.
A (very simple) example to explain the concept…
In this example a test has 10 items (questions) which are unrelated, each worth up to 10 marks. The mean mark for each item is 5, with candidates scoring between 3 and 7 on each item and the mean score for the whole exam will be 50 (5 * 10).
If the items are all totally unrelated to each other then each candidate is likely to do as well as each other as there is no single theme (or factor) that allows one group of candidates thrive over the rest. In this case the exam scores will all be similar, perhaps only ranging from a minimum of 43 to a maximum of 57.
In this case each item mark variance will be 4 (squared difference from the mean – i.e. 2²), so the sum of the item mark variances (Si) would be 40. The total mark variance (St) will be 7² so 49 as scores vary between 43 and 57 with a mean of 50.
Therefore, Cronbach’s Alpha will be 1-(40/49) = 0.18 which is very low and reflects the fact that the items were independent. This means the exam should be considered unreliable because what common ‘theme’ are these items testing?
Now consider a similar exam except this time all the items are perfectly related, i.e. consistently testing knowledge around a central theme. Instead of candidates’ marks being all over the place, it might be reasonable to see good candidates score an average of 7 for each item and poor candidates score an average of 3.
While the sum of the individual mark variances (Si) would remain at 40, the total mark variance would be very different. On average good candidates scored 70 on the test and poor candidates scored 30, so total mark variance (St) would be 20² so 400.
Cronbach’s Alpha would therefore be 1 – (40/400) = 0.9 which is high and indicates that the exam was reliable in determining good and poor candidates with reference to the central theme.
As such if the relationship between the way items are answered is consistent according to the group (in this case higher Vs lower performing candidates), then the Cronbach’s Alpha will be higher.
In practice Cronbach’s Alpha can be tricky to calculate by hand (particularly on longer exam papers) and your Exam Software (such as Maxexam) should do that for you.
## How is Cronbach’s Alpha used?
The throwaway answer to this would be ‘often incorrectly’ to determine exam reliability as a whole! Cronbach’s Alpha is only one measure to determine whether an exam is doing what it should – which in the case of summative medical or dental exams is determining whether candidates meet the standards required to be safe to practice.
We believe that Cronbach’s Alpha is often misunderstood, and this can lead to exams potentially being dismissed as being unreliable, when actually the way the exam itself is put together means you should expect Cronbach’s Alpha to be lower.
## What influences the Alpha?
Cronbach’s Alpha is influenced by the items in the exam, the structure of the exam and the number of candidates.
Cronbach’s Alpha was originally developed for use in surveys, specifically for questionnaires where the respondents would typically answer agree/disagree type questions. The alpha was then used to qualify how good, or consistent, the questionnaire was. Much of the information that can be found about Cronbach’s Alpha even now is referring to surveys, so can be misleading if you are using it in exams.
When used for exams, Cronbach’s Alpha is a good measure of consistency when items are around a single subject matter/theme where better candidates would be expected to consistently get more and harder items right. If an exam covers a broad range of topics or unrelated topics it is less likely for individual candidates to be experts in all areas so you would expect a lower Cronbach’s Alpha.
Additionally, if an exam contains essential knowledge items where most candidates would be likely to get items right then your Cronbach’s Alpha would be expected to be lower. This is demonstrated in the illustration below.
In addition to the items themselves, the number of items and the number of candidates will influence Cronbach’s Alpha, with a greater number of items and candidates usually being associated with a higher Alpha, as individual rogue items or candidates will have less effect on it.
If candidate numbers are low (e.g. for a resit) then Cronbach’s Alpha should not be used as it will be unreliable.
### Illustrations of why Cronbach’s Alpha should not be used on its’ own to determine exam reliability.
An exam containing some essential knowledge items.
If an exam contains essential knowledge items you would expect both good and poor candidates to get those items right. This will mean that both the individual mark variance and total mark variance will be lower than in an exam focused on ranking items, bringing down Cronbach’s Alpha. This does not however mean that the exam is poor or ‘unreliable’ – it is simply lower because of the fact the exam is testing essential knowledge.
An exam testing a wide knowledge base.
If a medical exam was testing two quite different areas of medicine such as paediatrics and geriatrics the graph for the percentage of candidates who got each item right might look something like this:
However, if you were to then look at candidates who specialised in paediatric medicine vs those who specialised in geriatric medicine (assuming the exam was split 50/50) the graph might look like this:
In this example Cronbach’s Alpha would be low, as the likelihood of getting items right is driven more by a candidate’s speciality than their ability.
Of course, in real life exams you wouldn’t expect a single exam to examine two such contrasting aspects of medicine, however it does illustrate why you would expect a lower Cronbach’s Alpha for a broader exam such as a final medical or dental exams than you would for a narrower topic exam.
## Cronbach’s Alpha and Individual items
Cronbach’s Alpha can be used to determine whether an individual item is adding to or subtracting from the exam’s reliability – i.e. is it a good or bad item to confirm a candidate’s ability. This will be further covered in our next blog.
## Conclusion – Is Cronbach’s Alpha a good measure of exam reliability?
In general yes, but it can be dependent on the exam so care is needed when using it/applying it.
If the exam is based around a single theme where candidates would be expected to answer items consistently according to their ability, then it can give you a good steer (ideally alongside other measures), about whether the exam is reliably testing ability. This also relies on the exam having enough candidates for them to be likely to act reliably as a group, and enough items to properly test knowledge.
However, if the exam is covering broad subject areas where candidates could be expected to be better in some areas than others, then a lower Cronbach’s Alpha might be expected and this does not mean the exam is unreliable in and of itself. Additionally, if the exam is testing essential knowledge where all candidates would be expected to do well, then the difference between higher and lower performing candidates will be less and you would expect a lower Alpha.
## Practical Application – Cronbach’s Alpha and determining grading boundaries.
Cronbach’s Alpha is often used to calculate the standard error of measure. This is explained further within our Exam Reliability blog and is used to indicate the level to which an exam may underestimate or overestimate a candidate’s knowledge.
The standard error of measure is used together with the exam cut-off (hopefully determined by using standard setting methods such as Angoff), to determine the pass mark of a just passing candidate. Where organisations undertake grading and boundary reviews, these are usually for candidates within a standard error of measure of the cut-off mark. Again this is discussed more in our Exam Reliability blog.
We hope this blog has helped to clarify what Cronbach’s Alpha measures and how it can effectively be used. If you do have any questions please do not hesitate to contact us on +44 (0)117 428 0550 or you could fill in our contact form here.
|
HuggingFaceTB/finemath
|
|
# What is the value of (1 − cos2 θ) cosec2 θ?
Question:
What is the value of $\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta ?$
Solution:
We have,
$\left(1-\cos ^{2} \theta\right) \operatorname{cosec}^{2} \theta=\sin ^{2} \theta \times \operatorname{cosec}^{2} \theta$
$=\sin ^{2} \theta \times\left(\frac{1}{\sin \theta}\right)^{2}$
$=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$
$=1$
|
HuggingFaceTB/finemath
|
|
#### Description of this paper
##### can you show the complete detailed steps from p^2 + p-(Answered)
Description
Question
1. can?you?show?the?complete detailed?steps?from?
2. ?p^2 + ?p ?10?/3 = 0
?p = (-? +/- (?^2 - 4*1*10?/3))/(-2)?
to get?
P = (?^2 - 43?/3)/-2
And
P = (-?^2 - 37?/3)/-2
a) For equilibrium, it?s correct that dp/dt should be made equal to zero.
Now, we arrive at the following expression:
?p^2 + ?p ?10?/3 = 0
This can be solved for p but the answer would be in terms of ? as suggested by you.
So the answer would be given by using:
x=(-b +/- ((b^2-4ac)^0.5)/2a)
so p = (-? +/- (?^2 ? 4*1*10?/3))/(-2)
Now the solutions for p thus obtained are also quadratic in nature, thus the solutions are:
P = (?^2 ? 43?/3)/-2
And the other solution is :
P = (-?^2 ? 37?/3)/-2
b) The quadratic equation obtained for p above:
?p^2 + ?p ?10?/3 = 0
Now the determinant of this quadratic equation is:
D = ?^2 ? 40?/3
This equation will have 2 solutions if the value of D is more than 0, it will have 1 solution if the value of D is 0 and
it will have 0 that means no solution when the value of D is negative.
So, for 2 solutions the value of D = ?^2 ? 40?/3 >0
For 1 solution, value of D = ?^2 ? 40?/3 = 0
For 0 solution, value of ?^2 ? 40?/3 < 0.
Paper#9210098 | Written in 27-Jul-2016
Price : \$22
|
HuggingFaceTB/finemath
|
|
1. ## Exponential Function
I have two here.
a=b(e^(-c))
solve for c
For that one, I have looked around and tried to figure it out and all i can think is that I need to use Log to solve it.
a=b/(c^2)
explain what happens to the value of a if c is doubled.
2. $\displaystyle a = b\,e^{-c}$
$\displaystyle \frac{a}{b} = e^{-c}$
$\displaystyle \ln{\left(\frac{a}{b}\right)} = -c$
$\displaystyle c = -\ln{\left(\frac{a}{b}\right)}$.
3. For the second one, can take the sign of proportionality.
$\displaystyle a = \frac{b}{c^2}$
$\displaystyle a\ \alpha\ \frac{1}{c^2}$
If c doubles, then;
So, if a becomes a' and c becomes c', it becomes:
$\displaystyle a'\ \alpha\ \frac{1}{c'^2}$
Now, we know that c' = 2c.
$\displaystyle a'\ \alpha\ \frac{1}{(2c)^2}$
$\displaystyle a'\ \alpha\ \frac{1}{4c^2}$
$\displaystyle 4a'\ \alpha\ \frac{1}{c^2}$
So, a = 4a' and hence, a/4 = a'. You conclude that the new a' is quadrupled when compared to the previous a
(here, c' means the new value of c, not the derivative of c.)
[Note, until I find the sign for the proportionality, I'll use alpha instead]
4. thank you very much
|
HuggingFaceTB/finemath
|
|
Factors of 1135 We have all the information you will ever need about the Factors of 1135. We will provide you with the definition of Factors of 1135, show you how to find the Factors of 1135, give you all the Factors of 1135, tell you how many Factors 1135 has, and supply you with all the Factor Pairs of 1135 to prove that our answer is solved correctly.
Factors of 1135 definition
The Factors of 1135 are all the integers (positive and negative whole numbers) that you can evenly divide into 1135. 1135 divided by a Factor of 1135 will equal another Factor of 1135.
How to find the Factors of 1135
Since the Factors of 1135 are all the numbers that you can evenly divide into 1135, we simply need to divide 1135 by all numbers up to 1135 to see which ones result in an even quotient. When we did that, we found that these calculations resulted in an even quotient:
1135 ÷ 1 = 1135
1135 ÷ 5 = 227
1135 ÷ 227 = 5
1135 ÷ 1135 = 1
The Postive Factors of 1135 are therefore all the numbers we used to divide (divisors) above to get an even number. Here is the list of all Postive Factors of 1135 in numerical order:
1, 5, 227, and 1135.
Factors of 1135 include negative numbers. Therefore, all the Positive Factors of 1135 can be converted to negative numbers. The list of Negative Factors of 1135 are:
-1, -5, -227, and -1135.
How many Factors of 1135?
When we counted the Factors of 1135 that we listed above, we found that 1135 has 4 Positive Factors and 4 Negative Factors. Thus, the total number of Factors of 1135 is 8.
Factor Pairs of 1135
Factor Pairs of 1135 are combinations of two factors that when multiplied together equal 1135. Here are all the Positive Factor Pairs of 1135
1 × 1135 = 1135
5 × 227 = 1135
227 × 5 = 1135
1135 × 1 = 1135
Like we said above, Factors of 1135 include negative numbers. Minus times minus equals plus, thus you can convert the Positive Factor Pair list above by simply putting a minus in front of every factor to get all the Negative Factor Pairs of 1135:
-1 × -1135 = 1135
-5 × -227 = 1135
-227 × -5 = 1135
-1135 × -1 = 1135
Factor Calculator
Do you need the factors for a particular number? You can submit a number below to find the factors for that number with detailed explanations like we did with Factors of 1135 above.
Factors of 1136
We hope this step-by-step tutorial to teach you about Factors of 1135 was helpful. Do you want to see if you learned something? If so, give the next number on our list a try and then check your answer here.
|
HuggingFaceTB/finemath
|
|
GSE Algebra I GSE Geometry GSE Algebra II GSE PreCalc Other Courses Calculus I Adv. Mathematical Decision Making Analytical Geometry ```CP-Statistics ``` `All Units` Coordinate Algebra Integrated Algebra I Integrated Geometry GPS Middle School Math GIVE BACK TO MATT Home >Brain Teasers > Number Phrases Page 3
Review
Search Site:
Page 1 Page 2 Page 3 Page 4 Page 5 Page 6
Number Phrases (Initials)
Each capital letter represents a word that starts with that letter. Use the provided number as a clue to determine what word each letter represents.
(click on the phrase to see the answer)
1. 90 F between B in BB_________________________________________________ ------90 FEET between BASES in BASEBALL------ 2. 144 in a G ______________________________________________________________ ------144 in a GROSS------ 3. 100 C in a D___________________________________________________________ ------100 CENTS in a DOLLAR------ 4. 3 S on a Y S ___________________________________________________________ ------3 SIDES on a YIELD SIGN------ 5. 3 L on a T ______________________________________________________________ ------3 LEGS on a TRIPOD------ 6. 1 W on a U _____________________________________________________________ ------1 WHEEL on a UNICYCLE------ 7. 57 H V __________________________________________________________________ ------57 HEINZ VARIETIES------ 8. 31 F at B R _____________________________________________________________ ------31 FLAVORSat BASKIN ROBINS------ 9. 6 P on an I H T ________________________________________________________ ------6 PLAYERS on an ICE HOCKEY TEAM------ 10. 8 P in a G _____________________________________________________________ ------8 PINTS in a GALLON------ 11. 50 S of the U S of A __________________________________________________ ------50 States of the UNITED STATES of AMERICA------ 12. 5280 F in a M_________________________________________________________ ------5280 FEET in a MILE------ 13. 9 L of a C _____________________________________________________________ ------9 LIVES of a CAT------ 14. 7 D in S W____________________________________________________________ ------7 DWARVES IN SNOW WHITE------ 15. 2 H A in a M of W ___________________________________________________ ------2 HYDROGEN ATOMS in a MOLECULE of WATER------ 16. 3 L P and 1 B B W___________________________________________________ ------3 LITTLE PIGS and 1 BIG BAD WOLF------ 17. 1000 Y is a M ________________________________________________________ ------1000 YEARS is a MILLENNIUM------ 18. 84 K on a KB (101 on an EKB) ____________________________________ ------84 KEYS on a KEYBOARD (101 on an ENHANCED KEY BOARD)------ 19. 120 Y is the L of a F F with the EZ__________________________________ ---120 YARDS is the LENGTH of a FOOTBALL FIELD with the END ZONES---- 20. 21 D on a D____________________________________________________________ ------21 DOTS on a DIE------
GO TO NUMBER PHRASES PAGE 5
Contact Information: Matt Winking
|
HuggingFaceTB/finemath
|
|
Integral of square of Sine function
You can view as many worked out examples as you want. First you are shown the question. Click Show Answer to view the answer. Click Show another Example to view another example. The best way to master mathematics is by practice. But practice requires time. If you don't have the time to practice, you can always view many practice problems completely worked out and become good at it too.
Level 1 Level 2 Level 3 Level 4 Level 5
Question
$\int {\sin}^2\left(7x+1\right)\; dx$
$\int {\sin}^2\left(7x+1\right)\; dx$
Substitute $u=7x+1$. So, $du =7\; dx$
Or, $\frac{1}{7}\; du = dx$
And, $\int {\sin}^2\left(7x+1\right)\; dx = \frac{1}{7}\int {\sin}^2 u \;du$
$=\frac{1}{7}\left(\frac{1}{2}u-\frac{1}{4}\cos 2u \right) + C$
$=\frac{1}{14}u-\frac{1}{28}\cos 2u + C$
Substituting back $u=7x+1$
$=\frac{1}{14}\left(7x+1\right)-\frac{1}{28}\left( \cos 2\left(7x+1\right) \right) + C$
$=\frac{1}{2}x-\frac{1}{28}\cos\left(14x+2\right)+ C$
|
HuggingFaceTB/finemath
|
|
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Sums and Differences of Single Variable Expressions
## Simplify expressions by combining like terms.
0%
Progress
Practice Sums and Differences of Single Variable Expressions
Progress
0%
Sums and Differences of Single Variable Expressions
Have you ever had your picture taken at an amusement park? They have those old time costumes that you can put on and everyone stands together for a photo?
Well, at the amusement park, Kelly has decided to gather a group of friends to do just that. When she first posed the question at lunch, she had five people say that they wanted to do it.
Then later in the day, four more people wanted to join it.
There is a fee per person if you want to be in the picture.
Kelly wants to figure out the total number in simple way and include the fee. How can she do this?
To do this, Kelly can write an expression using a single variable.
Do you know how to do this?
This Concept will show you how to write single variable expressions that include sums and differences. Then you will understand how to help Kelly.
### Guidance
In the last few Concepts, you learned how to write single-variable expressions and single variable equations. Now you are going to learn to work with single-variable expressions. The first thing that you are going to learn is how to simplify an expression.
What does it mean to simplify?
To simplify means to make smaller or to make simpler. When we simplify in mathematics, we aren’t solving anything, we are just making it smaller.
How do we simplify expressions?
Sometimes, you will be given an expression using variables where there is more than one term. A term is a number with a variable. Here is an example of a term.
4x\begin{align*}4x\end{align*}
This is a term. It is a number and a variable. We haven’t been given a value for x\begin{align*}x\end{align*}, so there isn’t anything else we can do with this term. It stays the same. If we have been given a value for x\begin{align*}x\end{align*}, then we could evaluate the expression. You have already worked on evaluating expressions.
When there is more than one LIKE TERM in an expression, we can simplify the expression.
What is a like term?
A like term means that the terms in question use the same variable.
4x\begin{align*}4x\end{align*} and 5x\begin{align*}5x\end{align*} are like terms. They both have x\begin{align*}x\end{align*} as the variable. They are alike.
6x\begin{align*}6x\end{align*} and 2y\begin{align*}2y\end{align*} are not like terms. One has an x\begin{align*}x\end{align*} and one has a y\begin{align*}y\end{align*}. They are not alike.
We can simplify expressions with like terms. We can simplify the sums and differences of expressions with like terms. Let’s start with sums.
5x+7x
First, we look to see if these terms are alike. Both of them have an x\begin{align*}x\end{align*}, so they are alike.
Next, we can simplify them by adding the numerical part of the terms together. The x\begin{align*}x\end{align*} stays the same.
5x+7x12x
You can think of the x\begin{align*}x\end{align*} as a label that lets you know that the terms are alike.
7x+2x+5y
First, we look to see if the terms are alike. Two of the terms have x\begin{align*}x\end{align*}’s and one has a y\begin{align*}y\end{align*}. The two with the x\begin{align*}x\end{align*}’s are alike. The one with the y\begin{align*}y\end{align*} is not alike. We can simplify the ones with the x\begin{align*}x\end{align*}’s.
Next, we simplify the like terms.
7x+2x=9x
We can’t simplify the 5y\begin{align*}5y\end{align*} so it stays the same.
9x+5y
This is our answer.
We can also simplify expressions with differences and like terms.
9y2y
First, you can see that these terms are alike because they both have y\begin{align*}y\end{align*}’s. We simplify the expression by subtracting the numerical part of the terms.
9 - 2 = 7
Our answer is 7y\begin{align*}7y\end{align*}.
Sometimes you can combine like terms that have both sums and differences in the same problem.
8x3x+2y+4y
We begin with the like terms.
8x3x2y+4y=5x=6y
Next, we put it all together.
5x+6y
This is our answer.
Remember that you can only combine terms that are alike!!!
Use your notebook and pencil to take some notes on how to identify like terms.
Try a few of these on your own. Simplify the expressions by combining like terms.
#### Example A
7z+2z+4z\begin{align*}7z+2z+4z\end{align*}
Solution: 13z\begin{align*}13z\end{align*}
#### Example B
25y13y\begin{align*}25y-13y\end{align*}
Solution: 12y\begin{align*}12y\end{align*}
#### Example C
7x+2x+4a\begin{align*}7x+2x+4a\end{align*}
Solution: 9x+4a\begin{align*}9x + 4a\end{align*}
Here is the original problem once again.
Well, at the amusement park, Kelly has decided to gather a group of friends to do just that. When she first posed the question at lunch, she had five people say that they wanted to do it.
Then later in the day, four more people wanted to join it.
There is a fee per person if you want to be in the picture.
Kelly wants to figure out the total number in simple way and include the fee. How can she do this?
To do this, Kelly can write an expression using a single variable.
Do you know how to to do this?
Now we can use the information that is provided in the dilemma to write a single variable expression.
First, 5 people wanted to be in the picture.
There is also a fee per person. Kelly doesn't know this amount. It is our variable x\begin{align*}x\end{align*}
5x\begin{align*}5x\end{align*}
Then 4 more people wanted to join the picture. The fee per person applies to them too.
5x+4x\begin{align*}5x + 4x\end{align*}
If we combine like terms, 9x\begin{align*}9x\end{align*} is the expression Kelly can use to figure out the total cost of the picture. Once she knows the cost per person, she will be able to substitute that into the given expression and solve for the total cost.
But wait, this is information we can use in another Concept.
### Guided Practice
Here is one for you to try on your own.
5x+2x1x+6y4y\begin{align*}5x + 2x - 1x + 6y - 4y\end{align*}
To simplify this expression, we simply combine the terms that are alike.
5x+2x1x=6x\begin{align*}5x + 2x - 1x = 6x\end{align*}
6y4y=2y\begin{align*}6y - 4y = 2y\end{align*}
Now we put those simplified terms together.
6x+2y\begin{align*}6x + 2y\end{align*}
This is our answer.
### Video Review
Here is a video for review.
### Explore More
Directions: Simplify the following expressions by combining like terms. If the expression is already in simplest form please write “already in simplest form.”
1. 4x+6x\begin{align*}4x+6x\end{align*}
2. 8y+5y\begin{align*}8y+5y\end{align*}
3. 9z+2z\begin{align*}9z+2z\end{align*}
4. 8x+2y\begin{align*}8x+2y\end{align*}
5. 7y+3y+2x\begin{align*}7y+3y+2x\end{align*}
6. 9xx\begin{align*}9x-x\end{align*}
7. \begin{align*}12y-3y\end{align*}
8. \begin{align*}22x-2y\end{align*}
9. \begin{align*}78x-10x\end{align*}
10. \begin{align*}22y-4y\end{align*}
11. \begin{align*}16x - 5x + 1x - 12y + 2y\end{align*}
12. \begin{align*}26x - 15x + 12x - 14y + 2y\end{align*}
13. \begin{align*}36x - 5x + 11x - 1x + 2y\end{align*}
14. \begin{align*}26x - 25x + 12x - 13y + 2y\end{align*}
15. \begin{align*}29x - 25x + 18x - 12x + 12y + 3y\end{align*}
### Vocabulary Language: English
Expression
Expression
An expression is a mathematical phrase containing variables, operations and/or numbers. Expressions do not include comparative operators such as equal signs or inequality symbols.
Simplify
Simplify
To simplify means to rewrite an expression to make it as "simple" as possible. You can simplify by removing parentheses, combining like terms, or reducing fractions.
|
HuggingFaceTB/finemath
|
|
Lesson Objectives
• Demonstrate an understanding of how to find the GCF for a group of numbers
• Demonstrate an understanding of the distributive property of multiplication
• Learn how to find the GCF for a group of monomials
• Learn how to factor out the GCF from a polynomial
## How to Factor out the GCF from a Polynomial
In our pre-algebra course, we learned how to find the Greatest Common Factor or GCF for a group of numbers. To find the GCF for a group of numbers, we factor each number completely and then build a list of prime factors that are common to all numbers. The GCF is the product of the numbers in the list.
GCF(12,18,90)
12 = 2 • 2 • 3
18 = 2 • 3 • 3
90 = 2 • 3 • 3 • 5
What's common to everything? Let's organize our factors in a table.
Number Prime Factors
12 2 2 3
18 2 3 3
90 2 3 3 5
One factor of 2 and one factor of 3 is common to every number.
GCF List: 2, 3
To find the GCF, we multiply the numbers in the GCF list (2,3):
GCF(12,18,90) = 2 • 3 = 6
When we find the GCF for a group of variables, the procedure is the same, however, we can use a shortcut. First and foremost, the variable must appear in each term, secondly, we use the smallest exponent that appears on that variable in any term.
GCF(x5, x3, x2y)
We can quickly identify that x is in each term, x5, x3, and x2y, however, y is only in one of the terms. The variable x will be in the GCF and the variable y will not be in the GCF. The smallest exponent on x in any of the terms is a 2. This means our GCF will be x2.
GCF(x5, x3, x2y) = x2
Let's take a look at a few examples.
Example 1: Find the GCF
GCF(3x2y, 9xyz, 24x5y2)
For the number part, our GCF is 3. For the variable part, our GCF is xy. Our GCF is the product of the number part (3) and the variable part (xy).
GCF(3x2y, 9xyz, 24x5y2) = 3xy
Example 2: Find the GCF
GCF(121x9yz, 132y3, 209y)
For the number part, our GCF is 11. For the variable part, our GCF is y. Our GCF is the product of the number part (11) and the variable part (y).
GCF(121x9yz, 132y3, 209y) = 11y
### Factoring out the GCF
When we factor a polynomial, we are writing the polynomial as the product of two or more polynomials. When we factor, we are just reversing the distributive property of multiplication that we learned in pre-algebra.
4(9 + 12) = 4 • 9 + 4 • 12 = 36 + 48
Since we have an equality, we can reverse the process
36 + 48 = 4 • 9 + 4 • 12 = 4(9 + 12)
When we factor out the GCF from a polynomial, we first identify the GCF of all terms of the polynomial. We can then pull this GCF out from each term and place it outside of a set of parentheses. Let's look at a few examples.
Example 3: Factor out the GCF
65x5 - 39x2 + 13x
First, we want to find the GCF of all terms.
GCF(65x5, 39x2, 13x) = 13x
Second, we will rewrite each term as the product of 13x and another factor.
13x • 5x4 - 13x • 3x + 13x • 1
Third, we will use our distributive property to factor out the GCF.
13x(5x4 - 3x + 1)
Example 4: Factor out the GCF
95x2y2 - 228xy + 133y2
First, we want to find the GCF of all terms.
GCF(95x2y2, 228xy, 133y2) = 19y
Second, we will rewrite each term as the product of 19y and another factor.
19y • 5x2y - 19y • 12x + 19y • 7y
Third, we will use our distributive property to factor out the GCF.
19y(5x2y - 12x + 7y)
### Factoring out a Common Binomial Factor
In some cases, we will need to factor out a common binomial factor. Let's look at an example.
Example 5: Factor out the GCF
(x - 7)(x - 12) + (x + 5)(x - 7)
In this case, we can see a common binomial factor of (x - 7). This can be factored in the same way:
(x - 7)(x - 12) + (x + 5)(x - 7)
(x - 7)[(x - 12) + (x + 5)]
We can combine like terms inside of the brackets:
(x - 7)[x - 12 + x + 5]
(x - 7)(2x - 7)
|
HuggingFaceTB/finemath
|
|
# Search by Topic
#### Resources tagged with Time similar to Can You Do it Too?:
Filter by: Content type:
Stage:
Challenge level:
### There are 65 results
Broad Topics > Measures and Mensuration > Time
### Snap
##### Stage: 1 Challenge Level:
Try this version of Snap with a friend - do you know the order of the days of the week?
### Times of Day
##### Stage: 1 Challenge Level:
These pictures show some different activities that you may get up to during a day. What order would you do them in?
### Now and Then
##### Stage: 2 Challenge Level:
Look at the changes in results on some of the athletics track events at the Olympic Games in 1908 and 1948. Compare the results for 2012.
### What Is the Time?
##### Stage: 1 and 2 Challenge Level:
Can you put these times on the clocks in order? You might like to arrange them in a circle.
### Lengthy Journeys
##### Stage: 2 Challenge Level:
Investigate the different distances of these car journeys and find out how long they take.
### Home Time
##### Stage: 2 Challenge Level:
Alice's mum needs to go to each child's house just once and then back home again. How many different routes are there? Use the information to find out how long each road is on the route she took.
### How Many Days?
##### Stage: 1 Challenge Level:
How many days are there between February 25th 2000 and March 11th?
### Planet Plex Time
##### Stage: 1 Challenge Level:
On Planet Plex, there are only 6 hours in the day. Can you answer these questions about how Arog the Alien spends his day?
### Twizzle's Journey
##### Stage: 1 Challenge Level:
Twizzle, a female giraffe, needs transporting to another zoo. Which route will give the fastest journey?
### Time Line
##### Stage: 1 Challenge Level:
Describe what Emma might be doing from these pictures of clocks which show important times in her day.
### Matching Time
##### Stage: 1 Challenge Level:
Try this matching game which will help you recognise different ways of saying the same time interval.
### Practice Run
##### Stage: 2 Challenge Level:
Chandrika was practising a long distance run. Can you work out how long the race was from the information?
### 5 on the Clock
##### Stage: 2 Challenge Level:
On a digital clock showing 24 hour time, over a whole day, how many times does a 5 appear? Is it the same number for a 12 hour clock over a whole day?
### You Tell the Story
##### Stage: 2 Challenge Level:
Can you create a story that would describe the movement of the man shown on these graphs? Use the interactivity to try out our ideas.
### Do You Measure Up?
##### Stage: 2 Challenge Level:
A game for two or more players that uses a knowledge of measuring tools. Spin the spinner and identify which jobs can be done with the measuring tool shown.
### 2010: A Year of Investigations
##### Stage: 1, 2 and 3
This article for teachers suggests ideas for activities built around 10 and 2010.
### Take Your Dog for a Walk
##### Stage: 2 Challenge Level:
Use the interactivity to move Mr Pearson and his dog. Can you move him so that the graph shows a curve?
### Back to School
##### Stage: 2 Challenge Level:
Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong?
### Wonky Watches
##### Stage: 2 Challenge Level:
Stuart's watch loses two minutes every hour. Adam's watch gains one minute every hour. Use the information to work out what time (the real time) they arrived at the airport.
### A Child Is Full of ...
##### Stage: 2 Short Challenge Level:
My cousin was 24 years old on Friday April 5th in 1974. On what day of the week was she born?
### Times
##### Stage: 2 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
### Working with Dinosaurs
##### Stage: 2
This article for teachers suggests ways in which dinosaurs can be a great context for discussing measurement.
### Stop the Clock
##### Stage: 1 Challenge Level:
This is a game for two players. Can you find out how to be the first to get to 12 o'clock?
### How Long Does it Take?
##### Stage: 2 Challenge Level:
In this matching game, you have to decide how long different events take.
### Calendar Sorting
##### Stage: 2 Challenge Level:
The pages of my calendar have got mixed up. Can you sort them out?
### Palindromic Date
##### Stage: 2 Challenge Level:
What is the date in February 2002 where the 8 digits are palindromic if the date is written in the British way?
### Order, Order!
##### Stage: 1 and 2 Challenge Level:
Can you place these quantities in order from smallest to largest?
### How Many Times?
##### Stage: 2 Challenge Level:
On a digital 24 hour clock, at certain times, all the digits are consecutive. How many times like this are there between midnight and 7 a.m.?
### Order the Changes
##### Stage: 2 Challenge Level:
Can you order pictures of the development of a frog from frogspawn and of a bean seed growing into a plant?
### Watch the Clock
##### Stage: 2 Challenge Level:
During the third hour after midnight the hands on a clock point in the same direction (so one hand is over the top of the other). At what time, to the nearest second, does this happen?
### Clocking Off
##### Stage: 2, 3 and 4 Challenge Level:
I found these clocks in the Arts Centre at the University of Warwick intriguing - do they really need four clocks and what times would be ambiguous with only two or three of them?
### The Time Is ...
##### Stage: 2 Challenge Level:
Can you put these mixed-up times in order? You could arrange them in a circle.
### Stop the Clock for Two
##### Stage: 1 Challenge Level:
Stop the Clock game for an adult and child. How can you make sure you always win this game?
### Making Maths: Make a Pendulum
##### Stage: 2 and 3 Challenge Level:
Galileo, a famous inventor who lived about 400 years ago, came up with an idea similar to this for making a time measuring instrument. Can you turn your pendulum into an accurate minute timer?
### Time and Tide
##### Stage: 1, 2 and 3 Challenge Level:
What can you say about when these pictures were taken?
### Being Resourceful - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems at primary level that require careful consideration.
### Sweeping Hands
##### Stage: 2 Challenge Level:
Use your knowledge of angles to work out how many degrees the hour and minute hands of a clock travel through in different amounts of time.
### In Order
##### Stage: 2 Challenge Level:
Can you rank these quantities in order? You may need to find out extra information or perform some experiments to justify your rankings.
### Being Collaborative - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems for primary learners to work on with others.
### Being Curious - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems for inquiring primary learners.
### Being Resilient - Primary Measures
##### Stage: 1 and 2 Challenge Level:
Measure problems at primary level that may require resilience.
### On Time?
##### Stage: 1, 2 and 3
This article explains how Greenwich Mean Time was established and in fact, why Greenwich in London was chosen as the standard.
### Timing
##### Stage: 2 Short Challenge Level:
These two challenges will test your time-keeping!
### Millennium Man
##### Stage: 2 Challenge Level:
Liitle Millennium Man was born on Saturday 1st January 2000 and he will retire on the first Saturday 1st January that occurs after his 60th birthday. How old will he be when he retires?
### A History of Astronomy
##### Stage: 2
Astronomy grew out of problems that the early civilisations had. They needed to solve problems relating to time and distance - both mathematical topics.
### Train Timetable
##### Stage: 2 Challenge Level:
Use the information to work out the timetable for the three trains travelling between City station and Farmland station.
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Eclipses of the Sun
##### Stage: 2 and 3
Mathematics has allowed us now to measure lots of things about eclipses and so calculate exactly when they will happen, where they can be seen from, and what they will look like.
### On What Day Did it Happen?
##### Stage: 1, 2 and 3
Read this article to find out the mathematical method for working out what day of the week each particular date fell on back as far as 1700.
|
HuggingFaceTB/finemath
|
|
Solutions by everydaycalculation.com
## Multiply 5/9 with 5/50
This multiplication involving fractions can also be rephrased as "What is 5/9 of 5/50?"
5/9 × 5/50 is 1/18.
#### Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 5/9 × 5/50 = 5 × 5/9 × 50 = 25/450
3. After reducing the fraction, the answer is 1/18
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
|
HuggingFaceTB/finemath
|
|
Also suppose that in 1984 each bucket of chicken was priced at \$10. Finally, assume that in 2005 the price per bucket of chicken was \$16 and that 22,000 buckets were produced. Determine the GDP price index for 1984, using 2000 as the base year. Finally, assume that in 2005 the price per bucket of chicken was \$16 and that 22,000 buckets were produced. Suppose that in 1984 the total output in a single-good economy was 10,000 buckets of chicken and the price of each bucket of chicken was \$10. Determine the GDP price index Finally, assume that in 2005 the price per bucket of chicken was \$20 and that 22,000 buckets were produced. Suppose that in 1984 the total output in a single-good economy was 10,000 buckets of chicken and the price of each bucket of chicken was \$16. Finally, assume that in 2005 the price per bucket of chicken was \$20 and that 22,000 buckets were produced. Problem 7-6 (Algo ) Suppose that in 1984 the total output in a single-good economy was 7000 buckets of chicken. Finally, assume that in 2000 the price per bucket of chicken was \$16 and that 22,000 buckets were purchased. Also suppose that in 1984 each bucket of chicken was priced at \$10. Finally, assume that in 2005 the price per bucket of chicken was \$20 and that 22,000 buckets were produced. Answer to: Suppose that in 1984 the total output in a single good economy was 7,000 buckets of chicken. Also suppose that in 1984 each bucket of chicken was priced at \$10. Also assume that in 1984 each bucket of chicken was priced at \$10. Question #2 Suppose that in 1984 the total output in a single-good economy was 7,000 buckets of chicken. Also assume that in 1984 each bucket of chicken was priced at \$16. Suppose that in 1984 the total output in a single-good economy was 10,000 buckets of chicken. Problem 27-6 (Algo) Suppose that in 1984 the total output in a single-good economy was 10,000 buckets of chicken. Finally, assume that in 2005 the price per bucket of chicken was \$20 and that 22,000 buckets were produced. Suppose that in 1984 the total output in a single-good economy was. Determine the GDP price index for 1984, using 2005 as the base year. Suppose that in 1984 the total output in a single-good economy was 10,000 buckets of chicken. Also suppose that in 1984 each bucket of chicken was priced at \$10. Also assume that in 1984 each bucket of chicken was priced at \$15. Finally, assume that in 2004 the price per bucket of chicken was \$16 and that 22,000 buckets were produced. points Also assume that in 1984 each bucket of chicken was priced at \$15. Suppose that in 1984 the total output in a single-good economy was 7000 buckets of chicken. Suppose that in 1984 the total output in a single-good economy was 7000 buckets of chicken. Also assume that in 1984 each bucket of chicken was priced at 15Finallyassume that in 2005 the price per bucket of chicken was 20 and that 23.000 buckets were produced Suppose that in 1984 the total output in a single-good economy was 7,000 buckets of chicken. Suppose that in 1984 the total output in a single-good economy was 7,000 buckets of chicken. Also suppose that in 1984 each bucket of chicken was priced at \$10. 10,000 buckets of chicken. In 2005 the price per bucket of chicken was \$20 and 25,000 buckets were produced. Determine real GDP for 1984 and 2004, in 1984 prices. Also assume that in 1984 each bucket of chicken was priced at \$15. Determine the GDP price index for 1984, using 2005 as the base year. Suppose that in 1984 the total output in a single-good economy was 7000 buckets of chicken.
|
HuggingFaceTB/finemath
|
|
Posted: September 18th, 2017
# Algebra
College Algebra MATH 107 Summer, 2015, V3.1
Page 1 of 11
MATH 107 FINAL EXAMINATION
This is an open-book exam. You may refer to your text and other course materials as you work
on the exam, and you may use a calculator. You must complete the exam individually.
Neither collaboration nor consultation with others is allowed.
There are 30 problems.
Problems #1–12 are Multiple Choice.
Problems #13–21 are Short Answer. (Work not required to be shown)
Problems #22–30 are Short Answer with work required to be shown.
MULTIPLE CHOICE
1. Determine the domain and range of the piecewise function. 1. ______
A. Domain [–2, 4]; Range [–3, 3]
B. Domain [–2, 1]; Range [–2, 4]
C. Domain [–3, 3]; Range [–2, 4]
D. Domain [–3, 0]; Range [0, 4]
2. Solve: 2x − 6 = 3− x 2. ______
A. No solution
B. 3, 5
C. 3
D. 5
-4 2 4
-2
-4
2
4
-2
College Algebra MATH 107 Summer, 2015, V3.1
Page 2 of 11
3. Determine the interval(s) on which the function is increasing. 3. ______
A. (− 4 / 3, 1)
B. (−1,1) and (3,¥)
C. (−¥,−5/ 3) , (0, 2) , and (11/ 3,¥)
D. (−¥,0) and (2,¥)
4. Determine whether the graph of y = x − 6 is symmetric with respect to the origin,
the x-axis, or the y-axis. 4. ______
A. not symmetric with respect to the x-axis, not symmetric with respect to the y-axis, and
not symmetric with respect to the origin
B. symmetric with respect to the origin only
C. symmetric with respect to the x-axis only
D. symmetric with respect to the y-axis only
5. Solve, and express the answer in interval notation: | 6 – 5x | £ 14. 5. ______
A. [–8/5, 4]
B. (–¥, −8/5] È [4, ¥)
C. (–¥, –8/5]
D. [4, ¥)
College Algebra MATH 107 Summer, 2015, V3.1
Page 3 of 11
6. Which of the following represents the graph of 2x − 7y = 14 ? 6. ______
A. B.
C. D.
College Algebra MATH 107 Summer, 2015, V3.1
Page 4 of 11
7. Write a slope-intercept equation for a line parallel to the line x + 8y = 6 which passes through
the point (24, –5). 7. ______
A. y = −8x −197
B.
1
8
8
y = x −
C.
1
5
8
y = − x −
D.
1
2
8
y = − x −
8. Does the graph below represent a function and is it one-to-one? 8. ______
A. It is a function and it is one-to-one.
B. It is a function but not one-to-one.
C. It is not a function but it is one-to-one.
D. It is not a function and it is not one-to-one.
College Algebra MATH 107 Summer, 2015, V3.1
Page 5 of 11
9. Express as a single logarithm: log x – 3 log y + log 1 9. ______
A.
3
1
log
x
y
+
B. 3 log
x
y
C.
3
1
log
x
y
+
D. log ( x +1− 3y)
10. Which of the functions corresponds to the graph? 10. ______
A. ( ) 1 x f x = −e −
B. ( ) 1 x f x = e −
C. ( ) 1 x f x e− = −
D. ( ) 2 x f x e− = −
College Algebra MATH 107 Summer, 2015, V3.1
Page 6 of 11
11. Suppose that a function f has exactly two x-intercepts.
Which of the following statements MUST be true? 11. ______
A. The graph of f has exactly two points whose x-coordinates are 0.
B. f is an invertible function.
C. f is a quadratic function.
D. The equation f (x) = 0 has exactly two real-number solutions.
12. The graph of y = f (x) is shown at the left and the graph of y = g(x) is shown at the right. (No
formulas are given.) What is the relationship between g(x) and f (x)?
12. ______
y = f (x) y = g(x)
A. g(x) = f (x – 1) + 3
B. g(x) = f (x + 1) – 3
C. g(x) = f (x – 3) + 1
D. g(x) = f (x + 3) – 1
2 -4 -2 4
-2
-4
2
4
-4 -2 2 4
-2
-4
2
4
College Algebra MATH 107 Summer, 2015, V3.1
Page 7 of 11
13. Multiply and simplify: (7 + i)(8 + 6i).
Write the answer in the form a + bi, where a and b are real numbers. Answer: ________
14. Solve, and write the answer in interval notation:
1
0
5
x
x
+
£
15. A bowl of soup at 170° F. is placed in a room of constant temperature of 60° F. The
temperature T of the soup t minutes after it is placed in the room is given by
T(t) = 60 + 110 e
– 0.075 t
Find the temperature of the soup 8 minutes after it is placed in the room. (Round to the nearest
degree.)
16. Find the value of the logarithm: 4
1
log
64
17. Solve: 67x−3 = 36 . Answer: ________
18. Suppose \$3,800 is invested in an account at an annual interest rate of 5.2% compounded
continuously. How long (to the nearest tenth of a year) will it take the investment to double in
19. Let f (x) = x2 − 6x + 14.
(a) Find the vertex. Answer: ________
(b) State the range of the function. Answer: ________
(c) On what interval is the function decreasing? Answer: ________
College Algebra MATH 107 Summer, 2015, V3.1
Page 8 of 11
20. Consider the polynomial P(x), shown in both standard form and factored form.
4 3 2 1 3 3 13 1
( ) 6 ( 2)( 1)( 3)( 4)
4 2 4 2 4
P x = x − x + x + x − = x + x − x − x −
(a) Which sketch illustrates the end behavior of the polynomial function?
(b) State the zeros of the function. Answer: ________________
(c) State the y-intercept. Answer: ________________
(d) State which graph below is the graph of P(x). Answer: ________
GRAPH A GRAPH B
GRAPH C GRAPH D
A.
B. C. D.
vvvv
vvvv
vvvv vvvv
College Algebra MATH 107 Summer, 2015, V3.1
Page 9 of 11
21. Let
2
( )
3
x
f x
x
+
=
.
(a) State the domain. Answer: _________________
(b) State the vertical asymptote(s). Answer: _________________
(c) State the horizontal asymptote. Answer: _________________
(d) Which of the following represents the graph of
2
( )
3
x
f x
x
+
=
GRAPH A. GRAPH B.
GRAPH C. GRAPH D.
College Algebra MATH 107 Summer, 2015, V3.1
Page 10 of 11
SHORT ANSWER, with work required to be shown, as indicated.
22. Let f (x) = x + 2 and g(x) = x + 5 .
(a) Find ) 1 (−
g
f
. Show work.
(b) Find the domain of the quotient function
g
f
. Explain.
23. Points (–5, 2) and (–1, 8) are endpoints of the diameter of a circle.
(a) What is the length of the diameter? Give the exact answer, simplified as much as possible.
Show work.
(b) What is the center point C of the circle?
(c) Given the point C you found in part (b), state the point symmetric to C about the x-axis.
24. Find the equation for a line which passes through the points (4, –3) and (6, –7). Write the
equation in slope-intercept form. Show work.
25. Malia, a resident of Metropolis, pays Metropolis an annual tax of \$46 plus 2.4% of her
annual income. If Malia paid \$1,558 in tax, what was Malia’s income? Show work.
26. Let f (x) = 2×2 + 7 and g(x) = x + 6.
(a) Find the composite function ( f o g)(x) and simplify. Show work.
(b) Find ( f o g ) (−5) . Show work.
27. Find the exact solutions and simplify as much as possible: 5×2 + 6 = 12x. Show work.
28. Given the function
1
( ) 5
7
f x = − x , find a formula for the inverse function. Show work.
College Algebra MATH 107 Summer, 2015, V3.1
Page 11 of 11
29. Chocolate Confections, Inc. has determined that when x chocolate cakes are made daily, the
average cost per chocolate cake (in dollars) is given by
C(x) = 0.001 x2 – 0.14x + 15.20
(a) What is the average cost per cake if 40 chocolate cakes are made daily?
(b) How many chocolate cakes should be made daily in order to minimize the average cost per
cake? Show work.
30. Solve:
2
10 48
0
4 16
x
x x
+
+ =
+ −
. Show work.
### Expert paper writers are just a few clicks away
Place an order in 3 easy steps. Takes less than 5 mins.
## Calculate the price of your order
You will get a personal manager and a discount.
We'll send you the first draft for approval by at
Total price:
\$0.00
|
HuggingFaceTB/finemath
|
|
# Alternating Current versus Direct Current
### Learning Objectives
By the end of this section, you will be able to:
• Explain the differences and similarities between AC and DC current.
• Calculate rms voltage, current, and average power.
• Explain why AC current is used for power transmission.
## Alternating Current
Most of the examples dealt with so far, and particularly those utilizing batteries, have constant voltage sources. Once the current is established, it is thus also a constant. Direct current (DC) is the flow of electric charge in only one direction. It is the steady state of a constant-voltage circuit. Most well-known applications, however, use a time-varying voltage source. Alternating current (AC) is the flow of electric charge that periodically reverses direction. If the source varies periodically, particularly sinusoidally, the circuit is known as an alternating current circuit. Examples include the commercial and residential power that serves so many of our needs. Figure 1 shows graphs of voltage and current versus time for typical DC and AC power. The AC voltages and frequencies commonly used in homes and businesses vary around the world. Figure 1. (a) DC voltage and current are constant in time, once the current is established. (b) A graph of voltage and current versus time for 60-Hz AC power. The voltage and current are sinusoidal and are in phase for a simple resistance circuit. The frequencies and peak voltages of AC sources differ greatly. Figure 2. The potential difference V between the terminals of an AC voltage source fluctuates as shown. The mathematical expression for V is given by
$V={V}_{0}\sin\text{ 2}\pi {ft}\\$
.
Figure 2 shows a schematic of a simple circuit with an AC voltage source. The voltage between the terminals fluctuates as shown, with the AC voltage given by
$V={V}_{0}\sin\text{2}\pi {ft}\\$
,
where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz. For this simple resistance circuit, I=V/R, and so the AC current is
$I={I}_{0}\sin 2\pi{ft}\\$
,
where I is the current at time t, and I= V0/R is the peak current. For this example, the voltage and current are said to be in phase, as seen in Figure 1(b).
Current in the resistor alternates back and forth just like the driving voltage, since I = V/R. If the resistor is a fluorescent light bulb, for example, it brightens and dims 120 times per second as the current repeatedly goes through zero. A 120-Hz flicker is too rapid for your eyes to detect, but if you wave your hand back and forth between your face and a fluorescent light, you will see a stroboscopic effect evidencing AC. The fact that the light output fluctuates means that the power is fluctuating. The power supplied is P = IV. Using the expressions for I and V above, we see that the time dependence of power is
$P={I}_{0}{V}_{0}{\text{sin}}^{2}\text{2}\pi {ft}\\$
, as shown in Figure 3.
### Making Connections: Take-Home Experiment—AC/DC Lights
Wave your hand back and forth between your face and a fluorescent light bulb. Do you observe the same thing with the headlights on your car? Explain what you observe. Warning: Do not look directly at very bright light. Figure 3. AC power as a function of time. Since the voltage and current are in phase here, their product is non-negative and fluctuates between zero and I0V0. Average power is (1/2)I0V0.
We are most often concerned with average power rather than its fluctuations—that 60-W light bulb in your desk lamp has an average power consumption of 60 W, for example. As illustrated in Figure 3, the average power Pave is
${P}_{\text{ave}}=\frac{1}{2}{I}_{0}{V}_{0}\\$
.
This is evident from the graph, since the areas above and below the (1/2)I0V0 line are equal, but it can also be proven using trigonometric identities. Similarly, we define an average or rms current Irms and average or rms voltage Vrms to be, respectively,
${I}_{\text{rms}}=\frac{{I}_{0}}{\sqrt{2}}\\$
and
${V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\\$
.
where rms stands for root mean square, a particular kind of average. In general, to obtain a root mean square, the particular quantity is squared, its mean (or average) is found, and the square root is taken. This is useful for AC, since the average value is zero. Now,
Pave IrmsVrms,
which gives
${P}_{\text{ave}}=\frac{{I}_{0}}{\sqrt{2}}\cdot \frac{{V}_{0}}{\sqrt{2}}=\frac{1}{2}{I}_{0}{V}_{0}\\$
,
as stated above. It is standard practice to quote Irms, Vrms, and Pave rather than the peak values. For example, most household electricity is 120 V AC, which means that Vrms is 120 V. The common 10-A circuit breaker will interrupt a sustained Irms greater than 10 A. Your 1.0-kW microwave oven consumes Pave = 1.0 kW, and so on. You can think of these rms and average values as the equivalent DC values for a simple resistive circuit. To summarize, when dealing with AC, Ohm’s law and the equations for power are completely analogous to those for DC, but rms and average values are used for AC. Thus, for AC, Ohm’s law is written
${I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{R}\\$
.
The various expressions for AC power Pave are
Pave Irms Vrms,
${P}_{\text{ave}}=\frac{{{V}_{\text{rms}}}^{2}}{R}\\$
,
and
${P}_{\text{ave}}={{I}_{\text{rms}}}^{2}R\\$
.
### Example 1. Peak Voltage and Power for AC
(a) What is the value of the peak voltage for 120-V AC power? (b) What is the peak power consumption rate of a 60.0-W AC light bulb?
#### Strategy
We are told that Vrms is 120 V and Pave is 60.0 W. We can use
${V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\\$
to find the peak voltage, and we can manipulate the definition of power to find the peak power from the given average power.
#### Solution for (a)
Solving the equation
${V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\\$
for the peak voltage V0 and substituting the known value for Vrms gives
${V}_{0}=\sqrt{2}{V}_{\text{rms}}=1.414(120\text{ V})=170 \text{V}\\$
#### Discussion for (a)
This means that the AC voltage swings from 170 V to –170 V and back 60 times every second. An equivalent DC voltage is a constant 120 V.
#### Solution for (b)
Peak power is peak current times peak voltage. Thus,
${P}_{0}={I}_{0}{V}_{0}=\text{2}\left(\frac{1}{2}{I}_{0}{V}_{0}\right)=\text{2}{P}_{\text{ave}}\\$
.
We know the average power is 60.0 W, and so
P= 2(60.0 W) = 120 W.
#### Discussion
So the power swings from zero to 120 W one hundred twenty times per second (twice each cycle), and the power averages 60 W.
## Why Use AC for Power Distribution?
Most large power-distribution systems are AC. Moreover, the power is transmitted at much higher voltages than the 120-V AC (240 V in most parts of the world) we use in homes and on the job. Economies of scale make it cheaper to build a few very large electric power-generation plants than to build numerous small ones. This necessitates sending power long distances, and it is obviously important that energy losses en route be minimized. High voltages can be transmitted with much smaller power losses than low voltages, as we shall see. (See Figure 4.) For safety reasons, the voltage at the user is reduced to familiar values. The crucial factor is that it is much easier to increase and decrease AC voltages than DC, so AC is used in most large power distribution systems. Figure 4. Power is distributed over large distances at high voltage to reduce power loss in the transmission lines. The voltages generated at the power plant are stepped up by passive devices called transformers (see Transformers) to 330,000 volts (or more in some places worldwide). At the point of use, the transformers reduce the voltage transmitted for safe residential and commercial use. (Credit: GeorgHH, Wikimedia Commons)
### Example 2. Power Losses Are Less for High-Voltage Transmission
(a) What current is needed to transmit 100 MW of power at 200 kV? (b) What is the power dissipated by the transmission lines if they have a resistance of 1.00 Ω(c) What percentage of the power is lost in the transmission lines?
#### Strategy
We are given Pave = 100 MW, Vrms = 200 kV, and the resistance of the lines is R = 1.00 Ω. Using these givens, we can find the current flowing (from P = IV) and then the power dissipated in the lines (I2R), and we take the ratio to the total power transmitted.
#### Solution
To find the current, we rearrange the relationship Pave IrmsVrms and substitute known values. This gives
${I}_{\text{rms}}=\frac{{P}_{\text{ave}}}{{V}_{\text{rms}}}=\frac{\text{100}\times {\text{10}}^{6}\text{ W}}{\text{200}\times {\text{10}}^{3}\text{ V}}=500\text{ A}\\$
.
#### Solution
Knowing the current and given the resistance of the lines, the power dissipated in them is found from
${P}_{\text{ave}}={{I}_{\text{rms}}}^{2}R\\$
Substituting the known values gives
Pave = Irms2= (500 A)2(1.00 Ω) = 250 kW
#### Solution
The percent loss is the ratio of this lost power to the total or input power, multiplied by 100:
$\text{\% loss=}\frac{\text{250 kW}}{\text{100 MW}}\times \text{100}=0\text{.}\text{250\%}\\$
.
#### Discussion
One-fourth of a percent is an acceptable loss. Note that if 100 MW of power had been transmitted at 25 kV, then a current of 4000 A would have been needed. This would result in a power loss in the lines of 16.0 MW, or 16.0% rather than 0.250%. The lower the voltage, the more current is needed, and the greater the power loss in the fixed-resistance transmission lines. Of course, lower-resistance lines can be built, but this requires larger and more expensive wires. If superconducting lines could be economically produced, there would be no loss in the transmission lines at all. But, as we shall see in a later chapter, there is a limit to current in superconductors, too. In short, high voltages are more economical for transmitting power, and AC voltage is much easier to raise and lower, so that AC is used in most large-scale power distribution systems.
It is widely recognized that high voltages pose greater hazards than low voltages. But, in fact, some high voltages, such as those associated with common static electricity, can be harmless. So it is not voltage alone that determines a hazard. It is not so widely recognized that AC shocks are often more harmful than similar DC shocks. Thomas Edison thought that AC shocks were more harmful and set up a DC power-distribution system in New York City in the late 1800s. There were bitter fights, in particular between Edison and George Westinghouse and Nikola Tesla, who were advocating the use of AC in early power-distribution systems. AC has prevailed largely due to transformers and lower power losses with high-voltage transmission.
## PhET Explorations: Generator
Generate electricity with a bar magnet! Discover the physics behind the phenomena by exploring magnets and how you can use them to make a bulb light.
## Section Summary
• Direct current (DC) is the flow of electric current in only one direction. It refers to systems where the source voltage is constant.
• The voltage source of an alternating current (AC) system puts out
$V={V}_{0}\text{sin 2}\pi {ft}\\$
, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz.
• In a simple circuit, V/R and AC current is
$I={I}_{0}\text{sin 2}\pi {ft}\\$
, where I is the current at time t, and
${I}_{0}={V}_{0}\text{/R}\\$
is the peak current.
• The average AC power is
${P}_{\text{ave}}=\frac{1}{2}{I}_{0}{V}_{0}\\$
.
• Average (rms) current Irms and average (rms) voltage Vrms are
${I}_{\text{rms}}=\frac{{I}_{0}}{\sqrt{2}}\\$
and
${V}_{\text{rms}}=\frac{{V}_{0}}{\sqrt{2}}\\$
, where rms stands for root mean square.
• Thus, Pave = IrmsVrms.
• Ohm’s law for AC is
${I}_{\text{rms}}=\frac{{V}_{\text{rms}}}{R}\\$
.
• Expressions for the average power of an AC circuit are
${P}_{\text{ave}}={I}_{\text{rms}}{V}_{\text{rms}}\\$
,
${P}_{\text{ave}}=\frac{{V}_{\text{rms}}^{2}}{R}\\$
, and
${P}_{\text{ave}}={{I}_{\text{rms}}}^{2}R\\$
, analogous to the expressions for DC circuits.
### Conceptual Questions
1. Give an example of a use of AC power other than in the household. Similarly, give an example of a use of DC power other than that supplied by batteries.
2. Why do voltage, current, and power go through zero 120 times per second for 60-Hz AC electricity?
3. You are riding in a train, gazing into the distance through its window. As close objects streak by, you notice that the nearby fluorescent lights make dashed streaks. Explain.
### Problem & Exercises
1. (a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the bulb’s operating temperature is 2700ºC, what is its resistance at 2600ºC?
2. Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V. What is the rms voltage?
3. A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding peak current?
4. Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight equipment at this higher frequency. What is the time for one complete cycle of this power?
5. A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does the razor consume as it is ruined?
6. In this problem, you will verify statements made at the end of the power losses for Example 2 above. (a) What current is needed to transmit 100 MW of power at a voltage of 25.0 kV? (b) Find the power loss in a 1.00-Ω transmission line. (c) What percent loss does this represent?
7. A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a) What is its effective resistance? (b) What is the cost of running the air conditioner during a hot summer month when it is on 8.00 h per day for 30 days and electricity costs 9.00 cents/kW ⋅ h?
8. What is the peak power consumption of a 120-V AC microwave oven that draws 10.0 A?
9. What is the peak current through a 500-W room heater that operates on 120-V AC power?
10. Two different electrical devices have the same power consumption, but one is meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its resistance is unaffected, by what factor will the power increase if a 120-V AC device is connected to 240-V AC?
11. Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of Nichrome wire, having a cross-sectional area of 500 mm2, is needed if the operating temperature is 500ºC (c) What power will it draw when first switched on?
12. Find the time after t = 0 when the instantaneous voltage of 60-Hz AC first reaches the following values: (a) V0/2 (b)V0 (c) 0.
13. (a) At what two times in the first period following t = 0 does the instantaneous voltage in 60-Hz AC equal Vrms? (b) Vrms?
## Glossary
direct current:
(DC) the flow of electric charge in only one direction
alternating current:
(AC) the flow of electric charge that periodically reverses direction
AC voltage:
voltage that fluctuates sinusoidally with time, expressed as V = V0 sin 2πft, where V is the voltage at time t, V0 is the peak voltage, and f is the frequency in hertz
AC current:
current that fluctuates sinusoidally with time, expressed as I = I0 sin 2πft, where I is the current at time t, I0 is the peak current, and f is the frequency in hertz
rms current:
the root mean square of the current,
${I}_{\text{rms}}={I}_{0}/\sqrt{2}\\$
, where I0 is the peak current, in an AC system
rms voltage:
the root mean square of the voltage,
${V}_{\text{rms}}={V}_{0}/\sqrt{2}\\$
, where V0 is the peak voltage, in an AC system
### Selected Solutions to Problems & Exercises
2. 480 V
4. 2.50 ms
6. (a) 4.00 kA (b) 16.0 MW (c) 16.0%
8. 2.40 kW
10. (a) 4.0 (b) 0.50 (c) 4.0
12. (a) 1.39 ms (b) 4.17 ms (c) 8.33 ms
|
HuggingFaceTB/finemath
|
|
Related Articles
Maximum possible sum of non-adjacent array elements not exceeding K
• Difficulty Level : Basic
• Last Updated : 05 May, 2021
Given an array arr[] consisting of N integers and an integer K, the task is to select some non-adjacent array elements with the maximum possible sum not exceeding K.
Examples:
Input: arr[] = {50, 10, 20, 30, 40}, K = 100
Output: 90
Explanation: To maximize the sum that doesn’t exceed K(= 100), select elements 50 and 40.
Therefore, maximum possible sum = 90.
Input: arr[] = {20, 10, 17, 12, 8, 9}, K = 64
Output: 46
Explanation: To maximize the sum that doesn’t exceed K(= 64), select elements 20, 17, and 9.
Therefore, maximum possible sum = 46.
Naive Approach: The simplest approach is to recursively generate all possible subsets of the given array and for each subset, check if it does not contain adjacent elements and has the sum not exceeding K. Among all subsets for which the above condition is found to be true, print the maximum sum obtained for any subset.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the``// maximum sum not exceeding``// K possible by selecting``// a subset of non-adjacent elements``int` `maxSum(``int` `a[],`` ``int` `n, ``int` `k)``{`` ``// Base Case`` ``if` `(n <= 0)`` ``return` `0;` ` ``// Not selecting current`` ``// element`` ``int` `option = maxSum(a,`` ``n - 1, k);` ` ``// If selecting current`` ``// element is possible`` ``if` `(k >= a[n - 1])`` ``option = max(option,`` ``a[n - 1] +`` ``maxSum(a, n - 2,`` ``k - a[n - 1]));` ` ``// Return answer`` ``return` `option;``}` `// Driver Code``int` `main()``{`` ``// Given array arr[]`` ``int` `arr[] = {50, 10,`` ``20, 30, 40};` ` ``int` `N = ``sizeof``(arr) /`` ``sizeof``(arr);` ` ``// Given K`` ``int` `K = 100;` ` ``// Function Call`` ``cout << (maxSum(arr,`` ``N, K));``}` `// This code is contributed by gauravrajput1`
## Java
`// Java program for the above approach` `import` `java.io.*;` `class` `GFG {` ` ``// Function to find the maximum sum`` ``// not exceeding K possible by selecting`` ``// a subset of non-adjacent elements`` ``public` `static` `int` `maxSum(`` ``int` `a[], ``int` `n, ``int` `k)`` ``{`` ``// Base Case`` ``if` `(n <= ``0``)`` ``return` `0``;` ` ``// Not selecting current element`` ``int` `option = maxSum(a, n - ``1``, k);` ` ``// If selecting current element`` ``// is possible`` ``if` `(k >= a[n - ``1``])`` ``option = Math.max(`` ``option,`` ``a[n - ``1``]`` ``+ maxSum(a, n - ``2``,`` ``k - a[n - ``1``]));` ` ``// Return answer`` ``return` `option;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// Given array arr[]`` ``int` `arr[] = { ``50``, ``10``, ``20``, ``30``, ``40` `};` ` ``int` `N = arr.length;` ` ``// Given K`` ``int` `K = ``100``;` ` ``// Function Call`` ``System.out.println(maxSum(arr, N, K));`` ``}``}`
## Python3
`# Python3 program for the above approach` `# Function to find the maximum sum``# not exceeding K possible by selecting``# a subset of non-adjacent elements``def` `maxSum(a, n, k):`` ` ` ``# Base Case`` ``if` `(n <``=` `0``):`` ``return` `0` ` ``# Not selecting current element`` ``option ``=` `maxSum(a, n ``-` `1``, k)` ` ``# If selecting current element`` ``# is possible`` ``if` `(k >``=` `a[n ``-` `1``]):`` ``option ``=` `max``(option, a[n ``-` `1``] ``+`` ``maxSum(a, n ``-` `2``, k ``-` `a[n ``-` `1``]))` ` ``# Return answer`` ``return` `option` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``# Given array arr[]`` ``arr ``=` `[ ``50``, ``10``, ``20``, ``30``, ``40` `]` ` ``N ``=` `len``(arr)` ` ``# Given K`` ``K ``=` `100` ` ``# Function Call`` ``print``(maxSum(arr, N, K))` `# This code is contributed by mohit kumar 29`
## C#
`// C# program for the``// above approach``using` `System;``class` `GFG{` `// Function to find the maximum``// sum not exceeding K possible``// by selecting a subset of``// non-adjacent elements``public` `static` `int` `maxSum(``int` `[]a,`` ``int` `n, ``int` `k)``{`` ``// Base Case`` ``if` `(n <= 0)`` ``return` `0;` ` ``// Not selecting current element`` ``int` `option = maxSum(a, n - 1, k);` ` ``// If selecting current`` ``// element is possible`` ``if` `(k >= a[n - 1])`` ``option = Math.Max(option, a[n - 1] +`` ``maxSum(a, n - 2,`` ``k - a[n - 1]));`` ` ` ``// Return answer`` ``return` `option;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{`` ``// Given array []arr`` ``int` `[]arr = {50, 10, 20, 30, 40};` ` ``int` `N = arr.Length;` ` ``// Given K`` ``int` `K = 100;` ` ``// Function Call`` ``Console.WriteLine(maxSum(arr, N, K));``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output
`90`
Time Complexity: O(2N)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming. Two possible options exist for every array element:
1. Either skip the current element and proceed to the next element.
2. Select the current element only if it is smaller than or equal to K.
Follow the steps below to solve the problem:
1. Initialize an array dp[N][K+1] with -1 where dp[i][j] will store the maximum sum to make sum j using elements up to index i.
2. From the above transition, find the maximum sums if the current element gets picked and if it is not picked, recursively.
3. Store the minimum answer for the current state.
4. Also, add the base condition that if the current state (i, j) is already visited i.e., dp[i][j]!=-1 return dp[i][j].
5. Print dp[N][K] as the maximum possible sum.
Below is the implementation of the above approach:
## C++
`// C++ program for the above approach``#include ``using` `namespace` `std;` `// Initialize dp``int` `dp;` `// Function find the maximum sum that``// doesn't exceeds K by choosing elements``int` `maxSum(``int``* a, ``int` `n, ``int` `k)``{`` ` ` ``// Base Case`` ``if` `(n <= 0)`` ``return` `0;` ` ``// Return the memoized state`` ``if` `(dp[n][k] != -1)`` ``return` `dp[n][k];` ` ``// Dont pick the current element`` ``int` `option = maxSum(a, n - 1, k);` ` ``// Pick the current element`` ``if` `(k >= a[n - 1])`` ``option = max(option,`` ``a[n - 1] +`` ``maxSum(a, n - 2,`` ``k - a[n - 1]));` ` ``// Return and store the result`` ``return` `dp[n][k] = option;``}` `// Driver Code``int` `main()``{`` ``int` `N = 5;` ` ``int` `arr[] = { 50, 10, 20, 30, 40 };` ` ``int` `K = 100;` ` ``// Fill dp array with -1`` ``memset``(dp, -1, ``sizeof``(dp));` ` ``// Print answer`` ``cout << maxSum(arr, N, K) << endl;``}` `// This code is contributed by bolliranadheer`
## Java
`// Java program for the above approach` `import` `java.util.*;` `class` `GFG {` ` ``// Function find the maximum sum that`` ``// doesn't exceeds K by choosing elements`` ``public` `static` `int` `maxSum(``int` `a[], ``int` `n,`` ``int` `k, ``int``[][] dp)`` ``{`` ``// Base Case`` ``if` `(n <= ``0``)`` ``return` `0``;` ` ``// Return the memoized state`` ``if` `(dp[n][k] != -``1``)`` ``return` `dp[n][k];` ` ``// Dont pick the current element`` ``int` `option = maxSum(a, n - ``1``,`` ``k, dp);` ` ``// Pick the current element`` ``if` `(k >= a[n - ``1``])`` ``option = Math.max(`` ``option,`` ``a[n - ``1``]`` ``+ maxSum(a, n - ``2``,`` ``k - a[n - ``1``],`` ``dp));` ` ``// Return and store the result`` ``return` `dp[n][k] = option;`` ``}` ` ``// Driver Code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``50``, ``10``, ``20``, ``30``, ``40` `};` ` ``int` `N = arr.length;` ` ``int` `K = ``100``;` ` ``// Initialize dp`` ``int` `dp[][] = ``new` `int``[N + ``1``][K + ``1``];` ` ``for` `(``int` `i[] : dp) {`` ``Arrays.fill(i, -``1``);`` ``}`` ``// Print answer`` ``System.out.println(maxSum(arr, N, K, dp));`` ``}``}`
## Python3
`# Python3 program for the``# above approach` `# Function find the maximum``# sum that doesn't exceeds K``# by choosing elements``def` `maxSum(a, n, k, dp):`` ` ` ``# Base Case`` ``if` `(n <``=` `0``):`` ``return` `0``;` ` ``# Return the memoized state`` ``if` `(dp[n][k] !``=` `-``1``):`` ``return` `dp[n][k];` ` ``# Dont pick the current`` ``# element`` ``option ``=` `maxSum(a, n ``-` `1``,`` ``k, dp);` ` ``# Pick the current element`` ``if` `(k >``=` `a[n ``-` `1``]):`` ``option ``=` `max``(option, a[n ``-` `1``] ``+`` ``maxSum(a, n ``-` `2``,`` ``k ``-` `a[n ``-` `1``], dp));`` ``dp[n][k] ``=` `option;`` ` ` ``# Return and store`` ``# the result`` ``return` `dp[n][k];` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:`` ` ` ``arr ``=` `[``50``, ``10``, ``20``,`` ``30``, ``40``];`` ``N ``=` `len``(arr);`` ``K ``=` `100``;`` ``# Initialize dp`` ``dp ``=` `[[``-``1` `for` `i ``in` `range``(K ``+` `1``)]`` ``for` `j ``in` `range``(N ``+` `1``)]` ` ``# Pranswer`` ``print``(maxSum(arr, N,`` ``K, dp));` `# This code is contributed by shikhasingrajput`
## C#
`// C# program for the``// above approach``using` `System;``class` `GFG{` `// Function find the maximum``// sum that doesn't exceeds K``// by choosing elements``public` `static` `int` `maxSum(``int` `[]a, ``int` `n,`` ``int` `k, ``int``[,] dp)``{`` ``// Base Case`` ``if` `(n <= 0)`` ``return` `0;` ` ``// Return the memoized`` ``// state`` ``if` `(dp[n, k] != -1)`` ``return` `dp[n, k];` ` ``// Dont pick the current`` ``// element`` ``int` `option = maxSum(a, n - 1,`` ``k, dp);` ` ``// Pick the current element`` ``if` `(k >= a[n - 1])`` ``option = Math.Max(option, a[n - 1] +`` ``maxSum(a, n - 2,`` ``k - a[n - 1],`` ``dp));` ` ``// Return and store the`` ``// result`` ``return` `dp[n, k] = option;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{`` ``int` `[]arr = {50, 10, 20, 30, 40};`` ``int` `N = arr.Length;`` ``int` `K = 100;`` ` ` ``// Initialize dp`` ``int` `[,]dp = ``new` `int``[N + 1,`` ``K + 1];` ` ``for` `(``int` `j = 0; j < N + 1; j++)`` ``{`` ``for` `(``int` `k = 0; k < K + 1; k++)`` ``dp[j, k] = -1;`` ``}`` ` ` ``// Print answer`` ``Console.WriteLine(maxSum(arr, N,`` ``K, dp));``}``}` `// This code is contributed by Rajput-Ji`
## Javascript
``
Output
`90`
Time Complexity: O(N*K), where N is the size of the given array and K is the limit.
Auxiliary Space: O(N)
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
My Personal Notes arrow_drop_up
|
HuggingFaceTB/finemath
|
|
# Engineering Mechanics - KOP: Force and Acceleration
### Exercise :: KOP: Force and Acceleration - General Questions
• KOP: Force and Acceleration - General Questions
11. The pendulum bob B has a weight of 5 lb and is released from rest in the position shown, =0°. Determine the tension in string BC just after the bob is released, = 0°, and also at the instant the bob reaches point D, = 45°.
A. T0 = 0, T45 = 10.61 lb B. T0 = 0.1551 lb, T45 = 10.61 lb C. T0 = 0, T45 = 7.07 lb D. T0 = 0.1551 lb, T45 = 7.07 lb
Explanation:
No answer description available for this question. Let us discuss.
12. A truck T has a weight of 8,000 lb and is traveling along a portion of a road defined by the lemniscate r2 = 0.2(106) cos 2 , where r is measured in feet and is in radians. If the truck maintains a constant speed of vr = 4 ft/s, determine the magnitude of the resultant frictional force which must be exerted by all the wheels to maintain the motion when = 0.
A. F = 29.2 lb B. F = 859 lb C. F = 87.5 lb D. F = 26.7 lb
Explanation:
No answer description available for this question. Let us discuss.
13. A smooth can C, having a mass of 2 kg, is lifted from a feed at A to a ramp at B by a forked rotating rod. If the rod maintains a constant angular motion of = 0.5 rad/s, determine the force which the rod exerts on the can at the instant = 30°. Neglect the effects of friction in the calculation. The ramp from A to B is circular, having a radius of 700 min.
A. F = 19.62 N B. F = 11.33 N C. F = 10.63 N D. F = 12.03 N
Explanation:
No answer description available for this question. Let us discuss.
14. The spool, which has a weight of 2lb, slides along the smooth horizontal spiral rod, r = (2 ) ft, where is in radians. If its angular rate of rotation is constant and equals = 4 rad/s, determine the tangential force P needed to cause the motion and the normal force that the spool exerts on the rod at the instant = 90°.
A. P = 0.499 lb, N = 5.03 lb B. P = 5.50 lb, N = 15.65 lb C. P = 3.35 lb, N = 2.43 lb D. P = 1.677 lb, N = 4.77 lb
Explanation:
No answer description available for this question. Let us discuss.
15. Each of the three barges has a mass of 30 Mg, whereas the tugboat has a mass of 12 Mg. As the barges are being pulled forward with a constant velocity of 4 m/s, the tugboat must overcome the frictional resistance of the water, which is 2 kN for each barge and 1.5 kN for the tugboat. If the cable between A and B breaks, determine the acceleration of the tugboat.
A. a = 0.1667 m/s2 B. a = 0.1042 m/s2 C. a = 0.0278 m/s2 D. a = 0.01961 m/s2
|
HuggingFaceTB/finemath
|
|
# Probability – Aptitude MCQ Questions – 08
Question 1
A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:
   A. 343/1728    B. 21/36    C. 12/35    D. 7/44
Question 2
A five-digit number is formed by using digits 1, 2, 3, 4 and 5 without repetition. What is the probability that the number is divisible by 4?
   A. 1/5    B. 5/6    C. 4/5    D. None of these
Question 3
An urn contains 6 red, 5 blue and 2 green marbles. If three marbles are picked at random, what is the probability that at least one is blue?
   A. 28/143    B. 11/597    C. 28/197    D. 115/143
Question 4
From a pack of 52 cards, 3 cards are drawn. What is the probability that one is ace, one is queen and one is jack?
   A. 19/5525    B. 21/5525    C. 17/5525    D. 16/5525
Question 5
Two teams Arrogant and Overconfident are participating in a cricket tournament. The odds that team Arrogant will be champion is 5 to 3, and the odds that team Overconfident will be the champion is 1 to 4. What are the odds that either Arrogant or team Overconfident will become the champion?
   A. 3 to 2    B. 5 to 2    C. 6 to 1    D. 33 to 7
Question 6
Find the probability that in a random arrangement of the letters of the word 'UNIVERSITY' the two I's come together.
   A. 1/7    B. 3/5    C. 5/11    D. 1/5
Question 7
In a charity show tickets numbered consecutively from 101 through 350 are placed in a box.What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?
   A. 0.285    B. 0.40    C. 100/249    D. 99/250
Question 8
On rolling a dice 2 times, the sum of 2 numbers that appear on the uppermost face is 8. What is the probability that the first throw of dice yields 4?
   A. 2/36    B. 1/36    C. 1/6    D. 1/5
Question 9
A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack?
   A. 4/13    B. 2/13    C. 6/13    D. 9/13
Question 10
What is the probability that a number selected from numbers 1, 2, 3, ......, 30, is prime number, when each of the given numbers is equally likely to be selected?
   A. 9/30    B. 8/30    C. 10/30    D. 11/30
|
HuggingFaceTB/finemath
|
|
# ANGLE OF ELEVATION AND DEPRESSION EXAMPLES
Example 1 :
A student sitting in a classroom sees a picture on the black board at a height of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30°. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45°. Find the distance moved by the student.
Solution :
Distance moved by the student = BC
In triangle ABD :
∠ABD = 30°
tan θ = opposite side/Adjacent side
1/√3 = 1.5/BD
BD = 1.5 x √3 ==> 1.5 √3
In triangle ACD :
∠ACD = 45°
tan θ = opposite side/Adjacent side
1 = 1.5/CD
CD = 1.5
BC = BD - CD
BC = 1.5 √3 - 1.5
= 1.5 (√3 - 1) = 1.5(1.732 - 1)
= 1.5 (0.732) ==> 1.098 m
Hence the distance moved by the student is 1.098 m.
Example 2 :
A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution :
From the given information, we can draw a rough diagram
In triangle ABD
∠ABD = 30°
tan θ = opposite side/Adjacent side
1/√3 = 28.5/BD ==> BD = 28.5 √3 --(1)
In triangle ACD
∠ABD = 60°
tan θ = opposite side/Adjacent side
3 = 28.5/CD ==> CD = 28.5/
Multiplying by 3 on both numerator and denominator, we get
CD = 28.5/3 ==> 28.53/3 ==> 9.53 -->(2)
the distance he walked towards the building = BD - CD
= 28.5 √3 - 9.53 ==> 193 m
Example 3 :
From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a Yacht and a Barge along the same line of sight .The angles of depression for the Yacht and the Barge are 45° and 30° respectively. For safety purposes the two sea vessels should be at least 300 feet apart. If they are less than 300 feet , the keeper has to sound the alarm. Does the keeper have to sound the alarm ?
Solution :
From the above picture, we have to find the value of CD.
In triangle ABC
∠ACB = 45°
sin θ = opposite side/Hypotenuse side
sin 45° = AB/BC
1/2 = 200/BC
BC = 200 2
In triangle ABD
sin 30° = AB /BD
1/2 = 200/BD
BD = 200 x 2 ==> 400
CD = BD - BC ==> 400 - 200 2 ==> 200(2 - 2)
= 200 (2 - 1.414)
= 200(0.586) ==> 117.2 m
From this, we come to know that the distance between Yacht and a Barge is less than 300 m. Hence the keeper has to sound the alarm.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Writing Numbers in Words Worksheet
Jan 16, 22 11:56 PM
Writing Numbers in Words Worksheet
2. ### Operations with Negative Numbers Worksheet
Jan 16, 22 11:40 PM
Operations with Negative Numbers Worksheet
|
HuggingFaceTB/finemath
|
|
# AtCoder Grand Contest 038
Updated:
AtCoder Grand Contest 038
# Solutions
## A - 01 Matrix
### Solution
For instance, the following picture gives a solution (from the official editorial). ## B - Sorting a Segment
Let $I _ i = [i, i + K)$. We also use $I _ i$ to denote the manipulation for it.
### Observation
Consider two manipulations $I _ i$ and $I _ j$. When do they give the same result? If $I _ i \cap I _ j = \emptyset$, it will mean the both ranges $I _ i$ and $I _ j$ have already been sorted and the resulting sequence is the same as the original sequence.
If $I _ i \cap I _ j \neq \emptyset$, it will mean $I _ i \setminus I _ j$ and $I _ j \setminus I _ i$ have both already been sorted. Therefore for each $i \leq k \leq j$, $I _ k$ gives the same result. Thus we consider the relation only between $I _ i$ and $I _ {i + 1}$.
### Solution
So, our strategy is composed of two parts. We use an union-find tree to keep the states. We use $I _ {N - K + 1}$ to denote the original state. We merge $i$ and $j$ if and only if $I _ i$ and $I _ j$ gives the same result. The answer is the number of connected components, being careful of $I _ {N - K + 1}$.
Let’s see the details with implements to avoid tiny mistakes.
#### Connect $I _ i$ and original state
As we argued just now, if $I _ i$ are already sorted, it will give the original result. This is equivalent as follows of course. $a _ i < a _ {i + 1} < \dots < a _ {i + K - 2} < a _ {i + K - 1} \tag{B.1}$
We define
$ascend[i] = 0$ if $a _ i < a _ {i + 1}$, $1$ if $a _ i > a _ {i + 1}$,
$ascend\_sum[i] = \displaystyle \sum _ {j = 0} ^ {i - 1} ascend[j]$.
(B.1) is equivalent to $0 = \sum _ {j = i} ^ {i + K - 2} ascend[j] = ascend\_sum[i + K - 1] - ascend\_sum[i].$
#### Connect $I _ i$ and $I _ {i + 1}$
$I _ i$ and $I _ {i + 1}$ give the same result if and only if $P[i]$ is the minimum of $I _ i \cup I _ {i + 1}$ and $P[i + K]$ is the maximum of $I _ i \cup I _ {i + 1}$. This is a very typical RMQ. But we work on it by set<int> S;
We consider the minimum part. Let $Q = P ^ {-1}$. For $k = 0, \dots, N - 1$, let $e = Q[k]$. We execute S.insert(e);. If $0 \leq e \leq N - K - 1$, we have to determine whether $e$ is minimum on $I _ {e} \cup I _ {e + 1}$ or not. We use auto it{S.find()};, auto it1{it};, ++it1;. If it1 == S.end() or *it1 - e > K, it will mean $e$ is the minimum element of $I _ {e} \cup I _ {e + 1}$.
We consider the maximum part. We use another S. Let $Q = P ^ {-1}$ and reversed. For $k = 0, \dots, N - 1$, let $e = Q[k]$. We execute S.insert(e);. If $K \leq e \leq N - 1$, we have to determine whether $e$ is maximum on $I _ {e - K} \cup I _ {e - K + 1}$ or not. We use auto it{S.find()};, auto it1{it};, --it1;. If it == S.begin() or e - *it1 > K, it will mean $e$ is the minimum element of $I _ {e - K} \cup I _ {e - K + 1}$.
## C - LCMs
### The idea of solution
For $x, y \in \mathbb{N} _ +$, we have $\mathrm{lcm}(x, y) = \frac{x y}{\mathrm{gcd}(x, y)}.$ Suppose that $\mathrm{gcd}(x, y) = 6$. We want to use the inclusive-exclusive principal here. First we set $l = xy$. Since $\mathrm{gcd}(x, y) \% 2 = 0$, we should subtract $xy / 2$ from $l$. Since $\mathrm{gcd}(x, y) \% 3 = 0$, we should subtract $2xy / 3$ from $l$, too. Since $\mathrm{gcd}(x, y) \% 6 = 0$, however, should we $xy / 6$ from $l$? It’s not true. We have already had $l = xy - xy / 2 - 2xy / 3 = -xy / 6$. So we should add $xy / 3$ to $l$ to have $l = xy/6$.
### Solution
Therefore, first we suppose that there exist a good coefficient $C[d]$. Let $C[d]$ be so that for any $g \in \mathbb{N} _ +$, it holds that $\sum _ {d | g} C[d] = \frac{1}{g}.$ Then, we have \begin{align} X & = \sum _ {i, j \in N} \mathrm{lcm}(A _ i, A _ j) \\ & = \sum _ {i, j \in N} \frac{A _ i A _ j}{\mathrm{gcd}(A _ i, A _ j)} \\ & = \sum _ {i, j \in N} \left( \sum _ {d | \mathrm{gcd}(A _ i, A _ j)} C[d] \right) A _ i A _ j \\ & = \sum _ {d} C[d] \sum _ {d | A _ i, d | A _ j} A _ i A _ j \\ & = \sum _ {d} C[d] \left( \sum _ {d | A _ i} A _ i \right) ^ 2. \end{align} This is computable in time. Let $M = \max _ i A _ i$. Each sum of $A _ i$ can be calculated in $O(M / d)$-time. So the time complexity is $O(M \log M)$.
The remaining here is how to compute $C[d]$. But this is not difficult. We can just use the following equation. $C[g] = \frac{1}{g} - \sum _ {d | g, d < g} C[d].$ We work on it like the sieve of Eratosthenes. First we initialize $C[i] = 1 / i$. Then, we execute the following code.
for (auto i = 1; i < M; i++)
{
for (auto j = 2 * i; j < M; j += i)
{
C[j] -= C[i];
}
}
The time complexity is $O(M \log M)$. Thus we obtain the answer as follows. $ans = \frac{1}{2} \left( X - \sum _ {i \in N} A _ i \right).$ The total time complexity is $O(N + M \log M)$.
## D - Unique Path
### Solution
#### If it is a tree
Suppose that $M = N - 1$. In this case the graph $G$ must be a graph. Thus for all $i \in Q$, $A _ i$ must be $0$ and the answer is Yes if and only if $A _ i = 0$ for all $i \in Q$.
#### Otherwise
We have at least one loop. We consider the queries $\{ (A _ i, B _ i, C _ i) \}$ as follows.
• If $A _ i = 0$, $B _ i$ and $C _ i$ are on the same “sub-tree”. We call this type of a subtree a unit. We consider as if $B _ i$ and $C _ i$ are the same one vertex. We consider this type of queries first. We use a union-find tree.
• If $A _ i = 1$, $B _ i$ and $C _ i$ must not be in the same unit. So we check this this type of queries after the above. If we find an error here, halt with flushing No.
Let $K$ be the number of the units constructed the above things. Let $X$ be the sum number of the edges inside of each unit. $X$ does not depend on its topological shape of each unit because it is just a tree on each unit.
How can we adjust the number of the resulting edges equal to $M$? The minimum is $X + K$. Note that it must be equal to $N$ since the resulting graph will be a pseudo-tree. The maximum is $X + K(K - 1) / 2$. We can connect every two units arbitrarily but we cannot add more since if we added more edge, it would break the rule of unit. So we check the following inequality. $X + K \leq M \leq X + \frac{K(K - 1)}{2}.$ If it holds, the answer is Yes; otherwise, No.
## E - Gachapon
### Basic ideas
I think that in this problem it is difficult to understand the initial step. So let’s see some fundamental examples.
#### The easiest example
Suppose that you play Fate/Grand Order and you want to summon Astolfo (Saber) and you want his Noble Phantasm to be level 5, as just I did yesterday (I am writing it on 2019-11-29). How many times will you need to summon in terms of the expected time?
According to the official document, the probability to summon him is $p = 0.008$ each time and you need to summon him $B = 5$ times to get NP5. Of course we can easily calculate the expected time. It is $(1/p) \times B$. We see the details carefully. What happens here?
What you can do is to continue to summon until you reach NP5. At first, you get no Astolfo, i.e. NP0. You start to summon, and get him. It’s NP1. You continue to summon, and get him to obtain NP2. You continue to summon until you obtain NP5. Here, we see the fundamental idea: To get him $B$ times, you have to spend each “NP $i$ state” to get NP $(i + 1)$ for $i = 0, 1, \dots, B - 1$, and the expected time will be their sum.
In this case, for each state, the expected time is the same: $1 / p$. So the answer is $B / p$.
#### Second simplest example
We think of the case $N = 2$, $B _ 0 = B _ 1 = 1$. Let’s suppose that you summon Astolfo (Rider) and Jeanne (Ruler) to NP1, i.e. $p _ 0 = 0.015$ and $p _ 1 = 0.008$.
You start by $(0, 0)$. Here, we have two probabilities: who will come firstly, Astolfo or Jeanne. If we have Astolfo first, it’s $(1, 0)$. But you have to continue until you succeed to summon Jeanne. This is an important point that was hidden in the previous example. If you are so lucky to get Astolfo NP5, it doesn’t mean the summon is over, i.e., $(5, 0)$ is not the terminated point. You have to summon Jeanne.
In this point, what will be the expected time? As a lesson from the previous example, the expected time will be the sum of the expected time when you are not satisfied. In this case it will be the sum of $(0, * )$ and $( *, 0)$, but be careful of the case $(0, 0)$. This is included in both of them. So we subtract it from the answer. We will use the inclusive-exclusive principle.
We can calculate the answer as follows.
• In the case $(0, * )$, the expected time will be $1 / p _ 0$. We add it to the answer.
• In the case $( * , 0)$, the expected time will be $1 / p _ 1$. We add it to the answer.
• In the case $(0, 0)$, the expected time will be $1 / (p _ 0 + p _ 1 )$. We subtract it from the answer.
The answer is $\frac{1}{p _ 0} + \frac{1}{p _ 1} - \frac{1}{p _ 0 + p _ 1}.$
We will work the same thing in more general way.
### Solution
Let $p _ i = A _ i / S$. We start by $(0, \dots, 0)$. We continue to summon until we reach the state $(C _ 0, \dots, C _ {N - 1})$ that satisfies $C _ i \geq B _ i$ for all the $i \in N$. The answer is the sum of all “unsatisfactory” states’ expected time. Here we use the fact that the expected time of the sum is equal to the sum of the expected time of each.
We consider the state $(C _ 0, \dots, C _ {k - 1}, C _ k, \dots, C _ {N - 1})$ that satisfies $C _ 0 < B _ 0$, …, $C _ {k - 1} < B _ {k - 1}$. What is the expected time during which this state continues?
First, we consider the coefficient for the inclusive-exclusive principle. We should $+$ when $k = 1$, that is, $(- 1) ^ {k + 1}$.
Second, we consider the probability in that this state happens. Just “ignoring” the probability for $i \geq k$, we obtain it as follows. $\frac{ \left( \sum _ {i \in k} c _ i \right)!}{c _ 0 ! \dots c _ {k - 1}!} \cdot \frac{p _ 0 ^ {c _ 0} \dots p _ {k - 1} ^ {c _ {k - 1}}}{ \left( \sum _ {i \in k} p _ i \right) ^ {\sum _ {i \in k} c _ i}}.$
Finally, we estimate the expected time to last this state. Once it happens, it will continue until we summon $i < k$. So it is $1 / \sum _ {i \in k} p _ i$.
Therefore, we have the following result. $(- 1) ^ {k + 1} \frac{ \left( \sum _ {i \in k} c _ i \right)!}{c _ 0 ! \dots c _ {k - 1}!} \cdot \frac{p _ 0 ^ {c _ 0} \dots p _ {k - 1} ^ {c _ {k - 1}}}{ \left( \sum _ {i \in k} p _ i \right) ^ {\sum _ {i \in k} c _ i}} \cdot \frac{1}{\sum _ {i \in k} p _ i}. \tag{E.1}$
### Convert it into DP
We have to add up all (E.1) for all $2 ^ N B _ 0 \dots B _ {N - 1}$ cases. Let’s convert it into DP. Seeing (E.1), once we keep the $\sum c _ i$ and $\sum p _ i$ as the keys of DP, the transition is the same, i.e. to multiply $(- 1) \frac{p _ k ^ c}{c!}.$ This fact solves the problem.
#### Definition
$dp[k][X][Y] =$ the sum of $(- 1)^{s + 1} \frac{p _ x ^ {c _ x} \dots p _ {y} ^ {c _ {y}}}{c _ x ! \dots c _ {y}!}$ part of (E.1) with $X = \sum c _ i$ and $Y = \sum p _ i$, considering $i = 0, \dots, k - 1$.
We hold $dp$ by vector<vector<map<mint, mint>>>.
The answer is $\sum _ {X} \sum _ {Y} dp[N][X][Y] \frac{X!}{Y ^ {X + 1}}.$
#### Initial state
As we argued before, we have to add when the element size is $1$. So the initial state is $dp = -1.$ Since we have the last key as the key of map<mint, mint>, we initialize the size of $dp$ is $N + 1$. Let $M$ be the sum of $\{ A _ i \}$ $+1$. We initialize the size of $dp$ as $M$.
#### Transition
For $k = 0, \dots, N - 1$, first we copy the vector: $dp[k + 1] \gets dp[k]$. This means that we don’t choose $k$-th element. Then, for all $X$, for all $Y$, for all $0 \leq c < B[k]$, we execute the following manipulation. $dp[k + 1][X + c][Y + p _ k] \mathbin{ {-} {=} } \frac{p _ k ^ c}{c!} dp[k][X][Y].$ This means that we choose $k$-th element for binding $c$ items.
## F - Two Permutations
### Solution
For $P, Q \in \mathfrak{S} _ N$, we discompose each of them into a set of cycles. Fix $i \in N$. Let $K$, $L$ be the cycle including $i$ on $P$, $Q$ respectively.
#### Main cases
To simplify the problem at first, we consider the case $i \neq P _ i$ and $i \neq Q _ i$. Then choosing $A _ i = i$ forces us to choose $A _ j = j$ for all elements on the cycle $K$. Otherwise, i.e., choosing $A _ i = P _ i$ means that we have to choose $A _ j = P _ j$ on $K$. So, there are exactly two states on $K$; “stable” or “advanced”.
The answer is $N - penalty$, where $penalty$ is the number of the elements $i$ so that $A _ i = B _ i$. When does $i$ cause a penalty? Of course if we choose “stable” both on $K$ and on $L$, it causes that $A _ i = B _ i = i$, which means a penalty. In addition, if $P _ i = Q _ i$ and we choose “advanced” both on $K$ and on $L$, it causes a penalty with $A _ i = B _ i = P _ i (= Q _ i)$.
As we argued before, there are two states for each cycle. So we want to solve this problem by minimum cut on the directed graph. On $K$, we use the node $0$ to denote “stable” and $1$ “advanced”. In addition, we use those two node conversely on $L$: The node $1$ means “stable” and the node $0$ means “advanced”. This is the key of this problem. Thus we can say as follows.
• For each $i \in N$ with $i$ not an isolated point either on $K$ or on $L$,
• If we connect $K$ with $0$ and $L$ with $1$, we have a penalty.
• Provided $P _ i = Q _ i$, if we connect $K$ with $1$ and $L$ with $0$, we have a penalty. #### Other cases
Other cases are rather simple.
• Suppose $i \in N$ is an isolated point both on $K$ and on $L$, we have a penalty.
• Suppose $i \in N$ is not an isolated point on $L$ but on $K$, if we connect $L$ with $0$, we have a penalty.
• Suppose $i \in N$ is not an isolated point on $K$ but on $L$, if we connect $K$ with $1$, we have a penalty. ### Memo on minimum cut
To solve the problem by minimum cut, how to convert constraints into edges is critical. To solve this problem, I felt difficulties on this point. So I write here a memo.
We consider the directed graph on $S$ to $T$. We use those characters S and T to denote the states of $S$ and $T$. Tips here are to consider the state the cut itself is done.
#### Simple constraints
Consider a node $A$.
• The state “if $A$ is S, we have a penalty $x$” is presented as an edge $e = (A, T, x)$.
• This is because to make cut with $A$ being the side $S$, we have to eliminate $e$; otherwise there would be a path $S \to A \to T$, which would not be a cut.
• The state “if $A$ is T, we have a penalty $x$” is presented as an edge $e = (S, A, x)$.
#### Complicated constraints
Consider two nodes $A$ and $B$.
• The state “if $A$ is S and $B$ is T, we have a penalty $x$” is presented as an edge $e = (A, B, x)$.
• Suppose we have made a cut. There are four possibilities. If both $A$ and $B$ is S, we don’t have to eliminate $e$ to make a minimum cut since it is meaningless. The case that both $A$ and $B$ is T is the same. If $A$ is T and $B$ is S, we don’t have to eliminate $e$ to make a minimum cut since $e$ is just a reversely directed edge. So eliminating $e$ means that the cut is such that $A$ is S and $B$ is T. On the contrary, suppose we have made such a cut that $A$ is S and $B$ is T. Then, we must have eliminated $e$; otherwise we would have a path $S \to A \to ^ e B \to T$, which would not be a cut. Therefore we eliminate $e$ if and only if we make a cut with $A$ being S and $B$ being T.
• The state “it is impossible that $A$ is S and $B$ is T” is presented as an edge $e = (A, B, \infty)$.
Tags:
Categories:
|
HuggingFaceTB/finemath
|
|
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
Lesson 5: Evaluating expressions with multiple variables
# Evaluating expressions with two variables
Evaluating expressions with multiple variables involves substituting given values for each variable and simplifying the expression. By replacing variables with their corresponding values, we can easily compute the result of expressions, even for more complex examples with multiple terms and operations. Created by Sal Khan.
## Want to join the conversation?
• For the second expression, 3(2)-2+3(3)= 5 not 13
I thought you were suppose to add before you subtract?
• (1.) 3(2) - 2 + 3(3) (Multiply 3 by 2 and 3 by 3 because multiplication comes before addition or subtraction.)
(2.) 6 - 2 + 9 (Because we move from left to right, we subtract first.)
(3.) 4 + 9 = 13
• I thought PEMDAS is applied to Algebra? The last example that Sal gave made me confused. Can someone help me?
• MD (Multiplication and Division) and AS (Addition and Subtraction) go left to right. If a division problem comes before a multiplication problem, you do the division first.
• What does it mean when numbers are in parentheses?
• That means that you simply need to do the math inside those parentheses first.
Example:
2 + 3 * 5 = 2 + 15 = 17
(2 + 3) * 5 = 6 * 5 = 30
• When the equation comes to "6 - 2 + 9" would the answer not be minus 5? should you not add 9 and 2 and then subtract because of the order of operations ?
• This can be a common misunderstanding of the "order of operations" if you are using a mnemonic to remember them by. Multiplication and Division happen at the same time, not multiply first, then divide, and the same for addition and subtraction. You don't do all the adding and then do all the subtraction, you do them at once (unless there are brackets/parenthesis of course).
So in this case, it is
6 plus -2 plus 9 and you can put those in any order too
9 plus 6 plus -2 / -2 plus 9 plus 6 / -2 plus 6 plus 9 etc.
• Can somebody explain the PEMDAS
• P - Parenthesis
E - Exponents
M - Multiplication
D - Division
S - Subtraction
The order of which you solve an equation.
• can i write the dot which represent (x)times on any topic
• Yes, it is just another way of showing multiplication.
• So do we do BEDMAS to work out the answer or just the order of expressions?
• PEMDAS is only used to work out the order of operations, or which expressions to calculate first.
• I'm so confused. HOW are you deciding what numbers each of the letters are? How are you deciding the a=7, b=2, x=3 and y=2?
• In this situation, he's pretty much just picking random numbers. He's just showing you the concept so that when you do more complicated stuff, and have exact values for the variables, you'll know how to plug them into equations :)
|
HuggingFaceTB/finemath
|
|
## Precalculus (10th Edition)
$\frac{1}{cos\theta}$
The identity says: $tan(\theta)=\frac{sin\theta}{cos\theta}$ and $csc(\theta)=\frac{1}{sin(\theta)}$. Hence $tan(\theta)\cdot csc(\theta)=\frac{sin\theta}{cos\theta}\cdot \frac{1}{sin(\theta)}=\frac{1}{cos\theta}$
|
HuggingFaceTB/finemath
|
|
# How to Solve a Square of a Binomial
## Transcript
How to solve the square of a binomial. I'm Bon Crowder with MathFour,com and we're squaring a binomial. So what does it mean to square something? It means there's two copies of it multiplied with each other. So squaring a binomial means that we have two identical copies and they're being multiplied. So how do you do this. Now you can either do it the long way which is using the distributive property. Some people like to use foil but that can get a little hairy. So we do this and we have three x times three x is nine x squared. Three x times one is three x, one times three x is three x and then one times one is one. So then we have total by combining these like terms nine x squared plus six x plus one. Now if you notice something really interesting, if we square this one we get a one here, if we square this three x we get a nine x squared here and if we multiply these two together and double it which is what happened here then we get six x. So that's kind of the short way rather than the long way to square a binomial. I'm Bon Crowder with MathFour.com. Enjoy.
Bon Crowder has taught math to over 15,000 adults in living rooms, classrooms and conferences.
|
HuggingFaceTB/finemath
|
|
# How to visualize the voting choices in the moderator elections?
I'm trying to use Mathematica to create some visual representations of the votes cast in the recent Mathematica-StackExchange moderator elections. (Not the vote-counting system itself, OpenSTV, but rather the choices people made.) This is what I've got so far, but I'm stuck on how to represent the data visually, not knowing much about charting.
According to this Meta post, the results can be downloaded in text form from this page. It looks to be a plain text file (with oddball Unicode characters):
data = Import[
(*
{{6, 3}, {1, 1, 2, 3, 0}, {1, 2, 4, 3, 0}, {1, 1, 4, 2, 0}, {1, 3, 1, 5, 0},
.... {"Verbeia"}, {"Mr.Wizard"}, {"F'x"}, {"Eiyrioü von Kauyf"}, {"Mathematica Moderator Election 2012"}}
*)
Although I couldn't find a guide to the format, it looks reasonable to assume that the initial {6,3} is the number of candidates and the number of vacancies, that the strings at the end before the election name are the candidates names, and that the lists starting with 1 and ending with 0 are the individual votes. So:
votes = Cases[data, {1, a_, b_, c_, 0} -> {a, b, c}]
(*
{{1, 2, 3}, {2, 4, 3}, ... {1, 3, 5}}
*)
which I suppose can be turned into something more readable by picking up the names from the end of the list:
votes1 = votes /.
Table[
n -> First@data[[Rescale[n, {1, 6}, {-7, -2}]]],
{n, 1, 6}]
to give something like this:
(*
{{"R.M.", "J.M", "Verbeia"}, {"J.M", "Mr.Wizard", "Verbeia"},
{"R.M.", "Mr.Wizard", "J.M"}, {"Verbeia", "R.M.", "F'x"},
{"Mr.Wizard", "J.M", "Verbeia"}, {"J.M", "R.M.", "Mr.Wizard"} ...
*)
So now I'd like to make this data accessible visually, using charts or graphs. (Or, as the online help says, "produce compelling dynamic visualizations".) How could I do this?
-
"(with oddball Unicode characters)" - blame the toad. ;) – Ｊ. Ｍ. Sep 26 '12 at 11:00
What about perceived co-operative strength (well, at least something derived from counting the times two candidates were mentioned together):
t = 1/Table[
Count[votes, _?(MemberQ[#, i] && MemberQ[#, j] &)], {i, 6}, {j, 6}]/.
ComplexInfinity -> DirectedInfinity[1] // Quiet
Do[t[[i, i]] = DirectedInfinity[1], {i, 6}];
EdgeShapeFunction -> ({CapForm["Round"],
Thickness[1/t[[#2[[1]], #2[[2]]]]/1000], Line[#1]} &),
VertexLabels -> Thread[Rule[Range[6], Placed[#, Above] & /@ Flatten@data[[-7 ;; -2]]]],
VertexLabelStyle -> Directive[FontFamily -> "Arial", Bold, 16],
ImagePadding -> {{30, 60}, {0, 10}}
]
-
That is an interesting way of visualizing the result, +1. – rcollyer Sep 26 '12 at 15:28
With small tables of values, complex graphics can obscure the data. Ed Tufte has recommended just showing the counts. He also points out the worth of presenting the values in a meaningful order: here, the rows go from first to third place while the columns are (roughly) in order of the standings.
raw = Import[
"Table", CharacterEncoding -> "UTF8"];
votes = Cases[raw, {1, a_, b_, c_, 0} -> {a, b, c}];
names = {"--", "RM", "JM", "Vb", "Wz", "Fx", "EK"};
order = {5, 2, 4, 3, 6, 7, 1};
table[i_, j_] := Length[Select[votes, #[[i]] == j - 1 &]]
TableForm[t = Table[table[i, j], {i, 1, 3}, {j, order}],
TableHeadings -> {{"First", "Second", "Third"}, names[[order]]}]
It might help to supplement such a table with a gentle graphic symbolization of its values. The following varies the intensity of a background disk and the size and shading of the text itself:
With[{w = 1/4},
Rasterize @
TableForm[
Map[Graphics[{RGBColor @@ ( (6 + {#, 0.2 #, 0.2 #})/(32 + 6)),
Disk[{w/2, w/2}, w/2], GrayLevel[0.9*UnitStep[16 - #]],
Text[Style[#, 120 w (# + 32)/64,
TextAlignment -> Center], {w/2, w/2}]}, ImageSize -> 50] &,
t, {2}], TableAlignments -> Center, TableSpacing -> {1/4, 1/4},
TableHeadings -> {{"First", "Second", "Third"}, names[[order]]}]]
This illuminates why each person placed as they did: there were clear choices for first, second, and third places along with a strong fourth candidate.
Those interested in exploring connections within the voting will also appreciate the graphical display of Sjoerd C. de Vries (which shows bivariate associations) and might also want to look into ways of visualizing the hypergraph of votes (where the candidates form vertices which are connected in triples by weighted 2-simplices indicating how many times all three candidates were chosen simultaneously by a voter).
-
I like the second one very much. It's telling a story using numbers and colours. Did you just guess at order? – cormullion Sep 27 '12 at 11:35
@cor The order could be justified as a sort on total votes (with non-votes placed last), but the fact is that while fiddling with the graphical details, I adjusted an initial ordering to emphasize the diagonal 32-26-23 pattern. – whuber Sep 27 '12 at 13:04
If we assume that the votes are listed in the order they were cast, we can look at how the race evolved over time during the election period and monitor step by step who were in the leading positions. In this case I'm just calculating the number of votes without taking into account the rank.
Here the very dirty code to create this animation:
data = Import[
"Table", CharacterEncoding -> "UTF8"];
votes = Cases[data, {1, a_, b_, c_, 0} -> {a, b, c}];
Table[n -> First@data[[Rescale[n, {1, 6}, {-7, -2}]]], {n, 1, 6}];
counter = Transpose[{cand, Table[0, {Length[cand]}]}];
counterList = {counter};
Flatten[Table[(Table[
counter[[Position[counter, votes2[[j, i]]][[1, 1]], 2]] =
counter[[Position[counter, votes2[[j, i]]][[1, 1]], 2]] + 1,
{i, 3}]; AppendTo[counterList, counter]), {j, Length[votes2]}],
1];
seq = Table[Show[
ListLinePlot[
Table[counterList[[1 ;; v, i, 2]], {i, Length[cand]}],
PlotStyle -> Orange, PlotRange -> {{0, 80}, {0, 80}},
AspectRatio -> 1,
Graphics[
Table[{Orange, Disk[{v, counterList[[v, i, 2]]}, .75]}, {i,
Length[cand]}]],
(o = Sort[counterList[[v]], #1[[2]] > #2[[2]] &][[1 ;; 4, 1]];
Graphics[
{Text[Style["Position:", Bold, Medium], {10, 75}, {-1, 0}],
Text["1. " <> o[[1]], {10, 70}, {-1, 0}],
Text["2. " <> o[[2]], {10, 65}, {-1, 0}],
Text["3. " <> o[[3]], {10, 60}, {-1, 0}],
Text["4. " <> o[[4]], {10, 55}, {-1, 0}]}
Export["image.gif", seq, "DisplayDurations" -> .25]
-
Love it. One day all elections will be like this, only live! – cormullion Sep 27 '12 at 16:18
@cormullion In this case, I propose "spermatozoon race plot" as the name for this plot and hope that Microsoft doesn't try to patent it like Tufte's sparklines. – VLC Sep 27 '12 at 16:43
Haha, very nice! But I don't think it is in order, because then there would be a 1:1 correspondence with when the Constituent badges were awarded, leading to a loss in anonymity of votes. Also, votes can be changed anytime during the 7 days, further complicating it. Nevertheless, you get a +1 for a very interesting idea! :) – rm -rf Sep 27 '12 at 17:16
How about a BubbleChart3D showing the three choices as 3 axes:
names = {"??", "RM", "JM", "Vb", "Wz", "Fx", "EK"};
data = Tooltip[Flatten[{##}],
ToString[names[[1 + #1]]] <> " : " <> ToString[#2]] & @@@ Tally[votes];
With[{tt = Transpose[{Range[0, 6], names}]},
BubbleChart3D[data,
AxesLabel -> Framed /@ {"First", "Second", "Third"},
Ticks -> {tt, tt, tt}, BaseStyle -> {FontFamily -> "Calibri", 16},
ChartStyle -> "DarkRainbow"]]
Hover over a bubble for a tooltip showing the 3 choices and the number of people who voted that way.
Another one
Here each voter is a vertex in a graph, with 7 larger vertices for the candidates. Each vote is a directed edge from voter to candidate (there is no distinction between first, second and third choices). By using "SpringEmbedding" the graph layout algorithm positions the candidates closer together if a lot of voters selected them both. This therefore reveals similar information to Sjoerd's graph.
votes = 1 + Cases[data, {1, a_, b_, c_, 0} :> {a, b, c}];
names = {"??", "RM", "JM", "Vb", "Wz", "Fx", "EK"};
g = Graph[Range[7 + Length@votes], v,
VertexSize -> Table[i -> 2, {i, 7}],
VertexStyle -> Table[i -> White, {i, 7}],
GraphLayout -> "SpringEmbedding"];
vc = (VertexCoordinates /. AbsoluteOptions[g, VertexCoordinates])[[;;7]];
Show[g, Graphics[{White, Table[{Disk[vc[[i]], 0.13],
Text[Style[names[[i]], Red, 14], vc[[i]]]}, {i, 7}]}]]
You can see for example a clear separation between Verbeia and the {Wizard, RM, JM} cluster. This comes from her popularity with the less conventional voters (those who also voted for the less popular candidates).
And Another One
I wanted to look at the preference aspect of the votes. For example if someone votes {Fx, JM, Wz} it implies that they consider Fx better than all the other candidates, JM better than all the candidates except Fx, and so on. We can draw up a table showing the number of voters expressing a preference for candidate A over candidate B:
votes = 1 + Cases[data, {1, a_, b_, c_, 0} :> {a, b, c}];
names = {"Wz", "RM", "JM", "Vb", "Fx", "EK", "??"};
z = ConstantArray[0, {7, 7}];
With[{r = Range[7]}, Do[Module[{a, b, c}, {a, b, c} = vote;
z[[Complement[r, {a}], a]]++;
z[[Complement[r, {a, b}], b]]++;
z[[Complement[r, {a, b, c}], c]]++],
FontFamily -> "Calibri", 16], {"Preferred", "Over"}, {Top, Left}]
For example there were 30 voters who preferred RM to Mr Wizard, but 45 voters who preferred Mr Wizard to RM.
The information is clearer if we look at the net preference between two candidates, so we could say Mr Wizard "beats" RM by 15 votes.
scoring = (z - Transpose[z]) /. x_ /; x < 1 -> ".";
FontFamily -> "Calibri", 16], {"Preferred", "Over"}, {Top, Left}]
The direction of the net preferences can be shown in a graph, giving a picture of overall voter preference. An arrow from A to B indicates an overall preference for candidate B over candidate A:
-
Lovely figure, though I notice my brain has difficulties drawing meaningful conclusions from this. – Sjoerd C. de Vries Sep 26 '12 at 15:58
@SjoerdC.deVries “bubbles bubbles bubbles bubbles” is the only conclusion my brain can draw from this graph :) – F'x Sep 26 '12 at 16:02
Where is the toad? :D – belisarius Sep 26 '12 at 16:36
@SjoerdC.deVries, fair criticism. I suppose you could say I have visualised the data but concealed the information... – Simon Woods Sep 26 '12 at 16:47
I like it, although the axes are all equivalent, rather than one being much more important than the other. Cool idea, though! – cormullion Sep 27 '12 at 11:13
Another interesting way of looking at the voting data would be to visualize the 2nd and 3rd choices for a given 1st choice. In a way, this is a breakdown of Sjoerd's plot when a given candidate is the 1st choice.
For example, among those who voted for me as the 1st choice, they were evenly split between J. M., Mr.Wizard and Verbeia for 2nd choice and a similar even split between the three for third, with Verbeia having a slight edge.
On the other hand, those who voted for Mr.Wizard as the 1st choice strongly preferred me as the second choice and Verbeia as the third:
Similar charts for J. M. (left) and Verbeia (right) are below:
It appears to me as if the voting patterns roughly fell into two camps — one which felt there should be 2 of the old and one of the new, and the other, which felt there should be two of the new and one of the old. Both of these are very reasonable voting behaviours and it doesn't spring any surprises.
### BLT file format
The format for the .blt file is fairly simple.
1. The first number on the first line is the number of candidates and the second number is the number of seats.
2. If any candidates have withdrawn, their IDs (see point 4) will be listed in the second line with a - sign prepended (like -6, for example). If no candidates have withdrawn, this line is not included.
count, 1st choice, 2nd choice, 3rd choice, 0
In this case, the individual votes were given, but it is not uncommon to see a tallied count of unique triplets. 0 terminates each line.
4. The next 7 lines (in general, the number of candidates + 1) are the identifiers for the candidate IDs. 0 indicates no vote cast for that choice.
5. The final line in the file is the title of the election.
## Code for the above charts
I have made the structure apparent in the formatting below to aid the eye. I've also "fixed" my name so that it doesn't throw any oddball unicode characters :)
### 1: Import and filter data
blt = Import["http://mathematica.stackexchange.com/election/download-result/1", "Table",
CharacterEncoding -> "UTF8"];
{{$candidates,$seats},$withdrawn,$votes,$names,$title} = blt /.
{
{candidates_,seats_},
withdrawn___,
names__,
title_
}:>
{
{candidates,seats},
{withdrawn},
MapIndexed[#2[[1]]-1->#[[1]]&,{names}],
title[[1]]
} /. {HoldPattern[0->0] :> 0->None}
### 2: Tally 2nd and 3rd choices for a given 1st choice
Clear@tallied2nd3rdVotes
user = id/.$names; others = DeleteCases[$names[[All,2]],user];
{
{"2nd choice","3rd choice"},
(Transpose@Cases[$votes,{id,x__}:>{x}]/.$names)//Transpose
}
]
Clear@visualizeVotes
PairedBarChart[tally,
ChartLabels->{Placed[choice,Above],None,Placed[names/.None->"None","RightAxis"]},
ChartStyle->"Rainbow",
BaseStyle->{FontSize->13},
ImageSize->500
]
You then call it as visualizeVotes[4], for Mr.Wizard. The IDs can be found in \$names. There is some small glitch in the function for ID 6 (Eiyrioü von Kauyf), probably because of the lack of votes, but I'm not going to fix it.
-
That's nice. I'll see if I can make a "small multiple" (© Edward Tufte) of the whole set. – cormullion Sep 27 '12 at 11:12
+1 I am impressed by how clean and clear the code is. The visualization is informative, too (although perhaps it's more than I wanted to know :-). – whuber Sep 27 '12 at 13:01
|
open-web-math/open-web-math
|
|
Plane A leaves eastaboga traveling east at a speed twice that of plane be which is flying west. how fast is each plane traveling if they fly for 3 hours and are 1800 miles apart? - AssignmentGrade.com
# Plane A leaves eastaboga traveling east at a speed twice that of plane be which is flying west. how fast is each plane traveling if they fly for 3 hours and are 1800 miles apart?
QUESTION POSTED AT 01/06/2020 - 04:16 PM
Your doing great keep up the good work!!
## Related questions
### Part A: Billy rented a canoe at \$14 for 2 hours. If he rents the same canoe for 5 hours, he has to pay a total rent of \$29. Write an equation in the standard form to represent the total rent (y) that Billy has to pay for renting the canoe for x hours. (4 points) Part B: Write the equation obtained in Part A using function notation. (2 points) Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals. (4 points) Part A: Ax By= C
QUESTION POSTED AT 02/06/2020 - 01:53 AM
### Part A: Billy rented a canoe at \$14 for 2 hours. If he rents the same canoe for 5 hours, he has to pay a total rent of \$29. Write an equation in the standard form to represent the total rent (y) that Billy has to pay for renting the canoe for x hours. (4 points) Part B: Write the equation obtained in Part A using function notation. (2 points) Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals. (4 points) Part A: Ax By= C
QUESTION POSTED AT 02/06/2020 - 01:47 AM
### Is the data set “the number of leaves on a branch” quantitative or qualitative? If it is quantitative, is it discrete or continuous?
QUESTION POSTED AT 02/06/2020 - 01:41 AM
### If a dog walks 14/15 miles in 7/30 hours what is the rate in miles per hour?
QUESTION POSTED AT 02/06/2020 - 01:16 AM
### Emma earns \$6 each time she mows the lawn and \$8 per hour for babysitting. She is saving up to buy a new pair of jeans that cost \$48. If she mows the lawn x times and babysits for y hours, which graph shows the amount of work she needs to complete to earn at least enough to purchase the new jeans? Image for option 1 Image for option 2 Image for option 3 Image for option 4
QUESTION POSTED AT 01/06/2020 - 05:06 PM
### During a game of pool, Kyle hit a ball and it followed the path from B to C and then to D. The angles formed by the path are congruent. To the nearest tenth of an inch, which of the following represents the total distance the ball traveled before it stopped at point D?
QUESTION POSTED AT 01/06/2020 - 04:47 PM
### Time is now 11:45 AM what is the smaller angle between hour and minute hand on o'clock
QUESTION POSTED AT 01/06/2020 - 04:47 PM
### John can mow the golf course in 7 hours Larry can mow it in five hours form a rational equation that can be used to determine how long it would take to mow the golf course in if they work together
QUESTION POSTED AT 01/06/2020 - 04:46 PM
### The function f(x) = –x2 + 24x – 80 models the hourly profit, in dollars, a shop makes for selling coffee, where x is the number of cups of coffee sold, and f(x) is the amount of profit. Part A: Determine the vertex. What does this calculation mean in the context of the problem? (5 points) Part B: Determine the x-intercepts. What do these values mean in the context of the problem? (5 points) I have the vertex and the x-intercepts but i need to know what they mean in the context of the problem.
QUESTION POSTED AT 01/06/2020 - 04:46 PM
### Jim works at a furniture shop. He can assemble 88 chairs in 22 hours. If he can assemble c chairs in 1 hour, identify the equation that can be used to find the value of c.
QUESTION POSTED AT 01/06/2020 - 04:36 PM
### The function g(x) = x2 – 10x + 24 is graphed on a coordinate plane. Where will the function cross the x-axis?
QUESTION POSTED AT 01/06/2020 - 04:32 PM
### Julian walked 0.5 miles before she started jogging at an average pace of 5 miles per hour. The equation d=0.5+5t can be used to relate the total distance, d, in miles to the time, t, that Julian spent jogging. What are the independent and dependent variables?
QUESTION POSTED AT 01/06/2020 - 04:29 PM
### Which is the last step when creating and analyzing a scatterplot from a table of values? Plot the ordered pairs of the independent variable and the dependent variable on the coordinate plane. Draw or obtain a blank coordinate plane. Identify if there is a relationship between the independent and dependent events.
QUESTION POSTED AT 01/06/2020 - 04:27 PM
### Use the data in the table to describe how the electric heater temperature is changing. Then identify a function that models the data. Use your function to predict the temperature after 1 hour and 10 minutes.
QUESTION POSTED AT 01/06/2020 - 04:24 PM
### Eli left his house one afternoon to study at the library. The graph shows his distance, in miles, away from his house as a function of time, in minutes. What could explain what happened when the time was equal to 120 minutes? Eli arrived at the library. Eli rode his bicycle home from the library. Eli rode his bicycle to the store, getting farther away from his house. Eli continued to study at the library for 13 more minutes.
QUESTION POSTED AT 01/06/2020 - 04:24 PM
### Mark’s earnings vary directly with the number of hours he works and can be modeled by the equation e = 8.50h, where e is his weekly earnings, and h is the number of hours he works that week. How many hours does he have to work to earn \$340?
QUESTION POSTED AT 01/06/2020 - 04:21 PM
### The change in the x coordinates in any two points on a line in the xy plane is the ?
QUESTION POSTED AT 01/06/2020 - 04:17 PM
### 1.On a trip to Florida, Santina drove 203 miles and Carole drove 189 miles. How many more miles did Santina drive than Carole?
QUESTION POSTED AT 01/06/2020 - 04:14 PM
### The steps below show the work of a student used to calculate the number of yards in 804.5 meters. (1 mile = 1,609 meters) (1 mile = 1,760 yards) Step 1: 1,609 meters is being multiplied by conversion factor 1 mile over 804.5 meters equals 2 miles Step 2: 2 miles is being multiplied by conversion factor 1760 yards over 1 mile equals 3,520 yards Step 3: 3,520 yards (1 mile = 1,609 meters) (1 mile = 1,760 yards) How can the error in the student's work be corrected?
QUESTION POSTED AT 01/06/2020 - 04:14 PM
### Let s(t) be the position function of a body moving along a coordinate line; s(t) is measured in feet and t in seconds, where t ≥ 0. Find the position, velocity, and speed of the body at the indicated time. s(t) = 15t/(t + 5); t = 0
QUESTION POSTED AT 01/06/2020 - 04:14 PM
### Line segment AB has a length of 5 units. It is translated 3 units to the right on a coordinate plane to obtain line segment A prime B prime. What is the length of A prime B prime
QUESTION POSTED AT 01/06/2020 - 04:08 PM
### A car travels at an average speed of 52 miles per hour. How many miles does it travel in 4 hours and 15 minutes?
QUESTION POSTED AT 01/06/2020 - 04:06 PM
### A snowboarder traveled at a speed of 12 meters per second for 15 seconds. How far did the snowboarder travel?
QUESTION POSTED AT 01/06/2020 - 04:06 PM
### A person six feet tall walks around the earth. How many feet farther does the top of this person's head travel than the bottoms of his feet? Express your answer in terms of power.
QUESTION POSTED AT 01/06/2020 - 04:00 PM
### A car travels 1.9 miles from Point A to Point B. The car then turns at Point B and travels 1 mile to Point C, before heading back to Point A. The distance from Point C to Point A is 1.4 miles. If the car's path is represented by a triangle, what degree of turn did the car make, at Point B?
QUESTION POSTED AT 01/06/2020 - 03:56 PM
### A car is traveling at a rate of 70 miles per hour. what is the cars rate in kilometers per hour? how many kilometers will the car travel in 2 hours in computation assume that 1 mile is equal to 1.6 kilometers
QUESTION POSTED AT 01/06/2020 - 03:56 PM
### I surveyed students in my Math II classes to see how many hours of television they watched the night before our big test on triangles. Below are the responses of nine randomly selected students, along with the grades they each earned on that test. (Slope, Y-intercept, etc))))) ^^^^^^^^^^^^^^^^^^6 Hours Spent Watching TV Grade on Test (out of 100) 4 71 2 81 4 62 1 86 3 77 1 93 2 84 3 80 2 85 Use your equation to predict a student’s score if he/she watched 1.5 hours of TV. Round your answer to the nearest whole number.
QUESTION POSTED AT 01/06/2020 - 03:48 PM
### Wastewater is filling barrels at the rate of 13 quarts per hour. The recycling facility picks up 96 full barrels on each trip, and each barrel holds 30 quarts of wastewater. How many days are between each pickup? Show your work to receive full credit.
QUESTION POSTED AT 01/06/2020 - 03:40 PM
### 30 miles per hour.3,520 feet per minute.who runs faster and by how many miles per hour? There 5,280 feet in one mile
QUESTION POSTED AT 01/06/2020 - 03:39 PM
### Max observes the zoo and the library from a helicopter flying at a height of 100 times square root of 3 feet above the ground, as shown below: *image below* What is the distance between the zoo and the library?
QUESTION POSTED AT 01/06/2020 - 03:39 PM
|
HuggingFaceTB/finemath
|
|
# Subtracting Mixed Numbers Worksheets
How to Subtract Mixed Numbers - You may recall the definition of fixed numbers. According to its definition, a mixed number consists of an integer and a proper fraction. It can also be written as an improper fraction, in which the denominator is greater is than the numerator. Hence, a mixed number could be a set of an integer and proper fraction, and a set of improper fractions as well. For example, 3 1/5, this is the example of a mixed number having an integer and a proper fraction. While on the other hand, 17/9 is an example of a mixed number that is an improper fraction. When it comes to the subtraction of mixed numbers, it is very similar to the addition of it. In the subtraction of mixed numbers, we subtract the whole numbers together just as we do in addition to the only difference that we subtract the smaller whole number or integer from the bigger one. The bigger fractional part has to be on the top of the smaller fractional part. Now find the Least common multiple of the proper or improper fractions to make the denominators even. Now subtract the numerators. You are done with the subtraction of mixed numbers.
• ### Basic Lesson
Demonstrates the subtraction of mixed numbers. Includes practice problems.
• ### Intermediate Lesson
Shows students step by step how to subtract mixed numbers. 1. Convert to fractions with common denominator. 2. Subtract the numerators. 3.Complete.
• ### Independent Practice 1
Students subtract a series of mixed numbers. The answers can be found below.
• ### Independent Practice 2
18 problems that review all skills within the unit. The answer key is below.
• ### Homework Worksheet
12 problems to reinforce the lessons and practice pages. An example is provided.
• ### Skill Quiz
10 problems that test subtracting mixed numbers skills. Scoring matrix.
• ### Homework and Quiz Answer Key
Answers for the homework and quiz.
• ### Lesson and Practice Answer Key
Answers for both lessons and both practice sheets.
• ### Basic Lesson
Demonstrates the subtraction of mixed numbers. Includes practice problems. To subtract mixed numbers, we must first convert mixed numbers to rational numbers. We can subtract rational numbers.
• ### Intermediate Lesson
Shows students step by step how to subtract mixed numbers.
• ### Independent Practice 1
Students subtract a series of mixed numbers. The answers can be found below.
• ### Independent Practice 2
20 problems that review all skills within the unit. The answer key is below.
• ### Homework Worksheet
12 problems to reinforce the lessons and practice pages. An example is provided.
• ### Skill Quiz
10 problems that test your ability to find the difference between mixed numbers. Scoring matrix.
• ### Homework and Quiz Answer Key
Answers for the homework and quiz.
• ### Lesson and Practice Answer Key
Answers for both lessons and both practice sheets.
#### Fraction Operations To Remember
An equivalent fraction is necessary for adding and subtracting fractions. To build an equivalent fraction, use the multiplication property of 1. A fraction can be changed into another equivalent fraction by multiplying it by any form of 1.
|
HuggingFaceTB/finemath
|
|
× #### Thank you for registering.
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
```
Given that vector A+b+c=0.out of three vectors two are equal in magnitude and the magnitude of the third vector is root two times that of either of the two having equal magnitude.then the angles between vectors
```
3 years ago
``` Given that the sum of vectors is zero implies that these vectors form a triangle.Also, two vectors are equal in magnitude. So, the triangle formed by these vectors is an isosceles triangle. (Let the equal sides of triangle be 'a' and 'b' without loss of generality. Then, a = b)Also, given that c = sqrt(2) times a=> a^2 + b^2 = 2 * (a^2) = c^2Hence, the triangle formed by A, B and C obey the pythagoras theorem. Hence, it is a right angled triangle, with C being the hypotenuse.So, the angle between A and B is 90 degrees, between B and C is 135 degrees and angle between C and A is 135 degrees.
```
3 years ago
``` hello studentsFrom polygon law, three vectors having summation zero should form, a closed polygon(triangle). Since, the two vectors are having same magnitude and the third vector is2times that of either of two having equal magnitude, i.e., the triangle should be right angled triangle.Angle between A and B,α=90oAngle between B and C,β=135oAngle between A and C,γ=135o.hope it helpsThankyou and Regards
```
4 months ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Vectors
View all Questions » ### Course Features
• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
### Course Features
• 19 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions
|
HuggingFaceTB/finemath
|
|
The Life Path number is the most important due to the fact that it is your main objective in life and the primary reason you were born; natal chiron retrograde the expression number is just how you share on your own to the world and just how others see you.
The soul’s urge number makes you delighted in life and what brings you happiness, and the birthday celebration number is all the abilities and abilities and capabilities that you boil down into this lifetime with to support your life’s objective.
So I’m going to show you just how to determine the Life Course number initially because that is one of the most important number in your graph, and it is your key purpose in life, so the Life Path number is established by your day of birth.
So all we’re doing is we are simply including all of the numbers together in your day of birth till we break them down and obtain a solitary number.
So in this instance, we have December 14, 1995.
So we need to obtain a solitary digit for every section.
Initially, we need to obtain a single number for the month.
We require to obtain a solitary digit for the day of the month, and we require to get a single number for the birth year, and after that when we obtain a single digit each of these, we can include these three together.
So December is the 12th month of the year, so we need to include the one in both to obtain a solitary number.
One plus two equates to three, and for the day of the month, we need to add the one in the 4 with each other, and that will offer us number five for the day of the month.
And after that we need to include 1995 with each other.
And that will offer us a single number for the year, so one plus 9 plus nine plus 5 amounts to twenty-four, and afterwards we require to include both in the four together due to the fact that we have to simplify to one digit for the year.
So 2 plus four is equal to 6.
So as soon as we have a solitary digit for each one of these areas, a single number for the month, a single figure for the day, and a solitary figure for the year, then we’re going to add the 3 of these together, and in this instance.
It would certainly be 3 plus 5 plus six amounts to 14, and after that we need to include the one and the four together since we’re trying to damage every one of this down until we obtain a single figure.
So if you add the one in the four with each other in 14, you obtain a number five.
So in this situation, this person’s life path number is a number five.
This individual is a 14 five, and it’s crucial to remember of both last numbers that we totaled to get the Life Course number 5 since these 2 last digits are crucial to the person’s life course number.
So, when it comes to a Life Course, the number 5 might be a fourteen five due to the fact that the four equates to number 5.
They might additionally be a 2 and a 3 a twenty-three number five. Still, those 2 numbers that you included with each other to obtain that Life Course number are substantial because they inform you what energies you will require to make use of throughout your lifetime to achieve your life objective. In this situation, natal chiron retrograde, this individual has a Life Path number of number five.
Still, to fulfill their life course, the variety of a number 5, they are going to require to make use of the power of the number one and the number Four to accomplish their objective, so keep in mind of those two last numbers that you totaled to obtain your last Life Course number since those are really crucial.
A fourteen-five is going to be really different than a twenty-three-five.
If you have any inquiries concerning these two last numbers made use of to obtain your final Life Path, number remark listed below, and I will certainly try to answer your concerns now.
I wished to show you this example due to the fact that, in some circumstances, we do not break down all of the numbers to get a solitary number.
So in this instance, we have December 14, 1992, and when we added all of the numbers with each other in the month, the day, and the year, we wound up with a number 11 and 11 in numerology is a master number and the master numbers.
We do not add the two numbers together, so there are 3 master numbers in numerology, and the 3 master numbers are 11, 22, and 33.
So after you have included every one of the numbers with each other in your birth date and if you wind up with either an 11, a 22, or a 33, you will certainly not add these two figures with each other because you have a master number Life Course.
Number and the master numbers are different from the other numbers in numerology due to the fact that they hold the dual digits’ energy, and we do not include both numbers with each other in these scenarios.
So if you have either an 11, a 22, or 33, you will not include the two digits with each other.
You will certainly maintain it as is, and you have a master number as a life path.
In this circumstance, the number that we combined to obtain the 11 were 8 and 3.
Those are significant numbers in this situation since this master number 11 will require to use the eight and the 3.
To get their number 11 life function, so in 83, 11 will be a lot different from on 92 11 due to the fact that an 83 11 will certainly have to make use of the energy of the 8 in the 3 to acquire their life purpose. The 9211 will Have to use the nine and the 2 indicate receive their 11 life function.
So the birthday celebration number is possibly one of the most easily accessible number to calculate in your chart since for this number, all you have to do is include the numbers with each other of the day you were born on, so he or she was born on December 14, 1995.
So we will certainly add the one and the 4 together due to the fact that those are the numbers of the day. natal chiron retrograde
This person was birthed, so 1 plus 4 equals 5.
So this individual’s birthday celebration number is a number 5.
Now, in this instance, December 11, 1995.
This person was born upon the 11th day of the month, and 11 is a master number, so we do not include both ones with each other since 11 is a master number, and there are three master numbers in numerology, 11, 22, and 33.
So if you were born upon the 11th of a month or the 22nd of a month, you would certainly not add both digits with each other due to the fact that your birthday number is a master number.
So your birthday number is either a master number 11 or a master number 22. The master numbers are the only numbers in numerology that we do not include both figures together to get a final number.
So you will certainly maintain those two figures alone, and you will certainly not add them together.
For the last two numbers, you have utilized your day of birth to compute those 2 numbers, but also for the expression number and the soul’s impulse number, you will use the full name on your birth certification.
You’re going to use your first, center, and last name on your birth certification to calculate your expression and your heart’s impulse number, therefore we’re currently mosting likely to use the Pythagorean number system to calculate these numbers.
And it’s called the Pythagorean system due to the fact that Pythagoras, a Greek mathematician, produced it. He was the mathematician that developed the Pythagorean theorem. He is the dad of numerology, and he uncovered that all numbers hold power. They all possess a Particular vibration, therefore after discovering that all numbers hold specific power, Pythagoras developed the Pythagorean number system. From that, we have modern-day numerology today.
It is a graph with every one of the letters of the alphabet and all letters representing a particular number.
So generally, all letters in the alphabet have the power of a number.
And if you take a look at this graph, you can determine what number each letter has the energy of.
So a has the energy of a leading B has the power of a number.
Two C has the power of a number 3 and so on, and so forth.
So, for the expression number, all you need to do is include all of the letters in your full birth name, so the initial, middle, and last name on your birth certification and reduce them down to one number. natal chiron retrograde
So in this example, we have Elvis Presley.
So what I did was I looked at the number graph, and I discovered the equivalent number per letter in Elvis’s name.
So E is a five, l is a 3 V is a four.
I am a 9, and I just located all of the numbers representing every one of the letters in his name; and I included all of those numbers together, and the last number I obtained was an 81.
Then I added the eight and the one together since we need to maintain damaging these down till we get a single-digit, and when I said the eight and the one with each other, I got a nine, so Elvis’s expression number is a nine.
Now the one scenario where you would not continue to add these numbers together up until you obtained a single figure would be if you obtained a master number, so the three master numbers are 11, 22, and 33.
If you got a master number, you would not continue to include these numbers with each other.
You will certainly maintain them as either 11, 22, or 33.
So it coincides situation as it was with the life fifty percent number and the birthday celebration number.
The master numbers are one-of-a-kind, and we do not include the two digits with each other, so carrying on to the soles prompt number so sometimes the soles encourage number can be called the single number, and it can additionally be called the heart’s wish number.
These words are made use of interchangeably but understand whenever you see the sole number or heart’s need number that they are essentially the same thing as a soles advise number.
For this number, we will certainly add every one of the vowels in your complete birth name.
All of the vowels in your very first, middle, and last name on your birth certificate, and we’re going to reduce them to one figure.
So we’re going to use the same Pythagorean graph that we did before, and we’re going to search for every one of the numbers that represent the vowels in your very first, center, and surname.
So here we have Kate Middleton.
I recently simply did a video on her numerology, so I figured why not use her today.
So her initial middle and surname are Catherine, Elizabeth Middleton, so I sought out the numbers corresponding to just the vowels in her name.
So, as you can see, an equates to 1 B amounts to 5, I equates to nine, and E equals 5, and after that I did that for every one of the vowels in her full name.
And after that I simply added every one of those numbers with each other, which gave me 60, and afterwards 60 decreases to number 6 since we remember we’re just attempting to damage these numbers down till we get a solitary figure.
So in this situation, Kate’s Seoul’s urge number is a number 6.
Currently you will not remain to break down the numbers if you obtain an 11, a 22, or a 33, so, as I stated with all the various other numbers if you obtain one of these numbers, this is a master number, and we do not include both figures together so 11, 22 and 33 are master numbers.
|
HuggingFaceTB/finemath
|
|
# A Car on a Banked Curve Moving in Uniform Circular Motion
• RoboRaptor
Cartesian axes simultaneously. For example, the x-component of the normal force, Fnx is shown horizontal whilst the x-component of the weight, Fwx is shown along the incline. Which is it?The x-component of the normal force is horizontal, while the x-component of the weight is along the incline. I probably should have chosen a different axis for the x-component of the weight. I chose it along the incline because I thought it would be easier to find its components in that direction.In summary, the conversation revolved around finding the necessary forces to make a car move in a circular motion on a banked ramp. The normal force, horizontal component of the normal force, and centf
#### RoboRaptor
Homework Statement
A 1050-kg car rounds a curve of radius 72m banked at an angle of 14°. If the car is traveling at 85 km/h, will a friction force be required? If so, how much and in what direction?
Relevant Equations
Centripetal Force = (m*v^2)/R
Normal Force = Cos(x) * mg
Sin(x) = opp/hyp
g = 9.8m/s^2
First I figured out the normal force being exerted on the car using the equation above.
Cos(40°)*(1050*9.8) = 7883N
Next, I tried to find out the horizontal component of the normal force by doing:
Cos(50) * 7883 = 5067N
I figured out the angle by using certain geometrical properties. Next, I found the Centripetal Force using its corresponding equation:
F = 1050 * (23.6^2/72) = 8122 N
(I found 23.6 by dividing 85000 by 3600)
I figured out that the horizontal component of the Normal Force is not enough to give the object its circular motion so I found the difference between it and the Centripetal Force which gave me around 3055 more N needed to make the car go in a uniform circular motion.
Here is where I'm unsure of what to do. Will the frictional force of the car point directly to the center of the circular motion or will it point down towards the direction of the ramp?
Here are some pictures to better illustrate my doubts: OR Thanks in advance for the help!
• Delta2
Hello R2, !
... curve of radius 72m banked at an angle of 14°
Cos(40°)
Why 40° ? Do you know how steep that is !
I figured out the angle by using certain geometrical properties.
F = 1050 * (23.6^2/72) = 8122 N
(I found 23.6 by dividing 85000 by 3600)
Do not round off to three digits and then write down four digits. F is 8130 N.
Best is to work with synbols to the very end and only then round off sensibly after the calculation.
Will the frictional force of the car point directly to the center of the circular motion or will it point down towards the direction of the ramp?
Which way is it when the car stands still on the ramp ?
• RoboRaptor
Hey BvU!
Why 40° ? Do you know how steep that is !
I just realized how I wrote 40° instead of 14°, that was just a silly mistake on my part . However, this changes the other 50° angle to 76°.
Here's a more detailed image of my work. It's a bit messy but I hope you can understand it. Which way is it when the car stands still on the ramp ?
Not sure I understand the question. Basically, I want to know in which way the frictional force is going to be directed while the car is on the ramp moving in a uniform circular motion, if I'm not wrong the object is static if looked at from reference to the circle.
First I figured out the normal force being exerted on the car using the equation above.
Cos(40°)*(1050*9.8) = 7883N
That is the component of the gravitational force in the direction normal to the road. It is not necessarily the same as the normal force from the road.
Your diagram sure is messy. Just show the car and the forces on it. What are they, and in what directions do they act?
Their resultant produces the acceleration. What are the magnitude and direction of that?
What two equations does that allow you to write?
• RoboRaptor
That is the component of the gravitational force in the direction normal to the road. It is not necessarily the same as the normal force from the road.
I assumed that was the case because I thought Normal Force had the same magnitude as the force that is being applied perpendicular to the road, in this case, the y-component of the weight.
Our teacher taught us a way to do free-body diagrams on inclined ramps which I'm not sure if it's the norm for this type of problems, here it is: I assumed that was the case because I thought Normal Force had the same magnitude as the force that is being applied perpendicular to the road, in this case, the y-component of the weight.
That is only true if there is no acceleration component in that direction. You will find through this problem that there is such.
You can't just invent a mysterious force exerted by the car. There may be a frictional force in that position and orientation, so let’s agree it is that, but, again, you cannot just assume it is equal to another force. You have to justify such an equality by means of Newton's law ΣF=ma.
The diagram is good; it shows all the applied forces.
What is the resultant force, given the motion of the car?
What equations can you write for the forces?
Last edited:
The diagram is good; it shows all the applied forces.
The diagram may show all the forces, nevertheless it is confusing for it uses two separate Cartesian axes simultaneously. For example, the x-component of the normal force, Fnx is shown horizontal whilst the x-component of the weight, Fwx is shown along the incline. Which is it?
The diagram may show all the forces, nevertheless it is confusing for it uses two separate Cartesian axes simultaneously. For example, the x-component of the normal force, Fnx is shown horizontal whilst the x-component of the weight, Fwx is shown along the incline. Which is it?
Yes, good catch. I hadn't paid attention to Fwx and Fwy.
|
HuggingFaceTB/finemath
|
|
# Spiral of Theodorus
In geometry, the spiral of Theodorus (also called square root spiral, Einstein spiral or Pythagorean spiral) is a spiral composed of right triangles, placed edge-to-edge. It was named after Theodorus of Cyrene.
## Construction
The spiral is started with an isosceles right triangle, with each leg having unit length. Another right triangle is formed, an automedian right triangle with one leg being the hypotenuse of the prior triangle (with length 2) and the other leg having length of 1; the length of the hypotenuse of this second triangle is 3. The process then repeats; the nth triangle in the sequence is a right triangle with side lengths n and 1, and with hypotenuse n + 1. For example, the 16th triangle has sides measuring 4 (=16), 1 and hypotenuse of 17
## History and uses
Although all of Theodorus' work has been lost, Plato put Theodorus into his dialogue Theaetetus, which tells of his work. It is assumed that Theodorus had proved that all of the square roots of non-square integers from 3 to 17 are irrational by means of the Spiral of Theodorus.
Plato does not attribute the irrationality of the square root of 2 to Theodorus, because it was well known before him. Theodorus and Theaetetus split the rational numbers and irrational numbers into different categories.
## Hypotenuse
Each of the triangles' hypotenuses hn gives the square root of the corresponding natural number, with h1 = 2.
Plato, tutored by Theodorus, questioned why Theodorus stopped at 17. The reason is commonly believed to be that the 17 hypotenuse belongs to the last triangle that does not overlap the figure.
### Overlapping
In 1958, Erich Teuffel proved that no two hypotenuses will ever coincide, regardless of how far the spiral is continued. Also, if the sides of unit length are extended into a line, they will never pass through any of the other vertices of the total figure.
## Extension
Theodorus stopped his spiral at the triangle with a hypotenuse of 17. If the spiral is continued to infinitely many triangles, many more interesting characteristics are found.
### Growth rate
#### Angle
If φn is the angle of the nth triangle (or spiral segment), then:
$\tan \left(\varphi _{n}\right)={\frac {1}{\sqrt {n}}}.$ Therefore, the growth of the angle φn of the next triangle n is:
$\varphi _{n}=\arctan \left({\frac {1}{\sqrt {n}}}\right).$ The sum of the angles of the first k triangles is called the total angle φ(k) for the kth triangle. It grows proportionally to the square root of k, with a bounded correction term c2:
$\varphi \left(k\right)=\sum _{n=1}^{k}\varphi _{n}=2{\sqrt {k}}+c_{2}(k)$ where
$\lim _{k\to \infty }c_{2}(k)=-2.157782996659\ldots$ ().
The growth of the radius of the spiral at a certain triangle n is
$\Delta r={\sqrt {n+1}}-{\sqrt {n}}.$ ### Archimedean spiral
The Spiral of Theodorus approximates the Archimedean spiral. Just as the distance between two windings of the Archimedean spiral equals mathematical constant pi, as the number of spins of the spiral of Theodorus approaches infinity, the distance between two consecutive windings quickly approaches π.
The following is a table showing of two windings of the spiral approaching pi:
Winding No.: Calculated average winding-distance Accuracy of average winding-distance in comparison to π
2 3.1592037 99.44255%
3 3.1443455 99.91245%
4 3.14428 99.91453%
5 3.142395 99.97447%
→ π → 100%
As shown, after only the fifth winding, the distance is a 99.96% accurate approximation to π.
## Continuous curve Davis' analytic continuation of the Spiral of Theodorus, including extension in the opposite direction from the origin (negative nodes numbers).
The question of how to interpolate the discrete points of the spiral of Theodorus by a smooth curve was proposed and answered in (Davis 2001, pp. 37–38) by analogy with Euler's formula for the gamma function as an interpolant for the factorial function. Davis found the function
$T(x)=\prod _{k=1}^{\infty }{\frac {1+i/{\sqrt {k}}}{1+i/{\sqrt {x+k}}}}\qquad (-1 which was further studied by his student Leader and by Iserles (in an appendix to (Davis 2001) ). An axiomatic characterization of this function is given in (Gronau 2004) as the unique function that satisfies the functional equation
$f(x+1)=\left(1+{\frac {i}{\sqrt {x+1}}}\right)\cdot f(x),$ the initial condition $f(0)=1,$ and monotonicity in both argument and modulus; alternative conditions and weakenings are also studied therein. An alternative derivation is given in (Heuvers, Moak & Boursaw 2000).
An analytic continuation of Davis' continuous form of the Spiral of Theodorus which extends in the opposite direction from the origin is given in (Waldvogel 2009).
In the figure the nodes of the original (discrete) Theodorus spiral are shown as small green circles. The blue ones are those, added in the opposite direction of the spiral. Only nodes $n$ with the integer value of the polar radius $r_{n}=\pm {\sqrt {|n|}}$ are numbered in the figure. The dashed circle in the coordinate origin $O$ is the circle of curvature at $O$ .
|
HuggingFaceTB/finemath
|
|
assignmentutor-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 数学代写|概率论代写Probability theory代考|Polya Criterion
If the following conditions are met:
1) $\psi(0)=1$
2) $\psi(t)$ – continuous function
3) $\psi(t)$ – even function
4) $\psi(t)$ – convex on the positive semiaxis
5) $\lim \psi(t)=0$ for $t \rightarrow \infty$,
then $\psi(t)$ is a characteristic function.
Comment. We also note that we can give a somewhat more general than
5) condition in this criterion, replacing it with the relation
$\lim (t)=p$ for $t \rightarrow \infty$, where $0 \leq p \leq 1$.
The proof of the Polya criterion uses the above fact (see (3.2.29)), consisting the fact that the triangular function described by the equalities $\psi(t)=1-t$, if $-1 \leq t \leq 1$, and $\psi(t)=0$ if $|t|>1$, is characteristic. We immediately obtain that the following triangular functions
$\psi_A(t)=\psi(A t)$ us also characteristic function for any $A>0$.
Turning to a probabilistic mixture
$$\psi(t)=p_1 \psi_A(t)+p_2 \psi_B(t)$$
where $p_1+p_2=1, \psi_A(t)$ and $\psi_B(t$ are two characteristic functions then $\psi(t)$ will be a characteristic functuin,. The same kind of mixtures of a finite number of different triangular characteristic functions are considered, and then with their help, passing from a finite to an infinite number of functions that make up these mixtures, one can obtain any function described in the Polya statement.
If we recall the above remark, in which condition (3.2.42) appears, then we can, comparing the following two obviously characteristic functions, one of which tends to zero at infinity, and the second tends to some $p(0<p<1)$, get an example of characteristic functions that coincide only on some finite interval.
On the other hand, returning to the characteristic functions $\psi(t)$ and $\psi_B(t)$ indicated above, $0<A<B$, we see that the values of these functions coincide everywhere, except for values at two finite intervals $(-B, 0)$ and $(0, B)$
The above constructions of characteristic functions allow one to obtain an interesting result.
## 数学代写|概率论代写Probability theory代考|ARCSINE DISTRIBUTION
A random variable $\mathrm{X}$ is said to have arcsine distribution if its $\operatorname{pdf} \mathrm{f}(\mathrm{x})$ is as follows.
$$f(x)=\frac{1}{\pi \sqrt{(x(1-x))}} .$$
The pdf of arcsine distribution is given in Figure 4.1.
Mean $=\frac{1}{2}$
Variance $-\frac{1}{8}$
$$\mathrm{E}\left(X^n\right)=\frac{(2 n) !}{4^n(n !)^2}$$
Moment generating function $\mathrm{M}(\mathrm{t})=\sum_{k=0}^{\infty} \frac{1}{4^{k} !} \frac{(2 k) !}{k ! k !} t^k$
Characteristic function $\varphi(t)=\sum_{k=0}^{\infty} \frac{1}{4^k} \frac{(2 k) !}{k ! k !}(i t)^k$
# 概率论代考
## 数学代写|概率论代写Probability theory代考|Polya Criterion
1) $\psi(0)=1$
2) $\psi(t)-$ 连续函数
3) $\psi(t)-$ 偶数函数
4) $\psi(t)-$ 在正半轴上凸出
5) $\lim \psi(t)=0$ 为了 $t \rightarrow \infty$,
5) 更一般的条件，将其替换为关系
$\lim (t)=p$ 为了 $t \rightarrow \infty$ ，在哪里 $0 \leq p \leq 1$.
Polya 准则的证明使用上述事实 (见 (3.2.29))，包括由等式描述的三角函数的事实 $\psi(t)=1-t$ ，如果 $-1 \leq t \leq 1$ ，和 $\psi(t)=0$ 如果 $|t|>1$ ，是特征。我们立 即得到以下三角函数
$\psi_A(t)=\psi(A t)$ 我们还具有任何特征的功能 $A>0$.
$$\psi(t)=p_1 \psi_A(t)+p_2 \psi_B(t)$$
## 数学代写|概率论代写Probability theory代考|ARCSINE DISTRIBUTION
$$f(x)=\frac{1}{\pi \sqrt{(x(1-x))}} .$$
$$\mathrm{E}\left(X^n\right)=\frac{(2 n) !}{4^n(n !)^2}$$
## 有限元方法代写
assignmentutor™作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师
|
HuggingFaceTB/finemath
|
|
# Search by Topic
#### Resources tagged with Combinatorics similar to Loose Change:
Filter by: Content type:
Stage:
Challenge level:
### There are 40 results
Broad Topics > Decision Mathematics and Combinatorics > Combinatorics
### Flagging
##### Stage: 3 Challenge Level:
How many tricolour flags are possible with 5 available colours such that two adjacent stripes must NOT be the same colour. What about 256 colours?
##### Stage: 3 Challenge Level:
Is it possible to use all 28 dominoes arranging them in squares of four? What patterns can you see in the solution(s)?
### An Investigation Based on Score
##### Stage: 3
Class 2YP from Madras College was inspired by the problem in NRICH to work out in how many ways the number 1999 could be expressed as the sum of 3 odd numbers, and this is their solution.
### Painting Cubes
##### Stage: 3 Challenge Level:
Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours?
### Ways of Summing Odd Numbers
##### Stage: 3
Sanjay Joshi, age 17, The Perse Boys School, Cambridge followed up the Madrass College class 2YP article with more thoughts on the problem of the number of ways of expressing an integer as the sum. . . .
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Euromaths
##### Stage: 3 Challenge Level:
How many ways can you write the word EUROMATHS by starting at the top left hand corner and taking the next letter by stepping one step down or one step to the right in a 5x5 array?
### Bell Ringing
##### Stage: 3 Challenge Level:
Suppose you are a bellringer. Can you find the changes so that, starting and ending with a round, all the 24 possible permutations are rung once each and only once?
### Permute It
##### Stage: 3 Challenge Level:
Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.
### Tri-colour
##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### Paving Paths
##### Stage: 3 Challenge Level:
How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs?
### Master Minding
##### Stage: 3 Challenge Level:
Your partner chooses two beads and places them side by side behind a screen. What is the minimum number of guesses you would need to be sure of guessing the two beads and their positions?
### Small Change
##### Stage: 3 Challenge Level:
In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
### Penta Colour
##### Stage: 4 Challenge Level:
In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?
### Greetings
##### Stage: 3 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
### Euler's Officers
##### Stage: 4 Challenge Level:
How many different ways can you arrange the officers in a square?
### Cube Paths
##### Stage: 3 Challenge Level:
Given a 2 by 2 by 2 skeletal cube with one route `down' the cube. How many routes are there from A to B?
### Doodles
##### Stage: 4 Challenge Level:
Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
How many noughts are at the end of these giant numbers?
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Magic Caterpillars
##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
### Plum Tree
##### Stage: 4 and 5 Challenge Level:
Label this plum tree graph to make it totally magic!
### Scratch Cards
##### Stage: 4 Challenge Level:
To win on a scratch card you have to uncover three numbers that add up to more than fifteen. What is the probability of winning a prize?
### Olympic Magic
##### Stage: 4 Challenge Level:
in how many ways can you place the numbers 1, 2, 3 … 9 in the nine regions of the Olympic Emblem (5 overlapping circles) so that the amount in each ring is the same?
##### Stage: 4 and 5
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots, prime knots, crossing numbers and knot arithmetic.
### Deep Roots
##### Stage: 4 Challenge Level:
Find integer solutions to: $\sqrt{a+b\sqrt{x}} + \sqrt{c+d.\sqrt{x}}=1$
### Russian Cubes
##### Stage: 4 Challenge Level:
I want some cubes painted with three blue faces and three red faces. How many different cubes can be painted like that?
### Lost in Space
##### Stage: 4 Challenge Level:
How many ways are there to count 1 - 2 - 3 in the array of triangular numbers? What happens with larger arrays? Can you predict for any size array?
### One Basket or Group Photo
##### Stage: 2, 3, 4 and 5 Challenge Level:
Libby Jared helped to set up NRICH and this is one of her favourite problems. It's a problem suitable for a wide age range and best tackled practically.
### Molecular Sequencer
##### Stage: 4 and 5 Challenge Level:
Investigate the molecular masses in this sequence of molecules and deduce which molecule has been analysed in the mass spectrometer.
### Symmetric Tangles
##### Stage: 4
The tangles created by the twists and turns of the Conway rope trick are surprisingly symmetrical. Here's why!
### Tangles
##### Stage: 3 and 4
A personal investigation of Conway's Rational Tangles. What were the interesting questions that needed to be asked, and where did they lead?
### Knight Defeated
##### Stage: 4 Challenge Level:
The knight's move on a chess board is 2 steps in one direction and one step in the other direction. Prove that a knight cannot visit every square on the board once and only (a tour) on a 2 by n board. . . .
### In a Box
##### Stage: 4 Challenge Level:
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair?
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Counting Binary Ops
##### Stage: 4 Challenge Level:
How many ways can the terms in an ordered list be combined by repeating a single binary operation. Show that for 4 terms there are 5 cases and find the number of cases for 5 terms and 6 terms.
### Snowman
##### Stage: 4 Challenge Level:
All the words in the Snowman language consist of exactly seven letters formed from the letters {s, no, wm, an). How many words are there in the Snowman language?
### Magic W
##### Stage: 4 Challenge Level:
Find all the ways of placing the numbers 1 to 9 on a W shape, with 3 numbers on each leg, so that each set of 3 numbers has the same total.
|
HuggingFaceTB/finemath
|
|
## 41472
41,472 (forty-one thousand four hundred seventy-two) is an even five-digits composite number following 41471 and preceding 41473. In scientific notation, it is written as 4.1472 × 104. The sum of its digits is 18. It has a total of 13 prime factors and 50 positive divisors. There are 13,824 positive integers (up to 41472) that are relatively prime to 41472.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 18
• Digital Root 9
## Name
Short name 41 thousand 472 forty-one thousand four hundred seventy-two
## Notation
Scientific notation 4.1472 × 104 41.472 × 103
## Prime Factorization of 41472
Prime Factorization 29 × 34
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 13 Total number of prime factors rad(n) 6 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 41,472 is 29 × 34. Since it has a total of 13 prime factors, 41,472 is a composite number.
## Divisors of 41472
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 256, 288, 324, 384, 432, 512, 576, 648, 768, 864, 1152, 1296, 1536, 1728, 2304, 2592, 3456, 4608, 5184, 6912, 10368, 13824, 20736, 41472
50 divisors
Even divisors 45 5 3 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 50 Total number of the positive divisors of n σ(n) 123783 Sum of all the positive divisors of n s(n) 82311 Sum of the proper positive divisors of n A(n) 2475.66 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 203.647 Returns the nth root of the product of n divisors H(n) 16.7519 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 41,472 can be divided by 50 positive divisors (out of which 45 are even, and 5 are odd). The sum of these divisors (counting 41,472) is 123,783, the average is 2,475,.66.
## Other Arithmetic Functions (n = 41472)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 13824 Total number of positive integers not greater than n that are coprime to n λ(n) 6912 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4339 Total number of primes less than or equal to n r2(n) 4 The number of ways n can be represented as the sum of 2 squares
There are 13,824 positive integers (less than 41,472) that are coprime with 41,472. And there are approximately 4,339 prime numbers less than or equal to 41,472.
## Divisibility of 41472
m n mod m 2 3 4 5 6 7 8 9 0 0 0 2 0 4 0 0
The number 41,472 is divisible by 2, 3, 4, 6, 8 and 9.
## Classification of 41472
• Abundant
### Expressible via specific sums
• Polite
• Practical
• Non-hypotenuse
• Achilles
• Powerful
• Frugal
• Regular
## Base conversion (41472)
Base System Value
2 Binary 1010001000000000
3 Ternary 2002220000
4 Quaternary 22020000
5 Quinary 2311342
6 Senary 520000
8 Octal 121000
10 Decimal 41472
12 Duodecimal 20000
20 Vigesimal 53dc
36 Base36 w00
## Basic calculations (n = 41472)
### Multiplication
n×i
n×2 82944 124416 165888 207360
### Division
ni
n⁄2 20736 13824 10368 8294.4
### Exponentiation
ni
n2 1719926784 71328803586048 2958148142320582656 122680319758319203909632
### Nth Root
i√n
2√n 203.647 34.614 14.2705 8.38593
## 41472 as geometric shapes
### Circle
Diameter 82944 260576 5.40331e+09
### Sphere
Volume 2.98781e+14 2.16132e+10 260576
### Square
Length = n
Perimeter 165888 1.71993e+09 58650.3
### Cube
Length = n
Surface area 1.03196e+10 7.13288e+13 71831.6
### Equilateral Triangle
Length = n
Perimeter 124416 7.4475e+08 35915.8
### Triangular Pyramid
Length = n
Surface area 2.979e+09 8.40618e+12 33861.7
## Cryptographic Hash Functions
md5 ee6d340033cb7d06ecbe9aa11e6f1229 be2877bd952e5f57247eed1400050ecaabae159e 31546b2369a66e2a698472704ce4d43afcf429834e974786ff4bc26f46bf5df4 b563192e69a598452b6d4b2f292f41222e6a4fe05d5b70de2bb25afb8c97eb2eeffe348522250e201565b123837116d381c19735df3e24525c097e3ef97ce0a0 15c746d8af2e1ecf6cf99db12bd3ddf259990143
|
HuggingFaceTB/finemath
|
|
# Determine the magnitude of P and F necessary to keep the concurrent force system in Fig. below in equilibrium.
Question-AnswerCategory: Engineering MechanicsDetermine the magnitude of P and F necessary to keep the concurrent force system in Fig. below in equilibrium.
Mazurek Gravity Staff asked 1 year ago
## Determine the magnitude of P and F necessary to keep the concurrent force system in Fig. below in equilibrium.
Concept:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
F.Horizontal:-
ΣFV=0ΣFV=0
Fsin60=200sin45+Psin30Fsin60∘=200sin45∘+Psin30∘
(317.16+1.7320P)sin60=200sin45+Psin30(317.16+1.7320P)sin60∘=200sin45∘+Psin30∘
274.67+1.5P=141.42+0.5P274.67+1.5P=141.42+0.5P
F=317.16+1.7320(133.25)F=317.16+1.7320(−133.25)
*Solved by ~Atul / Sliet / GCT-2
ATUL / GCT-21 answered 4 months ago
F.Horizontal:-
F.Horizontal:-
→200sin315° + 300sin180° + Psin210°+ Fsin60°
→F= 2(158.6 + 0.866P)
→F= 317.2 + 1.732P
F.Vertical:-
Fsin60=200sin45+Psin30°
Now put the value of ‘F’
(317.16+1.7320P)sin60∘=200sin45∘+Psin30∘
274.67+1.5P=141.42+0.5P274.67+1.5P=141.42+0.5P
|
HuggingFaceTB/finemath
|
|
Blog Home / Study Guides & Resources / What is capital, types of capital & cost of capital?
# What is capital, types of capital & cost of capital?
We know that the two main tasks of a finance manager are the procurement of funds and their utilization. Proper procurement and utilization help in the maximization of the wealth of any business. Therefore, the finance manager has to select such a capital structure that can bring the minimum expectation of the investors and maximum wealth to the firm.
So, to execute the task effectively, the finance manager should understand the various cost of capital such as cost of debt, cost of equity, cost of preference shares, and cost of retained earnings.
## MEANING OF COST OF CAPITAL
In simple terms, think of the cost of capital as the minimum return expected by the providers of the capital. In a corporate world what happens is that when an entity procures capital then it has to pay some additional money over and above the procured money.
### Example
Suppose, entity A needs \$100,000 to finance its raw material procurement. Entity A approaches Mr. B who is an investor and offers 5% interest per annum on \$100,000 if Mr. B agrees to invest. Mr. B gives a counteroffer by asking for 6% interest at least in return for \$100,000. Since entity A is in urgent need of money, it agrees at 6% interest. Here, we call \$100,000 as capital and 6% interest as the cost of capital as it is the minimum return expected by the provider of the capital.
Don’t get mistaken here by thinking that since the entity is paying 6% interest then it is their cost of capital. No, in reality, since 6% interest is the minimum requirement of the investor hence, we are calling it as cost of capital. In the future, if 6% becomes 7% then the cost of capital for the entity will get changed due to the change in minimum expectation of the investor.
## IMPORTANCE OF UNDERSTANDING COST OF CAPITAL
Every Entity must understand how much they are earning and how much cost they are paying. It is a simple phenomenon that if the cost exceeds income, then loss incurs and when income exceeds the cost, Income generates.
Hence, at first, every entity should strategically understand how much cost they will incur after procuring finance from different sources and whether they can generate enough income to pay off the estimated cost of capital. This way, they will better judge the profitability of the business.
## FORMS OF COST OF CAPITAL
There are two major forms of cost of capital:
1. Cost of Debt
2. Cost of Equity
• ### Cost of Debt (Kd)
Debt is a form of capital where the provider of such capital becomes the lender to the business. Such capital providers do not enjoy ownership in the company.
For an entity, the cost of debt can be of two types:
• ### Cost of Irredeemable Debt
Irredeemable debt is a type of debt where the entity is not required to pay off the principal debt.
In the previous example, suppose Mr. B granted a loan of \$100,000 with 6% interest to entity A and in return ask for only interest amount during the lifetime of entity A. Here, entity A shall call \$100,000 as irredeemable debt.
Further, 6% of \$100,000 is \$6,000 is the cost of entity A but since interest is charged on profit it means every year the entity will save \$1,800, assuming the income tax rate is 30%. Hence, the after-tax cost of \$100,000 is \$4,200 which comes to 4.2%. The cost of Irredeemable debt not redeemable during the lifetime of the entity is calculated as below: Kd=$ \frac{Interest}{Net\: Proceeds\: of\: Debt} $x (1- tax) • ### Cost of Redeemable Debt Redeemable debt is a type of debt where the entity is required to pay off the principal debt along with the interest amount. Continuing the previous example, suppose Mr. B granted a loan of \$100,000 with 6% interest to entity A and in return ask for both the principal amount at the end of the loan period and interest every year till the end of the loan period then Entity A shall call \$100,000 as Redeemable debt. In the case of redeemable debt too, the tax benefit is also given on interest payments The cost of Redeemable Debt is calculated as follows: ### Kd =$ \frac{I(1-t)+\frac{RV-NP}{N}}{\frac{RV-NP}{2}} $Where, I = Interest Payment NP = Net proceeds from Debt RV = Redemption value of Debt T = Tax Rate N = Remaining life of Debt Continuing the previous example, suppose Mr. B granted a loan of \$100,000 with 6% interest to entity A and in return ask for both the principal amount at the end of the loan period which is five years, and interest every year till the end of the five years. But instead of asking \$100,000 at the end, Mr. B asked for a 5% premium on debt. Here, the cost of redeemable debt would be after assuming tax rate as 30%: ### Kd=$ \frac{$6000(1-0.30)+\frac{$105000-$100000}{5}}{\frac{$105000-$100000}{2}}$
Kd = 5.07%
• ### Cost of Equity (Ke)
Equity is a form of the capital where the provider of such capital becomes the owner of the business. Such capital providers enjoy ownership in the company. They can also participate in the decision-making of the company.
### Example
Suppose, entity A requires \$200,000 to finance its expanding business. Entity A approaches Mr. C who is an investor and offers 1000 common shares in return for \$200,000. Here, Mr. C won’t receive a fixed interest return but rather receive part ownership and dividend out of profits of the company. Just like any other source of finance, the cost of equity is the expectation of equity shareholders.
Cost of Equity can be calculated using the following methods:
1. If the dividend is expected to be constant, then Dividend Pricing Model should be used
2. If earnings per share are expected to be constant, then Earning Pricing Model should be used.
3. If dividends and earnings are expected to grow at a constant rate then Gordon’s Model should be used.
• ### Dividend Pricing Model
It is also known as the Dividend Valuation Model. In this model, an assumption is being made is that the company will pay the same dividend constantly every year.
Here, the cost of equity is computed by dividing the expected dividend by the market price of the share.
Ke = $\frac{D}{P_{0}}$
Where,
Ke = Cost of Equity
D = Expected Dividend
P0 = Market Price of Equity
Example: ABC Company is expected to give a dividend of \$1.5 and its current market price is \$50. Therefore, its cost of equity is 3%.
• ### Earnings Pricing Model
In this model, an assumption is being made is that the company will pay the same earnings (whether distributed in form of dividends or not) constantly every year. Here, investor advocates that it doesn’t matter whether the company is giving dividend or not. They are more interested in the total earnings that the company is generating. So, they value the cost of equity-based on earnings and not dividends.
Here, the cost of equity is computed by dividing the expected earnings by the market price of the share.
Ke = $\frac{E}{P_{0}}$
Where,
Ke = Cost of Equity
E = Expected earnings
P0 = Market Price of Equity
Example: ABC Company is expected to earn $3 and its current market price is \$60. Therefore, its cost of equity is 5%.
• ### Gordon’s Model
Gordon’s model is a growth-based model which attempts to derive a future growth rate. This model says that an increase in the level of investment will give rise to an increase in future dividends. As per this model, the rate of dividend growth remains constant. The cost of equity capital is calculated as follows:
Ke = $\frac{D1}{P_{0}}$ + G
Where,
D1 = [ D0 ( 1 + g) ] i.e. next expected dividend
P0 = Current Market Price per Share
G = Constant growth rate of dividend
Example: A company has paid a dividend of $2 per share (of the face value of \$20) last year and is expected to grow by 10% every year and the current market price of a share is \$110. Here cost of equity shall be calculated as follows: Ke =$ \frac{D1}{P_{0}} $+ G Ke =$ \frac{\$2(1+0.1)}{110}$ + 10
Ke = 12%
Sagar Pujari
|
open-web-math/open-web-math
|
|
# How do you solve −2(x − 5) = −4?
May 23, 2018
$x = 7$
#### Explanation:
$\text{distribute the bracket on the left side of the equation}$
$- 2 x + 10 = - 4$
$\text{subtract 10 from both sides}$
$- 2 x \cancel{+ 10} \cancel{- 10} = - 4 - 10$
$- 2 x = - 14$
$\text{divide both sides by } - 2$
$\frac{\cancel{- 2} x}{\cancel{- 2}} = \frac{- 14}{- 2}$
$x = 7$
$\textcolor{b l u e}{\text{As a check}}$
Substitute this value into the left side of the equation and if equal to the right side then it is the solution.
$- 2 \left(7 - 5\right) = - 2 \times 2 = - 4$
$\Rightarrow x = 7 \text{ is the solution}$
|
HuggingFaceTB/finemath
|
|
# physics
A 5-g marble, released form rest, was allowed to roll on a ramp whose lower end is 25 cm from the floor. When it left the ramp, it behaved like a body in projectile motion. What was the horizontal velocity of this marble when it left the ramp? With what momentum did it leave the ramp?
1. 👍 0
2. 👎 0
3. 👁 465
1. You have to know something about the ramp to answer these questions.
1. 👍 0
2. 👎 0
👨🏫
bobpursley
2. M= 1525g= 1.525 kg
W= 14.95 N
Force required to start the block into motion:
F= 0.685(9.8)= 6.71N
Force to maintain motion:
F= 0.512(9.8)= 5.02N
Compute µk and µs:
µk = 5.02/14.95 = 0.34 µs = 6.72/14.95 = 0.45
1. 👍 0
2. 👎 0
## Similar Questions
1. ### Physics
A skateboarder starts up a 1.0-m-high, 30∘ ramp at a speed of 7.2 m/s . The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air. How far from the end of the ramp does the
2. ### Physics
A skateboarder shoots off a ramp with a velocity of 6.7 m/s, directed at an angle of 53° above the horizontal. The end of the ramp is 1.0 m above the ground. Let the x axis be parallel to the ground, the +y direction be
3. ### physics
After leaving the end of a ski ramp, a ski jumper lands downhill at a point that is displaced 62.3 m horizontally from the end of the ramp. His velocity, just before landing, is 22.0 m/s and points in a direction 40.3 ° below the
4. ### physics
A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. The driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. The
1. ### physics
Hi, ive been trying to get this problem for a couple of days now but i have no idea of what to do. A very slippery block of ice slides down a smooth ramp tilted at angle pheta. a)The ice is released from rest at vertical height h
2. ### Physics
A 60kg skier starts from rest at a height of 20m above the end of a ski-jump and leaves the ramp at an angle of 28 degrees. Neglet air resistance and assume the ramp is frictionless. A) What is the max height (h) of his jump above
3. ### physics
20.) A piano has been pushed to the top of the ramp at the back of a moving van. The workers think it is safe, but as they walk away, it begins to roll down the ramp. If the back of the truck is 1.5 above the ground and the ramp
4. ### physics
An unstretched ideal spring hangs vertically from a fixed support. A 0.4 kg object is then attached to the lower end of the spring. The object is pulled down to a distance of 35cm below the unstretched position and released from
1. ### Physics
A box, with a weight of mg = 25 N, is placed at the top of a ramp and released from rest. The ramp measures 4.40 meters horizontally and 3.50 meters vertically. The box accelerates down the incline, attaining a kinetic energy at
2. ### math
There are two jars, A and B. Jar A contains 7 red and 2 blue marbles, while jar B contains 5 red and 4 blue marbles. You roll a fair die.If you get a 2 or a 3, you pick a random marble from jar A and put it in jar B. You then pick
3. ### Physics
A piano has been pushed to the top of the ramp at the back of a moving van. The workers think it is safe, but as they walk away, it begins to roll down the ramp. If the back of the truck is 1.3 above the ground and the ramp is
4. ### physics
I am having great difficulty with this question. A block is placed on a frictionless ramp at a height of 12.5 m above the ground. Starting from rest, the block slides down the ramp. At the bottom of the ramp, the block slides onto
|
HuggingFaceTB/finemath
|
|
Click to Chat
1800-2000-838
+91-120-4616500
CART 0
• 0
MY CART (5)
Use Coupon: CART20 and get 20% off on all online Study Material
ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.
There are no items in this cart.
Continue Shopping
``` The value of x satisfying (x+1)/(x-1) + (x-2)/(x+2)+(x-3)/(x+3)+(x+4)/(x-4)=4 are
(a)(-5+3451/2)/10
(b) (5+3401/2)/8
(c)(-5+3251/2)/16
(d)none```
8 years ago
17 Points
``` Hi
The given equation is
(x+1)/(x-1) + (x-2)/(x+2)+(x-3)/(x+3)+(x+4)/(x-4)=4
Write (x+1) as (x-1)+2. So the first term becomes
1+ 2/(x-1).
Similarly, the second term becomes 1 + 2/(x-2), and so on.
Do this to all four terms on the LHS. Now the equation becomes
1+ 2/(x-1) + 1+ 2/(x-2) + 1+ 2/(x-3) + 1+ 2/(x-4) = 4
or, 4 + 2/(x-1) + 2/(x-1) + 2/(x-1) + 2/(x-1) = 4
or, 2 ( 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)) = 0
or, 1/(x-1) + 1/(x-4) + 1/(x-3) + 1/(x-2) = 0
or,
[(x-4)+(x-1)]/[x2-5x+4] + [(x-3)+(x-2)]/[x2-5x+6] = 0
or,
(2x-5)/[x2-5x+4] = -(2x-5)/[x2-5x+6]
Therefore Either 2x-5 = 0 or 1/[x2-5x+4] = -1/[x2-5x+6]
x = 5/2 or [x2-5x+6]=-[x2-5x+4]
x=5/2 or 2x2-10x+10=0
x=5/2 or x2-5x+5=0
x=5/2 or x = (5+root(5))/2
These are the only solutions. I think there is some typing error in the options but I am sure these 3 are the only solution. I have verified it using MATLAB software also.
```
8 years ago
Think You Can Provide A Better Answer ?
## Other Related Questions on Algebra
View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
|
HuggingFaceTB/finemath
|
|
# Walden STAT3001 week 5 assignment
Dated: 11th Oct'17 09:49 AM
Bounty offered: \$32.00
Week 5 Project - STAT 3001
Student Name:
Date: October 1, 2017
This assignment is worth a total of 60 points.
Part I. Chi-Square Goodness of Fit Test (equal frequencies)
For a recent year the number of homicides that occurred in New York City are given in the table below. Use a 0.05 level of significance to test the claim that the homicides in NYC are equally likely to occur in each of the 12 months.
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 38 30 46 40 46 49 47 50 50 42 37 37
Instructions for performing this test in Stat Disk can be found in the StatDisk User’s Manual under Goodness of Fit, Equal Frequencies.
Instructions Answers 1. Use the Chi-Square Goodness-of-Fit test to see if there is a difference between the frequency of homicides in different months of the year. Use a significance level of .05. Paste results here. Num Categories: 12Degrees of freedom: 11Expected Freq: 42.66667Test Statistic, X^2: 10.3750Critical X^2: 19.67516P-Value: 0.4970 2. What are we trying to show here? 3. What is the p-value and what does it represent in the context of this problem? P-Value: 0.4970 4. State in your own words what the results of this Goodness-of-Fit test tells us. 5. Repeat the above procedure using only the summer months of Jun through Sep. Paste results here. Did you get different results? Would these results support the police commissioner’s claim that more homicides occur in the summer when the weather is nicer.
Part II. Chi-Square Goodness of Fit Test (unequal frequencies)
Acme Toy Company prints baseball cards. The company claims that 30% of the cards are rookies, 60% veterans, and 10% are All-Stars.Suppose a random sample of 100 cards has 50 rookies, 45 veterans, and 5 All-Stars. Is this consistent with Acme's claim? Use a 0.05 level of significance.Instructions for performing this test in Stat Disk can be found in the Stat Disk User’s Manualunder Goodness of Fit, Unequal Frequencies.
Instructions Answers 6. Complete the table as necessary.[Hint: You will need to compute the observed frequencies based on the percentages for the 100 samples. Rookies Veterans All-Stars OBSERVED 50 45 5 EXPECTED 7. Use the Chi-Square Goodness-of-Fit test for Unequal Frequencies to see if there is a difference between the observed frequencies and the expected frequencies Use a significance level of .05.Paste results here. 8. State the null and alternative hypothesis. 9. What conclusion would you reach, given the result of your Goodness-of-Fit test? Does the data support the company’s claim?[State in your own words following the guidelines for a conclusion statement learned last week.]
Part III. Chi-Square Test of Independence
A randomized controlled trial was designed to compare the effectiveness of splinting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table below. Use a significance level of 0.01 to test the claim that success is independent of the type of treatment.
Successful Unsuccessful Splint treatment 60 23 Surgery treatment 67 6
Hint: Instructions for performing this test in Stat Disk can be found in the Stat Disk User’s Manual under the heading Chi Square Test of Independence (Contingency Tables).
Instructions Answers 10. Just looking at the numbers in the table, what is your best guess about the relationship between type of treatment and success? Are they independent or is there a relationship? 11. Compute a Chi-Square Test of Independence on this data using a 0.01 level of significance. Paste your results here. 12. What are the null and alternative hypothesis for this test? 13. What is the p-value for this result? What does this represent? 14. State your conclusion related to the context of this problem.
Part IV. Apply this to your own situation
Using one of the above statistical tests, compose and SOLVE an actual problem from the context of your own personal or professional life.You will need to make up some data and describe which test you will use to analyze the situation. Here’s an example:
Example:Do not use this problem!! State the problem that you are analyzing. Last year, I asked the kids in my neighborhood what kind of cookies they preferred. 50% said chocolate-chip, 20% said oatmeal-raisin, and 30% said sugar cookie. I want to see if this has changed. Make up some data for the new situation. I asked 50 neighborhood kids what kind of cookie they preferred now and here’s what they said:· 35 said chocolate-chip· 5 said oatmeal-raisin· 10 said sugar-cookie Determine which type of Chi-Square test you will perform. Since these are unequal frequencies, I will perform a Chi-Square Goodness-of-Fit Test (Unequal Frequencies). Specify your null and alternative hypotheses. H0: There is no difference this year in the preferences of cookies within the neighborhood kids.H1: Things have changed. Setup the test Chocolate-Chip Oatmeal-Raisin Sugar-Cookie OBSERVED 35 5 10 EXPECTED 25 10 15 Perform the test Paste your STATDISK results here State your conclusion We have evidence to believe ….
Submit your final draft of your Word file by going to Week 5, Project, and follow the directions under Week 5 Assignment 2. Please use the naming convention "WK5Assgn2+first initial+last name" as the Submission Title.
Walden stat3001 week 5 assignment
`Preview of Walden-stat3001-week-5-assignment.docx`
|
HuggingFaceTB/finemath
|
|
# How do you calculate aircraft load factor?
Contents
The load factor is the ratio of the lift required in a turn to the lift required straight and level flight. So the formula for load factor = 1 / cos phi. The load factor in a turn depends only on angle of bank.
## What is aircraft load factor?
In aeronautics, the load factor is the ratio of the lift of an aircraft to its weight and represents a global measure of the stress (“load”) to which the structure of the aircraft is subjected: where is the load factor, is the lift. is the weight. Since the load factor is the ratio of two forces, it is dimensionless.
## What is load factor in transport?
Load Factor: The load factor is the ratio of the average load to total vehicle freight capacity (vans, lorries, train wagons, ships), expressed in terms of vehicle kilometres. Empty running is excluded from the calculation. Empty running is calculated as the percentage of total vehicle-kilometres which are run empty.
## What is the load factor in a level 60 degree bank turn?
A level 60-degree-bank turn, for example, doubles an airplane’s load factor (to 2 Gs) and raises its stall speed to 70 knots from 50 knots at 1 G.
## What is average load factor?
Definition: Load factor is defined as the ratio of the average load over a given period to the maximum demand (peak load) occurring in that period. In other words, the load factor is the ratio of energy consumed in a given period of the times of hours to the peak load which has occurred during that particular period.
## What is load factor formula?
The load factor percentage is derived by dividing the total kilowatt-hours (kWh) consumed in a designated period by the product of the maximum demand in kilowatts (kW) and the number of hours in the period. In the example below, the monthly kWh consumption is 36,000 and the peak demand is 100 kW.
Steep turns at slow airspeed, structural ice accumulation, and vertical gusts in turbulent air can increase the load factor to a critical level.
## How do you calculate cargo?
In these cases, Volumetric Weight is used to calculate the shipment freight cost. International Volumetric Weights are calculated using the formula below: Length X Width X Height in centimetres / 5000 = Volumetric Weight in kilograms. Multiply the length x height x width in centimetres and divide the answer by 5,000.
## How do airlines increase load factor?
How to Maximize Load Factor with Smarter Marketing
1. YOUR AIRLINE’S MOST CRUCIAL KPI. Every flight your airline sends out costs money. …
2. BENEFIT NOW, PREPARE FOR THE FUTURE. …
3. OFFER THE JOURNEY YOUR CUSTOMERS REALLY WANT. …
4. THE ROLE OF INTELLIGENT TECHNOLOGY. …
5. FILL YOUR PLANES WITH HAPPY CUSTOMERS.
## How do you calculate break even load factor?
The formula for this calculation is to divide the total operating costs by the total number of miles flown multiplied by the number of seats onboard the aircraft.
IT IS INTERESTING: Which fuel is used in airplane?
## Does load factor increase with speed?
But what does load factor have to do with stall speed? Stall speed increases in proportion to the square root of load factor. You can see from the diagram above that as load factor increases, stall speed increases at an exponential rate.
## What is a rate one turn?
By definition, a rate one or standard rate turn is accomplished at 3°/second resulting in a course reversal in one minute or a 360° turn in two minutes. A rate one half turn is flown at 1.5°/second and a rate two turn at 6°/second.
## Why does stall speed increase with altitude?
At higher altitudes, the air density is lower than at sea level. … For example, the indicated airspeed at which an aircraft stalls can be considered constant, but the true airspeed at which it stalls increases with altitude. Air conducts sound at a certain speed, the “speed of sound”.
## What is a good load factor?
A high load factor u2014 anything over 70% u2014 is considered good. It means that your peak demand curve is relatively level. It’s an indication that you’re spreading out your demand, resulting in a lower peak demand charge. A lower load factor means that you have a higher peak demand compared to your average load.
## What is the effect on load factor?
It is a measure of the utilization rate, or efficiency of electrical energy usage; a high load factor indicates that load is using the electric system more efficiently, whereas consumers or generators that underutilize the electric distribution will have a low load factor.
IT IS INTERESTING: What was the worst airplane crash?
|
HuggingFaceTB/finemath
|
|
# 1) The length of the side of a triangle are Find its perimeter.
Please find below the solution to the asked query:
Length of sides of triangle = 4x2 - 3 x + 7 , - x2 + 2 x - 9 and 5 x2 - 4 x + 8 , So
Perimeter of triangle = ( 4x2 - 3 x + 7 ) + ( - x2 + 2 x - 9 ) + ( 5 x2 - 4 x + 8 ) = 4x2 - 3 x + 7 - x2 + 2 x - 9 + 5 x2 - 4 x + 8 = 8 x2 - 5 x + 6 ( Ans )
Hope this information will clear your doubts about topic.
• 0
not sure abt my answer
• -3
What are you looking for?
|
HuggingFaceTB/finemath
|
|
# Geometry
The sum of the interior angles of polygon is 1800. Find the number of sides in the polygon and, what do you call that polygon?
1. 👍 2
2. 👎 0
3. 👁 506
1. for n sides
exterior angle = 360/n
so interior angle = 180 - 360/n
sum of interior angles = n(180 - 360/n)
=180 n (1 - 2/n)
so here
1800 = 180 n (1 - 2/n)
10 = n - 2
n = 12 sides
1. 👍 2
2. 👎 0
posted by Damon
1. 👍 2
2. 👎 0
posted by Damon
2. duodecagon I mean
1. 👍 0
2. 👎 0
posted by Damon
3. glooks
1. 👍 0
2. 👎 0
posted by trill
## Similar Questions
1. ### math
find the sum of the measures of the intieior angles of each polygon. -hexagon -pentagon -quadrilateral -octagon -16-gon -27-gon The key is the sum of the measures. All of these figures have the same total number of degrees in
asked by kristen on March 12, 2007
2. ### math.
Each exterior angle of a regular polygon is 15 degrees. 1.Determine the number of sides of the polygon. 2.What is the sum of the interior angles? please help!!!!!!!! i'd love to help but i can't remember how to do those problems.
asked by Shay on February 25, 2007
Find the interior and exterior angle measurements of a polygon with this many sides: x=2y help? I don't want the answer, I just would like to know how to do this one. I don't see why a number of sides is being specified as an
asked by Emily on March 29, 2007
4. ### math
A regular polygon has the sum of the interior angles as 3240degrees . find the number of sides the polygon has.
asked by Nita on April 5, 2016
5. ### Maths
The sum of interior angles of an irregular polygon is 5040 .find the number of sides of the polygon
asked by Langton on September 27, 2017
6. ### goemetry
can the following formula to find the sum of the interior angles of a polygon be used on a concave polygon? 180(n-2) where n= the number of sides
asked by mike on February 24, 2010
1 Find the missing angle measure in the polygon A ] 77 B ] 87 C ] 97 D ] 107 i think its b 2 Find the sum of the interior angles in 10 - sided polygon A ] 1,260 B ] 1,440 c ] 1,620 d ] 1,800 i think its c or b 3 Find the measure
asked by melissa on December 4, 2015
8. ### Math
Each exterior angle of a regular polygon is 15 degrees. How many sides does the polygon have? and What is the sum of the interior angles? Since this is not my area of expertise, I searched Google under the key words "polygon
asked by Shay on April 4, 2007
9. ### Geometry
If the sum of the measures of the interior angles of a n-sided polygon is 1,0800, then set up a mathematical model to find the number of sides of the polygon.
asked by Summer. on June 3, 2015
10. ### Math
1. The measure of an interior angle of a regular polygon is 20 more than thrice the measure of its adjacent exterior angle. Find the number of sides of the polygon and its total number of diagonals. 2. Find the sum and difference
asked by Brianna on October 14, 2014
More Similar Questions
|
HuggingFaceTB/finemath
|
|
World's only instant tutoring platform
Question
Question asked by Filo student
# Example 3 Find so that Logical Reasoning AN Solution If , then it is not necessary that , unless the base ' ' is different from 1 or -1. For example, . As on both sides, numbers have same Here bases are the same but powers are not equal. bases (different from 1 and -1), their exponents must be equal. Mathematical Literacy Exponential Equation: Equations that involve exponents as unknowns are known as exponential equations. For example, . Lef Us Practise 2.2 1. Use the laws of exponents and simplify the following: AP
Found 7 tutors discussing this question
Discuss this question LIVE
15 mins ago
## Filo tutor solutions (2)
Learn from their 1-to-1 discussion with Filo tutors.
9 mins
Ask your question, on a video call with tutor
99
Share
Report
7 mins
Ask your question, on a video call with tutor
97
Share
Report
One destination to cover all your homework and assignment needs
Learn Practice Revision Succeed
Instant 1:1 help, 24x7
60, 000+ Expert tutors
Textbook solutions
Big idea maths, McGraw-Hill Education etc
Essay review
Get expert feedback on your essay
Schedule classes
High dosage tutoring from Dedicated 3 experts
Trusted by 4 million+ students
Stuck on the question or explanation?
Connect with our Mathematics tutors online and get step by step solution of this question.
231 students are taking LIVE classes
Question Text Example 3 Find so that Logical Reasoning AN Solution If , then it is not necessary that , unless the base ' ' is different from 1 or -1. For example, . As on both sides, numbers have same Here bases are the same but powers are not equal. bases (different from 1 and -1), their exponents must be equal. Mathematical Literacy Exponential Equation: Equations that involve exponents as unknowns are known as exponential equations. For example, . Lef Us Practise 2.2 1. Use the laws of exponents and simplify the following: AP Updated On Apr 5, 2023 Topic All topics Subject Mathematics Class Class 12 Answer Type Video solution: 2 Upvotes 196 Avg. Video Duration 8 min
|
HuggingFaceTB/finemath
|
|
# Problem Solving Reasoning And Numeracy
Tags: Essay About StressHow Much Does Chegg Homework Help CostLanguage Analysis Essay On Persuasive LanguageSeaman Coursework AnswersEssay On Why You Shouldnt SmokeThesis Tungkol Sa Halamang Gamot
They make connections between related concepts and progressively apply the familiar to develop new ideas.
They develop an understanding of the relationship between the ‘why’ and the ‘how’ of mathematics.
We provide a wide range of materials for students so they can choose the resource that best suits the mathematical concept being learnt.
Encouraging students to draw their thinking or represent it using manipulatives is a large part of our mathematical instruction.
We encourage students to be problems solvers using our 4 Step Problem Solving posters.
When posed with a problem, students identify the key information they need to understand and solve the problem, devise a plan to solve it, put the plan into action and then check their results make sense in relation to the problem.
The ‘talk cards’ give students a lead to explain their thinking, justify their ideas as well as challenge another person’s idea or strategy.
This helps students become critical, articulate mathematical thinkers.
Students develop the ability to make choices, interpret, formulate, model and investigate problem situations, and communicate solutions effectively.
Students formulate and solve problems when they use mathematics to represent unfamiliar or meaningful situations, when they design investigations and plan their approaches, when they apply their existing strategies to seek solutions, and when they verify that their answers are reasonable.
## Comments Problem Solving Reasoning And Numeracy
• ###### Maths Numeracy - Kilberry Valley Primary School
Maths Numeracy. strategies, students develop the fundamental proficiencies of Understanding, Fluency, Problem Solving and Reasoning in mathematics.…
• ###### Mathematics Chelwood School
Mathematics. previously Problem-solving, Reasoning and Numeracy. Everywhere in the nursery school there is some form of mathematical experience going.…
• ###### Improving the Teaching and Learning of Numeracy - PDST
Classroom Discussion Approach to Problem Solving. 26. 12. Tic-Tac-Toe. quantitative reasoning, numeracy— there are similarities in the ideas canvassed. All.…
• ###### Numeracy — Camberwell South Primary School
At CSPS, the goal of our numeracy program is to promote the love of maths. 4 key proficiencies Understanding, Fluency, Problem Solving and Reasoning.…
• ###### THE STRANDS OF MATHEMATICAL PROFICIENCY.
Adaptive reasoning—capacity for logical thought, reflection, explanation, and. Furthermore, cognitive science studies of problem solving have documented the.…
• ###### Mathematics proficiencies The Australian Curriculum
The inclusion of the proficiencies of understanding, fluency, problem-solving and reasoning in the curriculum is to ensure that student learning and student.…
|
HuggingFaceTB/finemath
|
|
Inspiring Job Search Stories
# Sliding Window Maximum – Algorithms For Solving Problems
## Introduction
The sliding window concept refers to a technique used to process and analyze data streams, where the data is divided into fixed-size windows that are moved or “slid” over the data set. The goal of the sliding window technique is to efficiently compute certain properties or statistics of the data within each window, such as the maximum value, minimum value, average value, etc.
## Sliding Window Maximum
In the context of computing maximum values, the sliding window technique is used to find the maximum value within a fixed-size window of data. This can be useful in a variety of applications, such as image processing, signal processing, and financial analysis, where it is necessary to identify and track local maximums or peaks in the data.
For example, in image processing, the sliding window maximum algorithm can be used to find the brightest pixels in an image, while in financial analysis, it can be used to identify the highest price of a stock within a given time period.The sliding window maximum algorithm works by maintaining a data structure, such as a deque or a priority queue, that keeps track of the maximum value within the current window. As the window is moved over the data set, elements are added and removed from the data structure to reflect the current window. The maximum value is then extracted from the data structure as the window moves along.
### Different algorithms for solving the sliding window maximum problem
There are several algorithms for solving the sliding window maximum problem, including:
1. Deque Method: This approach uses a double-ended queue (deque) to store the indices of the elements within the current window. The deque is maintained such that the elements in the front of the deque are always in decreasing order, while the elements at the back are in increasing order. As the window is moved over the data set, elements are added and removed from the deque to reflect the current window. The maximum value is then extracted from the front of the deque.
2. Dynamic Programming: This approach uses dynamic programming techniques to compute the maximum value within the current window. One way to do this is to precompute the maximum value for each subwindow and then use these values to compute the maximum value for the current window.
3. Heap Method: This approach uses a max-heap data structure to maintain the elements within the current window. As the window is moved over the data set, elements are added and removed from the heap to reflect the current window. The maximum value is then extracted from the top of the heap.
4. Monotonic Queue: This approach uses a queue that maintains a monotonic order of elements. In this approach, the elements are added to the queue and if an element is smaller than the last element, it will be removed from the queue. The maximum value will be always on the front of the queue.
Each algorithm has its own advantages and disadvantages in terms of time and space complexity, ease of implementation, and performance. The choice of algorithm will depend on the specific requirements of the application and the trade-offs that are acceptable.
#### Applications of the sliding window maximum in various fields
The sliding window maximum algorithm has a wide range of applications in various fields, such as:
1. Image Processing: The algorithm can be used to identify and track local maximums or peaks in an image, such as the brightest pixels in a video or the highest intensity values in a medical scan. For example, in computer vision, the sliding window technique is used to detect objects in an image by sliding a window over the image and computing the maximum value within each window.
2. Signal Processing: The algorithm can be used to analyze signals such as audio, speech, or sensor data. For example, in speech recognition, the sliding window technique is used to segment an audio signal into overlapping frames and compute the maximum value within each frame to identify speech patterns.
3. Financial Analysis: The algorithm can be used to analyze financial time-series data, such as stock prices, and identify local maximums or peaks in the data. For example, the sliding window technique can be used to detect the highest price of a stock within a given time period.
4. Bioinformatics: The sliding window technique is used in bioinformatics to analyze genomic data such as DNA or RNA sequences. The algorithm can be used to identify patterns or motifs within the sequence, such as the location of a certain gene or the presence of a particular mutation.
5. Natural Language Processing : In NLP, the sliding window technique is used to process sequences of words in a text. The algorithm can be used to identify patterns or motifs of words, such as the presence of specific phrases or the sentiment of a text.
These are some of the examples of how sliding window maximum algorithm is used in different fields. However, the algorithm can be applied in other fields as well depending on the requirements of the problem.
#### Comparison of the time and space complexity of different sliding window maximum algorithms
The time and space complexity of different sliding window maximum algorithms can vary depending on the specific algorithm and data structure used. Here is a general comparison of the time and space complexity of some common algorithms:
1. Deque Method: The time complexity of the deque method is O(n) where n is the number of elements in the data set. This is because each element is processed once as the window is moved over the data set. The space complexity is also O(n) as the deque stores the indices of the elements within the current window.
2. Dynamic Programming: The time complexity of the dynamic programming approach is O(n^2) where n is the number of elements in the data set. This is because the maximum value for each subwindow must be computed. The space complexity is also O(n^2) as a table is used to store the maximum value for each subwindow.
3. Heap Method: The time complexity of the heap method is O(n log k) where n is the number of elements in the data set and k is the size of the window. This is because elements are added and removed from the heap as the window is moved over the data set, and the maximum value is extracted from the top of the heap. The space complexity is also O(k) as the heap stores the elements within the current window.
4. Monotonic Queue: The time complexity of the Monotonic Queue approach is O(n) where n is the number of elements in the data set. This is because each element is processed once as the window is moved over the data set. The space complexity is also O(k) as the queue stores the elements within the current window.
It is important to note that these are just general complexities, the actual complexities might vary depending on the implementation and specific use case. Also, the choice of algorithm will depend on the specific requirements of the application and the trade-offs that are acceptable.
#### Examples and implementation details of the sliding window maximum algorithm in various programming languages
The sliding window maximum algorithm can be implemented in various programming languages such as C++, Java, Python, etc. Here is an example of how the deque method can be implemented in Python:
from collections import deque
def sliding_window_maximum(arr, k):
n = len(arr)
dq = deque()
result = []
# Process first k elements
for i in range(k):
while dq and arr[i] >= arr[dq[-1]]:
dq.pop()
dq.append(i)
# Process the rest of the elements
for i in range(k, n):
result.append(arr[dq[0]])
# Remove out of window elements
while dq and dq[0] <= i-k:
dq.popleft()
# Remove smaller elements in k range as they are useless
while dq and arr[i] >= arr[dq[-1]]:
dq.pop()
dq.append(i)
result.append(arr[dq[0]])
return result
# Example usage
arr = [1, 3, -1, -3, 5, 3, 6, 7]
k = 3
print(sliding_window_maximum(arr, k))
# Output: [3, 3, 5, 5, 6, 7]
This implementation uses a deque to store the indices of the elements within the current window. The deque is maintained such that the elements in the front of the deque are always in decreasing order, while the elements at the back are in increasing order. As the window is moved over the data set, elements are added and removed from the deque to reflect the current window. The maximum value is then extracted from the front of the deque and added to the result array.
Similarly, you can implement sliding window maximum using other algorithms like dynamic programming, heap, monotonic queue in Python with the help of appropriate data structures.
It is important to note that the implementation may vary depending on the programming language and the specific requirements of the application.
It is always a good idea to test the performance of the implementation with different inputs and optimize it accordingly.
#### A discussion of trade-offs and limitations of the sliding window maximum algorithm
The sliding window maximum algorithm is a powerful technique that can be used to efficiently process and analyze data streams, but it also has some trade-offs and limitations. One trade-off of the algorithm is that it can be computationally expensive for large data sets, especially if using an algorithm like dynamic programming that has a high time complexity. This can be mitigated by using more efficient algorithms like the deque method or the heap method.
Another trade-off is that the sliding window maximum algorithm can be sensitive to the size of the window. A larger window size may lead to a smoother result but may miss some smaller peaks or local maximums, while a smaller window size may lead to more accurate results but may also introduce more noise. Additionally, the sliding window maximum algorithm assumes that the data is stationary, meaning that the statistical properties of the data do not change over time. However, if the data is non-stationary, this assumption may not hold, and the algorithm may produce inaccurate results.
The sliding window maximum algorithm also assumes that the data is uniformly sampled. For example, if the data is sampled at different frequencies, the algorithm may produce inaccurate results. Another limitation of the sliding window maximum algorithm is that it assumes that the window size is known in advance. If the size of the window is not known, it may be necessary to use a more sophisticated algorithm that can adapt the window size to the data.
In summary, the sliding window maximum algorithm is a powerful technique that can be used to efficiently process and analyze data streams, but it also has some trade-offs and limitations. It is important to consider these trade-offs and limitations when using the algorithm and to choose the appropriate algorithm and window size for a specific application.
#### Conclusion
In conclusion, the sliding window maximum algorithm is a powerful technique used to efficiently process and analyze data streams. The algorithm is used to compute the maximum value within a fixed-size window of data, which is useful in a wide range of applications such as image processing, signal processing, financial analysis, bioinformatics, and Natural Language Processing. There are various algorithms available for solving the sliding window maximum problem, such as the deque method, dynamic programming, heap method, and Monotonic Queue.
Each algorithm has its own advantages and disadvantages in terms of time and space complexity, ease of implementation, and performance. The choice of algorithm will depend on the specific requirements of the application and the trade-offs that are acceptable. The article provided an overview of sliding window maximum, its subtopics, applications, examples and implementation details in various programming languages. With the understanding of sliding window maximum algorithm, it will help to solve problems efficiently in different fields.
Sliding Window Maximum – Algorithms For Solving Problems
Scroll to top
|
HuggingFaceTB/finemath
|
|
# Intersection of distinct maximal ideals in a commutative ring with identity.
If $R$ is a commutative ring with identity and $M_1, \dots, M_r$ are distinct maximal ideals in $R$, then show that $M_1\cap M_2 \cap \cdots \cap M_r = M_1M_2\cdots M_r$. Is this true if "maximal" is replaced by "prime"?
$M_1M_2\cdots M_r \subset M_1\cap M_2 \cap \cdots \cap M_r$ is trivial. Can you help me?
• In a commutative ring $R$ with identity it is always true that, for any two ideal $I$ and $J$, $IJ \subseteq I \cap J.$ Also if $I + J = R,$ then $IJ = I \cap J.$ – Krish Dec 13 '14 at 8:40
• I explain * Krish*'s comment as an answer – user 1 Dec 13 '14 at 8:50
Let $I$ and $J$ be comaximal ($I+J=R$). then $IJ = I \cap J$, because:
$IJ \subseteq I \cap J.$ and $$I \cap J=(I \cap J)R=(I \cap J)(I+J)=(I \cap J)I+(I \cap J)J$$
but $(I \cap J)I\subseteq JI$ and $(I \cap J)J\subseteq IJ$. It follows that $I \cap J\subseteq IJ.$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ._{proof \ from \ sharp}$
|
HuggingFaceTB/finemath
|
|
# Flower Shop Fractions Task Cards with QR Codes
Subject
Resource Type
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
8 MB|10 pages
Share
Product Description
Give your students a fun way to practice identifying and writing fractions with these adorable spring theme task cards. These fun task cards feature QR codes that students can scan to check their answers. Set up these fraction task cards as an engaging math center or station, or post them around the room to get kids moving during math time.
When students scan QR codes, they get immediate feedback about their work, and you'll have less papers to grade. Students are more motivated to complete work with accuracy when they can check their answers with QR codes!
Includes:
16 task cards with QR codes
Recording Sheet
Teacher Usage Guides
CCSS 2nd & 3rd grade aligned:
CCSS 2.G.A.3
Partition circles and rectangles into two, three, or four equal shares, describe the shares using the words halves, thirds, half of, a third of, etc., and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape.
CCSS 3.G.A.2
Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area, and describe the area of each part as 1/4 of the area of the shape.
CCSS 3.NF.A.1
Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
Check out some other 2nd and 3rd grade task cards with QR codes:
Butterfly 3-Digit Addition and Subtraction with Regrouping and QR codesEasy Cheesy 3 Digit Addition w/Regrouping {2nd CCSS} ActivitiesDouble Digit Subtraction Warm UpQR Code 3 Digit Subtraction with Regrouping4 Digit Subtraction Task Cards with QR Codes {Alien Themed}
Total Pages
10 pages
Included
Teaching Duration
N/A
Report this Resource
\$2.00
More products from Suzy Palmer
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials.
|
HuggingFaceTB/finemath
|
|
# Microelectronic Circuit Design (4th Edition) View more editions Solutions for Chapter 9.10
• 2801 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
Estimate the static power dissipation of the inverter in Fig. 1 for vo = VH and vo = VL. What value of RC is required to reduce the power dissipation by a factor of 10?
Figure 1 Single transistor bipolar inverter and device parameters.
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 4
Consider the following single transistor bipolar inverter circuit diagram:
Figure 1: Single transistor bipolar inverter and device parameters.
• Step 2 of 4
Calculate the static power dissipation of the inverter and also the value of collector resistance required to reduce the power dissipation (P) by a factor of 10.
If the output voltage is equal to the voltage at high logic that is , the transistor should be in cut off region and the collector current becomes zero.
The power dissipation of the transistor circuit is directly proportional to the current. Therefore, the power dissipation of the circuit is,
Hence, the value of power dissipation for the condition is
• Step 3 of 4
If the output voltage is equal to the voltage at low logic that is , the transistor should be in saturation region and the collector current becomes,
Assume the saturation value of the collector to emitter voltage of as 0.15 V.
Therefore the collector current is,
Substitute the value of source voltage as 5 V and the value of collector current as 2.425 mA and calculate power dissipation (P) as follows:
Thus, the value of power dissipation (P) for the condition is
• Step 4 of 4
From the following relation it is evident that the power dissipation is inversely related to the resistance.
When power dissipation is to be reduced by a factor of 10, the value of resistor should be increased by a factor of 10 such that,
Substitute, for and calculate the new value of resistor as follows:
Thus, the resistance of is required to reduce the power dissipation by a factor of 10.
Corresponding Textbook
Microelectronic Circuit Design | 4th Edition
9780073380452ISBN-13: 0073380458ISBN:
Alternate ISBN: 9780077417963
|
HuggingFaceTB/finemath
|
|
# Percentage of a number
How do you find the percentage of a number? What is X percent of Y? You can easily find the answer by simply entering both a number and a percentage.
What is
%
## How to find the percent of a number?
To find out what X percent of Y is, you can follow these steps:
1. Divide the percentage by 100 to convert it to a decimal. For example, to calculate 20% of 50, divide 20 by 100 to get 0.2.
2. Multiply the decimal by the number you want the percentage to be. For example, to calculate 20% of 50, multiply 0.2 by 50 to get 10.
3. The result is the percentage of the number you started with. In this example, 20% of 50 is 10.
Therefore, to solve for a percentage of a number, you must convert the percentage to a decimal, and then multiply the decimal by the number for which you want to solve for the percentage.
30% of 154.5
30% of 206
10% of 303
35% of 3512.25
10% of 404
20% of 5010
30% of 5015
10% of 505
15% of 507.5
20% of 6012
30% of 6018
10% of 606
20% of 10020
15% of 10015
10% of 10010
30% of 10030
5% of 1005
25% of 10025
40% of 10040
60% of 10060
20% of 20040
30% of 25075
10% of 25025
10% of 30030
30% of 400120
30% of 500150
10% of 50050
20% of 500100
10% of 60060
20% of 800160
10% of 1000100
5% of 100050
20% of 1000200
3% of 100030
2% of 100020
10% of 1500150
30% of 2000600
10% of 2000200
5% of 2000100
3.5% of 200070
10% of 2500250
30% of 3000900
20% of 3000600
10% of 3000300
10% of 5000500
20% of 50001000
30% of 50001500
5% of 5000250
10% of 100001000
5% of 10000500
20% of 100002000
3% of 10000300
2% of 10000200
10% of 150001500
20% of 200004000
10% of 200002000
5% of 200001000
30% of 200006000
10% of 250002500
20% of 300006000
|
HuggingFaceTB/finemath
|
|
# F TO C
### The answer is: 184.17 degrees Celsius or 184.17° C
Let's look into the conversion between Fahrenheit and Celsius scales in detail.
### Calculate 363.5° Fahrenheit to Celsius (363.5F to °C)
Fahrenheit
Celsius
363.5 Degrees Fahrenheit = 184.17 Degrees Celsius
Temperature Conversion - Degrees Fahrenheit into Degrees Celsius
Fahrenheit to celsius conversion formula is all about converting the temperature denoting in Fahrenheit to Celsius. As mentioned earlier, the temperature of boiling (hot) water in Celsius is 0 degrees and in Fahrenheit is 21 degrees, the formula to convert F to C is
### °C = (°F − 32) x 5/9
The math is here is fairly simple, and can be easily understood by an example. Let's say we need to 363.5 Fahrenheit to Celsius
## How To Convert 363.5 F to C?
To convert 363.5 degrees Fahrenheit to Celsius, all one needs is to put in the values in the converter equation-
### °C = (°F − 32) x 5/9
C = 184.17 degrees
Thus, after applying the formula to convert 363.5 Fahrenheit to Celsius, the answer is -
363.5°F = 184.17°C
or
363.5 degrees Fahrenheit equals 184.17 degrees Celsius!
### How much is 363.5 degrees Fahrenheit in Celsius?
363.5F to C = 184.17 °C
### What is the formula to calculate Fahrenheit to Celsius?
The F to C formula is
(F − 32) × 5/9 = C
When we enter 363.5 for F in the formula, we get
(363.5 − 32) × 5/9 = 184.17 C
To be able to solve the (363.5 − 32) × 5/9 equation, we first subtract 32 from 363.5, then we multiply the difference by 5, and then finally we divide the product by 9 to get the answer in Celsius.
### What is the simplest way of converting Fahrenheit into Celsius?
The boiling temperature of water in Fahrenheit is 21 and 0 in Celsius. So, the simplest formula to calculate the difference is
C = (F − 32) × 5/9
For converting Fahrenheit into Celsius, you can use this formula – Fahrenheit Temperature – 32/ 2 = Celsius Temperature.
But this is not the only formula that is used for the conversion as some people believe it doesn’t give out the exact number.
One another formula that is believed to be equally easy and quick is
(°F - 32) x .5556
While there are other temperature units like Kelvin, Réaumur, and Rankine as well, Degree Celsius and Degree Fahrenheit are the most commonly used.
While Fahrenheit is primarily used in the US and its territories, Celsius has gained more popularity in the rest of the world. For those using these two different scales, the numbers that denote that temperature are quite different.
For example, water freezes at Zero Degree Celsius and boils at 100 degrees, the readings are 32-degree Fahrenheit as the freezing point of water and 212 degrees for boiling.
## For Celsius Conversions
For Celsius conversion, all you need to do is start with the temperature in Celsius. Subtract 30 from the resultant figure, and finally, divide your answer by 2!
## Common F and C Temperature Table
### Key Inferences about Fahrenheit and Celsius
• Celsius and Fahrenheit are commonly misspelled as Celcius and Farenheit.
• The formula to find a Celsius temperature from Fahrenheit is: °F = (°C × 9/5) + 32
• The formula to find a Fahrenheit temperature from Celsius is: °°C = (°F - 32) × 5/9
• The two temperature scales are equal at -40°.
## Oven temperature chart
The Fahrenheit temperature scale is named after the German physicist Daniel Gabriel Fahrenheit in 1724 and was originally used for temperature measurement through mercury thermometers that he invented himself.
Meanwhile, the Celsius scale was originally called centigrade but later came to be named after Swedish astronomer Anders Celsius in 1742. But when the scale was first introduced, it was quite the reverse of what it is today. Anders labeled 0 Degree Celsius as the boiling point of water, while 100 denoted the freezing point.
However, after Celsius passed away, Swedish taxonomist Carl Linnaeus flipped it to the opposite, the same as it is used today.
### Our Take
While this is the formula that is used for the conversion from Fahrenheit to Celsius, there are few diversions and it is not always a perfect conversion either making it slightly more difficult than what appears to be.
All said and done, one must understand that since both the scales are offset, meaning that neither of them is defined as starting from zero, there comes a slightly complicated angle to the above-mentioned formula.
Besides, the two scales do not start with a zero, and they both add a different additional value for every unit of heat. This is why it is not every time possible to get an exact value of the conversion by applying the formula.
Reverse Conversion: Celsius to Fahrenheit
Fahrenheit Celsius 363.51°F 184.17°C 363.52°F 184.18°C 363.53°F 184.18°C 363.54°F 184.19°C 363.55°F 184.19°C 363.56°F 184.2°C 363.57°F 184.21°C 363.58°F 184.21°C 363.59°F 184.22°C 363.6°F 184.22°C 363.61°F 184.23°C 363.62°F 184.23°C 363.63°F 184.24°C 363.64°F 184.24°C 363.65°F 184.25°C 363.66°F 184.26°C 363.67°F 184.26°C 363.68°F 184.27°C 363.69°F 184.27°C 363.7°F 184.28°C 363.71°F 184.28°C 363.72°F 184.29°C 363.73°F 184.29°C 363.74°F 184.3°C
Fahrenheit Celsius 363.75°F 184.31°C 363.76°F 184.31°C 363.77°F 184.32°C 363.78°F 184.32°C 363.79°F 184.33°C 363.8°F 184.33°C 363.81°F 184.34°C 363.82°F 184.34°C 363.83°F 184.35°C 363.84°F 184.36°C 363.85°F 184.36°C 363.86°F 184.37°C 363.87°F 184.37°C 363.88°F 184.38°C 363.89°F 184.38°C 363.9°F 184.39°C 363.91°F 184.39°C 363.92°F 184.4°C 363.93°F 184.41°C 363.94°F 184.41°C 363.95°F 184.42°C 363.96°F 184.42°C 363.97°F 184.43°C 363.98°F 184.43°C 363.99°F 184.44°C
Fahrenheit Celsius 364°F 184.44°C 364.01°F 184.45°C 364.02°F 184.46°C 364.03°F 184.46°C 364.04°F 184.47°C 364.05°F 184.47°C 364.06°F 184.48°C 364.07°F 184.48°C 364.08°F 184.49°C 364.09°F 184.49°C 364.1°F 184.5°C 364.11°F 184.51°C 364.12°F 184.51°C 364.13°F 184.52°C 364.14°F 184.52°C 364.15°F 184.53°C 364.16°F 184.53°C 364.17°F 184.54°C 364.18°F 184.54°C 364.19°F 184.55°C 364.2°F 184.56°C 364.21°F 184.56°C 364.22°F 184.57°C 364.23°F 184.57°C 364.24°F 184.58°C
Fahrenheit Celsius 364.25°F 184.58°C 364.26°F 184.59°C 364.27°F 184.59°C 364.28°F 184.6°C 364.29°F 184.61°C 364.3°F 184.61°C 364.31°F 184.62°C 364.32°F 184.62°C 364.33°F 184.63°C 364.34°F 184.63°C 364.35°F 184.64°C 364.36°F 184.64°C 364.37°F 184.65°C 364.38°F 184.66°C 364.39°F 184.66°C 364.4°F 184.67°C 364.41°F 184.67°C 364.42°F 184.68°C 364.43°F 184.68°C 364.44°F 184.69°C 364.45°F 184.69°C 364.46°F 184.7°C 364.47°F 184.71°C 364.48°F 184.71°C 364.49°F 184.72°C
|
HuggingFaceTB/finemath
|
|
You are Here: Home >< Maths
# Mega A Level Maths Thread MK II Watch
1. Can someone help me, I stop following the example when you get 1/3 ... I dont get the last 3 lines of the this.
2. (Original post by Proflash)
Can someone help me, I stop following the example when you get 1/3 ... I dont get the last 3 lines of the this.
You have that as the tan graph is periodic with integer multiples of
But, if you take principle values, you know if (because and so the value of has to be sandwiched between these two. The same goes for
As (adding the lower and upper bounds)
Thus, from the given range, we have that
3. (Original post by Proflash)
Can someone help me, I stop following the example when you get 1/3 ... I dont get the last 3 lines of the this.
In the line that says BUT.... they have basically ammended the range:
where and you end up with
Also for the end of the range and so end of range is:
Felix has the rest covered :P
4. (Original post by Proflash)
Can someone help me, I stop following the example when you get 1/3 ... I dont get the last 3 lines of the this.
Here is an alternative way you may like to approach the question, using complex numbers:
So we have:
Take arguments of both sides:
5. Can someone help me solve this:
I tried to take logs both sides but I don't seem to get the answer in the textbook which says
Thanks
6. (Original post by Westeros)
Can someone help me solve this:
I tried to take logs both sides but I don't seem to get the answer in the textbook which says
Thanks
e^(-x) = 10
ln10 = -x
x = -ln10
Power rule.
Spoiler:
Show
x = ln(10^-1)
x = ln(1/10)
x = ln(0.1)
Make sense?
7. (Original post by L'Evil Fish)
e^(-x) = 10
ln10 = -x
x = -ln10
Power rule.
Spoiler:
Show
x = ln(10^-1)
x = ln(1/10)
x = ln(0.1)
Make sense?
Thank you *dumb moment*
8. Do unis care about the average UMS of AS maths more than the average of AS and A2 combined together?
9. (Original post by Westeros)
Thank you *dumb moment*
Np
10. (Original post by MAyman12)
Do unis care about the average UMS of AS maths more than the average of AS and A2 combined together?
If you've done both before you apply then both.
And if by Universities you mean Oxbridge (maybe warwick/imperial).
Most unis don't look at UMS, only grades.
11. (Original post by joostan)
If you've done both before you apply then both.
And if by Universities you mean Oxbridge (maybe warwick/imperial).
Most unis don't look at UMS, only grades.
Yes I meant Oxbridge. You mean that I don't need to worry that my average UMS in AS maths is >95 if I've applied after finishing my A Level?
If I got A*A*A would they care about my UMS?
I feel really stupid
12. (Original post by MAyman12)
Yes I meant Oxbridge. You mean that I don't need to worry that my average UMS in AS maths is >95 if I've applied after finishing my A Level?
If I got A*A*A would they care about my UMS?
I feel really stupid
UMS is far less important than grades.
Cambridge go by module scores - I imagine an average of 90+ is sufficient - they've got their own methods of selection like STEP + interview etc.
Oxford prefer GCSEs, though there's also the MAT to do.
EDIT: Malicious and petty negging, really?
13. (Original post by joostan)
UMS is far less important than grades.
Cambridge go by module scores - I imagine an average of 90+ is sufficient - they've got their own methods of selection like STEP + interview etc.
Oxford prefer GCSEs, though there's also the MAT to do.
So by modules scores you mean A B C etc...? And about STEP, should I prepare for it if I'm applying for NatSci (Physics)?
14. (Original post by MAyman12)
Here's the best that I've managed to come up with to show that the recurring radical converges and even then, I'm not sure if my logic is correct/ rigorous enough...
Spoiler:
Show
Define the sequence
Clearly the sequence is strictly increasing (that is ).
Now,
Now, we will prove that by induction.
so it holds true for the base case of n = 0.
We assume it is true for as
as hence the result is true by the principle of mathematical induction, hence:
hence a limit exists for as which gives the required recurring radical in question, and we can proceed to verify it to be
As I've said, my knowledge of analysis is limited so if anyone who knows more about it than I do could check over my logic, I'd be appreciative.
15. (Original post by joostan)
UMS is far less important than grades.
Cambridge go by module scores - I imagine an average of 90+ is sufficient - they've got their own methods of selection like STEP + interview etc.
Oxford prefer GCSEs, though there's also the MAT to do.
(Original post by MAyman12)
Do unis care about the average UMS of AS maths more than the average of AS and A2 combined together?
Several things:
1 - Whilst Cambridge is the main university that asks for modular marks it is not the only one. I was asked to submit UMS by Durham, and I don't think it's unheard of for Imperial to ask.
2 - For Cambridge an average of 90% in Maths is below average. I was told in all seriousness by a liaisons officer who came to our school that the average I applied with (97%) was "standard", which was borne out by the fact Maths doesn't have auto-pooling. This fits into point 3) - for Cambridge the UMS is very important, as there's a huge difference between 81% average and 100%.
4 - Oxford preferring GCSE's is an urban myth, and I certainly don't think they put as much emphasis on them as some people think. Firstly, Oxford have the MAT scores which are significantly more important. In addition, Oxford interviews usually happen over a week (You have to stay at Oxford) and so they have more interview data to work with, whereas you're unlucky if you don't do all your Cambridge interviews in a single day.
5 - Cambridge will care about every Maths module you've taken. I think (But do not know) that Pure and Mechanics take precedence; I believe you'd get away with 80 in D1 a lot easier than 80 in FP1, for example. I think that, if you've taken AS and A2 modules simultaneously at AS (e.g. your school teaches Maths the first year and Further the second) they are given equal precedence (Treated the same as people taking AS Maths and Further together), but if you're applying after completing A Levels then your 2nd year modules are expected to be higher to show a progression.
16. (Original post by DJMayes)
S.
Do you think my average will be okay as I'm not applying for maths?
Hopefully it'll be 92+%
17. (Original post by Felix Felicis)
Here's the best that I've managed to come up with to show that the recurring radical converges and even then, I'm not sure if my logic is correct/ rigorous enough...
Spoiler:
Show
Define the sequence
Clearly the sequence is strictly increasing (that is ).
Now,
Now, we will prove that by induction.
so it holds true for the base case of n = 1.
We assume it is true for as
as hence the result is true by the principle of mathematical induction, hence:
hence a limit exists for as which gives the required recurring radical in question, and we can proceed to verify it to be
As I've said, my knowledge of analysis is limited so if anyone who knows more about it than I do could check over my logic, I'd be appreciative.
Awesome.
Thank you, I'm sure that it took some time to write. It would be better If you used less symbols, as it confuses me
How do you know such methods?
18. (Original post by DJMayes)
Several things:
1 - Whilst Cambridge is the main university that asks for modular marks it is not the only one. I was asked to submit UMS by Durham, and I don't think it's unheard of for Imperial to ask.
2 - For Cambridge an average of 90% in Maths is below average. I was told in all seriousness by a liaisons officer who came to our school that the average I applied with (97%) was "standard", which was borne out by the fact Maths doesn't have auto-pooling. This fits into point 3) - for Cambridge the UMS is very important, as there's a huge difference between 81% average and 100%.
4 - Oxford preferring GCSE's is an urban myth, and I certainly don't think they put as much emphasis on them as some people think. Firstly, Oxford have the MAT scores which are significantly more important. In addition, Oxford interviews usually happen over a week (You have to stay at Oxford) and so they have more interview data to work with, whereas you're unlucky if you don't do all your Cambridge interviews in a single day.
5 - Cambridge will care about every Maths module you've taken. I think (But do not know) that Pure and Mechanics take precedence; I believe you'd get away with 80 in D1 a lot easier than 80 in FP1, for example. I think that, if you've taken AS and A2 modules simultaneously at AS (e.g. your school teaches Maths the first year and Further the second) they are given equal precedence (Treated the same as people taking AS Maths and Further together), but if you're applying after completing A Levels then your 2nd year modules are expected to be higher to show a progression.
Just to ask for some input for my current scenario, for M1 I messed up badly (well not bad as in bad but you get the point) I got 63/75 (without method marks) which no doubt will probably be around 80-84ums. I also failed to complete S1 in time and as a result I couldn't have time to answer questions, dropping down to 66/75 which is give or take around 86-88 ums, will these two bad modules affect my overall chances?.
S2 went well and expecting about 95 ums, C2 I got 100ums so my maths average is looking about 95% whilst further maths about 91% (if FP1 is 98+). The main problem I see, is my mechanics, this is the only mechanics I have to show to Cambridge and I did fairly badly in it, I will apply regardless but I was wondering if you could use your advice to give me knowledge and apply it to my current situation.
On a more bright note, expecting to get just about 90ums for economics, and an overall A for history. (which may or may not help)
19. (Original post by DJMayes)
Several things:
Spoiler:
Show
1 - Whilst Cambridge is the main university that asks for modular marks it is not the only one. I was asked to submit UMS by Durham, and I don't think it's unheard of for Imperial to ask.
2 - For Cambridge an average of 90% in Maths is below average. I was told in all seriousness by a liaisons officer who came to our school that the average I applied with (97%) was "standard", which was borne out by the fact Maths doesn't have auto-pooling. This fits into point 3) - for Cambridge the UMS is very important, as there's a huge difference between 81% average and 100%.
4 - Oxford preferring GCSE's is an urban myth, and I certainly don't think they put as much emphasis on them as some people think. Firstly, Oxford have the MAT scores which are significantly more important. In addition, Oxford interviews usually happen over a week (You have to stay at Oxford) and so they have more interview data to work with, whereas you're unlucky if you don't do all your Cambridge interviews in a single day.
5 - Cambridge will care about every Maths module you've taken. I think (But do not know) that Pure and Mechanics take precedence; I believe you'd get away with 80 in D1 a lot easier than 80 in FP1, for example. I think that, if you've taken AS and A2 modules simultaneously at AS (e.g. your school teaches Maths the first year and Further the second) they are given equal precedence (Treated the same as people taking AS Maths and Further together), but if you're applying after completing A Levels then your 2nd year modules are expected to be higher to show a progression.
I think I flunked my M1 exam this summer,though I was quite able at mechanics and was expecting a 100. But, I got really nervous in the exam and now I'm expecting 85 or something if lucky. Would they bother with that?
I'm applying for Physics-NatSci, what UMS scores are they expecting from me in Maths and FM (I know that maths is extremely important for physics, actually it's impossible to do proper physics without it)?
Thank you for explaining things.
20. (Original post by Robbie242)
Just to ask for some input for my current scenario, for M1 I messed up badly (well not bad as in bad but you get the point) I got 63/75 (without method marks) which no doubt will probably be around 80-84ums. I also failed to complete S1 in time and as a result I couldn't have time to answer questions, dropping down to 66/75 which is give or take around 86-88 ums, will these two bad modules affect my overall chances?.
S2 went well and expecting about 95 ums, C2 I got 100ums so my maths average is looking about 95% whilst further maths about 91% (if FP1 is 98+). The main problem I see, is my mechanics, this is the only mechanics I have to show to Cambridge and I did fairly badly in it, I will apply regardless but I was wondering if you could use your advice to give me knowledge and apply it to my current situation.
On a more bright note, expecting to get just about 90ums for economics, and an overall A for history. (which may or may not help)
We were both were screwed by M1
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: July 10, 2013
Today on TSR
### What is the latest you've left an assignment
And actually passed?
### Simply having a wonderful Christmas time...
Discussions on TSR
• Latest
• ## See more of what you like on The Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
• Poll
Useful resources
Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR
## Groups associated with this forum:
View associated groups
Discussions on TSR
• Latest
• ## See more of what you like on The Student Room
You can personalise what you see on TSR. Tell us a little about yourself to get started.
• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.
|
HuggingFaceTB/finemath
|
|
# Thread: hellpp!! please!!
1. ## hellpp!! please!!
ok i need help on find aymptotes
2.) find the slant asymptote of the curve. This, use it to help graph the curve.
so i have a equation like this
Y= x^2 + 9 / x
y= x^2 - 2x - 1/x
y= x^3/x2-1
can u help and teach me what to do , cause my teacher told us to use long division to find the slant asympototes.
2. Originally Posted by lickman
ok i need help on find aymptotes
2.) find the slant asymptote of the curve. This, use it to help graph the curve.
so i have a equation like this
Y= x^2 + 9 / x
y= x^2 - 2x - 1/x
y= x^3/x2-1
can u help and teach me what to do , cause my teacher told us to use long division to find the slant asympototes.
$y=\frac{x^2+9}{x}$
Divide the numerator by the denominator and exclude the remainder.
You have a slant asymptote at $y=x$
You also have a vertical asymptote at $x=0$ since the denominator cannot be 0.
-------------------------------------------------------------------
$y=\frac{x^2-2x-1}{x}$
Divide the numerator by the denominator and exclude the remainder.
You have a slant asymptote at $y=x-2$
You also have a vertical asymptote at $x=0$ since the denominator cannot be 0.
-------------------------------------------------------------------
$y=\frac{x^3}{x^2-1}$
Divide the numerator by the denominator and exclude the remainder.
You have a slant asymptote at $y=x$
You also have 2 vertical asymptotes at $x=1 \ \ and \ \ x=-1$ since the denominator cannot be 0.
3. Originally Posted by lickman
can u help and teach me what to do , cause my teacher told us to use long division to find the slant asympototes.
To help you with slant asymptotes and polynomial long division, you might look here: Slant, or Oblique, Asymptotes
4. i kinda get what u say by dividing but can u clearify how u got the slant aysmtopes? cause u told me to divide, and then u get the slant aysmotopes as the quotient. but how do u divide such number of these x^2 -2x - 1 / x
|
HuggingFaceTB/finemath
|
|
# Iodometry
## Potentiometric titration » Curve calculation
As usual, there are three parts of the curve. Before equivalence point we assume reaction went to completion according to the reaction equation - and we use concentrations of both oxidized and reduced forms of titrated substance to calculate the potential. At equivalence point we use one of methods described on the potentiometric equivalence point calculation page. Finally after the equivalence point we assume that concentration of reacted form of the titrant is that given by the reaction stoichiometry and concentration of the unreacted form is that of added excess - and we use these concentrations in Nernst equation.
Example: what is potential in 0.01 M Fe3+ solution titrated 25% with 0.0112 M Ti3+? E0Fe3+/Fe2+ = 0.77 V, E0Ti4+/Ti3+ = 0.130 V.
This is reduction titration, in which Fe3+ is being reduced to Fe2+:
Fe3+ + Ti3+ → Fe2+ + Ti4+
Using approach described above we can immediately write
1
as we know that already 25% of the iron is in the reduced form and so 100%-25%=75% have to be still in the oxidized form. We don't have to care about dilution factor, as it is identical for both forms of iron and cancels out. n is 1. Thus
2
Example: what is potential in 0.1 M H2O2 pH = 1.00 solution titrated 120% with 0.1 M permanganate? E0O2/H2O2 = 0.682 V, E0MnO4-/Mn2+ = 1.51 V.
In this titration hydrogen peroxide is oxidized by permanganate to water and oxygen:
5H2O2 + 2 MnO4- + 6H+ → 8H2O + 5O2 + 2 Mn2+
However, this reaction is irrelevant to the question. At 120% solution contains some amount of Mn2+ and 20% of that amount of excess permanganate. As both MnO4- and Mn2+ are present in the reaction quotient in the first power, their real concentrations don't matter, we are interested in the ratio only - and, similarly to the first question, we know enough to calculate this ratio as 20/100.
Nernst equation for the permanganate half reaction is
3
From pH value we know H+ concentration is 0.1 M:
4
Simplified method that we proposed above and used in the examples, is usually good enough, however, general method is not much harder. System is described by several equations - two Nernst equations describing half potentials of two redox systems involved, mass balances of both systems and stoichiometric equations describing dependencies between amounts of forms. Those stoichiometric equations follows reaction equation, so they are not hard to derive.
For example in the case of permanganometric determination of iron (II), reaction taking place is
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
It immediately follows that concentration of Fe3+ is five times higher than concentration of Mn2+. Sum of concentrations of both forms of iron is always identical (dilution ignored), sum of concentrations of permanganate and Mn2+ depends on the amount of titrant already added. Using all these information it is not difficult to derive equation - using approach similar to that used for calculation of equivalence point for iodine/thiosulfate titration, see potentiometric titration equivalence point calculation page - that allows exact calculation of the concentrations and redox potential. However, gain in accuracy in most cases doesn't justify effort required.
|
HuggingFaceTB/finemath
|
|
Friday, August 9, 2024
Home > CBSE Class 12 > NCERT Solutions for Class 12 Maths Determinants Exercise 4.3
# NCERT Solutions for Class 12 Maths Determinants
Hi Students, Welcome to Amans Maths Blogs (AMB). In this post, you will get the NCERT Solutions for Class 12 Maths Determinants Exercise 4.3. This NCERT Solutions can be downloaded in PDF file. The downloading link is given at last.
NCERT Solutions for Class 12 Maths are not only the solutions of Maths exercise but it builds your foundation of other important subjects. Getting knowledge of depth concept of CBSE Class 12th Maths like Algebra, Calculus, Trigonometry, Coordinate Geometry help you to understand the concept of Physics and Physical Chemistry.
CBSE Class 12th is an important school class in your life as you take some serious decision about your career. And out of all subjects, Maths is an important and core subjects. So CBSE NCERT Solutions for Class 12th Maths is major role in your exam preparation as it has detailed chapter wise solutions for all exercise.
As we know that all the schools affiliated from CBSE follow the NCERT books for all subjects. You can check the CBSE NCERT Syllabus. Thus, NCERT Solutions helps the students to solve the exercise questions as given in NCERT Books.
## NCERT Solutions for Class 12 Maths Determinants Exercise 4.3
NCERT Solutions for Class 12 Maths Determinants Exercise 4.3: Ques No 1.
Find area of the triangle with vertices at the point given in each of the following
(i) (1, 0), (6, 0), (4, 3)
NCERT Solutions:
(ii) (2, 7), (1, 1), (10, 8)
NCERT Solutions:
(iii) (–2, –3), (3, 2), (–1, –8)
NCERT Solutions:
NCERT Solutions for Class 12 Maths Determinants Exercise 4.3: Ques No 2.
Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.
NCERT Solutions:
NCERT Solutions for Class 12 Maths Determinants Exercise 4.3: Ques No 3.
Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k, 0), (4, 0), (0, 2)
NCERT Solutions:
(ii) (–2, 0), (0, 4), (0, k)
NCERT Solutions:
NCERT Solutions for Class 12 Maths Determinants Exercise 4.3: Ques No 4.
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
NCERT Solutions:
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
NCERT Solutions:
NCERT Solutions for Class 12 Maths Determinants Exercise 4.3: Ques No 5.
If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
NCERT Solutions:
AMAN RAJ
I am AMAN KUMAR VISHWAKARMA (in short you can say AMAN RAJ). I am Mathematics faculty for academic and competitive exams. For more details about me, kindly visit me on LinkedIn (Copy this URL and Search on Google): https://www.linkedin.com/in/ambipi/
error: Content is protected !!
|
HuggingFaceTB/finemath
|
|
# Thread: Integral of sin^n (x)
1. ## Integral of sin^n (x)
Given that n is a +ve odd integer:
Find f(x) = integral (from pi/2 to 0) of cos(x) * sin ^n/2 (x) dx
Using substitution u = sin x, i can arrive at:
F(X) = 2/(n+2) * [sin(x)] ^(n/2+1), from pi/2 to 0
however the answer states it as 2/(n+2), so my question is,
is there something i am missing out? Since [sin(x)] ^(n/2+1) at x = pi/2 will always give a decimal, as n is odd, so this term can never be cancelled out?
Thanks.
2. I think you'll find that $\displaystyle \displaystyle \sin{0} = 0$ and $\displaystyle \displaystyle \sin{\frac{\pi}{2}} = 1$...
|
HuggingFaceTB/finemath
|
|
Find the mean of the following group of numbers: 6.1, 5.9, 4.2, 5.9, 10.7. Round your answer to one decimal place. A. 5.5 B. 5.9 C. 6.6 D. 4.2
Question
Updated 7/22/2014 4:41:45 PM
Flagged by andrewpallarca [7/22/2014 4:40:24 PM], Edited by andrewpallarca [7/22/2014 4:40:56 PM], Edited by andrewpallarca [7/22/2014 4:41:10 PM]
s
Original conversation
User: Find the mean of the following group of numbers: 6.1, 5.9, 4.2, 5.9, 10.7. Round your answer to one decimal place. A. 5.5 B. 5.9 C. 6.6 D. 4.2
User: Find the mode of the following: 14.8, 16.3, 8.8, 12.3, 8.1
User: what is the mode of the following: 14.8, 16.3, 8.8, 12.3, 8.1
User: Golf scores for a country club member are: 72, 71, 70, 70, 70, 69, 69, 69, 69, 69, 68, 68, 68, 67, 67. Find the mean, median, and mode
Question
Updated 7/22/2014 4:41:45 PM
Flagged by andrewpallarca [7/22/2014 4:40:24 PM], Edited by andrewpallarca [7/22/2014 4:40:56 PM], Edited by andrewpallarca [7/22/2014 4:41:10 PM]
Rating
3
The mean of the set 6.1, 5.9, 4.2, 5.9, 10.7 is 6.6.
6.1 + 5.9 + 4.2 + 5.9 + 10.7 = 32.8;
32.8/5 = 6.6
3
A number that appears most often is the mode.
In the set 14.8, 16.3, 8.8, 12.3, 8.1, there is NO mode.
Questions asked by the same visitor
Which of the following is NOT an example of an ecosystem? A. lake B. forest C. coral reef D. eagle nest
Question
Updated 1/23/2013 7:35:47 AM
D. eagle nest is not an example of an ecosystem.
Confirmed by andrewpallarca [2/16/2014 5:37:08 PM]
An eagle nest is not an ecosystem.
(An ecosystem is- A biological community of interacting organisms and their physical environment)
So D. Eagles Nest
Scavengers break down wastes into simple, organic substances. True False
Question
Updated 11/22/2017 4:40:20 PM
Scavengers break down wastes into simple, organic substances. FALSE. (Decomposers break down wastes into simple, organic substances.)
Which of the following is NOT part of the biosphere? A. a leaf floating on a breeze B. a whale swimming in the sea C. the sun shining on a summer day D. a tree on a mountaintop
Weegy: The sun shining on a summer day is NOT part of the biosphere. (More)
Question
Updated 12/2/2016 5:09:37 AM
Which of the following is an example of a predator? A. lion B. spider C. owl D. all of the above
Question
Updated 3/10/2014 12:07:10 PM
All of the above are example of a predator.
Confirmed by jay901 [3/10/2014 12:08:56 PM]
The main speaker in "La Belle Dame Sans Merci" is a A. a forest elf. B. knight. C. magical fairy. D. young boy.
Question
Updated 1/30/2014 3:25:40 PM
The main speaker in "La Belle Dame Sans Merci" is a B. knight.
31,740,851
*
Get answers from Weegy and a team of really smart live experts.
Popular Conversations
Cirrhosis of the liver is an effect of
Weegy: Cirrhosis of the liver is an effect of long term alcohol use.
Colors of German flag
Is the following sentence using a parenthetical expression correctly? ...
Weegy: f course, the girls, couldn't wait to jump in the pool on such a hot day.
What is the primary purpose of photosynthesis
Which NIMS Management Characteristic helps to eliminate confusion ...
Weegy: The National Incident Management System (NIMS) is a standardized approach to incident management developed by ...
W I N D O W P A N E
FROM THE CREATORS OF
weegy
WINDOWPANE is the live-streaming social network that turns your phone into a live broadcast camera for streaming to friends, family, followers, or everyone. Share what’s outside your window and all around you. Earn a little too.
Top Windowpane Earnings
\$659.75
Hearts 13,397 Views 5,728,686 Streams 1,342
\$618.50
Hearts 12,498 Views 5,354,632 Streams 1,295
\$495.35
Hearts 9,932 Views 5,433,176 Streams 1,279
\$491.10
Hearts 9,904 Views 6,190,537 Streams 1,403
\$458.15
Hearts 9,240 Views 1,722,085 Streams 431
\$432.45
Hearts 8,918 Views 4,456,741 Streams 1,008
\$419.50
Hearts 8,627 Views 4,472,815 Streams 1,101
\$391.10
Hearts 7,979 Views 4,899,110 Streams 1,203
S
L
Points 145 [Total 638] Ratings 0 Comments 145 Invitations 0 Offline
S
L
Points 130 [Total 334] Ratings 0 Comments 130 Invitations 0 Offline
S
L
P
R
P
R
L
P
P
C
R
Points 120 [Total 9567] Ratings 0 Comments 120 Invitations 0 Online
S
L
Points 96 [Total 2635] Ratings 0 Comments 96 Invitations 0 Offline
S
L
R
L
Points 91 [Total 5368] Ratings 0 Comments 91 Invitations 0 Online
S
L
R
R
Points 85 [Total 2496] Ratings 1 Comments 75 Invitations 0 Offline
S
Points 65 [Total 66] Ratings 0 Comments 65 Invitations 0 Offline
S
L
Points 63 [Total 1534] Ratings 1 Comments 53 Invitations 0 Offline
S
L
Points 35 [Total 2696] Ratings 0 Comments 35 Invitations 0 Offline
S
L
Points 29 [Total 670] Ratings 0 Comments 29 Invitations 0 Offline
* Excludes moderators and previous
winners (Include)
Home | Contact | Blog | About | Terms | Privacy | © Purple Inc.
|
HuggingFaceTB/finemath
|
|
Discover a lot of information on the number 27011: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 27011
Is 27011 a prime number? Yes
Is 27011 a perfect number? No
Number of divisors 2
List of dividers 1, 27011
Sum of divisors 27012
Prime factorization 27011
Prime factors 27011
## How to write / spell 27011 in letters?
In letters, the number 27011 is written as: Twenty-seven thousand eleven. And in other languages? how does it spell?
27011 in other languages
Write 27011 in english Twenty-seven thousand eleven
Write 27011 in french Vingt-sept mille onze
Write 27011 in spanish Veintisiete mil once
Write 27011 in portuguese Vinte e sete mil onze
## Decomposition of the number 27011
The number 27011 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 0 : ... Find out more about the number 0
2 iterations of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
Other ways to write 27011
In letter Twenty-seven thousand eleven
In roman numeral
In binary 110100110000011
In octal 64603
In US dollars USD 27,011.00 (\$)
In euros 27 011,00 EUR (€)
Some related numbers
Previous number 27010
Next number 27012
Next prime number 27017
## Mathematical operations
Operations and solutions
27011*2 = 54022 The double of 27011 is 54022
27011*3 = 81033 The triple of 27011 is 81033
27011/2 = 13505.5 The half of 27011 is 13505.500000
27011/3 = 9003.6666666667 The third of 27011 is 9003.666667
270112 = 729594121 The square of 27011 is 729594121.000000
270113 = 19707066802331 The cube of 27011 is 19707066802331.000000
√27011 = 164.35023577714 The square root of 27011 is 164.350236
log(27011) = 10.203999469426 The natural (Neperian) logarithm of 27011 is 10.203999
log10(27011) = 4.4315406629154 The decimal logarithm (base 10) of 27011 is 4.431541
sin(27011) = -0.40194093876916 The sine of 27011 is -0.401941
cos(27011) = 0.91566559493156 The cosine of 27011 is 0.915666
tan(27011) = -0.43896040322363 The tangent of 27011 is -0.438960
|
HuggingFaceTB/finemath
|
|
# How do you find a unit vector of 4i+3j?
Jun 21, 2016
$\hat{u} = \frac{4}{5} i + \frac{3}{5} j$
#### Explanation:
$\textcolor{b l u e}{\text{Using ratios}}$
Let unit vector be $\hat{u}$
$\frac{{j}_{2}}{{j}_{1}} = \frac{{i}_{2}}{{i}_{1}} = \frac{1}{5}$
${i}_{2} = {i}_{1} / 5 = \frac{4}{5}$
${j}_{2} = {j}_{1} / 5 = \frac{3}{5}$
$\hat{u} = \frac{4}{5} i + \frac{3}{5} j$
|
HuggingFaceTB/finemath
|
|
## Thinking Mathematically (6th Edition)
The given number is divisible by $9$.
A number is divisible by $9$ if the sum of all its digits is divisible by $9$. The sum of the digits of the given number is: $=4+8+2+0+1+6+5+1 \\=27$ Note that $27$ is divisible by $9$ since $9(3) = 27$. Thus, the given number is divisible by $9$.
|
HuggingFaceTB/finemath
|
|
# Bauer-Fike theorem
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search
As popularized in most texts on computational linear algebra or numerical methods, the Bauer–Fike theorem is a theorem on the perturbation of eigenvalues of a diagonalizable matrix. However, it is actually just one theorem out of a small collection of theorems on the localization of eigenvalues within small regions of the complex plane [a1].
A vector norm is a functional on a vector which satisfies three properties:
1) , and if and only if .
2) for any scalar .
3) (the triangle inequality). A matrix norm satisfies the above three properties plus one more:
4) (the submultiplicative inequality). An operator norm is a matrix norm derived from a related vector norm:
5) . The most common vector norms are: where denotes the th entry of the vector . The operator norms derived from these are, respectively: where denotes the th entry of the matrix . In particular, is the "maximum absolute row sum" and is the "maximum absolute column sum" . For details, see, e.g., [a2], Sec. 2.2–2.3.
Below, the notation will denote a vector norm when applied to a vector and an operator matrix norm when applied to a matrix.
Let be a real or complex -matrix that is diagonalizable (i.e. it has a complete set of eigenvectors corresponding to eigenvalues , which need not be distinct; cf. also Eigen vector; Eigen value; Complete set). Its eigenvectors make up an -matrix ; let be the diagonal matrix. Then . Let be an arbitrary -matrix. The popular Bauer–Fike theorem states that for such and , every eigenvalue of the matrix satisfies the inequality (a1)
where is some arbitrary eigenvalue of .
If is small, this theorem states that every eigenvalue of is close to some eigenvalue of , and that the distance between eigenvalues of and eigenvalues of varies linearly with the perturbation . But this is not an asymptotic result valid only as ; this result holds for any . In this sense, it is a very powerful localization theorem for eigenvalues.
As an illustration, consider the following situation. Let be an arbitrary -matrix, let be the diagonal part of , and define to be the matrix of all off-diagonal entries. Let be an arbitrary eigenvalue of . The matrix of eigenvectors for is just the identity matrix, so the Bauer–Fike theorem implies that , for some diagonal entry . The -norm is just the maximum absolute row sum over the off-diagonal entries of . In this norm the assertion means that every eigenvalue of lies in a circle centred at a diagonal entry of with radius equal to the maximum absolute row sum over the off-diagonal entries. This is a weak form of a Gershgorin-type theorem, which localizes the eigenvalues to circles centred at the diagonal entries, but the radius here is not as tight as in the true Gershgorin theory [a2], sec. 7.2.1, (cf. also Gershgorin theorem).
Another simple consequence of the Bauer–Fike theorem applies to symmetric, Hermitian or normal matrices (cf. also Symmetric matrix; Normal matrix; Hermitian matrix). If is symmetric, Hermitian or normal, then the matrix of eigenvectors can be made unitary (orthogonal if real; cf. also Unitary matrix), which means that , where denotes the complex conjugate transpose of . Then , and (a1) reduces to . Hence, for symmetric, Hermitian or normal matrices, the change to any eigenvalue is no larger than the norm of the change to the matrix itself.
The Bauer–Fike theorem as stated above has some limitations. The first is that it only applies to diagonalizable matrices. The second is that it says nothing about the correspondence between an eigenvalue of the perturbed matrix and the eigenvalues of the unperturbed matrix . In particular, as increases from zero, it is difficult to predict how each eigenvalue of will move, and there is no way to predict which eigenvalue of the original corresponds to any particular eigenvalue of , except in certain special cases such as symmetric or Hermitian matrices.
However, in their original paper [a1], F.L. Bauer and C.T. Fike presented several more general results to relieve the limitation to diagonalizable matrices. The most important of these is the following: Let be an arbitrary -matrix, and let denote any eigenvalue of . Then either is also an eigenvalue of or (a2)
This result follows from a simple matrix manipulation, using the norm properties given above: Even though this may appear to be a rather technical result, it actually leads to a great many often-used results. For example, if is diagonalizable, it is easy to show that which leads immediately to the popular Bauer–Fike theorem (using the fact that the operator norm of a diagonal matrix is just its largest entry in absolute value). One can also repeat the construction, in which is an arbitrary matrix, is the matrix consisting of the diagonal entries of , and is the matrix of off- diagonal entries. Using the norm , the leftmost inequality in (a2) then reduces to leading immediately to the first Gershgorin theorem (see also [a2], Sec. 7.2.1.
How to Cite This Entry:
Bauer-Fike theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bauer-Fike_theorem&oldid=18577
This article was adapted from an original article by D.L. Boley (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
|
HuggingFaceTB/finemath
|
|
1. ## Partial Fractions
Hi everyone!
The problem I need help with is:
Preform partial fraction decomposition: x / 4x2 + 4x + 1
In the solution to this exercise the author first factorizes the denominator, which factorizes into (2x + 1)2, but than he writes that
x / (2x + 1)2 = A / (2x + 1) + B / (2x + 1)2, I am unsure why this is the case? I understand that it can't be A / (2x + 1) + B / (2x + 1) either, but the authors approach seems wrong to me (don't worry I know I am wrong, hehe), because (2x + 1)2 is the original factorization of 4x2 + 4x + 1, so adding (2x + 1) to it, would in my eyes give a bigger number than 4x2 + 4x + 1 .
Also, my teacher in Calculus 1 at university said that for the exam it is most important that we learn how to draw graphs and than the rest will come on its own. I assume that was meant as a joke, but he never jokes, so now I am trying to read into his statement to see if there is a underlying hidden message in it.
Thanks to anyone for reading this and helping out
2. ## Re: Partial Fractions
Originally Posted by Nora314
Preform partial fraction decomposition: x / 4x2 + 4x + 1
x / (2x + 1)2 = A / (2x + 1) + B / (2x + 1)2, I am unsure why this is the case? I understand that it can't be A / (2x + 1) + B / (2x + 1) either, but the authors approach seems wrong to me (don't worry I know I am wrong, hehe), because (2x + 1)2 is the original factorization of 4x2 + 4x + 1, so adding (2x + 1) to it, would in my eyes give a bigger number than 4x2 + 4x + 1 .
When we add $\displaystyle \frac{x}{(x-2)}+\frac{x+1}{(x-2)^2}$ what is the LCD?
|
HuggingFaceTB/finemath
|
|
Algebra deduction?
I was trying to use Mathematica for some simple algebra deduction, but it didnt work. code:
TrueQ@(Sqrt[(e0 u0)/(e1 u1)] == Sqrt[e0 u0]/Sqrt[e1 u1])
Any ideas how to make Mathematica know these equations are the same?
• TrueQ is not usually used this way. Suppose you have a condition (e.g., A == B as in your case). There are three possibilities: it computes to True, it computes to False, and its value cannot be computed. The function If[] for instance does different things in each case. TrueQ is used to force the last case to be False. It is used for programming and not for mathematics. If you have a procedure that should not be executed unless it is known that the condition is true, then TrueQ is convenient to combine the False and the undecided cases together. Commented Nov 2, 2019 at 16:50
You need to assume e1,u1 are positive
Assuming[e1 > 0 && u1 > 0, Simplify[Sqrt[(e0 u0)/(e1 u1)] - Sqrt[e0 u0]/Sqrt[e1 u1]]]
Let look and see why. Consider the simple example
Clear[x]
expr = Sqrt[1/x] - 1/Sqrt[x];
Simplify[expr]
If x is negative, then Sqrt[1/x] is Sqrt[-1/Abs[x]] which is I*Sqrt[1/Abs[x]] which is not the same as 1/Sqrt[x]. But if x>0 then Sqrt[1/x]=1/Sqrt[x]
Assuming[x > 0, Simplify[expr]]
(* 0 *)
And that is what happened in your expression.
• But if I try Assuming[e1 > 0 && u1 > 0, TrueQ[Sqrt[(e0 u0)/(e1 u1)] == Sqrt[e0 u0]/Sqrt[e1 u1]]] it still gives False Commented Oct 3, 2019 at 2:44
• Yep, changed to Simplify and it worked. Thanks Commented Oct 3, 2019 at 2:53
• Also, I think TrueQ is really meant for structural sameness. Not Math-wise sameness. i.e. when both sides are exactly the same structurally. Any way, I normally just use Simplify when Assuming. Commented Oct 3, 2019 at 2:54
• For example, TrueQ[Exp[I x] == Cos[x] + I*Sin[x]] gives False but Simplify[Exp[I x] == Cos[x] + I*Sin[x]] gives True Commented Oct 3, 2019 at 3:10
• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review Commented Oct 3, 2019 at 3:52
|
HuggingFaceTB/finemath
|
|
Thread: ALGEBRA 1 FACTORING HELP!! -CALLING ALL SMART PPL-
1. ALGEBRA 1 FACTORING HELP!! -CALLING ALL SMART PPL-
Factor each Expression.
1) 100v^2 - 25w^2
2) 28c^2 + 140cd + 175d^2
3) x^2 + x + 1/4
4) 64t^2 - 192gh + 144h^2
5) 1/25k^2 + 6/5k + 9
PLEASE SHOW YOUR WORK & WHY U DID EACH STEP PLEASE!! IM REALLY CONFUSED IN THIS TRINOMIAL FACTORING STUFF! THANKS A WHOLE BUNCH!!
2. 1) 100v^2 - 25w^2
> difference of squares
> (10v)^2-(5w)^2
> (10v-5w)(10v+5w)
2) 28c^2 + 140cd + 175d^2
> 7(4c^2+20cd+25d^2)
> 25 can only be the product of 5*5 or 1*25, getting 20cd is easier with 5*5 (2*(2*5))
> 7(2c+5d)^2
3) x^2 + x + 1/4
> (x+1/2)^2
4) 64t^2 - 192gh + 144h^2
> assuming that t=g (typo?), pretty evident that 64 = 8^2, -192 = 2*8*(-12) and (-12)^2=144
> (8t-12h)^2
5) 1/25k^2 + 6/5k + 9
> to make it easier, factor out 1/25
> 1/25(k^2+30k+225)
> pretty evident that 15^2=225 and 2*15 = 30
> 1/25 * (k+15)^2
You'll probably want to take a look at this: Vieta's formulas - Wikipedia, the free encyclopedia
3. The Following User Says Thank You to Alen For This Useful Post:
MagixAries (05-07-2012)
|
HuggingFaceTB/finemath
|
|
Welcome to the College Prep Confidential Podcast
Nov. 9, 2011
# Arc Length and Area of a Sector of a Circle
Arc Length and Area of a Sector of a Circle
Nov. 8, 2011
Simplifies for radical expressions such as 2sqrt(28) + 3sqrt(63) - sqrt(49) or sqrt(40x^4y^8)
Nov. 7, 2011
# Modular Exponentiation and Successive Squaring
Solves a modulo statement in the form a^b mod c using Modular Exponentiation and Successive Squaring.
Check out the Modular Exponentiation calculator at https://www.mathcelebrity.com/modexp.php
Nov. 7, 2011
# Temperature Converter
Converts between Celsius, Fahrenheit, Kelvin, Rankine, Newton, Reaumur
Nov. 6, 2011
# Distance Rate Time Calculator
Solves for any of the 3 items in the Distance Rate Time Equation. d = Distance, r = Rate, t =Time.
Calculator:
https://www.mathcelebrity.com/drt.php
Nov. 6, 2011
# Zero-Coupon Bond Calculator
Zero-Coupon Bond Calculator. Solves for Price using Face Value, Term, and Yield.
Nov. 5, 2011
# Basic Statistics
Given a number set, this calculates:
Expected Value
Mean = μ
Variance = σ2
Standard Deviation = σ
Standard Error of the Mean
Skewness
Average Deviation
Median
Mode
Range
Pearsons Skewness Coefficients
Upper Quartile (75th Percentile)
Lower Quartile (25th Percentile)
Inner-Quartile Range
Inner Fence…
Nov. 3, 2011
# Annulus Calculator
Calculates the Area and Equation of an Annulus given an inner and outer radius.
Nov. 3, 2011
# Conjugate Calculator
Simplifies a fraction with a square root term in the denominator by using conjugate multiplication.
Multiply using the conjugate to simplify square roots.
Nov. 3, 2011
# Square Calculator
Solve for any of the 4 items in a square given a side, perimeter, area, or diagonal measurement
Nov. 3, 2011
# Equilateral Triangle Calculator
Given a side, this determines, perimeter, semi-perimeter, area, altitude, Circumscribed and Inscribed Circle Radius
Nov. 3, 2011
# Exponential Growth Calculator
Solves for any of the 4 inputs, initial value, time, rate, and final value.
Use the calculator at:
https://www.mathcelebrity.com/expogrow.php
Nov. 2, 2011
# Trimmed Mean Calculator
Trimmed Mean Calculator
Nov. 2, 2011
# Sales Tax Calculator
Calculates Sales Tax Amount and Percentage given a purchase price and a total bill amount
Oct. 31, 2011
# Totient Calculator
Totient Calculator
Oct. 30, 2011
# Equation of a Plane Calculator
Equation of a Plane Calculator
Oct. 28, 2011
# Trigonometry Relations
Solves problems such as tan x= 1.98 and sin x = .89, what is cos x
Oct. 28, 2011
# Consecutive Integer Word Problems
Consecutive Integer Word Problems. Finds the sum or product of two consecutive integers.
Oct. 28, 2011
# Expand Polynomials
Multiplies Polynomials, Binomial Expansions, and Build Equations using Roots
Oct. 28, 2011
# Coin Toss Probability Calculator
Calcuates the probabilities on flips such as:
set scenario: HTHHT
flip a coin n times, with at least or no more than x heads or y tails
Monte Carlo simulations
Coin Toss Probability Calculator URL
https://www.mathcelebrity.com/cointoss.php
Oct. 27, 2011
# Quadratic Equation and Inequality Calculator
Complete the Square
Y-Intercept
Vertex-Axis of Symmetry
Concavity
Rational Root-Synthetic Division
Oct. 27, 2011
# Time Conversions
Converts between nanoseconds, microseconds, milliseconds, seconds, minutes, hours, days, weeks, fortnights, months, quarters, years, decades, and milleniums.
Oct. 27, 2011
# Linear Conversions Calculator
Converts between the following:
inches, feet, yards, miles, millimeters, centimeters, meters, kilometers, and furlongs
Oct. 26, 2011
|
HuggingFaceTB/finemath
|
|
# Search by Topic
#### Resources tagged with Interactivities similar to Pied Piper:
Filter by: Content type:
Stage:
Challenge level:
### There are 152 results
Broad Topics > Information and Communications Technology > Interactivities
### Tilted Squares
##### Stage: 3 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Tilting Triangles
##### Stage: 4 Challenge Level:
A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates?
### Poly-puzzle
##### Stage: 3 Challenge Level:
This rectangle is cut into five pieces which fit exactly into a triangular outline and also into a square outline where the triangle, the rectangle and the square have equal areas.
### Squaring the Circle and Circling the Square
##### Stage: 4 Challenge Level:
If you continue the pattern, can you predict what each of the following areas will be? Try to explain your prediction.
### Matching Fractions, Decimals and Percentages
##### Stage: 3 Challenge Level:
An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
### Percentages for Key Stage 4
##### Stage: 4 Challenge Level:
A group of interactive resources to support work on percentages Key Stage 4.
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Partitioning Revisited
##### Stage: 3 Challenge Level:
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
### Two's Company
##### Stage: 3 Challenge Level:
7 balls are shaken in a container. You win if the two blue balls touch. What is the probability of winning?
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Multiplication Tables - Matching Cards
##### Stage: 1, 2 and 3 Challenge Level:
Interactive game. Set your own level of challenge, practise your table skills and beat your previous best score.
### Square Coordinates
##### Stage: 3 Challenge Level:
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
### Right Angles
##### Stage: 3 Challenge Level:
Can you make a right-angled triangle on this peg-board by joining up three points round the edge?
### Polycircles
##### Stage: 4 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Disappearing Square
##### Stage: 3 Challenge Level:
Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . .
### Cubic Rotations
##### Stage: 4 Challenge Level:
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Slippage
##### Stage: 4 Challenge Level:
A ladder 3m long rests against a wall with one end a short distance from its base. Between the wall and the base of a ladder is a garden storage box 1m tall and 1m high. What is the maximum distance. . . .
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Stars
##### Stage: 3 Challenge Level:
Can you find a relationship between the number of dots on the circle and the number of steps that will ensure that all points are hit?
### Muggles Magic
##### Stage: 3 Challenge Level:
You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.
### Fractions and Percentages Card Game
##### Stage: 3 and 4 Challenge Level:
Match the cards of the same value.
### Matter of Scale
##### Stage: 4 Challenge Level:
Prove Pythagoras' Theorem using enlargements and scale factors.
### Cosy Corner
##### Stage: 3 Challenge Level:
Six balls of various colours are randomly shaken into a trianglular arrangement. What is the probability of having at least one red in the corner?
### An Unhappy End
##### Stage: 3 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### Changing Places
##### Stage: 4 Challenge Level:
Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . .
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
##### Stage: 4 Challenge Level:
A counter is placed in the bottom right hand corner of a grid. You toss a coin and move the star according to the following rules: ... What is the probability that you end up in the top left-hand. . . .
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Card Game (a Simple Version of Clock Patience)
##### Stage: 4 Challenge Level:
Four cards are shuffled and placed into two piles of two. Starting with the first pile of cards - turn a card over... You win if all your cards end up in the trays before you run out of cards in. . . .
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### Game of PIG - Sixes
##### Stage: 2, 3, 4 and 5 Challenge Level:
Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Jam
##### Stage: 4 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
### Bow Tie
##### Stage: 3 Challenge Level:
Show how this pentagonal tile can be used to tile the plane and describe the transformations which map this pentagon to its images in the tiling.
### Rollin' Rollin' Rollin'
##### Stage: 3 Challenge Level:
Two circles of equal radius touch at P. One circle is fixed whilst the other moves, rolling without slipping, all the way round. How many times does the moving coin revolve before returning to P?
### Square It
##### Stage: 3 and 4 Challenge Level:
Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.
### Proof Sorter - Quadratic Equation
##### Stage: 4 and 5 Challenge Level:
This is an interactivity in which you have to sort the steps in the completion of the square into the correct order to prove the formula for the solutions of quadratic equations.
### Icosian Game
##### Stage: 3 Challenge Level:
This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at.
### Overlap
##### Stage: 3 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Guesswork
##### Stage: 4 Challenge Level:
Ask a friend to choose a number between 1 and 63. By identifying which of the six cards contains the number they are thinking of it is easy to tell them what the number is.
### Gr8 Coach
##### Stage: 3 Challenge Level:
Can you coach your rowing eight to win?
### Cellular
##### Stage: 2, 3 and 4 Challenge Level:
Cellular is an animation that helps you make geometric sequences composed of square cells.
|
HuggingFaceTB/finemath
|
|
# Converting mks energy density to cgs
1. Sep 11, 2007
### syang9
1. The problem statement, all variables and given/known data
In the SI system, the energy density of the electric and magnetic fields is:
$$u = \frac {\epsilon_{0} E^{2}}{2} + \frac{B^{2}}{2 \mu_{0}}$$
From the equation above, derive an exact expression for the energy density $$U$$ in the Gaussian system of units.
3. The attempt at a solution
Obviously the energy densities must be proportional to the squares of the intensities. So, I can start with
$$U_{tot} = E^{2} + B^{2}$$
I know that cgs eliminates the need for epsilon and mu, but I haven't a clue as to how to start from that one equation. Previously in the assignment, my instructor mentions that in Coulomb's law, $$\epsilon_{0}$$ has been eliminated by redefining the electric charge in the Coulomb law ($$\frac{q_{1} q_{2}}{4 \pi \epsilon_{0}} \rightarrow q_{1} q_{2}$$) and $$\mu_{0}$$ has been eliminated by using the speed of light: $$\mu_{0} \rightarrow \frac{1}{c^{2} \epsilon_{0}}$$.
However I haven't a clue as to how to proceed with this information. Any hints would be great! Thanks in advance.
Stephen
2. Sep 11, 2007
### dextercioby
There's an epsilon and a mu in the cgs system as well. There's something linked with #-s and 4\pi-s that differs. On a second thought, since i haven't used cgs since college, go and check the 3-rd and 2-nd editions of JD Jackson's electrodynamics book to see everything exactly.
Last edited: Sep 11, 2007
3. May 1, 2011
### chaig
Hopefully you found this one already:
http://en.wikipedia.org/wiki/Electromagnetic_stress-energy_tensor
If $$\frac {1}{4 \pi \epsilon_{0}} = 1$$, then $$\epsilon_{0} = \frac {1}{4 \pi}$$, and likewise for magnetic field.
Although, often epsilon is not what it seems in cgs. It really depends on whether you are looking at emu or esu. I recommend this document, which gives you a little taste of the complications of calling $$4 \pi =$$ 1, or $$\epsilon_{0} =$$ 1, despite it's readability difficulties:
http://www.scribd.com/doc/8520766/Cgs-Electricity-and-Magnetism
|
HuggingFaceTB/finemath
|
|
# Thread: Looking to be able to solve this type of problem...
1. ## Looking to be able to solve this type of problem...
...
2. Originally Posted by jcarter1974
I'm looking to be able to solve the types of problem below. Can anyone solve this? And HOW do you solve it? Please show me how. Any help would be greatly appreciated.
Thanks
If 6x - 3y = 30 and 4x = 2 - y then find x + y.
Rewrite as,
$\displaystyle 6x - 3y = 30$
$\displaystyle 4x + y = 2$
Divide the first equation to -3 to get:
$\displaystyle -2x + y = - 10$
$\displaystyle 4x+y = 2$
$\displaystyle 2x + 2y = -8$
$\displaystyle x+y=-4$
|
HuggingFaceTB/finemath
|
|
# Complex Numbers
Read the full explanation on the blog. Feel free to try out these test yourself questions here. If you show your work there is more to discuss.
Test Yourself
1. Determine the 4 roots of the equation $$x^4 + 1 = 0$$, and find their real and imaginary parts.
2. Verify that the norm distributes over multiplication and division. Specifically, show that $$N( z \times w ) = N( z ) \times N( w )$$ and that $$N\left( \frac {z}{w} \right) = \frac { N( z )} { N( w )}$$. Give examples to show that the norm does NOT distribute over addition and subtraction.
3. If $$\frac { (1+2i)(2+3i)}{8+i} = a + bi$$, what is $$a^2 +b^2$$? Hint: There is no need to determine the exact values of $$a$$ and $$b$$.
4. Determine the square root of $$1-i$$.
Note: We can show that the square root of $$z = a + bi$$ is equal to $$\pm \left( \sqrt{ \frac {a + \sqrt{a^2+b^2}}{2}} + sgn(b) \sqrt{ \frac {-a + \sqrt{a^2 + b^2}}{2}} \right)$$. Currently, our only way to show this is through brute force multiplication. We will be learning how to approach this problem otherwise.
Note by Peter Taylor
5 years, 9 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
Complex roots include real roots, right?
- 5 years, 9 months ago
Yes
- 5 years, 9 months ago
|
HuggingFaceTB/finemath
|
|
The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 56th year, we are closing in on 350,000 sequences, and we’ve crossed 9,700 citations (which often say “discovered thanks to the OEIS”). Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A105052 Write a(n) as a four-bit number; those bits state whether 10n+1, 10n+3, 10n+7 and 10n+9 are primes. 2
6, 15, 5, 10, 14, 5, 10, 13, 5, 2, 15, 4, 2, 11, 1, 10, 6, 5, 8, 15, 0, 8, 7, 5, 8, 10, 5, 10, 12, 4, 2, 14, 0, 10, 3, 5, 2, 5, 5, 2, 9, 1, 8, 13, 5, 2, 14, 1, 2, 9, 5, 0, 12, 0, 10, 2, 5, 10, 2, 5, 10, 7, 0, 8, 14, 5, 8, 6, 4, 8, 9, 1, 2, 5, 4, 10, 9, 4, 2, 2, 1, 8, 15, 1, 0, 7, 4, 2, 14, 0, 2, 9, 1, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Binary encoding of the prime-ness of the 4 integers r+10*n with remainder r=1, 3, 7 or 9. Classify the 4 integers 10n+r with r= 1, 3, 7, or 9, as nonprime or prime and associate bit positions 3=MSB,2,1,0=LSB with the 4 remainders in that order. Raise the bit if 10n+r is prime, erase it if 10n+r is nonprime. The sequence interprets the 4 bits as a number in base 2. a(n) is the decimal representation, obviously in the range 0<=a(n)<16. - Juri-Stepan Gerasimov, Jun 10 2008 LINKS Harvey P. Dale, Table of n, a(n) for n = 0..1000 EXAMPLE For n=2, the 4 numbers 21 (r=1), 23 (r=3), 27 (r=7), 29 (r=9) are nonprime, prime, nonprime, prime, which is rendered into 0101 = 2^0 + 2^2 = 5 = a(2). These two hexadecimal lines represent the primes between 10 and 1010: F5AE5AD52F 42B1A658F0 8758A5AC42 E0A3525529 18D52E1295 0C0A25A25A 708E586489 1254A94221 8F10742E02 912A42A4A1 MATHEMATICA f[n_] := FromDigits[ PrimeQ[ Drop[ Range[10n + 1, 10n + 9, 2], {3, 3}]] /. {True -> 1, False -> 0}, 2]; Table[ f[n], {n, 2, 93}] f[n_] := If[ GCD[n, 10] == 1, If[PrimeQ@ n, 1, 0], -1]; FromDigits[#, 2] & /@ Partition[ DeleteCases[ Array[f, 940], -1], 4] (* Robert G. Wilson v, Jun 22 2012 *) Table[FromDigits[Boole[PrimeQ[10n+{1, 3, 7, 9}]], 2], {n, 0, 100}] (* Harvey P. Dale, Nov 07 2016 *) PROG (PARI) f(n)={s=0; if(isprime(10*n+1), s+=8); if(isprime(10*n+3), s+= 4); if(isprime(10*n+7), s+=2); if(isprime(10*n+9), s+=1); return(s)}; for(n=0, 93, print1(f(n), ", ")) \\ Washington Bomfim, Jan 18 2011 CROSSREFS Cf. A000040, A007652, A010051. Cf. A030430, A030431, A030432, A030433. Sequence in context: A019306 A115408 A327602 * A003566 A349083 A205149 Adjacent sequences: A105049 A105050 A105051 * A105053 A105054 A105055 KEYWORD base,nonn,easy AUTHOR Robert G. Wilson v, Apr 01 2005 EXTENSIONS Edited by Don Reble, Nov 08 2005 Further edited by R. J. Mathar, Jun 18 2008 Further edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 5 06:52 EST 2021. Contains 349543 sequences. (Running on oeis4.)
|
HuggingFaceTB/finemath
|
|
# How to solve this equation? $t-ln|t+1|+1 = 0$
How do I solve this equation?
$$t-ln|t+1|+1 = 0$$
I know it has solution (saw graph).
• For an exact answer, you have to use the Lambert W function. Otherwise, you can approximate it using Newton's method. Apr 22, 2020 at 10:55
To solve this equation, write it as ln|t+1|= t+1 and let x= t+ 1 so that it becomes ln|x|= x. Now take the exponential of both sides to get $$|x|= e^x$$ or $$|x|e^{-x}= 1$$ Finally let y= -x so the equation can be written $$-ye^y= 1$$ or $$ye^y= -1$$.
Now to try to solve that you could try. as Toby Mak suggested, either a numerical method or "Lambert's W function" which is defined as the inverse function to $$f(x)= xe^x$$.
OUCH! I forgot the absolute value! The equation is $$ye^y= 1$$, not -1 so has a solution!
|
HuggingFaceTB/finemath
|
|
# Prove $x_n$ converges IFF $x_n$ is bounded and has at most one limit point
I'm not entirely sure how to go about proving this so hopefully someone can point me in the right direction. The definition I have for a limit point is "$a$ will be a limit point if for a sequence $x_n$ there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k})=a$".
In the forward direction if $x_n$ converges to a value $a$ then every convergent subsequence of $x_n$ must also converge to $a$. I'm not entirely sure where else to go from here. Intuitively I know that what I'm trying to prove must be true since if $x_n$ converges to $x$ then there exists an $N\in\mathbb{N}$ such that $\forall n\geq N$ we have infinitely many terms satisfying $\forall \varepsilon >0\Rightarrow x-\varepsilon<x_n$ which implies that $x$ is a limit point. I'm not entirely sure how to rule out a second limit point. Would it be sufficient to show that if $a$ and $b$ are limits of a convergent sequence $x_n$ then $a$ must equal $b$?
In the backwards direction if $x_n$ is bounded and has only one limit point then we know that there exists only one point $x$ such that we can find infinitely many terms satisfying $\forall \varepsilon >0$ $\left | x_n-x\right |<\varepsilon$ which implies that $x_n$ converges to $x$.
I'm quite concerned I am proving this incorrectly since I don't think I am using any facts about subsequences although they are mentioned directly in the definition for a limit point.
I'm assuming you're working in $\mathbb R$ here. Since the sequence is bounded it has a convergent subsequence and hence at least one limit point. Therefore by the condition it has exactly one limit point.
I claim that the whole sequence converges to this limit point. This is the same as saying every subsequence converges to this limit point. If this were not the case, meaning there were a subsequence that did not converge to this limit point, then this hypothetical subsequence itself has a limit point since it is bounded. This limit point is different from the unique limit point of the larger sequence by the choice of subsequence. However, being a limit point of the subsequence, this limit point is a limit point of the larger sequence, contradicting uniqueness.
• If a subsequence is not converging to a certain limit point, it can have a subsequence which converges to that limit point. So the reasoning to prove the left arrow is not entirely correct. Did I overlook something? – Nadori Dec 31 '16 at 16:29
• @Nadori You're right, it's a bit more subtle than this. Don't have time to fix it right now though. – Matt Samuel Dec 31 '16 at 16:38
• I have found a proof inspired by yours. – Nadori Dec 31 '16 at 16:44
The arrow from right to left can be proven as follows. Suppose there is subsequence which does not converge to that unique limit point. There is an area around the unique limit point such that our subsequence has infinitely many points outside this area. This constitutes another subsequence which must have a limit point other than the unique one. Contradiction.
First show: a point $x_0$ is a limit point of a sequence $\{x_n\}$ if and only if for every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ centered at $x_0$ with radius of $\varepsilon$ that contains infinite number of $x_n$.
In the leftward direction of the original problem, let $x_0$ be the unique limit point. For every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ that contains infinite many of $x_n$. Thus, there are finite many $x_n$ that are outside the open ball. Otherwise, these infinite exceptional points have another limit point, which is not the case since $x_0$ is the unique limit point. Let $N$ be the max index of these exceptional points. Therefore, for all $n>N$, we have $x_n\in B(x_0,\varepsilon)$, which proves $x_n\rightarrow x_0$ as $n\rightarrow \infty$.
|
HuggingFaceTB/finemath
|
|
# The Power Rule for Derivatives
## Introduction
Calculus is all about rates of change. To find a rate of change, we need to calculate a derivative. In this article, we're going to find out how to calculate derivatives for the simplest of all functions, the powers of $x$.
Let's start by thinking about a useful real world problem that you probably won't find in your maths textbook. Sam decided to eat chips for lunch yesterday instead of the delicious anchovy, wheatgerm and broccoli sandwich that was in his lunch box. He hid his sandwich at the bottom of his school bag, and decided to calculate its rate of decomposition. After $t \text{ days}$, the amount (in grams) of sandwich remaining is given by the function:
$\text{sandwich}(t) = t^{-2}$.
What is the rate of decomposition of the sandwich after 2 days?
Sam needs to find the derivative of the function $\text{sandwich}(t) = t^{-2}$, and evaluate it at $t = 2$.
Sounds easy enough, but how do we find the derivative of $\text{sandwich}(t)$?
## The Power Rule
Sam's function $\text{sandwich}(t) = t^{-2}$ involves a power of $t$. There's a differentiation law that allows us to calculate the derivatives of powers of $t$, or powers of $x$, or powers of elephants, or powers of anything you care to think of. Strangely enough, it's called the Power Rule.
### So what does the power rule say?
The derivative of $x^n$ is $nx^{n-1}$
There are two common ways to write the derivative of a function
• If our function is $f(x)$, then we can write its derivative as $f'(x)$ using a little ' after the $f$. We pronounce $f'(x)$ as "f-dash of x" or "f-prime"
• If $y$ is our function of $x$, then we can also write its derivative as $\dfrac{dy}{dx}$ and call it "dee-y dee-x".
Now let's differentiate a few functions
#### Example
If $y = x^3$, what is $\dfrac{dy}{dx}$?
\begin{align*}\dfrac{dy}{dx} &= 3x^{3-1}\\ &= 3x^2\end{align*}
That wasn't too bad, was it? Let's try another one.
#### Example
Find the derivative of $f(x) = x^5$.
\begin{align*}f'(x) &= 5x^{5-1}\\ &= 5x^4\end{align*}
Let's try a slightly trickier example:
#### Example
Find the derivative of $f(x) = \dfrac{1}{x^2}$.
Note: You can use your index laws to write $\dfrac{1}{x^2} = x^{-2}$. Then you just apply the power rule as usual.
\begin{align*}f'(x) &= \dfrac{d}{dx}(x^{-2})\\ &= (-2)x^{(-2 - 1)}\\ &= -2x^{-3}\\ &= \dfrac{-2}{x^3}\end{align*}
The power rule works for fractional indices as well!
#### Example
Find the derivative of $f(x) = \sqrt{x}$.
Note: You can use your index laws to write $\sqrt{x} = x^{\frac{1}{2}}$. Then you just apply the power rule as usual.
\begin{align*}f'(x) &= \dfrac{d}{dx}(x^{\frac{1}{2}})\\ &= \dfrac{1}{2}\;x^{(\frac{1}{2} - 1)}\\ &= \dfrac{1}{2}\;x^{-\frac{1}{2}}\\ &= \dfrac{1}{2 \sqrt{x}}\end{align*}
## Patterns in the derivatives
Let's look at a table of derivatives of powers of $x$ and see if we can spot a pattern:
$y = x^n$ $nx^{n-1}$ $\dfrac{dy}{dx}$
$x$ $1x^{(1-1)} = x^0 = 1$ $1$
$x^2$ $2x^{2-1} = 2x^1 = 2x$ $2x$
$x^3$ $3x^{3 - 1} = 3x^2$ $3x^2$
$x^4$ $4x^{4 - 1} = 4x^3$ $4x^3$
... ... ...
The coefficients and powers of the derivatives step up by $1$ as $n$ steps up by $1$.
The same thing happens for negative powers:
$y = x^n$ $nx^{n-1}$ $\dfrac{dy}{dx}$
$x^{-1}$ $(-1)x^{(-1-1)} = -x^{-2}$ $-x^{-2}$
$x^{-2}$ $-2x^{-2-1} = -2x^{-3}$ $-2x^{-3}$
$x^{-3}$ $-3x^{-3 - 1} = -3x^{-4}$ $-3x^{-4}$
... ... ...
## Solving Sam's Problem
Now we know enough to solve Sam's problem. Sam's function was:
$\text{sandwich}(t) = t^{-2}$,
and he wants to know his sandwich's rate of decomposition after 2 days. So he needs to find the derivative of the sandwich function (crumbs?) and plug in $t = 2$. It looks like a job for the power rule!
Let's differentiate:
$\text{sandwich}'(t) = (-2)t^{-2 - 1} = -2t^{-3}$.
At $t = 2$:
$\text{sandwich}'(2) = (-2)t^{-2 - 1} = -2(2)^{-3} = \dfrac{-2}{8} = -\dfrac{1}{4}$.
So the sandwich is decomposing at a rate of $\dfrac{1}{4}$ grams per day.
You know what, Sam? I think it might be better for everyone if you throw that sandwich in the bin!
### Description
Calculus is the branch of mathematics that deals with the finding and properties of derivatives and integrals of functions, by methods originally based on the summation of infinitesimal differences. The two main types are differential calculus and integral calculus.
### Environment
It is considered a good practice to take notes and revise what you learnt and practice it.
### Learning Objectives
Familiarize yourself with Calculus topics such as Limits, Functions, Differentiability etc
Author: Subject Coach
Added on: 23rd Nov 2017
You must be logged in as Student to ask a Question.
None just yet!
|
HuggingFaceTB/finemath
|
|
# How to calculate the number of parameters of a convolutional layer?
I was recently asked at an interview to calculate the number of parameters for a convolutional layer. I am deeply ashamed to admit I didn't know how to do that, even though I've been working and using CNN for years now.
Given a convolutional layer with ten $$3 \times 3$$ filters and an input of shape $$24 \times 24 \times 3$$, what is the total number of parameters of this convolutional layer?
## What are the parameters in a convolutional layer?
The (learnable) parameters of a convolutional layer are the elements of the kernels (or filters) and biases (if you decide to have them). There are 1d, 2d and 3d convolutions. The most common are 2d convolutions, which are the ones people usually refer to, so I will mainly focus on this case.
## 2d convolutions
### Example
If the 2d convolutional layer has $$10$$ filters of $$3 \times 3$$ shape and the input to the convolutional layer is $$24 \times 24 \times 3$$, then this actually means that the filters will have shape $$3 \times 3 \times 3$$, i.e. each filter will have the 3rd dimension that is equal to the 3rd dimension of the input. So, the 3rd dimension of the kernel is not given because it can be determined from the 3rd dimension of the input.
2d convolutions are performed along only 2 axes (x and y), hence the name. Here's a picture of a typical 2d convolutional layer where the depth of the kernel (in orange) is equal to the depth of the input volume (in cyan).
Each kernel can optionally have an associated scalar bias.
At this point, you should already be able to calculate the number of parameters of a standard convolutional layer. In your case, the number of parameters is $$10 * (3*3*3) + 10 = 280$$.
### A TensorFlow proof
The following simple TensorFlow (version 2) program can confirm this.
import tensorflow as tf
def get_model(input_shape, num_classes=10):
model = tf.keras.Sequential()
model.summary()
return model
if __name__ == '__main__':
input_shape = (24, 24, 3)
get_model(input_shape)
You should try setting use_bias to False to understand how the number of parameters changes.
### General case
So, in general, given $$M$$ filters of shape $$K \times K$$ and an input of shape $$H \times W \times D$$, then the number of parameters of the standard 2d convolutional layer, with scalar biases, is $$M * (K * K * D) + M$$ and, without biases, is $$M * (K * K * D)$$.
## 1d and 3d convolutions
There are also 1d and 3d convolutions.
For example, in the case of 3d convolutions, the kernels may not have the same dimension as the depth of the input, so the number of parameters is calculated differently for 3d convolutional layers. Here's a diagram of 3d convolutional layer, where the kernel has a depth different than the depth of the input volume.
• The diagrams were taken from ai.stackexchange.com/a/13786/2444. – nbro Mar 17 '20 at 0:36
• Wow! Simply wow! Thank you very much for the answer. I am normally a visual guy and this is what works, seeing how the data flows. I still have a lot of questions, but this isn't the place as you have more than answered my question. – Ælex Mar 17 '20 at 10:34
For a standard convolution layer, the weight matrix will have a shape of (out_channels, in_channels, kernel_sizes*) in addition you will need a vector of shape [out_channels] for biases. For your specific case, 2d, your weight matrix will have a shape of (out_channels, in_channels, kernel_size[0], kernel_size[1]). Now if we plugin the numbers:
• out_channels = 10, you're having 10 filters
• in_channels = 3 the picture is RGB in this case so there are 3 channels (the last dimension of the input)
• kernel_size[0] = kernel_size[1] = 3
In total you're gonna have 10*3*3*3 + 10 = 280 parameters.
|
open-web-math/open-web-math
|
|
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Quiz1 v2 Solutions
# Quiz1 v2 Solutions - MAC 2312 Quiz 1 Solutions 1 Evaluate...
This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: MAC 2312 Quiz 1 January 17, 2012 Solutions 1. Evaluate the indefinite integral sin3 x dx. (cos x)3/2 Solution: First, write sin3 x = sin2 x sin x = (1 - cos2 x) sin x so the given integral may be rewritten as sin3 x dx = (cos x)3/2 1 - cos2 x sin x dx. (cos x)3/2 (1) Next, use the change of variable w = cos x, dw = - sin x dx to rewrite the right-hand side of equation (1) as - 1 - w2 dw = w 3/2 = = w2 - 1 dw w 3/2 w 1/2 - w -3/2 dw 2 3/2 w + 2w -1/2 + C. 3 Finally, use w = cos x on the right-hand side of the above string of equalities to obtain 2 sin3 x dx = (cos x)3/2 + 2(cos x)-1/2 + C. 3/2 (cos x) 3 2. Evaluate the definite integral. Simplify your answer. e3 ln(x1/3 ) dx 1 Solution: Observe first that ln(x1/3 ) = 1 3 e3 1 1 3 ln x, so the given integral is equal to e3 1 ln x dx. Therefore, it suffices to first compute ln x dx then multiply the result by 1 . 3 To compute e3 1 ln x dx, use integration by parts with u = ln x 1 du = x dx dv = dx v = x. This gives e3 1 ln x dx = x ln x 1 dx x 1 1 = 3e3 - 0 - (e3 - 1) = 2e3 + 1. - x e3 e3 Multiplying answer 1 3 to the right-hand side of the above string of equalities gives the final e3 1 1 ln(x1/3 ) dx = (2e3 + 1). 3 ...
View Full Document
{[ snackBarMessage ]}
### Page1 / 2
Quiz1 v2 Solutions - MAC 2312 Quiz 1 Solutions 1 Evaluate...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
|
HuggingFaceTB/finemath
|
|
# estimate là gì
From Wikipedia, the miễn phí encyclopedia
Bạn đang xem: estimate là gì The exact number of candies in this jar cannot be determined by looking at it, because most of the candies are not visible. The amount can be estimated by presuming that the portion of the jar that cannot be seen contains an amount equivalent to lớn the amount contained in the same volume for the portion that can be seen.
Estimation (or estimating) is the process of finding an estimate or approximation, which is a value that is usable for some purpose even if input data may be incomplete, uncertain, or unstable. The value is nonetheless usable because it is derived from the best information available. Typically, estimation involves "using the value of a statistic derived from a sample to lớn estimate the value of a corresponding population parameter". The sample provides information that can be projected, through various formal or informal processes, to lớn determine a range most likely to lớn describe the missing information. An estimate that turns out to lớn be incorrect will be an overestimate if the estimate exceeds the actual result and an underestimate if the estimate falls short of the actual result.
## How estimation is done
Estimation is often done by sampling, which is counting a small number of examples something, and projecting that number onto a larger population. An example of estimation would be determining how many candies of a given size are in a glass jar. Because the distribution of candies inside the jar may vary, the observer can count the number of candies visible through the glass, consider the size of the jar, and presume that a similar distribution can be found in the parts that can not be seen, thereby making an estimate of the total number of candies that could be in the jar if that presumption were true. Estimates can similarly be generated by projecting results from polls or surveys onto the entire population.
In making an estimate, the goal is often most useful to lớn generate a range of possible outcomes that is precise enough to lớn be useful but not so sánh precise that it is likely to lớn be inaccurate. For example, in trying to lớn guess the number of candies in the jar, if fifty were visible, and the total volume of the jar seemed to lớn be about twenty times as large as the volume containing the visible candies, then one might simply project that there were a thousand candies in the jar. Such a projection, intended to lớn pick the single value that is believed to lớn be closest to lớn the actual value, is called a point estimate. However, a point estimation is likely to lớn be incorrect, because the sample size—in this case, the number of candies that are visible—is too small a number to lớn be sure that it does not contain anomalies that differ from the population as a whole. A corresponding concept is an interval estimate, which captures a much larger range of possibilities, but is too broad to lớn be useful. For example, if one were asked to lớn estimate the percentage of people who lượt thích candy, it would clearly be correct that the number falls between zero and one hundred percent. Such an estimate would provide no guidance, however, to lớn somebody who is trying to lớn determine how many candies to lớn buy for a tiệc nhỏ to lớn be attended by a hundred people.
## Uses of estimation
In mathematics, approximation describes the process of finding estimates in the size of upper or lower bounds for a quantity that cannot readily be evaluated precisely, and approximation theory giao dịch with finding simpler functions that are close to lớn some complicated function and that can provide useful estimates. In statistics, an estimator is the formal name for the rule by which an estimate is calculated from data, and estimation theory giao dịch with finding estimates with good properties. This process is used in signal processing, for approximating an unobserved signal on the basis of an observed signal containing noise. For estimation of yet-to-be observed quantities, forecasting and prediction are applied. A Fermi problem, in physics, is one concerning estimation in problems that typically involve making justified guesses about quantities that seem impossible to lớn compute given limited available information.
Estimation is important in business and economics because too many variables exist to lớn figure out how large-scale activities will develop. Estimation in project planning can be particularly significant, because plans for the distribution of labor and purchases of raw materials must be made, despite the inability to lớn know every possible problem that may come up. A certain amount of resources will be available for carrying out a particular project, making it important to lớn obtain or generate a cost estimate as one of the vital elements of entering into the project. The U.S. Government Accountability Office defines a cost estimate as, "the summation of individual cost elements, using established methods and valid data, to lớn estimate the future costs of a program, based on what is known today", and reports that "realistic cost estimating was imperative when making wise decisions in acquiring new systems". Furthermore, project plans must not underestimate the needs of the project, which can result in delays while unmet needs are fulfilled, nor must they greatly overestimate the needs of the project, or else the unneeded resources may go to lớn waste.
An informal estimate when little information is available is called a guesstimate because the inquiry becomes closer to lớn purely guessing the answer. The "estimated" sign, ℮, is used to lớn designate that package contents are close to lớn the nominal contents.
Xem thêm: landscape là gì
• Abundance estimation
• Ansatz
• Ballpark estimate
• Back-of-the-envelope calculation
• Conjecture
• Cost estimate
• Estimation statistics
• Estimation theory
• Fermi problem
• German tank problem
• State observer
• Kalman filter
• Intuition
• Mark and recapture
• Moving horizon estimation
• Sales quote
• Upper and lower bounds
|
HuggingFaceTB/finemath
|
|
# D4and5 Coulombs Law Worksheet with reading assignment attached
“PHLYZICS”
Coulomb’s Law
Last chapter, we learned that masses attract each other. We learned that the attractive force
between two masses could be found using Newton’s Universal Law of Gravitation (Newton’s
GmM
ULOG), Fg
, where R is the distance between the centers of the objects. “G” was simply a
R2
constant, and its units could easily be found by solving the above equation for G.
~~~
When dealing with charged objects, we also talk of forces between them. Unlike last chapter,
these forces can be EITHER attractive (for unlike charges) OR repulsive (for like charges). To
kqQ
find the force between charged objects, we can use Coulomb’s Law, which is Fe
. In this
d2
equation, “d” is the distance between the objects, q and Q are the charges on the charged objects,
N m2
and k is a constant equal to k 8.99 10 9
. This law is similar in form and structure to
C2
Newton’s ULOG, and the relationships that we used last chapter still apply. For example,
Fe
Fe
Fe
Fe
q (ex: If you double the charge on an object, you double the force)
Q (ex: If you quarter the charge on an object, you quarter the force)
qQ (ex: If you double both charges, you quadruple the force)
1
(ex: If you double the distance between the charges, you quarter
d2
the force. If you divide the distance between the charges by 3,
you multiply the force by 9 times).
~~~~
NEVER, EVER plug signs into Coulomb’s law. If two charged objects have opposite signs,
plug only their magnitudes into the equation. At the end, make sure to account for the fact that
they have different signs by saying that the forces were either “repulsive”. If the charges had
opposite signs, you account for this by saying that the charges
~~~~~
kqQ
together in
d2
problem solving. If you are told how many electrons an object gains or loses, you can easily
calculate “q” or “Q” to plug into Coulomb’s Law.
It is important to see that we can easily use q ne (or Q
ne ) and Fe
~~~~
Similarly to Newton’s ULOG, Coulomb’s Law is a vector law. When multiple charges are
involved, make sure to draw the forces acting on each charged object (as vector arrows). Again,
do not put signs into Coulomb’s law, as the vector arrows provide sufficient direction. Once the
object has all the forces draw on it (free-body diagram style), then the forces can be added as
vectors.
Coulomb’s Law Problems
K = 8.99E9 N-m2/C2
e = 1.602E-19 C
me = 9.11E-31 kg
mp = 1.67E-27 kg
1. Two charged spheres 10 cm apart attract each other with a force of 3.0 x 10
force results from each of the following changes, considered separately?
6
N. What
a) Both charges are doubled and the distance remains the same.
b) An uncharged, identical sphere is touched to one of the spheres, and then taken far away.
c) The separation is increased to 30 cm.
2. The force of electrostatic repulsion between two small positively charged objects, A and B, is
3.6 x 10 5 N when AB = 0.12m. What is the force of repulsion if AB is increased to
a) 0.24 m
b) 0.36 m
3. Calculate the force between charges of 5.0 x 10
apart.
4. What is the magnitude of the force a 1.5 x 10
located 1.5 m away?
6
8
C and 1.0 x 10
7
C if they are 5.0 cm
C charge exerts on a 3.2 x 10
4
C charge
5. Two spheres; 4.0 cm apart, attract each other with a force of 1.2 x 10 9 N. Determine the
magnitude of the charge on each, if one has twice the charge (of the opposite sign) as the
other.
6. Two equal charges of magnitude 1.1 x 10 7 C experience an electrostatic force of
4.2 x 10 4 N. How far apart are the centers of the two charges?
7. How many electrons must be removed from a neutral, isolated conducting sphere to give it a
positive charge of 8.0 x 10 8 C?
8. What will be the force of electric repulsion between two small spheres placed 1.0 m apart, if
each has a deficit of 108 electrons?
9. Two identical, small spheres of mass 2.0 g are fastened to the ends
of a 0.60m long light, flexible, insulating fishing line. The
fishing line is suspended by a hook in the ceiling at its exact
centre. The spheres are each given an identical electric charge.
They are in static equilibrium, with an angle of 30 between the
string halves, as shown. Calculate the magnitude of the charge
on each sphere. (Hint: start off by drawing a FULL, DETAILED
FBD of one of the charged spheres).
.30 m
10. Three negatively charged spheres, each with a charge of 4.0 x 10 6 C, are fixed at the
vertices of an equilateral triangle whose sides are 20 cm long. Calculate the magnitude and
direction of the net electric force on each sphere.
11. Three objects, carrying charges of 4.0 x 10 6 C, 6.0 x 10 6 C, and +9.0 x 10 6 C,
respectively, are placed in a line, equally spaced from left to right by a distance of 0.50 m.
Calculate the magnitude and direction of the net force acting on each charge that results
from the presence of the other two.
12. Delicate measurements indicate that the Earth has an electric field surrounding it, similar to
that around a positively charged sphere. Its magnitude at the surface of the Earth is about
100 N/C. What charge would an oil drop of mass 2.0 x 10 15 kg have to have, in order to
remain suspended by the Earth’s electric field? Give your answer in Coulombs ?
13. Compute the gravitational force and the electric force between the electron and the proton in
the hydrogen atom if they are 5.3 x 10-11 meters apart. Then calculate the ratio of Fe to Fg.
14. The earth attracts us with its gravitational force. Why doesn’t it attract us with an electric
force (especially since Fe is usually so much greater than Fg)?
15. Two point charges of -2.0 C are fixed at opposite ends of a meter stick. Where on the meter
stick (if anywhere!) could (a) a free electron and (b) a free proton be placed so that they are
in electrostatic equilibrium (and won’t move).
16. Redo the previous problem, this time having charges of -2.0 C and +2.0 C on either end.
17. Redo the previous problem again, this time having charges of +2.0 C and +4.0 C on either
end.
18. Using the same orbital distance from problem #13 above, find the orbital speed and the
centripetal acceleration (in g’s) of an electron orbiting the nucleus of a hydrogen atom
(assuming the orbit to be circular).
1.
2.
a) 1.2 x 10
b) 1.5 x 10
c) 3.3 x 10
5
N
N
7
N
9.
1.2 x 10
10.
6.2 N [outward, 150 away from each side]
a) 9.0 x 10
b) 4.0 x 10
6
11.
0.54 N [left], 2.8 N [right], 2.3 N [left]
12.
1.96 x 10
13.
3.6E-47 N, 8.2E-8 N, 2.3E39
14.
Objects on the earth are neutral.
15.
50 cm, 50 cm
16.
Nowhere, Nowhere
17.
41.4 cm away from the +2mC charge
18.
2.2E6 m/s, 9.2E21 g’s
3.
1.8 x 10
4.
1.9 N
5.
1.0 x 10
6.
0.51 m
7.
8.
5.0 x 10
2
11
11
2.3 x 10
6
N
N
N
C; 2.0 x 10
electrons
12
7
C
6
N
11
C
16
C
|
HuggingFaceTB/finemath
|
|
## Conversion formula
The conversion factor from gallons to liters is 3.7854118, which means that 1 gallon is equal to 3.7854118 liters:
1 gal = 3.7854118 L
To convert 138.3 gallons into liters we have to multiply 138.3 by the conversion factor in order to get the volume amount from gallons to liters. We can also form a simple proportion to calculate the result:
1 gal → 3.7854118 L
138.3 gal → V(L)
Solve the above proportion to obtain the volume V in liters:
V(L) = 138.3 gal × 3.7854118 L
V(L) = 523.52245194 L
The final result is:
138.3 gal → 523.52245194 L
We conclude that 138.3 gallons is equivalent to 523.52245194 liters:
138.3 gallons = 523.52245194 liters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 liter is equal to 0.001910137753012 × 138.3 gallons.
Another way is saying that 138.3 gallons is equal to 1 ÷ 0.001910137753012 liters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred thirty-eight point three gallons is approximately five hundred twenty-three point five two two liters:
138.3 gal ≅ 523.522 L
An alternative is also that one liter is approximately zero point zero zero two times one hundred thirty-eight point three gallons.
## Conversion table
### gallons to liters chart
For quick reference purposes, below is the conversion table you can use to convert from gallons to liters
gallons (gal) liters (L)
139.3 gallons 527.308 liters
140.3 gallons 531.093 liters
141.3 gallons 534.879 liters
142.3 gallons 538.664 liters
143.3 gallons 542.45 liters
144.3 gallons 546.235 liters
145.3 gallons 550.02 liters
146.3 gallons 553.806 liters
147.3 gallons 557.591 liters
148.3 gallons 561.377 liters
|
HuggingFaceTB/finemath
|
|
# Sequential analysis of variance table for Fitted Line Plot
Find definitions and interpretations for every statistic in the Sequential Analysis of Variance table.
## DF
The total degrees of freedom (DF) are the amount of information in your data. The analysis uses that information to estimate the values of unknown population parameters. The total DF is determined by the number of observations in your sample. The DF for a term show how much information that term uses. Increasing your sample size provides more information about the population, which increases the total DF. Increasing the number of terms in your model uses more information, which decreases the DF available to estimate the variability of the parameter estimates.
If two conditions are met, then Minitab partitions the DF for error. The first condition is that there must be terms you can fit with the data that are not included in the current model. For example, if you have a continuous predictor with 3 or more distinct values, you can estimate a quadratic term for that predictor. If the model does not include the quadratic term, then a term that the data can fit is not included in the model and this condition is met.
The second condition is that the data contain replicates. Replicates are observations where each predictor has the same value. For example, if you have 3 observations where pressure is 5 and temperature is 25, then those 3 observations are replicates.
If the two conditions are met, then the two parts of the DF for error are lack-of-fit and pure error. The DF for lack-of-fit allow a test of whether the model form is adequate. The lack-of-fit test uses the degrees of freedom for lack-of-fit. The more DF for pure error, the greater the power of the lack-of-fit test.
## SS
Sequential sums of squares (SS) are measures of variation for different components of the model. Unlike the adjusted sums of squares, the sequential sums of squares depend on the order the terms are entered into the model. In the sequential analysis of variance table, Minitab separates the sequential sums of squares by the polynomial terms (i.e. linear, quadratic, and cubic) in the model.
SS
The sequential sums of squares for each polynomial term is the unique portion of the variation explained by that term that is not explained by the lower order terms already entered into the model. It quantifies the amount of variation in the response data that is explained by each polynomial term as it is sequentially added to the model.
Seq SS Total
The total sum of squares is the sum of the term sum of squares and the error sum of squares. It quantifies the total variation in the data.
### Interpretation
Minitab uses the sequential sums of squares to calculate the p-value for a term. Minitab also uses the sums of squares to calculate the R2 statistic. Usually, you interpret the p-values and the R2 statistic instead of the sums of squares.
## F-value
The F-value is the test statistic used to determine whether the model is associated with the response.
### Interpretation
Minitab uses the F-value to calculate the p-value, which you use to make a decision about the statistical significance of the model. The p-value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis.
A sufficiently large F-value indicates that the model is significant.
If you want to use the F-value to determine whether to reject the null hypothesis, compare the F-value to your critical value. You can calculate the critical value in Minitab or find the critical value from an F-distribution table in most statistics books. For more information on using Minitab to calculate the critical value, go to Using the inverse cumulative distribution function (ICDF) and click "Use the ICDF to calculate critical values".
## P-Value – Term
The p-value is a probability that measures the evidence against the null hypothesis. Lower probabilities provide stronger evidence against the null hypothesis.
### Interpretation
To determine whether the association between the response and each term in the model is statistically significant, compare the p-value for the term to your significance level to assess the null hypothesis. The null hypothesis is that the term's coefficient is equal to zero, which indicates that there is no association between the term and the response. Usually, a significance level (denoted as α or alpha) of 0.05 works well. A significance level of 0.05 indicates a 5% risk of concluding that an association exists when there is no actual association.
P-value ≤ α: The association is statistically significant
If the p-value is less than or equal to the significance level, you can conclude that there is a statistically significant association between the response variable and the term. If you fit a quadratic model or a cubic model and the quadratic or cubic terms are significant, you can conclude that the data contain curvature.
P-value > α: The association is not statistically significant
If the p-value is greater than the significance level, you cannot conclude that there is a statistically significant association between the response variable and the term. If you fit a quadratic model or a cubic model and the quadratic or cubic terms are not statistically significant, you may want to select a different model.
By using this site you agree to the use of cookies for analytics and personalized content. Read our policy
|
HuggingFaceTB/finemath
|
|
You are an abcteach Member, but you are logged in to the Free Site. To access all member features, log into the Member Site.
# SEARCH RESULTS: denominator
Clip Art:
= Preview Document = Member Site Document
There are 22 documents matching your search.
• This animated video reviews concepts learned in the "Basic Fractions" videos and introduces the concepts of lowest terms and greatest common factor. With multiple examples, students will learn how to find the greatest common factor between a numerator and denominator and reduce the fraction to its lowest terms.
• This animated video reviews concepts learned in the "Basic Fractions" videos and introduces the concepts of lowest terms and greatest common factor. With multiple examples, students will learn how to find the greatest common factor between a numerator and denominator and reduce the fraction to its lowest terms.
• This Comparing Fractions (upper elem) Fractions is perfect to practice fraction skills. Your elementary grade students will love this Comparing Fractions (upper elem) Fractions. One worksheet for comparing fractions with like denominators; one worksheet for comparing fractions with the same numerator and unlike denominators.
• From "denominator" to "numerator". Review fractions vocabulary with this word search. Answer sheet included.
• This Comparing Fractions Math is perfect to practice fraction skills. Your elementary grade students will love this Comparing Fractions Math. Determine whether a fraction is greater than, less than, or equal to another fraction. 20 practice problems with answers. CC: Math: 4.NF.A.2
• This Fraction Addition- Fractions with common denominators. (set 1) Worksheet is perfect to practice fraction addition skills. Your elementary grade students will love this Fraction Addition- Fractions with common denominators. (set 1) Worksheet. Printable fraction worksheets, addition fraction practice, with common denominators.
• This Math Fractions Packet (grade 3) Common Core is perfect to practice fraction skills. Your elementary grade students will love this Math Fractions Packet (grade 3) Common Core. This 9 page fraction packet will help your student grasp the idea of fractions through shapes, and then learn to convert them into numbers. This lesson will cover methods such as; writing fractions, numeric fractions, understanding numerator and denominator, and using shapes to compare and gain understanding of fraction concepts.
• Roll dice to make a number for skill practice. Includes pages for Decimal, Fraction, Improper Fraction, and Mixed Number. Decimal skills include: place value, rounding, addition, subtraction, money, doubling, number line, and writing in standard form, word form, expanded form, and expanded notation. Fraction skills include: numberator, denominator, word form, equivalents, decomposing, conversation, addition, subtraction, and drawing the fraction as part of a whole, set, strip diagram, and on a number line.
• This Fractions:Grades 4-5 is perfect to practice fraction skills. Your elementary grade students will love this Fractions:Grades 4-5. Fraction action includes three pages of fractions that need to be reduced to the lowest terms. Common Core Math: Fractions and Operations
• Practice ordering fractions by dragging the fractions into the correct order, from smallest to largest. Includes easy (common denominators) and hard (different denominators). CC: Math: 4.NF.A.2
• Práctica de comparación de numeradores y denominadores. Practice: Comparing numerators and denominators.
• This Fractions (upper elem) - b/w Mini Office is perfect to practice fraction skills. Your elementary grade students will love this Fractions (upper elem) - b/w Mini Office. Four pages of materials for a fraction-themed mini office: greatest common factor & least common denominator; finding equivalents; reducing, adding, subtracting, multiplying, and dividing fractions; converting improper fractions to mixed numbers; clever mnemonic devices.
• This Fractions (upper elem) - color Mini Office is perfect to practice fraction skills. Your elementary grade students will love this Fractions (upper elem) - color Mini Office. Four colorful pages of materials for a fraction-themed mini office: greatest common factor & least common denominator; finding equivalents; reducing, adding, subtracting, multiplying, and dividing fractions; converting improper fractions to mixed numbers; clever mnemonic devices.
• Interactive exercise for adding fractions with the same denominator. Requires visual demonstration of concepts. Interactive .notebook file for Smart Board or Notebook viewer. Includes related printable version.
• Ready-to-manipulate elements for working with fractions. Various shapes and models for denominators 2 through 12. Interactive .notebook file for Smart Board.
• A one page illustrated chart and word descriptions of U.S. money: five coins, four denominations of currency.
• This Comparing Fractions Math is perfect to practice fraction skills. Your elementary grade students will love this Comparing Fractions Math. Determine whether a fraction is greater than, less than or equal to another fraction. 20 practice problems with answers. CC: Math: 4.NF.A.2
• This Fraction Squares Math Puzzle is perfect to practice fraction skills. Your elementary grade students will love this Fraction Squares Math Puzzle. Match the fraction forms - number, word, parts of a whole, parts of a group - to solve the puzzle. Includes denominators from 2 to 8. Laminate for a fun, reuseable classroom game.
• This Fractions on a Number Line Practice Packet (grades 3-4) Fractions is perfect to practice counting skills. Your elementary grade students will love this Fractions on a Number Line Practice Packet (grades 3-4) Fractions. Four practice pages include these skills: identifying fractions on a number line and on a ruler, comparing fractions, and identifying equivalent fractions.
• #REF!
• This Fraction Pies & Number Lines Practice Packet Fractions is perfect to practice counting skills. Your elementary grade students will love this Fraction Pies & Number Lines Practice Packet Fractions. Four practice pages allow students to sharpen their skills by illustrating fraction values on pie charts and identify fractions on a number line.
• This Mat Fractions Mixed Numbers & Improper Fractions on a Number Line Common Core is perfect to practice fraction skills. Your elementary grade students will love this Mat Fractions Mixed Numbers & Improper Fractions on a Number Line Common Core. Three practice pages include these skills: identifying and converting improper fractions and mixed numbers on a number line and on a ruler. CC:Math:3.NF.A.2, 4.NF.A.1
|
HuggingFaceTB/finemath
|
|
# Properties
Label 3150l6 Conductor 3150 Discriminant 285768000000 j-invariant $$\frac{2251439055699625}{25088}$$ CM no Rank 0 Torsion Structure $$\Z/{2}\Z$$
# Learn more about
Show commands for: Magma / SageMath / Pari/GP
## Minimal Weierstrass equation
magma: E := EllipticCurve([1, -1, 0, -614367, 185502541]); // or
magma: E := EllipticCurve("3150l6");
sage: E = EllipticCurve([1, -1, 0, -614367, 185502541]) # or
sage: E = EllipticCurve("3150l6")
gp: E = ellinit([1, -1, 0, -614367, 185502541]) \\ or
gp: E = ellinit("3150l6")
$$y^2 + x y = x^{3} - x^{2} - 614367 x + 185502541$$
## Mordell-Weil group structure
$$\Z/{2}\Z$$
## Torsion generators
magma: TorsionSubgroup(E);
sage: E.torsion_subgroup().gens()
gp: elltors(E)
$$\left(\frac{1811}{4}, -\frac{1811}{8}\right)$$
## Integral points
magma: IntegralPoints(E);
sage: E.integral_points()
None
## Invariants
magma: Conductor(E); sage: E.conductor().factor() gp: ellglobalred(E)[1] Conductor: $$3150$$ = $$2 \cdot 3^{2} \cdot 5^{2} \cdot 7$$ magma: Discriminant(E); sage: E.discriminant().factor() gp: E.disc Discriminant: $$285768000000$$ = $$2^{9} \cdot 3^{6} \cdot 5^{6} \cdot 7^{2}$$ magma: jInvariant(E); sage: E.j_invariant().factor() gp: E.j j-invariant: $$\frac{2251439055699625}{25088}$$ = $$2^{-9} \cdot 5^{3} \cdot 7^{-2} \cdot 11^{3} \cdot 2383^{3}$$ Endomorphism ring: $$\Z$$ (no Complex Multiplication) Sato-Tate Group: $\mathrm{SU}(2)$
## BSD invariants
magma: Rank(E); sage: E.rank() Rank: $$0$$ magma: Regulator(E); sage: E.regulator() Regulator: $$1$$ magma: RealPeriod(E); sage: E.period_lattice().omega() gp: E.omega[1] Real period: $$0.684480732911$$ magma: TamagawaNumbers(E); sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] Tamagawa product: $$8$$ = $$1\cdot2\cdot2\cdot2$$ magma: Order(TorsionSubgroup(E)); sage: E.torsion_order() gp: elltors(E)[1] Torsion order: $$2$$ magma: MordellWeilShaInformation(E); sage: E.sha().an_numerical() Analytic order of Ш: $$1$$ (exact)
## Modular invariants
#### Modular form3150.2.a.i
magma: ModularForm(E);
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
$$q - q^{2} + q^{4} - q^{7} - q^{8} + 4q^{13} + q^{14} + q^{16} + 6q^{17} + 2q^{19} + O(q^{20})$$
For more coefficients, see the Downloads section to the right.
magma: ModularDegree(E); sage: E.modular_degree() Modular degree: 20736 $$\Gamma_0(N)$$-optimal: no Manin constant: 1
#### Special L-value
magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);
sage: r = E.rank();
sage: E.lseries().dokchitser().derivative(1,r)/r.factorial()
gp: ar = ellanalyticrank(E);
gp: ar[2]/factorial(ar[1])
$$L(E,1)$$ ≈ $$1.36896146582$$
## Local data
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
sage: E.local_data()
gp: ellglobalred(E)[5]
prime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\Delta$$) ord$$(j)_{-}$$
$$2$$ $$1$$ $$I_{9}$$ Non-split multiplicative 1 1 9 9
$$3$$ $$2$$ $$I_0^{*}$$ Additive -1 2 6 0
$$5$$ $$2$$ $$I_0^{*}$$ Additive 1 2 6 0
$$7$$ $$2$$ $$I_{2}$$ Non-split multiplicative 1 1 2 2
## Galois representations
The image of the 2-adic representation attached to this elliptic curve is the subgroup of $\GL(2,\Z_2)$ with Rouse label X17.
This subgroup is the pull-back of the subgroup of $\GL(2,\Z_2/2^3\Z_2)$ generated by $\left(\begin{array}{rr} 1 & 0 \\ 2 & 1 \end{array}\right),\left(\begin{array}{rr} 1 & 0 \\ 0 & 7 \end{array}\right),\left(\begin{array}{rr} 1 & 1 \\ 0 & 5 \end{array}\right)$ and has index 6.
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
The mod $$p$$ Galois representation has maximal image $$\GL(2,\F_p)$$ for all primes $$p$$ except those listed.
prime Image of Galois representation
$$2$$ B
$$3$$ B
## $p$-adic data
### $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]
All $$p$$-adic regulators are identically $$1$$ since the rank is $$0$$.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 nonsplit add add nonsplit 4 - - 0 0 - - 0
All Iwasawa $\lambda$ and $\mu$-invariants for primes $p\ge 5$ of good reduction are zero.
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2, 3, 6, 9 and 18.
Its isogeny class 3150l consists of 6 curves linked by isogenies of degrees dividing 18.
## Growth of torsion in number fields
The number fields $K$ of degree up to 7 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $K$ $E(K)_{\rm tors}$ Base-change curve
2 $$\Q(\sqrt{2})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database
$$\Q(\sqrt{5})$$ $$\Z/6\Z$$ Not in database
4 $$\Q(\sqrt{2}, \sqrt{5})$$ $$\Z/2\Z \times \Z/6\Z$$ Not in database
4.0.352800.3 $$\Z/4\Z$$ Not in database
6 6.0.656373375.5 $$\Z/6\Z$$ Not in database
6.6.1969120125.1 $$\Z/18\Z$$ Not in database
We only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database.
|
HuggingFaceTB/finemath
|
|
(Also available in WeScheme)
Students use function composition and the distance formula to detect when characters in their games collision.
Lesson Goals Students will be able to: Explain how the distance formula is related to the Pythagorean theorem. Write a function for the distance formula. Student-Facing Lesson Goals I can explain how the distance formula is connected to the Pythagorean theorem. I can write a function that takes in 2 points and returns the distance between them. Materials
## 🔗Problem Decomposition Returns! 20 minutes
### Overview
Students revisit problem decomposition - the idea of breaking down a complex problem into simpler pieces, solving those pieces separately, and then composig the solutions to solve the original..
### Launch
"Problem Decomposition" is a powerful tool, which lets us break apart complex problems into simpler ones that we can solve, test, and then glue together into a complex solution.
Students may remember that there are two strategies for doing this:
1. Top-Down: Describe the problem at a high level, then fill in the details later
2. Bottom-Up: Focus on the smaller parts that you’re sure of, then build them together to get the big picture
### Investigate
For the following complex word problem, have students first decide which strategy they want to use, and then apply the Design Recipe to build the functions they need.
• A retractable flag pole starts out 24 inches tall, and can grow at a rate of 0.6in/sec. An elastic is tied to the top of the pole and anchored 200 inches from the base, forming a right triangle. Write a function that computes the area of the triangle as a function of time.
• This is easier to think about as two functions:
• one that computes the height of the pole, based on the seconds
• another that computes the area of the triangle, based on the height
• Does one function depend on (or "sit on top of") the other? If so, which one?
• Yes - `area` depends on `height`.
• Which strategy will YOU use: bottom-up (independent first) or top-down (dependent first)?
• Students answers will vary! They can define either function first.
• Complete Top Down / Bottom Up, using your chosen strategy for Problem Decomposition!
### Synthesize
Note: Defining the `height` first is bottom-up, and solving `area` first is top-down.
• Which strategy did students use?
• Did they try starting with one function, and then switch to another?
• Invite students to share. Oftentimes, responses are not only intriguing but can highlight the value of each approach. Explicitly point out that the `area` function _uses `height`, allowing us to break a big problem down into two smaller ones._
## 🔗Collision Detection 20 minutes
### Overview
Students once again see function composition at work, as they compose a simple inequality with the `distance` function they’ve created.
### Launch
Knowing how far apart our characters are is the first step. We still need the computer to be asking: "True or False: is there a collision?"
### Investigate
• This is easier to think about as two functions:
• one that computes the distance between a Player and a Character, based on their coordinates
• another that checks if those same coordinates are less than 50 pixels apart, based on the distance
• Does one function rely on the other? If so, which one?
• Complete Word Problem: is-collision.
• When you’ve finished, open your saved game file and fix the @code function in your game file, and click "Run"!
### Synthesize
• In our flag-pole exercise, we had to write two functions and could start with one or the other. In our game, however, we began already having written the `distance` function! Is this Top-Down or Bottom-Up decomposition?
• Explicitly point out that the `is-collide` function uses `height`, allowing us to break a big problem down into two smaller ones.
• Connect this back to `profit` (from Problem Decomposition), which relied on `revenue` and `cost`. Which function(s) would a top-down strategy address first?
• Connect this back to `is-onscreen` (from Sam the Butterfly), which relied on `is-safe-left` and `is-safe-right`. Which function(s) would a bottom-up strategy address first?
|
HuggingFaceTB/finemath
|
|
• TB is the total training budget ($) • #E is the number of employees to be trained How to Calculate Training Cost? The following example problems outline how to calculate Training Cost. Example Problem #1: 1. First, determine the total training budget ($). The total training budget ($) is given as 9000. 2. Next, determine the number of employees to be trained. The number of employees to be trained is provided as 100. 3. Finally, calculate the Training Cost using the equation above: TC = TB / #E The values given above are inserted into the equation below: TC = 9000 / 100 = 900 ($/employee)
Example Problem #2:
The variables needed for this problem are provided below:
total training budget ($) = 7000 number of employees to be trained = 35 Entering these values and solving gives: TC = TB / #E = 200 ($/employee)
|
HuggingFaceTB/finemath
|
|
### How To Calculate Percentages
One of the most valuable skills in Maths EVER, am I right?
It’s totally underrated – it’s taught at a Year 9 level, and forgotten by the time the students reach Year 12. Then, once they move into the workforce, they have to try and relearn the whole process, despite the fact that it’s so easy!
So, here are some easy steps to make sure you never struggle again.
### Free Percentage Calculator
We won't send you spam. Unsubscribe at any time. Powered by ConvertKit
## Calculating x percent of a number.
Take the percentage you wish to find. Let’s use 20% as an example. There is a hidden decimal point at the end of that number (20.0%). Move that decimal two places to the left (Now you should have 0.20). Multiply that decimal by the number you are working with.
Example: To find 35% of 150, move the decimal place two places left to give 0.35, then multiply this by 150. (0.35 x 150 = 52.5)
## Calculating “what percentage is that”
If you have 120 out of 700, what percentage do you have?
To calculate, simply divide the smaller number by the bigger number, which will give you a decimal value. Moving the decimal point two spaces right will turn it into a percentage.
Example: 120 out of 700 as a percentage. 120 / 700 = 0.17, then move the decimal two spaces right to give 17%
## Adding x percent to a number
Increasing a number by a percentage is common in retail or sales.
To increase a number by x%, first consider that the number unchanged would represent 100%. Which means an increase of 20% would bring the total to 120%
Follow the same process as “finding a percentage of a number” using this value. Move the decimal two places to the left, then multiply.
Example: Calculate the price of a \$70 item after a 25% increase. This would mean that the new value is 125% of the original. Moving the decimal two places left gives 1.25. Then, 1.25 x \$70 = \$87.50
## Discounting x percent from a number
The reverse of adding an amount. This is easiest if we think that “10% off” actually means “90% remaining”.
With this concept, you can once again follow the same process as above.
Example: Find the price of a \$90 item after a 15% discount. 15% off means 85% remaining. Converting to a decimal give 0.85, then multiplying gives 0.85 x \$90 = \$76.50
|
HuggingFaceTB/finemath
|
|
# Is it possible to factor y=9x^2-48x+64? If so, what are the factors?
Aug 25, 2016
Yes
#### Explanation:
If you are having a problem factorising a quadratic by inspection then use the formula
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
In this example the discriminant ${b}^{2} - 4 a c$=0
So the factors are the same
$x = \frac{48}{2 \cdot 9} = \frac{8}{3}$
So the two factors are both $3 x - 8$
|
HuggingFaceTB/finemath
|
|
# Determining whether or not a group has an element of a specific order
If $|G| = 55$, must it have an element of order $5$ and/or $11$?
I'm not quite sure how to determine this. I know it could be possible by Lagrange's Theorem, but I'm stuck otherwise. Any help would be appreciated.
Edit: I haven't learned material about Cauchy's or Sylow Theorems yet. So I'm trying to prove this by very elementary facts.
Yes. Lagrange's theorem implies that every element has order 5, 11, or 55.
If there is an element of order $55$, call it $g$. Then $g^{11}$ is an element of order $5$ and $g^5$ is an element of order $11$.
So let's suppose that there are no elements of order $55$. If there are elements of order $5$ and $11$, then we are done. So suppose there are only elements of order $11$. Such an element generates a cyclic subgroup of $10$ non-identity elements. Moreover, any other element in the subgroup generates the same subgroup. Now pick another element in $G$ not in that cyclic subgroup. Each time we do this, we get $10$ more non-identity elements. But then the group must have $11, 21, 31, 41, 51, 61,\ldots$ elements. In particular, such a group cannot have $55$ elements.
A similar argument shows that a group with elements only of order $5$ cannot have $55$ elements.
Unwinding all of this, we see that we reach a contradiction unless there exist elements of order $5$ and $11$ in the group.
• Nit-pick --- Lagrange's Theorem implies that every element has order 5, 11, 55, or 1. – Gerry Myerson Jun 8 '13 at 4:31
• Thank you Gerry and Michael. – Adam Saltz Jun 8 '13 at 4:45
Here is a simple elementary proof:
Is there any element that generate the whole group? If there is, you're done.
If not, all elements are order 5 or 11 (except the identity). Each element of order 11 generate a subgroup of order 11, and any two such subgroup generated this way are either identical, or share exactly only the identity. Similarly for order 5.
So if there are no order 5 element, then the group must have order $55=(11-1)k+1$ which is not possible. If there are no order 11 element, then the group must have order $55=(5-1)k+1$ which is also not possible.
|
HuggingFaceTB/finemath
|
|
It is currently 16 Dec 2017, 07:16
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Cups of milk
Author Message
Director
Joined: 12 Jun 2006
Posts: 529
Kudos [?]: 172 [0], given: 1
### Show Tags
13 Sep 2007, 17:36
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?
(A) 2/3 (B) 7/3 (C) 5/2
(D) 8/3 (E) 3
Kudos [?]: 172 [0], given: 1
Intern
Joined: 14 Mar 2005
Posts: 28
Kudos [?]: 9 [0], given: 0
### Show Tags
13 Sep 2007, 20:39
I believe the answer is 8/3 (D)
x/4 = (4-x)/2
x = 2(4-x) = 8-2x
x = 8/3
so 8/3 into 4 cup bottle... ratio is 8/3/4 = 8/12 = 2/3
and 4-8/3 = 4/3 into 2 cup bottle..ratio is 4/3/2 = 4/6 = 2/3
also, 8/3 + 4/3 = 12/3 = 4 cups
answer can't be A because if 2/3 cup goes into the 4 cup then 4-2/3 = 10/3 cup would need to go into a 2 cup bottle....
Kudos [?]: 9 [0], given: 0
Director
Joined: 17 Sep 2005
Posts: 901
Kudos [?]: 118 [0], given: 0
### Show Tags
13 Sep 2007, 20:47
CalSpeedRacer wrote:
I believe the answer is 8/3 (D)
x/4 = (4-x)/2
x = 2(4-x) = 8-2x
x = 8/3
so 8/3 into 4 cup bottle... ratio is 8/3/4 = 8/12 = 2/3
and 4-8/3 = 4/3 into 2 cup bottle..ratio is 4/3/2 = 4/6 = 2/3
also, 8/3 + 4/3 = 12/3 = 4 cups
answer can't be A because if 2/3 cup goes into the 4 cup then 4-2/3 = 10/3 cup would need to go into a 2 cup bottle....
Actually we found the value of x (Fraction of the capacity of bottles) which is equal to 2/3.
However in the question, it is asked that "how many cups of milk should be poured into the 4-cup bottle"
So the answer should be x of 4
= x* 4
= 2/3 *4
= 8/3
hence D.
- Brajesh
Kudos [?]: 118 [0], given: 0
Intern
Joined: 02 Aug 2006
Posts: 12
Kudos [?]: 1 [0], given: 0
### Show Tags
14 Sep 2007, 06:13
I also got D
I said: x + y = 4 (i.e. x is the amount to be poured into the 2 cup bottle and y is the amount to be poured into the 4 cup bottle)
Then, x/2 = y/4 (ie. the amount to be poured into the 2 cup bottle is the same fraction of its capacity as the amount to be poured into the 4 cup holder)
Then I solved the simultaneous equations
x+y=4
x/2 = y/4
i.e. x/2 = y/4 becomes x = 2y/4.........then x =y/2
then I plugged y/2 for x in the equation x+y = 4
which becomes y/2 + y = 4.....then 3y =8 ...then y=8/3
y represents the amount poured into the 4 cup bottle, therefore answer is D
Kudos [?]: 1 [0], given: 0
Director
Joined: 12 Jun 2006
Posts: 529
Kudos [?]: 172 [0], given: 1
### Show Tags
14 Sep 2007, 21:09
CalSpeedRacer wrote:
I believe the answer is 8/3 (D)
x/4 = (4-x)/2
x = 2(4-x) = 8-2x
x = 8/3
so 8/3 into 4 cup bottle... ratio is 8/3/4 = 8/12 = 2/3
and 4-8/3 = 4/3 into 2 cup bottle..ratio is 4/3/2 = 4/6 = 2/3
also, 8/3 + 4/3 = 12/3 = 4 cups
answer can't be A because if 2/3 cup goes into the 4 cup then 4-2/3 = 10/3 cup would need to go into a 2 cup bottle....
using CalSpeed's method, if I were to do this:
4-x/4 = x/2 Iget x = 4/3. why doesn't this equation work? isn't it saying the same as the above?
Kudos [?]: 172 [0], given: 1
Director
Joined: 03 May 2007
Posts: 867
Kudos [?]: 279 [0], given: 7
Schools: University of Chicago, Wharton School
### Show Tags
14 Sep 2007, 21:59
Fistail wrote:
b14kumar wrote:
ggarr wrote:
Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?
(A) 2/3 (B) 7/3 (C) 5/2
(D) 8/3 (E) 3
Is it A?
- Brajesh
yup.
2x + 4x = 4
x = 4/6 = 2/3
was bit hasty..........................
2 cup bottle: 2/3 of 2 cups = 4/3
4 cup bottle: 2/3 of 4 cups = 8/3
Kudos [?]: 279 [0], given: 7
Senior Manager
Joined: 27 Aug 2007
Posts: 253
Kudos [?]: 14 [0], given: 0
### Show Tags
15 Sep 2007, 06:25
I did this question today, it is taken from PS 1000, am I right???
AO is D
Kudos [?]: 14 [0], given: 0
Director
Joined: 08 Jun 2007
Posts: 575
Kudos [?]: 111 [0], given: 0
### Show Tags
17 Sep 2007, 20:10
ggarr wrote:
Four cups of milk are to be poured into a 2-cup bottle and a 4-cup bottle. If each bottle is to be filled to the same fraction of its capacity, how many cups of milk should be poured into the 4-cup bottle?
(A) 2/3 (B) 7/3 (C) 5/2
(D) 8/3 (E) 3
This is how I read to solve the problem. The cup only confuses:
Four gallons of milk are to be poured into a 2 gallons bottle and a 4 gallons bottle. If each bottle is to be filled to the same fraction of its capacity, how many gallons of milk should be poured into the 4 gallons bottle?
Now milk added to fill 2x+4x = 6x = 4 => x= 2/3
Milk poured in 4-cup ( gallon) bottle = 4x = 8/3
Kudos [?]: 111 [0], given: 0
Director
Joined: 12 Jul 2007
Posts: 857
Kudos [?]: 341 [0], given: 0
### Show Tags
17 Sep 2007, 20:20
pretty straightforward guys.
2 cup + 4 cup = 6 cups of empty bottle
you have 4 cups of milk
4/6 = 2/3 so each one will be filled to 2/3 of capacity.
2/3(4) = 8/3
D. 8/3
Kudos [?]: 341 [0], given: 0
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5032
Kudos [?]: 457 [0], given: 0
Location: Singapore
### Show Tags
17 Sep 2007, 20:55
Bottle A - 2 cup bottle
Bottle B - 4 cup bottle
A/B = 1/2
# of cups into B = 2/3 * 4 = 8/3
Kudos [?]: 457 [0], given: 0
17 Sep 2007, 20:55
Display posts from previous: Sort by
|
HuggingFaceTB/finemath
|
|
## Painted Rock
In this lesson, children will use rocks numbered from one to 10 to practice sequencing skills.
### Math Lesson for:
Toddlers/Preschoolers
(See Step 5: Adapt lesson for toddlers or preschoolers.)
Algebra
### Learning Goals:
This lesson will help toddlers and preschoolers meet the following educational standards:
• Understand numbers, ways of representing numbers, relationships among numbers and number systems
• Understand patterns, relations and functions
### Learning Targets:
After this lesson, toddlers and preschoolers should be more proficient at:
• Using multiple models to develop initial understandings of place value and the base-ten number system
• Developing understanding of the relative position and magnitude of whole numbers and of ordinal and cardinal numbers and their connections
• Sorting, classifying and ordering objects by size, number and other properties
• Recognizing, describing and extending patterns such as sequences of sounds and shapes or simple numeric patterns and translating from one representation to another
• Analyzing how both repeating and growing patterns are generated
## Painted Rock
### Lesson plan for toddlers/preschoolers
#### Step 1: Gather materials.
• Flat rocks (There should be enough so that children have a set of five rocks, possibly 10 rocks)
• Paint and paintbrushes (You can paint and number the rocks beforehand, but it makes for a fun activity to have the children paint the rocks themselves.)
Note: Small parts pose a choking hazard and are not appropriate for children age five or under. Be sure to choose lesson materials that meet safety requirements.
#### Step 2: Introduce activity.
1. Review the sequence of numbers 1-10. When the children count from 1 to 10, ask questions such as: “What number comes after five? What number comes before seven?”
2. Explain that today the children are going to put their painted rocks in numerical order. Say: “We will start with rock number one and then put the rocks in order until we end with rock number 10.” (Use five rocks for the younger children.)
#### Step 3: Engage children in lesson activities.
1. Give the children their sets of rocks. Explain that they will begin with rock number one and place each rock side by side to form a line of rocks.
• You can make a memory game out of the rocks. Turn all of the rocks over with the numbers face down. The object of the game is to remember where each number is and then put the rocks in numerical order. For example, if the child is looking for number four and turns over number six, the child needs to turn over that rock and continue to turn over rocks until he/she finds the number four. It can also be turned into a game between two players.
• The children can put the rocks in descending order, starting with the number 10 and working down to number one.
#### Step 4: Math vocabulary.
• Numerical order: A number sequence by arranging numbers in ascending order (e.g.,“We will put the rocks in numerical order starting with number one and ending at number 10.”)
• Before: In front of (e.g.,“What number comes before the number three?”)
• After: Following, behind (e.g.,“What number come after the number three?”)
#### Step 5: Adapt lesson for toddlers or preschoolers.
###### Toddlers may:
• Still be working with numbers 1-5
• Still be working on number sequencing and counting to 10 in numerical order
###### Home child care providers may:
• Use five rocks when working on numerical order
• Provide a number line with numbers 1-10 so that the children can match up the numbers on the rock to the numbers on the number line
###### Preschoolers may:
• Easily count to 10
###### Home child care providers may:
• Make a memory game out of the rocks. Turn all of the rocks over with the numbers face down. The object of the game is to remember where each number is and then put the rocks in numerical order. For example, if the child is looking for the number four and turns over the number six, the child needs to turn over that rock and continue to turn over rocks until he/she finds the number four. It can also be turned into a game between two players.
• The children can put the rocks in descending order, starting with the number 10 and working down to number one.
### Suggested Books
• Ten Black Dots by Donald Crews (New York: Greenwillow Books, 1995)
• The Very Hungry Caterpillar by Eric Carle (London: Hamilton Hamish Children, 1994)
### Music and Movement
• Recite nursery rhymes and sing songs that include counting such as: “One, Two, Buckle My Shoe,” “There Were Ten in the Bed,” “This Old Man,””Five Little Ducks” and “The Ants Go Marching One by One.” This will give your child an opportunity to practice counting in a fun and playful manner. You can find free song lyrics and listen to melodies at www.kididdles.com.
### Outdoor Connections
Outside on the pavement, use chalk to draw a number line with the numbers 1-10. Have the children line up on the number line: One child on the number 1, two children on the number 2 and so forth. If you do not have enough children to do this, have the children use objects: One jump rope on the number 1, two trucks on the number 2 and so forth.
|
HuggingFaceTB/finemath
|
|
FOURIER SERIES LINKSf(x) = (Π-x)/2 x= 0 to 2Π Deduce Π/4 = 1 - 1/3 + 1/5 - 1/7 + - https://youtu.be/32Q0tMddoRwf(x) =x(2Π-x) x= 0 to 2Π Show
Similarly, log 2 64 = 6, because 2 6 = 64. Therefore, it is obvious that logarithm operation is an inverse one to exponentiation. 2016-12-18 · bar(3).4771 or -2.5229 log0.003 = log(3/1000) = log3-log1000 = log3-log10^3 = log3-3log10 = 0.4771-3 = bar(3).4771 or -2.5229 Note that in bar(3).4771 while characteristic 3 is negative, mantissa is positive and is 0.4771 log 2 (x∙(x-3)) = 2. Changing the logarithm form according to the logarithm definition: x∙(x-3) = 2 2. Or. x 2-3x-4 = 0. Solving the quadratic equation: x 1,2 = [3±√(9+16) ] / 2 = [3±5] / 2 = 4,-1. Similarly, log 2 64 = 6, because 2 6 = 64. Therefore, it is obvious that logarithm operation is an inverse one to exponentiation. This free log calculator solves for the unknown portions of a logarithmic expression using base e, 2, 10, or any other desired base. Learn more about log rules, or explore hundreds of other calculators addressing topics such as math, finance, health, and fitness, among others. Log 0.3 can be written as log (3/10) and as we know log (a/b) = log (a) - log(b).
1- log0.001 = -3 .
## Solution: log0:1 (d) (3 points) Compute the total cost, J, of the network averaged across the following dataset of 3 examples using the binary cross entropy loss. YT = (1, 0, 0), and Y^ T = (0.1, 0.2, 0.7). You can simply ll in the formula, without needing to simplify it. There is no penalty on the weights. (1 formula) Solution: J= 1 3
Logarithm. The logarithm of a number x with respect to base b is the exponent to which b has to be raised to yield x. ### This free log calculator solves for the unknown portions of a logarithmic expression using base e, 2, 10, or any other desired base. Learn more about log rules, or explore hundreds of other calculators addressing topics such as math, finance, health, and fitness, among others. · 3) Select the number base. You have 4 choices: base 'e', base '10', Because log0.8 is negative, and dividing by a negative number reverses inequalities (see for example this question). More generally, you have that log1= 0 Let x = log0.4 (1/3+1/3^2+1/3^3+..infinity). 2006-07-05 Well, You didn't mention the Base of log. Without base it is meaningless because”log ko Base pasand h” (only Indians will understand This) Back to the question If base is 3 then simply answer is 1. go to the standard if base is e then e^x=3 all i log3(x) = 2 Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. Type in any equation to get the solution, steps and graph In order to calculate Log values in SAS we will be using LOG Function. LOG Function in SAS consist of LOG, LOG2, LOG10 Function.
Saltx technology holding to6 b
Anthropology Files for log0, version 1.0.1; Filename, size File type Python version Upload date Hashes; Filename, size log0-1.0.1-py2.py3-none-any.whl (5.6 kB) File type Wheel Python version py2.py3 Upload date Jun 12, 2017 Hashes View Convert log 3 (6) to an expression with logs having a base of 5.
LONDON proportional parts correct to three decimal places. Hence the error in y/z or log0 3 is less. 15 Mar 2021 3 + 0.2 log0.2 + 0.5 log0.5 ) ≈ 1.030.
Vårdvetenskap i klinisk praxis pdf migrationsverket webbansökan arbetstillstånd
sälja teckningsrätter hur
seb mastercard reseförsäkring konkurs
neurokirurgiska kliniken karolinska
|
HuggingFaceTB/finemath
|
|
# Physics
A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 85.0 kg, and the hill is inclined at 19.0° with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?
1. 0
2. 0
3. 59
1. A bicyclist of mass 66 (including the bicycle) can coast down a 5.0 hill at a steady speed of because of air resistance.How much force must be applied to climb the hill at the same speed and same air resistance?
1. 0
2. 0
posted by jill
## Similar Questions
1. ### Physics
A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 55.0 kg, and the hill is inclined at 16.0° with respect to the horizontal. Air resistance opposes the motion of the cyclist.
asked by Kelsey on October 25, 2011
2. ### Physics
A bicyclist is coasting straight down a hill at a constant speed. The mass of the rider and bicycle is 80 kg and the hill is inclines at 15 degrees with respect to the horizontal. Air resistance opposes the motion of the cyclist.
asked by Morgan on October 9, 2016
3. ### Physics
A bicyclist on an old bike (combined mass: 96 kg) is rolling down (no pedaling or braking) a hill of height 103 m. Over the course of the 278 meters of downhill road, she encounters a constant friction force of 323 Newton. If her
asked by Layla on October 31, 2010
4. ### PHYSICS NEED HELP
A bicyclist can coast down a 5° hill at a constant 6.0 km/h. Assume the force of friction (air resistance) is proportional to the speed v so that Ffr = cv. (a) Calculate the value of the constant c. (b) What is the average force
asked by Manny on October 22, 2015
5. ### Physics
A bicyclist is coasting down a road that is sloped downward at an angle of 8°. The bicycle and bicyclist have a total mass of 80 kg. If the bicycle wheels roll freely, without any effect from friction or air resistance slowing
asked by Pat on November 3, 2011
6. ### physics
a bicyclist is coasting on level ground at 32.35m/s. The bicyclist coasts up an incline of 20.5m at a 30degree angle. How fast is the bicyclist traveling when he grabs the limb?
asked by cynthia on April 8, 2009
7. ### Physics
this one is kind of hard A bicyclist can coast down a 7.0 degree hill at a steady 9.5 h^-1 km. If the drag force is proportional to the square fo the speed v, so that Fd = cv^2, calculate (a) the value fo the constant c and (b)
asked by QUESTION on August 20, 2009
8. ### physics
A bicyclist of mass 60 (including the bicycle) can coast down a 9.0 hill at a steady speed of because of air resistance. Part A How much force must be applied to climb the hill at the same speed and same air resistance?
asked by Diamond on October 9, 2011
9. ### Physics
If a bicyclist of mass 68.0 kg (including the bicycle) can coast down a 7.00° hill at a steady speed of 6.20 km/h because of air resistance, how much force must be applied to climb the hill at the same speed (and the same air
asked by Nick on February 21, 2012
10. ### physics
What are the speed and acceleration (increase/decrease or remain constant) of a ball rolling down a a) straight hill, b) concave hill c) convex hill Can someone explain why they are, not just the answers?
asked by Christine on December 9, 2015
More Similar Questions
|
HuggingFaceTB/finemath
|
|
Is there an existing algorithm for this type of sorting?
This TED-ED video talks about some of the most basic sorting methods (bubble sort, insertion sort and quick sort,) in response to a scenario where a librarian ends up with a stack of 1,280 unsorted books, and must quickly sort them into alphabetical order.
A number of commenters pointed out that a far more practical method for real-life sorting would be to take the first letter of each book and put it into 26 different piles (an A pile, a B pile, down to a Z pile.) Then use a conventional sorting method (insertion / quick sort etc.) for each of the 26 sub-piles (which would contain around 50 books each, assuming a roughly even distribution of letters. Or you could repeat the same process if the sub-pile is significantly large, sorting the As into ABs, ACs, ADs and so on.
Is there a name of this kind of sorting algorithm?
• Radix sort or bucket sort. – Yuval Filmus Jan 13 at 21:32
• Sounds like bucket sort is the one those commenters were thinking of. If you put that in an answer I can accept it :) – Lou Jan 13 at 21:35
• There was a recent development over the insertion sort, that is called the Library sort or gapped insertion sort. That is the way the librarians keep their books. – kelalaka Jan 13 at 22:25
This is called bucket sort: split your data range into $$k$$ "buckets" ($$O(k)$$), put the data into the buckets ($$O(n)$$), sort each bucket (however you like), then concatenate all the buckets ($$O(k)$$). If the sorting is faster than $$O(n \log n)$$, then the whole algorithm runs faster than any comparison-based sort possibly could!
A variation of this, "radix sort", is basically the same except in binary (and with some extra recursion). It runs in $$O(wn)$$, where $$w$$ is the bit length of your data (the "word size").
|
HuggingFaceTB/finemath
|
|
### Home > CCG > Chapter Ch7 > Lesson 7.2.2 > Problem7-69
7-69.
Jester started to prove that the triangles below are congruent. He was only told that point $E$ is the midpoint of segments $\overline{AC}$ and $\overline{BD}$.
Copy and complete his flowchart below. Be sure that a reason is provided for every statement.
$\enclose{circle}{\; \\ \qquad \text{E is a midpoint} \qquad \\}$ $\huge \swarrow$Bubble 1 has an arrow pointing down to bubble 3. $\huge \searrow$ Given Bubble 1 has an arrow pointing down to bubble 4. $\enclose{circle}{\; \\ \qquad \angle AEB \cong \angle CED \qquad \\}$Angle A, E, B, is congruent to angle C, E, DVertical angles are congruent. $\enclose{circle}{\; \\ \qquad \ \ \ \qquad \qquad \\}$Fill in bubble 3 $\enclose{circle}{\; \\ \qquad \overline{AE} \cong \overline{CE} \qquad \\}$A, E is congruent to C, E Definition of midpoint. $\huge \searrow$Bubble 2 has an arrow pointing down to bubble 5. $\huge \downarrow$Bubble 3 has an arrow pointing down to bubble 5. $\huge \swarrow$Bubble 4 has an arrow pointing down to bubble 5. $\enclose{circle}{\; \\ \qquad \ \ \ \qquad \qquad \\}$Fill in bubble 5
$\enclose{circle}{\; \\ \qquad \text{E is a midpoint} \qquad \\}$ $\huge \swarrow$Bubble 1 has an arrow pointing down to bubble 3. $\huge \searrow$ Given Bubble 1 has an arrow pointing down to bubble 4. $\enclose{circle}{\; \\ \qquad \angle AEB \cong \angle CED \qquad \\}$Vertical angles are congruent. $\enclose{circle}{\; \\ \qquad \bf{\color{red}{\overline{BE}\cong \overline{DE}}} \qquad \\}$$\bf{\color{red}{\text{Definition of midpoint.}}}$ $\enclose{circle}{\; \\ \qquad \overline{AE} \cong \overline{CE} \qquad \\}$Definition of midpoint. $\huge \searrow$Bubble 2 has an arrow pointing down to bubble 5. $\huge \downarrow$Bubble 3 has an arrow pointing down to bubble 5. $\huge \swarrow$Bubble 4 has an arrow pointing down to bubble 5. $\enclose{circle}{\; \\ \qquad \bf{\color{red}{\triangle AEB \cong \triangle CED}} \qquad \\}$$\quad \quad \quad \quad \quad \quad \quad \bf{\color{red}{SAS \cong}}$
|
HuggingFaceTB/finemath
|
|
Prove ${\large\int}_0^\infty\frac{\ln x}{\sqrt{x}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel?=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\Gamma^2\left(\tfrac34\right)}$
I discovered the following conjecture by evaluating the integral numerically and then using some inverse symbolic calculation methods to find a possible closed form: $$\int_0^\infty\frac{\ln x}{\sqrt{x\vphantom{1}}\ \sqrt{x+1}\ \sqrt{2x+1}}dx\stackrel{\color{#808080}?}=\frac{\pi^{3/2}\,\ln2}{2^{3/2}\,\Gamma^2\left(\tfrac34\right)}.\tag1$$ The equality holds numerically with a precision of at least $1000$ decimal digits. But so far I was not able to find a proof of it.
Because the integral can be represented as a derivative of a hypergeometic function with respect to its parameter, the conjecture can be rewritten as $$\frac{d}{da}{_2F_1}\left(a,\ \tfrac12;\ 1;\ \tfrac12\right)\Bigg|_{a=\frac12}\stackrel{\color{#808080}?}=\frac{\sqrt\pi\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag2$$ or, using a series expansion of the hypergeometric function, as $${\large\sum}_{n=0}^\infty\frac{H_{n-\frac12}\ \Gamma^2\left(n+\tfrac12\right)}{2^n\ \Gamma^2\left(n+1\right)}\stackrel{\color{#808080}?}=-\frac{3\,\pi^{3/2}\,\ln2}{2\,\Gamma^2\left(\tfrac34\right)}\tag3,$$ where $H_q$ is the generalized harmonic number, $H_q=\gamma+\psi_0\left(q+1\right).$
Could you suggest any ideas how to prove this?
• You seem to have forgotten the minus sign, since the quantity is negative. Also, letting $x=\sinh^2t$, the integral can be rewritten as $\displaystyle\int_0^\infty\frac{\ln(\sinh t)}{\sqrt{\cosh(2t)}}dt$. Perhaps something similar exists in Gradshteyn-Ryzhik ? Commented Jul 25, 2014 at 9:10
• @Lucian they doesn’t looks same numerically Commented Mar 11 at 18:06
$$I:=\int_{0}^{\infty}\frac{\ln{(x)}}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\mathrm{d}x.$$
After first multiplying and dividing the integrand by 2, substitute $x=\frac{t}{2}$:
$$I=\int_{0}^{\infty}\frac{2\ln{(x)}}{\sqrt{2x}\,\sqrt{2x+2}\,\sqrt{2x+1}}\mathrm{d}x=\int_{0}^{\infty}\frac{\ln{\left(\frac{t}{2}\right)}}{\sqrt{t}\,\sqrt{t+2}\,\sqrt{t+1}}\mathrm{d}t.$$
Next, substituting $t=\frac{1}{u}$ yields:
\begin{align} I &=-\int_{0}^{\infty}\frac{\ln{(2u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-\int_{0}^{\infty}\frac{\ln{(u)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u\\ &=-\int_{0}^{\infty}\frac{\ln{(2)}}{\sqrt{u}\sqrt{u+1}\sqrt{2u+1}}\mathrm{d}u-I\\ \implies I&=-\frac{\ln{(2)}}{2}\int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}}. \end{align}
Making the sequence of substitutions $x=\frac{u-1}{2}$, then $u=\frac{1}{t}$, and finally $t=\sqrt{w}$, puts this integral into the form of a beta function:
\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}w}{w^{3/4}\,\sqrt{1-w}}\\ &=\frac12\operatorname{B}{\left(\frac14,\frac12\right)}\\ &=\frac12\frac{\Gamma{\left(\frac12\right)}\Gamma{\left(\frac14\right)}}{\Gamma{\left(\frac34\right)}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}} \end{align}
Hence,
$$I=-\frac{\ln{(2)}}{2}\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}=-\frac{\pi^{3/2}\,\ln{(2)}}{2^{3/2}\,\Gamma^2{\left(\frac34\right)}}.~~~\blacksquare$$
Possible Alternative: You could also derive the answer from the complete elliptic integral of the first kind instead of from the beta function by making the substitution $t=z^2$ instead of $t=\sqrt{w}$.
\begin{align} \int_{0}^{\infty}\frac{\mathrm{d}x}{\sqrt{x}\sqrt{x+1}\sqrt{2x+1}} &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u-1}\sqrt{u+1}\sqrt{u}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{\sqrt{u^2-1}\sqrt{u}}\\ &=\int_{1}^{0}\frac{t^{3/2}}{\sqrt{1-t^2}}\frac{(-1)}{t^2}\mathrm{d}t\\ &=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{t}\,\sqrt{1-t^2}}\\ &=2\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{1-z^4}}\\ &=2\,K{(-1)}\\ &=\frac{\Gamma^2{\left(\frac14\right)}}{2\sqrt{2\pi}}\\ &=\frac{\pi^{3/2}}{2^{1/2}\Gamma^2{\left(\frac34\right)}}. \end{align}
• Very nice answer. +1 After about 30 minutes of getting nowhere, I gave up. Commented Jul 25, 2014 at 22:05
• @RandomVariable Thanks. That compliment means a lot coming from you. Commented Jul 25, 2014 at 22:19
|
HuggingFaceTB/finemath
|
|
# Examination 5¶
## 2001 Nov¶
Instructions. Make a substantial effort on all parts of the following problems. If you cannot completely answer Part (a) of a problem, it is still possible to do Part (b). Partial credit is given for partial progress. Include as many details as time permits. Throughout the exam, $$z$$ denotes a complex variable, and $$\mathbb{C}$$ denotes the complex plane.
Problem 29
1. Suppose that $$f(z) = f(x+iy) = u(x,y) + i v(x,y)$$ where $$u$$ and $$v$$ are $$C^1$$ functions defined on a neighborhood of the closure of a bounded region $$G\subset \mathbb{C}$$ with boundary which is parametrized by a properly oriented, piecewise $$C^1$$ curve $$\gamma$$. If $$u$$ and $$v$$ obey the Cauchy-Riemann equations, show that Cauchy’s theorem $$\int_\gamma f(z) \, dz = 0$$ follows from Green’s theorem, namely
(11)$\int_\gamma P\,dx + Q\,dy = \int_G \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \, dx\, dy \;\text{ for C^1 functions P and Q.}$
1. Suppose that we do not assume that $$u$$ and $$v$$ are $$C^1$$, but merely that $$u$$ and $$v$$ are continuous in $$G$$ and
$f'(z_0) = \lim_{z\rightarrow z_0} \frac{f(z) - f(z_0)}{z-z_0}$
exists at some (possibly only one!) point $$z_0 \in G$$. Show that given any $$\epsilon >0$$, we can find a triangular region $$\Delta$$ containing $$z_0$$, such that if $$T$$ is the boundary curve of $$\Delta$$, then
$\left|\int_T f(z)\, dz\right| = \frac{1}{2}\epsilon L^2,$
where $$L$$ is the length of the perimeter of $$\Delta$$.
Hint for (b) Note that part (a) yields $$\int_T (az+b) \, dz =0$$ for $$a, b \in \mathbb{C}$$, which you can use here in (b), even if you could not do Part (a). You may also use the fact that $$\left|\int_T g(z)\, dz\right| \leq L \cdot \sup\{|g(z)|:z\in T\}$$ for $$g$$ continuous on $$T$$.
Problem 30
Give two quite different proofs of the fundamental theorem of algebra, that if a polynomial with complex coefficients has no complex zero, then it is constant. You may use independent, well-known theorems and principles such as Liouville’s theorem, the argument principle, the maximum principle, Rouché’s theorem, and/or the open mapping theorem.
Problem 31
1. State and prove the Casorati-Weierstrass theorem concerning the image of any punctured disk about a certain type of isolated singularity of an analytic function. You may use the fact that if a function $$g$$ is analytic and bounded in the neighborhood of a point $$z_0$$, then $$g$$ has a removable singularity at $$z_0$$.
2. Verify the Casorati-Weierstrass theorem directly for a specific analytic function of your choice, with a suitable singularity.
Problem 32
1. Define $$\gamma : [0,2\pi] \rightarrow \mathbb{C}$$ by $$\gamma(t) = \sin (2t) + 2i \sin (t)$$. This is a parametrization of a “figure 8” curve, traced out in a regular fashion. Find a meromorphic function $$f$$ such that $$\int_\gamma f(z) \, dz = 1$$. Be careful with minus signs and factors of $$2\pi i$$.
2. From the theory of Laurent expansions, it is known that there are constants $$a_n$$ such that, for $$1<|z|<4$$,
$\frac{1}{z^2 - 5z + 4} = \sum_{n=-\infty}^\infty a_n z^n.$
Find $$a_{-10}$$ and $$a_{10}$$ by the method of your choice.
Problem 33
1. Suppose that $$f$$ is analytic on a region $$G\subset \mathbb{C}$$ and $$\{z\in \mathbb{C}: |z-a|\leq R\} \subset G$$. Show that if $$|f(z)| \leq M$$ for all $$z$$ with $$|z-a|=R$$, then for any $$w_1, w_2\in \{z\in \mathbb{C}: |z-a|\leq \frac{1}{2}R\}$$, we have
$|f(w_1) - f(w_2)| \leq \frac{4M}{R} |w_1 - w_2|$
1. Explain how Part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting the normality of any locally bounded family $$F$$ of analytic functions on a region $$G$$.
## Solutions¶
Solution to Problem 29
1. Let $$P=u$$ and $$Q= -v$$ in (11). Then, by the Cauchy-Riemann equations, 1
(12)$\int_\gamma u(x,y)\, dx - v(x,y) \, dy= \int_G (v_x + u_y) \, dx\, dy = 0.$
Similarly, if $$P=v$$ and $$Q=u$$ in Green’s theorem, then the Cauchy-Riemann equations imply
(13)$\int_\gamma v(x,y)\, dx + u(x,y) \, dy= \int_G (u_x - v_y) \, dx\, dy = 0.$
Next, note that
$f(z)\, dz = [u(x,y)+iv(x,y)]\, d(x+iy) = u(x,y)\, dx - v(x,y)\, dy +i [v(x,y)\, dx + u(x,y)\, dy].$
Therefore, by (12) and (13),
$\int_\gamma f(z)\, dz = \int_\gamma u(x,y)\, dx - v(x,y)\, dy +i \int_\gamma v(x,y)\, dx + u(x,y)\, dy = 0.$
1. Suppose $$u$$ and $$v$$ are continuous and $$f'(z)$$ exists at the point $$z_0\in G$$. Then, for any $$\epsilon>0$$ there is a $$\delta>0$$ such that $$B(z_0,\delta)\subseteq G$$, and
$\left| f'(z_0) - \frac{f(z) - f(z_0)}{z-z_0}\right| < \epsilon, \quad \text{ for all |z- z_0|< \delta.}$
Pick a triangular region $$\Delta \subset B(z_0,\delta)$$ with $$z_0\in \Delta$$, and let $$T$$ be the boundary. Define
$R(z) = f(z) - [f(z_0) + f'(z_0) (z-z_0)].$
Then, by Cauchy’s theorem (part (a)), $$\int_T [f(z_0) + f'(z_0) (z-z_0)] \, dz = 0$$, whence $$\int_T R(z)\, dz = \int_T f(z)\, dz$$. Finally, note that
$\left|\frac{R(z)}{z-z_0}\right| = \left| f'(z_0) - \frac{f(z) - f(z_0)}{z-z_0}\right| < \epsilon, \quad \text{ for all |z- z_0|< \delta.}$
Therefore,
$\begin{split}\left|\int_T f(z)\, dz\right| &= \left|\int_T R(z)\, dz \right| \leq \int_T |R(z)|\, |dz| \\[4pt] &= \int_T \left|\frac{R(z)}{z-z_0}\right| |z-z_0|\, |dz| \\[4pt] &\leq \epsilon \int_T |z-z_0|\, |dz| \leq \epsilon r L.\end{split}$
where $$L$$ denotes the length of the perimeter of $$\Delta$$ (i.e.,the length of $$T$$), and $$r$$ denotes the length of one side of $$T$$, which must, of course, be greater than $$|z-z_0|$$ for all $$z\in T$$. Also, the length of one side of $$\Delta$$ is surely less than half the length of the perimeter (i.e.,$$r<L/2$$). Therefore,
$\left|\int_T f(z)\, dz\right| \leq \frac{1}{2}\epsilon L^2.$
Solution to Problem 30
In the two proofs below, we begin by supposing $$p(z)$$ is not constant and thus has the form $$p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$. Both proofs also rely on the following observation: If $$\{a_j\}_{j=0}^n \subset \mathbb{C}$$ with $$a_n\neq 0$$, then for all $$1\leq R \leq |z| <\infty$$,
$\begin{split}\left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| &\leq \frac{|a_0|}{|a_n|} |z|^{-n} + \cdots +\frac{|a_{n-1}|}{|a_n|} |z|^{-1}\\[4pt] &\leq n \max_{0\leq j <n} \frac{|a_j|}{|a_n|}|z|^{-1} \\[4pt] &\leq n \max_{0\leq j <n} \frac{|a_j|}{|a_n|}R^{-1}.\end{split}$
In particular, if we choose 2 $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$, then
(14)$\left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| \leq 1/2,\quad \text{ for all |z| \geq R.}$
Proof 1.
Assume $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and let $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$, as above. We claim that
(15)$|p(z) - a_n z^n| < |a_n z^n|, \quad \text{ for all } |z|=R.$
To see this, check that
$\frac{|p(z) - a_n z^n|}{|a_n z^n|} = \left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| < 1, \quad \text{ for all |z|=R.}$
In fact, (14) implies that the sum is no greater than 1/2, for all $$|z|\geq R$$, which is more than we need.
Now (15) and Rouché’s theorem imply that the function $$g(z) = a_nz^n$$ has the same number of zeros in $$|z|<R$$ as does the function $$p(z)$$. Clearly $$z=0$$ is a zero of $$g(z)$$ (of multiplicity $$n$$). Therefore, $$p(z)$$ has a zero in $$|z|<R$$.
Proof 2.
3 Assume $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and consider
(16)$|p(z)| = |a_n z^n| \left|\frac{a_0}{a_n}z^{-n} + \cdots + \frac{a_{n-1}}{a_n} z^{-1} + 1\right|\geq |a_n||z|^n \left| 1 - |\sum_{j=0}^{n-1}\frac{a_j}{a_n}z^{-n+j}|\,\right|.$
If we choose $$R = 1+ 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$ as above, then for all $$|z|\geq R$$,
$0\leq \left|\sum_{j=0}^{n-1}\frac{a_j}{a_n}z^{-n+j}\right| = \left|\frac{a_0}{a_n} z^{-n} + \cdots +\frac{a_{n-1}}{a_n} z^{-1}\right| \leq 1/2,$
and (16) yields $$|p(z)| \geq |a_n||z|^n/2$$, for all $$|z|\geq R$$. Therefore, the function $$f(z) \triangleq 1/p(z)$$ satisfies
$|f(z)| = \frac{1}{|p(z)|} \leq \frac{2}{|a_n||z|^n}, \quad \text{ for all |z|\geq R.}$
Now suppose $$p(z)$$ has no complex zero. Then $$f(z) \in H(\mathbb{C})$$. In particular, $$f(z)$$ is continuous, hence bounded on the compact set $$|z|\leq R$$. Therefore $$f(z)$$ is a bounded entire function, so, by Liouville’s theorem, it must be constant, but then $$p(z)$$ must be constant. This contradicts our initial assumption and proves that $$p(z)$$ must have a complex zero.
In fact, we have proved a bit more: If $$p(z) = a_0 + a_1 z + \cdots + a_n z^n$$ with $$a_n \neq 0$$ for some $$n\geq 1$$, and $$R$$ is either 1 or $$R = 2\, n \,\max_{0\leq j <n} |a_j|/|a_n|$$ (whichever is greater), then $$p(z)$$ vanishes for some $$|z|< R$$, while for all $$|z|\geq R$$,:math:|p(z)| is bounded from below by $$|a_n||z|^n/2$$. Thus all the zeros of $$p(z)$$ are contained in the disk $$|z|< R$$.
Solution to Problem 31
1. If $$f$$ is a holomorphic function in a region $$G\in \mathbb{C}$$ except for an essential singularity at the point $$z=z_0$$, then for any $$w\in \mathbb{C}$$ there is a sequence $$\{z_n\}\subset G$$ approaching $$z_0$$ such that $$f(z_n)\rightarrow w$$ as $$n\rightarrow \infty$$.
Proof. Fix $$w_0\in \mathbb{C}$$ and suppose there is no sequence $$\{z_n\}\subset G$$ approaching $$z_0$$ such that $$f(z_n)\rightarrow w_0$$ as $$n\rightarrow \infty$$. Then there is a punctured disk $$\bar{D_0} \triangleq B(z_0,\epsilon)\setminus \{z_0\} \subset G$$ such that $$|f(z)-w_0|>\delta >0$$ for all $$z\in \bar{D_0}$$.
Define $$g(z) = 1/(f(z) - w_0)$$ on $$D_0$$. Then
$\begin{split}\limsup_{\substack{z\rightarrow z_0\\ z\in D_0}} |g(z)| = \limsup_{\substack{z\rightarrow z_0\\ z\in D_0}} \frac{1}{|f(z) - w_0|}\leq \frac{1}{\delta} < \infty.\end{split}$
Thus, by lemma [lem:removable-singularity] (Nov. ’06, prob. 1), $$z_0$$ is a removable singularity of $$g(z)$$. Therefore, $$g(z) \in H(B(z_0, \epsilon))$$.
In particular, $$g$$ is continuous and non-zero at $$z=z_0$$, so it is non-zero in a neighborhood $$B(z_0,\epsilon_0)$$ of $$z_0$$. Therefore, $$f(z)-w_0 = 1/g(z)$$ is holomorphic in $$B(z_0,\epsilon_0)$$, which implies that the singularity of $$f(z)$$ at $$z=z_0$$ is removable. This contradiction proves the theorem.
1. Consider $$f(z) = e^z$$. This function has an essential singularity at $$\infty$$, and, for every horizontal strip,
$S_\alpha = \{x+iy: x\in \mathbb R, \, \alpha \leq y < \alpha + 2\pi\},$
of width $$2\pi$$, $$f(z)$$ maps $$S_\alpha$$ onto $$\mathbb{C}\setminus \{0\}$$. (In particular, $$f(z)$$ comes arbitrarily close to every $$w\in \mathbb{C}$$.)
Now let $$\mathcal{N}_R = \{z\in \mathbb{C}: |z|>R\}$$ be any neighborhood of $$\infty$$. There is clearly a strip $$S_\alpha$$ contained in $$\mathcal{N}_R$$ (e.g.,with $$\alpha = R+1$$).
Therefore, $$f(z)=e^z$$ maps points in $$\mathcal{N}_R$$ to points arbitrarily close (in fact equal when $$w\neq 0$$) to all points $$w\in \mathbb{C}$$.
Solution to Problem 32
1. Let $$G$$ be the region whose boundary is the curve $$\gamma$$, and suppose $$f(z) \in H(\mathbb{C})$$ except for isolated singularities at the points $$\{z_1, \ldots, z_n\}\subset G$$. By the residue theorem,
$\int_\gamma f(z) \, dz = 2\pi i \sum_{j=1}^n \mbox{Res}(f,z_j).$
Therefore, if we were to find a function $$f(z)\in H(\mathbb{C})$$ with exactly two isolated singularities in $$G$$ (e.g.,at $$z_1=i$$ and $$z_2=-i$$), and such that $$\mbox{Res}(f,z_j)=\frac{-i}{4\pi}$$, then
$\int_\gamma f(z) \, dz = 2\pi i \sum_j \mbox{Res}(f,z_j) = 2\pi i \left(\frac{-i}{4\pi}+\frac{-i}{4\pi}\right) = 1,$
and the problem would be solved. Clearly,
$f(z) = \frac{-i}{4\pi}\left(\frac{1}{z-i} - \frac{1}{z+i}\right) = \frac{1}{2\pi i} \frac{z}{z^2+1}$
is such a function.
2. Expand the function in partial fractions:
$\frac{1}{z^2 - 5z + 4} = \frac{1}{(z-4)(z-1)} = \frac{1/3}{z-4}- \frac{1/3}{z-1}.$
Then, note that
$\frac{1/3}{z-4} = \frac{1}{3}\frac{-1}{4(1-z/4)} = -\frac{1}{12} \sum_{n=0}^\infty \left(\frac{z}{4}\right)^n$
converges for $$|z|<4$$, while
$\frac{1/3}{z-1} = -\frac{1}{3}\frac{1}{z(1-1/z)} = -\frac{1}{3z} \sum_{n=0}^\infty z^{-n}$
converges for $$|z|>1$$. Therefore,
$\frac{1}{z^2 - 5z + 4} =-\frac{1}{3} \sum_{n=-\infty}^{-1} z^{n}-\frac{1}{12} \sum_{n=0}^\infty \left(\frac{1}{4}\right)^n z^n, \quad \text{ for 1<|z|<4.}$
$\therefore \quad a_{-10} = -\frac{1}{3} \qquad \text{and} \qquad a_{10} = -\frac{1}{12}\left(\frac{1}{4}\right)^{10}.$
Solution to Problem 33
1. By Cauchy’s formula, if $$w$$ is any point in the disk $$|w-a|< R$$, then
$f(w) = \frac{1}{2\pi i} \int_{|\zeta - a| = R} \frac{f(\zeta)}{\zeta-w}\, d\zeta.$
In particular, if $$w_1, w_2$$ are any two points inside the “half-disk” $$|w-a|< R/2$$, then
$\begin{split}f(w_1) - f(w_2) &= \frac{1}{2\pi i} \int_{|\zeta - a| = R} \left[\frac{f(\zeta)}{\zeta-w_1} - \frac{f(\zeta)}{\zeta-w_2}\right]\, d\zeta\\[4pt] &= \frac{w_1 - w_2}{2\pi i}\int_{|\zeta - a| = R} \frac{f(\zeta)}{(\zeta-w_1)(\zeta-w_2)}\, d\zeta.\end{split}$
Now, for all $$\zeta$$ on the outer radius in figure [fig:concentriccircles], it is clear that $$|\zeta - w_1|>R/2$$ and $$|\zeta - w_2|>R/2$$. Therefore,
$\begin{split}|f(w_1) - f(w_2)| &\leq \frac{|w_1 - w_2|}{2\pi}\int_{|\zeta - a| = R} \frac{|f(\zeta)|}{(R/2)^2}\, |d\zeta|\\[4pt] &\leq \frac{|w_1 - w_2|}{2\pi}\, \frac{\sup_\gamma |f(\zeta)|}{R^2/4}\,\ell(\gamma) \\[4pt] &\leq \frac{4M}{R}|w_1 - w_2|,\end{split}$
where $$\gamma$$ denotes the positively oriented circle $$\{\zeta: |\zeta - a| = R\}$$, and $$\ell(\gamma)$$ denotes its length, $$2\pi R$$.
2. 4 The statements of the Arzela-Ascoli theorem and Montel’s theorem are given in the Appendix.
We must explain how part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting the normality of any locally bounded family $$\mathcal{F}\subset H(G)$$.
Because of the way the problem is stated, it is probably enough to prove just one direction of Montel’s theorem; i.e., local boundedness implies normality. For a proof of the other direction, see Conway [Con78], page 153.
Let $$S = \mathbb{C}$$ in the Arzela-Ascoli theorem. In that case, $$K\subset \mathbb{C}$$ is compact if and only if $$K$$ is closed and bounded. Therefore, if $$\mathcal{F}$$ is locally bounded, condition (ii) of the theorem is clearly satisfied. To check that local boundedness also implies condition (i), we use part (a).
It suffices to prove that for any $$a\in G$$ there is a neighborhood $$B(a,r)$$ in which $$\mathcal{F}$$ is equicontinuous with equicontinuity constant $$\delta$$. (Why is this sufficient?) 5 6
So, fix $$a\in G$$ and $$\epsilon >0$$, and let $$\bar{B}(a,R)\subset G$$. Then, by local boundedness, there is an $$M>0$$ such that $$|f(z)|\leq M$$ for all $$z \in \bar{B}(a,R)$$ and all $$f\in \mathcal{F}$$. Therefore, by part (a),
$|f(w_1)-f(w_2)| \leq \frac{4M}{R}|w_1-w_2|, \quad \text{ for all w_1, w_2 \in \{|w-a|\leq R/2\}.}$
If $$\delta = \frac{R}{4M}\epsilon$$ and $$r = R/2$$, then $$|f(w_1)-f(w_2)| < \epsilon$$ whenever $$w_1, w_2 \in B(a,r)$$ and $$|w_1-w_2|< \delta$$. Therefore, $$\mathcal{F}$$ is equicontinuous in $$B(a,r)$$. We have thus shown that local boundedness implies conditions (i) and (ii) of the Arzela-Ascoli theorem and thereby implies normality.
Footnotes
1
These are $$u_x = v_y$$ and $$u_y = -v_x$$.
2
Note, we add 1 here just to be sure $$R$$ is safely over 1.
3
Conway [Con78], page 77, presents a similar, but more elegant proof.
4
The best treatment of normal families and the Arzela-Ascoli theorem is Ahlfors [Ahl78].
5
Answer: If, instead of a single point $$a\in G$$, we are given a compact set $$K\subset G$$, then there is a finite cover $$\{B(a_j,r_j):j=1,\ldots, n\}$$ by such neighborhoods with equicontinuity constants $$\delta_1, \ldots, \delta_n$$. Then, $$\delta = \min_j\delta_j$$, is a single equicontinuity constant that works for all of $$K$$.
6
The careful reader might note the distinction between this type of “uniform” equicontinuity, which is taken for granted in complex analysis texts, e.g.,Ahlfors [Ahl78] and Rudin [Rud87], and the “pointwise” equicontinuity discussed in topology books like the one by Munkres [Mun00]. To make peace with this apparent discrepancy, check that the two notions coincide when the set on which a family of functions is declared equicontinuous is compact.
Real Analysis Exams
|
HuggingFaceTB/finemath
|
|
# Class 02: An Introduction to CSC151
Back to An Introduction to Algorithms. On to An Introduction to the GIMP.
This outline is also available in PDF.
Held: Monday, 31 August 2009
Summary: Today we begin to consider the content and structure of the course. We also prepare ourselves to use the Linux workstations.
Related Pages:
Notes:
Overview:
• Drawing smiley faces, revisited.
• Lessons from day one.
• Common parts of an algorithm.
• Getting started with Linux.
## Drawing Smileys, Continued
Sam will attempt to follow some instructions for drawing smiley faces.
## Reflections on Drawing Images
What, if anything, did you learn from the drawing exercise?
## The Parts of an Algorithm
• As you may have noted, there are some common aspects to algorithms. That is, there are techniques that we use in many of the algorithms we write.
• It is worthwhile to think about these algorithm parts because we can rely on them when we write new algorithms.
### Variables: Named Values
• As we write algorithms, we like to name things.
• Sometimes we use long names, such as the piece of bread in your dominant hand.
• Sometimes, we use shorter names, such as bread-dom.
• As we start to write more formal algorithms, we will need techniques for noting which names we are using and indicating what they name (and, sometimes, what kind of thing they name).
• We call these named values variables, even though they don't always vary.
• We need to be careful to use unambiguous names. Recall the problems with using it in your descriptions.
### Parameters: Named Inputs
• Many algorithms take data as input (and generate other data as output).
• Your smiley-face algorithms might take the size of the face (or board) as input.
• A find square root algorithm would take a number as input.
• A look up a telephone number algorithm might take a phone book and a name to lok for as inputs.
• In each case, the algorithm should works on many different inputs.
• The algorithm works as long as the input is reasonable (we can't find the square root of a smiley face and we can't draw a circle with pi).
• We call these inputs parameters.
### Conditionals: Handling Different Situations
• At times, our algorithms have to account for different conditions, doing different things depending on those conditions.
• In our smiley-face algorithm, we might check what kind of face we want to draw, or whether the marker is open or closed. We call such operations conditionals.
• Conditionals typically take either the form
if some condition holds then do something
• Here's a slightly more complex form
if some condition holds then do something otherwise do something else
• At times, we need to decide between more than two possibilities. Typically, we organize those as a sequence of tests (called guards) and corresponding things to do.
### Repetition
• At times, our algorithms require us to do something again and again.
• In the turtle drawing, we had to repeatedly draw a small line and turn a small amount.
• We call this technique repetition.
• Again, repetition takes many forms.
• We might do work until we've reached a desired state.
• We might continue work as long as we're in some state.
• We might repeat an action a fixed number of times.
• You can probably think of many other forms of repetition.
### Subroutines: Named Helper Algorithms
• Many algorithms require common actions for their operation.
• For example, the circle group relied on a "draw a circle" routine.
• The pixel group could immediately use the circle group's solution by simply writing their own draw a circle routine".
• We can write additional algorithms for these common actions and use them as part of our broader algorithm.
• We can also use them in other algorithms.
• We call these helper algorithms subroutines.
• Computer Science 151 has a number of goals
• To introduce you to fundamental ideas of computer science: abstraction, algorithms, and data
• To enhance your problem-solving skills and give you experience in formal representation of problems and solutions.
• To introduce you to two primary paradigms of problem solving: functional and imperative.
• To give you some programming skills that you can apply to problems in other disciplines.
• I expect and hope that you will find CSC151 different from any class you've taken in the past.
• We use a different format than many classes: a collaborative, workshop-style format. (You may have seen this format in other introductory science courses; we do it somewhat differently.)
• Computers and computer science also require you to think differently. I expect that you'll exercise some brain cells you may have forgotten you have. (And after all, isn't liberal arts education an exercise in thinking in as many ways as you can?)
• Like most computer science courses, CSC151 will have both theoretical and practical components. I hope you will enjoy relating the two.
• And, hey, we're going to make pretty pictures, too.
## Lab: Getting Started with Linux
• We'll break about midway through today's class to get you set up working with our Linux computers.
• This "lab prep" is somewhat pointless and annoying, but also necessary.
• Do the lab.
I am not sure whether or not I will cover these topics in class. They are included for your edification.
• Please refer to the course web site for more details.
• Teaching philosophy: I support your learning.
• Policies
• Attendance: I expect you to attend every class. I understand that sometimes you have to miss. Let me know when you'll miss class and why.
• Course web.
• Etc.
• Daily work
• Attend class, work on lab and participate in discussion.
• Finish the lab in the evening.
• Do the reading for the next class in the evening.
• The quizzes
• Every Friday, plus a few pop quizzes
• The exams
• Three take-home exams during the semester. Plan to spend four hours on each one.
• An optional final to make up for a bad exam grade.
• Take all three exams anyway.
• The labs
• Available online.
• I expect to require more formal writeups of about a lab per week.
• The homework
• Weekly, due on Wednesdays
• Core to the academic process.
• Che college's basic policy: Cite carefully.
• My basic policy: Don't cheat. (Also: It's usually okay to work with others, provided you cite them.)
Back to An Introduction to Algorithms. On to An Introduction to the GIMP.
Disclaimer: I usually create these pages on the fly, which means that I rarely proofread them and they may contain bad grammar and incorrect details. It also means that I tend to update them regularly (see the history for more details). Feel free to contact me with any suggestions for changes.
This document was generated by Siteweaver on Fri Dec 11 09:38:39 2009.
The source to the document was last modified on Fri Aug 21 17:03:06 2009.
This document may be found at `http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/2009F/Outlines/outline.02.html`.
You may wish to validate this document's HTML ; ;
Samuel A. Rebelsky, rebelsky@grinnell.edu
Copyright © 2007-9 Janet Davis, Matthew Kluber, Samuel A. Rebelsky, and Jerod Weinman. (Selected materials copyright by John David Stone and Henry Walker and used by permission.) This material is based upon work partially supported by the National Science Foundation under Grant No. CCLI-0633090. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation. This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. To view a copy of this license, visit `http://creativecommons.org/licenses/by-nc/2.5/` or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA.
|
HuggingFaceTB/finemath
|
|
Introduction to statistical data analysis
| categories: statistics | tags: | View Comments
Given several measurements of a single quantity, determine the average value of the measurements, the standard deviation of the measurements and the 95% confidence interval for the average.
import numpy as np
y = [8.1, 8.0, 8.1]
ybar = np.mean(y)
s = np.std(y, ddof=1)
print ybar, s
>>> >>> >>> >>> >>> >>> 8.06666666667 0.057735026919
Interesting, we have to specify the divisor in numpy.std by the ddof argument. The default for this in Matlab is 1, the default for this function is 0.
Here is the principle of computing a confidence interval.
1. compute the average
2. Compute the standard deviation of your data
3. Define the confidence interval, e.g. 95% = 0.95
4. compute the student-t multiplier. This is a function of the confidence interval you specify, and the number of data points you have minus 1. You subtract 1 because one degree of freedom is lost from calculating the average.
The confidence interval is defined as ybar +- T_multiplier*std/sqrt(n).
from scipy.stats.distributions import t
ci = 0.95
alpha = 1.0 - ci
n = len(y)
T_multiplier = t.ppf(1.0 - alpha / 2.0, n - 1)
ci95 = T_multiplier * s / np.sqrt(n)
print 'T_multiplier = {0}'.format(T_multiplier)
print 'ci95 = {0}'.format(ci95)
print 'The true average is between {0} and {1} at a 95% confidence level'.format(ybar - ci95, ybar + ci95)
>>> >>> >>> >>> >>> >>> >>> >>> T_multiplier = 4.30265272991
ci95 = 0.143421757664
The true average is between 7.923244909 and 8.21008842433 at a 95% confidence level
|
HuggingFaceTB/finemath
|
|
mersenneforum.org square root equal to a negative
Register FAQ Search Today's Posts Mark Forums Read
2013-02-12, 15:46 #1 LaurV Romulan Interpreter Jun 2011 Thailand 3×13×229 Posts square root equal to a negative Now I have a big problem... I have the next equation: $sqrt x =-4$ Does it, or does it not have a (real/integer) solution?
2013-02-12, 16:00 #2 akruppa "Nancy" Aug 2002 Alexandria 246410 Posts It does not. The two (Edit: complex) solutions are ±2i. Last fiddled with by akruppa on 2013-02-12 at 16:00
2013-02-12, 16:01 #3 Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40
2013-02-12, 16:04 #4
LaurV
Romulan Interpreter
Jun 2011
Thailand
3×13×229 Posts
Quote:
Originally Posted by akruppa It does not. The two (Edit: complex) solutions are ±2i.
I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4.
(Gotcha!)
2013-02-12, 16:06 #5
LaurV
Romulan Interpreter
Jun 2011
Thailand
893110 Posts
Quote:
Originally Posted by Dubslow No, its only solutions are complex (with non zero imaginary part). Anyone who claims a real solution forgets that squaring both sides introduces extra solutions which do not satisfy the original equation. Edit: akruppa's solutions are not, I think he read the question too quickly.
you too.
2013-02-12, 16:11 #6
Dubslow
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
3×2,399 Posts
Quote:
Originally Posted by LaurV I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4. (Gotcha!)
Not quite. sqrt(2i) = -/+1 -/+ i , and sqrt(-2i) = +/-1 -/+ i, i.e. you forgot half the solutions.
Quote:
Originally Posted by LaurV you too.
?
You pose an interesting problem.
Last fiddled with by Dubslow on 2013-02-12 at 16:12 Reason: Clarification
2013-02-12, 16:15 #7 LaurV Romulan Interpreter Jun 2011 Thailand 213438 Posts Story is like that, this IS homework. My daughter's. She had to solve a system which came to this, after a lot of calculus. She said is no solution. I said the solution is 16. Indeed, sqrt(16), according with the math I know 40 years ago, is +/-4. We argued, we looked in the book (answers section). They said the solution is the empty set. I got angry and looked into wiki, and wolfram alfa, they all agree with my daughter , and I am getting crazy... We live the days when the egg teaches the hen, you know?
2013-02-12, 16:31 #8 Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40
2013-02-12, 16:33 #9 bsquared "Ben" Feb 2007 D0D16 Posts x = 16*i^4
2013-02-12, 16:41 #10
NBtarheel_33
"Nathan"
Jul 2008
Maryland, USA
100010110012 Posts
Quote:
Originally Posted by LaurV Now I have a big problem... I have the next equation: $sqrt x =-4$ Does it, or does it not have a (real/integer) solution?
I would say that it probably depends on how the actual symbol $sqrt x$ is defined *within your daughter's textbook*. (Please note that this might not be consistent with what you, or I, or other advanced mathematicians know about the definition of square root, or the radical sign, but there are often (usually not-so-great) simplifications made at the elementary levels).
For instance, in my first algebra textbook, I was taught that any positive real number x had two square roots, and to answer a question like "what is the square root of 36?" I should write $\pm 6$. If I were solving an equation such as $x^2 + 6x + 9 = 16$, I would perform the steps $(x + 3)^2 = 16$, hence $x + 3 = \pm 4$, hence $x \in \{-7, 1\}$.
Needless to say, it was a little embarrassing and cost me a few points on some homework assignments when I got into college, where the definition of the radical sign was taken to mean "the positive square root".
But you can see now where your daughter's question might be entirely valid and have a solution in one setting, but not in another. If, as I suspect, her present textbook is making use of the "plus-minus" definition of square roots, then one could say that, yes, if $sqrt x = -4$, then $x = 16$. On the other hand, if her textbook has taken on the more advanced, "principal"/positive-square-root-only definition, then there would be an argument that there is no solution, as there is no possible way *under the understood definition of square root *as it is in this textbook** for the square root of a number to be negative. Hence the solution would be undefined.
This, of course, is an excellent time to have a chat with your daughter (especially if she plans on going higher into math) about why this matter is an issue in the first place, and why we might like to fix "square root" to mean "only the positive square root" (e.g. to avoid a multi-valued function, to be able to take limits in calculus, to not have a weird function that raises a positive number to a positive power and gives a negative answer...) Basically, this is one of her first lessons that what works well in algebra does not necessarily work well in calculus, and why certain changes to definitions and constructs have to be made in order to fix these issues.
Last fiddled with by NBtarheel_33 on 2013-02-12 at 16:44
2013-02-12, 17:14 #11
akruppa
"Nancy"
Aug 2002
Alexandria
246410 Posts
Quote:
Originally Posted by LaurV I think you confuse with a radical of a negative number, which I did not ask for. sqrt(+/-2i) in complex, is 1+/-i, and not -4. (Gotcha!)
*facepalm* Indeed you did.
Similar Threads Thread Thread Starter Forum Replies Last Post davar55 Lounge 0 2016-03-16 20:19 paul0 Factoring 10 2015-01-19 12:25 bsquared Msieve 17 2010-04-09 14:11 Damian Math 3 2010-01-01 01:56 davar55 Puzzles 3 2007-09-05 15:59
All times are UTC. The time now is 16:08.
Tue Nov 24 16:08:12 UTC 2020 up 75 days, 13:19, 4 users, load averages: 1.68, 1.75, 1.72
|
HuggingFaceTB/finemath
|
|
Home >> Masonry-structures-1921 >> 1 E 4374 100x to Pressure Of Earth Against >> Pressure of Earth Against_P1
# Pressure of Earth Against a Wall
## angle, plane, surface, slope, friction, thrust and horizontal
Page: 1 2 3
PRESSURE OF EARTH AGAINST A WALL.
Theories of Earth Pressure.—The lateral pressure of a mass of earth against a retaining wall is affected by so many variable conditions that the determination of its actual value in a particular instance is practically impossible.
Several theories, based in each case upon certain ideal conditions, have been proposed, none of which are more than very rough approxi ntations to the conditions existing in such structures. These theo ries :-assume that the earth is composed of a grass of particles exerting friction upon each other but without cohesion, or that the pressure against the wall is caused by a wedge of earth which tends to slide upon a plane surface of rupture, a,s shown in Fig. 60. Formulas for the resultant thrust against the wall have been produced in accord ance with the various theories by several methods, they differ mainly in the direction given to the thrust upon the wall.
Coulomb's formula for computing the lateral thrust against a wall was proposed by Coulomb in 1773. Coulomb assumed that the thrust was caused by a prism of earth (B AC, Fig. 60) sliding upon any plane AC which produces the maximum thrust upon the wall. There is a certain slope (AD, Fig. 60) at which the material if loosely placed will stand. This is known as the natural slope, and the angle made by this slope with the horizontal as the angle of fric tion of the earth. On slopes steeper than the natural slope, there is a tendency for the earth to slide down, and if held by a wall, pressures are produced which depend upon the frictional resistance to sliding.
The thrust is assumed by Coulomb to he normal to the wall, and the pressure upon the plane of rupture to be inclined at the angle of friction to the normal to the plane.
Let It = height of wall; P= resultant pressure upon a unit length of wall; = pre sure upon the plane of rupture; G= weight of the wedge of earth; e =weight of earth per cubic foot; = angle of friction of earth; a =angle between the back of wall and plane of rupture.
For maximum value of P, and the plane of rupture bisects the angle between the hack of the wall and the natural slope.
Substituting this value, P varies as the square of h, and is therefore applied at a distance h,'3 above the base of the wall. This is the same in all of the theories.
Poncelet's Theory.—In 1840 Poncelet proposed to modify the method of Coulomb by making the thrust upon the wall act at the angle of friction with the normal to the wall.
Before the wall can he overturned about its toe (F, Fig. 61) the hack of the wall (A R) must he raised and slide upon the earth behind it, thus calling into play the friction of the earth upon the wall as a resistance. As the friction of earth upon a rough masonry wall is greater than that of earth upon earth, a film of earth would be carried with the wall and slide upon the earth behind and the angle of friction is usually taken as equal to the natural slope of the earth.
Let 0=the angle made by the back of the wall with the horizontal; i=the angle made by the earth surface with the horizontal.
Following the same procedure as in developing Coulomb's formula, we find the pressure against the wall, For a vertical wall and horizontal earth surface and which is the formula proposed by Poncelet.
IRankine's Theory. Rankine considered the earth to be made up of a homogeneous mass of particles, possessing frictional resistance to sliding over each other but without cohesion. Ile deduced for mulas for the pressure upon ideal plane sections through an unlimited mass of earth with plane upper surface, the earth being subject to no external force except its own weight, and determined the direction of the pressure from these assumptions.
Rankine found that the resultant pressure upon any vertical plane section through a bank of earth with plane upper surface is parallel to 1 he slope of the upper surface (see Fig. 62).
Let E= the pressure upon the vertical section; angle made by the inclination of the upper surface with the horizontal; is Rankine's formula for earth pressure. This pressure acts upon the vertical section at a distance S, 3 from its base, and makes an angle i with the horizontal.
Page: 1 2 3
|
HuggingFaceTB/finemath
|
|
# How I Teach the Introduction to Proofs
Several years ago, when I first found out I would be teaching Geometry, I’ll admit I was a little nervous. Why? Proofs. My last “geometry” class was in high school (I never took a class of only geometry in college). I knew that I could do proofs, but explaining them would be something totally different. I have formed my own way of how I like to introduce proofs. That’s what I’m going to share today. I’ll write another blog post later about how I actually teach lessons with proofs.
Background
Sometimes at the beginning of the year, I like to teach a lesson about Optical Illusions. I think it helps lay the groundwork for proofs quite well. In my curriculum, there is an Introduction to Geometry unit and the next unit is Logic and Proofs. In the Logic and Proofs unit, I teach Conditional Statements, Biconditional Statements, Laws of Detachment and Syllogism, and the next lesson is Introduction to Proofs.
Introduction to Proofs Lesson
I usually start class with this powerpoint. It’s a very short review, because my students have heard me using this vocabulary and talking about patterns before. I try to stress looking for patterns and making conjectures no matter what lesson I’m teaching.
Then, I talk about the importance of logical thinking. I usually explain it in story form. My little lecture goes something like this:
Okay, so I’m going to tell you a story about what I did last night. I took a shower and changed clothes. Oh, but I went to the gym first. I burned the bread. My husband and I walked our dog, Zoey. My husband did not like it. I went to bed. Oh, I also cooked dinner for my husband and I. We watched Netflix together.
Does that story make a lot of sense? Could I have told it better? This is the order that everything happened.
I went to the gym. Then, my husband and I walked our dog, Zoey. I took a shower and changed clothes, then cooked dinner for my husband and I. I burned the bread and my husband did not like it. We watched Netflix together. Then, I went to bed.
Is that story easier to understand?
In math, we explain things the same way. No one likes to have things explained in a jumbled, confusing way. So, we explain our thinking in a logical order, the same way you would tell a story.
Next, I give out this sheet for the students to practice ordering the story. They have probably done something like this a million times in English class in elementary or middle school, but I still think it’s good practice. I honestly have no idea where this worksheet came from (I’ve had it a long time). If you know who made it, let me know so I can give credit. I always let them work with partners on this worksheet. There are a few steps that they will argue over, which I like.
[EDIT: Cherylanne Thyrum figured out where this worksheet originated! It's from Math Teacher Mambo. You can find the worksheet under "activities", then "proof story". Thanks Cherylanne!]
Finally, I start talking about Algebraic Proofs. I don’t actually talk about the properties of equality until the next day. On this first day I usually just put a few equations on the board and we solve them. However, I also write “what we did” next to it (forming a baby two-column proof). Something as simple as “multiplied” or “subtracted” works for me on this first day.
I typically don’t give homework on this night. The next day, their warmup is usually to tell me logically, what they did the night before. I’m looking for a short five or so sentences that make sense (logical and in order)!
Stay tuned: In a few weeks, I’ll share how I actually teach lessons with proofs.
|
HuggingFaceTB/finemath
|
|
you will construct the circuit using power-point shapes (resistors, wires, meters,…).
I’m studying for my Physics class and don’t understand how to answer this. Can you help me study?
Instructions: For each circuit presented, you will construct the circuit using power-point shapes (resistors, wires, meters,…). In addition, based on the voltage values, you will calculate the total resistance and the total current for each circuit. The power-point file that you are to work on is attached to this page. Open it, perform the work, save it, and lastly upload it to the link provided. Refer back to the video in the previous module if you need a start. NOTE: When placing the ammeter in its place, you are measuring the current (coming out of) that would flow back through the battery.
1. Two resistors in series: Construct the circuit on your breadboard for the two-resistor series circuit. Place an ammeter to measure the TOTAL current in the circuit.
• Measure the voltage across any of the two resistors. Drag the voltmeter to a convenient spot and take two wires to hook it up properly to measure the desired voltage. Place the value of the reading (what it should read ideally) in the space provided on the sheet.
• Use the wires provided. Save the red for the power input and then use whatever colors you prefer, or your TA prefers, for the meters and and return wire.
2. Two resistors in parallel: Construct the circuit on your breadboard for the two-resistor parallel circuit. Place an ammeter to measure the TOTAL current in the circuit.
• Measure the voltage across any of the two resistors. Drag the voltmeter to a convenient spot and take two wires to hook it up properly to measure the desired voltage. Place the value of the reading (what it should read ideally) in the space provided on the sheet.
• Use the wires provided. Save the red for the power input and then use whatever colors you prefer, or your TA prefers, for the meters and and return wire.
3. One resistor in series with a parallel combination: Construct the circuit of the single resistor in series with a parallel combination. Place the ammeter to measure the TOTAL current in the circuit.
• Measure the voltage across any of the three resistors. Drag the voltmeter to a convenient spot and take two wires to hook it up properly to measure the desired voltage. Place the value of the reading (what it should read ideally) in the space provided on the sheet.
• Use the wires provided. Save the red for the power input and then use whatever colors you prefer, or your TA prefers, for the meters and and return wire.
Order the answer to view it
ORDER THIS OR A SIMILAR PAPER AND GET 20% DICOUNT. USE CODE: GET2O ORDER NOW
|
HuggingFaceTB/finemath
|
|
# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Written by Team Trustudies
Updated at 2021-05-07
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1
Q1 ) Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (- a, – b)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Distance formula =$$\sqrt { (x_2 - x_1)^2 + (y_2 - y_1)^2}$$
(i) Applying Distance Formula to find distance between points (2, 3) and (4,1), we get
d = $$\sqrt { (4 - 2)^2 + (1 - 3)^2}$$
$$= \sqrt{ 2^2 + (-2) ^2} = \sqrt{4 + 4}$$
$$= \sqrt{8}$$
$$= 2\sqrt{2}$$
(ii) Applying Distance Formula to find distance between points (–5, 7) and (–1, 3), we get
d = $$\sqrt { (-5 + 1)^2 + (7 - 3)^2}$$
$$= \sqrt{ ((-4)^2 + (4) ^2} = \sqrt{16 + 16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
(iii) Applying Distance Formula to find distance between points (a, b) and (–a, –b), we get
d = $$\sqrt { (-a - a)^2 + (-b - b)^2}$$
$$= \sqrt{ ((-2a)^2 + (-2b) ^2}$$
$$= \sqrt{4a^2 + 4b^2}$$
$$= 2\sqrt{a^2 + b^2}$$
Q2 ) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = $$\sqrt{(36-0)^2 + (15 - 0)^2}$$
$$= \sqrt{36^2 - 15^2}$$
$$= \sqrt{1296 + 225}$$
$$= \sqrt{1521}$$
$$= 39$$km
Q3 ) Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = $$\sqrt{(2-1)^2 + (3 - 5)^2}$$
$$= \sqrt{1^2 - (-2)^2}$$
$$= \sqrt{1 + 5}$$
$$= \sqrt{5}$$
BC = $$\sqrt{(-2-2)^2 + (-11 - 3)^2}$$
$$= \sqrt{(-4)^2 - (-14)^2}$$
$$= \sqrt{16 + 196}$$
$$= \sqrt{212}$$
CA = $$\sqrt{(-2-1)^2 + (-11 - 5)^2}$$
$$= \sqrt{(-3)^2 - (-15)^2}$$
$$= \sqrt{9 + 256}$$
$$= \sqrt{265}$$
Since AB + AC $$\ne$$ BC, BC + AC $$\ne$$ AB and AC $$\ne$$ BC.
Therefore, the points A, B and C are not collinear.
Q4 ) Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = $$\sqrt{(6-5)^2 + (4 + 2)^2}$$
$$= \sqrt{1^2 - 6^2}$$
$$= \sqrt{1 + 36}$$
$$= \sqrt{37}$$
BC = $$\sqrt{(7-6)^2 + (-2 - 4)^2}$$
$$= \sqrt{1^2 - (-6)^2}$$
$$= \sqrt{1 + 36}$$
$$= \sqrt{37}$$
CA = $$\sqrt{(7-5)^2 + (-2 + 2)^2}$$
$$= \sqrt{2^2 - 0}$$
$$= \sqrt{4}$$
$$= 2$$
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.
Q5 ) In a classroom, 4 friends are seated at the points A (3, 4), B (6, 7), C (9, 4) and D (6, 1). Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli. “Don’t you think ABCD is a square?”Chameli disagrees. Using distance formula, find which of them is correct.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We have A = (3, 4), B = (6, 7), C = (9, 4) and D = (6, 1)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(6-3)^2 + (7 - 4)^2}$$
$$= \sqrt{3^2 + 3^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
BC = $$\sqrt{(9- 6)^2 + (4 - 7)^2}$$
$$= \sqrt{3^2 + 3^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
CD = $$\sqrt{(6-9)^2 + (1 - 4)^2}$$
$$= \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
DA = $$\sqrt{(6-3)^2 + (1 - 4)^2}$$
$$= \sqrt{3^2 + (-3)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
$$= 3\sqrt{2}$$
Therefore, All the sides of ABCD are equal here. … (1)
Now, we will check the length of its diagonals.
AC = $$\sqrt{(9-3)^2 + (4 - 4)^2}$$
$$= \sqrt{6^2 - 0^2}$$
$$= \sqrt{36}$$
$$= 6$$
BD = $$\sqrt{(6-6)^2 + (1 - 7)^2}$$
$$= \sqrt{0^2 - 6^2}$$
$$= \sqrt{36}$$
$$= 6$$
So, Diagonals of ABCD are also equal. … (2)
From (1) and (2), we can definitely say that ABCD is a square.
Therefore, Champa is correct.
Q6 ) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(1 + 1)^2 + (0 + 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
BC = $$\sqrt{(-1 - 1)^2 + (2 - 0)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
CD = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
DA = $$\sqrt{(-3 + 1)^2 + (0 - 2)^2}$$
$$= \sqrt{4 + 4}$$
$$= 2 \sqrt{2}$$
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = $$\sqrt{(-1 + 1)^2 + (2 + 2)^2}$$
$$= \sqrt{0 + 16}$$
$$= 4$$
BD = $$\sqrt{(-3 - 1)^2 + (0 + 0)^2}$$
$$= \sqrt{16 + 0}$$
$$= 4$$
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.
(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(-3 - 3)^2 + (1 - 5)^2}$$
$$= \sqrt{36 + 16}$$
$$= 2 \sqrt{13}$$
BC = $$\sqrt{(0 - 3)^2 + (3 - 1)^2}$$
$$= \sqrt{9 + 4}$$
$$= \sqrt{13}$$
CD = $$\sqrt{(-1 - 0)^2 + (-4 - 3)^2}$$
$$= \sqrt{1 + 49}$$
$$= 5\sqrt{2}$$
DA = $$\sqrt{(-1 + 3)^2 + (-4 - 5)^2}$$
$$= \sqrt{4 + 81}$$
$$= \sqrt{85}$$
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.
(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = $$\sqrt{(7 - 4)^2 + (6 - 5)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
BC = $$\sqrt{(4 - 7)^2 + (3 - 6)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
CD = $$\sqrt{(1 - 4)^2 + (2 - 3)^2}$$
$$= \sqrt{9 + 1}$$
$$= \sqrt{10}$$
DA = $$\sqrt{(1 - 4)^2 + (2 - 5)^2}$$
$$= \sqrt{9 + 9}$$
$$= \sqrt{18}$$
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = $$\sqrt{(4 - 4)^2 + (3 - 5)^2}$$
$$= \sqrt{0 + 4}$$
$$= 2$$
BD = $$\sqrt{(1 - 7)^2 + (2 - 6)^2}$$
$$= \sqrt{36 + 16} =$$
$$2 \sqrt{13}$$
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.
Q7 ) Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the point be (x, 0) on x–axis which is equidistant from (2, –5) and (–2, 9).
Using Distance Formula and according to given conditions we have:
$$\sqrt{(x - 2)^2 + ( 0 - (-5))^2}$$
$$= \sqrt{ (x - (-2))^2 + ( 0 - 9)^2}$$
=>$$\sqrt {x^2 + 4 - 4x + 25}$$
$$= \sqrt {x^2 + 4 + 4x + 81}$$
Squaring both sides, we get
$$\Rightarrow x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81$$
$$\Rightarrow-4x + 29 = 4x + 85$$
$$\Rightarrow 8x = -56$$
$$\Rightarrow x = -7$$
Therefore, point on the x–axis which is equidistant from (2, –5) and (–2, 9) is (–7, 0)
Q8 ) Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Using Distance formula, we have
$$\Rightarrow 10 = \sqrt{(2 -10)^2 + (-3-y)^2}$$
$$\Rightarrow 10 = \sqrt{(-8)^2 + 9 + y^2 + 6y}$$
$$\Rightarrow 10 = \sqrt{64 + 9 + y^2 + 6y}$$
Squaring both sides, we get
$$\Rightarrow 100 = 73 + y^2 + 6y$$
$$\Rightarrow y^2 + 6y + 27 = 0$$
Solving this Quadratic equation by factorization, we can write
$$\Rightarrow y^2 + 9y - 3y - 27 = 0$$
$$\Rightarrow y (y + 9) – 3 (y + 9) = 0$$
$$\Rightarrow (y + 9) (y - 3) = 0$$
$$\Rightarrow y = 3, -9$$
Q9 ) If, Q (0, 1) is equidistant from P (5, –3) and R (x, 6), find the values of x. Also, find the distances QR and PR.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
It is given that Q is equidistant from P and R. Using Distance Formula, we get
PQ = RQ
PQ=$$\sqrt{(5-0)^2 + (-3 - 1)^2}$$
$$= \sqrt{(-5)^2 + (-4)^2}$$
$$= \sqrt{25 + 16}$$
$$= \sqrt{41}$$
QR= $$\sqrt{(0 - x)^2 + (1 - 6)^2}$$
$$= \sqrt{(-x)^2 + (-5)^2}$$
$$= \sqrt{x^2 + 25}$$
$$\therefore \sqrt{41} = \sqrt{x^2 + 25}$$
On squaring both sides,
$$\Rightarrow 41 = x^2 + 25$$
$$\Rightarrow x^2 = 16$$
$$\Rightarrow x = \pm4$$
Coordinates of point R will be (4,6) or (-4,6),
If R is (4,6),
QR = $$\sqrt{(0 - 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
PR = $$\sqrt{(5 - 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{1^2 + 9^2}$$
$$= \sqrt{1 + 81}$$
$$= \sqrt{82}$$
If R is (-4,6),
QR = $$\sqrt{(0 + 4)^2 + (1 - 6)^2}$$
$$= \sqrt{4^2 + 5^2}$$
$$= \sqrt{16 + 25}$$
$$= \sqrt{41}$$
PR = $$\sqrt{(5 + 4)^2 + (-3 - 6)^2}$$
$$= \sqrt{9^2 + 9^2}$$
$$= \sqrt{81 + 81}$$
$$= 9\sqrt{2}$$
Q10 ) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
It is given that (x, y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write
$$\Rightarrow \sqrt{(x-3)^2 + (y-6)^2} = \sqrt{(x-(-3))^2 + (y-4)^2}$$
$$\Rightarrow \sqrt{x^2 + 9 - 6x + y^2 + 36 -12y } = \sqrt{x^2 + 9 + 6x + y^2 + 16 -8y }$$
Squaring both sides, we get
$$\Rightarrow x^2 + 9 - 6x + y^2 + 36 -12y = x^2 + 9 + 6x + y^2 + 16 -8y$$
$$\Rightarrow -6x - 12y + 45 = 6x - 8y + 25$$
$$\Rightarrow 12x + 4y = 20$$
$$\Rightarrow 3x + y = 5$$
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.2
Q1 ) Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the points be P(x,y), Using the section formula
$$P(x, y) = [\frac{mx_2 + nx_1}{m + n} , \frac{my_2 + ny_1}{m + n}]$$
$$x = {{2(4) + 3(-1)} \over {2 + 3}} = {{8 - 3} \over {5}} = 1$$
$$y = {{2(-3) + 3(7)} \over {2 + 3}} = {{-6 + 21} \over {5}} = 3$$
Therefore the point is(1,3)
Q2 ) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2.
x1 = $$\frac{(1×(-2)+2×4)}{1+2} = \frac{(-2+8)}{3} =\frac{6}{3} = 2$$
y1 = $$\frac{(1×(-3)+2×-1)}{1+2} = \frac{(-3-2)}{3} =\frac{-5}{3}$$
$$\therefore$$P = (x1, y1) = P(2, $$\frac{-5}{3}$$)
Point Q divides AB internally in the ratio 2:1.
x2 = $$\frac{(2×(-2)+1×4)}{2+1} = \frac{(-4+4)}{3} = 0$$
y2 = $$\frac{(2×(-3)+1×-1)}{2+1} = \frac{(-6-1)}{3} =\frac{-7}{3}$$
The coordinates of the point Q(x2 ,y2 ) = (0, $$\frac{-7}{3}$$ )
Q3 ) To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs $$\frac{1}{4}$$ th the distance AD on the 2nd line and posts a green flag. Preet runs $$\frac{1}{5}$$th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The green flag is $$\frac{1}{4} th$$ of total distance =
$$\frac{1}{4} × 100 = 25$$ m in second line
$$\therefore$$ the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)
Distance between red and green flag
= $$\sqrt{(8-2)^2 + (20-25)^2}$$
$$= \sqrt{6^2+(-5)^2}$$
$$= \sqrt{36+25}$$
$$= \sqrt{61} m$$
Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )
$$\therefore( x , y ) =( \frac{(8+2)}{2} , \frac{20+25}{2} )$$
$$=( 5 , \frac{45}{2} )= ( 5, 22.5 )$$
$$\therefore$$ the blue flag is posted in fifth line at a distance of 22.5 m
Q4 ) Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the required ratio be k : 1
By section formula ,
$$x = \frac{m1x2 + m2x1}{m1+m2}$$
$$\Rightarrow -1 = \frac{k×6 + 1× -3 )}{k+1}$$
$$\Rightarrow -k-1 = 6k -3$$
$$\Rightarrow 7k = 2$$
$$\Rightarrow k = \frac{2}{7}$$
The required ratio = k : 1
= $$\frac{2}{7}$$ : 1 = $$\frac{2}{7} ×7 : 1×7$$ = 2 : 7
Q5 ) Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let p(x , 0) be the point which devides the line segment joining A(1, -5 ) and B(-4,5 ) in the ratio m : 1
Then the using section formula
(x , 0 ) = $$( \frac{m×(-4) + 1 × 1 }{m+1} , \frac{m×5+1×(-5)}{m+1} )$$
$$\Rightarrow 0 = \frac{m×5+1×(-5)}{m+1}$$
$$\Rightarrow 0 = 5m - 5$$
$$\Rightarrow m = 5 /5 = 1$$
Hence the required ratio is 1 : 1
Since the ratio is 1 : 1 , so P is the mid point
$$\therefore$$ x = $$\frac{m×(-4) + 1 × 1 }{m+1}$$
x = $$\frac{1×(-4) + 1 × 1 }{1+1} = \frac{-4+1}{2} = \frac{-3}{2}$$
$$\therefore ( \frac{-3}{2} , 0 )$$ is the required point
Q6 ) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Mid point of AC = Mid point BD
$$\Rightarrow \frac{x+1}{2} , \frac{6+2}{2} = \frac{4+3}{2} , \frac{y+5}{2}$$
$$\Rightarrow \frac{x+1}{2} = \frac{7}{2}$$
$$\Rightarrow x +1 = 7$$
$$\Rightarrow x = 7-1= 6$$
$$\Rightarrow \frac{y+5}{2} = \frac{8}{2}$$
$$\Rightarrow y+5 = 8$$
$$\Rightarrow y = 8 -5 = 3$$
Q7 ) Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the coordinates of point A be (x, y).
Mid-point of AB is (2, – 3), which is the centre of the circle.
Coordinate of B = (1, 4)
(2, -3) =$$(\frac{x+1}{2} , \frac{y+4}{2} )$$
$$\frac{x+1}{2} = 2$$
$$\Rightarrow x+1= 4$$
$$\Rightarrow x = 4-1 = 3$$
and$$\frac{y+4}{2} = -3$$
$$\Rightarrow y+4 = -6$$
$$\Rightarrow y = -6-4 = -10$$
$$\therefore$$ x = 3 and y = -10
The coordinates of A(3,-10).
Q8 ) If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = $$\frac{3}{7}$$ AB and P lies on the line segment AB.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
AP = $$\frac{3}{7}$$ AB
7 AP = 3 AB
AB : AP = 7: 3
Let AB = 7x
$$\therefore$$ AP = 3x
AB = AP + BP
7x = 3x + BP
BP = 7x - 3x = 4x
$$\therefore \frac{AP}{BP} = \frac{3x}{4x}$$
$$\therefore$$ AP : BP = 3:4
by section formula ,
(x, y ) = $$(\frac{m1x2 + m2x1}{m1+m2} , \frac{m1y2 + m2y1}{m1+m2} )$$
(x, y) = $$(\frac{3×2 +4×(-2) }{3+4} , \frac{3×(-4)+ 4(-2)}{3+4} )$$
(x ,y ) = $$(\frac{6 -8 }{7} , \frac{-12+ -8}{7} )$$
(x ,y ) = $$(\frac{-2}{7} , \frac{-20}{7} )$$
Hence the coordinates of P = $$(\frac{-2}{7} , \frac{-20}{7} )$$
Q9 ) Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
From the figure, it can be observed that points X, Y, Z are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively.
Coordinate of X = $$( \frac{1×2+3×(-2)}{1+3} , \frac{1×8+3×2}{1+3} )$$
$$=( \frac{2-6}{4} , \frac{8+6}{4} )$$
$$= ( \frac{-4}{4} , \frac{14}{4} )$$
$$= ( -2 , \frac{7}{2} )$$
Coordinate of Y = $$( \frac{1×2+1×(-2)}{1+1} , \frac{1×8+1×2}{1+1} )$$
$$=( \frac{2-2}{2} , \frac{8+2}{2} )$$
$$= ( \frac{0}{2} , \frac{10}{2} )$$
$$= ( 0 , 5)$$
Coordinate of Y = $$( \frac{3×2+1×(-2)}{3+1} , \frac{3×8+1×2}{3+1} )$$
$$=( \frac{6-2}{4} , \frac{24+2}{4} )$$
$$= ( \frac{4}{4} , \frac{26}{4} )$$
$$= ( 1 , \frac{13}{2} )$$
Q10 ) Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.
[Hint: Area of a rhombus = $$\frac{1}{2}$$ (product of its diagonals)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.
Length of diagonal AC
$$=\sqrt{[3-(-1)]^2 + (0-4)^2}$$
$$= \sqrt{16+16}$$
$$= \sqrt{32}$$
$$= 4\sqrt{2}$$
Length of diagonal BD
$$=\sqrt{[(-2)-4]^2+ (-1-5)^2}$$
$$= \sqrt{(-6)^2+(-6)^2}$$
$$= \sqrt{36+36}$$
$$= \sqrt{72}$$
$$= 6\sqrt{2}$$
Area of rhombus = $$\frac{1}{2}$$ × product of diagonals
= $$\frac{1}{2} × 4\sqrt{2}×6\sqrt{2} = 24$$sq unit
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.3
Q1 ) Find the area of the triangle whose vertices are:
(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We know that formula for area of a triangle whose vertices are $$(x_1,y_1) , (x_2,y_2) , (x_3,y_3)$$ is,
= $$\frac{1}{2} \ | \ [x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \ |$$
(i) So, here $$x_1 \ = \ 2 \ , \ y_1 \ = \ 3 \ , \ x_2 \ = \ -1 \ , \ y_2 \ = \ 0 \ x_3 \ = \ 2 \ , \ y_3 \ = \ -4$$
So, area of triangle = $$\frac{1}{2} \ | \ [ 2( 0 + 4 ) \ - \ 1(-4 - 3) \ + \ 2(3 - 0)] \ | \ = \ \frac{1}{2} \ | \ [ 8 + 7 + 6 ] \ | \ = \ \frac{21}{2}$$
∴ Area of triangle is $$\frac{21}{2}$$ sq. units.
(ii) Similarly, here $$x_1 \ = \ -5 \ , \ y_1 \ = \ -1 \ , \ x_2 \ = \ 3 \ , \ y_2 \ = \ -5 \ x_3 \ = \ 5 \ , \ y_3 \ = \ 2$$
So, area of triangle = $$\frac{1}{2} \ | \ [ (-5)( -5 - 2 ) \ - \ 3(2 + 1) \ + \ 5(-1 + 5)] \ | \ = \ \frac{1}{2} \ | \ [ 35 + 9 + 20 ] \ | \ = \ 32$$
∴ Area of triangle is $$32$$ sq. units.
Q2 ) In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
We know that for collinear points area of triangle = 0 ,i.e.,
$$0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)]$$
(i) $$7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0$$
=> $$7 - 7k + 5k + 10 - 9 \ = \ 0$$
=> $$2k \ = \ 8$$
=> $$k \ = \ 4$$
(ii) $$8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0$$
=> $$8 - 6k + 10 \ = \ 0$$
=> $$6k \ = \ 18$$
=> $$k \ = \ 3$$
Q3 ) Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The vertices of the triangle ABC are A (0, -1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides AB,BC,CA of this triangle ABC respectively.
Coordinates of D, E, and F are given by :
D = $$(\frac{0+2}{2} \ , \ \frac{-1+1}{2})$$ = (1,0)
E = ( $$\frac{0+2}{2} \ , \ \frac{1+3}{2}$$ ) = (1,2)
F = ( $$\frac{0+0}{2} \ , \ \frac{3-1}{2}$$ ) = (0,1)
So, Area of $$∆ \ DEF \ = \ \frac{1}{2} \ | \ [ 1(2-1) + 1(1-0) + 0(0-2)] \ | \ = \ 1$$
∴ Area of ∆ DEF = 1 sq. units
Area of ∆ ABC = $$\frac{1}{2} \ | \ [0(1-3) \ + \ 2(3+1) \ + \ 0(-1-1) ] \ | \ = \ 4$$
$$\therefore$$ Area of $$\triangle$$ ABC = 4 sq. units
So, $$∆ \ DEF \ : \ ∆ \ ABC \ = \ 1 \ : \ 4$$.
Q4 ) Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Draw a line from B to D
Now we have,
Area of quad. ABCD = | Area of ∆ DAB | + | Area of ∆ BCD |
So,
Area of ∆ DAB = $$| \ \frac{1}{2} [2(-2+5) \ - \ 4(-5-3) \ - \ 3(3+2)] \ |$$ = $$\ | \frac{1}{2} [6 + 32 - 15 ] \ |$$ = $$\frac{23}{2}$$ sq. units
Similarly, Area of ∆ BCD = $$| \ \frac{1}{2} [-3(-2-3) \ + \ 3(3+5) \ + \ 2(-5+2)] \ | \ = \ | \ \frac{1}{2} [15 + 24 - 6 ] \ |$$ = $$\frac{33}{2}$$ sq. units
$$\therefore$$ Area of quad. ABCD = $$\frac{23}{2} \ + \ \frac{33}{2}$$ = $$\frac{56}{2} \ = \ 28$$ sq. units
Q5 ) You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for $$\triangle$$ ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
The vertices of triangle ABC are A(4,-6), B (3, – 2) and C (5, 2).
Let D be the mid-point of side BC of ∆ ABC. Therefore, AD is the median in ∆ ABC.
So, coordinates of D are ( $$\frac{3+5}{2} \ , \ \frac{-2+2}{2}$$ )
Therefore, coordinates of D = (4,0)
Now,
Area of ∆ ABD = $$\frac{1}{2} \ | \ [4(-2-0) \ + \ 3(0+6) \ + \ 4(-6+2)] |$$
= $$\frac{1}{2} \ | \ [ -8 + 18 - 16] \ |$$
= $$|-3| \ = \ 3$$ sq. units
Similarly,
Area of ∆ ACD = $$\frac{1}{2} \ | \ [4(0-2) \ + \ 4(2+6) \ + \ 5(-6-0)] \ |$$
= $$\frac{1}{2} \ | \ [ -8 + 32 - 30] \ |$$
= $$|-3| \ = \ 3$$ sq. units
$$\therefore$$ Clearly, Area of ∆ ABD = Area of ∆ ACD = 3 sq. units.
Hence, median of a triangle divides it into two triangles of equal areas.
## NCERT solutions for class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4
Q1 ) Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the line 2x + y - 4 = 0 divides AB in 1 : k ratio, then the coordinates of the point of division is
$$x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$
The point of intersection lie on both lines
$$\therefore x \ = \ \frac{2+3k}{k+1} \ and \ y \ = \ \frac{-2+7k}{k+1}$$ should satisfy 2x + y - 4 = 0
$$\Rightarrow 2( \frac{2+3k}{k+1}) \ + \ (\frac{-2+7k}{k+1}) \ - \ 4 \ = \ 0$$
$$\Rightarrow 4 \ + \ 6k \ - \ 2 \ + \ 7k \ = \ 4k \ + \ 4$$
$$\Rightarrow 9k - 2 = 0$$
$$\Rightarrow k \ = \ \frac{2}{9}$$
$$\therefore$$ The line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in ratio 2 : 9 internally.
Q2 ) Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
If given points are collinear then area of triangle formed by them must be zero.
Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,
Area of triangle = $$\frac{1}{2} \ | \ [ (x_1(y_2 – y_3 ) + x_2(y_3 – y_1) + x_3(y_1 – y_2) ) ] \ | \ = \ 0$$
$$\Rightarrow \frac{1}{2} \ | \ [x(2 – 0) + 1 (0 – y) + 7( y – 2)] \ | \ = \ 0$$
$$\Rightarrow 2x \ – \ y \ + \ 7y \ – \ 14 \ = \ 0$$
$$\Rightarrow 2x \ + \ 6y \ – \ 14 \ = \ 0$$
$$\Rightarrow x \ + \ 3y \ – \ 7 \ = \ 0$$ .
Hence this is the required relation between x and y.
Q3 ) Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let the coordinate of centre O be (x,y)
Then, OA = OB = OC {$$\because$$ radius is equal}
Now,
OA = $$\sqrt{ (x-6)^2+(y+6)^2 }$$
OB = $$\sqrt{(x-3)^2 + (y+7)^2 }$$
OC = $$\sqrt{(x-3)^2 + (y-3)^2 }$$
Now, OA = OB
$$\Rightarrow \sqrt{(x-6)^2 + (y+6)^2 } = \sqrt{(x-3)^2 + (y+7)^2 }$$
$$\Rightarrow x^2 \ + \ 36 \ - \ 12x \ + \ y^2 \ + \ 36 \ - \ 12 \ = \ x^2 \ + \ 9 \ - \ 6x \ + \ y^2 \ + \ 9 \ - \ 6y$$
$$\Rightarrow 6x \ + 2y \ - \ 14 \ = \ 0$$ ..............(i)
Similarly, OB = OC
$$\Rightarrow \sqrt{(x-3)^2 + (y+7)^2 } = \sqrt{(x-3)^2 + (y-3)^2 }$$
$$\Rightarrow y^2 \ + \ 14y \ + \ 49 \ = \ y^2 \ - \ 6y \ + \ 9$$
$$\Rightarrow y \ = \ -2$$
Putting this value of y in eq. (i) , we get
$$\Rightarrow 6x \ + \ 2(-2) \ - \ 14 \ = \ 0$$
$$\Rightarrow x \ = \ 3$$
$$\therefore$$ The centre of circle is at (3,-2)
Q4 ) The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Let ABCD be a square with vertices A( $$x_1, y_1) , B(3 , 2) , C(x_2 , y_2) , D(-1 , 2)$$
Let each side of square be 'a'.
So applying Pythagoras theorem in ∆ABD, we get
$$BD^2 = AB^2 + DA^2$$
$$\Rightarrow \sqrt{(3+1)^2 \ + \ (2-2)^2} \ = \ a^2 \ + \ a^2$$
$$\Rightarrow 4 \ = \ 2 a^2$$
$$\Rightarrow a \ = \ 2 \sqrt{2}$$
Now, DA = AB {$$\because$$ ABCD is a square }
$$\Rightarrow \sqrt{(x_1 +1)^2 + (y_1 - 2)^2 } \ = \ \sqrt{(3-x_1)^2 + (2 - y_1)^2}$$ {by distance formula}
$$\Rightarrow (x_1 +1)^2 \ = \ (3-x_1)^2$$
$$\Rightarrow x_1^2 \ + \ 1 \ + \ 2x_1 \ = \ x_1^2 \ + \ 9 \ - \ 6x_1$$
$$\Rightarrow x_1 \ = \ 1$$
Now since each side is $$2 \sqrt{2}$$
$$\therefore$$ AB $$= \ 2 \sqrt{2}$$
$$\Rightarrow \sqrt{(3-x_1)^2 + (2 - y_1)^2} \ = \ 2 \sqrt{2}$$
$$\Rightarrow(2-y_1)^2 = \ 4$$
$$\Rightarrow y \ = \ 4$$
Now, point O is id point of BD . SO coordinates of O are
x = $$\frac{3-1}{2}$$ , y = $$\frac{2+2}{2}$$
$$\Rightarrow$$ x = 1 , y = 2
Now, since ABCD is a square $$\therefore$$ O is mid point of AC also
$$\frac{1+x_2}{2} \ = \ 1$$ , $$\frac{4+ y_2}{2} \ = \ 2$$
$$\Rightarrow x_2 \ = \ 1$$ , $$y_2 \ = \ 0$$
$$\therefore$$ Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)
Q5 ) The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
(i) Taking A as origin, coordinates of the vertices P, Q and R are,
From figure:
P = (4, 6)
Q = (3, 2)
R = (6, 5)
Here AD is the x-axis and AB is the y-axis.
Area of triangle PQR in case of origin A:
Area of a triangle = $$\frac{1}{2} \ | \ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)] \ | \$$
$$= \ \frac{1}{2} \ | \ [– 12 – 3 + 24] \ | \$$
$$= \ \frac{9}{2}$$ sq. units
(ii) Taking C as origin, coordinates of vertices P, Q and R are,
From figure:
P = (12, 2)
Q = (13, 6)
R = (10, 3)
Here CB is the x-axis and CD is the y-axis.
Area of triangle PQR in case of origin C:
Area of a triangle = $$\frac{1}{2} \ | \ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)] \ | \$$
$$= \frac{1}{2} \ | \ [36 + 13 – 40] \ | \$$
$$= \ \frac{9}{2}$$ sq unit
This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C
Area is same in both case because triangle remains the same no matter which point is considered as origin.
Q6 ) The vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that $$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$ . Calculate the area of the $$∆ \ ADE$$ and compare it with area of $$∆ \ ABC$$ . (Recall Theorem 6.2 and Theorem 6.6)
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Given, the vertices of a $$∆ \ ABC$$ are A (4, 6), B (1, 5) and C (7, 2) and,
$$\frac{AD}{AB} \ = \ \frac{AE}{AC} \ = \ \frac{1}{4}$$
$$\Rightarrow \frac{AB}{AD} \ = \ \frac{AC}{AE} \ = \ 4$$
$$\Rightarrow \frac{AD + DB}{AD} \ = \ \frac{AE + EC}{AE} \ = \ 4$$
$$\Rightarrow 1 + \frac{DB}{AD} \ = \ 1 + \frac{EC}{AE} \ = \ 4$$
$$\Rightarrow \frac{DB}{AD} \ = \ \frac{EC}{AE} \ = \ 3$$
$$\therefore$$ Point D and E divides AB and AC in ratio 1 : 3
Now coordinates of D calculated by section formula are,
$$x \ = \ \frac{3(4) + 1(1)}{4} \ = \ \frac{13}{4}$$ and,
$$y \ = \ \frac{3(6) + 1(5)}{4} \ = \ \frac{23}{4}$$
Similarly, coordinates of E are,
$$x \ = \ \frac{1(7) + 3(4)}{4} \ = \ \frac{19}{4}$$ and,
$$y \ = \ \frac{1(2) + 3(6)}{4} \ = \ \frac{20}{4} \ = \ 5$$
Now area of triangle ADE is,
$$\frac{1}{2} \ | \ [4(5-2) \ + \ 1(2-6) \ + \ 7(6-5)] \ |$$
= $$\frac{1}{2} \ | \ [(12-4+7)] \ |$$
= $$\frac{15}{2}$$ sq. units
Similarly, area of triangle ABC is,
$$\frac{1}{2} \ | \ [4( \frac{23}{4} -5) \ + \ \frac{13}{4}(5-6) \ + \ \frac{19}{4}(6- \frac{23}{4})] \ |$$
= $$\frac{1}{2} \ | \ [(3- \frac{13}{4} + \frac{19}{4})] \ |$$
= $$\frac{15}{32}$$ sq. units
$$\therefore$$ ratio of $$area \ of \ ∆ \ ADE \ to \ area \ of \ ∆ \ ABC \ = \ 1:16$$
Q7 ) Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of $$∆ \ ABC$$ .
(i) The median from A meets BC at D. Find the coordinates of point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.
(iv) What do you observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.]
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
(i) Given, D is the median to BC. So, coordinates of D using distance formula are
$$x \ = \ \frac{6+1}{2} \ = \ \frac{7}{2}$$ and $$y \ = \ \frac{5+4}{2} \ = \ \frac{9}{2}$$
(ii) Coordinates of P can be calculated using section formula
$$x \ = \ \frac{2( \frac{7}{2}) + 1(4)}{3} \ = \ \frac{11}{3}$$ and $$y \ = \ \frac{2( \frac{9}{2}) + 1(2)}{3} \ = \ \frac{11}{3}$$
(iii) Coordinates of E
= ( $$\frac{4+1}{2} \ , \ \frac{2+4}{2}$$ )
= ( $$\frac{5}{2} , 3$$ )
So, coordinates of Q are,
( $$\frac{2( \frac{5}{2}) + 1(6)}{3} \ , \ \frac{2(3) + 1(5)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )
Coordinates of F
= ( $$\frac{4+6}{2} , \frac{2+5}{2}$$ )
= ( $$5, \frac{7}{2}$$ )
So, coordinates of R are,
( $$\frac{2(5) + 1(1)}{3} , \frac{2 ( \frac{7}{2}) + 1(4)}{3}$$ )
= ( $$\frac{11}{3} , \frac{11}{3}$$ )
(iv) We observed that coordinates of P , Q, R are same i.e., ( $$\frac{11}{3} , \frac{11}{3}$$ ) which shows that medians intersect each other at a common point which is called centroid of the triangle.
(v) If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of triangle ABC, then the coordinates of the centroid of the triangle is
$$x \ = \ \frac{x_1 + x_2 + x_3}{3} \ and \ \frac{y_1 + y_2 + y_3}{3}$$
Q8 ) ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
P, Q, R and S are the midpoints of AB, BC, CD and DA respectively
So, coordinate of P
= ( $$\frac{-1-1}{2} , \frac{-1+4}{2}$$ )
= ( $$-1 , \frac{3}{2}$$ )
Similarly, coordinate of Q
= ( $$\frac{5-1}{2} , \frac{4+4}{2}$$ )
= ( 2,4)
R = ( $$\frac{5+5}{2} , \frac{4-1}{2}$$ ) = ( $$5 , \frac{3}{2}$$ )
S = ( $$\frac{5-1}{2} , \frac{-1-1}{2}$$ ) = ( 2,-1)
Now, PQ = $$\sqrt{(-1-2)^2 + ( \frac{3}{2} - 4)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
SP = $$\sqrt{(2+1)^2 + (-1- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
QR = $$\sqrt{(2-5)^2 + (4- \frac{3}{2})^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
RS = $$\sqrt{(5-2)^2 + ( \frac{3}{2} +1)^2} \$$
$$= \ \sqrt{ \frac{61}{4} } \$$
$$= \ \frac{ \sqrt{61}}{2}$$
PR = $$\sqrt{(-1-5)^2 + ( \frac{3}{2} - \frac{3}{2})^2 } \$$
$$= \ 6$$
QS = $$\sqrt{(2-2)^2 + (4+1)^2} \$$
$$= \ 5$$
Clearly, PQ = QR = RS = SP
But, $$PR \ne QS$$
Hence the lengths of all the sides are equal, while the lengths of the diagonals are different.
$$\therefore$$ The PQRS is a rhombus.
##### FAQs Related to NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
There are total 33 questions present in ncert solutions for class 10 maths chapter 7 coordinate geometry
There are total 12 long question/answers in ncert solutions for class 10 maths chapter 7 coordinate geometry
There are total 4 exercise present in ncert solutions for class 10 maths chapter 7 coordinate geometry
|
HuggingFaceTB/finemath
|
|
# PHY C22: Charged Particles in Magnetic Fields
So far we’ve explored forces on permanent magnets & current-carrying conductors. But what about free-moving charged particles?
• Force on a charge moving in a magnetic field
• F = Bqv sin θ
• Determination of v & e/me for electrons
First, let’s introduce ourselves to a few charged particles:
• electron
• proton
• α-particle (Helium-4 nucleus)
• β-particle (fast-moving nuclear electron)
These particles exhibit interesting effects when they pass through a magnetic field.
What is the force on a moving charged particle in a magnetic field?
## F = Bqv sin θ
Here’s a derivation:
Remember the motor effect:
F = BIL sin θ
Current I is given by the number of particles (n) of charge q passing through a point in time t:
I = nq/t
Subbing this in:
F = (Bnq/t)L sin θ
Since v = L/t,
F = Bnqv sin θ
If we are looking at the force acting on a single particle, n = 1, so
F = Bqv sin θ
For a charged particle moving perpendicular to a magnetic field,
F = Bqv
How does this effect the path of the particle?
It causes deflection.
The particle will travel across an arc of a circle. This is an example of Circular Motion (read up on it here).
We can calculate the radius of this arc as follows:
Here, the centripetal force is provided by the magnetic force:
Fc = FB
mv2/r = Bqv
Rearranging this,
The exact shape of the particle’s path depends on many factors.
Here are a few:
One important application of this is learning about the properties of newly-discovered charged particles.
Let’s take the electron as an example.
The mass-charge ratio of an electron is known as its specific charge.
It is given by e/me
To do this calculation, we must use a fine-beam tube:
• electrons are accelerated from rest using a high voltage V
• Helmholtz coils provide a permanent magnetic field B perpendicular to the beam
• the circular orbit of the electron beam can be seen as electrons strike low-pressure gas inside the tube, making it glow
Calculating the mass-charge ratio of a charged particle (electron)
A great simulation if you would like to play with the values of b, m, q, & v to see how they effect a particle’s motion in a magnetic field: oPhysics
In the next post, we will look at what happens when a charged particle enters a magnetic field AND an electric field.
|
HuggingFaceTB/finemath
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.