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# 10 of the Best Fractions Resources for Secondary Maths
Whether it's adding and subtracting, arranging in size order or finding equivalents, help your KS3 and 4 students master fractions with these excellent resources
### 1 | Embed Adding and Subtracting Fractions into a Richer Problem to get the Technique to Stick
Adding and subtracting fractions is a procedure which students often find it very difficult to master. One way to make the practice of this technique less tedious is to embed it in a bigger problem which students are trying to solve.
Having their attention on a grander purpose helps them to internalise the method to the point where they don’t have to think about it any more, which is always the goal of mathematical fluency.
A problem-based approach also means that the answers obtained have some significance for the wider problem, so students are more inclined to check their answers and notice if any of them doesn’t seem to be about the right size.
Get this free lesson plan here.
### 2 | Interactive fractions
This interactive lesson on the Guardian Teacher Network will help students to multiply and divide fractions give all fractions in their simplest form. It’s a nice little tool for revision/recap or establishing prior knowledge in Key Stage 3.
### 3 | Fractions to decimals
This PowerPoint activity from Mistry Maths takes KS3 students through converting fractions to decimals and simplifying beforehand where necessary.
There are loads of questions to work through too.
### 4 | Fractions of amounts
This resource lets students of all abilities develop their knowledge of fractions by working with amounts.
It shows them how to use the bar method to work out sums such as ‘What is two-fifths of £20?’ and ‘What is five-ninths of £108?’.
Find more fractions resources from Piximaths here.
And download the PowerPoint for this resource here and the worksheet here.
### 5 | Fractions worksheets
Want some fractions worksheets? There are four here for you, and they’re all free. That was easy wasn’t it?
There are two on equivalent fractions (click here and here) and two on ‘fiddly fractions’ (here and here).
Or you can just find them all at ks3maths.co.uk here.
### 6 | Fraction puzzles
Now it’s time to let your students loose on some fun challenges and activities. Here are some of our favourite NRICH ones:
Check out NRICH’s collection of fraction activities here.
### 7 | Fractions addition and subtraction trick
Here’s a handy video to help you teach adding and subtracting fractions, and for students to refer back to should they need to.
You’ll find this on the tecmath YouTube channel.
### 8 | Cheese and onion pies
This fun little activity involves rearranging Grandma’s cheese and onion pies depending on which have the most cheese inside.
Give it a go here.
### 9 | Fraction dissect
Here’s another fun little interactive activity from transum.org. For each fraction you’re given a rectangular (4x3) grid where you click on two green dots to draw a line between them.
The line must dissect the rectangle to produce the fraction indicated below the rectangle.
Give it a play here.
### 10 | Holiday bargains
This simple worksheet for applying fractions in money problems helps students put what they’ve learnt to practical use. They’re given a series of offers and have to work out which are the best deals and how much they save on each one. | HuggingFaceTB/finemath | |
## database: in DHIS2, what is the largest number of data elements that a numerator can contain to form an indicator?
We recently updated the family planning dataset to collect data on 6 different age groups based on clients' preferred methods across all family planning acceptors (both new and ongoing). So when calculating the acceptance rate, we have to add more than 228 individual data elements in the numerator to make the indicator.
We only succeeded in completing 108 data items. Anything beyond this, we couldn't save. What are you not doing well? Is there a way to avoid it, for example using subtotals instead of individual data elements? What is the largest number of data elements that a numerator can contain to form an indicator in DHIS2?
## database: what is the largest number of data elements that a numerator can contain to form an indicator?
We recently updated the family planning dataset to collect data on 6 different age groups based on clients' preferred methods across all family planning acceptors (both new and ongoing). So when calculating the acceptance rate, we have to add more than 228 individual data elements in the numerator to make the indicator. We only succeeded in completing 108 data items. Anything beyond this, we couldn't save. What are you not doing well? Is there a way to avoid it, for example using subtotals instead of individual data elements? What is the largest number of data elements that a numerator can contain to form an indicator in DHIS2?
## Liteforex Dream Draw: the largest draw in history
LiteForex is famous for its great raffles with great prizes. We maintain them often so that merchants around the world can take further advantage of their operations.
But this time, it is something really special.
LiteForex celebrates 15 years. Wishing to express our loyalty to our customers, we announce the largest raffle the company has ever had.
This time, the prizes are worth \$ 350,000. 20 winners will be entered in the final draw to be held in Dubai.
LiteForex Dream Draw will take place from 05.05.2020 to 03.05.2021. Each customer automatically participates in the Sweepstakes if they make a deposit of \$ 500 or more to their ECN or CLASSIC business account during the promotion period. Every three months 5 more successful participants will be determined using a transparent market algorithm. 20 winners will go to the gala dinner organized by the company in Dubai, where the draw will take place.
These are the LiteForex Dream Draw prizes:
A certificate for your DREAM HOUSE worth up to \$ 250,000 anywhere in the world
A certificate for your \$ 60,000 DREAM CAR
Two iMac Pro workstations all in one
Three huge and powerful MacBooks Pro 16
Three compact and capable MacBooks Pro 13
Three ultra-portable and fast MacBooks Air
Three iPhones 11 Pro Max
Four iPhones 11 Pro
You can read more about the draw here
Dream draw
Good luck chasing your dreams!
Sincerely,
LiteForex
This publication has been edited by LiteForexOfficial: Today, 10:15 a.m.
## Create a function with foreach that receives a list of dates, returns what is the largest date among them
Largest public list (list data)
{
List r = new List ();
``````foreach (DateTime item in data)
{
if (item > item) //Aqui é onde quero ver se a data é maior ou não
{
}
}
``````
}
## Reference Request – Dependency Range: What is the size of the largest sub-collection of random variables that is statistically independent?
Leave $$X_1, ldots, X_p$$ be random variables in the same space. Define your range of dependency, denoted $$range (X_1, ldots, X_p)$$ as the largest non-negative integer $$k$$ such that there is a subcollection of $$k$$ outside $$p$$ random variables that are statistically independent by pairs (that is, their joint distribution function factors as a product of their individual distribution functions). Of course, each singleton $${X_i }$$ forms an independent collection so $$1 le range (X_1, ldots, X_p) le p$$, and the upper limit is reached if $$X_1, ldots, X_p$$ they are independent to begin with.
Question. Is the above notion of rank defined above being studied in the literature? Is there any interesting quantity property? Are they the known representatives of this concept?
## What is an algorithm for finding the largest white space rectangle in an image?
Subtract $$254.9999999$$ of all pixel values, so black corresponds to $$-254.9999999$$ and white to $$0.0000001$$.
Find the 2D subrectangle with maximum sum. There is standard $$O (n ^ 3)$$time algorithms for this problem (for example https://stackoverflow.com/q/19064133/781723, https://www.geeksforgeeks.org/maximum-sum-rectangle-in-a-2d-matrix-dp- 27 /). As long as you have no more than 10,000,000 pixels in the image, this will be the solution to your problem (since each white pixel contributes another $$0.0000001$$ to the sum).
Proof that this finds the optimal solution: Only all-white subrectangles have a positive sum (if you have a single non-white pixel, then you contribute a negative value of $$-0.9999999$$ or less than the sum, which cannot be exceeded by any number of white pixels); and the larger the completely white subrectangle, the greater its sum.
## algorithms – Find the largest sum of elements \$ k \$ below the threshold
You can use the encounter function in the middle to reduce the execution time to $$O (n ^ { lceil k / 2 rceil})$$.
For simplicity let me assume that $$k$$ even.
The idea is the following:
• Divide $$A$$ in two parts
• For each part, compute an ordered list of sums of $$k / 2$$ part elements
• For each $$k / 2$$-sum in the first part, use binary search to find the best $$k / 2$$-sum in the second part that conforms to the restrictions.
As stated, there are several problems with the idea:
1. The optimal solution is assumed to contain exactly $$k / 2$$ addends of each part.
2. Run on time $$O (n ^ {k / 2} log n)$$.
3. Uses $$O (n ^ {/ 2})$$ memory.
To handle the first difficulty, there are several options. We could repeat the whole algorithm $$sqrt {k}$$ times. Alternatively, there may be deterministic ways to achieve the same goal. Here is a hybrid solution:
• Random partition $$A$$ inside $$k ^ 3$$ parts $$P_1, ldots, P_ {k ^ 3}$$ (something substantially larger than $$k ^ 2$$ would work). With high probability, each element of the optimal solution is in its own part.
• Consider all possible partitions of $$A$$ in two parts of the form $$P_1, ldots, P_i$$ Y $$P_ {i + 1}, ldots, P_ {k ^ 3}$$. One of these will contain exactly $$k / 2$$ elements of the optimal solution.
To handle the second difficulty, we need to be a little more careful in implementation. By merging (the mergesort family subroutine), it should be possible to calculate the $$k / 2$$-sums of each part in $$O (n ^ {/ 2})$$. The final step can be implemented in $$O (n ^ {/ 2})$$ using a classic two-pointer technique (the first pointer goes up in the first half, the second pointer goes down in the second half).
There are tricks to reduce the required memory of $$O (n ^ {/ 2})$$ to $$O (n ^ {/ 4})$$: Divide each part into two sub parts. You can easily review everything $$k / 2$$-sum in the first part considering all pairs of $$k / 4$$-sums in its subparts. The binary search in the second part can be implemented using the two pointer technique.
## Probable Probability: The probability that the third largest of 39 nine-sided dice equals 8?
How can I calculate the probability that the third greatest of 39 nine-sided dice equals 8?
I only see three results:
(two nines, one eight, the rest is less than 8)
(one nine, two eights, the rest is less than 8)
(three eights, the rest is less than 8)
How would this be done mathematically?
## Encryption signals, daily tips and updates from Bitmex in our largest encryption community!
Visit the link and verify it yourself, our success rate is incredible: https://t.me/btctradingclub | open-web-math/open-web-math | |
$18,227 in 1966 is worth$78,927.41 in 1992
$18,227 in 1966 has the same purchasing power as$78,927.41 in 1992. Over the 26 years this is a change of $60,700.41. The average inflation rate of the dollar between 1966 and 1992 was 5.73% per year. The cumulative price increase of the dollar over this time was 333.02%. The value of$18,227 from 1966 to 1992
So what does this data mean? It means that the prices in 1992 are 789.27 higher than the average prices since 1966. A dollar in 1992 can buy 23.09% of what it could buy in 1966.
We can look at the buying power equivalent for $18,227 in 1966 to see how much you would need to adjust for in order to beat inflation. For 1966 to 1992, if you started with$18,227 in 1966, you would need to have $78,927.41 in 1966 to keep up with inflation rates. So if we are saying that$18,227 is equivalent to $78,927.41 over time, you can see the core concept of inflation in action. The "real value" of a single dollar decreases over time. It will pay for fewer items at the store than it did previously. In the chart below you can see how the value of the dollar is worth less over 26 years. Value of$18,227 Over Time
In the table below we can see the value of the US Dollar over time. According to the BLS, each of these amounts are equivalent in terms of what that amount could purchase at the time.
Year Dollar Value Inflation Rate
1966 $18,227.00 2.86% 1967$18,789.56 3.09%
1968 $19,577.15 4.19% 1969$20,646.02 5.46%
1970 $21,827.40 5.72% 1971$22,783.75 4.38%
1972 $23,515.08 3.21% 1973$24,977.74 6.22%
1974 $27,734.29 11.04% 1975$30,265.82 9.13%
1976 $32,009.76 5.76% 1977$34,091.24 6.50%
1978 $36,679.02 7.59% 1979$40,841.98 11.35%
1980 $46,355.09 13.50% 1981$51,136.86 10.32%
1982 $54,287.21 6.16% 1983$56,031.15 3.21%
1984 $58,450.16 4.32% 1985$60,531.64 3.56%
1986 $61,656.77 1.86% 1987$63,907.01 3.65%
1988 $66,551.05 4.14% 1989$69,757.65 4.82%
1990 $73,526.82 5.40% 1991$76,620.91 4.21% | HuggingFaceTB/finemath | |
# Math Vocab Words- Chapter 8
## 33 terms
### angle
figure formed by two rays with a common endpoint
### vertex
common endpoint of two rays
### acute angle
angle measure of less than 90 degrees
### right angle
the 90 degree angle between two perpendicular lines
### obtuse angle
An angle that measures more than 90 degrees but less than 180 degrees
### straight angle
an angle of 180 degrees
### congruent
Having the same size, shape, and measure
### vertical angle
either of two equal and opposite angles formed by the intersection of two straight lines
2 angles that have the same vertex and a common ray but no interior points in common
### complementary angle
two angles that add up to 90 degrees
### supplementary angle
two angles whose measures have a sum of 180 degrees
### point
an exact location
### line
a straight path that goes on forever in both directions
### plane
a flat surface that extends without end in all directions
### line segment
part of a line with two endpoints
### ray
a straight line extending from a point that goes on forever in one direction
### parallel lines
Lines in the same plane that do not intersect
### perpendicular lines
Two lines that intersect to form right angles
### skew lines
lines that lie in different planes. they are neither parallel or intersecting
### acute triangle
a triangle that has three acute angles
### obtuse triangle
a triangle with one obtuse angle
### right triangle
a triangle with one right angle
a plane figure with 4 sides and 4 angles
### parallelogram
a quadrilateral whose opposite sides are both parallel and equal in length
### rectangle
a parallelogram with four right angles
### rhombus
a parallelogram with 4 congruent sides
### square
a rectangle with four equal sides and four right angles
### trapezoid
a quadrilateral with two parallel sides
### line symmetry
a property that a shape can have. means that when you fold the figure in half , the 2 halves match exactly.
### line of symmetry
a line that divides a figure into 2 halves that match exactly when the figure is folded on that line
### equilateral triangle
a triangle with three equal sides
### isosceles triangle
a triangle with two congruent sides
### scalene triangle
a triangle with no congruent sides | HuggingFaceTB/finemath | |
# How Many Hearts Beat in Sync With Yours?
The idea of two hearts beating in unison has long been a powerful image in poetry, music, and romance. Let us see what the realm of mathematics has to say about this idea. The question we will answer is a natural one, but one which will require careful analysis:
On average, how many hearts beat in sync with yours?
Of course, our first question should be:
What does it mean for two hearts to beat in sync?
We will define two basic notions in answering this question.
Heart rate: The number of times a heart beats per minute measured in beats per minute (BPM)
Offset: Given that two hearts share a heart rate, the time delay between their beats measured in seconds
Thus, we say that two hearts beat in sync if they have the same heart rate and same offset.
We will be using a half-empirical, half-theoretical approach to answering our question about how many hearts beat with yours.
First the statistics!
From Statistics Canada, Canada’s National Statistics Agency, we gather data on the distribution of resting heart rates for people ages 6 to 79. This data gives the 5th, 10th, 25th, 50th, 75th, 90th and 95th percentile for resting heart rate.
Since we wish to have finer grained data, we will need to make an assumption here.
Assumption 1: The distribution of heart rates between the provided percentiles is uniformly distributed.
This assumption is surely not fully correct but it will serve us well in the absence of finer grained data. We will also assume that our minimum resting heart rate is 40 BPM and the maximum is 99 BPM. These numbers come from an article about normal heart rates.
Given our assumption so far, we can generate a histogram of the heart rates.
Time for some math!
First we’ll define
If we index each person in the world as $i=1,2,3 … , N$ where you are person $i=N$, then we can define
What we are after is the mean of $S$, also called the expected value of $S$ and denoted as $\mathbf{E}(S)$. Note that
Why? Well $S$ is the number of people in the world whose hearts beat with yours and $B_{i}$ is 1 if and only if person i’s heart beats in sync with yours. So, summing up all the $B_{i}$’s we will get exactly the count of how many people whose hearts are in sync with yours.
So we want:
using in the second equality that the expected value of a sum is the sum of the expected values and in the last equality the fact that the mean of an indicator variable is the probability that this variable is 1.
We can go one step further if we make another assumption:
Assumption 2: Heart rates are independent between people.
This is likely much easier to swallow than Assumption 1 but of course is still not completely true as people who work out together, attend the same sporting event, etc. will have more similar heart rates. Still, on a global level, we should be safe with this assumption.
Then, our independence assumption allows us to reduce the result to:
so that we need only to find $\mathbf{P}(B_{i}=1)$.
Exiting the math-o-sphere for a moment, we only need to find the probability that someone’s heart is in sync with yours.
Now, under what conditions would someone’s heart be in sync with yours? As noted earlier, we need only that both your heart rates are the same and your offsets are the same. For example, both your heart rates could be 55 BPM and you have the same offset, or perhaps your heart rates are 77 BPM and you have the same offset, or perhaps you both share a heart rate of 90 BPM with the same offset.
In fact, there are infinitely many heart rates that you can share! How do we simplify this? Well, we will make another assumption here.
Assumption 3: We will treat the continuous range of heart rates as a range of integer values.
That is, if we are considering heart rates between 40 BPM and 52 BPM, we choose to split this interval up into 12 allowable integer heartbeats. Why do we do this? Looking at the way heart rate is typically reported by medical equipment, it is given in integer valued beats per minute.
Now, given that two people have the same heart rate, how do we measure whether they have the same offset?
Let’s think about this with an example. Say both you and your friend have a heart rate of 40 BPM which is the same as 40 beats per 60 seconds which implies that your hearts beat once every 1.5 seconds.
Now, the probability that your friend’s heart beats exactly when yours does is 0 because there are infinitely many possible offsets in a finite amount of time. Thus, we will need to make a final assumption in order to conduct a meaningful analysis going forward.
Assumption 4: Two hearts with the same heart rate will be considered as having the same offset if their beats are within some error tolerance of one another.
Essentially, if your friend’s heart beats just a tiny fraction of a second before or after yours does, we will consider the offsets to be the same. We will denote the error tolerance as $\varepsilon$. Going back to our 40 BPM example, we allow that your friend’s heart beats in any of the green regions in the figure below. Note that the probability that your friend’s heart beats in the green region is $\frac{2\varepsilon}{1.5}$ since we have an allowable range of $2\varepsilon$ in a total range of $1.5$.
So, if two hearts beat at $K$ Beats per Minute, they each beat once every $\frac{60}{K}$ seconds and the probability that they have the same offset is $\frac{2\varepsilon}{\frac{60}{K}} = \frac{K\varepsilon}{30}$.
We will let $\varepsilon$ be 1% of $\frac{60}{K}$ going forward which implies that the probability that two hearts with the same heart rate share an offset is $\frac{0.01 \times K \times \frac{60}{K}}{30} = 0.02$ regardless of the heart rate.
Now, we are ready to finish our calculation!
Recall that we are looking for $\mathbf{P}(B_{i}=1)$, the probability that someone’s heart beats in sync with yours.
Note that:
by the independence assumption between two people’s heart rates and the fact that the probability of a matching offset is always 0.02.
Now, finding the probability that a heart beats with heart rate $i$ is as simple as taking the percentile range that the heart rate falls into and dividing by the number of integers in that percentile range.
To be more clear suppose we are trying to find $\mathbf{P}(Heart Rate = 85)$. We note that 85 BPM falls into the 83-88 BPM range, which contains 5 integer values and comprises 5% of the total range of heart rates.
Thus, using our Assumption 1 about uniform distribution, $\mathbf{P}(Heart Rate = 85) = \frac{0.05}{5} = 0.01$ (We can also just read this 0.01 as the height of the 85 BPM bar in our histogram above). We use the same procedure with all other heart rates.
After doing all appropriate calculations, we end up with:
That is, the probability that your heart beats in sync with another heart is 0.053%, very small indeed.
Using the fact that the current world population is around 7.5 billion,
So, under our four assumptions, the expected number hearts that beat in sync with yours is around:
Wow!
In the interest of transparency, let us see what happens to this 4 million if we relax each of our four assumptions.
To relax Assumption 1, we would need to find finer grained data than we do now. Given the true distribution of heart rates between the given percentiles, our estimate of 4 million may rise or fall.
Assumption 2 may be the most difficult to relax since we would need some underlying model of how one person’s heart rate affects that of others in the world, an extremely challenging task. This also has an indeterminate effect on our final result.
If we relax Assumption 3, we will no longer treat heart rates as only being able to take integer values but rather consider them to take continuous values in the range 40 to 100. This will cause our final result of how many hearts beat with yours to be driven to 0.
Why would that be? Well, without even going into the calculations, if we allow any real valued heart rate between 40 and 100, the probability that your heart rate exactly matches someone else’s is very very slim, actually it’s 0. It’s a consequence of the fact that there are an infinite amount of heart rates to choose from, something that isn’t true with discrete integer values.
Last, if we relax Assumption 4, essentially forcing $\varepsilon$ to be 0, we will also drive the final result to 0. This is basically the same reason as for Assumption 3: two offsets being exactly the same is a probability 0 event.
Note also that our final result is quite dependent on the choice of $\varepsilon$. That is, if we let $\varepsilon$ be 2% of $\frac{60}{K}$, then the probability that two hearts with the same heart rate share an offset is $\frac{0.02 \times K \times \frac{60}{K}}{30} = 0.04$, which will eventually double our 4 million to around 8 million.
Finally note that there are some implicit assumptions that were made such as that the Canadian population’s heart beats are representative of those in the whole world and that we consider those outside the range 6 to 79 to have the same distribution of heart rates as those in that range.
The latter of these is definitely not true since those outside the range are young children and elderly people, but again, this assumption serves us well in the absence of the appropriate data.
All that said, the above analysis, complete with its assumptions, is an interesting way to think about how many of us around the world are walking around with our tickers pounding in rhythm. | open-web-math/open-web-math | |
### Theory:
Basic properties of Exponents.
$\begin{array}{l}{a}^{n}\cdot {a}^{m}={a}^{n+m};\\ \\ {a}^{n}:{a}^{m}={a}^{n-m},\phantom{\rule{0.147em}{0ex}}n>m,\phantom{\rule{0.147em}{0ex}}a\ne 0;\\ \\ {\left({a}^{n}\right)}^{m}={a}^{n\cdot m}\end{array}$
Where n and $$m$$ are integers.
In practice, several properties are often applied simultaneously.
Example:
Calculate$\frac{{\left({5}^{4}\cdot 5\right)}^{3}}{{5}^{7}\cdot {5}^{6}}$
Perform actions in the numerator:
1.${5}^{4}\cdot 5={5}^{4}\cdot {5}^{1}={5}^{4+1}={5}^{5}$ — applied property ${a}^{n}\cdot {a}^{m}={a}^{n+m}$.
2. ${\left({5}^{5}\right)}^{3}={5}^{5\cdot 3}={5}^{15}$ — applied property ${\left({a}^{n}\right)}^{m}={a}^{n\cdot m}$.
Perform the multiplication in the denominator:
3. ${5}^{7}\cdot {5}^{6}={5}^{7+6}={5}^{13}$ — applied property ${a}^{n}\cdot {a}^{m}={a}^{n+m}$.
Replace the fraction line with division:
4.${5}^{15}:{5}^{13}={5}^{15-13}={5}^{2}$ — applied property ${a}^{n}:{a}^{m}={a}^{n-m}$.
5. ${5}^{2}=25$.
Answer: $$25$$. | HuggingFaceTB/finemath | |
Related Articles
Rotations of a Binary String with Odd Value
• Last Updated : 11 May, 2021
Given a binary string. We are allowed to do circular rotation of the string without changing the relative order of the bits in the string.
For Example, all possible circular rotation of string “011001” are:
```101100
010110
001011
100101
110010```
We are required to tell total number of distinct odd decimal equivalent possible of binary string, by doing circular rotation.
Examples:
```Input : 011001
Output : 3
Explanation:
All odd possible binary representations are:
["011001", "001011", "100101"]
Input : 11011
Output : 4
Explanation:
All odd possible binary representations are:
["11011", "01111", "10111", "11101"]```
Concept:
It can be observed that a binary string can only be odd if it’s last bit is 1, because the value of last bit is 2^0.Hence, since we are doing circular rotation.
## C++
`// CPP program to find count of rotations``// with odd value.``#include ``using` `namespace` `std;` `// function to calculate total odd decimal``// equivalent``int` `oddEquivalent(string s, ``int` `n)``{`` ``int` `count = 0;`` ``for` `(``int` `i = 0; i < n; i++) {`` ``if` `(s[i] == ``'1'``)`` ``count++;`` ``}`` ``return` `count;``}` `// Driver code``int` `main()``{`` ``string s = ``"1011011"``;`` ``int` `n = s.length();`` ``cout << oddEquivalent(s, n);`` ``return` `0;``}`
## Java
`// Java program to find count of rotations``// with odd value.` `class` `solution``{``static` `int` `oddEquivalent(String s, ``int` `n)``{` `int` `count = ``0``;``// function to calculate total odd decimal``// equivalent``for` `(``int` `i = ``0``; i < n; i++)`` ``{`` ``if``(s.charAt(i) == ``'1'``)`` ``count++;`` ``}`` ``return` `count;``}` `// Driver code``public` `static` `void` `main(String ar[])``{` `String s = ``"1011011"``;``int` `n = s.length();``System.out.println(oddEquivalent(s, n));` `}``}``//This code is contributed``//By Surendra_Gangwar`
## Python3
`# Python3 program to find count``# of rotations with odd value` `#function to calculate total odd equivalent``def` `oddEquivalent(s, n):`` ``count``=``0`` ``for` `i ``in` `range``(``0``,n):`` ``if` `(s[i] ``=``=` `'1'``):`` ``count ``=` `count ``+` `1`` ``return` `count`` ` `#Driver code``if` `__name__``=``=``'__main__'``:`` ``s ``=` `"1011011"`` ``n ``=` `len``(s)`` ``print``(oddEquivalent(s, n))` `# this code is contributed by Shashank_Sharma`
## C#
`// C# program to find count of``// rotations with odd value.``using` `System;` `class` `GFG``{``static` `int` `oddEquivalent(String s, ``int` `n)``{`` ``int` `count = 0;`` ` ` ``// function to calculate total`` ``// odd decimal equivalent`` ``for` `(``int` `i = 0; i < n; i++)`` ``{`` ``if``(s[i] == ``'1'``)`` ``count++;`` ``}`` ``return` `count;``}` `// Driver code``public` `static` `void` `Main()``{`` ``String s = ``"1011011"``;`` ``int` `n = s.Length;`` ``Console.WriteLine(oddEquivalent(s, n));``}``}` `// This code is contributed``// by Subhadeep`
## PHP
``
## Javascript
` `
Output:
`5`
My Personal Notes arrow_drop_up | HuggingFaceTB/finemath | |
ahh
Te Arapiki Ako
"Towards better teaching & learning"
# Understanding volume Add to your favourites Remove from your favourites Add a note on this item Recommend to a friend Comment on this item Send to printer Request a reminder of this item Cancel a reminder of this item
Last updated 26 October 2012 15:28 by NZTecAdmin
Understanding volume (PDF, 49 KB)
Measurement progression, 2nd–3rd steps
In this activity, the learners develop an understanding of volume as a description of the number of cubic units needed to fill a solid shape. They do this by finding the volume of rectangular solids, initially by counting cubic units and then by forming a mental image of the number of cubic units that fit into the shape.
## The teaching points
• Volume is the amount of space occupied by a solid expressed in cubic units.
• A cube has equal sides and equal square faces. • For a rectangular solid, working out how many squares fit on one level and how many levels there are is a faster way of calculating the volume than counting each cube.
• Metric units for volume are cubic centimetres (cubes with sides of 1 centimetre) with the symbol of cm3 and cubic metres (cubes with sides of 1 metre) with the symbol of m3.
• m3 means m x m x m and is said as metres cubed.
• Discuss with the learners the situations in which they might want to measure volumes.
## Resources
• Nets with squares of 1 centimetre to build solid shapes (templates below).
• Lots of centimetre cubes.
• Rulers or measuring tapes that can measure 1 metre.
• Scissors.
• Sellotape.
## The guided teaching and learning sequence
1. Ask the learners to build a solid from the net below (made from 1 centimetre squares) and to discuss how they might measure the space inside it. Listen for and encourage responses that suggest finding out how much fits in the solid. 2. Give each learner a 1 centimetre cube and ask them to describe its features. You may need to prompt them to measure the sides. Listen for and reinforce the fact that the length of the sides are all 1 centimetre and the faces are equal squares. Discuss the name “centimetre cube”.
3. Ask the learners to predict how many cubes will fit into the solid. Record their predictions.
4. Ask them to check their predictions by filling the solid with cubic centimetres and share their results. Discuss the fact that the number of cubes that fill the space inside the solid is a way of measuring that space/volume and that, in this example, the volume is 16 cubes or 16 cubic centimetres or 16 cm3.
5. Ask the learners to compare the results with their predictions. Encourage those with more accurate predictions to share their methods. Listen for and encourage, or prompt if necessary, comments like “8 cubes fit on the bottom, and they are stacked 2 high so there must be 16 cubes”.
6. Ask the learners to make a solid from a second net, predict its volume and, if necessary, check the prediction by filling the shape with centimetre cubes (volume equals 36 cm3). 7. Ask the learners to estimate the number of cubic centimetres that would fit into a freight container and whether cubic centimetres are an appropriate unit to measure its volume. Ask them to suggest an alternative unit. Listen for “cubic metres”.
8. Discuss with the learners what a cubic metre might look like and how they might demonstrate it with newspaper. Using square metres of newspaper (as they made in the activity “Understanding area”), ask four learners to make a cubic metre. The learners will each hold a side face of a cubic metre. The bottom and top faces are contained within other faces. Emphasise that a cubic metre is a cube with sides of 1 metre with the symbol m3, from 1 m x 1 m x 1 m. Emphasise that it is the unit used to describe the volume of larger shapes.
9. Ask the learners to estimate the volume of the room and other objects in the room (large cardboard box, carry bag, etc). You may need to discuss the use of fractions of a metre for objects with a volume smaller than 1 cubic metre.
## Follow-up activity
Ask the learners to choose small and large solid objects in the room, decide which metric unit of measurement they would use to describe the volume and give an estimation of the volume of those objects. | HuggingFaceTB/finemath | |
# Addition and Subtraction of Integers
Mathematics is a subject that deals with numbers. An arithmetic operation is an elementary branch of mathematics. Arithmetical operations include addition, subtraction, multiplication and division. It helps us to find the sum (total) or difference (how more or less) of something. Not only sum or difference, it also helps one to compare and divide things equally. Arithmetic operations are applicable to all real numbers including integers. OPERATIONS OF INTEGERS
Integers are a special group of numbers that are positive, negative and zero, which are not fractions. Rules for addition and subtraction are same for all, whether it is natural number or an integer, because natural numbers are itself integers. We just extend the rule and apply it for integers as well.
Addition of Integers: Addition of integers means there are three possibilities. They are:
• Addition between two positive numbers,
• Addition between two negative numbers; and
• Addition between a positive number and a negative number.
Type of Numbers Operation Result Example Positive + Positive Add Positive (+) 10 + 15 = 25 Negative + Negative Add Negative (-) (-10) + (-15) = -25 Positive + Negative* Subtract Positive (+) (-10) + 15 =5 Negative + Positive* Subtract Negative (-) 10 + (-15)= -5
Whenever a positive number and a negative number are added, sign of the greater number will decide the operation and sign of the result. In above example 10 + (-15) = -5 and (-10) + 15 =5; here, without sign 15 is greater than 10 hence numbers will be subtracted and answer will give the sign of greater number.
Alternatively, to find the sum of a positive and a negative integer, take the absolute value (“absolute value” means to remove any negative sign of a number, and make the number positive) of each integer and then subtract these values. Take above example, 10 + (-15); absolute value of 10 is 10 and -15 is 15.
⇒ 10 – 15 = -5
Thus we can conclude above table as follow:
• Addition oftwo positive integers always gives a positive sum.
• Additiontwo negative integers always give a negative sum.
• Addition of a positive and a negativeinteger give either a positive or negative sum depending on the value of the given numbers.
Note: The sum of an integer and its opposite is always zero. (For example, -5 + 5= 0)
Subtraction of Integers: Like addition, subtraction of integers also has three possibilities. They are:
• Subtraction between two positive numbers,
• Subtraction between two negative numbers; and
• Subtraction between a positive number and a negative number.
For the ease of calculation, we need to renovate subtraction problems into addition problems. There are two steps for this:
1. After converting the sign, take the inverse of the number which comes after the sign.
Once the transformation is done, follow the rules of addition given above.
For example, find the value of: (-5) – (7)
⇒ (-5) + (7)
Step 2: Take the inverse of the number which comes after the sign
5 + (-7) (opposite of 7 is -7)
5 + (-7) = -12 [Add and put the sign of greater number]
Solve: Evaluate the following:
1. (-5 )+ 9
2. (-1) – ( -2)
Solution:
1. (-5 )+ 9 = 4 [Subtract and put the sign of greater number]
2. (-1) – ( -2)= 1
(-1) + (-2) [Transform subtraction problems into addition problems]
(-1) + (2) [Subtract and put the sign of greater number] | HuggingFaceTB/finemath | |
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## 7 posts in this topic
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This is an easy one. A 2x4 is cut into three pieces,
The first piece is 3 inches longer then the third piece.
The seconded piece is 2 inches shorter than the third piece.
The three pieces together measure 1 inch over one yard.
How long is each piece?
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Algebra says:
Let x = 3rd piece. Therefore...
x+3 = 1st piece, and
x-2 = 2nd piece. So...
x + x+3 + x-2 = 37
3x + 1 = 37
3x = 36
x = 12
x+3 = 15
x-2 = 10
So the length of the three pieces are 15, 10, and 12 inches.
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Posted · Report post
This is an easy one. A 2x4 is cut into three pieces,
The first piece is 3 inches longer then the third piece.
The seconded piece is 2 inches shorter than the third piece.
The three pieces together measure 1 inch over one yard.
How long is each piece?
Omg why don't just state a number in stead of saying "1 inch over one yard" , inches / yards / feet don't make sence to everyone. The puzzle is easy, I just don't know what number to use for "1 inch over 1 yard".
[b]Spoiler[/b] for [i]solution[/i]:
x + x + 3 + x - 2 = y
3x + 1 = y
piece 3 = x = (y-1)/3
piece 1 = x + 3 = ((y-1)/3)+3
piece 2 = x - 2 = ((y-1)/3)-2
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Hmm i can't even edit my own reply, strange. So here is my addition to my previous post.
edit: What i want to say is: '1 inch over a yard' isn't really working as a number, also i could transform this to a distance in lightyears if I wanted to. You didn't say how you want your answer, metrics system, feet / inches. I guess that makes it safe to use Y in stead of a number.
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Hmm i can't even edit my own reply, strange. So here is my addition to my previous post.
everybody can edit his/her own posts within 10 minutes (so that nobody changes anything which was subject to further reactions)
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Posted (edited) · Report post
First piece = 14
Second piece = 12
Third piece = 11
Edited by ShijeFace
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Posted · Report post
First piece = 14
Second piece = 12
Third piece = 11
The second piece is 2 in. shorter than the first peace.
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Register a new account
Followers 0 | HuggingFaceTB/finemath | |
Video: Understanding the Medians of a Triangle
In β³πππ, where π΄ is the midpoint of line ππ, what name is given to line π΄π? [A] base [B] height [C] hypotenuse [D] median
03:54
Video Transcript
In triangle πππ, where π΄ is the midpoint of line ππ, what is the name given to the line π΄π? A) Base, B) height, C) hypotenuse, or D) median.
First, letβs sketch a triangle, given these conditions. Hereβs a triangle. If we make this side π, weβll follow the naming convention such that we have π and then π. And this is our triangle πππ. If π΄ is the midpoint of line ππ, then π΄ is halfway between π and π. This also means that the segments ππ΄ and π΄π are equal in length because the midpoint divides the line ππ in half.
But weβre interested in what we would call the line π΄π. We know that the base can be any one of the three sides of the triangle. But the segment π΄π is not one of the original sides of the triangle. Therefore, it cannot be the base. What about the height of a triangle? The height of the triangle depends on which base youβre using. If we let line ππ be the base, then this would be the height because the height is the perpendicular distance from the base to the vertex opposite that base. But remember, weβve just sketched this triangle. We donβt know that thatβs exactly what the triangle looks like. So, letβs leave this information here and keep going.
If we consider the word hypotenuse, that is the longest side of a right triangle. We donβt know if triangle πππ is a right triangle. Even if triangle πππ was a right triangle and π΄ was halfway between π and π, the line ππ΄ would still not be the hypotenuse because it is a line segment inside the triangle and therefore would not be the hypotenuse.
So now, we should consider the definition of a median of a triangle. The median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. We know that point π΄ is a midpoint and π is the vertex opposite the line ππ. This means we can say line segment π΄π is a median. Itβs worth noting that there are triangles in which the median and the height are the same line. Remember that the height needs to be perpendicular to the vertex opposite it, if we drew a line that was perpendicular to the midpoint π΄ and the π vertex fell on that line.
Here is a triangle πππ, where ππ is the base. Line segment π΄π is still the median because point π΄ is the midpoint of ππ. But because the line segment π΄π is perpendicular to the base ππ, in this case, π΄π is also the height. This fact where the median and height are the same value is true in isosceles triangles. But since we havenβt been told whether or not triangle πππ is isosceles, the only thing we can say for sure is that π΄π is the median. We cannot tell if it is the height or not, which makes option D median the best answer. | HuggingFaceTB/finemath | |
amo
6
# A school debate team has 4 girls and 6 boys. A total of 3 of the team members will be chosen to participate in the district debate. What is the probability that 1 girl and 2 boys will be selected? A. 1/4 B. 1/2 C. 3/4 D. 3/10
The probability that 1 girl and 2 boys will be selected is given by: $P(1\ boy,\ 2\ girls)=\frac{4C1\times6C2}{10C3}=\frac{1}{2}$ The correct answer choice is B. 1/2. | HuggingFaceTB/finemath | |
# Is it possible for rigid objects to slip before friction is saturated everywhere?
A common technique in statics problems is to assume that static friction is saturated everywhere. Consider, for example, the following problem:
A thin, uniform bar of mass $$m$$ lies on a uniform floor, with coefficient of static friction $$\mu$$. What is the minimum force required to make the object start moving, if the force can only be applied horizontally? Assume the normal pressure on the floor remains uniform.
The minimum force required comes from pushing perpendicular to the bar at its edge. If we assume the friction has the maximum magnitude everywhere, that is $$df = \mu g dm$$ for every infinitesimal segment of mass $$dm$$, and it points either directly along or against the applied force, we then have enough information to solve the problem. The friction will oppose the applied force on the segment of length $$l$$ closest to the applied force, while the friction will align with the applied force for the remaining $$L-l$$ segment, where $$L$$ is the total length of the bar. Balancing forces yields
$$F = \mu m g \left(\frac{l}{L} - \frac{L-l}{L} \right)$$
While balancing torques about where the applied force acts yields
$$\mu m g \frac{l}{L} \left(\frac{l}{2} \right) = \mu m g \frac{L-l}{L} \left(\frac{L+l}{2} \right)$$
Solving the set of equations yields $$l=L/\sqrt{2}$$ and $$\boxed{F=(\sqrt{2}-1)\mu mg}$$.
My question is two-fold:
1. In the above example, in order for the bar to be right at the threshold between motion and being static, friction must be maximal in at least one location along the rod. Intuitively, why must it be maximal at all locations along the rod?
2. Is there an example of a statics problem/situation in which when at the threshold between motion and being static, friction does not need to be maximal along all locations? (other than trivial situations such as one foot sliding and the other foot being static—such a situation has essentially independent sources of friction).
• Is the minimum force to be applied assumed to only act on a single point? Jan 20 at 5:36
• Yes, the minimum force is applied at a single point. Jan 20 at 6:31
• Does this answer your question? Force required to move bar at one end Jan 20 at 14:36
• No, this doesn't answer my question. First, the answer you have linked does not address part 2 of my question. Second, the answer linked does not adequately explain part 1 of my question; the only portion of the answer that seems to address why the assumption is true is the fact that the bar experiences "uniform contact pressure," but again that reasoning is vague and doesn't explain how one part of the bar could have friction in one direction and the other part could have friction in the other direction. Jan 20 at 23:05
• Can you draw schematics of what actually happening, because I cannot follow your strange scriptures. You put together some letters about forces without even making a sketch of what goes where. Jan 24 at 2:21 | HuggingFaceTB/finemath | |
Significant figures practice worksheet
Comments Off on Significant figures practice worksheet
You may select the numbers to be whole, these Significant Figures Worksheets are great for testing students in their ability on identifying and working with significant figures practice worksheet digits. The “Significant Figures Rules Handout Worksheet” is great for reinforcing the rules in determining the correct number of significant digits in a number. You may select the problems to be addition, use the “Identify Significant Digits Worksheet” to measure their ability to correctly identify the number of significant digits in a number.
You may select the problems to be multiplication – the “Adding and Subtracting” and Multiplying and Dividing” with Significant Figures Worksheets are great for solving problems with significant digits and rounding to the correct answer. You may select the numbers to be whole, click here for a Detailed Description of all the Significant Figures Worksheets.
Click the image to be taken to that Significant Figures Worksheet. Find Any Errors, this Significant Figures Worksheet is a great handout for reinforcing the rules of significant figures. Introducing Multiplication KS1Flip, these Significant Figures Worksheets are great for testing the students in their ability to determine the number of significant digits for a given number.
It is important for students to be able to identify significant digits using a few basic rules, or all three. The Identifying Significant Figures Worksheet includes up to 30 randomly generated whole numbers, creates up to 30 unique problems that can include whole numbers, these Significant Figures Worksheets will produce twenty problems per worksheet. Introducing Multiplication KS1Flip, these Significant Figures Worksheets are great for solving addition and subtraction problems with significant figures and correctly rounding to the correct answer. Your calculator doesn’t know the difference, in this tutorial, these Significant Figures Worksheets will produce twenty problems per worksheet.
We’ll attempt to explain them — these Significant Figures Worksheets are great for solving multiplication and division problems with significant figures and correctly rounding to the correct answer. For some reason, these Significant Figures Worksheets will produce twenty problems per worksheet. They may know what significant figures are and how to find them, please forward this error screen to 162. In this case, this Significant Figures Worksheet is great for testing the students in their ability to determine the number of significant digits for a given number.
When working with significant figures, or all three. Never write too many because if you do, this Significant Figures Worksheet will produce twenty problems per worksheet.
Your data will show that you’ve got amazing super, now you are ready to create your Significant Figures Worksheet by pressing the Create Button. And if you don’t write enough, choose the correct answer for each question. The biggest rule of science is to tell the truth; please forward this error screen to 162. If an instrument is precise, a worksheet with a series of whole numbers and decimals to round to a given number of significant figures.
You need to be logged in to save a worksheet. If something weighs 150 grams and you measure the weight as 130 grams, rounding significant figures numbers worksheets to 1sf 2. If an answer isn’t reproducible – rounding significant figures up to 2sf sheet 1.
If you find that the mass of a textbook is 4300 kilograms; rounding significant figures up to 2sf sheet 2. Quiz worksheet significant figures and scientific notation print worksheet. It’s easy to make mistakes, rounding significant figures numbers worksheets to 1sf 2. Before I want to go any further, rounding significant figures up to 2sf sheet 2. | HuggingFaceTB/finemath | |
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#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 7 - Mathematics4.49 Special Products and Expansions - Formulas
Formulas:
(a+b)2 = (a+b)(a+b) =a2+2ab+b2
(a-b)2 = (a-b)(a-b) =a2-2ab+b2
(a+b)3 = (a+b)(a+b)(a+b) = (a+b)(a2+2ab+b2) =a3+3ab2+3a2b+b3
(a-b)3 = (a-b)(a-b)(a-b) = (a-b)(a2-2ab+b2) =a3+3ab2-3a2b+b3
Sum of squares: a2+b2 =(a+b)2- 2ab a2+b2 =(a-b)2+ 2ab Difference of squares: a2-b2 = (a+b)(a-b) Sum of cubes: a3 + b3 = (a2 - ab + b2)(a + b) Difference of cubes: a3 - b3 = (a2 + ab + b2)(a - b)
Directions:Solve the following problems. Also write two examples for each formula above.
Q 1: If x=2, then find the value of 8x2+6x+7.514947 Q 2: Multiply (7x2-6x+7) with 6x.42x3+36x2+42x42x3-36x2-42x42x3-36x2+42x Q 3: Multiply (x-7) with (x2+1).x3-7x2-x+7x3-7x2+x-7x3-7x2+x+7 Q 4: Multiply (6x+7) with (7x+3).42x2+65x+2142x2+67x+2142x2+67x+24 Q 5: Multiply (6x2+7x-1) with x2.6x4-7x3-x26x4+7x3+x26x4+7x3-x2 Q 6: Multiply (7x+7y) with (6x+3).42x2+21x+42xy+21y42x2+15x+42xy+21y42x2+21x+46xy+21y Q 7: Multiply (3x+2) with (3x-5).9x2-9x+109x2+9x-109x2-9x-10 Q 8: Multiply (6x+7) with (2y+3).12xy+18y+14x+2112xy+18x+14y+2112xy+18x+14y+20 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
#### Subscription to kwizNET Learning System offers the following benefits:
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• Unlimited practice with auto-generated 'WIZ MATH' quizzes | HuggingFaceTB/finemath | |
# Discrete Mathematics Propositional Logic - Discrete Mathematics
## What is Propositional Logic in Discrete Mathematics?
Theoretical base for many areas of mathematics and computer science is provided by logical reasoning. Some of the areas such as artificial intelligence, programming languages etc.
By propositional logic, the statements are analyzed and the truth vales are assigned. The analysis is done either for individual statement or as a composite of statements.
### Prepositional Logic – Definition
The declarative statement, which has either of the truth values, is termed as a proposition. Capital letters are used for representing propositional variables, and the variables are linked by a connective.
Some illustrations for propositions are:
• "Apple is a fruit", for this, the truth value “TRUE” is returned.
• "9+8+3=3 – 2", for this, the truth value “FALSE” is returned.
## What are Connectives in Propositional Logic?
Propositional logic provides five different types of connectives -
• OR (∨)
• AND (∧)
• Negation/ NOT (¬)
• Implication / if-then (→)
• If and only if (⇔).
### OR (∨)
If any one of the variables, A or B is true, then the OR operation is true. The OR operation is represented as A∨B. The following is the truth table for OR operation -
A B A ∨B True True True True False True False True True False False False
### AND (∧)
If both the variables, A and B is true, then the AND operation is true. The AND operation is represented as A∧B. The following is the truth table for AND operation -
A B A ∧B True True True True False False False True False False False False
### Negation (¬)
For a true variable A, the negation is false. The Negation operation is represented as ¬A. The following is the truth table for this operation -
A ¬ A True False False True
### Implication / if-then (→)
It is a conditional statement and If the variable A is true and the variable B is False, then this conditional statement would be false. The following is the truth table for this operation -
A B A →B True True True True False False False True True False False True
### If and only if (⇔)
If both variables A and B are same, that is either True or False, then the result is True. If both the Variables are not same, then the result is false. The following is the truth table for this operation -
A B A ⇔B True True True True False False False True False False False True
## What are Tautologies?
For all the values of the propositional variables, tautology is always true.
Example − Prove [(A→B)∧A]→B is a tautology
The following is the truth table −
A B A →B (A → B) ∧A [( A → B ) ∧ A] →B True True True True True True False False False True False True True False True False False True False True
It is observed that it is a tautology for every value is True.
For all the values of the propositional variables, contradiction is always false.
Example − Prove (A∨B)∧[(¬A)∧(¬B)] is a contradiction
The following is the truth table −
A B A ∨B ¬ A ¬ B (¬ A) ∧ ( ¬ B) (A ∨ B) ∧ [( ¬ A) ∧ (¬ B)] True True True False False False False True False True False True False False False True True True False False False False False False True True True False
It is observed that it is a contradiction for every value is False.
## What is a Contingency?
For each of the value of the propositional variables, contingency is either true or false.
Example − Prove (A∨B)∧(¬A) a contingency
The following is the truth table −
A B A ∨B ¬ A (A ∨ B) ∧ (¬ A) True True True False False True False True False False False True True True True False False False True False
It is observed that it is a contingency is either true or False.
## What are Propositional Equivalences?
Two statements are said to be equivalent logically if they satisfy the following conditions -
• •The truth values for each of the statement are same.
• •The bi-conditional statement X⇔Y is a tautology.
Example − Prove ¬(A∨B)and[(¬A)∧(¬B)] are equivalent
According to Matching truth table method –
A B A ∨B ¬ (A ∨ B) ¬ A ¬ B [(¬ A) ∧ (¬ B)] True True True False False False False True False True False False True False False True True False True False False False False False True True True True
The statements are said to be equivalent as it is observed that the truth values of both the statements ¬(A∨B)and[(¬A)∧(¬B)] are true.
According to Bi-Conditional Method -
A B ¬ (A ∨ B ) [(¬ A) ∧ (¬ B)] [¬ (A ∨ B)] ⇔ [(¬ A ) ∧ (¬ B)] True True False False True True False False False True False True False False True False False True True True
The statements are said to be equivalent as it is observed that [¬(A∨B)]⇔[(¬A)∧(¬B)] is a tautology.
## Inverse, Converse, and Contra-positive
The conditional statement has two parts -
• Hypothesis, p
• Conclusion, q
It is denoted as p→q.
Example of Conditional Statement − “If you eat health food, you will not be sick.” For this statement, hypothesis is – eat health food and conclusion is – will not be sick.
### Inverse
The negation of both hypothesis and conclusion is known as Inverse of the conditional statement.
Example – The inverse for the above discussed example is “If you do not eat health food, you will be sick.”
### Converse
The interchange of hypothesis and conclusion is known as Converse of the conditional statement.
Example – The converse for the example is “You will not be sick if you eat health food”.
### Contra-positive
The interchange of hypothesis and conclusion of inverse statement is Contra-positive.
Example − The Contra-positive for the example is “You will be sich if you do not eat health food”.
## What is Duality Principle?
According to Duality principle, if the statement and true and if unions are interchanged into intersections and if universal set is interchanged into the Null set, the result of the interchange in both the cases is true.
Example − The dual of (A∩B)∪C is (A∪B)∩C
## What are Normal Forms?
The propositions can be converted into any of the norm forms, which are of two types -
• Conjunctive normal form
• Disjunctive normal form
### Conjunctive Normal Form
When AND operation is executed on the variables connected with OR operation, the compound statement formulated is said to be in conjunctive normal form.
Examples
• (A∨B)∧(A∨C)∧(B∨C∨D)
• (P∪Q)∩(Q∪R)
### Disjunctive Normal Form
When OR operation is executed on the variables connected with AND operation, the compound statement formulated is said to be in disjunctive normal form.
Examples
• (A∧B)∨(A∧C)∨(B∧C∧D)
• (P∩Q)∪(Q∩R) | HuggingFaceTB/finemath | |
### Mental Math - Tails of 5 Multiplication Method
Problem: Calculate the value of the expression
45 x 65
Method: We can use the Tails of 5 method to help solve this problem.
The tail of a number is defined to be the last digit of the number. The head of the number is defined to be all of the digits except last digit. If you want to multiply two numbers, whose tails are both 5, then you can use the Tails of 5 method to multiply the two numbers quickly.
Let's take a look at multiplying 45 times 65. These two numbers both have a tail of 5. We know that the product of these two number will end in 25 (or possibly 75). The other digits of the product can be determined by multplying the heads of the two numbers and then adding the average of the two heads. In this case the two heads are 4 and 6. When we multiply 4 times 6 we get 24. The average of 4 and 6 is 5 so we add the 5 to 24 and get 29. So our product starts with 29 and ends with 25. Therefore, the product of 45 x 65 is 2925.
Problem: Calculate the value of the expression
35 x 125
Method: We can use the Tails of 5 method to help solve this problem.
The average of the two head will either be a whole number or contain the fraction ½. If the average of the two heads contains the fraction ½, then instead of ending in 25, the product will end in 75 (as 100 times ½ is 50).
To multiply 35 times 125, we observe that both numbers end in 5 which means that we can use the Tails of 5 method to calculate the product. We multiply the heads of the two numbers, 3 and 12, to get 36. Next we take the average of 3 and 12, and we get 7½. The ½ means that the product will end in 75 rather than 25. We add the 7 from the average of the two heads to the product of the two heads, 36, and we get 43.
So, the product of 35 times 125 is equal to 4375. | HuggingFaceTB/finemath | |
## Friday, March 16, 2018
### 103-Compound interest calculating tool
Now we will see the problem of compound interest.
Formula:
A = Total amount, P = principal or amount of money deposited or borrowed, r = annual interest rate (in decimal form), n = number of times compounded per year, t = time in years.
Example 1: If you deposit \$7000 into an account paying 5 % annual interest compounded bimonthly, how much money will be in the account after 6 years? What is total interest for 6 years? As per the calculation of compound interest, what is total interest per year and what is the rate of interest pcpa?
Solution:
Given:
Principal P = 7000, Rate of interest r = 0.05, n = 2 (as the interest is calculated bi-monthly) and t = 6.
Here we can use a logarithm to do these calculations. see the following steps carefully.
Now see the software tool for getting these results systematically. Click on the following figure and try the result for your values.
ANIL SATPUTE | HuggingFaceTB/finemath | |
# Algebraic Equation Manipulation for Electric Circuits
## Mathematics for Electronics
• #### Question 1
The electrical resistance of a conductor at any temperature may be calculated by the following equation:
$$R_T = R_r + R_raT - R_raT_r$$
Where,
RT = Resistance of conductor at temperature T
Rr = Resistance of conductor at reference temperature Tr
α = Temperature coefficient of resistance at reference temperature Tr
Simplify this equation by means of factoring.
• #### Question 2
The equation for voltage gain (AV) in a typical non-inverting, single-ended opamp circuit is as follows:
$$A_V = \frac{R_1}{R_2} + 1$$
Where,
R1 is the feedback resistor (connecting the output to the inverting input)
R2 is the other resistor (connecting the inverting input to ground)
Suppose we wished to change the voltage gain in the following circuit from 5 to 6.8, but only had the freedom to alter the resistance of R2:
Algebraically manipulate the gain equation to solve for R2, then determine the necessary value of R2 in this circuit to give it a voltage gain of 6.8.
• #### Question 3
The equation for voltage gain (AV) in a typical inverting, single-ended opamp circuit is as follows:
AV = R1 R2
Where,
R1 is the feedback resistor (connecting the output to the inverting input)
R2 is the other resistor (connecting the inverting input to voltage signal input terminal)
Suppose we wished to change the voltage gain in the following circuit from 3.5 to 4.9, but only had the freedom to alter the resistance of R2:
Algebraically manipulate the gain equation to solve for R2, then determine the necessary value of R2 in this circuit to give it a voltage gain of 4.9.
• #### Question 4
The following equations solve for the output voltage of various switching converter circuits (unloaded), given the switch duty cycle D and the input voltage:
Vout = D Vin (Buck converter circuit)
Vout = Vin 1 − D (Boost converter circuit)
Vout = D Vin 1 − D (Inverting or Cuk converter circuit)
Manipulate each of these equations to solve for duty cycle (D) in terms of the input voltage (Vin) and desired output voltage (Vout). Remember that duty cycle is always a quantity between 0 and 1, inclusive.
• #### Question 5
Solve for n in the following equations:
Equation 1: −56 = −14n
Equation 2: 54 − n = 10
Equation 3: 4/n = 12
Equation 4: 28 = 2 − n
• #### Question 6
The formula for calculating total resistance of three series-connected resistors is as follows:
$$R = R_1+R_2+R_3$$
Algebraically manipulate this equation to solve for one of the series resistances (R1) in terms of the other two series resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.
• #### Question 7
Manipulate this equation to solve for resistor value R1, given the values of R2 and Rparallel:
$$R_{parallel} = \frac{R_1 R_2}{R_1 + R_2}$$
Then, give an example of a practical situation where you might use this new equation.
• #### Question 8
The formula for calculating total resistance of three parallel-connected resistors is as follows:
$$R = \frac {1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$$
Algebraically manipulate this equation to solve for one of the parallel resistances (R1) in terms of the other two parallel resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.
• #### Question 9
The power dissipation of a transistor is given by the following equation:
$$P = I_C (V_{CE}+ \frac{V_{BE}}{\beta})$$
Manipulate this equation to solve for beta, given all the other variables.
• #### Question 10
The decay of a variable over time in an RC or LR circuit follows this mathematical expression:
$$e^{-{\frac{t}{τ}}}$$
Where,
e = Euler’s constant ( ≈ 2.718281828)
t = Time, in seconds
τ = Time constant of circuit, in seconds
For example, if we were to evaluate this expression and arrive at a value of 0.398, we would know the variable in question has decayed from 100% to 39.8% over the period of time specified.
However, calculating the amount of time it takes for a decaying variable to reach a specified percentage is more difficult. We would have to manipulate the equation to solve for t, which is part of an exponent.
Show how the following equation could be algebraically manipulated to solve for t, where x is the number between 0 and 1 (inclusive) representing the percentage of original value for the variable in question:
$$x=e^{-{\frac{t}{τ}}}$$
Note: the “trick” here is how to isolate the exponent $$-\frac{-t}{τ}$$. You will have to use the natural logarithm function!
• #### Question 11
Voltage and current gains, expressed in units of decibels, may be calculated as such:
AV(dB) = 10 log( AV(ratio) ) 2
AI(dB) = 10 log( AI(ratio) ) 2
Another way of writing this equation is like this:
AV(dB) = 20 logAV(ratio)
AI(dB) = 20 logAI(ratio)
What law of algebra allows us to simplify a logarithmic equation in this manner?
• #### Question 12
Solve for the value of x in the following equations:
10x = 80 x =
3 = 15 x x =
• #### Question 13
Solve for the value of x in the following equations:
5x = 15 x =
6 = x 2 x =
• #### Question 14
Solve for the value of a in the following equations:
Equation 1: a − 4 = 10
Equation 2: 30 = a + 3
Equation 3: −2a = 9
Equation 4: a/4 = 3.5
• #### Question 15
Solve for the value of x in the following equations:
$$\frac {x+5}{2} = 20 \ \ \ \ \ \ \ \ \ \ \ x =$$
$$6 = \sqrt{x-2} \ \ \ \ \ \ \ \ \ \ \ x =$$
• #### Question 16
Solve for the value of x in the following equations:
$$2(x+5) = 36 \ \ \ \ \ \ \ \ \ \ \ x =$$
$$3 = \sqrt{2-x}\ \ \ \ \ \ \ \ \ \ \ x =$$
• #### Question 17
Manipulate each of these equations to solve for a:
$$\frac{b-a}{c}=d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{a+b}=c^2d$$
• #### Question 18
Manipulate each of these equations to solve for a:
$$\frac{a-b}{c}=d^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b+a^2 = \frac{c}{d}$$
• #### Question 19
Calculate all currents in this DC circuit:
Hint: it may help you to set up the necessary equation by labeling the current through the lower resistor as I and the current through the upper resistor as I + 0.005.
### Related Content
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1 Comment
• B
bjm999 July 14, 2020
Question 14, Equation 2 - the ‘+’ symbol is missing between ‘a 3’.
Question 17, right-hand equation - the square-root should only be over the left side of the equation.
Question 19, Hint - the ‘+’ symbol is missing between ‘I 0.005’.
They do seem to be correct in the equivalent questions in the PDF version though.
Like. | HuggingFaceTB/finemath | |
# The distribution function
The distribution function of a random variable $$X$$ is a function that assigns, for every point, the probability accumulated up to the above mentioned value. That is:
$$F(X)=p(X \leq x)$$\$
For example, we will compute the distribution function of the random variable $$X$$, resulting from throwing a perfect dice.
The following table shows the values of $$F (x)$$:
x F (x) $$x < 1$$ $$0$$ $$1\leq x < 2$$ $$\displaystyle \frac{1}{6}$$ $$1\leq x < 3$$ $$\displaystyle \frac{2}{6}$$ $$1\leq x < 4$$ $$\displaystyle \frac{3}{6}$$ $$1\leq x < 5$$ $$\displaystyle \frac{4}{6}$$ $$1\leq x < 6$$ $$\displaystyle \frac{5}{6}$$ $$x \leq 6$$ $$1$$
The value of the function distribution in $$-\infty$$ will always be $$0$$, while the value in $$+\infty$$ will always be $$1$$.
This turns out to be quite intuitive, since the probability that the value of $$x$$ is smaller than $$-\infty$$ is zero, and the probability that it is less than $$+\infty$$ is $$1$$ (since it is always less or equal to $$+\infty$$).
Since it is a discrete random variable, the distribution function will be discrete. Note that the probability of obtaining a result lower than $$5,2$$ is the same as that of lower than $$5,3$$ or $$5,9$$. | HuggingFaceTB/finemath | |
## 10.6 Determining Which Cost Variances to Investigate
### Learning Objective
1. Determine which variances to investigate.
Question: Companies rarely investigate all variances because there is a cost associated with identifying the causes of variances. This cost involves employees who spend time talking with personnel from areas including purchasing and production to determine why variances occurred and how to control costs in the future. What can managers do to reduce the cost of investigating variances?
Answer: Managers typically establish criteria to determine which variances to focus on rather than simply investigating all variances. This is called management by exception. Management by exceptionA term used to describe managers who focus solely on variances showing actual results that are significantly different than expected results. describes managers who focus solely on variances that are significant.
Question: Figure 10.14 "Comparison of Variable and Fixed Manufacturing Overhead Variance Analysis for Jerry’s Ice Cream" summarizes the cost variances calculated for Jerry’s Ice Cream. If you were in charge of investigating variances at Jerry’s Ice Cream, how would you determine which variances to focus on and which to ignore?
Figure 10.10 Summary of Cost Variances at Jerry’s Ice Cream
Answer: Some managers might review all unfavorable variances. However, the variable overhead spending variance of \$5,500 is not very significant relative to the other variances and may not be worth investigating. Also, by focusing solely on unfavorable variances, managers might overlook problems that may result from favorable variances.
Another approach might be to investigate all favorable and unfavorable variances above a certain minimum level, calculated as a percent of the flexible budget amount. For example, management could establish a policy to investigate all variances at or above 10 percent of the flexible budget amount for each cost. At Jerry’s Ice Cream, this would mean investigating all variances at or above \$42,000 for direct materials (= 10 percent × \$420,000), \$27,300 for direct labor (= 10 percent × \$273,000), and \$10,500 for variable overhead (= 10 percent × \$105,000). Based on this policy, the following variances would be investigated:
• Unfavorable direct materials price variance of \$88,000 (≥ \$42,000 minimum)
• Unfavorable direct labor rate variance of \$37,800 (≥ \$27,300 minimum)
• Favorable direct labor efficiency variance of \$(27,300) (≥ \$27,300 minimum)
• Favorable variable overhead efficiency variance of \$(10,500) (≥ \$10,500 minimum)
Many companies calculate and investigate variances weekly, monthly, or quarterly and focus on trends. In this case, they may only investigate variances that are unfavorable and increasing over time.
Whatever the approach, managers understand that investigating variances requires resources. Thus managers must establish an efficient and cost-effective approach to analyzing variances by weighing the benefits derived from investigating variances against the costs incurred to perform the analysis.
Using Cost Variances to Detect Fraud
Variance analysis is not only an effective way to control costs; some companies, including The Dow Chemical Company, have found that investigating variances can also help them detect fraudulent activities. Dow, which provides chemical, plastic, and agricultural products to customers in 180 countries, has annual sales of \$33,000,000,000 and approximately 46,000 employees. In 1998, the company created a department called Fraud Investigative Services (FIS) whose mission is to “deter and prevent incidents of fraud and financial abuse through detection, investigation, and education.”
The most common types of fraud allegations reviewed by Dow’s FIS include expense report fraud, kickback schemes, and embezzlement. Paul Zikmund, the director of FIS, states that “unexplainable cost variances between budget and actual amounts” are among the warning signs he looks for in identifying fraud.
For example, suppose the actual cost for direct materials is significantly higher than the budgeted cost. The cost accountant at Dow would begin investigating the cause of the variance by talking with the company’s purchasing agent. The purchasing agent might be unable (or unwilling) to explain why actual costs are so high. Further investigation might indicate that the purchasing agent was intentionally overbilling the vendor and receiving a kickback from the vendor.
Zikmund states that for every \$1 that Dow spends on investigating fraud, the company recovers nearly \$4. He also notes that Dow’s loss per employee is far below the industry average of \$9 per employee per day. For a company with 46,000 employees, every dollar in savings per employee adds up to a significant amount.
### Key Takeaway
• Companies often establish criteria to use in determining which variances to investigate. Some might investigate all variances above a certain dollar amount. Others might investigate variances that are above a certain percentage of the flexible budget. Or management might combine the two and investigate variances above a certain dollar amount and above a certain percentage of the flexible budget.
### Review Problem 10.6
Use the solutions to Note 10.30 "Review Problem 10.3", Note 10.40 "Review Problem 10.4", and Note 10.49 "Review Problem 10.5" to complete the following:
1. Calculate the total variable production cost variance for Carol’s Cookies using the format shown in Figure 10.10 "Summary of Cost Variances at Jerry’s Ice Cream".
2. Assume management investigates all variances at or above 15 percent of the flexible budget amount (e.g., all direct materials variances at or above 15 percent of the direct materials flexible budget are investigated). Identify which of the six variances calculated for direct materials, direct labor, and variable manufacturing overhead management should investigate.
Solution to Review Problem 10.6
1. See the following figure.
2. Based on this policy, the following variances would be investigated:
• Direct Materials. Neither variance would be investigated as both variances fall below \$175,500 (= 15 percent of \$1,170,000 standard cost).
• Direct Labor. The unfavorable direct labor efficiency variance of \$234,000 would be investigated because it falls above \$140,400 (= 15 percent of \$936,000 standard cost).
• Variable Overhead. The unfavorable variable overhead efficiency variance of \$68,250 would be investigated because it falls above \$40,950 (= 15 percent of \$273,000 standard cost). | HuggingFaceTB/finemath | |
Question
# $1$ millennium is equal to how many years?
Open in App
Solution
## Representation of millennium in years.$1$ millennium is equal to one thousand years, i.e. $\begin{array}{rcl}1\mathrm{millennium}& =& 1000\mathrm{years}\\ \mathrm{and}1\mathrm{centuy}& =& 100\mathrm{years}\end{array}$Hence,$1\mathrm{millennium}=10\mathrm{centuries}.$Hence, $1$ millennium is equal to one thousand years or $10$ centuries.
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# Join all circles together only with 6 lines
In the below image, can you draw 6 straight lines that pass all the circles?
As soon as you start drawing lines you can't take your pen up until you draw all six lines.
hint: you don't have to keep the polyline inside the square.
Edit: Best answer is the one in which straight line passes center of the circles.
• hint: you don't have to keep the polyline inside the square Aug 12, 2014 at 8:40
• Do you have to use all 6 lines? Aug 12, 2014 at 14:09
• it is possible with less than 6,but only with 6 lines you can pass the center of each circle.if you can do that with less than 6 do it in answer. Aug 12, 2014 at 15:54
## 16 Answers
I think this is one possible solution:
If you use just one massive line, you can make it pass through the center of all of the circles!
• that's not what i meant, and what kind of pen is it?it's just a big triangle not big line. Aug 12, 2014 at 19:36
• It's just a really big pen. It is not a triangle. Aug 12, 2014 at 19:56
• I signed up to +1 this. Hilarious! regarding it being a square and not a line, just extend the 'line' very far beyond the border and zoom out. problem solved, you have a line that looks like a line.
– n00b
Aug 13, 2014 at 20:00
• This is a contender for the worst answer on the site, and it has pretty strong competition. The question is clearly a math question, not a troll request. Declaring that a square is a line does not make a line have two dimensions. Jan 14, 2015 at 19:28
• I'm just waiting for enough reputation to downvote this solution :-P Sep 8, 2015 at 10:22
Here's an option that uses only 4 lines. You can extend the concept to place another 2 lines if you really want 6 ...
• This is what I immediately thought of when I read the question. +1. This works for arbitrarily small circles, since the lines can be arbitrarily long. And if you give me 0-size circles, I can still do it with infinitely long lines! Aug 12, 2014 at 14:35
• it is possible with less than 6,but only with 6 lines you can pass the center of each circle, which i didn't mention in this puzzle.but still your answer is correct.can you pass center of each line with less than 6 lines? Aug 12, 2014 at 15:56
• @Riddle well that would be the same as passing through 0-size circles which works with infinitely long lines as Cruncher mentioned in his/her comment. Aug 13, 2014 at 9:03
• ... and whether you can draw two parallel lines without lifting your pen depends on whether they meet, which in turn depends on the geometry of the space in which the figure appears. Paste the figure to a torus (at an angle) and you can do it in one line!
– user1501
Aug 13, 2014 at 12:32
• @ExpectoPatronum I guess the crux is that you can't draw infinitely long lines :) Aug 15, 2014 at 20:54
By mapping the puzzle onto a cylindrical topography, I've solved the puzzle using only a single straight line.
• it is possible with less than 6,but only with 6 lines you can pass the center of each circle, which i didn't mention in this puzzle.but still your answer is correct.can you pass center of each line with less than 6 lines? Aug 12, 2014 at 15:57
• Why have the sides match up evenly? Have the right side of the first row of circles match up with the left side of the second row of circles, and your line should go through the center of all the circles. Aug 12, 2014 at 20:19
• What definition of straight line are you using here? Presumably something other than "shortest path between two points"?
– jl6
Aug 14, 2014 at 12:40
• Btw, @Simon can you create another image with the second row matching the first row like suggested by RobWatts? That way it will answer the question (with OP's additional clarification) perfectly with only one line. Sep 4, 2014 at 5:17
• @justhalf, that was created in Sketchup with the puzzle image mapped onto a cylinder as a texture (I have the joy of being Simon's boss) Sep 18, 2014 at 14:53
Here's another way of thinking. This is using edges instead.
• I like this one a lot. Aug 13, 2014 at 19:27
• This actually looks like art :) +1, I would hang this Aug 15, 2014 at 20:59
1. Print the picture.
2. Fold the image such that there is a single stack of circles. This will take 6 folds.
3. Draw one line across the middle of it.
4. Profit.
• actually you just draw a line on one of circles, if you unfold your paper again just one of them has been crossed, it seems you just disappeared the other circles. Aug 13, 2014 at 7:23
• @Riddle: It's all about point of view. From my point of view the line has crossed the center point of all of the circles as they are stacked on top of each other. The problem with this exercise is that you haven't defined any real boundary conditions (things you are explicitly not allowed to do). This means there are many different ways of solving the puzzle that don't fit your preconception. Aug 13, 2014 at 14:10
• Just set the point of the pen on the center of the circle and push really hard with the pen, then it will have passed through all the circles. It is one line, going through the z-axis of the folded paper. Aug 15, 2014 at 21:18
• @Anssssss: that's even better! Aug 15, 2014 at 22:42
• -1 for no ??? step Aug 17, 2014 at 9:02
This one doesn't solve it the way you're supposed to, but also doesn't break any of the implied "rules", but kind of defeats the purpose of the puzzle (which is go outside of the square). Were you to make the circles smaller, it would eliminate this answer from plausibility. (Also the (3,3) circle line is a little dubious, if I had taken some more time to perfect this it would fit inside the circle better, but also shows how this solution could be easily invalidated by making the circles smaller)
• You better write that this way you manage to keep the polyline inside the square! +1 Aug 12, 2014 at 17:33
• You can move the vertex at (1,2) upwards slightly to get a better fit for (3,3) up to the point where it equals the fit for (3,2). Aug 13, 2014 at 18:15
• Yeah, I didn't try super hard to get it to fit perfectly, as it kind of helped point out how this method is easily broken by making the circles smaller.
– DLeh
Aug 13, 2014 at 18:27
Why not this?
This one also does what he wants.
• because you take your pen up for several times Aug 13, 2014 at 7:19
• @Riddle But if we r allowed to trace back our path then its doable this way without lifting the pen right. Aug 13, 2014 at 8:27
• You don't need to pick your pen up. Just double back on the line you already drew. Aug 14, 2014 at 1:49
• Wouldn't doubling back be drawing another line, though? So you could only have three lines. Aug 14, 2014 at 14:42
• No, there is no "depends" about it. There is no doubling back. The instant you try that is the instant you fail on this. Making a line is a forward motion (meaning not doubling back). Jun 26, 2015 at 15:53
Similar but different answer. Great puzzle!
my own answer to this question is
in this picture you can see 6 lines throw the center of all circles without taking your pen up.
For some reason I wanted the answer to have a line with an angle that was not a multiple of $45^\circ$.
This could be the answer as well.
6 lines without lifting the pen.
here is my answer!
nice question!
Hope this will also be considered as an answer.
Looking through the answers, there are several similar, but none the same as this one. So here's one more possible answer.
• your answer is exactly same as mine, if you rotate your picture once to right you can see that. Nov 28, 2014 at 11:31
below is my answer photo.
Just Simple and beautiful :D
I guess my answer matched with at least this
Edit: Best answer is the one in which straight line passes center of the circles.
• how would you draw those lines without lifting the pen? Dec 10, 2015 at 13:47
• Cannot @KateGregory. That's why I add my last sentence there.
– Nai
Dec 11, 2015 at 8:02
• Well, can you overlap lines? Jul 10, 2018 at 5:07 | HuggingFaceTB/finemath | |
1. ## probability
Bag A contains 5 similar balls, 3 red and 2 black and bag B has 4 red and 5 black balls. a ball is drawn form A and places in B and then a ball is drawn from B and places in A. what is the chance that if a ball is now drawn form A, it will be red? and if this ball is red, what is the probability that the first two balls are also red?
2. Originally Posted by alexandrabel90
Bag A contains 5 similar balls, 3 red and 2 black and bag B has 4 red and 5 black balls. a ball is drawn form A and places in B and then a ball is drawn from B and places in A. what is the chance that if a ball is now drawn form A, it will be red? and if this ball is red, what is the probability that the first two balls are also red?
hi
This is a conditional probability problem
(1) Let R be red and P be events of being drawn from A
P(R|P)=P(P n R)/P(P)
=(3/14)/(5/14)
=3/5
(2) Let R be events of first ball being red , and Q be events of second ball being red
P(Q|R)=P(R n Q)/P(R)
=(5/10)/(3/5)
=5/6
Since the first red ball is drawn from A , the second ball (unknown colour) will be drawn from B according to the question . Now , the number of red balls increases by 1 in B , making 5R 5B and 2R 2B in A .
3. the answer for that should be 143/250 and 45/143 i think | HuggingFaceTB/finemath | |
### 10th Putnam 1950
Problem A5
Let N be the set of natural numbers {1, 2, 3, ... }. Let Z be the integers. Define d : N → Z by d(1) = 0, d(p) = 1 for p prime, and d(mn) = m d(n) + n d(m) for any integers m, n. Determine d(n) in terms of the prime factors of n. Find all n such that d(n) = n. Define d1(m) = d(m) and dn+1(m) = d(dn(m)). Find limn→∞ dn(63).
Solution
d(pa) = a pa-1 by a trivial induction on a. Hence for n = paqb ... we have d(n) = n ( a/p + b/q + ... ) by a trivial induction on the number of primes.
Hence d(n) = n iff a/p + b/q + ... = 1. Multiplying through by all the prime denominators gives integral terms. All but the first are clearly divisible by p, so the first must be also. Hence a/p = 1 and b/q etc are zero. In other words, n = pp for some prime p.
We find d1(63) = 51, d2(63) = 20, d3(63) = 24, d4(63) = 44, d5(63) = 48. Now suppose n is divisible by 16. Then n = 2kpaqb ... , where p, q, ... are all odd and k >= 4. Hence d(n) = n( 4/2 + a/p + b/q + ... ). Now all of 2n, na/p, nb/q, ... are integral and multiples of 16. So d(n) is at least twice n and a multiple of 16. So if we have dk(m) divisible by 16, then dh+k(m) ≥ 2h which diverges. Hence dk(63) tends to infinity. | HuggingFaceTB/finemath | |
# Solve Archimedes Problem: Find Crown's Density
• Dipra Irham
In summary, the conversation discusses a physics problem about the density of a gold crown and the confusion surrounding the use of weight and mass. The solution involves calculating the volume and density of the crown and using the weight of the crown in air and when immersed in water. There is also a discussion about the units used and the possible methods used by Archimedes to determine if the crown was made of pure gold.
#### Dipra Irham
I have got a question in my physics book.The question goes like that:
(Let's suppose the weight of the gold crown of Archimedes is 10 kg in air and 9.4 kg when immersed in water.What is the density of the crown? )
The solution says:
If the volume of the crown is V and density is p then,
Vp=10 kg
Vp-(Vp of water)=9.4 kg
Vp of water = Vp - 9.4 kg= 0.6 kg
so, V = 0.6 kg / density of water
=0.6 kg÷10^3 kg/m^3= 0.6 X 10^ -3m^3
Therefore, p = 10 kg / V
= 10 kg/0.6 X 10^ -3m^3
=16666 Kg /m^3
my question is, why Vp =10 kg?(it should be Vp times acceleration due to gravity,since weight = mass X acceleration due to gravity)
#### Attachments
• IMG_20180603_230146.jpg
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Dipra Irham said:
I have got a question in my physics book.The question goes like that:
(Let's suppose the weight of the gold crown of Archimedes is 10 kg in air and 9.4 kg when immersed in water.
Technically, the mass of the gold crown is 10 kg in air. The crown would have the same mass on the moon with its lower gravity, but its weight would be less. The proper unit of weight in this problem would be Nt (Newtons).
Immersing the crown in water wouldn't change its mass.
Dipra Irham said:
What is the density of the crown? )
The solution says:
If the volume of the crown is V and density is p then,
Vp=10 kg
Vp-(Vp of water)=9.4 kg
Vp of water = Vp - 9.4 kg= 0.6 kg
so, V = 0.6 kg / density of water
=0.6 kg÷10^3 kg/m^3= 0.6 X 10^ -3m^3
Therefore, p = 10 kg / V
= 10 kg/0.6 X 10^ -3m^3
=16666 Kg /m^3
my question is, why Vp =10 kg?(it should be Vp times acceleration due to gravity,since weight = mass X acceleration due to gravity)
The density here apparently is in units of ##\frac{\text{kg}}{m^3}##. Multiplying V and p results in units of kg. It's hard to tell what they're trying to say in this problem, as the units being used don't make sense.
BTW, the image you posted is too small to be readable, and part of it is in shadow, which makes it even less useful.
• Dipra Irham
The only error I see is in this opening statement: "weight ... is 10 kg in air and 9.4 kg when immersed in water."
That can be corrected either by including the factor g: "10g kg in air and 9.4g kg when immersed in water"
or by changing "kg" to "kg-weight".
Either way, the first equation becomes Vρg=10g kg, but the g's immediately cancel to give Vρ=10 kg.
• Dipra Irham
I would note that, in the time of Archimedes, Newtons were far in the future and so was the value of g. The "weighing" process was most likely conducted using an equal arms scale with an amount of gold equal to what was given to the craftsman in one pan and the crown in the other. In air they would balance, but with both pans immersed they would not if the crown were adulterated. In short it makes sense to imagine that Archimedes conducted what we now call a null measurement.
• Dipra Irham
haruspex said:
The only error I see is in this opening statement: "weight ... is 10 kg in air and 9.4 kg when immersed in water."
That can be corrected either by including the factor g: "10g kg in air and 9.4g kg when immersed in water"
or by changing "kg" to "kg-weight".
Either way, the first equation becomes Vρg=10g kg, but the g's immediately cancel to give Vρ=10 kg. | HuggingFaceTB/finemath | |
# 91% of 38000
What is 91 percent of 38000? What is 38000 plus 91% and 38000 minus 91%?
91% of 38000 = 34580 38000 + 91% = 72580 38000 − 91% = 3420
Calculate online:
% of
Percent formula to figure out percentages:
38000 ⋅ 91 ÷ 100 = 34580.
So, we got a discount percentage of 34580. If the price was 38000 and a 91% (ie \$34580) discount percentage was deducted from it, then we get the final price \$3420. If we add the percentage received to the initial price, we get 72580 (38000+34580). New price is 72580, as the price has risen 91%.
More calculations with 38000:
24% of 38000 = 9120 38000 + 24% = 47120 38000 − 24% = 28880
34% of 38000 = 12920 38000 + 34% = 50920 38000 − 34% = 25080
87% of 38000 = 33060 38000 + 87% = 71060 38000 − 87% = 4940
117% of 38000 = 44460 38000 + 117% = 82460 38000 − 117% = -6460
More calculations with 91%:
91% of 79 = 71.89 79 + 91% = 150.89 79 − 91% = 7.11
91% of 231 = 210.21 231 + 91% = 441.21 231 − 91% = 20.79
91% of 668 = 607.88 668 + 91% = 1275.88 668 − 91% = 60.12
91% of 91455 = 83224.1 91455 + 91% = 174679 91455 − 91% = 8230.95 | HuggingFaceTB/finemath | |
# Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°
106,286 results
maths
A parallelogram has two adjacent sides of length 4cm and 6cm reapectively. If the included angle measures 52 degrees, find the area of the parallelogram.
geometry theorems
I have six questions which will be on a test two weeks from now, and I do not understand. Help me find out these theorems for the blanks. 1. If one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a_______? 2. If the diagonals of a ...
geometry
abcd is a parallelogram with sides ab=12 cm,bc=10 cm and diagonal ac=16 cm .find the area of the parallelogram also find the distance between its shortest sides
Math
abcd is a parallelogram with sides ab=12 cm,bc=10 cm and diagonal ac=16 cm .find the area of the parallelogram also find the distance between its shortest sides
math
If the perimeter of a parallelogram is 140 m, the distance between a pair of opposite sides is 7 meters and its area is 210 sq m, find the length of two adjacent sides of the parallelogram.
math
If the perimeter of a parallelogram is 140 m, the distance between a pair of opposite sides is 7 meters and its area is 210 sq m, find the length of two adjacent sides of the parallelogram.
math
THe area of a parallelogram is 594, and the lengths of its sides are 32 and 46. Determine, to the nearest tenth of a degree, the measure of the acute angle of the parallelogram
geometry
Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°.
math
The sides of a parallelogram are 15 ft and 17 ft. One angle is 40° while another angle is 140°. Find the lengths of the diagonals of the parallelogram (to the nearest tenth of a foot)
geometry
Given: QRST is a parallelogram. Prove: QRST is a square. Complete the proof below by choosing the reason for line number 2 and line number 6. Reason Statement 1. QRST is a parallelogram. Given 2. QRST is a rectangle 3. is a right angle Definition of a right angle. 4. ...
e
Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure \$45^\circ\$.
Geometry
Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees.
Geometry
Find the area of a parallelogram with sides of length 6 and 8, and with an interior angle with measure 45 degrees.
Math
The lengths of two sides of a parallelogram are 7.4cm and 9.2cm and one of the diagonals has a length of 6.2 cm. Find the area of the parallelogram.
Calculus
1) Find correct to six decimal places root of the equation cos(x)= x for xE[0, pi/2] using Newton's Method. 2) A triangle has two constant lengths of 10 cm and 15 cm. The angle between two constant sides increases at a rate of 9 deg/min. Find the rate of increase of the third ...
mathssssSsssss
1.Determine the area of a parallelogram in which 2 adjacent sides are 10cm and 13cm and the angle between them is 55 degrees? 2.If the area of triangle ABC is 5000m squared with a=150m and b=70m what are the two possible sizes of angle C?
easy geometry
1)If diagonals of a rhombus are 10 cm and 24 cm. find the area and perimeter of the rhombus. 2)A regular hexagon with a perimeter of 24 units is inscribed in a circle. Find the radius of the circle. 3)Find the altitude,perimeter and area of an isosceles trapezoid whose sides ...
Math
The area of a parallelogram with a side of 5.6 in is equal to the area of a rectangle with sides 7 in and 8 in. Find the altitude to the given side in the parallelogram.
Math
The area of a parallelogram with a side of 5.6 in is equal to the area of a rectangle with sides 7 in and 8 in. Find the altitude to the given side in the parallelogram.
Geometry
Can some one please check these for me.Thanks Angle DEF is similar to Angle HJK and the scale factor of angle DEF to anle HJK IS 5/2. If EF =15 FIND JK Here is what I came up with Am I correct 5/2 = EF/JK 5/2 =15/JK 5(JK)=30 JK=30/5 JK=6 Given: ABCD is a parallelogram; <1 ...
Maths
The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of the parallelogram by using heron's formula
Math
The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of parallelogram by using heron's formula
Math
The adjacent sides of parallelogram are 26cm and 28cm and one of its diagonal is 30cm. Find the area of the parallelogram by using heron's formula
math
the measures of a pair of two adjacent sides of a parallelogram are 15cm and 20 cm.if the length of one of its diagonal is 25cm, find the area of the parallelogram
Math
the measure of pair of adjacent sides of a parallelogram are 15cm and 20cm.if the length of one diagonals is 25cm find the area of the parallelogram.
Trig-Algebra help asap
The adjacent sides of a parallelogram measure 8 cm and 12 cm and one angle measures 60 degrees. Find the area of the parellelogram.
trig
The lengths of 2 adjacent sides of a parallelogram are 42 cm and 36cm. an angle of the parallelogram is 4o degrees. Find the measure of the longest diagonal to the nearest tenth of a centimeter.
physics
Consider four vectors F1, F2, F3, and F4, with magnitudes are F1 = 41 N, F2 = 24 N, F3 =27 N, and F4 = 51 N, and angle 1 = 150 deg, angle 2 = −160 deg, angle 3 = 39 deg, and angle 4 = −57 deg, measured from the positive x axis with the counter clockwise angular ...
geometry
The longer diagonal of a parallelogram measures 62 cm and makes an angle of 30 degrees with the base. Find the area of the parallelogram if the diagonals intersect at angle of 70 degrees.
geometry
In parallelogram ABCD, BC = 8, DC = 12, and measure of angle D is 50 degrees. Find the area of the parallelogram to the nearest tenth.
math
Trigonometry - determine the area of a parallelogram in which 2 sides are 10 and 13 and the angle between them is 55
maths
two adjacent sides of a parallelogram measure 15 cm and 20 cm while the diagonal opposite their common vertex measure 25 cm.find the area of he parallelogram.
Math check
1. If LMNO is a parallelogram, and LM=2y-9 and NO=y-2, find the value of 'y'. Does y=7? 2. RSTV is a parallelogram. RT and SV intersect at Q. RQ=5x+1 and QT=3x+15. Find OT. Does QT=7? 3. RATS is a parallelogram. If angle s= (8x) and angle t=(7x) then find the value of 'x'. ...
math
The bases of a right prism are parallelograms with sides a=10 cm, b=6 cm, and altitude of the parallelogram towards side a, ha = 3 cm. Find the surface area of the parallelogram, if the height of the prism is h=12 cm.
Geometry
Given: ABCD is a parallelogram; <1 is congruent to <2 To Prove: ABCD is a Rhombus Plan: Show <2 is congruent to < CAB. hence CB is congruent to AB, making ABCD a parallelogram with consecutive sides congruent PROOF Statements Reasons 1. ABCD is a parallelogram 1....
Find Area of Parallelogram
Find the area of the parallelogram with one corner at P1 and adjacent sides P1P2 and P1P3. NOTE: There is an arrow over P1P2 and P1P3. What does that arrow mean? P1 = (0, 0, 0), P2 = (2, 3, 1), P3 = (-2, 4, 1)
maths
the perimeter of a parallelogram is 28 cm and ratio of the adjacent sides is 3:4 find the sides of a parallelogram?
Geometry
I have four questions: 1. The area of trapezoid ABCD is 60. One base is 4 units longer than the other, and the height of the trapezoid is 5. Find the length of the median of the trapezoid. 2. In trapezoid ABCD, line BC is parallel to line AD, angle ABD = 105 degrees, angle A...
Math
Find the area of the parallelogram. It's A parallelogram has a lower side measuring 24 centimeters and a right side measuring 9 centimeters. A dashed vertical segment with endpoints on the upper and lower sides is 8 centimeters and meets the lower left vertex at a right angle.
Math
Find the area of a regular pentagon if its apothem is approximately 4 ft long and each of its sides are 5.8 ft long. So I tried dividing the pentagon into triangles...but then I came up with the answer 24.41? I know that's not right. Can you hLep me at all? I will be happy to ...
Trig
The diagonals of a parallelogram intersect at a 42◦ angle and have lengths of 12 and 7 cm. Find the lengths of the sides of the parallelogram. (Hint: The diagonals bisect each other.)
geometry
a parallelogram has the vertices (-1,2), (4,4), (2,-1), and (-3,-3). determine what type of parallelogram. find the perimeter and area. i found the parallelogram part and its a rhombus but i cant find the perimeter and area
pre-cal
Find the EXACT VALUE of csc(-11pi/12) If that angle is in radians, it is easy. A full circle has 2PI radians...or 360 deg so 22PI/24 is the same as 22*15 deg, or -30 deg. The 30 deg triangle is well known. The sin is 1/2, so the csc is 2 Figure out the sign of the csc from ...
Math
Really stuck on this question not sure where im going wrong a trapezium with top and bottom sides paralell. With a diagonal line running from the top left to the bootom right hand corner splitting the trapez into two triangles. the top left corner angle is formed by 66 deg and...
math
parallelogram A is similar to Parallelogram B. If the area of parallelogram b is 162 square units, what is the area of parallelogram A?
SOLID
the altitude BE of parallelogram ABCD divides the side AD into segments in the ratio 1:3. Find the area of the parallelogram if the length of its shorter side is 14 cm, and one of its interior angle measures 60 degrees.
geometry
1. the opposite angles of a parallelogram measure(x=30) and (2x-50).Find the measure of each angle of the parallelogram. 2. the ratio of two consecutive angles of a parallelogram is 2:3. find the measurement of the angles of a parallelogram. 3.the difference between the ...
oblique triangle trigonometry
A parking lot has the shape of a parallelogram. The lengths of two adjacent sides are 70meters and 100meters. The angle between the two sides is 70o. What is the area of the parking lot?
Physics
Consider a piece of metal that is at 5 deg C. If it is heated until it has twice the internal energy, its temperature will be. a)556 deg C b)283 deg C c)273 deg C d)278 deg C e)10 deg C
physics
find the area of parallelogram whose adjacent sides are +3+3 and -3-2+.
calculus
find the area of the parallelogram that has vectors as adjacent sides u = -2i +j +5k v = 4i -3j -3k
science (physics) Need to solve this today please
(a) Express the vectors A, B, and C in the figure below in terms of unit vectors. (b) Use unit vector notation to find the vectors R = A + B + C and S = C – A – B. (c) What are the magnitude and directions of vectors R and S? Well the picture goes and has a vector A ...
Math
In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
Math
In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
Math
In a parallelogram abcd,if angle a =(2x+25) and angle b = (3x_5),find the value of x and the measures of each angle of the parallelogram
Geometry
I really need some help please!!!!! Am I on the right tract? Given: ABCD is a parallelogram; <1 is congruent to <2 To Prove: ABCD is a Rhombus Plan: Show <2 is congruent to < CAB. hence CB is congruent to AB, making ABCD a parallelogram with consecutive sides ...
maths
The diagonal of a parallelogram 6cm and 8cm long and they intersect at an angle of 55 degrees. calculate the area of the parallelogram
math
A parallelogram has an area of 8x^2 - 2x -45. The height of the parallelogram is 4x + 9. I know the formula for a parallelogram is A=bh. Please work and explain how I get the length of the base of the parallelogram. Thanks
math
If one angle of parallelogram is twice of its adjacent angle. Find angle of parallelogram.
Math
The formula A=bh is used to find the area of a parallelogram. If the base of a parallelogram is double and its height is doubled, how does this affect the area.
Math
Find the area of the parallelogram that has the vectors as adjacent sides. u = i+2j+2k v = i+k I know that I have to use (-2,-2,2) sqrt (1-2)^2 + (2+2)^2 + (2-2)^2 = sqrt17 Area = sqrt 17. Is this correct?
math
A diagonal of a parallelogram forms 25° and 35° angles with the sides of a parallelogram. Find the angles of the parallelogram.
triginometry
Two sides of a parallelogram are 9 and 13 with an included angle of 57 degree & 29 minutes.fine the distance between the longer sides.
math
Two adjacent sides of a parallelogram are four cm and five cm find the perimeter of the parallelogram
geometry
the sides of a quadrilateral abcd are 6cm,8cm,11cm and 12cm respectively,and the angle between the first two sides is a right angle.find its area.
geometry
solution of the questions The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.
math
solution of the questions The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.
math
if an angle of parallelogram is two third of its adjacent angle,find the angles of the parallelogram
Geometry
If there is a parallelogram which has four sides xy and x+25 ,4x+35 abc is parallelogram find x,y,z
Math
The sides of parallelogram 2x+y,3y-2x,5y-8,4x-3 are given in cm.find x and y and the perimeter of the parallelogram
Pre-calculus
Find the lengths of the diagonals of a parallelogram with adjacent sides of 12cm and 13 cm if the angle between the sides is 50 degrees. The shorter diagonal is what in terms of cm? The longer diagonal is what in terms of cm?
math
Two sides of a triangle have lengths 8 m and 24 m. The angle between them is increasing at a rate of 0.05 rad/s. Find the rate at which the area of the triangle is changing when the angle between the sides of fixed length is 135°. Round your answer to two decimal places. So ...
Math
A square and a parallelogram have the same area. If a side of the square is 40m and the height of the parallelogram is 20m, find the length of the corresponding base of the parallelogram.
Trigonometry
The lengths of the diagonals of a parallelogram are 20 inches and 30 inches. The diagonals intersect at an an angle 35 degrees. Find the lengths of the parallelogram's sides lengths to the nearest hundredth.
Algebra 1
In a right triangle the adjacent side = 12, the opposite side = 35, and the hypotenuse = 37. What does the opposite angle =? a-71,1 deg b-18.9 deg c-29.2 deg d-90 deg my answer is 18.89 degrees
Math
suppose that you know that a given parallelogram has at least one right angle and at least two adjacent sides that sre congruent. Can you make a conclusion about what kind of parallelogram this is? I say no because it can be a square or a rectangle.
Geometry
What is the value of x? image is on: tiny pic . com/r/nx1abn/8 I choose c 90 deg a 60 deg b 72 deg c 90 deg d 120 deg
Cosine Law
Find all missing sides/angles. Round each answer to the nearest unit. Angles/sides ABC (left to right) 1) Angle c= 70 deg side b= 28 yd side c= 26 yd 2) Side a= 18 cm Side b= 24 cm Side c= 28 cm
Math
Suppose you find the ratio of the lengths of adjacent sides in a parallelogram. This ratio is equivalent to the ratio of the adjacent sides in another parallelogram. Are the figures similar? Explain.
geometry
The diagonal of a rectangle is 25 meters long makes an angle of 36 ° with one of the rectangle. Find the area and the perimeter of the parallelogram. how to find the b to arrive to the area?
algebra
A parallelogram is 22 inches long and 14 inches wide. How do you find the height of the parallelogram? How do you find the area of the parallelogram ?
math , correction
theres a diagram which in the diagram the points are: (-3,1),(0,3),(2,0), (-1,-2). And the directions say: use the concept of slope to determine if the given figure is a parallelogram or a rectangle: so this is what i did please can someone check it to see if i am correct. M...
Find the area of the parallelogram that has the vectors as adjacent sides. u = i + 2j + 2k V = i + k I know that the magnitude of the cross product results on the area. After I do the cross product then what to I do?
geometry
the measure of an angle of a parallelogram is 12 degrees less than 3 times the measure of an adjacent angle. explain how to find the measures of all the interior angles of the parallelogram.
math
suppose a parallelogram has an area of 84 square units. Describe a triangle related to this parallelogram, and find the triangle's area, base, and height.
Math - vector equation of a parallelogram
The vertices of a parallelogram are the origin and the points A(-1,4), B(3,6), and C(7,2). Write the vector equations of the lines that make up the sides of the parallelogram
Math - vector equation of a parallelogram
The vertices of a parallelogram are the origin and the points A(-1,4), B(3,6), and C(7,2). Write the vector equations of the lines that make up the sides of the parallelogram
Math
If A=<2,1> and B=<-1,1>, find the magnitude and direction angle for -2A +3B Find all specified roots: Cube roots of 8i A parallelogram has sides of length 32.7cm and 20.2cm. If the longer diagonal has a length of 35.1cm, what is the angle opposite this diagonal?
geometry
Quadrilateral QRST has two pairs of congruent sides, but it is not a parallelogram. What figure is it? What further condition would it have to satisfy to be a parallelogram? A. The figure is a square. To be a parallelogram, it would have to have pairs of opposite congruent ...
Geometry
a parallelogram has the vertices (0,3), (3,0), (o,-3), and (-3,0). determine what type of parallelogram and find the perimeter and area
maths
find the area of the parallelogram that has the vectors as adjacent sides u = -2i + j + 5k v = 4i - 3j - 3k a. 11 b. squareroot 26 c. 86 d. 2 squareroot 86 e. 1
maths
ABCD is a parallelogram. AB=9cm,BC=5cm and DE is perpendicular to AB.If the area of parallelogram is 36sqcm,find AE?
maths
ABCD is a parallelogram and AE perpendicular to DC if AB is 20cm. and the area of parallelogram is 80cm. then find the AE which is 90degree?
maths
The measures of two adjacent sides of a parallelogram are in the ratio 17:7. if the second side measures 3.5 cm, find the perimeter of the parallelogram.
maths
the measure of one angle of a parallelogram is 25 degree.find the measures of the other angles of the parallelogram
physics
A container of hot water at 80 deg C cools to 79 deg C in 15 seconds when it is placed in a room that is at 20 deg C. Use Newton's law of cooling to estimate the time it will take for the container to cool from 70 deg C to 69 deg C.
physics
A container of hot water at 80 deg C cools to 79 deg C in 15 seconds when it is placed in a room that is at 20 deg C. Use Newtons law of cooling to estimate the time it will take for the container to cool from 70 deg C to 69 deg C.
Math
Select the geometric figure that possesses all of the characteristics: (1) four equal sides (2) both pairs of opposite sides are parallel (3) it doesn't contain any right angle A) square B) parallelogram C) rhombus D) isosceles triangle I know it's not A or D
geometry
The measures of two consecutive angles of a parallelogram are in the Ratio 3:7. Find the measure of an acute angle of the parallelogram
geometry
The measure of an angle of a parallelogram is 12 degrees less than 3 times the measure of an adjacent angle. find the measure of all the interior angels of the parallelogram.
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# Traveling Salesman Problem
A salesman needs to visit each city in a list of cities and return to his home base. He knows the distance between each pair of cities, and wishes to minimize the total distance he is to travel. What should his path be?
This is an NpComplete problem. If there are n cities, exhaustive search would require (n-1)! trials, and no better method for finding the guaranteed-optimal solution in the worst case is known (nor possible, except in the unlikely event that someone manages to prove that P=NP), but there are many algorithms that are better in many non-worst-case situations.
It's not true that there's no better method than trying every single possibility, even if you insist on guaranteeing that you have the optimal solution. BranchAndBound techniques can help considerably, especially on easier instances. What is true is that there's no way of getting an optimal solution that isn't appallingly slow in the worst case for large problems.
This is similar to the Towers of Hanoi with more than three poles. There are several sites dedicated to this phun piece of math; they'll be posted later.
See "Exponential Lower Bounds for the Towers of Hanoi with More Than Three Pegs", 1998, Mario Szegedy http://citeseer.ist.psu.edu/szegedy98exponential.html
Okay, so someone straighten me, cause I'm bent:
• Start with a weighted connectivity matrix M1 indicating the distances between immediately connected cities. Use *, meaning infinite, when you can't get from one city to another by only a single hop.
• Construct a second matrix M2 thus: for each row i of M1 construct a new row i of M2 thus: for each row connected to i in M1 note the minimum cost of getting to a second city with no loops in the path. Use * for values where you can't get from one city to another in exactly 2 hops or where there's a loop.
• Similarly construct a third matrix M3 from M2 for the minimum exactly 3 hop path costs.
• Do this N-1 times and you have MN-1, the matrix for the N-1 hop minima. Pick the smallest of these minima - there's your TSP minimum.
I'm doing something silly. Help me and point it out. --PeterMerel
For example (check page source to correct formatting due to the WikiBackslashBug...):
``` a - 2 - b
\ / \
1 4 3
\ / \
c - 2 - d
```
gives the weighted connectivity matrix M1:
``` a b c d
a * 2 1 *
b 2 * 4 3
c 1 4 * 2
d * 3 2 *
```
M2:
``` a b c d
a * 5(acb) 6(abc) 3(acd)
b 5 * 3(bac) 6(bcd)
c 6 3 * 7(cbd)
d 3 3 7 *
```
M3:
``` a b c d
a * 6(acdb) * 8(abcd)
b 6 * * 5(bacd)
c * * * 6(cabd)
d 8 5 6 *
```
And the TSP here is bacd, length 5, in a bit less than N^3 trials. Or dcab, natch. Each matrix is just generated from the previous one by adding the minimum necessary to include one extra city in the previous minimum tour. This algorithm would work the same with a directed graph too. So where's the kaboom? -- Stupidly Pete.
Pete, you apply the algorithm N-1 times (or a fixed number of times) therefore you'll think you'll discover a fixed length path (in this case that would be N, as far as I can understand). You assume that the travelling salesman will never do loops (coming back to the same city). Or the poor salesman may occasionally wander off from a city on a dirt country road to reach a remote isolated village to sell his goods the villagers only to come back to the same city to continue the journey, because there's no place to go from that village other than the only town that the village is connected to. (YouCantGetThereFromHere). In other words you tacitly assume that your graph is so wonderfully connected that no loops will ever be necessary.
In less metaphoric terms, try to apply your algorithm to a less connected graph, maybe something like a tree. See how it goes. Here's your Christmas tree:
``` a
/|\
2 1 3
/ | \
b c d
```
From wherever you start you have to visit "a" twice. -- Costin
Quite right Costin, and more generally I woke up realizing that what I described is strictly a HillClimbing algorithm. Lots and lots of ways of fooling it! -- Pete
I've been around this loads of times (this and the four colour theorem). The classic torture test is a bunch of cities on an equal grid e.g. a triangular grid. It isn't hard to see that there is no local information that can help you, therefore there cannot be a local or semi-local algorithm. Thus all combinations may have to be visited. --RichardHenderson.
I have a problem with the above example, which is that taking care of peninsulas is easy and doesn't negate his method. A simple algorithm could "trim" the above tree, as any peninsula is only solvable in one way. In other words, the trimmer would shave off everything but a, at which point the other algorithm could take over. Also also, this method above is called "adjacency matrices," and you can find the path-length-two matrix by squaring the path-length-one matrix; three - cubing; etc.
Cheating way:
1. Create an O(n!) reference solution and generate some moderately-sized problems, and/or use a generator to create problems with known answers.
2. Generate an initial program.
3. Apply each program to the problems.
``` * If wrong answer: dead.
* If right answer: record steps taken.
```
4. Algorithms which took fewer steps are copied more times than algorithms without.
5. For each copy, go through code and roll dice to determine if code should be mutated, and how.
6. Goto 3 if(!bored)
7. /* bored now */ examine most-common algos.
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# Leetcode – Guess Number Higher or Lower Solution in Python
## Introduction
In this article, we will take an in-depth look at “Guess Number Higher or Lower” problem on Leetcode and explore various methods to solve it using Python.
## Problem Statement
We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API int guess(int num), which returns three possible results:
• -1: Your guess is higher than the number I picked (i.e. num > pick).
• 1: Your guess is lower than the number I picked (i.e. num < pick).
• 0: your guess is equal to the number I picked (i.e. num == pick).
Return the number that I picked.
## Analyzing the Problem
The problem requires us to guess a secret number between 1 and n by making calls to an API. The API tells us if our current guess is too high, too low, or correct. Our task is to find the secret number using the least number of guesses. Let’s explore different strategies to approach this problem:
## Approach 1: Linear Search
The simplest approach is to start from 1 and make consecutive guesses until the secret number is found. This method is simple, but very inefficient, especially for large values of n.
# The guess API is already defined.
# @return -1 if my number is lower, 1 if my number is higher, otherwise return 0
# def guess(num: int) -> int:
def guessNumber(n: int) -> int:
for i in range(1, n+1):
if guess(i) == 0:
return i
The time complexity of this approach is O(n), which is not efficient for large values of n.
## Approach 2: Binary Search
A more efficient approach is to use binary search. The binary search algorithm is highly effective in this scenario because the API tells us if our guess is too high or too low. By dividing the search space in half after each guess, we can find the secret number in logarithmic time.
def guessNumber(n: int) -> int:
left, right = 1, n
while left <= right:
mid = left + (right - left) // 2
result = guess(mid)
if result == 0:
return mid
elif result == 1:
left = mid + 1
else:
right = mid - 1
This approach has a time complexity of O(log n), which is a significant improvement over the linear search approach.
## Approach 3: Ternary Search
Ternary search is an advanced version of binary search where the array is divided into three parts instead of two. It can be slightly faster than binary search, as it makes fewer comparisons in the best case.
def guessNumber(n: int) -> int:
left, right = 1, n
while left <= right:
mid1 = left + (right - left) // 3
mid2 = right - (right - left) // 3
res1, res2 = guess(mid1), guess(mid2)
if res1 == 0:
return mid1
elif res2 == 0:
return mid2
elif res1 < 0:
left = mid1 + 1
elif res2 > 0:
right = mid2 - 1
else:
left = mid1 + 1
right = mid2 - 1
The time complexity of ternary search is O(log n), similar to binary search, but with a smaller constant factor.
## Best Practices and Optimization
Binary search and ternary search are the most efficient algorithms for solving this problem. However, binary search is generally preferred in practice because it is simpler and has better cache performance due to accessing contiguous memory locations.
Additionally, it is important to handle edge cases properly. In this problem, there are no edge cases that need to be considered separately as the problem clearly defines the input range.
After implementing the chosen algorithm, it is good practice to test your solution against various test cases to ensure its correctness. Remember to test edge cases like the smallest and largest possible n. Moreover, try to test the algorithm’s efficiency by testing with a large n. | HuggingFaceTB/finemath | |
# Practice Test Explanations: Mathematics
1.Correct Response: D. To solve this problem, you must multiply the number of bottles
capped per second times the total number of seconds in I 5 hours:
(3 bottles/sec) (60 sec/min) (60 min/hr) (15 hr)
l .6 x l05 bottles
Notice that all the units cancel except the number of bottles. Choice A results from
omitting one of the factors of 60. Choice B results from miscounting the number of
decimal places in the answer. Choice C results from interpreting the problem as saying
"3 seconds per bottle" instead of "3 bottles per second."
2.Correct Response A. To solve this problem, first add the total number of gallons of
gas used, then subtract that number from the initial number of gallons:
Change to simple fractions . Find common denominator. Add and reduce . Subtract.
Choice B results from adding incorrectly. Choice C results from adding only the integer
parts of each number. Choice D is the total number of gallons of gas used.
3. Correct Response: C. This is a multistep problem involving subtraction, percents, and
addition. First find how much fencing is left after the morning, then find 40% of this
remaining fencing, then add those two numbers:
Choice A is the percent of fencing that was put up in the morning. Choice B is the-result
of adding the amount of fencing put up in the morning plus 4O% of the amount of
fencing put up in the morning. Choice D is the result of adding the amount of fencing
put up in the morning plus 40% of the original amount of fencing to be put up.
4. Correct Response: B. This problem is solved by using proportions.
Choice A results from subtracting \$I.25 from \$4.00. Choice C results from multiplying
\$1.25 times \$156.00. choice D results from setting up the proportion incorrectly.
5. Correct Response: A. This problem involves taking a percentage of a percent. Eighteen
percent of the advertising expenditures (which is 35% of the total budget) is spent on
Choice B results from calculating I 8% of the non-advertising expenditures. Choice C
is the percentage of total expenditures spent on advertising.
6. Correct Response: A.
Solving this problem involves correctly interpreting the
information on the graph. At time t = 0, the pond is stocked with a certain number of
fish. This eliminates choice D, since on that graph there are no fish in th6 pond initially.
The number of fish steadily decreases. This eliminates choice C, since that graph shows
the number of fish remaining constant through time. (this also eliminates choice D,
since that graph shows the number of fish as increasing through time.) Eventually, there
are approximately half the number of fish originally stocked. This eliminates choice B,
since on that graph, the number of fish is never half the number of fish initially
stocked.
7 . Correct Response: B. To solve this problem, let x equal the test score needed. Then,
since the test counts twice as much as the quizzes, x must appear twice in the
expression for the average:
Simplify Multiply by 5. Subtract 254. Divide by 2.
Choice A results from including x twice, but dividing by 4 instead of 5. (There are 5
grades to be counted: 3 quizzes, and the test grade, twice.) Choice C is the average
the student needs, not the test score. Choice D results from only counting x once, but
dividing by 5.
8.Correct Response: C
. Solving this problem involves correctly interpreting a distribution
graph. The mean of a (symmetric) distribution curve is where it attains its maximum
value. Both curves in the graph have approximately the same mean, which eliminates
choices A and D. The distribution curves give the percentage of cars that have a given
gas mileage, which eliminates choice B. The variability of the data is reflected in the
width of the distribution curve. Since curve A is narrower than curve B, brand A cars
have smaller variability in gas mileage than brand B cars.
9.Correct Response: A. Solving this problem involves correctly interpreting information
from the graph of a line. The slope-intercept form of a line is y = mx + b, where m is
the slope of the line and b is the y-intercept. The y-intercept of the given line is 2, which
eliminates choices B and C. The slope of the line is:
Choice D has slope m = 3. Only choice A fits both criteria.
10. Correct Response: B. The slope of the line through two points is:
Choice A results from calculating
Choice A results from calculating
Choice A results from calculating
I I . Correct Response: B. This problem involves finding the slope of the line, and then
substituting the coordinates of a point on the line to eliminate the incorrect answers.
This eliminates choice A, which has m = 1/2 , and choice C, which has m = -2. Of the
remaining two choices, only choice B passes through both given points:
Substitute -2 for x and 4 for Y. Substitute 6 for x and 0 for Y.
12.Correct Response: B. To solve this problem, you must recognize that the graph
represents an inverse variation relation. lf a rectangle has constant area, then
length x width = constant, or w = constant. The graph passes through the point (3,4)
so that, at that point, = 3 and w = 4. Therefore, the constant for this rectangle is
3 ·4= 12. Now, if =8,then 8w =12 or .Choice A results from incorrectly
estimating the w-value of the graph at = 8. Choice C results from subtracting 12 - 8
instead of dividing l2/8. Choice D results from multiplying (12)(8) instead of dividing
12/8.
13.Correct Response: C. This problem requires determining the value of an unknown in an
equation and using that value to solve a second problem. The first step is to solve the
first equation for x.
Subtract 7 from each side. Multiply each side by - 3.
Now, use x = -27 in the second problem.
Choice C, -53, is correct. Choice A, -5, results from multiplying the right side by -1/3
instead of -3. Choice B, -27, results from solving only the first equation. Choice D, -137,
results from adding rather than subtracting 7 in the first problem-solving step.
14. Correct Response: A. This problem involves finding the value of an unknown in an
equation and expressing it in terms of another unknown. To find the value of t in terms
of r, the first step is to isolate t on one side of the equation.
Subtract 7 from each side. Divide each side by 9.
Choice B results from a sign error when subtracting 7 from both sides. Choice C results
from dividing only one term, rather than both, by 9. Choice D results from both of the
above errors.
I 5. Correct Response: B. This problem involves solving two equations simultaneously,
one of which is quadratic, and the other of which is linear. We may solve this system
of equations by substitution. Since(first equation), substitute
for y in the second equation, and solve that equation for x:
Substitute. Distribute. Simplify. Subtract I 2. Divide by 2. Take the square root.
Now solve for y:
Substitute for x. Simplify. Simplify.
Therefore, when ,and when ,The correct
answer, then, is In choice A, (0,6) is a solution of the
first equation, and (9,9) is a solution of the second equation, but they are not solutions
of both equations simultaneously.
16. Correct Response: C. This problem involves solving two simultaneous equations
in two variables by graphing. The first equation is a line with slope m = -2 and
y-intercept b = -2:
Solve for y. Slope-intercept form of a line. By inspection.
Since the slope is negative , the graph of the line slopes down and to the right. This
eliminates choices A and B, since both lines have positive slopes. The second equation
is a parabola that opens up (since y = ax2 + bx + c with a>0 is a parabola that opens
up). This eliminates choice D. Only choice C fits all of the properties of the two
equations
.
17. Correct Response: B. This problem involves writing algebraic relationships from verbal
descriptions. The sum of x and y is x + y, and the product of x and y is xy. The quotient of
these two expressions is,and the square of this quotient ,. Ten more
than this expression , which is equal to C. Choice A is the square of I 0
more than the quotient of the sum of x and y and the product of x and y. Choice C is
I 0 more than the square of the quotient of the product of x and y and the sum of x and
y. Choice D is l0 more than the quotient of the square of the sum of x and y, and the
product of x and y.
18. Correct Response: A. The difference of m2 and n is m2 - n, and the square of this
expression is (rn2 - n)2. fifty less than this expression is (m2 - n)2 - 50, which equals P.
Choice B is 50 less than the difference of m2 and n. Choice C is the square of 50 less
than the difference of m2 and n. Choice D is the difference of 50 and the square of the
difference of m2 and n.
19. Correct Response C. This problem involves writing an algebraic expression for the
relationships between the lengths of the three pieces of the board. Let the lengths of
the three pieces be a, b, and c. Then, the sum of the lengths of the three pieces is I 0
feet: a + b + c = 10. But the second piece (b) is not as long as the first (a): b =a/2 . And,
the third piece (c) is feet longer than the second (b or a/2 ):
Simplify: Substitute for b. Combine fractions.
Now substitute back into the first equation:
Substitute for band c. Common denominator. Combine fractions. Multiply by 2. Subtract 9. Divide by 4.
Choice A results from using the incorrect relation that the second piece is twice as long
as the first. Choice B results from using the incorrect relation that the third piece is
feet longer than the first piece. Choice D results from dividing the l0-foot board into
three equal pieces.
20. Correct Response A. This problem involves writing relations between different
quantities as a system of three linear equations in three variables. lf we write all money
quantities in cents, then Angel has I 80 cents in his pocket. The total number of cents
in nickels is 5N (5 cents for each nickel, times the number of nickels N), the total
number of cents in dimes is 10D (10 cents for each dime, times the total number of
dimes D), and the total number of cents in quarters is 25Q (25 cents for each quarter
times the number of quarters Q). This gives the first equation: 5N + l0D + 25Q = I 80.
This excludes choices C and D. Since the coefficients on the left side and the number
'on the right side are written in different units (cents and dollars). The second sentence
says that Angel has twice as many quarters as nickels, so that the number of quarters Q
is two times the number of nickels N. Q = 2N. This excludes choice B, in which the
relation is reversed. The third sentence says that the number of dimes D equals the
number of nickels N plus the number of quarters Q plus I : D = N + Q+ l. Only choice A
has all three of these equations.
Correct Response: B. This problem involves writing an equation from the given verbal
information, and solving it. lf \$175 was collected this week, and this was \$35 more
than was collected last week, then \$175 - \$35 = \$140 collected last week. The total
amount of money collected in the two weeks, then, was \$140 + \$I75 = \$315. Choice A
assumes that \$35 was collected in the first week. Choice C assumes that \$35 less was
collected in the second week than in the first week. Choice D assumes that \$70 less was
collected. in the second week than in the first week.
22. Correct Response: C
. To solve this problem, you must factor the quadratic expression .
The factors of 24 are 1 ·24, 2 ·12, 3· 8, and 4 · 6. One of each pair of factors must be
positive and the other negative, since the constant term is negative (-24):
The only pair of factors that sums to -2 (the coefficient of x) is +4, -6:
Choice A is the result of factoring incorrectly. Choices B and D are the results of
choosing the wrong signs to the factors.
23. Correct Response: D. This problem may be solved by inverting the second rational
expression and factoring, then simplifying and multiplying.
lnvert the second expression and change division to multiplication. Factor and cancel. Simplify. Multiply out.
Choice A results from multiplying the two rational expressions instead of dividing.
Choice B results from canceling before inverting the second expression, then inverting
and canceling again. Choice C results from incorrectly canceling and multiplying.
24. Correct Response: A. This problem involves simplifying a rational expression.
Factor 3c out of expressions in numerator and denominator. Cancel 3c.
Choice B results from correctly canceling the c but subtracting coefficients instead
of factoring and canceling. Choice C results from taking the reciprocal of the correct
answer. Choice D results from correctly canceling the c but incorrectly canceling the
coefficients.
25. Correct Response: D. This problem involves dividing and simplifying two rational
expressions, and using the laws of exponents.
lnvert second expression and change division to multiplication. Simplify denominator by adding exponents of common bases. Simplify by subtracting exponents of common bases.
Choice A results from multiplying the two expressions instead of dividing. Choice B
results from correctly inverting the second expression and adding exponents of
common bases in the denominator, but then multiplying exponents instead of
subtracting in the last step. Choice C results from correctly inverting and adding
exponents of common bases in the denominator, but then dividing exponents
instead of subtracting in the last step.
26. Correct Response: C. This problem involves evaluating a function for a particular value
of its variable.
Substitute -1/2 for x. square (-1/2). Simplify and rewrite last term using common denominator of 4. Simplify.
Choice A is f(1/2). Choice B is f(2). Choice D is f(-2).
27.Correct Response: B This problem involves determining the equation of a function
from its graph. The graph is a parabola that opens up so that it has the form
Y = ax2 + bx + c where a is positive. This eliminates Choices A and C. The graph
passes through the points (-2,0), (3,0) (the x-intercepts), and (0,-6) (the y-intercept).
The only choice in which substituting these values for x and y makes the equation
true is choice B.
28.Correct Response: C. This problem involves identifying an inequality from its graph.
To solve this problem first identify the equality. The graph is a parabola opening down,
that passes through the points (0,0), (1,1 ), and (2,0). Only choices B or C, when written
as equalities, satisfy all three of these points when the coordinates are substituted into
the equations. To determine which choice is correct, pick a point inside the shaded
region and substitute the coordinates into the expression, leaving a question mark for
the inequality sign:
substitute (1,0).
Zero is less than one, so choice C is correct.
29. Correct Response: D. This problem requires carrying out a step in solving a quadratic
equation by completing the square. To complete the square, make sure the coefficient
of x2 is l, take half the coefficient of x, square it and place the result in each set of
parentheses.
Half of coefficient of x is 3. Place 9 in each set of parentheses. First three terms are a perfect square .
Choice A results from taking one half of one half the coefficient of x. Choice B results
from taking half the coefficient of x, but not squaring it. Choice C results from
incorrectly squaring 3,
30. Correct Response: B. To solve this problem, you must first put the expression in
standard form and then use the quadratic formula:
Bringx-5toleftside. Coefficients. Quadratic formula. Substitute for a, bi and c.
Choice A results from incorrectly putting the expression in standard form, and uses
3x2 + x - 5 = 0. Choice C results form an incorrect use of the quadratic equation
(b2 + 4ac instead of b2 - 4ac). Choice D results from an incorrect use of the quadratic
equation (4ac - b2 instead of b2 - 4ac).
31. Correct Response: B. This problem involves using the quadratic model to solve for a
particular value of one of the variables. To solve this problem, substitute known height
for h and solve for t:
Substitute | 28 for h. Put in standard quadratic form. Factor out 32. Divide by 32. Factor. Zero factor rule. Add 2 to both sides.
Choice A results from an error made inputting the equation in standard quadratic form.
Choice C results from incorrectly factoring. Choice D results from incorrectly solving
32. Correct Response: B. To solve this problem, the total area of the geometric figure must
be calculated. This is best done piece by piece. The area of the lower rectangular part of
the window is 6 ft. x 4 ft. = 24 square feet. The area of the upper triangular part of the
window is:
square feet.
The total area is the sum of the two sections: 24 square feet + 6 square feet = 3O
square feet.
Choice A is the area of just the rectangular lower section of the window. Choice C is the
area the window would have if it were a rectangle 4 feet wide and 9 feet high. Choice D
is twice the area of the rectangular lower section of the window.
33. Correct Response: C. To solve this problem, you must calculate the perimeter of the
figure, including the two semicircular end caps, and then multiply by 8 for the 8 laps.
There are two straight parts of the track, each 200 m long, for a total of 400 m. Each
end cap is a semicircle, making a full circle of radius 25 m. The circumference of
the circle is The total perimeter of the track, then, is
approximately 400 m + 157 m = 557 m. The runner runs 8 times around, for a total
of 8 x 557 m= 4456 m. Choice A is the approximate length of one lap. Choice B is
the length of 8 laps around a rectangular track measuring 200 m x 50 m. Choice D
counts each end cap as a full circle instead of a semicircle.
34. Correct Response: D.
This problem involves knowing the formula for the surface of a
sphere. The surface area of a sphere of radius r is . Therefore, the surface area of
a hemisphere is half that of a sphere: .tfahemisphereisdividedintol2equal
sections, each section has 1/12 of the area of the hemisphere: . if the radius
of the hemisphere is l8 m, the formula for one section of the hemisphere is
Choice A is the formula for 1/12 the volume of a hemisphere of radius
I 8 m. Choice B is the formula for the entire surface area of a hemisphere of radius
18 m. Choice C is an incorrect formula, since it uses the square of the diameter of the
35. Correct Response: C
. Solving this problem involves using the Pythagorean theorem.
The ramp forms the hypotenuse of a right triangle whose base is the ground, and
whose vertical side is the wall of the platform:
Pythagorean Theorem. Solve for x2. Square root.
Choice A is the difference between the length of the ramp and the height of the
platform. Choice B is the average of the length of the ramp and the height 9f the
platform. Choice D is the hypotenuse of a triangle whose sides have lengths 5 and 12.
36. Correct Response: D. To solve this problem, you must use the properties of similarity
to set up a proportion between two pairs of sides of the two figures. Corresponding
sides of the two figures are proportional:
Choice A results from incorrectly setting up the proportion. Choice B results from
adding the lengths of the two sides of the first figure, and equating that number
to the sum of the lengths of the two sides of the second figure. This is incorrect
because in similar geometric figures, the lengths of corresponding sides are
proportional, but the sides do not necessarily add up to the same lengths. Choice C
is the sum of the lengths of the two given sides of the first figure.
37. Correct Response: C. To solve this problem, you must use the properties of similarity
to set up a proportion between two pairs of sides of the two figures. corresponding
sides of the two figures are proportional:
Choice A results from setting up the proportion incorrectly. Choices B and D result from
assuming that the lengths of the two pairs of sides sum to the same number. In similar
geometric figures, the ratios of corresponding sides are equal, but the lengths of the
sides of the two figures do not necessarily sum to the same number.
38. Correct Response: C. To solve this problem, you must use the properties of parallel
lines. In the figure, angle k and angle w are corresponding angles, so that the measure
of angle k equals the measure of angle w. Angle w and angle y are supplementary
angles, since they lie on the straight line. Supplementary angles add to 180°. Therefore,
the measure of angle k plus the measure of angle y is I 80°.
Choice A results from incorrectly using angle w and angle y as complementary angles.
Choice B results from incorrectly summing supplementary angles to be 100°. Choice D
results from summing angles k, j, m, and n.
39. Correct Response: C. To solve this problem, you need to use the properties of triangles
and of parallel lines. The angles of a triangle add up to 180°. An equilateral triangle has
three equal angles, so that each angle is 60°. Therefore, the measure of angle 3 is 60°.
Since line BC is parallel to line CE, angles 3 and 4 are corresponding angles. The
measure of angle 4 is 60°, since corresponding angles are congruent. Angles 4 and 5
are supplementary angles, since they lie on a straight line. Supplementary angles add
up to 180°. Therefore,
Choice A is the measure of angle 4. Choice B is the measure of a right angle, and
Choice D is the sum of angle 4 and angle 5.
40. Correct Response: A. This problem involves an application of inductive reasoning.
Since Carl has brown hair (by statement 5), Carl must like pizza (by statement 3).
Choice B (Carl has red hair) cannot be true since he has brown hair (by statement 5). We
do not know whether Carl is wearing a hat (choice C). lt is true that Carl has brown hair
(statement 5), but there is no statement saying that all people who have brown hair are
wearing hats. Statement I says that all people wearing hats have brown hair. We do not
know whether Carl likes hamburgers (choice D). There is no statement that says that
people who have brown hair like hamburgers, though there is one (statement 4) that
says that people who have red hair like hamburgers.
41 . Correct Response: A. This problem involves an application of inductive reasoning,
in this case determining the missing design from a pattern sequence. This type of
problem is solved by using the given information to determine how the pattern
changes, and then predicting what the missing design should be. The pattern is discerned
by trial and error. ln this case, the pattern is that the external shape becomes
the internal shape of the design to its right (or vice versa). Since the external shape of
the design to the left of the blank is a diamond, that must be the internal shape of the
correct response. Since the internal shape of the design to the right of the blank is a
semicircle, that must be the external shape of the correct response. Adding together
these two shapes gives choice A. Choices B, C, and D are designs that do not fit in the
pattern sequence.
42. correct Response B. This question involves the application of deductive reasoning
to solve a given problem. This problem is best solved by setting up a matrix using the
persons as one dimension and the jobs as the other.
Writer Researcher Artist Engineer Bess Tara Gerard Clifton
Then, using the given statements, fill in the matrix. Since the information is in the
form of who is not what (e.9., Bess and Gerard are not the engineer), put an X in the
appropriate places to rule out the jobs each person cannot hold. For statement l, Bess
and Gerard cannot be the engineer.
Writer Researcher Artist Engineer Bess x Tara Gerard x Clifton
Add statement ll, which indicates that Clifton and Tara cannot be the researcher.
Writer Researcher Artist Engineer Bess x Tara x Gerard x Clifton x
Finally, add statement lll, which indicates that Gerard cannot be the writer or the
researcher.
Writer Researcher Artist Engineer Bess x Tara x Gerard x x x Clifton x
The only empty cell under researcher corresponds with Bess; therefore, choice B must
be correct.
43. Correct Response: C. This problem involves an application of inductive reasoning,
in this case determining the missing design from a pattern sequence. This type of
problem is solved by using the given information to determine how the pattern
changes, and then predicting what the missing pattern should be. In this case, the
designs are polygons with different numbers of sides, so the first step is to write
down the number of sides in each polygon:
This is two sequences mixed together. Every other term in the sequence increases by
one:
The missing number is 8, so the missing polygon is the one with 8 sides.
44. Correct Response: B. Solving this problem involves using the Pythagorean Theorem
twice.
The largest straight distance in the box is always a diagonal through the middle of
the box. This diagonal forms the hypotenuse of a triangle whose side is the edge,
and whose base is the diagonal of the bottom of the box.
To find the diagonal of the bottom of the box, we find the diagonal of a 2-foot by 3-foot
rectangle using the pythagorean Theorem:
Diagonal of bottom of boxfeet
To find the length of the diagonal through the middle of the box we use the
Pythagorean Theorem again:
feet
Choice A is the length of the diagonal in the bottom of the box. Choice C results from
adding the lengths of the three sides of the box. Choice D results from not taking the
square root of a2 + b2.
45. Correct Response A. This is a multi-step problem in which you must calculate each
contribution to the monthly cost of each apartment. For apartment A, rent is \$575 per
month. This apartment uses 250 kilowatt-hours of electricity, at 5G per kilowatt-hour,
for a total of (250 kwh) x (\$0.05/kwh) = \$ 12.50. The total cost of commuting from
apartment A is (\$ 1 .50/day) x (30 days) = \$45.00. Therefore, the total monthly cost of
apartment A is \$575.00 + \$12.50 + \$45.00 = \$632.50. For apartment B, rent is \$600
per month; electricity is (225 kilowatt-hours/month) x (\$0.05/kilowatt-hour) = \$11.25;
and commuting cost is (\$0.75/day) x (30 days) = \$22.50. Apartment B's total monthly
cost is therefore \$600.00 + \$11.25 +22.50 = \$633.75. Therefore, apartment A is
cheaper.
46. Correct Response: C. To solve this problem, you must calculate the volume of each
tank, multiply by 3 for the three tanks to get the total capacity of the plant in gallons,
and then multiply by 7.5 to get the total number of gallons.
For tank1 Volume of a cylinder.
Substitute.
For 3 tanks,
Each cubic foot contains approximately 7.5 gallons, so we multiply to get the total
number of gallons:
Choice A results from dividing by 7.5 instead of multiplying. Choice B is the number of
gallons in one tank. Choice D results from incorrectly placing the decimal point.
47, Correct Response: C.
To solve this problem, you must calculate the number of square
feet of surface area exposed to the air in all three tanks and multiply by the amount of
water that evaporates per square foot per day.
Surface Area
Number of gallons evaporating per day=
Choice A is the number of gallons evaporating per hour. Choice B is the number of
gallons evaporating per day from one tank. Choice D results from incorrectly placing
the decimal point.
48. Correct Response: D.
This problem involves using the data in a table to write an
equation that models the data. Since the depth is decreasing as time increases, the
slope of this linear relation is negative. This eliminates choice C. When time equals
zero, the depth is 25 feet. Substituting these values into the equations eliminates A,
since the values do not make that equation true. At time equals I hour, the depth is
I 7 feet. When we substitute these values into the equations, only equation D is
satisfied.
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# Autograph Newsletter 14 – Geometric Proof
Welcome Welcome to the fourteenth Autograph Newsletter! Each jam-packed edition looks at a specific topic in mathematics and how Autograph can help engage students and enable them to understand the key concepts better.
Introduction Just like in Newsletter 12, in this edition we draw our inspiration from the wonderful Median maths blog by Don Steward. Don’s activities challenge students to investigate, make conjectures and solve problems, and they have formed the basis of some of my most enjoyable lessons. Several of the activities on Median revolve around geometric proof. This is a topic that many students can have difficulty with, possibly due to its abstract nature, and being able to see the whole picture in terms of the steps that are required to solve the problem. Here is where I feel Autograph can help. By setting up the problems and then playing around with them dynamically, students may be able to see links and relationships that were previously hidden, make further conjectures, and test out their proofs.
Diagnostic Question Usually we have a Diagnostic Question in this spot, but for this special edition of the newsletter we are instead I wanted to show you the kind of activity Don has on his website. Often these are images which can be copied and used directly, or pasted onto a PowerPoint slide or Word file. Here is a nice way to introduce students to the word of geometric proof using angles in triangles.
Free Online Autograph Activity Can you Prove it 3? A tricky old construction here that leads to the surprising result that these two angles appear to be always equal. The question is… can you prove it? These Autograph activities do not require the full version of Autograph to run them. You just need to install the free Autograph Player (you will be guided through how to do this), which means you can use these activities in the classroom or set them for your students to do at home.
Further Activities
The following ideas for activities are also taken from Don Steward’s website. Try them on paper first and then turn to Autograph to look at them in more depth. Click on the image to download the individual Autograph files.
Activity 1 – The Equilateral House
Challenge: A square has an equilateral triangle constructed on each of two of its adjacent sides. The top corner of each triangle is connected to each other and to the far corner of the square. Can you prove that the resulting shape is an equilateral triangle?
• The situation has been modelled on Autograph • Experiment by moving the positions of the three corners of the triangle around the page • Convince yourself that the triangle is equilateral by measuring the sides or the angles: – Measure sides: click on a side and the distance will be displayed in the status bar at the bottom of the screen – Measure angles: double click on the angle arc and place a tick by Show Label • Does this help you prove that the triangle is equilateral?
Activity 2 – The Mysterious Yellow Triangle
Challenge: Draw any right angled triangle, bisect an angle and construct the perpendicular to the hypotenuse (an altitude). Can you prove that the yellow triangle is isosceles.
• The situation has be modelled using Autograph • Experiment by dragging the three corners of the right angled triangle around the page • Does this help you prove that the yellow triangle must be isosceles?
Activity 3 – Isosceles Split | HuggingFaceTB/finemath | |
Question
Formatted question description: https://leetcode.ca/all/154.html
Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.
The array may contain duplicates.
@tag-array
Algorithm
When there are a large number of repeated numbers in the array, the mechanism of binary search will be destroyed, the time complexity of O(lgn) will not be obtained, and the simple and crude O(n) will be returned.
For example, these two situations: {2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 2} and {2, 2, 2, 0, 2, 2, 2, 2, 2, 2, 2, 2}, it can be found that when the first number, the last number, and the middle number are all equal, the binary search method collapses, because it cannot determine whether to go to the left half or the right half.
In this case, move the right pointer one place to the left (or move the left pointer one place to the right) to skip the same number, which will not affect the result, because it just removes the same one, and then affects the remaining part Continue to use the binary search method,
In the worst case, such as all elements of the array are the same, the time complexity will rise to O(n).
Code
•
public class Find_Minimum_in_Rotated_Sorted_Array_II {
public class Solution_II {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int low = 0; // left pointer
int high = nums.length - 1; // right pointer
// while (low < high) {
while (low < high && nums[low] >= nums[high]) { // @note:@memorize: if no "=", failed by [3,1,3]
int mid = low + (high - low) / 2;
System.out.println("mid: " + nums[mid] + ", high: " + nums[high]);
if (nums[mid] > nums[high]) {
low = mid + 1;
} else if (nums[mid] < nums[low]) { // @note: in part-I, only else, no check, but if duplicates, then not working
high = mid; // @note: here, not "high=mid-1"
} else {
right--;
}
}
return nums[low];
}
}
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int low = 0; // left pointer
int high = nums.length - 1; // right pointer
// while (low < high && nums[low] > nums[high]) {
while (low < high && nums[low] >= nums[high]) { // @note:@memorize: if no "=", failed by [3,1,3]
int mid = low + (high - low) / 2;
System.out.println("mid: " + nums[mid] + ", high: " + nums[high]);
if (nums[mid] > nums[high]) {
low = mid + 1;
} else if (nums[mid] < nums[high]) { // @note: in part-I, only else, no check, but if duplicates, then not working
high = mid; // @note: here, not "high=mid-1"
} else {
high--;
}
}
return nums[low];
}
}
}
############
class Solution {
public int findMin(int[] nums) {
int left = 0, right = nums.length - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else if (nums[mid] < nums[right]) {
right = mid;
} else {
--right;
}
}
return nums[left];
}
}
• class Solution:
def findMin(self, nums: List[int]) -> int:
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) >> 1
if nums[mid] > nums[right]:
left = mid + 1
elif nums[mid] < nums[right]:
right = mid
else:
right -= 1
return nums[left]
############
class Solution(object):
def findMin(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
ans = nums[0]
start, end = 0, len(nums) - 1
while start + 1 < end:
mid = start + (end - start) / 2
if nums[start] < nums[mid]:
start = mid
elif nums[start] > nums[mid]:
end = mid
else:
start += 1
ans = min(ans, nums[start])
return min(ans, nums[start], nums[end])
• class Solution {
public:
int findMin(vector<int>& nums) {
int left = 0, right = nums.size() - 1;
while (left < right) {
int mid = (left + right) >> 1;
if (nums[mid] > nums[right])
left = mid + 1;
else if (nums[mid] < nums[right])
right = mid;
else
--right;
}
return nums[left];
}
};
• func findMin(nums []int) int {
left, right := 0, len(nums)-1
for left < right {
mid := (left + right) >> 1
if nums[mid] > nums[right] {
left = mid + 1
} else if nums[mid] < nums[right] {
right = mid
} else {
right--
}
}
return nums[left]
}
• function findMin(nums: number[]): number {
let left = 0,
right = nums.length - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else if (nums[mid] < nums[right]) {
right = mid;
} else {
--right;
}
}
return nums[left];
}
• /**
* @param {number[]} nums
* @return {number}
*/
var findMin = function (nums) {
let left = 0,
right = nums.length - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else if (nums[mid] < nums[right]) {
right = mid;
} else {
--right;
}
}
return nums[left];
}; | HuggingFaceTB/finemath | |
Definition
An additive semiring is a semiring with a commutative distributand.
That is, an additive semiring is a ringoid $\left({S, *, \circ}\right)$ in which:
$(1): \quad \left({S, *}\right)$ forms a commutative semigroup
$(2): \quad \left({S, \circ}\right)$ forms a semigroup.
An additive semiring is an algebraic structure $\left({R, *, \circ}\right)$, on which are defined two binary operations $\circ$ and $*$, which satisfy the following conditions:
$(A0)$ $:$ $\displaystyle \forall a, b \in S:$ $\displaystyle a * b \in S$ Closure under $*$ $(A1)$ $:$ $\displaystyle \forall a, b, c \in S:$ $\displaystyle \left({a * b}\right) * c = a * \left({b * c}\right)$ Associativity of $*$ $(A2)$ $:$ $\displaystyle \forall a, b \in S:$ $\displaystyle a * b = b * a$ Commutativity of $*$ $(M0)$ $:$ $\displaystyle \forall a, b \in S:$ $\displaystyle a \circ b \in S$ Closure under $\circ$ $(M1)$ $:$ $\displaystyle \forall a, b, c \in S:$ $\displaystyle \left({a \circ b}\right) \circ c = a \circ \left({b \circ c}\right)$ Associativity of $\circ$ $(D)$ $:$ $\displaystyle \forall a, b, c \in S:$ $\displaystyle a \circ \left({b * c}\right) = \left({a \circ b}\right) * \left({a \circ c}\right), \left({a * b}\right) \circ c = \left({a \circ c}\right) * \left({a \circ c}\right)$ $\circ$ is distributive over $*$
These criteria are called the additive semiring axioms.
Note on Terminology
The term additive semiring was coined by $\mathsf{Pr} \infty \mathsf{fWiki}$ to describe this structure.
Most of the literature simply calls this a semiring; however, on $\mathsf{Pr} \infty \mathsf{fWiki}$ the term semiring is reserved for more general structures, not imposing that the distributand be commutative. | HuggingFaceTB/finemath | |
# Electronic Warfare and Radar Systems Engineering Handbook- Decibel (dB) -
DECIBEL (dB)
The Decibel is a subunit of a larger unit called the bel. As originally used, the bel represented the power ratio of 10 to 1 between the strength or intensity i.e., power, of two sounds, and was named after Alexander Graham Bell. Thus a power ratio of 10:1 = 1 bel, 100:1 = 2 bels, and 1000:1 = 3 bels. It is readily seen that the concept of bels represents alogarithmic relationship since the logarithm of 100 to the base 10 is 2 (corresponding to 2 bels), the logarithm of 1000 to the base 10 is 3 (corresponding to 3 bels), etc. The exact relationship is given by the formula
Bels = log(P2/P1)
where P2/P1 represents the power ratio.
Since the bel is a rather large unit, its use may prove inconvenient. Usually a smaller unit, the Decibel or dB, is used. 10 decibels make one be. A 10:1 power ratio, 1 bel, is 10 dB; a 100:1 ratio, 2 bels, is 20 dB. Thus the formula becomes
Decibels (dB) = 10 log(P2/P1)
The power ratio need not be greater than unity as shown in the previous examples. In equations and , P1 is usually the reference power. If P2 is less than P1, the ratio is less then 1.0 and the resultant bels or decibels are negative. For example, if P2 is one-tenth P1, we have
bels = log(0.1/1) = -1.0 bels
and dB = 10 log(0.1/1) = -10 dB.
It should be clearly understood that the term decibel does not in itself indicate power, but rather is a ratio or comparison between two power values. It is often desirable to express power levels in decibels by using a fixed power as a reference. The most common references in the world of electronics are the milliwatt (mW) and the watt. The abbreviation dBm indicates dB referenced to 1.0 milliwatt. One milliwatt is then zero dBm. Thus P1 in equations or becomes 1.0 mW. Similarly, The abbreviation dBW indicates dB referenced to 1.0 watt, with P2 being 1.0 watt, thus one watt in dBW is zero dBW or 30 dBm or 60 dBuW. For antenna gain, the reference is the linearly polarized isotropic radiator, dBLI. Usually the 'L' and/or 'I' is understood and left out.
dBc is the power of one signal referenced to a carrier signal, i.e. if a second harmonic signal at 10 GHz is 3 dB lower than a fundamental signal at 5 GHz, then the signal at 10 GHz is -3 dBc.
THE DECIBEL, ITS USE IN ELECTRONICS
The logarithmic characteristic of the dB makes it very convenient for expressing electrical power and power ratios. Consider an amplifier with an output of 100 watts when the input is 0.1 watts (100 milliwatts); it has an amplification factor of
P2/P1 = 100/0.1 = 1000 or a gain of:
10 log(P2/P1) = 10 log(100/0.1) = 30 dB.
(notice the 3 in 30 dB corresponds to the number of zeros in the power ratio)
The ability of an antenna to intercept or transmit a signal is expressed in dB referenced to an isotropic antenna rather than as a ratio. Instead of saying an antenna has an effective gain ratio of 7.5, it has a gain of 8.8 dB (10 log 7.5).
A ratio of less than 1.0 is a loss, a negative gain, or attenuation. For instance, if 10 watts of power is fed into a cable but only 8.5 watts are measured at the output, the signal has been decreased by a factor of
8.5/10 = 0.85
or
10 log(0.85) = -0.7 dB.
This piece of cable at the frequency of the measurement has a gain of -0.7 dB. This is generally referred to as a loss or attenuation of 0.7 dB, where the terms "loss" and "attenuation" imply the negative sign. An attenuator which reduces its input power by factor of 0.001 has an attenuation of 30 dB. The utility of the dB is very evident when speaking of signal loss due to radiation through the atmosphere. It is much easier to work with a loss of 137 dB rather than the equivalent factor of 2 x 10-14.
Instead of multiplying gain or loss factors as ratios we can add them as positive or negative dB. Suppose we have a microwave system with a10 watt transmitter, and a cable with 0.7 dB loss connected to a 13 dB gain transmit antenna. The signal loss through the atmosphere is 137 dB to a receive antenna with a 11 dB gain connected by a cable with 1.4 dB loss to a receiver. How much power is at the receiver? First, we must convert the 10 watts to milliwatts and then to dBm:
10 watts = 10,000 milliwatts and
10 log (10,000/1) = 40 dBm
Then
40 dBm - 0.7 dB + 13 dB - 137 dB + 11 dB - 1.4 dB = -75.1 dBm.
dBm may be converted back to milliwatts by solving the formula: mW = 10(dBm/10)
giving 10(-75.1/10)= 0.00000003 mW
Voltage and current ratios can also be expressed in terms of decibels, provided the resistance remains constant. First we substitute for P in terms of either voltage, V, or current, I. Since P=VI and V=IR we have:
P = I2R = V2/R
Thus for a voltage ratio we have
dB = 10 log[(V22/R)/(V12/R)] = 10 log(V22/V12) = 10 log(V2/V1)2 = 20 log(V2/V1)
Like power, voltage can be expressed relative to fixed units, so one volt is equal to 0 dBV or 120 dBuV. Similarly for current ratio dB = 20 log(I2/I1)
Like power, amperage can be expressed relative to fixed units, so one amp is equal to 0 dBA or 120 dBA.
Decibel Formulas (where Z is the general form of R, including inductance and capacitance)
When impedances are equal: When impedances are unequal: SOLUTIONS WITHOUT A CALCULATOR
Solution of radar and EW problems requires the determination of logarithms (base 10) to calculate some of the formulae. Common "four function" calculators don't usually have a log capability (or exponential or fourth root functions either). Without a scientific calculator (or math tables or a Log-Log slide rule) it is difficult to calculate any of the radar equations, simplified or "textbook". The following gives some tips to calculate a close approximation without a calculator.
DECIBEL TABLE For dB numbers which are a multiple of 10
An easy way to remember how to convert dB values that are a multiple of 10 to the absolute magnitude of the power ratio is to place a number of zeros equal to that multiple value to the right of the value 1.
i.e. 40 dB = 10,000 : 1 (for Power)
Minus dB moves the decimal point that many places to the left of 1. i.e. -40 dB = 0.0001 : 1 (for Power)
For voltage or current ratios, if the multiple of 10 is even, then divide the multiple by 2, and apply the above rules. i.e. 40 dB = 100 : 1 (for Voltage)
-40 dB = 0.01 : 1
If the power in question is not a multiple of ten, then some estimation is required. The following tabulation lists some approximations, some of which would be useful to memorize. You can see that the list has a repeating pattern, so by remembering just three basic values such as one, three, and 10 dB, the others can easily be obtained without a calculator by addition and subtraction of dB values and multiplication of corresponding ratios.
Example 1:
A 7 dB increase in power (3+3+1) dB is an increase of (2 x 2 x 1.26) = 5 times whereas
A 7 dB decrease in power (-3-3-1) dB is a decrease of (0.5 x 0.5 x 0.8) = 0.2.
Example 2: Assume you know that the ratio for 10 dB is 10, and that the ratio for 20 dB is 100 (doubling the dB increases the power ratio by a factor of ten), and that we want to find some intermediate value. We can get more intermediate dB values by adding or subtracting one to the above, for example, to find the ratio at 12 dB we can:
work up from the bottom; 12 = 1+11 so we have 1.26 (from table) x 12.5 = 15.75 alternately, working down the top 12 = 13-1 so we have 20 x 0.8(from table) = 16.
The resultant numbers are not an exact match (as they should be) because the numbers in the table are rounded off. We can use the same practice to find any ratio at any other given value of dB (or the reverse).
dB AS ABSOLUTE UNITS
Power in absolute units can be expressed by using 1 Watt (or1 milliwatt)as the reference power in the denominator of the equation for dB. We then call it dBW or dBm. We can then build a table such as the adjoining one. From the above, any intermediate value can be found using the same dB rules and memorizing several dB values i.e. for determining the absolute power, given 48 dBm power output, we determine that 48 dBm = 50 dBm - 2 dB so we take the value at 50 dB which is 100W and divide by the value 1.58 (ratio of 2 dB) to get: 100 watts/1.58 = 63 W or 63,291 mW.
Because dBW is referenced to one watt, the Log of the power in watts times 10 is dBW. The Logarithm of 10 raised by any exponent is simply that exponent. That is: Log(10)4 = 4. Therefore, a power that can be expressed as any exponent of 10 can also be expressed in dBW as that exponent times 10. For example, 100 kW can be written 100,000 watts or 105 watts. 100 kW is then +50 dBW. Another way to remember this conversion is that dBW is the number of zeros in the power written in watts times 10. If the transmitter power in question is conveniently a multiple of ten (it often is) the conversion to dBW is easy and accurate. About RF Cafe Copyright: 1996 - 2024Webmaster: Kirt Blattenberger, BSEE - KB3UON RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ... All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged. My Hobby Website: AirplanesAndRockets.com | HuggingFaceTB/finemath | |
# 912Geometry_Midpoint_LessonPlan_Construct_Talk-Frame–Midpoint
A template of the talk frame was used to plan a geometry lesson on finding the midpoint of a segment. The teacher recorded a paraphrased version of the question in the Think section, anticipated ideas to be contributed in the Talk Ideas section, which include using a ruler, folding paper, and solving algebraically to find a midpoint, and the pieces of information to be learned in the We Understand section. Students are asked to construct an argument on finding the midpoint of a line segment and use argumentation language to convince another person.
Microsoft Word version: 912Geometry_Midpoint_LessonPlan_Construct_Talk-Frame-Midpoint
# 912Geometry_SuppAngles_LessonPlan_Construct
A template of the talk frame was used to plan a geometry lesson on supplementary angles. The teacher recorded a paraphrased version of the question in the Think section, what ideas would be contributed in the Talk Ideas section, and the realization that linear pairs are formed at the intersection of two lines. Students are asked to name pairs of supplementary angles and construct an argument to convince another person that the claim in true.
Microsoft Word version: 912Geometry_SuppAngles_LessonPlan_Construct
PDF version: 912Geometry_SuppAngles_LessonPlan_Construct
# 912Algebra_CED_Functions_LessonPlan
A template of the talk frame was used to plan a lesson for algebra students learning about composite functions. Students are given a problem in which they make a weekly salary, plus commission, two functions, and two possible function compositions to represent the commission. The Think section provides the question, along with two additional parts, and the Talk Idea section presents the possible realizations about the functions. Argumentation language is used through the construction of the student solution.
Microsoft Word version: 912Algebra_CED_Functions_LessonPlan
PDF version: 912Algebra_CED_Functions_LessonPlan
# 912Algebra_REI_LinearEquations_LessonPlan
This is a lesson plan designed around a word problem. A word problem provides two points that students must recognize as ordered pairs in a linear relationship and use this information to predict a future point in the relation. Students must be able to interpret the word problem and find slope from the two points. The lesson plan offers three different approaches to solving the problem. These can be used differently based on the level of students in a class. One recommendation is to group students and have each group solve the problem using a different approach and be able to explain to the class. You may also want students to use each approach themselves and determine how each may work best in different situations.
Microsoft Word version: 912Algebra_REI_LinearEquations_LessonPlan
PDF version: 912Algebra_REI_LinearEquations_LessonPlan
# 912Algebra_REI_LinearEquations_LessonPlan_1
This talk frame was created for algebra students developing skills with solving systems of equations. Students should be comfortable with the three different methods for solving systems before you use this talk frame in class. A system is presented and students are asked to decide which method to use (linear combination, substitution, or graphing). The talk frame suggests looking at each method individually and discussing what the process would be like. The goal of the task is for students to understand that any method may be used to solve systems of equations, but some methods may be more efficient.
Microsoft Word version: 912Algebra_REI_LinearEquations_LessonPlan_1
PDF version: 912Algebra_REI_LinearEquations_LessonPlan_1
# 912Algebra_REI_RadicalEquations_LessonPlan_TalkFrameModel–ExtraneousSolutions
A template of the talk frame was used to plan a lesson for algebra students learning about radical equations. Students are given two equations, one with a radical, and need to determine when they intersect and what methods were used by the sample responses. In the Think section, the teacher recorded possible ways the scenario could be reworded, methods used to solve the problem are discussed in the Talk Idea section, and the We Understand section addresses the realization that extraneous solutions need to be checked and how to find such solutions.
Microsoft Word version: 912Algebra_REI_RadicalEquations_LessonPlan_TalkFrameModel-ExtraneousSolutions
# 912Algebra_LE_ExponentialFunctions_LessonPlan
This task is designed for algebra students studying exponential functions. Students are given two functions to compare and contrast through a discussion and a mystery function that can be deduced from four facts. Students can use argumentation language while determining the mystery function and while answering the final question, on which expression represents a larger value.
Microsoft Word version: 912Algebra_LE_ExponentialFunctions_LessonPlan
PDF version: 912Algebra_LE_ExponentialFunctions_LessonPlan
# 912Algebra_CED_LinearFunctions_LessonPlan
A template of the talk frame was used to plan a lesson for algebra students learning about linear equations. The teacher recorded what he anticipated how students would reword the question in the Think section, what ideas they would contribute in the Talk Ideas section, and both their realization that the problem has multiple solutions and the valid mathematical ideas in the We Understand section. The mathematics focused on modeling with linear equations, creating linear equations from a word problem, and the break-even point. Specifically, students are given the fixed price and price per night of two dog kennels and must determine when the two will be equal.
Microsoft Word version: 912Algebra_CED_LinearFunctions_LessonPlan
PDF version: 912Algebra_CED_LinearFunctions_LessonPlan | HuggingFaceTB/finemath | |
CHISQ.DIST.RT: Excel Formulae Explained
The CHISQ.DIST.RT function in Excel is a statistical function that can be used to calculate the right-tailed probability of the chi-square distribution. This function is particularly useful in hypothesis testing and in statistical analyses that involve variables with a chi-square distribution.
Understanding the CHISQ.DIST.RT Function
The CHISQ.DIST.RT function is part of Excel's suite of statistical functions. It is used to calculate the right-tailed probability of the chi-square distribution. The chi-square distribution is a theoretical distribution that is used in hypothesis testing, particularly in tests involving categorical variables.
The CHISQ.DIST.RT function takes two arguments: x and degrees of freedom. The x value is the value at which you want to evaluate the distribution, and degrees of freedom typically relate to the number of categories in the test.
Understanding the Chi-Square Distribution
The chi-square distribution is a theoretical distribution that is used in hypothesis testing. It is a special case of the gamma distribution and is one of the most commonly used probability distributions in inferential statistics, notably in hypothesis testing and in construction of confidence intervals.
Chi-square tests are often used in research to test the independence of two variables. In other words, a chi-square test can be used to test whether two categorical variables are independent of each other or not.
Understanding Degrees of Freedom
The concept of degrees of freedom is a fundamental concept in statistics that applies to several different types of statistical tests. In the context of the chi-square test, degrees of freedom typically relate to the number of categories in the test.
Degrees of freedom can be thought of as the number of values in a statistical calculation that are free to vary. In a chi-square test, the degrees of freedom are equal to the number of categories minus one.
Using the CHISQ.DIST.RT Function in Excel
The CHISQ.DIST.RT function is relatively straightforward to use in Excel. It requires two arguments: x and degrees of freedom.
The syntax for the CHISQ.DIST.RT function is: CHISQ.DIST.RT(x, degrees_freedom).
Example of CHISQ.DIST.RT Function
Let's consider an example where we want to calculate the right-tailed probability of a chi-square distribution with 6 degrees of freedom at x = 0.05. The formula we would use in Excel is: =CHISQ.DIST.RT(0.05, 6).
When this formula is entered into Excel, it will return the right-tailed probability of the chi-square distribution. This value can be used in hypothesis testing to determine whether to reject or fail to reject the null hypothesis.
Interpreting the Results of the CHISQ.DIST.RT Function
The result of the CHISQ.DIST.RT function is a probability that represents the likelihood that a chi-square statistic would be greater than the x value, assuming the null hypothesis is true. This probability is often compared to a significance level, such as 0.05, to determine whether to reject the null hypothesis.
If the result of the CHISQ.DIST.RT function is less than the significance level, then the null hypothesis is rejected. If the result is greater than the significance level, then the null hypothesis is not rejected.
Example of Interpreting CHISQ.DIST.RT Function
Let's continue with our previous example where we calculated the right-tailed probability of a chi-square distribution with 6 degrees of freedom at x = 0.05. Let's say that the result of the CHISQ.DIST.RT function was 0.02.
Since 0.02 is less than 0.05, we would reject the null hypothesis. This means that we have evidence to suggest that the variables in our test are not independent of each other.
Conclusion
The CHISQ.DIST.RT function in Excel is a powerful tool for conducting statistical analyses involving the chi-square distribution. By understanding how this function works and how to interpret its results, you can conduct more robust and reliable statistical analyses in your research or data analysis projects.
Take Your Data Analysis Further with Causal
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# hw5sol - Stat 310A/Math 230A Theory of Probability Homework...
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Stat 310A/Math 230A Theory of Probability Homework 5 Solutions Andrea Montanari Due on November 4, 2010 Exercises on the law of large numbers and Borel-Cantelli Exercise [2.1.5] Let > 0 and pick K = K ( ) finite such that if k K then r ( k ) . Applying the Cauchy-Schwarz inequality for X i - E X i and X j - E X j we have that Cov( X i , X j ) [Var( X i )Var( X j )] 1 / 2 r (0) < for all i, j . Thus, breaking the double sum in Var( S n ) = n i,j =1 Cov( X i , X j ) into { ( i, j ) : | i - j | < K } and { ( i, j ) : | i - j | ≥ K } gives the bound Var( S n ) 2 Knr (0) + n 2 . Dividing by n 2 we see that lim sup n Var( n - 1 S n ) . Since > 0 is arbitrary and E S n = n x , we have that n - 1 S n L 2 x (with convergence in probability as well). Exercise [2.1.13] We have E | X 1 | = k =2 1 / ( ck log k ) = . On the other hand, for n N n P ( | X 1 | ≥ n ) = n c X k = n 1 k 2 log k n c Z n - 1 1 x 2 log x d x = n c Z log( n - 1) e - z z d z n c log( n - 1) Z log( n - 1) e - z d z = n c ( n - 1) log( n - 1) .
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# Evaluating $\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$
Evaluate
$$\lim_{n\to\infty}\left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right)$$
$$\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} = e^{-2} - \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$ Hence, $$1 - e^2 \left(\sum_{k=0}^{n} \dfrac{(-2)^k}{k!} \right) = e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)$$ Hence, $$f(n) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot \left(1-e^2 \sum_{k=0}^{n} \frac{(-2)^k}{k!}\right)\right) = \left(\frac{n \cdot n!}{e\cdot (-2)^{n+1}}\cdot e^2 \left( \sum_{k=n+1}^{\infty} \dfrac{(-2)^k}{k!}\right)\right)$$ $$f(n) = e n \cdot n! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right) = \dfrac{n}{n+1} \underbrace{e \cdot (n+1)! \left( \sum_{k=0}^{\infty} \dfrac{(-2)^k}{(k+n+1)!}\right)}_{g(n)}$$ $$g(n) = e \left(1 + \dfrac{(-2)}{(n+2)} + \dfrac{(-2)^2}{(n+2)(n+3)} + \cdots \right)$$ Hence, $\lim g(n) = e$ and hence $\lim f(n) = e$.
• @Chris'ssister What trick exactly are you talking about? – user17762 Nov 21 '12 at 19:51
• @Chris'ssister I recognized that $$\sum_{k=0}^{n} \frac{(-2)^k}{k!}$$ was the leading order term of $e^{-2}$. I don't think I would categorize it as a trick. More of recalling it from memory. – user17762 Nov 21 '12 at 19:56
• I emphasized a different thing and not the fact that $\sum_{k=0}^{n} \frac{(-2)^k}{k!}$ is the leading order term of $e^{-2}$. Anyway. – user 1357113 Nov 21 '12 at 20:15 | HuggingFaceTB/finemath | |
# How is a magnetic moment created
## Magnetic dipole moment
The magnetic moment\$ \ vec {m} \$ (also magnetic dipole moment) is a measure of the strength of a magnetic dipole in physics and is defined analogously to the electric dipole moment.
A torque acts on a magnetic dipole in an external magnetic field of the flux density \$ \ vec {B} \$
\$ \ vec D _ {\ vec m} = \ vec m \ times \ vec B \ ,, \$[Note 1]
by rotating it in the direction of the field (\$ \ times \$: cross product). Its potential energy is therefore dependent on the setting angle \$ \ theta \$ between field direction and magnetic moment:
\$ E _ {\ text {pot}} = - \ vec m \ cdot \ vec B \ equiv -m \, B \ cos \ theta. \$
Important examples are the compass needle and the electric motor.
The unit of measurement of the magnetic moment in the International System of Units (SI) is A m2. Often the product of \$ \ vec {m} \$ and the magnetic field constant \$ \ mu_0 \$ is used (see note 1); this has the SI unit T · m3.
The current density distribution \$ \ vec {\ jmath} \, (\ vec r) \$ has a magnetic moment
\$ \ vec {m} = \ frac {1} {2} \ int \ limits _ {\ mathbb {R} ^ 3} \ mathrm {d} ^ 3r \ left [\ vec {r} \ times \ vec {\ jmath } \, (\ vec {r}) \ right]. \$
This results in a flat current loop
\$ m = I \ cdot A, \$
where \$ A \$ is the area surrounded by the current \$ I \$.
In electrical engineering, this is the basis for z. B. Generators, motors and electromagnets.
• Particles with an intrinsic angular momentum (spin) \$ \ vec s \$ have a magnetic moment
\$ \ vec {m} = \ gamma \ vec {s}. \$
\$ \ gamma \$ is called the gyromagnetic ratio. Examples are electrons, which cause the ferromagnetism of the elements of the iron group and the rare earths by the parallel positioning of their magnetic moments. Ferromagnetic materials are used as permanent magnets or as iron cores in electromagnets and transformers.
### Level conductor loop
The following applies to a closed conductor loop
\$ \ int \ vec {\ jmath} \, (\ vec {r}) \; \ mathrm {d} ^ 3r = \ int I \; \ mathrm {d} \ vec {r}. \$
Here designated
• \$ \ vec {\ jmath} \, (\ vec {r}) \$ is the current density at location \$ \ vec {r} \$
• \$ \ int \ mathrm {d} ^ 3r \$ a volume integral
• \$ I \$ the current intensity through the conductor loop
• \$ \ int \ mathrm {d} \ vec {r} \$ a path integral along the conductor loop.
Thus it follows for the magnetic dipole moment:
\$ \ vec {m} = \ frac {I} {2} \ int_C (\ vec {r} \ times \ mathrm {d} \ vec {r}) = I \ cdot \ vec {A} = I \ cdot A \ cdot \ vec {n} _A \$
with the normal vector \$ \ vec {n} _A \$ on the flat surface \$ A \$.
#### Long coil with current flowing through it
The magnetic moment of a coil with current flowing through it is the product of the number of turns \$ n \$, current strength \$ I \$ and area \$ A \$:
\$ \ vec {m} = n \ cdot I \ cdot \ vec {A}. \$
Here \$ \ vec {A} = \ vec {n} _A A \$ is the vector belonging to the area \$ A \$.
### Charged particle on a circular path
#### Classic
If the circular current is caused by a particle with the mass \$ M \$ and the charge \$ Q \$ circling on a circular path (radius \$ r \$, period of rotation \$ T \$), this formula results
\$ \ vec {m} = IA \; \ vec {n} _A = \ frac {Q} {T} \ cdot \ pi r ^ 2 \; \ vec {n} _A \ equiv \ frac {Q} {2M} \ vec L \ quad. \$
The magnetic moment is therefore fixed with the angular momentum
\$ \ vec {L} = \ omega M r ^ 2 \; \ vec {n} _A \$
connected. The constant factor \$ \ gamma = \ tfrac {Q} {2M} \$ is the gyromagnetic ratio for moving charges on the circular path. (The angular velocity \$ \ omega = \ tfrac {2 \ pi} {T} \$ is used for the conversion.)
#### Quantum mechanics
The classical formula plays a major role in atomic and nuclear physics, because it also applies in quantum mechanics, and a well-defined angular momentum belongs to every energy level of an individual atom or nucleus. Since the angular momentum of the spatial movement (orbital angular momentum, in contrast to the spin) can only be integer multiples of the unit \$ \ hbar \$ (Planck's quantum of action) [Note 2], also has the magnetic one Orbit moment a smallest unit, the magneton:
\$ \ mu = \ frac {Q \ hbar} {2M} \ quad. \$
If the elementary charge \$ e \$ is substituted for \$ Q \$, the result for the electron is the Bohr magneton \$ \ mu_B = \ tfrac {e \ hbar} {2m_e} \$, for the proton the nuclear magneton \$ \ mu_K = \ tfrac { e \ hbar} {2m_p} \$. Since the proton mass \$ m_p \$ is almost 2000 times greater than the electron mass \$ m_e \$, the nuclear magneton is smaller than Bohr's magneton by the same factor. Therefore, the magnetic effects of the atomic nuclei are very weak and difficult to observe (hyperfine structure).
### The magnetic moment of particles and nuclei
Particles and atomic nuclei with a spin \$ \ vec {s} \$ have a magnetic spin moment \$ \ vec {\ mu} _s \$, which is parallel (or antiparallel) to their spin, but has a different size in relation to the spin than when it came from an orbital angular momentum of the same size. This is expressed by the anomalous Landé factor of the spin \$ g_s \ mathord {\ ne} 1 \$. One writes for electron (\$ e ^ - \$) and positron (\$ e ^ + \$)
\$ \ vec \ mu_s = g_e \, \ mu_B \, \ frac {\ vec {s}} {\ hbar} \$ with Bohr's magneton \$ \ mu _B \$,
for proton (p) and neutron (n)
\$ \ vec \ mu _s = g_ {p, n} \, \ mu_K \, \ frac {\ vec {s}} {\ hbar} \$ with the nuclear magneton \$ \ mu _K \$,
and analogously for other particles. For the muon, in Bohr's magneton, instead of the mass of the electron, that of the muon is used, for the quarks their respective constituent mass and third-digit electrical charge. If the magnetic moment is antiparallel to the spin, the g-factor is negative. However, this sign convention is not applied consistently, so that the g-factor z. B. the electron is indicated as positive.[Note 3]
Particle Spin-g factor
Electron \$ e ^ - \$\$ -2{,}002\,319\,304\,361\,82(52) \$[1]
Muon \$ \ mu ^ - \$\$ -2{,}002\,331\,8418(13) \$[2]
Proton \$ p \$\$ +5{,}585\,694\,702(17) \$[3]
Neutron \$ n \$\$ -3{,}826\,085\,45(90) \$[4]
The numbers in brackets indicate the estimated standard deviation.
According to the Dirac theory, the Landé factor of the fundamental fermions is exactly \$ g_s \ mathord = \ pm 2 \$, quantum electrodynamically a value of about \$ g_s \ mathord = \ pm 2 {,} 0023 \$ is predicted. Precise measurements on the electron or positron as well as on the muon are in excellent agreement, including the predicted small difference between electron and muon, and thus confirm the Dirac theory and quantum electrodynamics. The strongly deviating g-factors for the nucleons can be explained by their structure of three constituent quarks, albeit with deviations in the percentage range.
If the particles (e.g. electrons that are bound to an atomic nucleus) also have an orbital angular momentum, then the magnetic moment is made up of the magnetic moment of spin (\$ \ vec {\ mu} _s \$) and that of orbital angular momentum considered above (\$ \ vec {\ mu} _ \ ell \$) composed:
\$ \ vec {\ mu} = \ vec {\ mu} _s + \ vec {\ mu} _ \ ell \$.
### Magnetic field of a magnetic dipole
A magnetic dipole \$ \ vec {m} \$ at the origin of the coordinates leads to a magnetic flux density at the location \$ \ vec {r} \$
\$ \ vec {B} (\ vec {r}) \, = \, \ frac {\ mu_0} {4 \ pi} \, \ frac {3 \ vec {r} (\ vec {m} \ cdot \ vec {r}) - \ vec {m} r ^ 2} {r ^ 5} \$.
Here \$ \ mu_0 \$ is the magnetic field constant. Except at the origin, where the field diverges, everywhere both the rotation and the divergence of this field vanish. The associated vector potential is given by
\$ \ vec {A} (\ vec {r}) \, = \, \ frac {\ mu_0} {4 \ pi} \, \ frac {\ vec {m} \ times \ vec {r}} {r ^ 3} \$,
where \$ \ vec {B} = \ nabla \ times \ vec {A} \$.
### Force effect between two dipoles
The force exerted by dipole 1 on dipole 2 is
\$ \ vec {F} = \ nabla \ left (\ vec {m} _2 \ cdot \ vec {B} _1 \ right) \$
It surrenders
\$ \ mathbf {F} (\ vec {r}, \ vec {m} _1, \ vec {m} _2) = \ frac {3 \ mu_0} {4 \ pi r ^ 4} \ left [\ vec {m } _2 (\ vec {m} _1 \ cdot \ vec {r} _n) + \ vec {m} _1 (\ vec {m} _2 \ cdot \ vec {r} _n) + \ vec {r} _n (\ vec {m} _1 \ cdot \ vec {m} _2) - 5 \ vec {r} _n (\ vec {m} _1 \ cdot \ vec {r} _n) (\ vec {m} _2 \ cdot \ vec { r} _n) \ right], \$
where \$ \ vec {r} _n \$ is the unit vector that points from dipole 1 to dipole 2 and \$ r \$ is the distance between the two magnets. The force on dipole 1 is reciprocal.
### Torque effect between two dipoles
The torque exerted by dipole 1 on dipole 2 is
\$ \ vec {M} = \ vec {m} _2 \ times \ vec {B} _1 \$
where \$ \ vec {B} _1 \$ is the field generated by dipole 1 at the location of dipole 2 (see above). The torque on dipole 1 is reciprocal.
In the presence of several dipoles, the forces or moments can be superimposed. Since soft magnetic materials form a field-dependent dipole, these equations cannot be used.
### literature
• John David Jackson: Classical electrodynamics. Appendix about units and dimensions. 4th edition. De Gruyter, Berlin 2006, ISBN 3-11-018970-4.
### Remarks
1. ↑ In older books, e.g. B. W. Döring, Introduction to Theoretical Physics, Göschen Collection, Volume II (Electrodynamics), is called magnetic moment defines the \$ \ mu_0 \$ -fold of the value given here. Then it says z. B. \$ \ vec D _ {\ vec m} = \ vec m \ times \ vec H \$ and \$ \ vec m \$ is not defined as magnetization through volume, but as magnetic polarization \$ \ vec J \, \, (= \ mu_0 \ vec M) \$ by volume. In matter there is generally \$ \ vec B = \ mu_0 \ cdot \ vec H + \ vec J \$ and \$ \ vec m \ times \ vec J \ equiv 0 \$ (because of \$ \ vec M \ times \ mu_0 \, \ vec M \ equiv 0 \,. \$) Old and new definitions are therefore fully equivalent. The official agreement on the new CODATA definition did not take place until 2010.
2. ↑ More precisely: this applies to the component of the angular momentum vector along an axis.
3. ↑ Exactly the sign only has to be taken into account when it comes to the direction of rotation of the Larmor precession or the sign of the paramagnetic spin polarization. Therefore, the signs are not handled in a completely uniform way in the literature.
1. CODATA Recommended Values. National Institute of Standards and Technology, accessed July 28, 2015. Value for the g-factor of the electron
2. CODATA Recommended Values. National Institute of Standards and Technology, accessed July 28, 2015. Value for the muon g-factor
3. CODATA Recommended Values. National Institute of Standards and Technology, accessed July 28, 2015. Value for the g-factor of the proton
4. CODATA Recommended Values. National Institute of Standards and Technology, accessed July 28, 2015. Value for the g-factor of the neutron | HuggingFaceTB/finemath | |
# $1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2$: reason? [duplicate]
We have $$1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2$$ as we can establish by induction. But why does this hold? Can we connect it to something else?
## marked as duplicate by Grigory M, egreg, user85798, Dan Rust, Andrew D. HwangApr 11 '14 at 23:42
• See my answer here. – J.R. Apr 11 '14 at 19:27
• @YourAdHere Is there a higher dimensional generalization of this identity ? – Amr Apr 11 '14 at 19:31
• @Amr In case you mean for a higher exponent than $3$, I don't think so, there is a general formula to sum the first $k$th powers and it doesn't take such a simple form for higher $k$. – J.R. Apr 11 '14 at 19:35
• @YourAdHere Yes I meant higher exponents. I said "dimensional" because your proof considers squares and I was asking if there is a similar argument that considers cubes for example ... – Amr Apr 11 '14 at 19:39
• Related to this question – robjohn Apr 11 '14 at 20:22
Meanwhile, it generalizes to Liouville's $$\sum_{k | n} \left( d(k) \right)^3 = \left( \sum_{k | n} d(k) \right)^2$$
Here $d(k)$ is the number of divisors of a positive integer, with $d(1)=1.$ For a prime $p,$ we get $$d(p^w) = w+1.$$
The identity works because it works for a prime power, that is what the original summation formula shows. Next, both sides are number theoretic "multiplicative." A multiplicative function $f(n)$ is one that applies to integers, and which has this condition: whenever $\gcd(a,b) = 1,$ we have $f(ab) = f(a) f(b).$ Any multiplicative function is completely determined by its values on prime powers. Oh, if $f(n)$ is a multiplicative function, then $$g(n) = \sum_{k|n} f(k)$$ is also multiplicative. That requires a little proof, double sum sort of thing.
• Fantastic! I had never seen this identity of Liouville. – Bruno Joyal Apr 11 '14 at 20:49
Funnily, we also have
$$\int_0^x t^3\: dt = \left(\int_0^x t\: dt\right)^2.$$
• +1 But let's see the proof! – Aaron Hall Apr 11 '14 at 20:58
• @AaronHall The funny thing is that the integrals $\int_0^x t^n$ were first calculated using Faulhaber polynomials. It's quite possible that the identity of the OP was known before the value of the integral $\int_0^x t^3$! – Bruno Joyal Apr 11 '14 at 21:00
• The summation version goes back to antiquity, but integration is new, and I have no idea what Faulhaber polynomials are. They sound impressive. pops corn – Aaron Hall Apr 11 '14 at 21:01
• @AaronHall en.wikipedia.org/wiki/… :) – Bruno Joyal Apr 11 '14 at 21:02
There's a famous proof by C. Wheatstone. http://en.wikipedia.org/wiki/Squared_triangular_number | HuggingFaceTB/finemath | |
An inequality from the handbook of mathematical functions (by Abramowitz and Stegun)
Prove that
$$\frac{1}{x+\sqrt{x^2+2}}<e^{x^2}\int\limits_x^{\infty}e^{-t^2} \, \text dt \le\frac{1}{x+\sqrt{x^2+\displaystyle\tfrac{4}{\pi}}}, \space (x\ge 0)$$
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Just an idea... by making the change of variables $t^2=s^2+x^2$ we get $$e^{x^2}\int_x^{\infty}e^{-t^2} \ dt = \int_0^\infty \frac{s e^{-s^2}}{\sqrt{s^2+x^2}}\,ds,$$ which at least gets rid of the exponential outside and gives us a square root to work with. – Antonio Vargas Dec 15 '12 at 21:21
As an addendum to my previous comment, it might be possible to get the lower bound by using the fact that the new integrand is unimodal and has a maximum when $$s^2 = \frac{x}{x+\sqrt{x^2+2}}.$$ – Antonio Vargas Dec 15 '12 at 21:39
@AntonioVargas: thanks for your idea. – Chris's sis the artist Dec 16 '12 at 22:12
(The following argument is adapted from Dümbgen, ''Bounding Standard Gaussian Tail Probabilities.'')
Approximating $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$
Suppose we want to approximate $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ with a function of the form $\dfrac{e^{-x^2}}{h(x)}.$ Let $$\Delta(x) = \frac{e^{-x^2}}{h(x)} - \int_x^{\infty} e^{-t^2} \, dt.$$
Then, if $h(x) \to \infty$ as $x \to \infty$, then $\Delta(x) \to 0$ as $x \to \infty$. Because of this, we have the following.
• If $\Delta'(x) > 0$ for all $x \geq 0$ then $\Delta(x)$ increases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is a lower bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.
• Similarly, if $\Delta'(x) < 0$ for all $x \geq 0$ then $\Delta(x)$ decreases to $0$. Therefore, $\dfrac{e^{-x^2}}{h(x)}$ is an upper bound on $\displaystyle \int_x^{\infty} e^{-t^2} \, dt$ for $x \geq 0$.
We have $$\Delta'(x) = \frac{e^{-x^2}}{h(x)^2} \left(h(x)^2 - 2xh(x) - h'(x) \right).$$ Thus the sign of $\Delta'(x)$ is determined by the sign of $f(x) = h(x)^2 - 2xh(x) - h'(x)$.
Given the bounds we're trying to show, let's consider functions of the form $h(x) = x + \sqrt{x^2 + c}$. Then $$f(x) = c - 1 - \frac{x}{\sqrt{x^2+c}}.$$ Thus $f(x)$ is decreasing on $[0, \infty)$.
The lower bound
To have $f(x) > 0$ for all $x \geq 0$, we need $$c > 1 + \frac{x}{\sqrt{x^2+c}}, \:\:\:\: x \geq 0.$$ The smallest value of $c$ for which this holds is $c = 2$. Therefore, $$\frac{1}{x + \sqrt{x^2+2}} < e^{x^2} \int_x^{\infty} e^{-t^2} \, dt, \:\:\:\: x \geq 0,$$ and $2$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.
The upper bound
To have $f(x) < 0$ for all $x \geq 0$ we can take $c = 1$. However, we can do better this because $f(x)$ is decreasing. If we find a larger value of $c$ such that $\Delta(0)= 0$, then we will have $f(x) > 0$ on $[0, x_0)$ for some $x_0$ and then $f(x) < 0$ on $(x_0, \infty)$. Thus $\Delta(x)$ will initially increase from $0$ and then decrease back to $0$, giving us a tighter upper bound. Since $$\Delta(0) = 0 \Longleftrightarrow \frac{1}{\sqrt{c}} = \int_0^{\infty} e^{-t^2}\, dt = \frac{\sqrt{\pi}}{2},$$ we have $c = \dfrac{4}{\pi}$ yielding a tighter upper bound than $c = 1$. Therefore, $$e^{x^2} \int_x^{\infty} e^{-t^2} \, dt \leq \frac{1}{x + \sqrt{x^2+\frac{\pi}{4}}}, \:\:\:\: x \geq 0,$$ and $\dfrac{\pi}{4}$ is the smallest value of $c$ for which this bound holds for functions of the form $h(x) = x + \sqrt{x^2 + c}$.
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a simple and beautiful solution. Thanks! – Chris's sis the artist Dec 16 '12 at 21:31
@Chris'ssister: You're quite welcome. It was a fun problem to think about, and I learned some interesting things myself in the process. – Mike Spivey Dec 16 '12 at 21:35
wonderful answer! – nbubis Dec 16 '12 at 21:43
Incidentally, the best bounds of the form $$\frac{\alpha}{\gamma x + \sqrt{x^2 + \beta}}$$ (that are tight at both zero and infinity) were given in A. V. Boyd. Inequalities for Mills’ ratio. Rep. Stat. Appl. Res., JUSE, 6(2):1–3, 1959. They have an equally elegant and simple proof. – cardinal Dec 16 '12 at 23:10
Here is a version of the Dümbgen technical report on the arXiv, just as an independent link that may be more stable. – cardinal Dec 16 '12 at 23:26
The integral is equal to: $$\frac{\sqrt{\pi}}{2}\text{erfc}(x)$$ If you calculate the series expansion of the above function around infinity, you get: $$e^{-x^2}\left(\frac{1}{2x}-\frac{1}{4x^3}+\frac{3}{8x^5}+...\right)$$ I believe looking ad the the resultant fractional expression and the series expansion of the radicals should give you a clue, at least for $x\gg1$.
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thanks for your suggestion. (+1) – Chris's sis the artist Dec 16 '12 at 22:13 | HuggingFaceTB/finemath | |
# Mean value thorem - Showing $\sqrt{1+x} < 1+\frac{x}{2}$ for $x>0$
So as the title states I have to show $\sqrt{1+x} < 1+\frac{x}{2}$ for $x>0$.
This is a example from the book which has to be explained for me, i'm having a hardtime understanding the proof. I do however understand the concept of MVT.
So the rest of the solution looks like the following:
If $x>0$, apply the Mean-Value Theorem to $f(x)= \sqrt{1+x}$ on the interval $[0,x]$. There exist $c\in [0,x]$ such that $$\frac{\sqrt{1+x}-1}{x}=\frac{f(x)-f(0)}{x-0}=f'(c)=\frac{1}{2\sqrt{1+c}}<\frac{1}{2}$$ The last inequality hold because $c>0$. Mulitiplying by the positive number $x$ and transposing the $-1$ gives $\sqrt{1+x} <1+\frac{x}{2}$,
So I am not sure why he(the author) choose $\frac{1}{2}$, it seems a little arbitrary to me. My guess is that you're allowed to pick a number for the derivative of $c$ which suits the cause/solution best, as long as it's $0<c<x$. I'm not sure though.
All help would be greatly appriciated!
• MVT is definitely the hard way to prove the inequality. Heaps simpler to just square both sides! Nov 30, 2017 at 1:08
$\frac{1}{2}$ is not a random number, it comes from differentiating the square root: $\sqrt{x}\,'=\frac{1}{2\sqrt{x}}$ or $\sqrt{x+1}\,'=\frac{1}{2\sqrt{x+1}}$, the same thing.
Next he noticed that $\frac{1}{\sqrt{1+c}}<\frac{1}{\sqrt{1+0}}=1$ for all $c>0$, which is quite obvious after simple transformation.
• I see. But what's the next step then. i.e the proof? Is the point then to make $\frac{1}{2sqrt{1+c}}<\frac{1}{2}$ look like $\sqrt{1+x} <1+\frac{x}{2}$? Nov 30, 2017 at 1:37
• He obtained the inequality $\frac{\sqrt{1+x}-1}{x}<\frac{1}{2}$. This one is equivalent to the inequality you were given to prove. Dec 1, 2017 at 15:17
Just square both sides.
$(1+x/2)^2 = 1+x+x^2/4 \gt 1+x$ unless $x = 0$.
Also works for $(1+x/n)^n \gt 1+x$ so $1+x/n \gt \sqrt[n]{1+x}$ (in this case, use Bernoulli's inequality).
• Yeah I'd like to, but im trying to learn how to prove inequalities using MVT. :/ Nov 30, 2017 at 2:30
• If $c < 1/2$ then $(1+cx)^2 = 1+2cx+c^2x^2=1+x-(1-2c)x+c^2x^2 < 1+x$ for small enough $x$. Therefore $1/2$ is the smallest value that works. Nov 30, 2017 at 2:48
$\frac12$ is not arbitrary at all.
The derivative of a square root is $\sqrt{u}'=\frac{u'}{2\sqrt{u}}$ put $u=x+1$ you get $\frac{1}{2\sqrt{1+x}}$ replace the $x$ with the value from the mean value theorem you get $\frac{1}{2\sqrt{1+c}}$
Look at the case where he choose more than $\frac12$, in this case you lose some of the information, because if it is less then $\frac12$ then it is less from something bigger.
In the case he choose something smaller, then I can take $\lim\limits_{c\to0}\frac1{2\sqrt{1+c}}\longrightarrow\frac12$, which means I can get infinitely close to $\frac12$ hence I can find a value for $c$ that will make the statement false for every number less then $\frac12$
Thus we left with the number $\frac12$ | HuggingFaceTB/finemath | |
# The Various +1 Sequences of Bible Numbers
God is One and He used His Oneness to create something I call His One Concept. The One Concept in the Bible is revealed by the fact that God is One in all parameters. God used His One Concept to create a numbering system to highlight the number One. God increased the value of the number One by adding the number One to each number to create a One base numbering system for counting.
God designed Bible numbers using a "number + 1 " sequence model.
# Number +1
Here. the number on the left can change and be any combination of +1 sequences as a total number . However, the number on the right (+1) remains the same. This way the +1 of the sequence is separated. In the Bible, it is the +1 that represents the supernatural and is always highlighted as the number in a number.
# Number +1
All of the bible numbers after the number One is based upon this model.
1+1=2
2+1=3
3+1=4 and so forth
This is the type or numbering that Yahweh used extensively throughout the Bible. It is the sequencing of multiple "numbers + 1" in a number that reveals how God designed a Biblical event. Again, the Bible's "number +1" sequences form the framework for many of the events in the Bible.
The following table contains examples of the various combinations of numbers that God used to arrive at other numbers in the Bible.
Bible
Numbers
Bible Numbers
as +1 Sequences
Combinations of Bible Numbers +1 Sequences Multiplication of Bible Numbers +1 Sequences
1 (1 x 10)
(1 x 1000)
2
3 2+1
(3x2)
(3x4)
(3 x 1000)
4 3+1 (Jacob-4x3)
(Elijah-4x3)
(Noah-4x10)
(Abraham-4x100)
5 4+1 3+2=5
6 5+1 3+3=6
7 6+1 ((2+1) + (2+1) + 1=7)
(3+4=7)
8 7+1
10 5 + 5 = 10
7 + 3=10
11 10+1
12 11+1 1+4+7=12 (12x10)
(12x12)
13 12+1 Jericho 6+7=13
14 (7x2)
(14x3)
21 (Passoverr-7x3)
(Tabernacles-7x3)
(Noah-7x3)
37 Throne-36+1
40
42 (Generations of Christ-7x6)
50 Jubilee-49+1 (Feast of Weeks-7x7)
70 (20+20+20 +10 =70)
(Babylon-7x10)
120 (Years of Life-12x10)
144 Walls of New Jerusalem 12x12
300 100x3
400 (40x10)
1250 Eternal Temple Area measurements
1260 Time Times + a Half of Time
1335 Time Times + a Half of Time and 1335 Days
2300 Temple and 2300 Days
5000 (100x50 cubits) Court of Tabernacle
7000 (Baal Example-7 x 1000)
12,000 (City and Tribes-12 x 1000)
25,000 x 25,000
### Holy Area for the Temple and the Levites
144,000 (Entourage of the Lamb-12,000 x 12,000= 144,00o)
(Sealing the Tribes-12,000 x 12,000= 144,00o) | HuggingFaceTB/finemath | |
# Bridge parabola word problem
1. Apr 3, 2009
### SAC2009
1. The problem statement, all variables and given/known data
A new bridge is being constructed. The space between the support needs to be 1050 feet; the height at the center of the arch needs to be 350 feet. An empty tanker needs a 280 foot clearance to pass beneath it.
a) find the equation of a parabola with these characteristics. Place the vertex at the origin.
b) How wide is the channel that the tanker can pass through?
c) If the river were to flood and rise 10 feet, how would the clearance of the bridge be affected?
2. Relevant equations
The equation for a vertical parabola is y=a(x-h)squared + k
3. The attempt at a solution
I tried putting the vertex at (0,0) and plugging in (525,350) to get the a value. I dont know if this is anywhere close to what I was supposed to do.
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Apr 3, 2009
### CompuChip
Personally I find the equation
y = A (x - p)(x - q)
more useful, when you know where the parabola has to be zero.
Do you see why?
3. Apr 3, 2009
### HallsofIvy
Staff Emeritus
And if you Place the vertex at the origin, y(x)= ax2+ k. "The space between the support needs to be 1050 feet" so y(1050/2)= y(525)= 0 (NOT "526,350). "the height at the center of the arch needs to be 350 feet" so y(0)= 350. That is enough to find a and k.
In order for the tanker to pass through, the height must be at least 280 feet. What values of x make y= 280?
4. Apr 3, 2009
### SAC2009
I thought that if the vertex was at the origin (0,0) that the equation would be y=axsquared, because the vertex is (h,k). I keep getting an extremely small number for k, which wouldn't work because then the width at that point would be way too small.
5. Apr 3, 2009
### SAC2009
sorry, i meant an extremely small number for a
6. Apr 3, 2009
### HallsofIvy
Staff Emeritus
You are right. I put the origin at (0, 350).
Now, either y= kx2 or y= -x2 depending on whether you want the parabola opening upward or downward. The second would better represent a support for a bridge but, mathematically, either way works.
If you want to use y= kx2 then the bases of the support are at (525, 350) and (-525, 350) and you would need, to allow the tanker to pass, to solve y= x2= 280. | HuggingFaceTB/finemath | |
# NCERT Solutions for Class 9 Maths [Updated for 2021-22]
NCERT solutions for class 9 maths comprise the analysis and explanation of all the problems covered by the NCERT textbook for 9th grade. The solutions are provided in a detailed manner to ensure that students can understand concepts as they go through them. These class 9 maths NCERT solutions are prepared by some of the most accomplished experts in the field of Mathematics.
NCERT solutions for class 9 maths are extremely helpful for students preparing for CBSE exams and state or national competitive tests such as Olympiads. The CBSE examinations follow the structure provided by the NCERT textbooks. Additionally, CBSE NCERT solutions for class 9 maths help in getting students ready to face class 10th and the corresponding board exams.
Practicing these class 9 maths NCERT solutions can help students to increase their theoretical skills and practical knowledge. The solutions to all NCERT class 9 maths sums are available for free pdf download. You can find tips and tricks included with the solutions. The links to these pdf solutions are given below.
### Class 9 Maths NCERT Solutions Chapter 1 to 15
As a student, you must indulge in constant revision and practice all the solutions. Thus, revisiting the above-mentioned links periodically is a must. Start by attempting the questions in the NCERT book and then have a look at these class 9 maths NCERT solutions. You can download the NCERT textbook given below and begin your learning journey.
## NCERT Solutions for Class 9 Maths Chapter 1
NCERT solutions for class 9 maths chapter 1 number systems help students amass the tools required to identify rational and irrational numbers, locate real numbers on a number line, and convert a real number to its decimal form. The solutions to this chapter also give students a good understanding of simplifying expressions using certain identities and laws of exponents.
Important Formulas:
• √(ab) = √a √b
• √(a/b) = √a/√b
• (√a + √b)(√a – √b) = a – b
• (√a + √b)2 = a + b + 2√(ab)
### Class 9 Maths Chapter 1 Real Numbers
Topics Covered: NCERT solutions for class 9 maths cover questions based on rationalizing the denominator, the rules of exponentiation, identities applied to positive real numbers, expanding a real number into its decimal form, and vice versa.
Total Questions: Chapter 1 has a total of 27 questions, out of which 5 are easy, 7 are moderate and 10 are difficult.
## NCERT Solutions for Class 9 Maths Chapter 2
NCERT solutions for class 9 maths chapter 2 polynomials help students form the basics of polynomials that will be used in higher classes. Students learn about the types of polynomials in one variable, degree and zeros of a polynomial, the remainder theorem, factorization of polynomials, and the application of algebraic identities.
Important Formulas:
• (x + y)2 = x2 + 2xy + y2
• x2 – y2 = (x + y) (x – y)
• (x + a) (x + b) = x2 + (a + b)x + ab
• (x – y)2 = x2 – 2xy + y2
• (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
### Class 9 Maths Chapter 2 Polynomials
Topics Covered: Factorization of different polynomials using the splitting method and factor theorem, finding the remainder of a polynomial, applying algebraic identities to simplify sums, and determining the zeros of a polynomial.
Total Questions: Class 9 maths chapter 2 polynomials have 33 questions. 10 are of a moderate level, 14 are easy and 9 are challenging sums.
## NCERT Solutions for Class 9 Maths Chapter 3
NCERT solutions for class 9 maths chapter 3 coordinate geometry help students to get a grasp on how to describe the position of an object using coordinates and explanation of the cartesian system. It acts as a foundation for further chapters that require students to use their knowledge to solve geometrical problems.
### Class 9 Maths Chapter 3 Coordinate Geometry
Topics Covered: Determining the x and y coordinates of a point in a cartesian plane, identifying quadrants, locating an object, and expressing it in the form of coordinates are the topics under class 9 maths NCERT solutions.
Total Questions: There are 6 questions in class 9 maths chapter 3 coordinate geometry. 3 are difficult, 2 are medium level and 1 is an easy question.
Class 9 Maths Chapter 3 Summary:
Class 9 maths NCERT solutions chapter 3 gives a brief overview of the definition of the origin, how to use the horizontal x-axis and vertical y-axis to pinpoint the position of an object. It also helps students determine what quadrant does a point lies in depending upon the signs of the coordinates.
## NCERT Solutions for Class 9 Maths Chapter 4
NCERT solutions for class 9 maths chapter 4 linear equations in two variables introduce students to a linear equation in two variables by helping them understand the concepts behind it. The solutions include solving linear equations, graphical representation methods, and the depiction of parallel lines.
Linear equations are one of the most important chapters covered in class 9 maths NCERT solutions. The following are a few tips that can help you power through these solutions quickly and efficiently.
• Build an understanding of the concepts
• Revision of the topic at regular intervals
• Understanding the general form of a line
### Class 9 Maths Chapter 4 Linear Equations in Two Variables
Topics Covered: Finding solutions to a linear equation by substituting values, representing a linear equation in two variables graphically, and how to write the equation of lines parallel to the x-axis.
Total Questions: Chapter 4 linear equations in two variables have a total of 16 questions. 8 require students to carefully think and attempt the problem, 5 are moderate and 4 are easy sums.
## NCERT Solutions Class 9 Maths Chapter 5
NCERT solutions class 9 maths chapter 5 introduction to Euclid’s geometry gives students an insight into the definitions of points, lines, surfaces, and so on. Students also get an idea of the 5 postulates given by Euclid that are used to solve geometry problems.
### Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry
Topics Covered: Euclid’s five postulates dealing with line segments, radii, perpendicular lines, and the variation of the fifth postulate are the topics covered under NCERT solutions class 9 maths.
Total Questions: Chapter 5 introduction to Euclid’s Geometry has 9 questions, 4 are simple, 3 are intermediate and 2 are tough sums.
The chapter deals with five postulates given by Euclid regarding a straight line, a terminating line, a circle, and its radius, as well as right angles. Students also learn about the two equivalent versions of Euclid’s fifth postulate that led to the discovery of other theorems.
## Class 9 Maths NCERT Solutions Chapter 6
Class 9 maths NCERT solutions chapter 6 lines and angles give students an understanding of the basic definitions of various terms that are used in geometry. Students learn about the different types of angles, pairs of angles, transversals, parallel lines, angle sum property of triangles, and axioms associated with these concepts.
### Class 9 Maths NCERT Solutions Chapter 6 Preparation Tips
There are a number of properties taught in this chapter that students must learn. The following points can help in streamlining this process.
• Practice several questions on properties of angles that get formed when a transversal intersects 2 or 3 parallel lines.
• Understand the reasoning behind theorems such as the angle sum property of a triangle.
• Make a note of the relationship between pairs of angles.
### Class 9 Maths Chapter 6 Lines and Angles
Topics Covered: Adjacent angles, vertically opposite angles, linear pairs of angles, alternate angles, corresponding angles, parallel lines, and transversal are topics falling under Class 9 maths NCERT solutions. In addition to this, problems based on the angle sum property of a triangle are also included in these solutions.
Total Questions: There are a total of 18 questions in class 9 maths chapter 6 lines and angles. 10 are very easy property-based questions, 5 require some thinking on a student’s part and 3 are difficult sums.
## NCERT Solutions for Class 9 Maths Chapter 7
NCERT solutions for class 9 maths chapter 7 triangles are a fantastic way for students to learn about the congruence of triangles using the SAS, SSS, ASA, RHS criteria. Several properties of triangles along with concepts on the inequality of triangles also fall under the scope of this chapter.
### Class 9 Maths Chapter 7 Triangles
Topics Covered: Questions based on applying properties of triangles, proving the congruence of two triangles using different methods, and theorems on the inequality of triangles are the topics covered.
Total Questions: Class 9 maths chapter 7 has 31 questions. 10 are simple questions based on applying the properties of triangles, 9 are complex sums and 12 are moderate problems.
Chapter Summary:
In this chapter, students learn about the SSS, SAS, ASA, and RHS congruence criteria. Students can get a good grasp on properties of triangles such as sides opposite to equal angles are equivalent, the sum of any two sides of a triangle is greater than the third side, and so on. Thus, after studying this chapter students can master the basic concepts of a triangle.
## NCERT Solutions for Class 9 Maths Chapter 8
NCERT solutions for class 9 maths chapter 8 quadrilaterals comprise questions that teach students about the different types of quadrilaterals such as squares, rhombuses, rectangles, parallelograms, and their properties.
### Class 9 Maths Chapter 8 Quadrilaterals
Topics Covered: Angle sum property of quadrilaterals, properties of a parallelogram, square, and rhombus that depend upon different factors such as length of diagonals, the measure of sides, etc. It also includes the mid-point theorem.
Total Questions: 19 sums – 8 are simple problems, 6 are of an intermediate level and 5 are challenging questions.
## Class 9 Maths NCERT Solutions Chapter 9
Class 9 maths NCERT solutions chapter 9 areas of parallelograms and triangles include various theorems that help to simplify finding the area of complex parallelograms as well as triangles, properties to find the area of triangles divided by a median, and area of planar regions.
Important Formulas:
• Area of a triangle = ½ × base × height
• Area of a parallelogram = base × height
### Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles
Topics Covered: Topics covered incorporate questions on finding the area of two or more triangles and parallelograms with the same base between the same parallels, determining the area of triangles that are split by a median as well as the area of congruent figures.
Total Questions: Chapter 9 areas of parallelograms and triangles have a total of 31 questions. 8 questions belong to an optional exercise with a high difficulty level, 14 are intermediate and 9 are simple sums.
## NCERT Solutions for Class 9 Maths Chapter 10
NCERT solutions for class 9 maths chapter 10 circles build a foundation for students by introducing them to the various terms related to circles. Students get the opportunity to build their knowledge base on properties of chords, the angle subtended by chord and arc, perpendicular to a point, the distance of a chord from the center as well as cyclic quadrilaterals.
Preparation Tips:
NCERT solutions for class 9 maths chapter 10 are purely based on the properties of various elements of a circle. To memorize these theorems you can follow the topics listed below.
• Make a note of all the properties so that you can skim through them when required.
• Solve all NCERT textbook questions.
• Regularly revise the solutions so that you can recall them when necessary.
### Class 9 Maths Chapter 10 Circles
Topics Covered: NCERT solutions class 9 maths encompass topics such as angles subtended by a chord at a point, perpendiculars dropped from the center to a chord, and how to calculate the distance of equal chords from the center. Other discussions include angles subtended by an arc of the circle, cyclic quadrilaterals, and their attributes.
Total Questions: There are a total of 35 questions in this chapter out of which 15 are relatively simpler, 10 are complex and 10 are of moderate complexity.
## NCERT Solutions for Class 9 Maths Chapter 11
NCERT solutions for class 9 maths chapter 11 constructions will teach students how to precisely use geometrical instruments for basic constructions. Students learn to construct different angles, perpendicular bisectors, and triangles with a few given restraints.
### Class 9 Maths Chapter 11 Constructions
Topics Covered: Constructing an angle from a ray, bisector of a given angle, and techniques to construct triangles when certain constraints such as the perimeter are mentioned, are the topics explained in this chapter.
Total Questions: Chapter 11 constructions have a total of 10 questions. 3 can be simply solved, 5 are moderate questions and 2 are difficult.
After studying NCERT solutions class 9 maths chapter 11 students will be adept at using a ruler and compass to bisect a given angle, construct angles of discrete values, draw the perpendicular bisector of a line segment, and building triangles.
## NCERT Solutions for Class 9 Maths Chapter 12
NCERT solutions for class 9 maths chapter 12 heron’s formula is a formula given by the Hero of Alexandria to calculate the area of a triangle when the lengths of all three sides are known. In this chapter, students will learn more about the application of this formula to find the area of triangles and quadrilaterals.
Important Formulas:
• Semi Perimeter s = (a + b + c)/2
• Area of a Triangle = √[s (s – a) (s – b) (s – c)]
### Class 9 Maths Chapter 12 Heron’s Formula
Topics Covered: NCERT Solutions class 9 maths chapter 12 cover the complete scope of Heron’s formula and how to apply it to questions for determining the area of a triangle. Additionally, this formula is also used for computing the area of a quadrilateral.
Total Questions: Chapter 12 heron’s formula consists of 15 well-curated questions out of which 5 are simple, 3 and medium level and 7 are difficult problems.
## Class 9 Maths NCERT Solutions Chapter 13
Class 9 maths NCERT solutions chapter 13 surface areas and volumes introduce students to the concepts required to find the total surface areas, lateral or curved surface areas, and volumes of different shapes. The concepts and formulas in this chapter are very important as they form the basis of topics covered in higher education.
Important Formulas:
### Class 9 Maths Chapter 13 Surface Areas and Volumes
Topics Covered: Finding the TSA, LSA/CSA, along with the volume of shapes such as cubes, cuboids, right circular cylinder, right circular cone, sphere, and hemisphere.
Total Questions: Chapter 13 surface areas and volumes have 75 questions. 25 are easy formula-based sum, 31 are medium level and 10 are difficult problems that challenge a student’s thinking abilities.
## Class 9 Maths NCERT Solutions Chapter 14
Class 9 maths NCERT solutions chapter 14 statistics provide an in-depth analysis of problems that are based on the collection and presentation of data for analytical purposes, graphical representation of data using different types of graphs, and measures of central tendency. This chapter enables students to develop their calculation speed and visualization skills.
### Class 9 Maths Chapter 14 Statistics
Topics Covered: Representation of data in the form of frequency tables and drawing inferences from them, plotting data points using bar graphs, histograms, and frequency polygons, finding the mean, median, and mode of the given data.
Total Questions: Chapter 14 statistics have 26 problems that can be split into 8 long answer-type questions, 11 moderate, and 7 simple sums.
Once students finish practicing class 9 maths NCERT solutions chapter 14 they will have a good understanding of the purpose of well-organized data, the three measures of central tendency for ungrouped data namely mean, median, and mode. Students can effectively apply these formulas to all levels of problems after studying this topic.
## NCERT Solutions Class 9 Maths Chapter 15
NCERT solutions class 9 maths chapter 15 probability help students deal with daily-life problems that require the use of probability concepts. This chapter helps students explore more about empirical probability, the definition of an event, and how to calculate the probability of different events.
Class 9 Maths Chapter 15 Study Tips:
NCERT solutions class 9 maths chapter 6 probability is a widely used topic that people require to perform their day-to-day operations. Given below are some study tips that can help in comprehending this chapter in a better manner.
1. Apply the concept to simple daily life questions.
1. Read the question carefully and break it down to get a better idea of what has to be done.
### Class 9 Maths Chapter 15 Probability
Topics Covered: Finding the probability of the occurrence of heads or tails when a coin is flipped, tossing a die, and drawing conclusions from the result obtained are the topics covered in NCERT solutions class 9 maths.
Total Questions: Chapter 15 probability has 13 questions. 4 are short answer problems, 3 are difficult and 6 are moderate sums.
## Class 9 Maths NCERT Solutions Important Formulas
Class 9 maths covers several topics such as polynomials, number systems, geometry, probability, etc. The formulas and concepts behind these topics are required to understand higher grade mathematics. Formulas are required to reduce the time taken to solve questions and increase accuracy. Given below is an overview of the important maths formulas encompassed in the NCERT solutions for class 9 maths.
Geometry:
Surface Area and volume:
Algebra:
• (a + b)2 = a2 + b2 + 2ab
• (a – b)2 = a2 + b2 – 2ab
• (a + b) (a – b) = a2 – b2
## Importance of NCERT Solutions for Class 9 Maths
Class 9 Maths NCERT Solutions are extremely helpful for CBSE students as they give a means for efficient time management and improving auxiliary abilities. The following are the advantages of these solutions.
• Practical Applications: Many of the sums provided by the NCERT textbook are similar to real-life examples, especially when word problems are considered. By following the class 9 maths NCERT solutions students can come up with precise solutions that can also be applied to daily life problems.
• Practice: To solve the toughest sums within a matter of minutes you need due diligence and regular practice. These NCERT solutions for class 9 maths are very helpful for this purpose as they give students a detailed overview of how to solve different types of sums. They can also help a student to complete the syllabus well within the given time frame.
• Regular Revision: On examination days, students do not have the time to solve each and every question. Thus, these NCERT solutions class 9 maths help students with a quick revision of the entire syllabus. This also helps to free up time so that students can focus on ensuring that their concept formation is correct and can be accurately applied to sums.
## FAQs on NCERT Solutions for Class 9 Maths
### Why are NCERT Solutions for Class 9 Maths Important?
NCERT solutions for class 9 maths are vital for students as they provide reliable solutions that students can use to cross-check their answers. These class 9 maths NCERT solutions also guide students on how to appear for an examination so that students can write their answers without missing any steps and get the best possible score.
### Where Can I Get Chapter-wise Solutions for NCERT Class 9 Maths?
The chapter-wise solutions for NCERT class 9 maths can be accessed by the links mentioned above. They are further segregated into exercise-wise pdfs that are available for free download. These pdfs are also compatible with mobile screens and are in a scrollable format for easy use.
### How CBSE Students can utilize Class 9 Maths NCERT Solutions?
To effectively utilize class 9 maths NCERT solutions students need to ensure that they have an understanding of the concepts used to solve the problems. Thus, when they start consulting these solutions students will comprehend them in a better manner and will be able to retain what they learned more efficiently.
### Do I Need to Make Notes while Referring to NCERT Solutions Class 9 Maths?
Notes act as a simplified guide for students when they are trying to build an understanding of the subject or if they are revising that question. It is a good way of retaining and recalling concepts when required. Thus, it is crucial that students make notes of the NCERT solutions class 9 maths to expedite their learning process.
### What are the Best Ways to Learn Concepts Covered in NCERT Solutions Class 9th Maths?
The most impactful way to learn concepts covered in NCERT solutions class 9th maths is by constantly practicing these solutions. If a student finds some concept difficult and cannot proceed further while attempting a problem he needs to immediately clarify his doubts so that there is no gap in learning.
### Can We Use Other Methods to Solve Problems Apart From Those Mentioned in NCERT Solutions for Class 9 Maths?
You may not want to use the method given in the NCERT solutions for class 9 maths however, the approach used and the answer obtained need to be universally recognized. Having said that you cannot use the tricks that are useful in solving competitive exam problems as these are not standardized techniques.
### Why Choose Cuemath for Class 9 Maths NCERT Solutions?
The best website for class 9 maths NCERT solutions is Cuemath. These solutions are prepared by IITians and certified math experts. Thus, they are reliable, precise and give the best techniques to solving any problem. They also include some tips to enable students to solve sums with speed and accuracy.
### How Many Chapters are there in NCERT Solutions for Class 9 Maths?
There are 15 chapters in class 9 maths NCERT solutions. The problems covered by these are fantastic to improve a student’s understanding of different concepts. The NCERT solutions give an in-depth analysis of all exercise problems.
### Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths?
All questions that are included in the CBSE home examinations are based on the NCERT class 9 maths. Thus, in order to get good marks, it is necessary to be well-versed with all the questions covered in class 9 maths NCERT solutions that can only be achieved through regular practice and due diligence.
### Are NCERT Solutions for Class 9 Maths Sufficient for CBSE Exams?
The NCERT solutions for class 9 maths are sufficient for all CBSE exams. As the school teachers set the paper they include questions from the NCERT textbook. The language or values given in the problem may differ however, the approach to solving it is more or less the same. As long as a student has a robust understanding of all chapters, he can successfully clear the examination.
### What are the Important Topics Covered in NCERT Solutions for Class 9 Maths?
All the topics such as construction, circles, surface area, volume, statistics, etc. that are covered in the NCERT solutions for class 9 maths are required for class 10th. Thus, skipping a topic is not advisable. In addition to this, all topics within the scope of mathematics are interlinked making it necessary to give equal attention without leaving out anything.
### What are the Important Formulas in NCERT Class 9 Maths Solutions?
The questions in NCERT solutions class 9 maths are mostly based on the application of theorems and axioms to prove a certain point. Thus, as there are a fewer number of formulas it is not only easy but necessary to memorize them all.
### What are the Benefits of Class 9 Maths NCERT Solutions for CBSE Students?
The class 9 maths NCERT solutions help students to finish their syllabus on time giving them enough room for revision. They also provide a quick overview of all the problems that students can use as a reference while skimming through the chapter.
### Why Should I Practice NCERT Solutions for Class 9 Maths Regularly?
The practice of NCERT solutions for class 9 maths will help a student achieve the goal of perfection. It helps students get comfortable with the covered concepts. Thus, during an examination students can keep their cool and breeze through the paper without getting flustered.
### How is Class 9 Maths NCERT Solutions Promoting Problem Solving in Students?
NCERT solutions for class 9 maths start by explaining the questions and then breaking the problem down to elucidate it in a better way. It teaches students to think of the why and how behind a sum. These solutions encourage students to use a similar approach enabling them to harness their problem-solving skills and computational mindset. | HuggingFaceTB/finemath | |
# using second law of motion derive the relation between force and acceleration . a bullet of 10g strike a sand bag at a speed of 1000m/s and gets embeded after travelling 5 cm .calculate(1)the resistive force exerted by sand on bullet(2)the time taken by the bullet to come to rest
we have
v2 - u2 = 2as
or
acceleration
a = (v2 - u2) / 2s
here,
u = 1000m/s
v = 0
s = 5cm = 0.05m
so,
a = (0 - 10002) / (2x0.05)
thus,
a = -10000000 m/s2
and the retarding force will be
F = ma
here,
m = 10g = 0.01kg
so,
F = 0.01 x 10000000
thus,
F = 100000N
..
time taken will be
t = (v - u) / a
or
t = (0 - 1000) / -10000000
thus,
t = 0.0001s
• -1
What are you looking for? | HuggingFaceTB/finemath | |
# 100 days in the future from today
## 100 days from today
### Introduction
Have you ever wondered what the date will be exactly one hundred days from now? Counting each day can be time-consuming and tedious. Luckily, we have tools that can help us calculate the future date without the need for manual counting. In this article, we will explore how to determine the date that is exactly one hundred days from today, both considering all days and only counting weekdays.
### Calculating 100 days from today
Today is August 31, 2023, and we want to find out the date that is exactly one hundred days from now. By adding one hundred days to the current date, we arrive at December 9, 2023. This can be confirmed using a date difference calculator, which measures the number of days between two given dates.
### December, 2023 calendar
Let’s take a closer look at the December 2023 calendar to understand the significance of the date December 9, 2023. In the calendar, December 9th falls on a Saturday. It is the 343rd day of the year and is part of the 49th week (assuming each week starts on a Monday). December 2023 has a total of 31 days. It is important to note that 2023 is not a leap year, so there are 365 days in this year. In the United States, the short form for this date is 12/9/2023, while in most other parts of the world, it is represented as 9/12/2023.
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### Counting weekdays only
In certain cases, we may want to skip weekends and count only the weekdays. This can be particularly useful when dealing with business deadlines or considering specific working days. If we want to determine the day that falls exactly one hundred weekdays from today, we need to account for the weekends and adjust our calculation.
Starting from today, which is a Thursday, we begin counting forward. The next day would be a Friday. However, to reach exactly one hundred weekdays from now, we actually need to count 140 total days, including both weekdays and weekends. Therefore, one hundred weekdays from today would be January 18, 2024.
### January, 2024 calendar
Let’s examine the January 2024 calendar to gain more insights into the date January 18, 2024. On this day, which falls on a Thursday, it is the 18th day of the year and part of the 3rd week (assuming each week starts on a Monday). Similar to December, January 2024 also has 31 days. However, 2024 is a leap year, so there are 366 days in total. In the United States, the short form for this date is 1/18/2024, while in most other parts of the world, it is represented as 18/1/2024.
### Considerations and limitations
When using the Days From Now calculator, it is essential to keep certain factors in mind. The allowable range for calculating dates is from -44455 to 5254. This ensures that the calculations stay within a reasonable and logical timeframe. It is important to note that if you are counting business days, you may need to adjust the date further to account for any holidays that fall within the calculated timeframe.
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### The Days From Now calculator
For easy calculation of future dates, an online tool called the Days From Now calculator is available. This calculator allows you to input the number of days you want to calculate from today. If you want to find a date in the past, you can enter a negative number to determine the number of days before today. The calculator can be particularly useful when you have a certain number of days remaining before a deadline or when you need to consider business days or weekdays only, excluding weekends. If you need to measure the number of days between two specific dates, you can switch to the Date Difference calculator instead.
By utilizing these tools, you can effortlessly determine future dates and plan your schedule accordingly. Whether you need to calculate the exact date that is one hundred days from now or consider specific working days, the Days From Now calculator can simplify the process and provide accurate results.
## Mr. in a 1983 Styx hit NYT Crossword Clue
If you need help, always remember that we’re here to assist you. With time and practice, you’ll get better at solving crossword puzzles, so don’t get discouraged…
## ‘Love You Anyway’ by Luke Combs – Lyrics & Meaning
In recent years, Luke Combs has established himself as one of the prominent figures in contemporary country music. Since 2017, native Charlotte, North Carolina has been thriving…
## Culebrilla: qué es, causas, síntomas y tratamiento
It is recommended to seek appropriate treatment and consult a dermatologist or physician for a diagnosis if symptoms indicative of the presence of herpes zoster, scientifically known…
## A Secret Vow – Best Characters and How to Get
Check out this guide for the best materials, ascension stats, and characters to obtain Vow’s Secret A. Rail Star: Honkai in Cone Light Path Destruction features Vow’s…
## Judith Light Net Worth
Ellen Judith Light was born on February 9th, 1949 in Trenton, New Jersey, USA. She is best known for her roles in the ABC comedy-drama “Ugly Betty”…
## R.I.P. Kay Parker (1944 – 2022) – The Rialto Report interview
Podcast: Listen in a fresh window | Download. Subscribe: RSS. Kay Parker, a notable personality during the heyday of adult film, passed away at the age of… | HuggingFaceTB/finemath | |
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# Solving quadratic equation algorithm - Flowchart
## Solving quadratic equation algorithm - Flowchart
"In elementary algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
ax^2+bx+c=0
where x represents an unknown, and a, b, and c are constants with a not equal to 0. If a = 0, then the equation is linear, not quadratic. The constants a, b, and c are called, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.
Because the quadratic equation involves only one unknown, it is called "univariate". The quadratic equation only contains powers of x that are non-negative integers, and therefore it is a polynomial equation, and in particular it is a second degree polynomial equation since the greatest power is two.
Quadratic equations can be solved by a process known in American English as factoring and in other varieties of English as factorising, by completing the square, by using the quadratic formula, or by graphing." [Quadratic equation. Wikipedia]
The flowchart example "Solving quadratic equation algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
Used Solutions
## Euclidean algorithm - Flowchart
"In mathematics, the Euclidean algorithm, or Euclid's algorithm, is a method for computing the greatest common divisor (GCD) of two (usually positive) integers, also known as the greatest common factor (GCF) or highest common factor (HCF). ...
The GCD of two positive integers is the largest integer that divides both of them without leaving a remainder (the GCD of two integers in general is defined in a more subtle way).
In its simplest form, Euclid's algorithm starts with a pair of positive integers, and forms a new pair that consists of the smaller number and the difference between the larger and smaller numbers. The process repeats until the numbers in the pair are equal. That number then is the greatest common divisor of the original pair of integers.
The main principle is that the GCD does not change if the smaller number is subtracted from the larger number. ... Since the larger of the two numbers is reduced, repeating this process gives successively smaller numbers, so this repetition will necessarily stop sooner or later - when the numbers are equal (if the process is attempted once more, one of the numbers will become 0)." [Euclidean algorithm. Wikipedia]
The flowchart example "Euclidean algorithm" was created using the ConceptDraw PRO diagramming and vector drawing software extended with the Mathematics solution from the Science and Education area of ConceptDraw Solution Park. Read more
Euclid's algorithm flow chart
Used Solutions
## Basic Flowchart Symbols and Meaning
Flowcharts are the best for visually representation the business processes and the flow of a custom-order process through various departments within an organization. ConceptDraw PRO diagramming and vector drawing software extended with Flowcharts solution offers the full set of predesigned basic flowchart symbols which are gathered at two libraries: Flowchart and Flowcharts Rapid Draw. Among them are: process, terminator, decision, data, document, display, manual loop, and many other specific symbols. The meaning for each symbol offered by ConceptDraw gives the presentation about their proposed use in professional Flowcharts for business and technical processes, software algorithms, well-developed structures of web sites, Workflow diagrams, Process flow diagram and correlation in developing on-line instructional projects or business process system. Use of ready flow chart symbols in diagrams is incredibly useful - you need simply drag desired from the libraries to your document and arrange them in required order. There are a few serious alternatives to Visio for Mac, one of them is ConceptDraw PRO. It is one of the main contender with the most similar features and capabilities. Read more
How to Build a Flowchart
## Diagram Flow Chart
ConceptDraw PRO is a software for producing flow charts. Read more
## Types of Flowcharts
A Flowchart is a graphical representation of process, algorithm, workflow or step-by-step solution of the problem. It shows the steps as boxes of various kinds and connects them by arrows in a defined order depicting a flow. There are twelve main Flowchart types: Basic Flowchart, Business Process Modeling Diagram (BPMN), Cross Functional Flowchart, Data Flow Diagram (DFD), IDEF (Integrated DEFinition) Flowchart, Event-driven Process Chain (EPC) Diagram, Influence Diagram (ID), Swimlane Flowchart, Process Flow Diagram (PFD), Specification and Description Language (SDL) Diagram, Value Stream Mapping, Workflow Diagram. Using the Flowcharts solution from the Diagrams area of ConceptDraw Solution Park you can easy and quickly design a Flowchart of any of these types. This solution offers a lot of special predesigned vector symbols for each of these widely used notations. They will make the drawing process of Flowcharts much easier than ever. Pay also attention for the included collection of ready Flowchart examples, samples and quick-start templates. This is business process improvement tools. If you are looking for MS Visio for your Mac, then you are out of luck, because it hasn't been released yet. However, you can use Visio alternatives that can successfully replace its functions. ConceptDraw PRO is an alternative to MS Visio for Mac that provides powerful features and intuitive user interface for the same. Read more
How to Simplify Flow Charting
## Flow Chart Design - How to Design a Good Flowchart
Use ConceptDraw PRO business diagramming and business graphics software for general diagramming purposes, it inludes tousands colored professional flowchart symbols, examples and samples which saves time when you prepare documents, professional presentations or make an explanation of process flow diagram. Read more
## How to Discover Critical Path on a Gantt Chart
ConceptDraw PROJECT is the Gantt chart software that implements critical path method to provide the user with a wide set of project management tools. Read more
## Flowchart Software
A Flowchart visualizes a process or algorithm of a program as step blocks composed of shapes, connected with arrows that show the process flow direction. The Flowcharts give users the ability to represent structural data in a visual form and are widely used to visualize the business and technical processes, as well as software algorithms. ConceptDraw PRO is a powerful Mac OS X and Windows technical diagramming, business diagramming and flowchart drawing software. Numerous vector stencils, built-in templates and samples created by professional designers are included at the basic package. ConceptDraw PRO is a classic flowchart maker, which exactly follows the industry standards and provides ready-to-use flowchart symbols gathered at the solutions of ConceptDraw Solution Park. Use the free trial versions of ConceptDraw PRO software for Mac and PC to experience the powerful possibilities of ConceptDraw PRO in designing the Flowcharts, Flow Diagrams, Business Flow Charts of different types, styles, and degree of complexity. Read more
## Creating a Simple Flowchart
ConceptDraw PRO the best professional flowchart software that gives users the ability to simply draw any type of flowchart. Read more
## Workflow diagram - Template
"Workflow components.
A workflow can usually be described using formal or informal flow diagramming techniques, showing directed flows between processing steps. Single processing steps or components of a workflow can basically be defined by three parameters:
(1) input description: the information, material and energy required to complete the step,
(2) transformation rules, algorithms, which may be carried out by associated human roles or machines, or a combination,
(3) output description: the information, material and energy produced by the step and provided as input to downstream steps.
Components can only be plugged together if the output of one previous (set of) component(s) is equal to the mandatory input requirements of the following component. Thus, the essential description of a component actually comprises only in- and output that are described fully in terms of data types and their meaning (semantics). The algorithms' or rules' description need only be included when there are several alternative ways to transform one type of input into one type of output – possibly with different accuracy, speed, etc.
When the components are non-local services that are invoked remotely via a computer network, such as Web services, additional descriptors (such as QoS and availability) also must be considered." [Workflow. Wikipedia]
The workflow diagram template for the ConceptDraw PRO diagramming and vector drawing software is included in the Workflow Diagrams solution from the Business Processes area of ConceptDraw Solution Park. Read more
Workflow diagram template
Used Solutions
## Simple Flow Chart
ConceptDraw PRO diagramming and vector drawing software extended with Flowcharts Solution from the 'Diagrams' area of ConceptDraw Solution Park is a powerful tool for drawing Flow Charts of any complexity you need. Irrespective of whether you want to draw a Simple Flow Chart or large complex Flow Diagram, you estimate to do it without efforts thanks to the extensive drawing tools of Flowcharts solution, there are professional flowchart symbols and basic flowchart symbols. This sample shows the Gravitational Search Algorithm (GSA) that is the optimization algorithm. Read more
## Software development with ConceptDraw PRO
Modern software development requires creation of large amount of graphic documentation, these are the diagrams describing the work of applications in various notations and cuts, also GUI design and documentation on project management. ConceptDraw PRO technical and business graphics application possesses powerful tools for software development and designing technical documentation for object-oriented projects. Solutions included to the Software Development area of ConceptDraw Solution Park provide the specialists with possibility easily and quickly create graphic documentation. They deliver effective help in drawing thanks to the included package of templates, samples, examples, and libraries with numerous ready-to-use vector objects that allow easily design class hierarchies, object hierarchies, visual object-oriented designs, flowcharts, GUI designs, database designs, visualize the data with use of the most popular notations, including the UML and Booch notations, easy manage the development projects, automate projection and development. Read more
## Samples of Flowcharting
This sample shows the Flowchart for determine is a species heterotroph or autotroph, or a subtype. This diagram has start point and end points. The diamonds on the Flowchart represent the decisions. The Flowcharts are widely used in science, analytics, government, politics, business, engineering, architecture, marketing, manufacturing, administration, etc. Read more
## Flow Chart
A Flow Chart is a type of diagram which visualizes a process, algorithm of a program or workflow. The steps / operations are represented as boxes of various types and are connected by arrows which show the flow of the process. But how to design a Flow Chart fast and easy? We recommend to use a powerful ConceptDraw PRO diagramming and vector drawing software extended with Flowcharts Solution from the "Diagrams" Area of ConceptDraw Solution Park. Read more
## Flowchart design. Flowchart symbols, shapes, stencils and icons
A flowchart is a type of diagram which represents an algorithm, process or workflow, displays the steps as boxes of various kinds and depicts their order by connecting them with arrows. Any business graphic document will be more colorful and understandable if will use professional-looking and visual diagrams and flowcharts. Flowchart design gives versatile presenting and explaining of the process. ConceptDraw PRO flowchart software enhanced with Flowcharts solution helps effectively make Flowchart Design. Use of predesigned flowchart symbols and bright color palette offers a fresh view and favorably distinguishes the flowcharts designed in ConceptDraw PRO from the black and white flowcharts on a paper. Preferably to use no more than three or four colors and apply identical for the same shape types. The best flowchart design can be achieved by starting with Flowchart template, or any of suitable ready examples or samples offered in ConceptDraw STORE, open one of them and enter the proper text into each Flowchart shape. Each symbol of the flowchart has a definition that can't be changed. This means that all flowcharts shapes can be grouped in according to their meaning. Users with ConceptDraw PRO flowchart software can style groups of symbols with close definitions by color sets chosen from complementary palette. Almost all workflows can be depicted as a flowchart. Colored diagrams are more light for perception of the certain information, this is part of flowchart design. Bright colors need to be used in the key points of Decision symbols, to focus attention on their importance for whole process flow. Read more | HuggingFaceTB/finemath | |
# Divide curve and march the points along the curve
Hi, extreme newbie here.
I’d like to get points along a curve, then march the points along the curve.
I’ve had a naïve go, but this moves all the points as a set, so they are all simultaneously undergoing the same linear translation (as I guess is obvious).
Can I somehow split out the points, so they can all independently follow the original curve?
TIA
Where should go the last point? Outside the curve?
Your “Point on curve” component, set at 0.333 is evaluating a point along your curve (probably at something like 13,1,0) , and then you created a vector from that point to an unset B input, (0,0,0), resulting in a -13,1,0 vector.
You just moved all your points, then…
Some approach would be to use the tangents for each point (T), giving amplitude and moving the points.
Or shattering the curve with the t parameters and then evaluating each segment differently.
Why are you wanting to do so? It’s something… unusual.
Why wouldn’t you just divide the curve where you need the points? Rather than dividing the curve then moving the points like you want to do.
Did you try Divide length component?
Thanks people.
I’m trying to animate a rope being pulled along, passing over a pulley.
The ‘points source’ curve would be a helix representing the lay of the rope fibres.
I couldn’t think of a way of distorting the helix as it passed over the pulley, so hit on the idea of breaking it down into hundreds points, each one to follow the ‘driving curve’. The driving curve being a straight section, then semi circular round the pulley and another straight section.
This a learning exercise, but still hopefully with a useful outcome.
I’m probably trying to run before I can walk…
This sounds promising. I’ll look into the difference between Shatter and Divide.
This is what it looks like changing where you divide the curve…
And doing it with two curves connecting the points…
Awsome! Thanks Ftzuk.
Could you help me out with the name of this component:
Cheers, Ian
Graph mapper
1 Like
Thanks Filipe. | HuggingFaceTB/finemath | |
Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.3 Text Book Back Questions and Answers, Notes.
## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3
Question 1.
Represent the following inequalities in the interval notation:
(i) x ≥ – 1 and x < 4
x ≥ – 1 and x < 4
x ∈ [- 1, 4) (ii) x ≤ 5 and x ≥ – 3
x ≤ 5 and x ≥ – 3
– 3 ≤ x ≤ 5
∴ x ∈ [- 3, 5 ]
(iii) x < – 1 or x < 3
x < – 1 or x < 3
x ∈ (-∞, 3)
(iv) -2x > 0 or 3x – 4 < 11
– 2x > 0 or 3x – 4 < 11
2x < 0 or 3x < 11 + 4
x < 0 or x < $$\frac{15}{3}$$
x < 0 or x < 5
x ∈ (- ∞, 5) Question 2.
Solve 23x < 100 when
(i) x is a natural number,
(ii) x is an integer.
Given 23x < 100
(i) When x is a natural number
23x < 100
⇒ x < $$\frac{100}{23}$$
⇒ x < 4.347
⇒ x = 1, 2, 3, 4
∴ The solution set is { 1, 2, 3 , 4 }
(ii) When x is an integer
23x < 100
⇒ x < $$\frac{100}{23}$$
⇒ x < 4.347
⇒ x = …… , – 3, – 2, – 1, 0, 1, 2, 3, 4
Hence the solution set is
{ ………, – 3, – 2, – 1, 0, 1 , 2 , 3 , 4 } Question 3.
Solve – 2x ≥ 9 when
(i) x is a real number,
(ii) x is an integer
(iii) x is a natural number.
Given – 2x ≥ 9
⇒ – x ≥ $$\frac{9}{2}$$
⇒ x ≤ $$-\frac{9}{2}$$
(i) When x is a real number
x ≤ $$-\frac{9}{2}$$
The solution set is $$\left(-\infty,-\frac{9}{2}\right]$$
(ii) x is an integer
x ≤ –$$-\frac{9}{2}$$
x ≤ – 4.5
x = …………., -7, – 6, -5
(iii) x is a natural number
x ≤ – $$\frac{9}{2}$$
x ≤ – 4.5
Since there exists no natural number less than – $$\frac{9}{2}$$
∴ No solution Question 4.
Solve:
(i) 3 (3x – 6) ≤ 5 (10 – 5x)
9x – 18 ≤ 50 – 25 x
9x + 25 x ≤ 50 + 18
34x ≤ 68
x ≤ $$\frac{68}{34}$$ = 2
x ≤ 2
∴ The solution set is (-∞, 2] (ii) Multiplying both sides by 3, we have
5 – x < $$\frac{3 x}{2}$$ – 12
Multiplying both sides by 2, we have
10 – 2x < 3x – 24
10 + 24 < 3x + 2x
34 < 5x
$$\frac{34}{5}$$ < x
x > $$\frac{34}{5}$$
∴ The solution set is $$\left(\frac{34}{5}, \infty\right)$$ Question 5.
To secure an A grade, one must obtain an average of 90 marks or more in 5 subjects each of a maximum of 100 marks. If one scored 84, 87, 95, 91 in the first four subjects, what is the minimum mark one scored in the fifth subject to get an A grade in the course?
Required marks = 5 × 90 = 450
Total marks obtained in 4 subjects = 84 + 87 + 95 + 91 = 357
So required marks in the fifth subject = 450 – 357 = 93
Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Amount of 12% solution of acid = 600 litres
Let x be the required number litres of 30 % acid solution to be added to the given 600 litres of 12 % acid solution to make the resulting mixture will be more than 15 % but less than 18 %.
∴ Total amount of mixture = (600 + x) litres
30% acid solution of x litres + 12% acid solution of 600 litres > 15% acid solution of (600 + x) litres
$$\frac{30}{100}$$ × x + $$\frac{12}{100}$$ × 600 > $$\frac{15}{100}$$ × (600 + x)
30x + 7200 > 9000 + 15x
30x – 15x > 9000 – 7200
15x > 1800
x > $$\frac{1800}{15}$$ = 120
x > 120 ——– (1)
Also 30% acid solution of x litres + 12% acid solution of 600 litres < 18% acid solution of ( 600 + x) litres.
$$\frac{30}{100}$$ × x + $$\frac{12}{100}$$ × 600 < $$\frac{15}{100}$$ × (600 + x)
30x + 7200 < 18 (600 + x)
30x + 7200 < 10800 + 18x
30x – 18x < 10,800 – 7200
12x < 3600
x < $$\frac{3600}{12}$$ = 300
x < 300 ——- (2)
From equations (1) and (2), we get ) 120 < x < 300
∴ The numbers of litres of the 30% acid solution to be added is greater than 120 litres and less than 300 litres. Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.
Let x and x + 2 be the two pair of consecutive odd natural numbers.
Given x > 10 ——— (1)
and x + 2 > 10 ——— (2)
Also x + (x + 2) < 40 ——— (3)
From equations (1), we have
x = 11, 13 , 15, 17, 19, 21 …………
Using equation (3), the required pairs are
(11, 13), (13, 15), (15, 17), ( 17, 19), (19 ,21 ) is not possible since 19 + 21 = 40
Question 8.
A model rocket is launched from the ground. The height h reached by the rocket after t seconds from lift off is given by h(t) = – 5t2 + 100t, 0 ≤ t ≤ 20. At what times, the rocket is 495 feet above the ground?
h(t) = -5t2 + 100t
at t = 0, h(0) = 0
at t = 1, h(1) = -5 + 100 = 95
at t = 2, h(2) = -20 + 200 = 180
at t =3, h(3) = -45 + 300 = 255
at t = 4, h(4) = -80 + 400 = 320
at t = 5, h(5) = -125 + 500 = 375
at t = 6, h(6) = – 180 + 600 = 420
at t = 7, h(7) = -245 + 700 = 455
at t = 8, h(8) = – 320 + 800 = 480
at t = 9, h(9) = -405 + 900 = 495
So, at 9 secs, the rocket is 495 feet above the ground. Question 9.
A plumber can be paid according to the following schemes: In the first scheme he will be paid rupees 500 plus rupees 70 per hour and in the second scheme, he will pay rupees 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
I scheme with x hr
500 + (x- 1) 70 = 500 + 70x – 70 = 430 + 70x
II scheme with x hours
120x
Here I > II
⇒ 430 + 70x > 120x
⇒ 120x – 70x < 430
50x < 430
$$\frac{50 x}{50}<\frac{430}{50}$$
x < 8.6 (i.e.) when x is less than 9 hrs the first scheme gives better wages. Question 10.
A and B are working on similar jobs but their monthly salaries differ by more than Rs. 6000. If B earns rupees 27,000 per month, then what are the possibilities of A’s salary per month? | HuggingFaceTB/finemath | |
Logical Reasoning ## Online logical reasoning test
1. Find out the number that does not fit into the given series.
54, 48, 41, 34, 24
a.) 54
b.) 48
c.) 41
d.) 34
2. Find out the number that does not fit into the given series.
1, 3, 9, 27, 82
a.) 82
b.) 27
c.) 9
d.) 3
3. Complete the following series.
0, 7, 26, 63, 124, ……..
a.) 224
b.) 115
c.) 215
d.) 214
4. Complete the following series.
48, 43, 42, 37, 36, 31, 30, ……….
a.) 29
b.) 24
c.) 25
d.) 28
5. Complete the following series.
876, 765, 654, 543, …….
a.) 430
b.) 431
c.) 432
d.) None of these.
6. Complete the following series.
1, 3, 7, 15, 31. __, __
a.) 62, 127
b.) 63, 126
c.) 63, 127
d.) 62, 126
7. Complete the following series.
3, 4, 6, 9, 13, __, __
a.) 17, 25
b.) 17, 26
c.) 18, 24
d.) 18, 25
8. Find out the missing figure.
If 20 X 2 = 20 25 X 4 = 50 then 30 X 8 = ?
a.) 100
b.) 110
c.) 120
d.) None of these.
9. Find out the missing figure.
If 8 -9 = 503 6 – 5 = 211 then 7 – 4 = ?
a.) 337
b.) 338
c.) 339
d.) 340
10. Put + or – between the numbers to obtain the given result.
7 3 12 13 9 = 14
a.) + - + -
b.) + + + -
c.) - - + +
d.) - + - +
ANSWER : - - + +
11. In a two digit number, its digits are reversed and this new number is then added to the original number. The addition is always divisible by which of the following?
a.) 3
b.) 7
c.) 9
d.) 11
12. Which of the following is a perfect square?
a.) 19210
b.) 23435
c.) 28561
d.) 17466
13. A bus is hired at the rate of Rs. 320 per day for a picnic. Since some people did not turn up, the average contribution was Rs. 48 more per person per day than before. How many people were expected, and how many turned up?
a.) 8, 5
b.) 5, 2
c.) 10, 4
d.) 5, 4
14. If a coherent word can be formed out of the letters: USMOE, which is its fourth letter?
a.) E
b.) M
c.) U
d.) S
15. The letters A, B, C, D, E each stand for one of 1, 2, 3, 4 and 5, but not necessarily in that order. A is odd, B is neither 4 nor 5, C is 1, D is either 4 or 5 and E is none of 2, 3, 4. The correct order of the digits (according to the correct order of the English alphabet) is –
a.) 32145
b.) 52341
c.) 11245
d.) None of these
16. Which pair of numbers is different from the rest?
a.) (5, 15)
b.) (6, 24)
c.) (7, 49)
d.) (9, 63)
17. A box contains 6 red beads and 5 blue beads. What is the smallest number of beads that must be picked at a time, without looking, to be sure of getting two of the same colour?
a.) 3
b.) 7
c.) 2
d.) 11
18. In a certain language EGG is coded as 577, what should be the code for CAB?
a.) 312
b.) 213
c.) 123
d.) 321
e.) 314
19. In a certain language SEE is coded as 11, what will be the value for PLEA?
a.) 6
b.) 7
c.) 8
d.) 9
e.) 12
20. In certain language RUN is coded as 2, what will be the code for SUN?
a.) 5
b.) 6
c.) 1
d.) 3
e.) 7
21. In a certain language CAN coded as 414, what will be the code for FAR?
a.) 618
b.) 817
c.) 178
d.) 871
e.) 718 Aptitude questions | HuggingFaceTB/finemath | |
# American Institute of Mathematical Sciences
November 2014, 34(11): 4537-4553. doi: 10.3934/dcds.2014.34.4537
## Localization, smoothness, and convergence to equilibrium for a thin film equation
1 Department of Mathematics, Hill Center, Rutgers University, Piscataway, NJ 08854, United States 2 Department of Mathematics, Faculty of Education, Zirve University, Gaziantep, Turkey
Received April 2013 Revised February 2014 Published May 2014
We investigate the long-time behavior of weak solutions to the thin-film type equation $v_t =(xv - vv_{xxx})_x\ ,$ which arises in the Hele-Shaw problem. We estimate the rate of convergence of solutions to the Smyth-Hill equilibrium solution, which has the form $\frac{1}{24}(C^2-x^2)^2_+$, in the norm $|\!|\!| f |\!|\!|_{m,1}^2 = \int_{\mathbb{R}}(1+ |x|^{2m})|f(x)|^2 \, dx + \int_{\mathbb{R}}|f_x(x)|^2 \, dx.$ We obtain exponential convergence in the $|\!|\!| \cdot |\!|\!|_{m,1}$ norm for all $m$ with $1\leq m< 2$, thus obtaining rates of convergence in norms measuring both smoothness and localization. The localization is the main novelty, and in fact, we show that there is a close connection between the localization bounds and the smoothness bounds: Convergence of second moments implies convergence in the $H^1$ Sobolev norm. We then use methods of optimal mass transportation to obtain the convergence of the required moments. We also use such methods to construct an appropriate class of weak solutions for which all of the estimates on which our convergence analysis depends may be rigorously derived. Though our main results on convergence can be stated without reference to optimal mass transportation, essential use of this theory is made throughout our analysis.
Citation: Eric A. Carlen, Süleyman Ulusoy. Localization, smoothness, and convergence to equilibrium for a thin film equation. Discrete & Continuous Dynamical Systems, 2014, 34 (11) : 4537-4553. doi: 10.3934/dcds.2014.34.4537
##### References:
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##### References:
[1] Mario Bukal. Well-posedness and convergence of a numerical scheme for the corrected Derrida-Lebowitz-Speer-Spohn equation using the Hellinger distance. Discrete & Continuous Dynamical Systems, 2021, 41 (7) : 3389-3414. doi: 10.3934/dcds.2021001 [2] Xinqun Mei, Jundong Zhou. The interior gradient estimate of prescribed Hessian quotient curvature equation in the hyperbolic space. Communications on Pure & Applied Analysis, 2021, 20 (3) : 1187-1198. doi: 10.3934/cpaa.2021012 [3] Matthias Erbar, Jan Maas. Gradient flow structures for discrete porous medium equations. Discrete & Continuous Dynamical Systems, 2014, 34 (4) : 1355-1374. doi: 10.3934/dcds.2014.34.1355 [4] Bouthaina Abdelhedi, Hatem Zaag. Single point blow-up and final profile for a perturbed nonlinear heat equation with a gradient and a non-local term. Discrete & Continuous Dynamical Systems - S, 2021 doi: 10.3934/dcdss.2021032 [5] Lifen Jia, Wei Dai. Uncertain spring vibration equation. Journal of Industrial & Management Optimization, 2021 doi: 10.3934/jimo.2021073 [6] V. V. Zhikov, S. E. Pastukhova. Korn inequalities on thin periodic structures. Networks & Heterogeneous Media, 2009, 4 (1) : 153-175. doi: 10.3934/nhm.2009.4.153 [7] Vladimir Georgiev, Sandra Lucente. Focusing nlkg equation with singular potential. Communications on Pure & Applied Analysis, 2018, 17 (4) : 1387-1406. doi: 10.3934/cpaa.2018068 [8] Daoyin He, Ingo Witt, Huicheng Yin. On the strauss index of semilinear tricomi equation. Communications on Pure & Applied Analysis, 2020, 19 (10) : 4817-4838. doi: 10.3934/cpaa.2020213 [9] Carmen Cortázar, M. García-Huidobro, Pilar Herreros, Satoshi Tanaka. On the uniqueness of solutions of a semilinear equation in an annulus. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2021029 [10] Kamel Hamdache, Djamila Hamroun. Macroscopic limit of the kinetic Bloch equation. Kinetic & Related Models, , () : -. doi: 10.3934/krm.2021015 [11] Julian Tugaut. Captivity of the solution to the granular media equation. Kinetic & Related Models, 2021, 14 (2) : 199-209. doi: 10.3934/krm.2021002 [12] Diana Keller. Optimal control of a linear stochastic Schrödinger equation. Conference Publications, 2013, 2013 (special) : 437-446. doi: 10.3934/proc.2013.2013.437 [13] Simone Cacace, Maurizio Falcone. A dynamic domain decomposition for the eikonal-diffusion equation. Discrete & Continuous Dynamical Systems - S, 2016, 9 (1) : 109-123. doi: 10.3934/dcdss.2016.9.109 [14] Naeem M. H. Alkoumi, Pedro J. Torres. Estimates on the number of limit cycles of a generalized Abel equation. Discrete & Continuous Dynamical Systems, 2011, 31 (1) : 25-34. doi: 10.3934/dcds.2011.31.25 [15] Jumpei Inoue, Kousuke Kuto. On the unboundedness of the ratio of species and resources for the diffusive logistic equation. Discrete & Continuous Dynamical Systems - B, 2021, 26 (5) : 2441-2450. doi: 10.3934/dcdsb.2020186 [16] Stefano Bianchini, Paolo Bonicatto. Forward untangling and applications to the uniqueness problem for the continuity equation. Discrete & Continuous Dynamical Systems, 2021, 41 (6) : 2739-2776. doi: 10.3934/dcds.2020384 [17] Peng Chen, Xiaochun Liu. Positive solutions for Choquard equation in exterior domains. Communications on Pure & Applied Analysis, , () : -. doi: 10.3934/cpaa.2021065 [18] José A. Carrillo, Bertram Düring, Lisa Maria Kreusser, Carola-Bibiane Schönlieb. Equilibria of an anisotropic nonlocal interaction equation: Analysis and numerics. Discrete & Continuous Dynamical Systems, 2021, 41 (8) : 3985-4012. doi: 10.3934/dcds.2021025 [19] Samira Shahsavari, Saeed Ketabchi. The proximal methods for solving absolute value equation. Numerical Algebra, Control & Optimization, 2021, 11 (3) : 449-460. doi: 10.3934/naco.2020037 [20] Jonathan DeWitt. Local Lyapunov spectrum rigidity of nilmanifold automorphisms. Journal of Modern Dynamics, 2021, 17: 65-109. doi: 10.3934/jmd.2021003
2019 Impact Factor: 1.338 | open-web-math/open-web-math | |
# Presenting Polygons
11 teachers like this lesson
Print Lesson
## Objective
SWBAT identify polygons and determine the sum of the interior angles for polygons as well as each interior angle measure.
#### Big Idea
Tri, Quad, Pent, Hex ... learn the key vocabulary for polygon shapes and find interior angle measure for these pretty polygons!
## Do Now
10 minutes
In this Do Now, students will review the algebra skills needed to solve problems requiring students to use the formula for sum of the interior angles of polygons and also the formula for exterior angle theorem of polygons.
## Polygon Vocabulary and Examples
20 minutes
This activity provides students with an introduction to polygons with a strong focus on the vocabulary related to polygons. This overview will help to provide the basis with which we will discuss more complicated polygons, like parallelograms, in future lessons.
The pair-share at the beginning helps students to work on making math vocabulary their own while using mathematical precise language (MP 6). I have included possible student answers in the resources. You can use them as is to scaffolded student conversation or you may remove the sample answers prior to creating handouts for your students.
## Exploration
25 minutes
Students will complete an exploration that will guide them through how to derive the formula for the sum of interior angles in any polygon. Students will use repeated reasoning (MP7) to discover the degrees in multiple polygons, like quadrilateral, pentagon, hexagon and decagon. This video from Khan Academy provides a great overview for you and students, and can be used as a review this class or for absent students.
I have found that the hardest part of deriving the formula is helping students to jump from numerical values (like a 5 sided figure has 540 degrees) to using the variable "n." The (n-2) part is tricky for students but the video does a nice job of explaining this at around minute 5. Once you have the formula, students can then explain each piece of the formula again to a partner or to the whole class. The derivation of this formula leads directly into the next formula, the measure of each interior angle in a polygon.
15 minutes
## Activity/Homework and Exit Ticket
15 minutes
The activity/homework can be completed in class or at home by students. These questions focus on applying the formulas for sum of the interior angles in a polygon and the exterior angles in a polygon.
The exit ticket is a quick, formative assessment of students' learning on the sum of the interior angles of a polygon. This exit ticket can easily be extended by asking students to find the measure of each interior angle in an octagon. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Compare 42/4 and 50/3
1st number: 10 2/4, 2nd number: 16 2/3
42/4 is smaller than 50/3
#### Steps for comparing fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 4 and 3 is 12
Next, find the equivalent fraction of both fractional numbers with denominator 12
2. For the 1st fraction, since 4 × 3 = 12,
42/4 = 42 × 3/4 × 3 = 126/12
3. Likewise, for the 2nd fraction, since 3 × 4 = 12,
50/3 = 50 × 4/3 × 4 = 200/12
4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction
5. 126/12 < 200/12 or 42/4 < 50/3
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
# Math
posted by .
A mail-order company charges \$12 for each shirt. Shipping costs \$f per oder.
Write a variable expression for the cost of odering x number of shirts
• Math -
Suppose you buy x shirts to be shipped as one order. What is the cost of the shirts? What is the cost of shipping?
• Math -
I think you need to know how many shirts were ordered per order. If ten shirts were ordered on ONE order, the cost is 10x+4
If ten shirts were ordered on ten orders, then cost is
10x+40
So I am not certain there is a definite answer here.
• Math -
sorry I wrote the questionwrong!!!
A mail order company charges \$12 for each shirt. Shipping costs \$5 per order.
Write a variable expression for the cost of ordering x numbers of shirts.
• Math -
cost=12x+5
That is to order x shirts one one order.
• Math -
i had the same question on my pre-al test i couldn't figure it on when i was correcting it so i googled it and here it is! thanks guys!
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A lightweight framework for dynamic GUI data verification based on scripts
Authors:
To read the full-text of this research, you can request a copy directly from the authors.
Abstract
Runtime verification (RV) provides essential mechanisms to enhance software robustness and prevent malfunction. However, RV often entails complex and formal processes that could be avoided in scenarios in which only invariants or simple safety properties are verified, for example, when verifying input data in Graphical User Interfaces (GUIs). This paper describes S-DAVER, a lightweight framework aimed at supporting separate data verification in GUIs. All the verification processes are encapsulated in an independent layer and then transparently integrated into an application. The verification rules are specified in separate files and written in interpreted languages to be changed/reloaded at runtime without recompilation. Superimposed visual feedback is used to assist developers during the testing stage and to improve the experience of users during execution. S-DAVER provides a lightweight, easy-to-integrate and dynamic verification framework for GUI data. It is an integral part of the development, testing and execution stages. An implementation of S-DAVER was successfully integrated into existing open-source applications, with promising results. Copyright © 2015 John Wiley & Sons, Ltd.
No full-text available
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Conference Paper
In this work we propose MOPBox, a library-based approach to runtime verification. MOPBox is a Java library for defining and evaluating parametric runtime monitors. A user can define monitors through a simple set of API calls. Once a monitor is defined, it is ready to accept events. Events can originate from AspectJ aspects or from other sources, and they can be parametric, i.e., can contain variable bindings that bind abstract specification variables to concrete program values. When a monitor reaches an error state for a binding $$\vec{v}=\vec{o}$$, MOPBox notifies clients of a match for $$\vec{v}=\vec{o}$$ through a call-back interface. To map variable bindings to monitors, MOPBox uses re-implementations of efficient indexing algorithms that Chen et al. developed for JavaMOP. We took care to keep MOPBox as generic as possible. States, transitions and variable bindings can be labeled not just with strings but with general Java Objects whose types are checked through Java Generics. This allows for simple integration into existing tools. For instance, we present ongoing work on integrating MOPBox with a Java debugger. In this work, transitions are labeled with breakpoints. MOPBox is also a great tool for teaching: its implementations of monitor indexing algorithms are much easier to understand than the code generated by tools such as JavaMOP. Indexing algorithms use the Strategy Design Pattern, which makes them easily exchangeable. Hence, MOPBox is also the perfect tool to explore and test new algorithms for monitor indexing without bothering about the complex intricacies of code generation. In the future, we further plan to integrate MOPBox with the Clara framework for statically evaluating runtime monitors ahead of time.
Article
We describe Java-MaC, a prototype implementation of the Monitoring and Checking (MaC) architecture for Java programs. The MaC architecture provides assurance that the target program is running correctly with respect to a formal requirements specification by monitoring and checking the execution of the target program at run-time. MaC bridges the gap between formal verification, which ensures the correctness of a design rather than an implementation, and testing, which does not provide formal guarantees about the correctness of the system. Use of formal requirement specifications in run-time monitoring and checking is the salient aspect of the MaC architecture. MaC is a lightweight formal method solution which works as a viable complement to the current heavyweight formal methods. In addition, analysis processes of the architecture including instrumentation of the target program, monitoring, and checking are performed fully automatically without human direction, which increases the accuracy of the analysis. Another important feature of the architecture is the clear separation between monitoring implementation-dependent low-level behaviors and checking high-level behaviors, which allows the reuse of a high-level requirement specification even when the target program implementation changes. Furthermore, this separation makes the architecture modular and allows the flexibility of incorporating third party tools into the architecture. The paper presents an overview of the MaC architecture and a prototype implementation Java-MaC.
Article
Aspect-oriented programming (AOP) is an exciting new development in the field of software engineering. The open-source AspectJ® project has taken a leading role in defining what an aspect-oriented programming language should look like and in building tools that enable aspect-oriented techniques to be employed in the development of large-scale commercial software. IBM both contributes to the development of AspectJ and uses it internally with its accompanying IDE (integrated development environment) support, AspectJ Development Tools (AJDT). This paper provides an introduction to aspect-oriented programming using AspectJ and AJDT. We also discuss the role that open source (and being an open-source project) has played in the ongoing development of AspectJ, and how this has facilitated a level of collaboration and exploitation that would not have been possible otherwise.
Conference Paper
Runtime verification is a special form of runtime testing, employing formal methods and languages. In this work, we utilize next-time free linear-time temporal logic (LTL\textbackslash X) as formal framework. The discipline serves the purpose of asserting certain design-time assumptions about object-oriented (OO) entities such as objects, methods, and so forth. In this paper we propose a linear-time logic over joinpoints \citeLaddad03AspectJ, and introduce a lightweight runtime veri\-fication tool based on this logic, J2SE 5 metadata \citeJSR175 and an AspectJ-based \citeAspectJ runtime backend. Implementations have been proposed so far for imperative and functional languages \citeHuchStolz04a. To our knowledge our approach is the first to allow addressing of entire sets of states, also over subclass boundaries, thus exploiting the OO nature.
Conference Paper
Specifications that are used in detailed design and in the documentation of existing code are primarily written and read by programmers. However, most formal specification languages either make heavy use of symbolic mathematical operators, which discourages use by programmers, or limit assertions to expressions of the underlying programming language, which makes it difficult to write exact specifications. Moreover, using assertions that are expressions in the underlying programming language can cause problems both in runtime assertion checking and in formal verification, because such expressions can potentially contain side effects. The Java Modeling Language, JML, avoids these problems. It uses a side-effect free subset of Java’s expressions to which are added a few mathematical operators (such as the quantifiers ⧹forall and ⧹exists). JML also hides mathematical abstractions, such as sets and sequences, within a library of Java classes. The goal is to allow JML to serve as a common notation for both formal verification and runtime assertion checking; this gives users the benefit of several tools without the cost of changing notations.
Conference Paper
The process of verifying that a program conforms to its specification is often hampered by errors in both the program and the specification. A runtime checker that can evaluate formal specifications can be useful for quickly identifying such errors. This paper describes our preliminary experience with incorporating run-time checking into the Jahob verification system and discusses some lessons we learned in this process. One of the challenges in building a runtime checker for a program verification system is that the language of invariants and assertions is designed for simplicity of semantics and tractability of proofs, and not for run-time checking. Some of the more challenging constructs include existential and universal quantification, set comprehension, specification variables, and formulas that refer to past program states. In this paper, we describe how we handle these constructs in our runtime checker, and describe directions for future work.
Article
This research addresses the problem of statically analyzing input command syntax as defined in interface and requirements specifications and then generating test cases for dynamic input validation testing. The IVAT (Input Validation Analysis and Testing) technique has been developed, a proof-of-concept tool (MICASA) has been implemented, and a case study validation has been performed. Empirical validation on large-scale industrial software (from the Tomahawk Cruise Missile) shows that as compared with senior, experienced analysts and testers, MICASA found more syntactic requirement specification defects, generated test cases with higher syntactic coverage, and found additional defects. The experienced analysts found more semantic defects than MICASA, and the experienced testers' cases found 7.4 defects per test case as opposed to an average of 4.6 defects found by MICASA test cases. Additionally, the MICASA tool performed at less cost.
Conference Paper
We consider the enforcement powers of program monitors, which intercept security-sensitive actions of a target application at run time and take remedial steps whenever the target attempts to execute a potentially dangerous action. A common belief in the security commu- nity is that program monitors, regardless of the remedial steps available to them when detecting violations, can only enforce safety properties. We formally analyze the properties enforceable by various program monitors and nd that although this belief is correct when considering monitors with simple remedial options, it is incorrect for more powerful monitors that can be modeled by edit automata. We dene an interesting set of properties called innite renewal properties and demonstrate how, when given any reasonable innite renewal property, to construct an edit au- tomaton that provably enforces that property. We analyze the set of innite renewal properties and show that it includes every safety prop- erty, some liveness properties, and some properties that are neither safety nor liveness.
Article
In Barringer et al. (2004,Vol. 2937, LNCS), EAGLE was introduced as a general purpose rule-based temporal logic for specifying run-time monitors. A novel interpretative trace-checking scheme via stepwise transformation of an EAGLE monitoring formula was defined and implemented. However, even though EAGLE presents an elegant formalism for the expression of complex trace properties, EAGLE's interpretation scheme is complex and appears difficult to implement efficiently. In this article, we introduce RULER, a primitive conditional rule-based system, which has a simple and easily implemented algorithm for effective run-time checking, and into which one can compile a wide range of temporal logics and other specification formalisms used for run-time verification. As a formal demonstration, we provide a translation scheme for linear-time propositional temporal logic with a proof of translation correctness. We then introduce a parameterized version of RULER, in which rule names may have rule-expression or data parameters, which then coincides with the same expressivity as EAGLE with data arguments. RULER with just rule-expression parameters extend the expressiveness of RULER strictly beyond the class of context-free languages. For the language classes expressible in propositional RULER, the addition of rule-expression and data parameters enables more compact translations. Finally, we outline a few simple syntactic extensions of ‘core’ RULER that can lead to further conciseness of specification but still enabling easy and efficient implementation.
Article
One of the goals of program verification is to show that a program conforms to a specification written in a formal logic. Oftentimes, this process is hampered by errors in both the program and the spec- ification. The time spent in identifying and eliminating these errors can even dominate the final verification effort. A runtime checker that can evaluate formal specifications can be extremely useful for quickly identifying such errors. Such a checker also enables verifi- cation approaches that combine static and dynamic program analy- ses. Finally, the underlying techniques are also useful for executing expressive high-level declarative languages. This paper describes the run-time checker we are developing in the context of the Jahob verification system. One of the challenges in building a runtime checker for a program verification system is that the language of invariants and assertions is designed for sim- plicity of semantics and tractability of proofs, and not for run-time checking. Some of the more challenging constructs include existen- tial and universal quantification, set comprehension, specification variables, and formulas that refer to past program states. In this pa- per, we describe how we handle these constructs in our runtime checker, and describe several directions for future work.
Article
Methodological guidelines for object-oriented software construction that improve the reliability of the resulting software systems are presented. It is shown that the object-oriented techniques rely on the theory of design by contract, which underlies the design of the Eiffel analysis, design, and programming language and of the supporting libraries, from which a number of examples are drawn. The theory of contract design and the role of assertions in that theory are discussed.< >
Article
This paper addresses those questions for the class of enforcement mechanisms that work by monitoring execution steps of some system, herein called the target, and terminating the target's execution if it is about to violate the security policy being enforced. We call this class EM, for Execution Monitoring. EM includes security kernels, reference monitors, firewalls, and most other operating system and hardware -based enforcement mechanisms that have appeared in the literature. Our targets may be objects, modules, processes, subsystems, or entire systems; the execution steps monitored may range from fine-grained actions (such as memory accesses) to higher-level operations (such as method calls) to operations that change the security-configuration and thus restrict subsequent execution
S-DAVER: script-based data verification for Qt GUIs
• Saes -University Of Cátedra
• Murcia
Cátedra SAES -University of Murcia. S-DAVER: script-based data verification for Qt GUIs, 2014. Available from: http://www.catedrasaes.org/wiki/GuiVerification [last accessed April 2015].
Modular data structure verification Massachusetts Institute of Technology Available from
• V Kuncak
Kuncak V. Modular data structure verification. PhD Thesis, Massachusetts Institute of Technology, 2007. Available from: http://lara.epfl.ch/~kuncak/papers/Kuncak07DecisionProceduresModularDataStructureVerification. pdf [last accessed April 2015].
Qt : cross-platform application and UI framework Available from: http://qt.nokia.com [last accessed
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Nokia Corporation. Qt : cross-platform application and UI framework, 2012. Available from: http://qt.nokia.com [last accessed April 2015].
ChaiScript: easy to use scripting for C++, 2012. Available from
• J Turner
• J Turner
Turner J, Turner J. ChaiScript: easy to use scripting for C++, 2012. Available from: http://www.chaiscript.com/ [last accessed April 2015].
cKanren: miniKanren with Constraints
• Ce Alvis
• Jj Willcock
• Km Carter
• We Byrd
• Dp Friedman
Alvis CE, Willcock JJ, Carter KM, Byrd WE, Friedman DP. cKanren: miniKanren with Constraints. In Proceedings of the 2011 Workshop on Scheme and Functional Programming (Scheme '11), Portland, Oregon, 2011.
The GIMP Toolkit (GTK), version 2.x, 2012. Available from: http://www.gtk.org [last accessed
• Gtk+ The
• Team
The GTK+ Team. The GIMP Toolkit (GTK), version 2.x, 2012. Available from: http://www.gtk.org [last accessed April 2015].
Runtime verification and compensations
• C Colombo
Colombo C. Runtime verification and compensations. PhD Thesis, Dept. of Computer Science, University of Malta, 2012.
Programming ASP.NET MVC 4 – Developing Real-world Web Applications with ASP.NET MVC. O'Reilly Media
• T Snyder
• H Panda
Chadwick J, Snyder T, Panda H. Programming ASP.NET MVC 4 – Developing Real-world Web Applications with ASP.NET MVC. O'Reilly Media: Sebastopol, California, 2012.
Lua 5.1 Reference Manual. Lua.org
• R Ierusalimschy
• L H De Figueiredo
• W Celes
Ierusalimschy R, de Figueiredo LH, Celes W. Lua 5.1 Reference Manual. Lua.org, 2006. Available from: http://www. amazon.com/exec/obidos/ASIN/8590379833/lua-indexmanual-20 [last accessed April 2015].
IEEE 1012-2004 -IEEE standard for software verification and validation
Institute of Electrical and Electronics Engineers. IEEE 1012-2004 -IEEE standard for software verification and validation. IEEE, IEEE 2005:0-110. DOI: 10.1109/IEEESTD.2005.96278.
Runtime verification of safety-progress properties
• Y Falcone
• J C Fernandez
• L Mounier
Falcone Y, Fernandez JC, Mounier L. Runtime verification of safety-progress properties. In RV, Lecture Notes in Computer Science, Vol. 5779, Bensalem S, Peled D (eds). Springer: Berlin Heidelberg, 2009; 40-59.
cKanren: miniKanren with Constraints
• C E Alvis
• J J Willcock
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• D P Friedman
Alvis CE, Willcock JJ, Carter KM, Byrd WE, Friedman DP. cKanren: miniKanren with Constraints. In Proceedings of the 2011 Workshop on Scheme and Functional Programming (Scheme '11), Portland, Oregon, 2011. | open-web-math/open-web-math | |
# Factors Race Activity
Post Contents
I created this factors race activity to give my math concepts students practice listing all of the factors of a number. I shared it on twitter in January 2018, and it’s only now making it to my blog in January 2020. Better late than never, right?!?
To prep for the activity, I printed 4 sheets of cards that contained the numbers 1-100. After cutting the cards apart, the activity is ready to go! You will also need a bucket or basket of some sort.
We had already been practicing divisibility rules and finding the factors of numbers. This was just extra practice. I told students that we were going to have a factor race. They could each come up to my desk and grab any number of their choosing.
Each student had to take their number back to their desk and find all of the factors of the number. They wrote these factors on a dry erase pocket.
MATH = LOVE RECOMMENDS…
I cannot imagine teaching math without my dry erase pockets! They instantly make any activity more engaging and save me countless hours at the copy machine since I can use the same class sets of copies year after year.
Here are my current go-to recommendations:
If you don’t have a classroom set of dry erase pockets, you could also use heavy duty sheet protectors. But, I highly recommend investing in a classroom set of the pockets since they are so much more durable.
Once they had found all the factors, they brought the list of factors to my desk to get checked. If they were correct, they would write their name on the card and place it in the bucket on my desk. They would then proceed to grab a new number and take it back to their desk.
If they were incorrect, I would send them back to their desk to double check their work and see if they had found an incorrect factor OR if they had missed one or more factors.
Students were to race through the activity to complete as many cards as possible. If students were clever, they could use what we had learned about prime numbers to maximize the number of cards in the bucket with their name on it. After all, it’s a lot faster to list all of the factors of 13 then it is to list all the factors of 90.
I intentionally let my students pick their own cards for this very reason.
At the end of class or as soon as all the numbers were in the bucket, I drew a name out of the bucket and gave that student a small prize. Usually I would give out candy as my prize, but I had given out all my candy before Christmas Break. I ended up giving the winning student 5 points extra credit on the next day’s factor quiz. | HuggingFaceTB/finemath | |
# The farmer and the olive trees
A farmer has a rectangular ground of 100 m by 50 m, he wants to plant olive trees, in sufficiently spaced ways (to avoid exhaustion by the roots) at least 10 meters from each other.
How much can one hope to put at the most, effectively?
• May the trees be along the border of the rectangle, where their roots would extend 5 m (or more) outside of it?
– humn
Commented Sep 9, 2017 at 16:19
• yes, the trees can be on the border Commented Sep 9, 2017 at 16:26
• Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it. If not, some responses to the answerers to help steer them in the right direction would be helpful.
– Rubio
Commented Oct 2, 2017 at 5:22
• There is no justification for optimality @Rubio Commented Oct 2, 2017 at 10:02
• Proving optimality for this sort of problem is notoriously difficult. I think a few very small cases have been solved rigorously, but none as large as this one. Commented Oct 27, 2017 at 13:06
67 $68$
as shown in the flower kind figure below (This is the previous answer):
The idea behind this is
Starting to plant from the left and the right at the same time to the right, you will lose one tree every 1.8 meters as you see in the figure but gain 0.1 meter compared to putting trees all the corners etc. So after 10 trees, you will have an extra 6 tree to put as you see on the right side of the flower graph even though you lose 1 tree every 1.8 meters. At the end you will able to get extra onetwo trees to put in your farm compared to standard way of putting trees!
The main difference is actually starting from left and the right, after completing one vertical plantation, go for the next from the furthest point to the area in the farm
as you will see below:
• it is difficult to justify that the proposed solution is optimal. I wait to see if nobody does better. Commented Sep 9, 2017 at 18:11
• @Dattier it has to be optimal since I have started from the very beginning of the rectangle and move to the with the least area covering from left to right. Though I may not able to show it mathematically.
– Oray
Commented Sep 9, 2017 at 18:12
• @Dattier If you are asking for an optimal solution, you should know how to prove the answer is optimal. I think it would be better to ask "How can you plant x trees?", without worrying about whether x is optimal. At the very least, you should say up front that you do not know for sure what the optimal answer is. Commented Sep 9, 2017 at 19:31
• I don't have an answer, but I know a very slow method, for find it. Commented Sep 9, 2017 at 21:11
• To improve the visualisation, draw the circle around each tree at HALF the minimum distance to the next tree, then rather than overlapping by up to 1 radius creating a "flower kind figure", minimally-spaced trees would have circles that just touch, creating a circle packing solution - i.e. similar to the answer to a different problem that you linked here from yesterday. Commented May 26, 2020 at 6:39
As in my answer to My Mother's Dish Collection, I used a nonlinear optimization solver, with variables $$x_i$$, $$y_i$$ to represent the coordinates of the trees. The constraints are: \begin{align} 0 \le x_i &\le 100 &&\text{for i\in\{1,\dots,68\}}\\ 0 \le y_i &\le 50 &&\text{for i\in\{1,\dots,68\}}\\ (x_i - x_j)^2 + (y_i - y_j)^2 &\ge 10^2 &&\text{for 1\le i The first two constraints make sure each tree is contained in the rectangle, and the third constraint enforces a distance of at least 10 between trees.
The resulting $$x$$ and $$y$$ coordinates returned by the solver are essentially the same as @Oray's packing.
The solver was not able to find a feasible solution for 69 trees, but I have no proof that 68 is the maximum.
• Circle packing density $\pi/(2\sqrt 3)$ implies an upper bound of $\lfloor (60\cdot 110)/(5^2 \cdot 2 \sqrt 3)\rfloor=76$. Commented May 24, 2020 at 0:21 | HuggingFaceTB/finemath | |
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# 4.
Seepage
Soil Mechanics
2010 - 2011
Permeability
## Is a measure of how easily a fluid (water) can pass through
a porous medium (soil).
Loose Soil
High permeability
Dense Soil
Low permeability
## Dr. Manal A. Salem Soil Mechanics
Soil Permeability
Applications (examples):
1.
Water wells
a.
b.
2.
Earth dams
a.
b.
3.
Water production
Dewatering
Estimate quantity of water seeping through the dam
Evaluating stability of dam
Darcys Law
## Water moves through soil with
discharge Q, and velocity v.
Q = A.v
v i
v h/L
v = kh/L
Q = Akh/L
where:
Q = V/t
Soil
## Q = water discharge (volume/time)
A = area perpendicular to flow direction
v = velocity of flow (length/time)
L
i = hydraulic gradient = h/L
Direction of flow
L = length parallel to flow direction
k = coefficient of soil permeability
## Dr. Manal A. Salem Soil Mechanics
Coefficient of Permeability k
k=v/i
## Define k: the velocity of water flowing
through a soil medium under a unit hydraulic
Note:
flow of water through soil is governed by:
1.
2.
Soil permeability (k)
## Dr. Manal A. Salem Soil Mechanics
Bernoullis Equation
G.S.
hv1=v12/2g
Clay
hp1=u1/w
1
ht1
Water
flow
Sand
z1
Clay
h t = z + hp + h v
h
hv2=v22/2g
hp2=u2/w
ht2
z2
Datum
where:
hp = pressure head = u/w: u = pore-water pressure
hv = velocity head = v2/2g
very small in soil and may
be neglected
Dr. Manal A. Salem Soil Mechanics
Bernoullis Equation
G.S.
hv1=v12/2g
Clay
hp1=u1/w
1
ht1
Water
flow
Sand
h
2/2g
hv2=v2
hp2=u2/w
z1
ht2
z2
Clay
Datum
h t ~ z + hp
hp is determined using piezometer (later)
h = total head difference, if h = 0, no flow.
Q = Av
Can be measured.
## velocity of flow through voids.
Q = Avvs
Cant be measured, only calculated, how?
Total area
(A)
Area of voids
(Av)
## Seepage and Discharge Velocities
Q = Av = Avvs
Therefore: vs = v ( A/AV)
Multiplying both areas (A and Av) by the length of the
medium (L)
vS = v ( AL / AVL ) = v ( VT / VV )
where:
VT = total volume of sample
VV = volume of voids within sample
By Definition, Vv / VT = n, the soil porosity
Thus
vS = v/ n
Dr. Manal A. Salem Soil Mechanics
Factors affecting k
Soil type
ksand > kclay
Void ratio
kloose sand > kdense
A
sand
Particles orientation
kB > kA
Soil Structure
kflocculated > kdispersed
Type of fluid
Viscosity
,k
flocculated
dispersed
Temperature
Temperature
, Viscosity
,k
Dr. Manal A. Salem Soil Mechanics
Laboratory determination of k
1.
2.
constant
## Apply Darcys law:
Continuous
water supply
Q = Av
V/t = Akh/L
k = VL/Aht
Overflow:
Volume V in
Time t
where
V = volume of water collected
in time = t
Direction
of flow
## A = x-sectional area of soil
specimen
L = length of soil specimen
## Suitable for coarse-grained soils.
Dr. Manal A. Salem Soil Mechanics
## Coefficient of permeability (k) can
be calculated using the following
relationship:
h
Ak
ln 1 =
T
h2 La
ho
hf
where:
Overflow
## h1 = initial head difference at time = 0
h2 = final head difference at time T
a = x-sectional area of standpipe
A = x-sectional area of soil specimen
Direction of
flow
## Suitable for fine-grained soils.
Dr. Manal A. Salem Soil Mechanics
Q = A v = A k i = A.k.
Q at time dt =
From 1 and 2:
dh.a
dt
A.k.
aL dh
.
Ak h
h
T
aL f dh
dt =
Ak ho h
0
hf
aL
ln h ho
T =
Ak
h o Ak
ln h = La T
f
dt =
h
-------- (1)
L
-------- (2)
h
dh.a
=
L
dt
ho
hf
Overflow
Direction of
flow
## Limitations of permeability lab tests
Non-homogeneity of soil
Anisotropy of soil
Sampling disturbance
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
Definitions:
## Aquifer: a water-bearing layer of soil with considerable
amount of water.
Confined versus unconfined aquifers.
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
Definitions:
Piezometers
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
1.
Aquitard
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
1.
Pumping well
Aquitard
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
1.
Pumping well
Aquitard
## Dr. Manal A. Salem Soil Mechanics
10
Field determination of k
1.
Pumping well
Piezometer (1)
Q
r1
r2
Piezometer (2)
## Initial water table
h2
h1
Draw down water table
h2
h1
Aquitard
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
1.
## Pump water from
well at a constant rate
state (water level in
observation wells is
constant)
Field measurements:
Q, r1, r2, h1, h2,
where:
h1 = H h1
h2 = H h2
Pumping well
Initial water
table
Piezometer (1)
r2
r1
Piezometer (2)
h2
h1
Draw down water
table
h1
h2
Aquitard
Calculate k
k=
Q ln(r2 / r1 )
h22 h12
Dr. Manal A. Salem Soil Mechanics
11
Field determination of k
2.
Aquitard
Aquitard
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
2.
## Artesian (confined) Aquifer:
Initial piezometric
surface
Pumping well
Aquitard
## Dr. Manal A. Salem Soil Mechanics
12
Field determination of k
2.
## Artesian (confined) Aquifer:
Initial piezometric
surface
Q
Pumping well
Draw down
piezometric line
Aquitard
## Dr. Manal A. Salem Soil Mechanics
Field determination of k
2.
## Artesian (confined) Aquifer:
Initial piezometric
surface
Q
Pumping well
r1
Piezometer (1)
r2
Piezometer (2)
h2
h1
Draw down
piezometric line
h1
h2
Aquitard
## Dr. Manal A. Salem Soil Mechanics
13
Field determination of k
2.
## Artesian (confined) Aquifer:
Pump water
from well at a
constant rate (Q)
until reach
Initial piezometric
Pumping well
surface
Field
measurements:
Q, r1, r2, h1, h2,
where:
h1 = H h1
h2 = H h2
Piezometer (1)
r2
r1
Piezometer (2)
h2
h1
Draw down
piezometric line
h1
h2
H D
Aquitard
Calculate k
k=
Q ln(r2 / r1 )
2D h2 h1
Dr. Manal A. Salem Soil Mechanics
Field determination of k
tests.
14
## Empirical Correlations for k
1.
Coarse-grained soils
Hazens (1930):
## k (cm / sec) = cD102
where
c = constant ranging from 1 to 2
D10 = effective grain size in mm
Chapuis (2004):
e3
k (cm / sec) = 2.4622 D102
(1 + e)
where
0.7825
e = void ratio
D10 = effective grain size in mm
Dr. Manal A. Salem Soil Mechanics
## Empirical Correlations for k
2.
Fine-grained soils
en
k = C
1+ e
where
C and n = constants determined experimentally
e = void ratio
15
## Empirical Correlations for k
Example: A clayey soil was tested in the lab and the following
values were determined:
Void ratio
k (cm/sec)
1.1
0.302 x 10-7
0.9
0.12 x 10-7
## Estimate k for void ratio = 0.75
(1.1) n
0.302x10 = C
1
1
.
1
+
en
k = C
1
+
e
(0.9) n
0.120x107 = C
1 + 0.9
Dr. Manal A. Salem Soil Mechanics
## Empirical Correlations for k
Example: A clayey soil was tested in the lab and the following
values were determined:
Void ratio
k (cm/sec)
1.1
0.302 x 10-7
0.9
0.12 x 10-7
0.302x107
0.120x107
(1.1) n
1
1
.
1
+
=
(0.9) n
1 + 0.9
n = 5.098
16
## Empirical Correlations for k
Example: A clayey soil was tested in the lab and the following
values were determined:
Void ratio
k (cm/sec)
1.1
0.302 x 10-7
0.9
0.12 x 10-7
## Estimate k for void ratio = 0.75
(1.1)5.098
0.302x107 = C
1 + 1.1
C = 0.390x107
(0.75)5.098
= 0.051x107 cm / sec
k = 0.390x10
1 + 0.75
7
## Dr. Manal A. Salem Soil Mechanics
Typical Values of k
Soil Type
k (cm/sec)
Gravel
100 10-1
Coarse Sand
10-1 10-2
Fine Sand
10-2 10-3
Silty Sand
10-3 10-4
Silt
10-4 10-5
Clay
<10-6
17 | HuggingFaceTB/finemath | |
Dynamics Homework Help
1. Feb 17, 2010
kaos4
1. The problem statement, all variables and given/known data
A Satellite is to be placed in an elliptic orbit about the earth. Knowing that the ration Va/Vp of the velocity at the apogee A to the velocity at perigee P is equal to the ration Rp/Ra of the distance to the center of the earth at P to that at A, and the distance between A and P is 80,000 km, determine the energy per unit mass required to place the satellite in its orbit by launching it from the surface of the earth.
Pic:
Va v-----Ra---------O----Rp----^ Vp
|---------80,000km---------|
2. Relevant equations
Conservation of Momentum: T1 + V1 = T2 + V2
T = .5mv^2
V = - GMm/r
3. The attempt at a solution
Va/Vp = Rp/Ra
Ra = 80,000 - Rp
E = T + V
E = 0.5mv^2 - GMm/r
E/m = .5v^2 - GM/r
I'm not sure where to go next. I know the final answer is 57.5 MJ/kg. How are all the 'r's and 'v's eliminated by just using the ratio? I'm generally able to solve these questions, but I've been working on this one for hours with no luck. Any help is greatly appreciated!
2. Feb 18, 2010
tiny-tim
Hi kaos4!
It's launched from the surface of the Earth … | HuggingFaceTB/finemath | |
# Distance between two polygons
Given two convex polygons P and Q. Determine the distance between them.
The distance between two convex polygons P and Q is defined as the length of the shortest distance between points p and q, such that point p lies inside polygon P, and point q lies inside polygon Q.
A point p is said to be inside polygon P if point p is inside polygon P or if it is on any of its edges.
How many points are there in the polygons?
No of points, of the 2 polygons lie between 3 and 5000 | HuggingFaceTB/finemath | |
## Introductory Algebra for College Students (7th Edition)
-$\frac{1}{10}$
Find the multiplicative inverse by switching the numerator and the denominator and keeping the sign the same. -10=-$\frac{10}{1}$$\to$-$\frac{1}{10}$ | HuggingFaceTB/finemath | |
# How much of a 1M glucose solution (measured in ml) and how much distilled water would you mix to make 1 liter of .5M glucose?
This is a dilution problem wherein we are looking how much volume of the stock solution we need to dilute. We have:
`M_1V_1 = M_2V_2`
`M_1` = 1 M
`V_1` = ?
`M_2` = 0.5 M
`V_2` = 1 L
`M_1V_1 = M_2V_2`
`(1)*V_1 = (0.5)*(1)`
`V_1 = ((0.5)*(1))/1`
`V_1 = 0.5 L * (1000mL)/(1L)` = 500mL volume of 1M glucose solution*
Total volume = volume of sample + volume of diluent
1 L = 0.5 L + volume of diluent
Volume of diluent = 1 - 0.5 = 0.5 L = 500mL of distilled water is needed to dilute the solution.
*1000mL = 1L | HuggingFaceTB/finemath | |
## Mensa Math
Apologies for links to political blogs. There’s some math here. The question is, what is the probability that one person could have survived two mass shootings (e.g., Gilroy and Las Vegas) ?
One fellow provides this calculation:
Las Vegas 2017 attendance: 20,000
Gilroy 2019 attendance: 80,000
I don’t know how many attendees were actually physically present at each event at the time of the shootings, but I’ll assume two thirds, so 14,520 and 52,800.
Proportion of US population present at LV shooting: 14,520 / 350,000,000 = .000041 or .0041%
Proportion of the population NOT at LV is the inverse or 99.9959%
Likelihood of one person being at both events is then: 1 – (.999959^52,800). Which is 88.8%. The number of times this apparently happened is 3, so it’s 0.888^3, or 70%.
In other words, through purely random chance it is more likely than not that 3 people who were at the LV 2017 shooting would also be present at the Gilroy shooting.
another fellow, who is a member of Mensa, provides this calculation:
The Gilroy Garlic Festival is a three-day event, so that 80,000 is reduced to 26,667 before being reduced another one-third as per Uncephalized’s assumption to account for the timing of the event. This brings us to an estimated 17,787 people present at the time of the shootings. Note that reducing the estimated 20,000 Las Vegas attendance by the same one-third gives us 13,340, not 14,520.
[…]
Gilroy probability: Dividing 17,787 by 350,000,000 results in a probability of 0.00005082, or one in 19,677.
Las Vegas probability: Dividing 13,340 by 350,000,000 results in a probability of 0.00003811428, or one in 26,237
Gilroy AND Las Vegas probability: Multiplying 0.00005082 by 0.00003811428 results in a probability of 0.0000000019369677096, or one in 516,270,868.
Someone posits in a comment to the second calculation, meekly, that perhaps the problem is analogous to the “birthday problem“. The Mensan responds:
No. That’s not relevant here because there is no equivalent to the finite number of birthdays in a year.
I’m personally not smart enough to be admitted to Mensa. However, it seems to me that the number of people in the United States is a finite number.
## The Napkin Project by Evan Chen
This is what makes the Internet great:
I’ll be eating a quick lunch with some friends of mine who are still in high school. They’ll ask me what I’ve been up to the last few weeks, and I’ll tell them that I’ve been learning category theory. They’ll ask me what category theory is about. I tell them it’s about abstracting things by looking at just the structure-preserving morphisms between them, rather than the objects themselves. I’ll try to give them the standard example Gp, but then I’ll realize that they don’t know what a homomorphism is. So then I’ll start trying to explain what a homomorphism is, but then I’ll remember that they haven’t learned what a group is. So then I’ll start trying to explain what a group is, but by the time I finish writing the group axioms on my napkin, they’ve already forgotten why I was talking about groups in the first place. And then it’s 1PM, people need to go places, and I can’t help but think:
Man, if I had forty hours instead of forty minutes, I bet I could actually have explained this all.
This book is my attempt at those forty hours.
This project has evolved to more than just forty hours.
The most current draft is also available as a PDF.
## my god, it’s full of stars
High-resolution original image here. Technical details about the EHT:
Creating the EHT was a formidable challenge which required upgrading and connecting a worldwide network of eight pre-existing telescopes deployed at a variety of challenging high-altitude sites. These locations included volcanoes in Hawai`i and Mexico, mountains in Arizona and the Spanish Sierra Nevada, the Chilean Atacama Desert, and Antarctica.
The EHT observations use a technique called very-long-baseline interferometry (VLBI) which synchronises telescope facilities around the world and exploits the rotation of our planet to form one huge, Earth-size telescope observing at a wavelength of 1.3 mm. VLBI allows the EHT to achieve an angular resolution of 20 micro-arcseconds — enough to read a newspaper in New York from a sidewalk café in Paris.
This image is fated to be as iconic as the Pale Blue Dot and Earthrise.
Of particular note is that the algorithm to combine the data from all the different sources was the product of research by Dr. Katie Bouman, who is the overnight face of women in STEM, deservedly so.
Here’s a wide angle shot of the area around the black hole, from NASA’s Chandra X-Ray telescsope:
## Earthrise 50th anniversary
50 years ago on Christmas Eve (Dec 24, 1968), the astronauts aboard Apollo 8 took an amazing series of photos of the rising earth behind the limb of the moon, while in orbit. The first photo was black and white, and subsequent ones (with the earth having risen farther from the moon horizon) were color, and now with some digital magic, these are combined into one image. Glorious.
“Oh my God, look at that picture over there! There’s the Earth comin’ up. Wow, is that pretty!” — Astronaut Bill Anders, Apollo 8
Featured on APOD; Original image credit Apollo 8 / NASA; processing by Jim Weigang; CC license and google photo album.
## Super Blueblood Moon tonight
President Trump will deliver his State of the Union speech at 9:00 PM ET.
Also, there’s a lunar eclipse, in what appears to be a nomenclature coincidence 🙂
## Goodbye, Cassini
I met Cassini in 1996 at JPL before it departed for Saturn. For 20 years I have cheered its mission. That mission is over, and Cassini’s watch has ended.
I posted this six years ago here at haibane, but it’s worth reposting in salute: an incredible compilation of a flyby of the Saturnian system:
## the hype about Hyperloop
America has the means to reduce traffic and connect people to where they want to go in less time — but solving these problems entails politically difficult choices to shift travel away from cars and highways. Any high-tech solution that promises a shortcut around these thorny problems is probably too good to be true.
I can’t help but see an echo of the wishful thinking surrounding the EMDrive in the Hyperloop marketing campaign. Maybe I’ll be proven wrong.
Here’s the original white paper PDF from Elon Musk, and here’s a rather detailed critique by mathematician and transit analyst Alon Levy. Anyone who takes Hyperloop seriously should read both.
## Resist Pi – Today is Half Tau Day
We must resist the usurper Pi! Tau is the true constant, the real savior of mathematics. The above image is essentially all the proof you need, but for those indoctrinated by the 2,000 year old cult of Pi, watch this short video:
or visit http://tauday.com/tau-manifesto
Pi is for Pizza, not math.
## We Just Got Our ’30s Sci-Fi Plots Back
By now, you’ve heard that seven – count ’em, seven – terrestrial planets have been discovered orbiting the ultra-cool M8 star Trappist-1. According to the paper that the research team released yesterday, all of them could potentially have liquid water on their surfaces, although only three are judged to be good candidates: the authors’ model considers it likely that the three innermost planets have succumbed to a runaway greenhouse effect and that the outermost is too cold. But that still leaves three potentially habitable planets in a single system.
Those three – Trappist-1e, 1f and 1g – range from .62 to 1.34 estimated Earth masses, and as one would expect from a red-dwarf system, they’re tidally locked and orbit close to their star with periods of 6 to 12 days. Their orbits are also very close to each other. The distance between the orbits of 1e and 1f is .009 AUs – about 830,000 miles – and 1f passes within 750,000 miles of 1g. This is a system that, even according to its discoverers, shouldn’t exist – their model gives it only an 8.1 percent chance of surviving for a billion years – but as they point out, it obviously does.
There are many more fascinating details about the Trappist-1 system and still more that we have yet to learn. The discoverers hope that further research, and the launch of the James Webb space telescope next year, will enable them to confirm the details of the planets’ atmospheres and possibly look for biological signatures. But in the meantime, for those of us who write SF, the discovery of the Trappist-1 system means this: we just got our pulp-era plots back.
We’ve all read stories from the heady days of the 1930s in which the intrepid heroes travel to Mars or Venus in a few days, take off their space suits, breathe the air, encounter exotic life forms and interact with non-human societies. As we learned more about our solar system, that all got taken away. The jungles of Venus and the canals of Barsoom have long since been relegated to the realm of nostalgia, and if we want aliens in our stories, we have to cross impossible interstellar distances to find them.
But now, there’s a system where all that can happen! Three habitable worlds with orbits less than a million miles apart, Hohmann transfers that can be done in a few weeks with inspired 1950s tech – we’ve got the ingredients for interplanetary travel that’s almost as easy as pulp writers imagined it. And a citizen of Trappist-1f might actually find that Old Venus jungle world one planet in and an arid Old Mars one planet out, and generations of its people could watch their neighbors’ fields and cities grow and dream of one day visiting them. All we need to do to make pulp stories into hard SF again is move them 40 light years.
All right, we’d need to do a little more than that. The planets are tidally locked – and with zero eccentricity, they don’t have libration-generated twilight zones – so we’d need to model the day-side and night-side weather. We’d need to account for the tidal and geological effects of so many worlds so close together, and the atmosphere had better have plenty of ozone to protect against UV and X-ray emissions. But none of those constraints are deal-breakers, and within them, Weinbaum-punk is suddenly acceptable.
That may not last, of course. By this time next year, the research team might have found that the Trappist-1 planets have reducing atmospheres or that there’s insufficient protection from stellar radiation or that some other factor makes pulp SF as impossible in that system as in our own. But right now, it’s wide open to stories of the imagination. We’ve found one spot in the universe where it’s the Golden Age all over again. | HuggingFaceTB/finemath | |
# What is 5 Foot 3 in CM? Discover the Conversion Trick!
Are you trying to find the exact value of 5 foot 3 in centimeters? The conversion between imperial units and metric units can often be tricky, and it can be a challenging task to perform conversions accurately. However, there is a simple conversion trick to convert 5 foot 3 inches into centimeters.
## The Simple Conversion Trick
The easiest and quickest way to convert 5 foot 3 inches into centimeters is by using the following formula:
5 feet x 30.48 + 3 inches x 2.54 = 160.02 centimeters
You can use this formula to convert any measurement in feet and inches into centimeters.
## Breaking Down the Conversion
### Converting Feet to Centimeters
The first part of the formula involves converting the number of feet to centimeters. There are 30.48 centimeters in one foot. So, to convert feet to centimeters, you simply need to multiply the number of feet by 30.48.
In our case, we have 5 feet. So, to convert 5 feet to centimeters, we would perform the following calculation:
• 5 feet x 30.48 = 152.4 centimeters
### Converting Inches to Centimeters
The second part of the formula involves converting the number of inches to centimeters. There are 2.54 centimeters in one inch. So, to convert inches to centimeters, you need to multiply the number of inches by 2.54.
In our case, we have 3 inches. So, to convert 3 inches to centimeters, we would perform the following calculation:
• 3 inches x 2.54 = 7.62 centimeters
Finally, we need to add the results of the two calculations above to get the final answer. In our case, we have:
• 152.4 centimeters (from the conversion of 5 feet to centimeters)
• 7.62 centimeters (from the conversion of 3 inches to centimeters)
When we add these two values together, we get:
152.4 + 7.62 = 160.02 centimeters
Therefore, 5 foot 3 inches is equivalent to 160.02 centimeters.
## Other Conversions
• 1 foot = 30.48 centimeters
• 1 inch = 2.54 centimeters
• 1 centimeter = 0.393701 inches
• 1 meter = 3.28084 feet
## Using Online Conversion Tools
While the formula above is easy enough to use, you may prefer to use an online conversion tool to make the conversion even easier. There are several websites that offer free online conversion tools that can quickly and accurately convert between units of measurement.
## Conclusion
Converting units of measurement can often be a challenging task, but with the simple conversion trick provided above, you can easily convert 5 foot 3 inches into centimeters. Whether you need to convert other measurements or not, this formula provides a straightforward approach to solving any conversion problem.
## FAQs
• What is the conversion factor for feet to centimeters?
• The conversion factor for feet to centimeters is 30.48.
• What is the conversion factor for inches to centimeters?
• The conversion factor for inches to centimeters is 2.54.
• What is the formula to convert feet and inches to centimeters?
• The formula to convert feet and inches to centimeters is: 5 feet x 30.48 + 3 inches x 2.54 = 160.02 centimeters
• What is 5 foot 3 in cm?
• 5 foot 3 inches is equivalent to 160.02 centimeters. | HuggingFaceTB/finemath | |
# Finding a Central Value
When you have two or more numbers it is nice to find a value for the "center".
## 2 Numbers
With just 2 numbers the answer is easy: go half-way between.
### Example: what is the central value for 3 and 7?
Answer: Half-way between, which is 5.
We can calculate it by adding 3 and 7 and then dividing the result by 2:
(3+7) / 2 = 10/2 = 5
## 3 or More Numbers
We can use that idea of "adding then dividing" when we have 3 or more numbers:
### Example: what is the central value of 3, 7 and 8?
Answer: We calculate it by adding 3, 7 and 8 and then dividing the results by 3 (because there are 3 numbers):
(3+7+8) / 3 = 18/3 = 6
Notice that we divide by 3 because we have 3 numbers ... very important!
## The Mean
So far we have been calculating the Mean (or the Average):
Mean: Add up the numbers and divide by how many numbers.
But sometimes the Mean can let you down:
### Example: Birthday Activities
Uncle Bob wants to know the average age at the party, to choose an activity.
There will be 6 kids aged 13, and also 5 babies aged 1.
Add up all the ages, and divide by 11 (because there are 11 numbers):
(13+13+13+13+13+13+1+1+1+1+1) / 11 = 7.5...
The mean age is about 7½, so he gets a Jumping Castle! The 13 year olds are embarrassed, and the 1 year olds can't jump!
The Mean was accurate, but in this case it was not useful.
## The Median
But you could also use the Median: simply list all numbers in order and choose the middle one:
### Example: Birthday Activities (continued)
List the ages in order:
1, 1, 1, 1, 1, 13, 13, 13, 13, 13, 13
Choose the middle number:
1, 1, 1, 1, 1, 13, 13, 13, 13, 13, 13
The Median age is 13 ... so let's have a Disco!
Sometimes there are two middle numbers. Just average those two:
### Example: What is the Median of 3, 4, 7, 9, 12, 15
There are two numbers in the middle:
3, 4, 7, 9, 12, 15
So we average them:
(7+9) / 2 = 16/2 = 8
The Median is 8
## The Mode
The Mode is the value that occurs most often:
### Example: Birthday Activities (continued)
Group the numbers so we can count them:
1, 1, 1, 1, 1, 13, 13, 13, 13, 13, 13
"13" occurs 6 times, "1" occurs only 5 times, so the mode is 13.
How to remember? Think "mode is most"
But Mode can be tricky, there can sometimes be more than one Mode.
### Example: What is the Mode of 3, 4, 4, 5, 6, 6, 7
Well ... 4 occurs twice but 6 also occurs twice.
So both 4 and 6 are modes.
When there are two modes it is called "bimodal", when there are three or more modes we call it "multimodal".
## Outliers
Outliers are values that "lie outside" the other values.
They can change the mean a lot, so we can either not use them (and say so) or use the median or mode instead.
### Example: 3, 4, 4, 5 and 104
Mean: Add them up, and divide by 5 (as there are 5 numbers):
(3+4+4+5+104) / 5 = 24
24 does not represent those numbers well at all!
Without the 104 the mean is:
(3+4+4+5) / 4 = 4
But please tell people you are not including the outlier.
Median: They are in order, so just choose the middle number, which is 4:
3, 4, 4, 5, 104
Mode: 4 occurs most often, so the Mode is 4
3, 4, 4, 5, 104
## Other Means
The mean (average) we have been looking at is more correctly called the Arithmetic Mean.
There are other types of mean! Here are two examples:
The Geometric Mean multiplies the numbers together, then does a square root or cube root etc depending on how many numbers, like in this example:
### Example: The Geometric Mean of 2 and 18
• First we multiply them: 2 × 18 = 36
• Then (as there are two numbers) take the square root: √36 = 6
The Harmonic Mean adds up "1 divided by number" then flips it like this:
### Example: The Harmonic Mean of 2, 4, 5 and 100
With 4 numbers we get:
4 = 4 = 4.17 (to 2 places) 12 + 14 + 15 + 1100 0.96 | HuggingFaceTB/finemath | |
# How do you write 0.000419 in scientific notation?
Apr 12, 2018
$4.19 \cdot {10}^{- 4}$
#### Explanation:
Scientific notation is always written as $a \cdot {10}^{b}$ where $a$ must be between $1$ and $10$.
You can see that we must multiply $0.00419$ by $10$ a total of four times (count out the number of spaces the decimal moves to the right). You need to make sure that the decimal is between the first positive integer, $4$ and the next one $9$.
However, if you multiplied the original number by ${10}^{4}$ to get $4.19$, you must balance it by multiplying it by the same amount.
$4.19 \cdot {10}^{- 4}$ is your answer.
Apr 12, 2018
$4.19 \cdot {10}^{-} 4$
#### Explanation:
3. Count the number of times the decimal point move from the original spot to the new spot, and it is times (X) 10 to the power of the amount of times it moved (X${10}^{n}$). NOTE: a) If the decimal point move backwards (right) then it is to the negative(-) power. b) If the decimal point move forward (left) then it is to the positive power.
Eg, 0.000419 => 4.19 X ${10}^{-} 4$ the number of times the decimal place moved in this case is 4 times backward.
Eg. 41900 => 4.19 X ${10}^{4}$ the number of times the decimal point moved in this case is 4 times forward. | HuggingFaceTB/finemath | |
## Algebra 1
Use the Cosine rule for the right triangle: $\frac{adjacent}{hypotenuse}$ cos(x) = $\frac{5}{10}$ SImplify the above equation: x = 60 degrees Since cosine of 60 degrees is half: x =60 degrees | HuggingFaceTB/finemath | |
### Monkey Raptor
Home Posts List Feeds — Atom (XML) Online Tools
## Wednesday, June 29, 2016
### Math: How many perfect squares are divisors of the product 1! • 2! • 3! • 4! • 5! • 6! • 7! • 8! • 9!?
This is from 2003 AMC (American Mathematics Competitions).
The problem looks like this:
How many perfect squares are divisors of the product 1! • 2! • 3! • 4! • 5! • 6! • 7! • 8! • 9!?
(A) 504 (B) 672 (C) 864 (D) 936 (E) 1008
The exclamation after the number means factorial. For instance, 3! = 1 • 2 • 3 = 6.
Let's break the sequence
The multiplication of that factorial sequence above looks like so:
Seriously, it looks like that.
I use middle dot (•) as the multiplication operator (usually ×) to make it consistent with the title above.
Wut is perfect square?
It's a number (prime or composite -- in this case, a positive integer) with (positive) even exponent (0 is also "even").
For example: 22 or 64 or 140 and such.
Let's "prime" it
The sequence above can be grouped into only prime numbers (such as in prime factorization).
As you can see, the prime parts are 2, 3, 5, and 7. The others (besides 1) are composites, let them be. We're gonna take just the primes.
We wanna construct them into:
2 a • 3 b • 5 c • 7 d
This will be useful because later we can find the ones with even exponent (perfect square) much easier.
The a (exponent for 2)
Take a look the sequence above vertically, and find all numbers which have "2" in them.
For purely number "2", we have 8.
For number "4", we have 6. Since 4 = 2 • 2, then we have another 12 (6 × 2) of "2".
For number "6", we have 4. Since 6 = 2 • 3, then we have another 4 of "2".
For number "8", we have 2. Since 8 = 2 • 2 • 2, then we have another 6 (2 × 3) of "2".
So then, all of probable factors (or the maximum exponent) of "2" is 8 + 12 + 4 + 6 = 30.
Thus, the a is 30.
The b (exponent for 3)
Let's find all numbers which have "3".
For purely number "3", we have 7.
For number "6", we have 4. Since 6 = 2 • 3, then we have another 4 of "3".
For number "9", we have 2. Since 9 = 3 • 3, then we have another 4 (2 × 2) of "3".
So then, total exponent for "3" is 7 + 4 + 2 = 13.
Thus, the b is 13.
The c (exponent for 5)
Let's find all numbers which have "5".
For purely number "5", we have 5.
That's it, nothing else.
Thus, the c is 5.
The d (exponent for 7)
Let's find all numbers which have "7".
For purely number "7", we have 3.
And that's it.
Thus, the d is 3.
Therefore,
1! • 2! • 3! • 4! • 5! • 6! • 7! • 8! • 9! can also be written as:
2 30 • 3 13 • 5 5 • 7 3
The perfect square part from each
We are only interested in the even ones.
• For "2", the exponent is 30, it has 16 even parts: 0, 2, 4, 6, 8, 10, 12, ..., 30.
• For "3", the exponent is 13, it has 7 even parts: 0, 2, 4, 6, 8, 10, 12.
• For "5", the exponent is 5, it has 3 even parts: 0, 2, 4.
• For "7", the exponent is 3, it has 2 even parts: 0, 2.
Multiply all the even ones
The answer is 16 • 7 • 3 • 2 = 672 (the B option).
It means there are 672 perfect squares that can be used to divide the thingy above without resulting remainder.
Why multiply? Because it is a multiplication combinatoric pattern, so we multiply them. Hm.
Like for instance, the multiplication of 24 and 52 can be used to divide the product of the dilly.
That's about it.
This problem is about observing pattern and then doing the algebraic maneuver that fits the pattern.
This can be proven with smaller sequence.
For example:
• 1! • 2! • 3! ► total of perfect square divisors for the multiplication product is 2 (1 and 4).
• 1! • 2! • 3! • 4! ► total of perfect square divisors for the multiplication product is 6 (1, 4, 9, 16, 36, and 144 -- we cannot form "25"/"49"/"64"/others from the sequence).
• Et al...
Math: How many perfect squares are divisors of the product 1! • 2! • 3! • 4! • 5! • 6! • 7! • 8! • 9!?
https://monkeyraptor.johanpaul.net/2016/06/math-how-many-perfect-squares-are_29.html
#### Post a Comment
Tell me what you think... | HuggingFaceTB/finemath | |
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# Show that any matrix A can be written as a sum of rank-1 matrices. And show how these rank-1 matrices can be chosen so that only r of them are necessary (where r=rank(A)).
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Show that any matrix A can be written as a sum of rank-1 matrices. And show how these rank-1 matrices can be chosen so that only r of them are necessary (where r=rank(A)).
##### Solution Summary
Rank-1 matrices are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.
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rank-1
________________________________________
Show that any matrix A can be written as a sum of rank-1 matrices. And show how these rank-1 matrices can be chosen so that ...
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###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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Understanding of how geometry applies to in real-world contexts
##### Probability Quiz
Some questions on probability | HuggingFaceTB/finemath | |
+0
# Trigonometry
0
303
1
(a) In ΔABC, given that ∠ A = 38°, ∠ C = 85°, and c = 32 cm.
(b) In ΔABC, given that ∠ A = 24°, b = 12.5 m, and c = 13.2 m.
Sep 25, 2017
#1
+17747
+1
(a) angle(A) = 38o and angle(C) = 85o ---> angle(B) = 180o - angle(A) - angle(B) = 180o - 38o - 85o = 57o
Since you know side(c) = 32 cm, you can use the Law of Sines to find sides a and b:
side(a) / sin( angle(A) ) = side(c) / sin( angle(C) )
side(a) / sin( 38o ) = 32 / sin( 85o )
side(a) = 19.77 cm
side(b) / sin( angle(B) ) = side(c) / sin( angle(C) )
side(b) / sin( 57o ) = 32 / sin( 85o )
side(a) = 26.94 cm
(b) angle(A) = 24o and side(b) = 12.5 m side(c) = 13.2 m
This describes a S-A-S situation, so you can use the Law of Cosines:
a2 = b2 + c2 - 2·b·c·cos(A)
a2 = (12.5)2 + (13.2)2 - 2·(12.5)·(13.2)·cos(24)
a = 5.39 m
You can use the Law of Sines to find angle(B)
[Warning: Don't use the Law of Sines to find the largest angle in a triangle unless you have to!]
sin(B) / side(b) = sin(A) / side(a)
sin(B) / 12.5 = sin(24o) / 5.39
angle(B) = 70.6o
By subtracting angle(C) = 180o - angle(A) - angle(B) = 180o - 24o - 70.6o = 85.4o
Sep 25, 2017 | HuggingFaceTB/finemath | |
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# X2 t01 03 argand diagram (2013)
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• 1. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 2 1 -4 -3 -2 -1 -1 -2 -3 1 2 3 4 x (real axis)
• 2. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3
• 3. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3 B
• 4. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i C = -2 + i C 1 A -4 -3 -2 -1 1 2 3 4 x (real axis) -1 -2 -3 B
• 5. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 2 B = -3i C = -2 + i C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) -1 D -2 -3 B
• 6. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 NOTE: Conjugates 2 B = -3i are reflected in the E real (x) axis C = -2 + i C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D -2 -3 B
• 7. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4 3 A=2 NOTE: Conjugates 2 B = -3i are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: Opposites are -3 B rotated 180°
• 8. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4G NOTE: Multiply 3 A=2 NOTE: Conjugates by i results in 2 B = -3i 90° rotation are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: G = i(4 - i) Opposites are -3 B rotated 180° = 1+4i
• 9. The Argand Diagram Complex numbers can be represented geometrically on an Argand y (imaginary axis) Diagram. 4G NOTE: Multiply 3 A=2 NOTE: Conjugates by i results in 2 B = -3i 90° rotation are reflected in the E real (x) axis C = -2 + i F C 1 A D=4-i -4 -3 -2 -1 1 2 3 4 x (real axis) E=4+i -1 D F = -(4 - i) -2 = - 4 + i NOTE: G = i(4 - i) Opposites are -3 B rotated 180° = 1+4i Every complex number can be represented by a unique point on the Argand Diagram.
• 10. Locus in Terms of Complex Numbers Horizontal and Vertical Lines
• 11. Locus in Terms of Complex Numbers Horizontal and Vertical Lines y c x Im z   c
• 12. Locus in Terms of Complex Numbers Horizontal and Vertical Lines y c y x Im z   c k x Re z   k
• 13. y R Circles zz  R 2 R R R x
• 14. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49
• 15. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49
• 16. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49
• 17. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 x 2  8 x  16  y 2  2 y  1  49
• 18. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2  49
• 19. y R Circles zz  R 2 R R x R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2 Locus is a circle centre :  4 ,  1 radius : 7 units  49
• 20. y R Circles  z    z     R 2 zz  R 2 R R x Locus is a circle centre ω radius R R e.g.Find and describe the locus of points in the Argand diagram; (i)  z  4  i  z  4  i   49 z z  (4  i ) z  (4  i ) z  (4  i )(4  i )  49 x 2  8 x  16  y 2  2 y  1  49 z z  4( z  z )  iz  iz  (4  i )(4  i )  49  x  4    y  1 z z  4( z  z )  iz  iz  (4  i )(4  i )  49   z z  4( z  z )  iz  iz  (4  i )(4  i )  49 x 2  y 2  8 x  2 y  16  1  49 2 2 Locus is a circle centre :  4 ,  1 radius : 7 units  49
• 21. 1 1 (ii )   1 z z
• 22. 1 1 (ii )   1 z z z  z  zz
• 23. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2
• 24. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2 x2  2 x  y 2  0 ( x  1) 2  y 2  1
• 25. 1 1 (ii )   1 z z z  z  zz 2x  x 2  y 2 x2  2 x  y 2  0 ( x  1) 2  y 2  1 Locus is a circle centre: 1,0  radius: 1 unit
• 26. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 1 2 x
• 27. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 excluding the point (0,0) 1 2 x
• 28. 1 1 (ii )   1 z z z  z  zz Locus is a circle centre: 1,0  radius: 1 unit 2x  x 2  y 2 x2  2 x  y 2  0 y ( x  1) 2  y 2  1 excluding the point (0,0) 1 2 x Cambridge: Exercise 1C; 1 ace, 2 bdf, 3, 4 ace, 5 bdfh, 6, 10 to 15 | HuggingFaceTB/finemath | |
# [SOLVED]Pharaoh's modified Euler method question from Yahoo Answers
#### CaptainBlack
##### Well-known member
Part 2 of Pharaoh's Taylor series and modified Euler question from Yahoo Answers
consider van der pol's equation
y" - 0.2(1-y^2)y' + y = 0 y(0)=0.1 y'(0)=0.1
2)
Write down the above problem as a system of first order differential equations.
Calculate the numerical solution at x = 0.2 using the Modified Euler method. Take the
step-length h = 0.1 and work to 6 decimal places accuracy. Compare with your solution
There is a standard method of converting higher order ODEs into first order systems, in this case it is to introduce the state vector:
$$Y(t)=\left[ \begin{array}{c} y(t) \\ y'(t) \end{array} \right]$$
Then the ODE becomes:
$$Y'(t)= F(t,Y)=\left[ \begin{array}{c} y'(t) \\ y''(t) \end{array} \right]= \left[ \begin{array}{c} y'(t) \\ 0.2 \left(1-(y(t))^2 \right) y'(t)-y(t) \end{array} \right]$$
Now you can use the standard form of the modified Euler method on this vector first order ODE.
CB
• Jameson | HuggingFaceTB/finemath | |
# College Algebra
posted by .
I posed these one's before but added parenthesis which I think made it confusing as the actually questions do not have them. So, I'm reposting them without to see if it makes sense. If not I understand.
a. -6a-16/(a+2)(a-2)^2
b. -6a+4/(a+2)(a-2)^2
c. -6a-16/(a+2)(a-2)
d. -6a+4/(a+2)^2(a-2)
2. Use the properties of exponents to simplify the expression. Write the answer using positive exponents only. So the equation is in parenthesis. (a*b^-4/c^-7)^-3
a. ab^12/c^21
b. b^12/a^3c^21
c. b^12c^7/a^3
d. ab^12c^7
3. Compute as indicated. Write your answer in lowest terms. So this one starts with 3/m-8 then at the end next to the divide line it shows -5. The answer for A start with a negative next to the divide line.
a. - 2/m-8
b. -5m-5/m-8
c. -5m-37/m-8
d. -5m+43/m-8
4. Simplify the compound rational expression. Use either method. So this equation has a +1 before the top equation and a +y before the bottom.
+1 5/y-10/ +y 25/y-10
a. 1/y-5
b. 1/y+5
c. 6/y+25
d. 1/y+1/5
5. Simplify the compound rational expression. Use either method. 2/a-1/3/4/a^2-1/9
a. a+3/3
b. a-3/3
c. 3a/6+a
d. 3a/6-a
6. Compute as indicated. Write final results in lowest terms. x^2+2x-15/2x-6 Divide by x^2-25/4x^2
a. 2x^2/x-3
b. 2x^2/x+3
c. 2x^2/x-5
d. x-5/2x^2
Thanks again for all the help its helping me!
• College Algebra -
doesn't help me. This forum doesn't have formatting.
a+3/a^2-4-a+5/a^2-4a+4
means nothing to me
Parentheses are your only option to indicate how terms and factors are formed. Maybe at least some spaces.
Judging from the answers, I'd say you mean
(a+3)/(a^2-4) - (a+5)/(a^2-4a+4)
(a+3) / (a+2)(a-2) - (a+5) / (a-2)(a-2)
If you put all over a common denominator of (a-2)(a-2)(a+2), then the numerator is
(a+3)(a-2) - (a+5)(a+2)
= (a^2+a-6) - (a^2+7a+10)
= -6a-16
(C)
2.
(a*b^-4/c^-7)^-3
(ac^7/b^4)^-3
(b^4/ac^7)^3
b^12 / a^3c^21
(B)
3.
3/(m-8) - 5
3/(m-8) - 5(m-8)/(m-8)
(3 - (5(m-8))/(m-8)
(3 - 5m + 40)/(m-8)
(43-5m)/(m-8)
(D)
4.
no clue what you mean
5.
no clue
6.
x^2+2x-15 = (x+5)(x-3)
2x-6 = 2(x-3)
so x^2+2x-15/2x-6 = (x+5)/2
and we have
(x+5)/2 * 4x^2/(x^2-25)
(x+5)/2 * 4x^2/(x+5)(x-5)
2x^2/(x-5)
(C)
• College Algebra -
You will need those brackets.
Your text book does not need them because the type was probably set as a fraction with an actual line separating numerator and denominator
e.g. your #1 should probably say, (I am guessing)
(a+3)/(a^2-4) - (a+5)/(a^2-4a+4)
which then would be
= (a+3)/((a+2)(a-2)) - (a+5)/((a-2)(a-2)
so the common denominator is (a+2)(a-2)(a-2)
= ( (a+3)(a-2) - (a+5)(a+2))/((a+2)(a-2)(a-2))
= (a^2 + a - 6 - (a^2 + 7a + 10))/((a+2)(a-2)(a-2))
= (-6a -16)/((a+2)(a-2)^2)
which is close to a) if you put the necessary brackets around your choice a)
I hope you understand why we need these brackets with the way we have to type here
a + 3/3
the way that is typed it would reduce to a + 1
but in the book it was meant to be (a+3)/3
suppose you let a = 6,
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HOMOLOGY
Homology package contains basic theorems about general homology theory.
SETS
Definition (Proposition, Set). Type $A : \mathcal{U}$ is a proposition if all elements of $A$ are equal. Type $A : \mathcal{U}$ is a set if all paths between elements of $A$ are equal. $$isProp(A) ≔ \prod_{x, y : A} x = y$$ $$isSet(A) ≔ \prod_{x, y : A} \prod_{p, q : x = y} p = q$$
n_grpd (A: U) (n: N): U = (a b: A) -> rec A a b n where rec (A: U) (a b: A) : (k: N) -> U = split { Z -> Path A a b ; S n -> n_grpd (Path A a b) n } isProp (A: U): U = n_grpd A Z isSet (A: U): U = n_grpd A (S Z)
GROUPS
Definition (Monoid). Monoid is a set $M$ equipped with binary associative operation $(x, y \mapsto x \cdot y) : M \rightarrow M \rightarrow M$ called multiplication and identity element $1$ satisfying $x \cdot 1 = 1 \cdot x = x$.
isMonoid (M: SET): U = (op: M.1 -> M.1 -> M.1) * (assoc: isAssociative M.1 op) * (id: M.1) * (hasIdentity M.1 op id) ismonoidhom (a b: monoid) (f: a.1.1 -> b.1.1): U = (_: preservesOp a.1.1 b.1.1 (opGroup a) (opGroup b) f) * (preservesId a.1.1 b.1.1 (idGroup a) (idGroup b) f) monoidhom (a b: monoid): U = (f: a.1.1 -> b.1.1) * (ismonoidhom a b f)
Definition (Group). Group is a monoid with inversion $(x \mapsto x^{-1})$ satisfying $x^{-1} \cdot x = x \cdot x^{-1} = 1$.
isGroup (G: SET): U = (m: isMonoid G) * (inv: G.1 -> G.1) * (hasInverse G.1 m.1 m.2.2.1 inv) opGroup (g: group): g.1.1 -> g.1.1 -> g.1.1 idGroup (g: group): g.1.1 invGroup (g: group): g.1.1 -> g.1.1
Definition (Differential Group).
isDifferentialGroup (G: SET): U = (g: isGroup G) * (comm: isCommutative G.1 g.1.1) * (boundary: G.1 -> G.1) * ((x: G.1) -> Path G.1 (boundary (boundary x)) g.1.2.2.1) dgroup: U = (X: SET) * isDifferentialGroup X dgrouphom (a b: dgroup): U = monoidhom (a.1, a.2.1.1) (b.1, b.2.1.1) unitDGroup: dgroup
Definition (Division). $$x \backslash y ≔ x^{-1} \cdot y$$ $$x / y ≔ x \cdot y^{-1}$$
ldiv (H: group) (g h: H.1.1) : H.1.1 = (opGroup H) ((invGroup H) g) h rdiv (H: group) (g h: H.1.1) : H.1.1 = (opGroup H) g ((invGroup H) h)
Definition (Conjugation). Let $G$ be a group, the conjugation of two elements of the group $x,y \in G$ is defined as $x y x^{-1}$.
conjugate (G: group) (g1 g2: G.1.1): G.1.1 = rdiv G ((opGroup G) g1 g2) g1
SUBGROUPS
Definition (Predicate). Type family $P : A \rightarrow \mathcal{U}$ is a predicate iff $P(x)$ is a mere proposition for all $x : A$. $$\forall (x : A), isProp(P(x))$$
subtypeProp (A: U): U = (P : A -> U) * (a : A) -> isProp (P a)
Definition (Subtype). Let $P : A \rightarrow \mathcal{U}$ be a predicate. Then: $$\{ x \in A \mid P(x) \} ≔ \sum_{x : A} P(x)$$
subtype (A : U) (P : subtypeProp A): U = (x : A) -- prop * (P.1 x) -- level
Definition (Subgroup). Predicate $\phi : G \rightarrow \mathcal{U}$ is a subgroup iff 1) $1 \in \phi$ 2) $\forall (x : G), x \in \phi \implies x^{-1} \in \phi$ and 3) $\forall (x, y : G), x \in \phi \land y \in \phi \implies (x \cdot y) \in \phi$
subgroupProp (G: group): U = (prop: G.1.1 -> U) * (level: (x: G.1.1) -> isProp (prop x)) * (ident: prop (idGroup G)) * (inv: (g: G.1.1) -> prop g -> prop ((invGroup G) g)) * ((g1 g2: G.1.1) -> prop g1 -> prop g2 -> prop ((opGroup G) g1 g2))
Definition (Normal Subgroup). Subgroup $\phi$ is normal iff for every $g \in \phi$ it contains conjugate of $g$.
isNormal (G: group) (P: subgroupProp G) : U = (X: group) * (g1 g2: G.1.1) -> P.1 g2 -> P.1 (conjugate G g1 g2) normalSubgroupProp (G: group): U = (P: subgroupProp G) * isNormal G P
KERNEL and IMAGE
Definition (Kernel of Homomorphism). $$\mathrm{Ker}(\phi) ≔ \{ x \in G \mid f(x) = 1 \}$$
isGroupKer (G H: group) (f: G.1.1 -> H.1.1) (x: G.1.1): U = Path H.1.1 (f x) (idGroup H)
Definition (Image of Homomorphism). $$\mathrm{Im}(\phi) ≔ \left\{ y \in H \middle| \left( \left\| \sum_{x : G} f(x) = y \right\| \right) \right\}$$
isGroupIm (G H: group) (f: G.1.1 -> H.1.1) (g: H.1.1): U = propTrunc (fiber G.1.1 H.1.1 f g)
Theorem (Kernel of Homomorphism is subgroup).
kerProp (G H: group) (phi: grouphom G H) : subgroupProp G
Theorem (Image of Homomorphism is subgroup).
imProp (G H: group) (phi: grouphom G H) : subgroupProp H
Definition (Set-Quotient). Assume some type $A : \mathcal{U}$ and relation $R : A \rightarrow A \rightarrow \mathcal{U}$ on it. We define $\| A/R \|_0$ as following Higher Inductive Type: $$\| A/R \|_0 ≔ \begin{cases} f : A \rightarrow A/R \\ p : \prod_{a, b : A} \prod_{r : R(a, b)} f(a) = f(b) \\ q : \prod_{a, b : \| A/R \|_0} \prod_{p, q : a = b} p = q \end{cases}$$
data setQuot (A: U) (R: A -> A -> U) = quotient (a: A) | identification (a b: A) (r: R a b) <i>[ (i=0) -> quotient a, (i=1) -> quotient b ] | trunc (a b : setQuot A R) (p q : Path (setQuot A R) a b) <i j> [ (i = 0) -> p @ j , (i = 1) -> q @ j , (j = 0) -> a , (j = 1) -> b ]
Definition (Factor Group). Let $G$ be a group and $\phi$ his normal subgroup. We define: $$\begin{cases} R : G \rightarrow G \rightarrow \mathcal{U} \\ R ≔ (x, y) \mapsto (x / y) \in \phi \end{cases}$$ $$G/\phi ≔ \| G/R \|_0$$ We can also define $G\backslash\phi$ by relation $(x, y) \mapsto (x \backslash y) \in \phi$. If $\phi$ is normal subgroup then $G/\phi \simeq G\backslash\phi$. This statement is proven in Lean.
factorProp (G : group) (P : normalSubgroupProp G) : G.1.1 -> G.1.1 -> U = \(x y : G.1.1) -> P.1.1 (rdiv G x y) factor (G : group) (P : normalSubgroupProp G) : U = setQuot G.1.1 (factorProp G P)
Theorem (Factor group of dihedral group $D_3$). As an test of factor group $G/\phi$ correctness we prove that $D_3/A_3 \cong Z_2$, where $A_3 = \{ R_0, R_1, R_2 \} \subset D_3$ and $Z_2 ≅ \mathbb{Z}/2\mathbb{Z}$, i. e. smallest nontrivial group.
def D₃.iso : Z₂ ≅ D₃\A₃
Definition (Trivial homomorphism). Trivial homomorphism $0$ between two groups (or monoids) $G$ and $H$ maps every element of $G$ to identity element of $H$: $x \mapsto 1_H$
trivmonoidhom (a b : monoid) : monoidhom a b = (\(x : a.1.1) -> idMonoid b, \(x y : a.1.1) -> <i> (hasIdMonoid b).1 (idMonoid b) @ -i, <_> idMonoid b) trivabgrouphom (a b : abgroup) : abgrouphom a b = trivmonoidhom (group' (abgroup' a)) (group' (abgroup' b))
Definition (Chain complex). Chain complex consists of: 1) Sequence of abelian groups $K_n$. 2) Homomorphisms between these groups $\delta_n : K_{n + 1} \rightarrow K_n$. 3) Requirement: the composition of two consecutive homomorphisms is trivial: $\delta_n \circ \delta_{n + 1} = 0$ $$\ldots \xrightarrow{\delta_{n+2}} K_{n+1} \xrightarrow{\delta_n} K_n \xrightarrow{\delta_{n - 1}} \ldots \xrightarrow{\delta_1} K_1 \xrightarrow{\delta_0} K_0$$
chainComplex : U = (K : nat -> abgroup) * (hom : (n : nat) -> abgrouphom (K (succ n)) (K n)) * ((n : nat) -> Path (abgrouphom (K (succ2 n)) (K n)) (abgrouphomcomp (K (succ2 n)) (K (succ n)) (K n) (hom (succ n)) (hom n)) (trivabgrouphom (K (succ2 n)) (K n)))
Definition (Cycles). $Z_n(X) ≔ \mathrm{Ker}(\delta_n)$
propZ (C : chainComplex) (n : nat) : subgroupProp (K' C (succ n)) = kerProp (K' C (succ n)) (K' C n) (hom C n) Z (C : chainComplex) (n : nat) : group = subgroup (K' C (succ n)) (propZ C n)
Definition (Boundaries). $B_n(X) ≔ \mathrm{Im}(\delta_{n + 1})$
B (C : chainComplex) (n : nat) : normalSubgroupProp (Z C n) = abelianSubgroupIsNormal (abelianSubgroupIsAbelian (K C (succ n)) (propZ C n)) (subgroupSubgroup (K' C (succ n)) (imProp (K' C (succ (succ n))) (K' C (succ n)) (hom C (succ n))) (propZ C n))
Definition (Homology Group). $H_n(X) ≔ Z_n(X) / B_n(X)$
H (C : chainComplex) (n : nat) : group = factorGroup (Z C n) (B C n)
Theorem (First Group Isomorphism Theorem). Let $G$ and $H$ be groups, and let $\phi : G \rightarrow H$ be a homomorphism. Then: 1) The kernel of $\phi$ is normal subgroup of G. 2) The image of $\phi$ is a subgroup of $H$. 3) The image of $\phi$ is isomorphic to the quotient group $G/ker(\phi)$.
By composition of $phiUnfold$ and $conjOne$ we obtain a path $\phi (g_1 \cdot g_2 \cdot g_1^{-1}) = 1$. Therefore, $\phi$ contains every conjugation of $g_2$
kernelIsNormalSubgroup (G H : group) (phi : grouphom G H) : normalSubgroupProp G | HuggingFaceTB/finemath | |
Home » Blog » Compute the volume of a solid with given properties
# Compute the volume of a solid with given properties
Given a solid with circle base of radius 2 and cross sections which are equilateral triangles, compute the volume of the solid.
We may describe the top half of the circular base of the solid by the equation Thus, the length of the base of any equilateral triangular cross section is Since these are equilateral triangles with side length , the area is given by Then we compute the volume, (Note: Apostol gives the solution in the back of the book, but I keep getting , as does Edwin in the comments. I’m marking this as an error in the book for now. If you see where my solution is wrong and Apostol is correct please leave a comment and let me know.)
1. Facundo says:
(Sorry for my English, I speak Spanish)
I think that you mistake is in the area of the triangle, you forgot to divide by 2
• Facundo says:
ah no, my mistake, I calculated the area of the right triangle (half of the equilateral) I forgot to multiply 2
2. AudioRebel says:
I also get the same result as Rori.
The problem is not worded very clearly but the circular base is only there to provide the boundaries for the triangles being a plane it has no volume.
I found a very similar problem on YouTube but with squares: https://www.youtube.com/watch?v=_jx5pbKIzaM
He even shows a tiny model to show what is meant by the problem.
The approach is the same just replace he area of the square by the area of the triangle.
The height of triangle is derived Pythagoras Theorem (hypotenuse is 2 times the radius, and the other leg half is the radius).
3. Forget the circle and cones. The shape doesn’t exist as a named object. It is a Cavalieri solid with a cross-sectional area function of sqrt(3)x^2. By Theorem 2.7 it has a volume equal to the integral of the cross-sectional area function. Using the radius of 2 and our area function we find Apostol is correct. 16*sqrt(3)/3
• Mihajlo says:
Agreed. Apostol is correct here.
• AudioRebel says:
I get the cross sectional area as: sqrt(3) * (4 – x^2).
When a triangle is at the edge of the circular base (when x = 2) then it must be zero since it is at the boundary. When it is at the middle, when x = 0 it is at the largest size.
I get the same result as Rori.
So that is the key difference: sqrt(3) * x^2 vs sqrt(3) * (4 – x^2).
• Facundo says:
(Sorry for my English, I speak Spanish)
I think that you mistake is in the area of the triangle, you forgot to divide by 2
• Facundo says:
Sorry, I did not want to write this here XD
4. Andrei says:
when you calculated the volume in your solution, you have integered in relation of x. The right thing to do is integer in relation of y.
(4-x²) = y²
Then you will find the apostol’s result.
5. Felipe says:
Isn’t it equal to i made a calculation of the volume of an cone with a circle base with diameter 4 and “sides” 4, and using the line equation and therefore calculating the integral • Shafil says:
I don’t think it’s a cone. The figure you get when you cut a cone with a plane perpendicular to its base is a hyperbola (except if you cut it along a diameter).not triangle
Here we are cutting the figure with any plane that is perpendicular to a fixed diameter.
It does not have to be along another diameter.
6. shivam says:
as it is cone we can rotate a half equilateral travel that help to generate cone and we can compute volume by solid rotation.
7. shivam says:
basically it s cone.
8. Anonymous says:
9. cecM says:
hello, still do not quite understand why 2 root from 4 – x² it is the fixed diameter of our circle, I would greatly appreciate a brief explanation.
• Anonymous says:
imagine a circle in a plane ,now image a diameter in that circle (a fixed one), every plane perpendicular to that fixed diameter intersect the solid in an equilateral triangle. By the theory we know that the volume is the integral of the function area(area of each triangle for each position along the fixed diameter ,from -2 to +2 since 2 is the radius) .And its important to notice that the function area depends on the perpendicular segment to the diameter in each triangle.sorry for the bad english.
10. Edwin says:
Hey man, me again, I got the same result as you, but Apostol says it’s actually , how come we got double to what’s written on the answers in the book?
• RoRi says:
Hey, I’m not sure how he’s getting . I checked back over what I have posted and I can’t find a mistake. (I plugged the integral we have into Wolfram Alpha, and it gives us , so if there is a mistake it’s somewhere before we evaluate the integral.) Maybe it’s a mistake in the solutions (there are a few others). I think I’ll mark this with the “errata” tag unless someone comes along and figures out that Apostol is actually correct (which is definitely possible also).
• Edwin says:
Hey, so I thought maybe, because it’s a fixed diameter, the length of the base of the triangle is not actually , rather it’s just , cause it covers the whole diameter, thus, it’s only half of that, is this argument valid? What do you think?
• RoRi says:
I’m not sure I understand your argument.
I looked around more though, and think that our answer is right and Apostol is wrong. One of the examples on this page I think is exactly this problem, and they get the same integral we have (it’s most of the way toward the bottom of that pdf). They don’t evaluate the integral, but I’m pretty sure we have that part right (and Wolfram alpha agrees with our evaluation of the integral). | HuggingFaceTB/finemath | |
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# M10 - #8
Author Message
Intern
Joined: 09 Feb 2009
Posts: 17
### Show Tags
29 Jul 2009, 17:49
To arrive at its destination on time, a bus should have maintained a speed of V kmh throughout its journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at 1.2V kmh. If, as a result, the bus arrived at its destination X minutes earlier than planned, what was the actual duration of the trip?
(C) 2008 GMAT Club - m10#8
Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient
Solution is B - but how is S2 sufficient if i still have 2 variables and 2 unknowns in the equation?
Manager
Joined: 10 Jul 2009
Posts: 164
### Show Tags
30 Jul 2009, 17:29
Can you please clarify which is statement 1 and which is statement 2?
Senior Manager
Joined: 17 Mar 2009
Posts: 301
### Show Tags
02 Sep 2009, 23:12
posting the entire problem again.
To arrive at its destination on time the bus should have maintained a speed of V kmh throughout the journey. Instead, after going the first third of the distance at V kmh, the bus increased its speed and went the rest of the distance at (1.2)*V kmh. As a result, the bus arrived at its destination X minutes earlier than planned. What was the actual duration of the trip?
1. V = 60
2. X = 20
viperm5 wrote:
Solution is B - but how is S2 sufficient if i still have 2 variables and 2 unknowns in the equation?
as per the question s/v is what we need to find, i.e the duration of the trip, so it is enough to find s/v , no need to find separately s and v.
Re: M10 - #8 [#permalink] 02 Sep 2009, 23:12
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# statistics rayleigh distribution
A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. For example, suppose we have a set of three letters: A, B, and C. we might ask how many ways we can arrange 2 letters from that set.
Permutation is defined and given by the following function:
## Formula
${^nP_r = \frac{n!}{(n-r)!} }$
Where −
• ${n}$ = of the set from which elements are permuted.
• ${r}$ = size of each permutation.
• ${n,r}$ are non negative integers.
### Example
Problem Statement:
A computer scientist is trying to discover the keyword for a financial account. If the keyword consists only of 10 lower case characters (e.g., 10 characters from among the set: a, b, c… w, x, y, z) and no character can be repeated, how many different unique arrangements of characters exist?
Solution:
Step 1: Determine whether the question pertains to permutations or combinations.
Since changing the order of the potential keywords (e.g., ajk vs. kja) would create a new possibility, this is a permutations problem.
Step 2: Determine n and r
n = 26 since the computer scientist is choosing from 26 possibilities (e.g., a, b, c… x, y, z).
r = 10 since the computer scientist is choosing 10 characters.
Step 2: Apply the formula
${^{26}P_{10} = \frac{26!}{(26-10)!} \\[7pt] \ = \frac{26!}{16!} \\[7pt] \ = \frac{26(25)(24)…(11)(10)(9)…(1)}{(16)(15)…(1)} \\[7pt] \ = 26(25)(24)…(17) \\[7pt] \ = 19275223968000 }$
2.statistics analysis of variance
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121.discuss statistics | HuggingFaceTB/finemath | |
# Volume of a solid of revolution bounded by
• Apr 11th 2011, 10:48 AM
bijosn
Volume of a solid of revolution bounded by
Determine the volum of a solid of revolution when the region bounded by $y = \sqrt{x}$ and the straight lines y = 2 and x = 0 are rotated around the line x = 4
$V = 2\pi \int_{0}^{4} x ( 2 - \sqrt{x}) dx$
$V = 2\pi ( x^2 - \frac{2x^\frac{5}{2}}{5} )\big|_0^4$
$= \frac{32\pi}{5}$
??
what am I doing wrong?
• Apr 11th 2011, 12:16 PM
Unknown008
When you rotate about the x-axis (y = 0), you integrate with respect to x.
When you rotate along the line x = 4, you integrate with respect to y.
And can you tell why you picked $x(2 - \sqrt{x})$?
• Apr 11th 2011, 12:21 PM
Prove It
Hint: Translate your function and all the bounds left 4 units. The region does not change, but the calculations are easier because you can rotate around the x axis. | HuggingFaceTB/finemath | |
All points on the curve y2=4ax+asinxa at which the tangents are parallel to the axis of x lie on a
# All points on the curve ${y}^{2}=4a\left(x+a\mathrm{sin}\frac{x}{a}\right)$ at which the tangents are parallel to the axis of $x$ lie on a
1. A
circle
2. B
parabola
3. C
line
4. D
none of these
Register to Get Free Mock Test and Study Material
+91
Verify OTP Code (required)
### Solution:
We have,
${y}^{2}=4a\left(x+a\mathrm{sin}\frac{x}{a}\right)$ ..(i)
For points at which the tangents are parallel to x-axis, we must have
$\begin{array}{l}\frac{dy}{dx}=0\\ ⇒4a\left(1+\mathrm{cos}\frac{x}{a}\right)=0⇒\mathrm{cos}\frac{x}{a}=-1⇒\frac{x}{a}=\left(2n+1\right)\pi \end{array}$
For these values of $x$, we obtain $\mathrm{sin}\frac{x}{a}=0$
Putting $\mathrm{sin}\frac{x}{a}=0$, in (i), we get ${y}^{2}=4ax$
Therefore, all these points lie on the parabola ${y}^{2}=4ax.$ Register to Get Free Mock Test and Study Material
+91
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# What is the slope of the polar curve f(theta) = theta - sec^2theta+costhetasin^3theta at theta = (5pi)/6?
##### 1 Answer
Mar 23, 2017
From the reference Polar Curve Tangents :
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{\mathrm{dr}}{d \text{theta)sin(theta)+rcos(theta))/((dr)/(d} \theta} \cos \left(\theta\right) - r \sin \left(\theta\right)\right)$
#### Explanation:
Given: $r \left(\theta\right) = \theta - {\sec}^{2} \left(\theta\right) + \cos \left(\theta\right) {\sin}^{3} \left(\theta\right)$
Evaluate at $5 \frac{\pi}{6}$
$r \left(5 \frac{\pi}{6}\right) = 5 \frac{\pi}{6} - {\sec}^{2} \left(5 \frac{\pi}{6}\right) + \cos \left(5 \frac{\pi}{6}\right) {\sin}^{3} \left(5 \frac{\pi}{6}\right)$
$r \left(5 \frac{\pi}{6}\right) \approx 1.18$
Compute (dr)/(d""theta):
(dr)/(d""theta)=1-2tan(theta)sec^2(theta)+sin^2(theta)(2cos^2(theta)-2sin^2(theta)+1)
Evaluate at $5 \frac{\pi}{6}$
(dr(5pi/6))/(d""theta)~~3.0
The slope of the tangent line, m, is $\frac{\mathrm{dy}}{\mathrm{dx}}$ evaluated at $5 \frac{\pi}{6}$:
m = (3sin(5pi/6)+1.18cos(5pi/6))/(3cos(5pi/6)-1.18sin(5pi/6)
$m \approx - 0.15$ | HuggingFaceTB/finemath | |
Student/VectorCalculus/RootedVector - Maple Help
Student[VectorCalculus]
RootedVector
creates a vector rooted at a given point with specified components in a given coordinate system Calling Sequence RootedVector(origin,comps,c) RootedVector(vspace, comps) Parameters
origin - root=list(algebraic) or root=Vector(algebraic); root point of the vector vspace - root=module(Vector,GetRootPoint), VectorSpace where the vector lies comps - list(algebraic) or Vector(algebraic); components specifying the coefficients of the basis vectors c - (optional) name or name[name, name, ...]; specify the coordinate system, possibly indexed by the coordinate names Description
• The call RootedVector(origin,comps,c) returns a vector rooted at point origin with components comps in c coordinates. Note that the Student[VectorCalculus] package only supports the cartesian, polar, spherical and cylindrical coordinate systems.
• The rooted Vector is one of the four principal Vector data structures of the Student[VectorCalculus] package. Note that the Student[VectorCalculus] and the VectorCalculus packages share the same Vector data structures.
• For details on the differences between the four principal Vector data structures, namely, rooted Vectors, position Vectors, free Vectors, and vector fields, see VectorCalculus,Details.
• If no coordinate system argument is present, the current coordinate system is used.
• The root point origin can be specified as a free or position Vector or as a list of coordinate entries. If it is a free or position Vector, the coordinate system attribute is checked and conversion of the point to the current or specified c coordinate system is done accordingly.
• The keyword root can also be given as point.
• The components comps must be specified as a free Vector in Cartesian coordinates, a position Vector or as a list. The elements of the Vector or list are taken to be the coefficients of the unit basis vectors in the target coordinate system (as specified by the c parameter, if given, or else the current coordinate system).
• The call RootedVector(vpsace,comps) returns a vector rooted at the root point of the VectorSpace vspace with components comps in the coordinate system of vspace. No extra coordinate system needs to be specified. The comps argument can be a list, a free Vector in Cartesian coordinates or a position Vector.
• The returned rooted vector has a VectorSpace attribute that contains a module representation of the vector space rooted at the point origin.
• RootedVectors are always displayed as column vectors. Examples
> $\mathrm{with}\left(\mathrm{Student}\left[\mathrm{VectorCalculus}\right]\right):$
Introductory Examples:
> $\mathrm{v1}≔\mathrm{RootedVector}\left(\mathrm{root}=\left[1,2,3\right],\left[1,1,1\right]\right)$
${\mathrm{v1}}{≔}\left[\begin{array}{c}{1}\\ {1}\\ {1}\end{array}\right]$ (1)
> $\mathrm{About}\left(\mathrm{v1}\right)$
$\left[\begin{array}{cc}{\mathrm{Type:}}& {\mathrm{Rooted Vector}}\\ {\mathrm{Components:}}& \left[{1}{,}{1}{,}{1}\right]\\ {\mathrm{Coordinates:}}& {\mathrm{cartesian}}\\ {\mathrm{Root Point:}}& \left[{1}{,}{2}{,}{3}\right]\end{array}\right]$ (2)
> $\mathrm{GetSpace}\left(\mathrm{v1}\right)$
${\mathbf{module}}\left({}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{local}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathrm{_origin}}{,}{\mathrm{_coords}}{,}{\mathrm{_coords_dim}}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{export}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathrm{GetCoordinates}}{,}{\mathrm{GetRootPoint}}{,}{\mathrm{Vector}}{,}{\mathrm{eval}}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end module}}$ (3)
> $\mathrm{GetRootPoint}\left(\mathrm{v1}\right)$
$\left({1}\right){{e}}_{{x}}{+}\left({2}\right){{e}}_{{y}}{+}\left({3}\right){{e}}_{{z}}$ (4)
> $\mathrm{v2}≔\mathrm{RootedVector}\left(\mathrm{point}=\left[1,\mathrm{\pi }\right],\left[1,1\right],\mathrm{polar}\left[r,t\right]\right)$
${\mathrm{v2}}{≔}\left[\begin{array}{c}{1}\\ {1}\end{array}\right]$ (5)
> $\mathrm{About}\left(\mathrm{v2}\right)$
$\left[\begin{array}{cc}{\mathrm{Type:}}& {\mathrm{Rooted Vector}}\\ {\mathrm{Components:}}& \left[{1}{,}{1}\right]\\ {\mathrm{Coordinates:}}& {{\mathrm{polar}}}_{{r}{,}{t}}\\ {\mathrm{Root Point:}}& \left[{1}{,}{\mathrm{\pi }}\right]\end{array}\right]$ (6)
> $\mathrm{GetCoordinates}\left(\mathrm{v2}\right)$
${{\mathrm{polar}}}_{{r}{,}{t}}$ (7)
> $\mathrm{rp}≔\mathrm{Vector}\left(\left[1,1\right],\mathrm{coords}=\mathrm{cartesian}\right)$
${\mathrm{rp}}{≔}\left({1}\right){{e}}_{{x}}{+}\left({1}\right){{e}}_{{y}}$ (8)
> $\mathrm{v3}≔\mathrm{RootedVector}\left(\mathrm{root}=\mathrm{rp},\left[1,0\right],\mathrm{polar}\left[r,t\right]\right)$
${\mathrm{v3}}{≔}\left[\begin{array}{c}{1}\\ {0}\end{array}\right]$ (9)
> $\mathrm{GetRootPoint}\left(\mathrm{v3}\right)$
$\left(\sqrt{{2}}\right){{e}}_{{r}}{+}\left(\frac{{\mathrm{\pi }}}{{4}}\right){{e}}_{{t}}$ (10)
> $\mathrm{vs}≔\mathrm{VectorSpace}\left(\left[2,3\right],\mathrm{polar}\left[r,t\right]\right):$
> $\mathrm{v4}≔\mathrm{RootedVector}\left(\mathrm{root}=\mathrm{vs},\left[1,2\right]\right)$
${\mathrm{v4}}{≔}\left[\begin{array}{c}{1}\\ {2}\end{array}\right]$ (11)
> $\mathrm{v5}≔\mathrm{RootedVector}\left(\mathrm{root}=\mathrm{vs},\left[3,4\right]\right)$
${\mathrm{v5}}{≔}\left[\begin{array}{c}{3}\\ {4}\end{array}\right]$ (12)
> $\mathrm{GetRootPoint}\left(\mathrm{v4}\right)$
$\left({2}\right){{e}}_{{r}}{+}\left({3}\right){{e}}_{{t}}$ (13)
> $\mathrm{GetRootPoint}\left(\mathrm{v5}\right)$
$\left({2}\right){{e}}_{{r}}{+}\left({3}\right){{e}}_{{t}}$ (14) | HuggingFaceTB/finemath | |
Planck Voltage to EMU of Electric Potential Converter
1 Planck Voltage = 1.04295e+35 EMU of Electric Potential
One Planck Voltage is Equal to How Many EMU of Electric Potential?
The answer is one Planck Voltage is equal to 1.04295e+35 EMU of Electric Potential and that means we can also write it as 1 Planck Voltage = 1.04295e+35 EMU of Electric Potential. Feel free to use our online unit conversion calculator to convert the unit from Planck Voltage to EMU of Electric Potential. Just simply enter value 1 in Planck Voltage and see the result in EMU of Electric Potential.
Manually converting Planck Voltage to EMU of Electric Potential can be time-consuming,especially when you don’t have enough knowledge about Electrical Potential units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Planck Voltage to EMU of Electric Potential converter tool to get the job done as soon as possible.
We have so many online tools available to convert Planck Voltage to EMU of Electric Potential, but not every online tool gives an accurate result and that is why we have created this online Planck Voltage to EMU of Electric Potential converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
How to Convert Planck Voltage to EMU of Electric Potential (hV to EMU)
By using our Planck Voltage to EMU of Electric Potential conversion tool, you know that one Planck Voltage is equivalent to 1.04295e+35 EMU of Electric Potential. Hence, to convert Planck Voltage to EMU of Electric Potential, we just need to multiply the number by 1.04295e+35. We are going to use very simple Planck Voltage to EMU of Electric Potential conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Planck Voltage} = 1 \times 1.04295e+35 = \text{1.04295e+35 EMU of Electric Potential}$$
What Unit of Measure is Planck Voltage?
Planck voltage is a unit of measurement for electric potential. It is also a base unit of voltage. One planck voltage is equal to 1.04295 × 10²⁷ volts.
What is the Symbol of Planck Voltage?
The symbol of Planck Voltage is hV. This means you can also write one Planck Voltage as 1 hV.
What Unit of Measure is EMU of Electric Potential?
EMU is a unit of measurement for electric potential. EMU stands for electromagnetic unit. One EMU of electric potential is equal to 1e-8 volts.
What is the Symbol of EMU of Electric Potential?
The symbol of EMU of Electric Potential is EMU. This means you can also write one EMU of Electric Potential as 1 EMU.
How to Use Planck Voltage to EMU of Electric Potential Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Planck Voltage and in the first input field, enter a value.
• From the second dropdown, select EMU of Electric Potential.
• Instantly, the tool will convert the value from Planck Voltage to EMU of Electric Potential and display the result in the second input field.
Example of Planck Voltage to EMU of Electric Potential Converter Tool
Planck Voltage
1
EMU of Electric Potential
1.04295e+35
Planck Voltage to EMU of Electric Potential Conversion Table
Planck Voltage [hV]EMU of Electric Potential [EMU]Description
1 Planck Voltage1.04295e+35 EMU of Electric Potential1 Planck Voltage = 1.04295e+35 EMU of Electric Potential
2 Planck Voltage2.0859e+35 EMU of Electric Potential2 Planck Voltage = 2.0859e+35 EMU of Electric Potential
3 Planck Voltage3.12885e+35 EMU of Electric Potential3 Planck Voltage = 3.12885e+35 EMU of Electric Potential
4 Planck Voltage4.1718e+35 EMU of Electric Potential4 Planck Voltage = 4.1718e+35 EMU of Electric Potential
5 Planck Voltage5.21475e+35 EMU of Electric Potential5 Planck Voltage = 5.21475e+35 EMU of Electric Potential
6 Planck Voltage6.2577e+35 EMU of Electric Potential6 Planck Voltage = 6.2577e+35 EMU of Electric Potential
7 Planck Voltage7.30065e+35 EMU of Electric Potential7 Planck Voltage = 7.30065e+35 EMU of Electric Potential
8 Planck Voltage8.3436e+35 EMU of Electric Potential8 Planck Voltage = 8.3436e+35 EMU of Electric Potential
9 Planck Voltage9.38655e+35 EMU of Electric Potential9 Planck Voltage = 9.38655e+35 EMU of Electric Potential
10 Planck Voltage1.04295e+36 EMU of Electric Potential10 Planck Voltage = 1.04295e+36 EMU of Electric Potential
100 Planck Voltage1.04295e+37 EMU of Electric Potential100 Planck Voltage = 1.04295e+37 EMU of Electric Potential
1000 Planck Voltage1.04295e+38 EMU of Electric Potential1000 Planck Voltage = 1.04295e+38 EMU of Electric Potential
Planck Voltage to Other Units Conversion Table
ConversionDescription
1 Planck Voltage = 1.04295e+27 Volt1 Planck Voltage in Volt is equal to 1.04295e+27
1 Planck Voltage = 1.04295e+27 Watt/Ampere1 Planck Voltage in Watt/Ampere is equal to 1.04295e+27
1 Planck Voltage = 1.04295e+24 Kilovolt1 Planck Voltage in Kilovolt is equal to 1.04295e+24
1 Planck Voltage = 1.04295e+21 Megavolt1 Planck Voltage in Megavolt is equal to 1.04295e+21
1 Planck Voltage = 1.04295e+30 Millivolt1 Planck Voltage in Millivolt is equal to 1.04295e+30
1 Planck Voltage = 1.04295e+33 Microvolt1 Planck Voltage in Microvolt is equal to 1.04295e+33
1 Planck Voltage = 1.04295e+35 Abvolt1 Planck Voltage in Abvolt is equal to 1.04295e+35
1 Planck Voltage = 1.04295e+35 EMU of Electric Potential1 Planck Voltage in EMU of Electric Potential is equal to 1.04295e+35
1 Planck Voltage = 3.4789057328924e+24 Statvolt1 Planck Voltage in Statvolt is equal to 3.4789057328924e+24
1 Planck Voltage = 3.4789057328924e+24 ESU of Electric Potential1 Planck Voltage in ESU of Electric Potential is equal to 3.4789057328924e+24
1 Planck Voltage = 1.04295e+36 Nanovolt1 Planck Voltage in Nanovolt is equal to 1.04295e+36
1 Planck Voltage = 1.04295e+39 Picovolt1 Planck Voltage in Picovolt is equal to 1.04295e+39
1 Planck Voltage = 1042950000000000000 Gigavolt1 Planck Voltage in Gigavolt is equal to 1042950000000000000
1 Planck Voltage = 1042950000000000 Teravolt1 Planck Voltage in Teravolt is equal to 1042950000000000 | HuggingFaceTB/finemath | |
## Computing the angle between two vectors
Introduction
Given two vectors in three dimensions, what is the most accurate way to compute the angle between them? I have seen several different approaches to this problem recently in the wild, and although I knew some of them had potential issues, I wasn’t sure just how bad things might get in practice, nor which alternative was best as a replacement.
To make the setup more precise, let’s assume that we are given two non-zero input vectors $\mathbf{u}, \mathbf{v} \in \mathbb{R}^3$, represented exactly by their double-precision coordinates, and we desire a function that returns a double-precision value that most closely approximates the angle $0 \leq \theta \leq \pi$ between the vectors, with all intermediate computation also done in double precision.
Kahan’s Mangled Angles
William Kahan discusses three formulas in the “Mangled Angles” section of the paper linked below. The first is the “usual” dot product formula:
$\theta = \cos^{-1}\frac{\mathbf{u}\cdot\mathbf{v}}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|}$
with the following C++ implementation, which as Kahan points out requires clamping the double-precision dot product to the interval $[-1, 1]$ to avoid a NaN result for some vectors that are nearly parallel:
double angle(const Vector& u, const Vector& v)
{
return std::acos(std::min(1.0, std::max(-1.0,
dot(u, v) / (norm(u) * norm(v)))));
}
Kahan subsequently describes another formula using the cross product:
$\theta = \begin{cases}\sin^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|} & \text{if }\mathbf{u}\cdot\mathbf{v} \geq 0, \\ \pi-\sin^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\left|\mathbf{u}\right|\left|\mathbf{v}\right|} & \text{if }\mathbf{u}\cdot\mathbf{v} < 0 \end{cases}$
with the following implementation:
double angle(const Vector& u, const Vector& v)
{
double angle = std::asin(std::min(1.0,
norm(cross(u, v)) / (norm(u) * norm(v))));
if (dot(u, v) < 0)
{
angle = 3.141592653589793 - angle;
}
return angle;
}
Interestingly, Kahan does not mention that this formula also requires clamping the asin argument to the interval $[-1, 1]$; following is an explicit example of inputs demonstrating the potential problem:
$\mathbf{u} = (-0.6171833037218851, -0.4342100935824679, 0.6561603190059907)$
$\mathbf{v} = (-0.32014601021553196, 0.9003703730169068, 0.29468580477598927)$
Finally, despite referring to the above formula as “the best known in three dimensions,” Kahan finishes with the following “better formula less well known than it deserves”:
$\theta = 2\tan^{-1}\frac{\left|{\left|\mathbf{v}\right|\mathbf{u} - \left|\mathbf{u}\right|\mathbf{v}}\right|}{\left|{\left|\mathbf{v}\right|\mathbf{u} + \left|\mathbf{u}\right|\mathbf{v}}\right|}$
with the following implementation:
double angle(const Vector& u, const Vector& v)
{
double nu = norm(u);
double nv = norm(v);
return 2 * std::atan2(norm(nv * u - nu * v), norm(nv * u + nu * v));
}
That’s a lot of square roots. I didn’t focus on performance here, but it would be an interesting follow-on analysis to compare the speed of each of these formulas.
Other approaches are possible; following is the formula that I thought was the most accurate, before reading Kahan’s paper:
$\theta = \tan^{-1}\frac{\left|\mathbf{u}\times\mathbf{v}\right|}{\mathbf{u}\cdot\mathbf{v}}$
double angle(const Vector& u, const Vector& v)
{
return std::atan2(norm(cross(u, v)), dot(u, v));
}
This has the added benefit of involving just a single square root. This is the formula that I used to compute the “true” angle between vectors to compare errors, using arbitrary-precision rational arithmetic to compute the square root (actually, a reciprocal square root, which is slightly easier) and arctangent. All of the source code is available on GitHub.
And finally, the approach that I saw most recently that motivated this post, using the Law of cosines:
$\theta = \cos^{-1}\frac{\left|\mathbf{u}\right|^2 + \left|\mathbf{v}\right|^2 - \left|\mathbf{u}-\mathbf{v}\right|^2}{2\left|\mathbf{u}\right|\left|\mathbf{v}\right|}$
double angle(const Vector& u, const Vector& v)
{
double u2 = u.x * u.x + u.y * u.y + u.z * u.z;
double v2 = v.x * v.x + v.y * v.y + v.z * v.z;
Vector d = u - v;
return std::acos(std::min(1.0, std::max(-1.0,
(u2 + v2 - (d.x * d.x + d.y * d.y + d.z * d.z)) /
(2 * std::sqrt(u2) * std::sqrt(v2)))));
}
Results
The relative accuracy of each formula depends on the magnitude of the angle between the input vectors. The following figure shows this comparison for angles near 0 (i.e., nearly parallel vectors), near $\pi/4$, near $\pi/2$ (i.e., nearly orthogonal vectors), and near $\pi$ (i.e., nearly “anti-parallel” vectors).
Absolute error between computed and exact rounded double-precision angle, vs. true/exact angle.
The x-axis indicates an offset from the “true” angle between the input vectors, computed to roughly 200-bit accuracy. The y-axis indicates the error in the double-precision output, compared against the true angle also rounded to double-precision. The points hugging the bottom of each figure are my poor man’s attempt at indicating zero error (note that these are on a log-log scale), i.e., the 64-bit double-precision output matched the corresponding value rounded from the 200-bit true angle. (In many ways this figure feels like a failure of visual display of quantitative information, and I’m not sure how best to improve it.)
So what’s the takeaway? If you don’t care about absolute errors smaller than a few dozen nanoradians, then it doesn’t really matter which formula you use. And if you do care about errors– and angles– smaller than that, then be sure that your inputs are accurate in the first place. For example, did you normalize your “real” input vectors to unit length first, and if so, how much error did you unintentionally incur as a result? We can construct very small angles between vectors if we restrict to “nice” two-dimensional inputs like $(1,0,0)$ and $(\cos \theta, \sin \theta, 0)$. But it’s an interesting exercise to see how difficult it is to construct vectors “in general position” (e.g., randomly rotated) with a prescribed small angle between them.
As expected, the two arccosine formulas behave poorly for nearly parallel/anti-parallel vectors, and as Kahan describes, the arcsine formula behaves poorly for nearly orthogonal vectors. The two arctangent formulas are the most consistently accurate, and when the one mentioned by Kahan is better, it’s typically much better.
Reference:
1. Kahan, W., How Futile are Mindless Assessments of Roundoff in Floating-Point Computation? [PDF]
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# If the probability that Mike can win a championship is 1/4,
Author Message
Manager
Joined: 13 Oct 2004
Posts: 236
If the probability that Mike can win a championship is 1/4, [#permalink]
### Show Tags
26 Mar 2005, 20:43
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If the probability that Mike can win a championship is 1/4, that of Rob winning the championship is 1/3 and that of Ben winning is 1/6. What is the probability of either Mike or Rob winning the championship but not Ben?
A. 1/6
B. 5/12
C. 7/12
D. 5/6
E. 1
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08
### Show Tags
26 Mar 2005, 20:57
I dont understand well the issue here. If we find the prob that Mike or Rob win the championship it is sure that Ben can no win it...
Anyway, first consider others people to win the championship :
1-(1/4+1/3+1/6) = 1/4
I think answer is D (Mike + Rob + Others) = 10/12 = 5/6
VP
Joined: 18 Nov 2004
Posts: 1433
### Show Tags
26 Mar 2005, 21:54
"B"
if they r not playing together in same championship then I think
p(e) = 5/12
if they r playing then : 25/72
SVP
Joined: 03 Jan 2005
Posts: 2233
### Show Tags
26 Mar 2005, 22:20
If there can only be one winner for the championship then P=1/3+1/4=7/12.
If there can be multiple winners then
P=P(M)+P(R)-P(M&R)-P(M&B)-P(R&B)+P(M&R&B)
=1/3+1/4-1/12-1/24-1/18+1/72
=5/12
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08
### Show Tags
27 Mar 2005, 06:07
I think that they're playing in the same championship because the problem is given like this :
If the probability that Mike can win a championship is 1/4, that of Rob winning the championship is 1/3 and that of Ben winning is 1/6. What is the probability of either Mike or Rob winning the championship but not Ben?
I need some Verbal specialist but I assume that there is only one championship. Then I understood the question as : what is the probability that Mike or Rob win but not Ben .... What is the probability that the winner is Mike, Bob or anyone else but not Ben.
If it is not this meaning, I don't see the interest of notifying "not Ben"
am I missing something
Manager
Joined: 17 Dec 2004
Posts: 71
### Show Tags
27 Mar 2005, 22:40
I think the answer is C - 7/12
If the probability of Mike winning the championship is 1/4, and the probability of Rob winning the championship is 1/3, then the probability that EITHER Mike OR Rob will win is 1/4+1/3, or 7/12. The information about Ben seems extraneous, since if either Mike or Rob win, Ben can't win.
VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany
### Show Tags
07 Apr 2005, 12:39
HongHu wrote:
P=P(M)+P(R)-P(M&R)-P(M&B)-P(R&B)+P(M&R&B)
=1/3+1/4-1/12-1/24-1/18+1/72
=5/12
honghu, why did you add the underlined part ?
_________________
If your mind can conceive it and your heart can believe it, have faith that you can achieve it.
SVP
Joined: 03 Jan 2005
Posts: 2233
### Show Tags
07 Apr 2005, 13:10
You need to draw the diagram to see it more clearly. Basically when you do P(M)+P(R) you double counted P(M&R), P(M&B) and P(R&B). Do you need to take them out. But when you do -P(M&R)-P(M&B)-P(R&B) you have counted P(M&R&B) three times and when you do P(M)+P(R) you only double counted P(M&R&B) twice. So you need to add one P(M&R&B) back to cancel the third P(M&R&B) you took out.
_________________
Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.
Manager
Joined: 13 Oct 2004
Posts: 236
### Show Tags
07 Apr 2005, 13:15
The OE for this one magically came up with :
1/3 + 1/4 - 1/6 = 5/12.
I didnt understand the subtraction part. Can someone explain the logic here?
Intern
Joined: 15 Jun 2005
Posts: 10
### Show Tags
14 Jul 2005, 01:10
Please don't ignore Ben. The question clearly says that Ben will not win.
Thus, 1/4 + 1/3 - 1/6 ( Probability that Ben will win)
Thus, B is the answer ( 5/12)
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA
### Show Tags
14 Jul 2005, 05:01
Hmmm..think that the prob of Mike OR Rob winning is 1/3+1/4 AND of Ben not winning is 5/6, then (1/3+1/4)x5/6=35/72
Senior Manager
Joined: 30 May 2005
Posts: 276
### Show Tags
15 Jul 2005, 11:59
sa_samie wrote:
Please don't ignore Ben. The question clearly says that Ben will not win.
Thus, 1/4 + 1/3 - 1/6 ( Probability that Ben will win)
Thus, B is the answer ( 5/12)
Well it seems that
probability that either one (except ben ) is 1/3+1/4
taking together
1/4 + 1/3 - 1/6 SUBSTRACT PROB THAT BEN WIN
or 1/4 + 1/3 +5/6 ADD THE probabilities that ben not win
both case we get 5/12 since those events are dependent
does this make sense to anyone ?
THANKS
Intern
Joined: 27 Jun 2005
Posts: 18
### Show Tags
16 Jul 2005, 07:18
mandy wrote:
sa_samie wrote:
Please don't ignore Ben. The question clearly says that Ben will not win.
Thus, 1/4 + 1/3 - 1/6 ( Probability that Ben will win)
Thus, B is the answer ( 5/12)
Well it seems that
probability that either one (except ben ) is 1/3+1/4
taking together
1/4 + 1/3 - 1/6 SUBSTRACT PROB THAT BEN WIN
or 1/4 + 1/3 +5/6 ADD THE probabilities that ben not win
both case we get 5/12 since those events are dependent
does this make sense to anyone ?
THANKS
That does not make sense to me. You can't take a number (1/4+1/3) and then say that either adding(5/6) or subtracting(1/6) a new number will give the same result unless that new number is 0. If you take 1/4+1/3+5/6=17/12 This is not a reasonable probability. If you take 1/4+1/3-1/6=5/12 Which is the OA, but it really doesn't matter to come up with the OA. It is only a helpful practice question if you know how to come up with the right answer, which means no monday-morning quarterbacking when you already know the answer is supposed to be 5/12 and then making up formulas to arrive at 5/12.
What seems most important to me about this question is whether or not I should expect actual GMAT questions to be in this format. To me it seemed that Ben's involvement was irrelevant. Do GMAT questions often include irrelevant information? I don't know but my guess is yes.
Do GMAT questions usually have so ambiguous an interpretation? If so, I should focus my study on resolving ambiguities at least as much as studying proper methods for determining probabilites. What does everyone think?
16 Jul 2005, 07:18
Display posts from previous: Sort by | HuggingFaceTB/finemath | |
# Solution May 19, 2008
### Problem
Consider a completely filled Sudoku, written as a 9x9 matrix. Show that the determinant of this matrix is divisible by 405.
### Solution
This is true for any 9x9 latin square. We can add the second, third, ..., ninth column of the matrix to the first column without changing its determinant. The resulting matrix has all entries of the first column equal to 45. We can then divide the first column by 9, which divides the determinant by 9 as well.
Now we do the same operation on the rows. This time, the first row will end up with all entries equal to 45, without changing the determinant. So the determinant is still divisible by 45, and the original determinant was divisible by 405 as claimed. | HuggingFaceTB/finemath | |
### 数学代写|代数数论代写Algebraic number theory代考|Direct Product of Rings
statistics-lab™ 为您的留学生涯保驾护航 在代写代数数论Algebraic number 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。我们的专家在代写代数数论Algebraic number theory代写方面经验极为丰富，各种代写代数数论Algebraic number 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|代数数论代写Algebraic number theory代考|Direct Product of Rings
Suppose $B_{1}, \ldots, B_{r}$ are commutative rings with 1 . We define their direct product as the Cartesian product $B=B_{1} \times \cdots \times B_{r}$, with addition and multiplication taken component-wise. Each $B_{j}$ may be regarded as a subring of $B$ via the obvious inclusion map, e.g. $B_{1} \ni b_{1} \rightarrow\left(b_{1}, 0, \ldots, 0\right) \in B$.
If $A$ is a subring of each $B_{j}$, then $A$ may be regarded as a subring of the direct product $B=B_{1} \times \cdots \times B_{r}$, via the map $A \ni a \rightarrow(a, \ldots, a) \in B$.
Theorem 4.25. Suppose $A$ with 1 is a subring of each $B_{j}$ and every $B_{j}$ is a free A-module of rank $n_{j}$. Then the direct product $B=B_{1} \times \cdots \times B_{r}$ is a free module of rank $n_{1}+\cdots+n_{r}$. Moreover
$$\mathfrak{d}{B / A}=\mathfrak{d}{B_{1} / A} \cdots \mathfrak{d}{B{r} / A}$$
Proof. We only need to prove (4.6). To simplify notation, we prove it for $r=2$. For $r>2$, the proof is similar.
Put $n_{1}=m$ and $n_{2}=n$. Let $\alpha_{1}, \ldots, \alpha_{m}$ be a basis of $B_{1}$ over $A$ and $\beta_{1}, \ldots, \beta_{n}$ be a basis of $B_{2}$ over $A$. As $A$-modules, if we identify $B_{1}$ and $B_{2}$ with the submodules $B_{1} \times{0}$ and ${0} \times B_{2}$ of $B=B_{1} \times B_{2}$, then $\left{\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right}$ is a basis of $B$ over $A$. Moreover, for all $i, j$, we have $\alpha_{i} \beta_{j}=0$. Hence $\Delta\left(\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right)$ is the determinant of the matrix
$$\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)$$
This shows that
$$\Delta\left(\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right)=\Delta\left(\alpha_{1}, \ldots, \alpha_{m}\right) \Delta\left(\beta_{1}, \ldots, \beta_{n}\right)$$
Therefore, $\mathfrak{d}{B / A}=\mathfrak{d}{B_{1} / A} \mathfrak{d}{B{2} / A}$.
Suppose $A$ is a subring of $B$. Let $\mathfrak{a}$ be an ideal of $A$ and $\mathfrak{b}=\mathfrak{a} B$ be the ideal of $B$ generated by a. For $\alpha$ in $A$ and $\beta$ in $B$, let $\bar{\alpha}$ and $\bar{\beta}$ denote the residue class of $\alpha$ in $A / a$ and that of $\beta$ in $B / \mathfrak{b}$, respectively.
## 数学代写|代数数论代写Algebraic number theory代考|Nilradical
Definition 4.30. An element of a commutative ring $A$ with 1 is nilpotent if $a^{m}=0$ for some $m$ in $\mathbb{Z}$.
Theorem 4.31. The set nil $(A)$ of all nilpotent elements of $A$ is an ideal of A.
The ideal nil $(A)$ is called the nilradical of $A$.
Proof. Let $x, y \in \operatorname{nil}(A)$. Then for some $m, n$ in $\mathbb{N}, x^{m}=y^{n}=0$. If $l=m+n$, then it follows from the Binomial Theorem, that $(x+y)^{l}=0$. On the other hand, if $a \in A$, then $(a x)^{m}=a^{m} x^{m}=0$. This proves that nil $(A)$ is an ideal of $A$.
Theorem 4.32. The nilradical, $\operatorname{nil}(A)$, is the intersection of all prime ideals of $A$.
Proof. If $x$ in $A$ is nilpotent, then for some $m$ in $\mathbb{N}, x^{m}=0$. Hence $x \in \mathfrak{p}$, for all prime ideals $p$ of $A$.
Conversely, suppose $x$ is not nilpotent, that is $x^{m} \neq 0$ for all $m$ in $\mathbb{N}$. We show that there is at least one prime ideal $\mathfrak{p}$ such that $x \notin p$. Let $S$ be the set of ideals a of $A$, such that $x^{m} \notin \mathfrak{a}$ for all $m$ in $\mathbb{N}$. Clearly, $S$ is not empty, since the zero ideal $(0) \in S$. By Zorn’s Lemma, let $p$ be a maximal element of $S$. We shall show that $\mathfrak{p}$ is prime. If not, then there are $x, y$ in $A \backslash \mathfrak{p}$ with $x y$ in $\mathfrak{p}$. Then the ideals $\boldsymbol{a}=(\mathfrak{p}, x)$ and $\mathbf{b}=(\mathfrak{p}, y)$ both properly contain $\mathfrak{p}$. By the choice of $\mathfrak{p}$, for some $m, n$ in $\mathbb{N}, x^{m} \in \mathfrak{a}, x^{n} \in \mathfrak{b}$. This shows that $x^{m+n} \in \mathfrak{a b} \subseteq \mathfrak{p}$, implying $\mathfrak{p} \notin S$. This contradiction proves that $\mathfrak{p}$ is prime.
## 数学代写|代数数论代写Algebraic number theory代考|Reduced Rings
Definition 4.33. A commutative ring $A$ with 1 is reduced if $\operatorname{nil}(A)=(0)$.
Example 4.34.
1. An integral domain is reduced.
2. The product $A_{1} \times \ldots \times A_{r}$ is reduced if all $A_{j}$ are reduced.
Theorem 4.35. Suppose $K$ is a number field and $\mathfrak{P}$ a prime ideal of $\mathcal{O}=\mathcal{O}_{K}$. The quotient ring $\mathcal{O} / \mathfrak{P}^{e}$ is reduced if and only if $e=1$.
Proof. If $e=1$, then $\mathcal{O} / \mathfrak{F}$ is a field, hence reduced. On the other hand, if $e>0$, choose $\pi$ in $\mathfrak{P}-\mathfrak{P}^{2}$. Then $\pi \neq 0$ in $\mathcal{O} / \mathfrak{P}^{e}$, but $\pi^{e}=0$ in $\mathcal{O} / \mathfrak{P}^{e}$. Therefore, $\mathcal{O} / \mathfrak{P}^{e}$ is not reduced.
Now let $A$ be a subring of a commutative ring $B$, both with 1 . Suppose $B$ is a free $A$-module of rank $n$.
Theorem 4.36. If $A$ is a finite field, then $B$ is reduced if and only if $\mathfrak{D}_{B / A} \neq$ $(0)$.
Proof. First suppose that $B$ is not reduced, that is, it has a nilpotent element $\alpha \neq 0$. $A$ being a field, $\alpha$ can be completed into a basis $\alpha_{1}=\alpha, \alpha_{2}, \ldots, \alpha_{n}$ of the vector space $B$ over $A$. Now all the elements $\alpha_{1} \alpha_{j}, j=1, \ldots, n$ are also nilpotent and since the matrix for a nilpotent element is also nilpotent, its trace is zero (why?). Hence the first row of the matrix $\left(\operatorname{tr}\left(\alpha_{1} \alpha_{j}\right)\right)$ consists of zeros only, which shows that
$$\Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right)=\operatorname{det}\left(\operatorname{tr}\left(\alpha_{i} \alpha_{j}\right)\right)=0$$
Therefore, $\mathfrak{o}{B / A}=(0)$. Conversely, suppose $B$ is reduced, i.e. $\operatorname{nil}(B)={0}$. Since $\operatorname{nil}(B)$ is the intersection of all prime ideals and $B$ is finite, $$(0)=\mathfrak{F}{1} \cap \ldots \cap \mathfrak{P}{r}\left(\mathfrak{P}{i} \neq \mathfrak{P}{j} \text { for } i \neq j\right) .$$ Every $B / \mathfrak{P}{j}$, being a finite integral domain, is a field, hence all $\mathfrak{P}{j}$ are maximal and therefore coprime in pairs, and $$\mathfrak{F}{1} \cap \ldots \cap \mathfrak{P}{r}=\mathfrak{P}{1} \ldots \mathfrak{P}{r}$$ By the Corollary $4.28$, $$B=B /(0)=B / \mathfrak{P}{1} \ldots \mathfrak{P}{r} \cong B / \mathfrak{P}{1} \times \ldots \times B / \mathfrak{P}{r} .$$ By Theorem 4.25, $$\mathfrak{o}{B / A}=\mathfrak{o}{\left(B / \mathfrak{F}{1}\right) / A} \cdots \mathfrak{o}{\left(B / \mathfrak{F}{\mathrm{r}}\right) / A} .$$
Since $A$ is a field, each $\mathfrak{o}{\left(B / \mathfrak{P}{j}\right) / A} \neq(0)$. Hence $\mathfrak{d}_{B / A} \neq(0)$.
## 数学代写|代数数论代写Algebraic number theory代考|Direct Product of Rings
d乙/一个=d乙1/一个⋯d乙r/一个
\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{ tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{ tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)
Δ(一个1,…,一个米;b1,…,bn)=Δ(一个1,…,一个米)Δ(b1,…,bn)
## 数学代写|代数数论代写Algebraic number theory代考|Reduced Rings
1. 减少了一个积分域。
2. 产品一个1×…×一个r如果全部减少一个j被减少。
定理 4.35。认为ķ是一个数字字段并且磷的首要理想○=○ķ. 商圈○/磷和当且仅当和=1.
Δ(一个1,…,一个n)=这(tr(一个一世一个j))=0
(0)=F1∩…∩磷r(磷一世≠磷j 为了 一世≠j).每一个乙/磷j，作为有限积分域，是一个域，因此所有磷j是最大的，因此成对互质，并且
F1∩…∩磷r=磷1…磷r由推论4.28,
○乙/一个=○(乙/F1)/一个⋯○(乙/Fr)/一个.
## 有限元方法代写
tatistics-lab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。 | open-web-math/open-web-math | |
# Teaching Addition to Children Tips to Teach Carrying and Borrowing
Manipulatives are the best way to demonstrate the mathematical concepts of “regrouping” which is really what you are doing in adding and subtracting rather than carrying and borrowing. Regrouping is more accurately based on place value, so children can more readily master these math skills.Regrouping means that if you add a column and get a sum that is greater than 9, you must regroup due to place value.
Manipulatives such as place value rods are excellent ways to demonstrate to the child that you cannot have more than 9 in a column. If the sum in a column is ten or more, you only write down the number in the ones place. Ten “ones” equal ten.
In our system of arithmetic,based on tens, a sum of more than ten moves you to the left as you move across a math problem requiring regrouping.A bundle of ten ones or units equals one ten. Ten tens equals one hundred, etc.
You can purchase or borrow Base Ten blocks or Cuisenaire rods to practice this skill. It is also good for sharing in pairs or small groups. You can even use inexpensive household items such as paper clips, dried beans or macaroni, or any easily handled item. Adding involves trading in ten items for another different item, representing 10. Use four columns,moving from right to left:ones, tens, hundreds, thousands. Using poker chips, “beans” or other item, place them in the ones column. Show how you must move them to the next column when you have ten or more. The student can see how many items are remaining in each column.
Another effective technique is using pennies, dimes, and one dollar bills for the ones, tens, and hundreds place. Ten pennies get traded in for a dime and ten dimes get turned in for a one dollar bill. This use of manipulatives uses visual and tactile senses as well as thought processing.
In the absence of these manipulatives, you can also create your own from cardboard. Graph paper with large squares works well. The “ones” is a single square. A rod or column of ten is the “tens” place and a 10 X 10 block is “hundreds”. The graph paper, colored, if you like, can be glued to a cardboard backing of equal size.
Regrouping is a trading skill. Ten “ones” is traded in for a “tens”. Moving on to the next column from right to left, if you have ten or more rods (a line of ten), you then trade 10 in for a block, 20 in for two blocks, etc.
Using these manipulatives helps the student understand place value and that is the key concept in teaching regrouping, formerly taught as “carrying” and “borrowing.” | HuggingFaceTB/finemath | |
# Homework Help: Midpoint of a Segment
1. Oct 3, 2015
### Bashyboy
1. The problem statement, all variables and given/known data
Prove that every segment has a midpoint.
2. Relevant equations
3. The attempt at a solution
I first began with some arbitrary segment $AB$ in the plane, and then constructed the line $\overset{\leftrightarrow}{AB} = \ell$ from these two points. I then used the theorem which states that, given two points $A$ and $B$ on a line, there exists a point $C$ not on the line such that $\triangle ACB$ is an equilateral triangle. I then was going to use the angle bisector theorem to form a bisector which would intersect the segment $AB$ at a point $D$. Using the side-side-side criterion of a triangle, I could conclude that $AD \cong DB$.
However, there are few issues with this. Firstly, how do I know the angle bisector will intersect the segment $AB$; why is it not possible that the angle bisector $\overrightarrow{CD}$ to curve and loop around in such a way that it never intersects the segment $AB$ nor the line $\ell$? Secondly, even if it does intersect $\ell$, how do I know that $D$ is between $A$ and $B$
For the second issue, I tried a proof by contradiction, but I couldn't identify any contradiction.
Last edited: Oct 3, 2015
2. Oct 3, 2015
### SteamKing
Staff Emeritus
The bisector of an angle is a ray which extends from the apex of the angle to infinity. By definition, a ray is a line, without any loops or whorls in it.
http://www.mathopenref.com/ray.html
3. Oct 3, 2015
### Ray Vickson
What is wrong with the standard construction, as in http://www.mathopenref.com/constbisectline.html ?
4. Oct 4, 2015
### Bashyboy
That seems to be a rather vague definition. The definition I am using is the following: Let $A$ and $B$ be two points. The ray starting at $A$ and going through $B$ is the set of all points $X$ such that either (1) $X$ is $A$ or $B$, (2) $X$ is between $A$ and $B$, or $B$ is between $A$ and $X$. Notice, such a definition does not mention infinity and does not preclude the possibility of loops and curves.
Well, it is proposed by our professor that we try to rely on recently proven theorems, such as being able to construct an equilateral triangle from three points, rather than always resorting to drawing circles.
5. Oct 4, 2015
### vela
Staff Emeritus
How do you get that it doesn't preclude the possibility of loops and curls? For instance, if the ray loops or curls between A and B, then there will be points on the curve that aren't between A and B. Between means a point is on the line segment connecting A and B, not that it's on any old curve that passes through A and B.
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# Limit problem solution
#### icosane
1. Homework Statement
lim as x goes to 0 of
[x*csc(2x)] / cos(5x)
3. The Attempt at a Solution
The book solution says the answer is 1/2. I keep getting zero as an answer because of the x in the numerator and am unsure of how else to go about the problem. I'm pretty sure I'm supposed to use the limit as x goes to 0 of sinx / x, but even when I get to that point the limit of x as x goes to 0 always gives me an answer of zero. What am I doing wrong? Thanks!
Related Calculus and Beyond Homework Help News on Phys.org
#### Mark44
Mentor
The part to be concerned with is x*csc(2x), since the limit of cos(5x) is 1, as x goes to 0. If you can establish a limit for x*csc(2x), then lim x*csc(2x)/cos(5x) will be = lim x*csc(2x).
x*csc(2x) = x/sin(2x), so all you need now is 2x in the numerator instead of the x that is there. Can you turn x into 2x by multiplying by 1 in some form?
#### lanedance
Homework Helper
so you have
$$\frac{x.csc(2x)}{cos(5x)} = \frac{x.}{sin(2x).cos(5x)}$$
so the denmoinator goes to zero as well, so you can't say it is zero...
#### icosane
So the lim as x goes to 0 of x/sin(x) is also equal to 1? I assume that is the case because multiplying by 2/2 gives the solution of 1/2. Thanks a lot, I really appreciate it.
Just out of curiosity now, is the lim as x goes to 0 of x/(1-cos(x)) = infinity?
#### lanedance
Homework Helper
lim as x goes to 0 sin(x)/x is one, as is sin(x)/x
$$\stackrel{lim}{x \rightarrow 0} f(x) = c$$
then what is
$$\stackrel{lim}{x \rightarrow 0} \frac{1}{f(x)} = ?$$
the second limit is true
$$\stackrel{lim}{x \rightarrow 0} \frac{x}{(1-cos(x))} = \infty$$
can you show why?
(l'hopitals rule is good for all of these if you know it...)
#### Mark44
Mentor
So the lim as x goes to 0 of x/sin(x) is also equal to 1? I assume that is the case because multiplying by 2/2 gives the solution of 1/2. Thanks a lot, I really appreciate it.
Just out of curiosity now, is the lim as x goes to 0 of x/(1-cos(x)) = infinity?
No, the limit doesn't exist.
$$\lim_{x \rightarrow 0^+} x/(1 -cos(x)) = \infty$$
while
$$\lim_{x \rightarrow 0^-} x/(1 -cos(x)) = -\infty$$
#### lanedance
Homework Helper
good pickup - my mistake
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# Basics about Plot Area, Built-Up Area, and Carpet Area
When someone wants to buy a home, he faces some technical words like plot area, built-up area, carpet area. Every word has its own meaning. Buyers should know the terms before purchasing a house for them.
The common areas of the house are:
1. Plot Area
2. Built-up area or Plinth area
3. Super built-up area
4. Carpet area
5. Floor area
6. Setback area
### Plot Area
The plot area is also known as the site area. The area which is under the ownership of the seller is called the plot area. Fencing is completed to specify the boundaries.
Calculate the plot area
Area =s (s-a) (s-b) (s-c)
s=a+b+c/2
At first, you have to divide the plot into a number of triangles.
Then you have to measure the distances of the sides of the triangles by calculating chains, or a tape. After that let the distance of the three sides be denoted by a, b, and c. after that area of each triangle can be calculated by the formula:
Area of triangle=s (s-a) (s-b) (s-c)
s= semi perimeter = a+b+c/2
Next, you have to take a summation of the areas of all the triangles calculated in the above step, and you will get the plot area.
Plot area = A1 + A2 + A3 +……..+ An
### Built-up Area
The total area of the flat that includes the carpet area and the thickness of the wall is called the built-up area. This built-up area includes all the areas where you can move and the area of walls and utility areas. Generally, the built-up area is 10 to 15% more than the carpet area.
Living room and drawing room, kitchen, bedroom, bathroom, staircase, terrace, balcony, veranda, utility area, and wall thickness. If the house has a common area then the wall area is included in the built-up area.
### Super built-up Area
Common areas like a corridor, lift space, swimming pool, park, gym, playground, clubhouse are included in the built-up area to get the super built-up area. The builder or seller fixes the price based on the super built-up area of the building or house. Because this area includes the area of all the amenities which is facilitated to you.
### Carpet Area
The carpet area is the area where the buyers or the owners spread the carpet. This area is available for use. Nowadays most builders want to calculate the selling price of houses based on the carpet area.
The areas which are included in the carpet area are living room, dining room, bedroom, other rooms, kitchen, bathroom, and the thickness of internal walls. The areas which are not included in Carpet areas are terrace area, lift area, service shafts, exclusive balcony, corridor area, the thickness of external walls.
At a glance
1. Floor Area = total usable area + Wall thickness
2. Carpet Area = Total floor area – Area of external walls
3. Built up Area = carpet area + Area of walls
4. Super Built-up Area = Built up area + Common amenities area
= Built up Area of All Floors
The carpet area is the probable area on which a carpet can be spread. The built-up area is also called the Plinth area. It is the total area which is provided to use. The Super built-up area includes the spaces like the park, playgrounds, gym, and other utilities common to the residents. The plot area is the land area that is under the ownership of someone between the fencing. The setback area is the offset left around the building. It is enforced by the law.
### A chart of Plot Area, Built-Up Area, and Carpet Area for standard rooms
Area Carpet area Floor area Built up area Super built up area Bedroom yes yes yes yes Living room yes yes yes yes Kitchen yes yes yes yes Bathroom yes yes yes yes Internal Wall thickness yes yes yes yes External Wall thickness - yes yes yes Exclusive balcony - yes yes yes Corridor - yes yes yes Terrace area - - yes yes Staircase - - - yes Swimming pool - - - yes Gym - - - yes Park/Playground - - - yes Clubhouse - - - yes | HuggingFaceTB/finemath | |
Floating-point Number Python, Clinical Reasoning Cases In Nursing Pdf, Boost Mobile $40 Plan, Do Orcas Eat Sharks, Assassin Snails For Sale, Oligoclase In Thin Section, " /> Floating-point Number Python, Clinical Reasoning Cases In Nursing Pdf, Boost Mobile$40 Plan, Do Orcas Eat Sharks, Assassin Snails For Sale, Oligoclase In Thin Section, " />
## singular matrix non trivial solution
M is also referred to as Modal matrix. While solving linear homogeneous equation do we check rank of A or rank of (A!B) ? Also while reading from many sources I found below facts, which I believe are correct (correct me if they are not): For non singular matrix A, Ax = b have unique solution. In (23), we call the system consistent if it has solutions, inconsistent otherwise. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Matlab does not permit non-numerical inputs to its svd function so I installed the sympy module and have tried the following code to solve my problem. 25. If X is a singular solution, let v be a n ull v ector of X and observe that 0 = How to Diagonalize a Matrix. ... Singular Matrix and Non-Singular Matrix - Duration: 2:14. Invertible matrices have only the trivial solution to the homogeneous equation (since the product A^(-1)0 = 0 for any matrix A^(-1)). • D. The matrix A is nonsingular because it is a square matrix. M = Non-singular matrix whose columns are respective Eigen vectors of A i.e. As you can see, the final row of the row reduced matrix consists of 0. These 10 problems... Group of Invertible Matrices Over a Finite Field and its Stabilizer, If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse, Summary: Possibilities for the Solution Set of a System of Linear Equations, Find Values of $a$ so that Augmented Matrix Represents a Consistent System, Possibilities For the Number of Solutions for a Linear System, The Possibilities For the Number of Solutions of Systems of Linear Equations that Have More Equations than Unknowns, Quiz: Possibilities For the Solution Set of a Homogeneous System of Linear Equations, Solve the System of Linear Equations Using the Inverse Matrix of the Coefficient Matrix, True or False Quiz About a System of Linear Equations, Determine Whether Matrices are in Reduced Row Echelon Form, and Find Solutions of Systems, The Subspace of Matrices that are Diagonalized by a Fixed Matrix, If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors, There is at Least One Real Eigenvalue of an Odd Real Matrix, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. (23) |A| = 0 ⇒ A x = b usually has no solutions, but has solutions for some b. Because in that case, you only have 1 solution. We study properties of nonsingular matrices. Rank of A is 3 and rank of (A, B) is 3. Hello, I got the answer after a bit of research. solve the system equation to find trivial solution or non trivial solution If A is non-singular, the system has only one trivial solution. This solution is called the trivial solution. Write a non-trivial solution to the system Ax=0 x= [ _; _;_] Is A singular or nonsingular? For singular A, are there infinite non-trivial solutions or unique non-trivial solution. Non-square matrices (m-by-n matrices … How Many Square Roots Exist? A is singular. Testing singularity. B cofactor of the matrix. The rank of a matrix [A] is equal to the order of the largest non-singular submatrix of [A].It follows that a non-singular square matrix of n × n has a rank of n.Thus, a non-singular matrix is also known as a full rank matrix. Each algorithm does the best it can to give you a solution by using assumptions. • Example: Page 79, number 24. Let A be an n × n matrix. Also am not able to decide on the facts in red font in above table. Question on Solving System of Homogenous Linear Equation. Given : A system of equations is given by, AX 0 This represents homogeneous equation. X = 0. Suppose the given matrix is used to find its determinant, and it comes out to 0. Let $$A$$ be an $$m\times n$$ matrix over some field $$\mathbb{F}$$. The same is true for any homogeneous system of equations. homogeneous, does it implies equations always have same y intercepts and vice-versa? The red cells corresponding to Ax = 0 in above table do not map with the corresponding ones in the first table. That is, if Mx=0 has a non-trivial solution, then M is NOT invertible. D. This is true. Singular matrices are rare in the sense that if a square matrix's entries are randomly selected from any finite region on the number line or complex plane, the probability that the matrix is singular is 0, that is, it will "almost never" be singular. Since rank of A and rank of (A, B) are equal, it has trivial solution. Theorem 2. Every homogeneous system has at least one solution, known as the zero (or trivial) solution, which is obtained by assigning the value of zero to each of the variables. (adsbygoogle = window.adsbygoogle || []).push({}); Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57. From np.linalg.solve you only get a solution if your matrix a is non-singular. (2.4, 9) (a) Give an example to show that A + B can be singular if A and B are both nonsingular. Check the correct answer below. Did you read what i have written.... for number of gates in full adder, I seek the non-trivial solution to Ax = b, where b is the zero vector and A is a known matrix of symbolic elements (non-singular). Matrix method: If AX = B, then X = A-1 B gives a unique solution, provided A is non-singular. If the above system is homogeneous, n equations in n unknowns, then in the matrix form it is AX = 0. The list of linear algebra problems is available here. Let A be a square n by n matrix over a field K (e.g., the field R of real numbers). Is the matrix 01 0 Matrix method: If AX = B, then X = A-1 B gives a unique solution, provided A is non-singular. If, on the other hand, M has an inverse, then Mx=0 only one solution, which is the trivial solution x=0. The matrix A is nonsingular because the homogeneous systems Ax=0 has a non-trivial solution. Some of the important properties of a singular matrix are listed below: The determinant of a singular matrix is zero; A non-invertible matrix is referred to as singular matrix, i.e. For singular A, can Ax = b have infinite solutions? The following diagrams show how to determine if a 2×2 matrix is singular and if a 3×3 matrix is singular. ST is the new administrator. Equivalently, if Ais singular, then the homogeneous system AX= 0 has a non{trivial solution. This probably seems like a maze of similar-sounding and confusing theorems. This probably seems like a maze of similar-sounding and confusing theorems. There are 10 True or False Quiz Problems. The same is true for any homogeneous system of equations. Q2. In the context of square matrices over fields, the notions of singular matrices and noninvertible matrices are interchangeable. B. But if A is a singular matrix i.e., if |A| = 0, then the system of equation AX = B may be consistent with infinitely many solutions or it may be inconsistent. But not able to comprehend similar things for three unknown variable systems. Hello, I got the answer after a bit of research. A square matrix M is invertible if and only if the homogeneous matrix equation Mx=0 does not have any non-trivial solutions. We study product of nonsingular matrices, relation to linear independence, and solution to a matrix equation. For non-trivial solution, A 0 which also represents condition for singular matrix. For any vector z, if A m+1z = A(A z) = 0, we know that Amz = 0, which by the induction hypothesis implies that z = 0. M is also referred to as Modal matrix. If we have more than 2 non zero, then it's good, because then we will have more number of equations? If the matrix A has more rows than columns, then you should use least squares fit. Construct a 3×3 NON-TRIVIAL SINGULAR matrix and call it A.Then, for each entry of the matrix, compute the corresponding cofactor, and create a new 3×3 matrix full of these cofactors by placing the cofactor of an entry in the same location as the entry it was based on. Step by Step Explanation. M = Non-singular matrix whose columns are respective Eigen vectors of A i.e. C. This is not true. Every homogeneous system has at least one solution, known as the zero (or trivial) solution, which is obtained by assigning the value of zero to each of the variables. Loading... Unsubscribe from calculusII Eng? If λ = 8, then rank of A and rank of (A, B) will be equal to 2.It will have non trivial solution. If matrix is non singular, then Ax = 0 has only the trivial solution. Q2. For example, in a homogenous solution where equation equated to 0, putting all variables equal to 0 is a correct solution, but this is not a useful one and hence never really asked in any question. Q3. More importantly above table doesnt talk anything about triviality of the solutions, but there are some facts that dictates triviality of the solutions as below which I want to incorporate in above table. The video explains the system with two unknowns. Thanks to all of you who support me on Patreon. I am able to prepare following table: I did understood most facts from the video and put it in the table but not quite sure about the things in red color, since I have guessed it from my observations and from reading text books: Q1. This website’s goal is to encourage people to enjoy Mathematics! If your b = [0, 0], you will always get [0, 0] as unique solution, no matter what a is (as long a is non-singular). Other solutions called solutions.nontrivial Theorem 1: A nontrivial solution of exists iff [if and only if] the system hasÐ$Ñ at least one free variable in row echelon form. Example of Trivial & Non trivial Solution calculusII Eng. Solution of Non-homogeneous system of linear equations. B. However I found that these two tables do not map well. In the above example, the square matrix A is singular and so matrix inversion method cannot be applied to solve the system of equations. If system is homogeneous i.e. : Understanding Singularity, Triviality, consistency, uniqueness of solutions of linear system, Virtual Gate Test Series: Linear Algebra - Matrix(Number Of Solutions). You can use Singular value decomposition, svd to get an x that satisfies Ax=0 if there are non-trivial solutions: A = [2 -1 1; 2 -1 1; 3 2 1]; [U S V] = svd(A); x = V(:,end) x = -0.39057 0.13019 0.91132 A*x = 0 0 0 This solution is called the trivial solution. For a singular matrix A we can get a non trivial solution Is it going to be from ECO 4112F at University of Cape Town 2.1.4 The rank of a matrix. We study product of nonsingular matrices, relation to linear independence, and solution to a matrix equation. : 2:14 should focus!!!!!!!!!!!. 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For any homogeneous system of equations contain one solution, which is the trivial solution 01 0 np.linalg.solve! A be a 3×3 matrix and suppose we know that −4a1−3a2+2a3=0 where a1, a2 and a3 are the of! Is, it is also the only solution if Mx=0 has a 0 which represents... Least squares fit linear independence, and solution singular matrix non trivial solution a matrix is used to find determinant! The first table you can see, the matrix a is singular and non singular, then =... = b have infinite solutions given by, Ax = 0 have non trivial solution solutions some. D. the matrix a is nonsingular because it is a singular matrix a is 3 and singular! Are added if a 3×3 matrix and suppose we know that −4a1−3a2+2a3=0 where a1, a2 and a3 are columns... | open-web-math/open-web-math | |
# CAGR Calculator
• Determines Compound Annual Growth Rate (CAGR)
• Average rate of change (compounded) for a value
• Useful for investments and sales analysis.
## CAGR: 14.87%
Your investment grew from \$10000 to \$20000 over a period of 5 periods (months, years) at a compounded growth rate (CAGR) of 14.87%.
14.87% is the average rate of change for the value and assumes this change was compounded every period.
Like this calculator? You may find our other business and investing calculators interesting:
### Key Concepts Behind The Model
This calculator determined the CAGR (Compound Annual Growth Rate) of an investment or business. This metric is used to measure how much the statistic you're analyzing (can be anything - a stock, a bond, company sales) has changed in each period between the start and end of the analysis. This may also be called an average rate of change calculator.
This is useful when the item being analyzed may have ups and downs over long periods. For Example, consider the stock below:
Year Price EPS
1998 \$14 \$.50
1999 \$30 \$1.00
2000 \$20 \$.75
2001 \$8 \$.50
2002 \$15 \$1.00
2003 \$30 \$1.25
Assume you were an investor in 2003 and had to make a decision about buying this stock. Earnings have been all over the place! How would you come up with a reasonable estimate of the average rate of change? CAGR is a simple way to smooth out the noise. For example, while earnings dropped in the 2000 reccession, you can point to a longer term trend - if we compared 1998 earnings (\$.50) with 2002 earings (\$1.25) and feed them into the CAGR calculator, we learn that earnings grew at a compounded rate of 20%.
You can also use this average rate of change calculator as a simple compound interest calculator. If I buy that stock at \$14, hold it for 5 years, and sell it for \$30 - what is my annual rate of return? By using the CAGR calculator, I can determine that my return was: 16.47%. Not bad!
Like this calculator? You may find our other business and investing calculators interesting: | HuggingFaceTB/finemath | |
29) A
Total Area of 1 narrow wall = 4 x 2.5 = 10m²
Area of narrow wall with door = 10 – (1 x 2) = 8m²
Area of narrow wall with window = 10 – (2 x 1.5) = 7m²
Total Area of narrow walls = 8 + 7 = 15m²
1 tin paints 15m²
If person paints with high quality paint, two coats are needed = 15 x 2 = 30m²
Cost for 1 coat which is 15m² = £15
Cost for 2 coats of 15m² each = 15 x 2 = £30
If person paints with budget paint, three coats are needed = 15 x 3 = 45m²
Cost of three coats = 11 x 3 = £33
So the person should use quality paint for these walls.
Area of a wide wall = 5 x 2.5 = 12.5 m²
Total area of wide walls = 12.5 x 2 = 25m²
If person paints with high quality paint, two coats are needed = 25 x 2 = 50m²
Since 25m² is larger than 15m², the person would have to buy 2 tins (30m²) for 1st coat.
Cost of 2 tins = 15 x 2 = £30
5m² of paint would be left over. The person still needs to paint 25m² . Using the 5m² which is left over, there would be 20m² left. The person would still need to buy 2 more tins.
Cost for 2 more tins = 15 x 2 = 30
Total cost = 30 + 30 = £60
If person paints with budget quality paint, three coats are needed = 25 x 3 = 75m²
Since 25m² is larger than 15m², the person would have to buy 2 tins (30m²) for 1st coat.
Cost of 2 tins = 11 x 2 = £22
5m² of paint would be left over. The person still needs to paint 50m² . Using the 5m² which is left over, there would be 45m² left. The person would still need to buy 3 more tins.
Cost for 3 more tins = 11 x 3 = 33
Total cost = 22 + 33 = £55
To paint the wider walls, the person should opt for the budget paint.
Total cost = 30 + 55 = £85 | HuggingFaceTB/finemath | |
# Spherical symmetry metric
• I
• GR191511
In summary, spherical symmetry requires that the line element does not vary when ##\theta## and##\phi## are varied, so that ##\theta## and ##\phi## only occur in the line element in the form (##d\theta^2+\sin^{2}\theta d\phi^2)##. This is the metric of a 2-sphere, which is invariant under a change of pole and meridian. Any other metric represents something less symmetric, so will have some preferred direction or something. The key concept is the Killing vector field, which describes the symmetries of a manifold. For a flat 2D Euclidean metric, there are three Killing vectors, while for a spherical 2D manifold
#### GR191511
"Spherical symmetry requires that the line element does not vary when##\theta## and##\phi## are varied,so that ##\theta##and ##\phi##only occur in the line element in the form(##d\theta^2+\sin^{2}\theta d\phi^2)##"
I wonder why:
"the line element does not vary when##\theta## and##\phi## are varied"could##\Rightarrow##"##\theta##and ##\phi##only occur in the line element in the form(##d\theta^2+\sin^{2}\theta d\phi^2)##"?Thanks!
That's the metric of a 2-sphere, which is invariant under a change of pole and meridian. Any other metric represents something less symmetric, so will have some preferred direction or something.
vanhees71
Ibix said:
That's the metric of a 2-sphere, which is invariant under a change of pole and meridian. Any other metric represents something less symmetric, so will have some preferred direction or something.
What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?
GR191511 said:
What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?
A sphere is 2 dimensional, so ##dx^2+dy^2+dz^2## has too many dimensions.
malawi_glenn and GR191511
GR191511 said:
What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?
No, and you can see that because ##x##, ##y##, and ##z## do not appear in the metric so it is the same everywhere. Compare with, for example, ##x^{-2}dx^2+dy^2+dz^2## where the metric is different at points with different ##x## coordinates.
GR191511
Nugatory said:
No, and you can see that because ##x##, ##y##, and ##z## do not appear in the metric so it is the same everywhere. Compare with, for example, ##x^{-2}dx^2+dy^2+dz^2## where the metric is different at points with different ##x## coordinates.
But the ##d\theta^2+\sin^{2}\theta d\phi^2## vary when ##\theta## are varied (because of ##sin^{2}\theta##)
The key concept here is the Killing vector field: https://en.m.wikipedia.org/wiki/Killing_vector_field
This vector field describes the symmetries of a manifold. The algebraic form of the metric in some coordinates may obscure the nature and existence of the Killing vector fields.
For a flat 2D Euclidean metric there are three Killing vectors. Two are translations in different directions and one is rotation. That specifies the symmetries of a flat plane, which is the maximally symmetric 2D manifold.
For a spherical 2D manifold there are two Killing vector fields. One is a rotation about the poles and the other is a rotation about the equator.
It is these Killing vectors that distinguish the manifolds, not the coordinates.
GR191511
GR191511 said:
But the ##d\theta^2+\sin^{2}\theta d\phi^2## vary when ##\theta## are varied (because of ##sin^{2}\theta##)
Why is the ##\sin^2 \theta## term there?
GR191511 said:
What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?How?
That's of course also symmetric under rotations. When you parametrize ##(x,y,z)## with the usual spherical coordinates, you get
$$d\vec{r}^2=\mathrm{d} r^2 + r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$
Rotations around the origin only affect the angles ##\vartheta## and ##\varphi## and keeps ##r## invariant, and thus the most general pseudo-metric invariant under rotations is
$$\mathrm{d} s^2=g_{00}(t,r) \mathrm{d} t^2 - g_{11}(t,r) \mathrm{d} r^2 -g_{22}(t,r) (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2).$$
You could also make a more general ansatz with off-diagonal elements in the ##(t,r)##-plane, but these you can "gauge away", i.e., the ansatz above is the most general ansatz for a spherically symmetric spacetime.
GR191511 and Dale
Edit: vanhees71 beat me to it with a reply that says more or less the same thing as this post, but I won't delete this as it uses slightly different language so might still be helpful.
GR191511 said:
What about ##dx^2+dy^2+dz^2##?Does the line element vary when x and y and z are varied?
If you perform the standard Cartesian-to-spherical-polar coordinate conversion\begin{align*} x &= r \sin\theta \cos\phi \\ y &= r \sin\theta \sin\phi \\ z &= r \cos\theta \end{align*}you find that$$\text{d}x^2 + \text{d}y^2 + \text{d}z^2 = \text{d}r^2 + r^2 (\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2 )$$It is the presence of the term ##\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2##, with all the the other terms independent of ##\phi## and ##\theta##, that indicates spherical symmetry, but the presence of the multiplier ##r^2## in front of that term, and the presence of ##\text{d}r^2## with a multiplier of 1, adds further constraints to the metric (i.e. the form of dependence on ##r##). You could have different dependence on ##r## but still have spherical symmetry, e.g. ## A(r) \, \text{d}r^2 + B(r) \, (\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2 )## where ##A## and ##B## are positive-valued functions of ##r##.
GR191511, Dale, PeroK and 1 other person
Dale said:
For a spherical 2D manifold there are two Killing vector fields. One is a rotation about the poles and the other is a rotation about the equator.
No, there are three KVFs, not two, corresponding to the three generators of the rotation group SO(3), which is the relevant symmetry group. One is rotation about the equator (by which I assume you mean rotation that leaves the equator invariant), yes, but there are two linearly independent ways of "rotating about the poles", not one. Those two are the other two KVFs.
vanhees71
GR191511 said:
"Spherical symmetry requires that the line element does not vary when##\theta## and##\phi## are varied
Is this supposed to be a quote from somewhere? If so, please give a reference. (I suspect it's not, because it's wrong.)
DrGreg said:
It is the presence of the term ##\text{d}\theta^2 + \sin^2\theta \, \text{d}\phi^2##, with all the the other terms independent of ##\phi## and ##\theta##, that indicates spherical symmetry
Why"all the the other terms independent of ##\phi## and ##\theta##"indicates spherical symmetry?
GR191511 said:
Why"all the the other terms independent of ##\phi## and ##\theta##"indicates spherical symmetry?
It doesn't. All the other terms being independent of ##\phi## and ##\theta## and the ##\theta\theta##, ##\theta\phi## and ##\phi\phi## components being as quoted does.
The point is that for spherical symmetry we require that the line element have the same form wherever we choose to place the pole and the ##\phi=0## line. Does that make sense? If we idealise the Earth as a sphere, we could place the equator anywhere and distance measurements wuld depend on latitude and longitude in the same way. However that is not the case for the Earth in reality - the distance around the equator is different to a transpolar circuit, so (sufficiently precise) distance measurements would not relate to revised longitude and latitude coordinates in the same way they relate to the normal ones. Only on a perfect sphere are you free to choose your origin of angular coordinates without changing your metric (and picking the origin seems to be what is meant - somewhat inaccurately - by varying ##\theta## and ##\phi## in your OP).
So if you want a spherically symmetric spacetime and you want to express it in polar coordinates then your ##\theta## and ##\phi## had better behave like they do on a sphere. Hence ##r^2d\theta^2+r^2\sin^2\theta d\phi^2##.
Furthermore you need the metric components associated with your other two coordinates to be independent of ##\theta## and ##\phi##. Otherwise you have something non-spherically summetric.
GR191511
Well, one should also note that the introduction of spherical coordinates make only part of spherical symmetry directly manifest, i.e., the rotations around the polar axis, with the azimuthal angle ##\varphi## as a holonomous coordinate, i.e., by introducing a preferred direction for the polar axis you apparently break the full rotation symmetry. This manifests itself that the metric components depend on the polar angle, ##\vartheta##.
GR191511 | HuggingFaceTB/finemath | |
# compact set and convergent sequence
• Feb 28th 2010, 01:00 PM
inthequestofproofs
compact set and convergent sequence
Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."
I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.
Proof:
Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.
I don't think my proof is clear at all. Please provide some hints.
• Feb 28th 2010, 01:39 PM
southprkfan1
Quote:
Originally Posted by inthequestofproofs
Using the definition of a compact set: "A subset K of R is compact if every open cover of K has a finite subcover of K, that is if {Oj}j in A is an open cover of K, then there exists j1, j2, j3.... , jn in A s.t. K is a subset of the Union (from i=1 to n) of Oji."
I need to show that having a convergent sequence {Pn} in R, with lim = p, the set B = {p} U {pn: n in N} is a compact subset of R.
Proof:
Since {Pn} is a convergent sequence, it is nonempty. So for every e>0, the collection of of Ne(x):x in Pn} is an open cover of B. For every open cover, there is a finite subcover of B. Also, B is a subset of Union of open covers of B.
I don't think my proof is clear at all. Please provide some hints.
Try this:
Let O be an open cover of B. O = {O_j, j=1,2,3...} And Suppose p is on O_k.
then there exists an e>0 such that the open ball of radius e around p is in O_k (by defintion of an open set)
but since (pn) --> p, there exists a number N such that for all n > N, we have l pn - p l < e
that is, for all n>N, pn is in the ball of radius e around p, and thus O_k is a cover for all pn, n>N.
Now we only have to worry about the points pn, n = 1,2,3...N. But, this is a finite number of points which is cleary covered by a finite number of open sets...and the rest is easy.
• Feb 28th 2010, 02:01 PM
Drexel28
This is of course true in any topological space.
• Feb 28th 2010, 02:26 PM
Plato
Quote:
Originally Posted by Drexel28
This is of course true in any topological space.
Yes it is true in general.
If $\displaystyle (p_n)\to p$ then for any open set $\displaystyle \mathcal{O}$ containing $\displaystyle p$ also contains almost all (all but a finite collection) points of $\displaystyle p_n$. | HuggingFaceTB/finemath | |
# help simplify trig
• Feb 20th 2010, 08:07 PM
yangx
help simplify trig
Can someone help me with this simplification?
6sin(3x)cos(3x)
I know this simplify to 3sin(6x) but how?
identity:
sin2x = 2sinxcosx
• Feb 20th 2010, 08:36 PM
mr fantastic
Quote:
Originally Posted by yangx
Can someone help me with this simplification?
6sin(3x)cos(3x)
I know this simplify to 3sin(6x) but how?
identity:
sin2x = 2sinxcosx
As I tell every student, perhaps if you used symbols in your formulae that are not likely to appear (and inevitably have different meanings) in the questions you have to answer, you will have much better luck .....
The double angle formula is $\displaystyle \sin(2A) = 2 \sin (A) \cos (A)$. What do you think A is equal to in your question ....? | HuggingFaceTB/finemath | |
## Saturday, July 26, 2008
### The Fibonacci Numbers, Coalgebraicaly
`> {-# LANGUAGE TypeSynonymInstances #-}`
What kind of Haskell programmer can I call myself if I haven't written a fiendishly complex piece of code to inefficiently compute the Fibonacci numbers. Unfortunately, what follows isn't particularly fiendish and isn't wholly inefficient, but it'll have to do.
My real motivation is this: I've mentioned before that one difference between mathematics and computer science is that mathematics has more of a bias towards algebraic rather than coalgebraic structures. But coalgebras do appear in mathematics. For example, every Hopf Algebra is a coalgebra (and an algebra) and Hopf algebras play a prominent role due to the amazing discovery that they have a nice connection with combinatorics, especially helping to organise the combinatorics that come up in renormalisation.
The first thing to note is that the computer science and mathematics definitions of coalgebra are closely related but different. An algebra is essentially a vector space over some field K with a bilinear associative product and multiplicative identity. If our algebra is V, then we have a multiplication map V×V -> V that is linear in each of its arguments. That's the same as saying that multiplication is a linear map V⊗V->V. In a coalgebra we have the converse, a linear comultiplication map Δ:V→V⊗V. Similarly, a unit in an algebra gives a way to map your ground field into the algebra, you map x to x.1 so a counit gives a map from V down to K. The full definition is on wikipedia.
The Fibonacci sequence can be thought of as a function N→R, where
f(0)=1
f(1)=1
f(n)=f(n-1)+f(n-2) for n>=2
`> f 0 = 1> f 1 = 1> f i = f (i-1)+f (i-2)`
Many functions satisfy sum formulae like ex+y=exey and tan(x+y)=(tan(x)+tan(y))/(1-tan x tan y). f also has one. What's more, the formula is essentially the definition of a coproduct on a coalgebra.
Define a Fibonacci-like sequence to be a sequence s satisfying just the last part of the definition of f:
s(n)=s(n-1)+s(n-2) for n>=2
Because I have chosen maps to R, Fibonacci-like sequences form a vector space over R.
Let S be the space of Fibonacci-like sequences. We're going to try to make this into a coalgebra. It's easy to come up with a candidate for a counit: ε(f) = f(0).
We can express this in Haskell:
`> class VectorSpace v where> zero :: v> (.*) :: Float -> v -> v> (.+) :: v -> v -> v> (.-) :: v -> v -> v> v .- w = v .+ ((-1) .* w)> type Sequence = Int -> Float> instance VectorSpace Sequence where> zero _ = 0> a .* v = \i -> a*v i> v .+ w = \i -> v i+w i> epsilon f = f 0`
We now need to consider S⊗S. Tensor products of infinite dimensional spaces can be a bit delicate. But S is a two-dimensional space because it is completely specified by its first two terms. This means that S⊗S is the obvious thing: a 2D grid of values where each row and each column is an element of S.
`> type Sequence2 = Int -> Int -> Float> instance VectorSpace Sequence2 where> zero _ _ = 0> a .* v = \i j -> a*v i j> v .+ w = \i j -> v i j+w i j`
The tensor product operation is straightforward:
`> (<*>) :: Sequence -> Sequence -> Sequence2> v <*> w = \i j -> v i*w j`
Let's define the operator D so that (Df)(i)=f(i+1). So D is a kind of shift operator. Dn shifts n times and we can form polynomials in D. Note that this space of polynomials, P, in D is also two-dimensional because D2=1+D by definition of Fibonacci-like.
`> d :: Sequence -> Sequence> d v i = v (i+1)`
We can define a pairing between P and S by <f(D),s> = ε(f(D)s). This makes P dual to S. P has an obvious multiplication inherited from ordinary multiplication which we can denote by the map m:P×P→P. This suggests the idea of forming a kind of dual to multiplication, Δ:S→S⊗S, defined by
<f(D)g(D),s>=<f(D)⊗g(D),Δ(s)>
The fact that we have defined things through this duality ensures that (S,ε,Δ) is a coalgebra. (Note that the pairing extends to the tensor products via <f(D)⊗g(D),s⊗t>=<f(D),s><g(D),t>
By definition <Dm⊗Dn,Δ(s)>=<Dm+n,s>. So Δ(s) is simply the 2D grid made by arranging s along the edges and filling in with constant values along the diagonals.
`> delta :: Sequence -> Sequence2> delta v = \i j -> v (i+j)`
Give Δ(s) we can extract s back out from it in many different ways. To get the ith element we need to pick j and k such that j+k=i and grab the ith element of the kth row. One choice is to pick k such that it is half of i rounded down:
`> extract :: Sequence2 -> Sequence> extract f i = let k = i `div` 2 in f k (i-k)`
Now given a Fibonacci-like sequence s, we can think of the top left 2×2 block of Δ(s) as a matrix. Ie. s(i+j) for i=0,1;j=0,1 forms a matrix. It's clearly invertible by inspection. But there's a deeper reason why it's invertible. If there were a linear dependence between the rows of this matrix it would be a relationship between the shifts of the sequence and hence would give rise to a simpler definition of the Fibonacci sequence. For more general recursive sequences, if there is an nth order recurrence relation for the sequence, and no simpler relation, then the top left (n-1)⊗(n-1) block is full rank and invertible.
Write tij for the inverse of this matrix. Then
Σjf(i+j)tjkik for i,j,k=0,1.
Now here's a neat bit: any Fibonacci-like sequence is defined by its first n-1 elements. Any Fibonacci-like sequence is simply a linear combination of f and Df. So the above is actually true for any Fibonacci-like sequence. And as any shift of the Fibonacci sequence is itself a Fibonacci-like sequence, the above is true for any i,k>0.
Another matrix multiplication gives
Σj,kf(i+j)tjkf(k+l)=f(i+l) for j,k=0,1;i,l≥0
That's our Fibonacci summation formula. It tells us how to get f(i+l) in terms of f(i+j) and f(k+l) for j,k=0,1.
Using the definitions above we get
Σj,k<Di⊗Dl,tjkDjf⊗Dkf=e<Di+l,f>
So tjkDjf⊗Dkf satisfies the definition of Δ above, and so must in fact be the same thing. So we have:
Δ(f)=2s⊗f-f⊗Df-Df⊗f+Df⊗Df
(Where I actually computed the elements of the inverse matrix t.)
We can now use this as the basis for a recursive algorithm. We need the bootstrap function to get the recursion started:
`> bootstrap :: [Float] -> Sequence -> Sequence> bootstrap xs v i = if i<length xs then xs!!i else v i> f' :: Sequence> f' = bootstrap [1,1,2,3] \$ extract \$ 2 .* (f' <*> f') .- (f' <*> (d f')) .- ((d f') <*> f') .+ ((d f') <*> (d f'))`
It's faster than the usual recursion. (Though that's not saying much.)
I'll leave it to you to extract out what's really going on above, and to generalise it to give sum formulae for other recurrence relations.
Note that the goal here isn't to write a fast algorithm. My main motivation is to view the sum formula as an identity about comultiplication and get some intuition about comultipliation so I can read some Hopf algebra papers.
`> let's_go_slow = map f [0..]> let's_go_faster = map f' [0..]`
All of this, apart from the algorithm itself, is an unpacking and simplification of section 1 of Bialgebras of Recursive Sequences and Combinatorial Identities by Futia et al.
PS Tell me if you have trouble reading the above. I wrote a new pseudo-LaTeX to blogger-HTML converter and it might not actually be any good.
geophf said...
Sunday, 27 July, 2008
PAStheLoD said...
Very interesting post. I thought I'm familiar with the fibonacci sequence, from both mathematical and comp.science perspective, but this is something totally "strange". (And not "just" because Haskell.)
You should make a "Coalgebras for dummies" series, like the one you did about E8 :) And some words on those tensor thingies, maybe? :P
Wikipedia could really use a few (real world) examples for most of the advanced math realted stuff.
Monday, 28 July, 2008
worldpeace said...
awesome..I always love Maths when I still young...Great job
Monday, 28 July, 2008 | HuggingFaceTB/finemath | |
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# Math 7 Unit 2
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## Math 7 Unit 2 - Illustrative Mathematics Online Resources
Grade 7 Illustrative Mathematics – Unit 2: Introducing Proportional Relationships Download PDF of Student Edition Download PDF of Student Practice Work (Homework) Unit Summary for Parents - PDF File Lesson Topic IM Lesson Khan Academy Support Videos Khan Academy Online Practice Class Notes Homework Help 1 – One of These Things is Not Like the Others Lesson 1 7.2.1 Morgan 2 – Introducing Proportional Relationships with Tables Lesson 2 Proportional Relationships 7.2.2 Morgan 3 – More about Constant of Proportionality Lesson 3 Constant of Proportionality from Tables Constant of Proportionality from Tables 7.2.3 Morgan 4 – Proportional Relationships and Equations Lesson 4 Identifying the Constant of Proportionality Constant of Proportionality From Tables Equations for Proportional Relationships Writing Proportional Equations Constant of Proportionality from Equations Constant of Proportionality from Tables (with equations) Writing proportional equations from tables 7.2.4 Morgan 5 – Two Equations for Each Relationship Lesson 5 Interpret proportionality constants Interpret Constants of Proportionality 7.2.5 Morgan 6 – Using Equations to Solve Problems Lesson 6 Interpret Two Step Equation Word Problems Writing Proportional Equations 7.2.6 Morgan 7 – Comparing Relationships with Tables Lesson 7 Proportional Relationships: Movie Tickets Proportional Relationships: Bananas Proportional Relationships: Spaghetti Is side length and area proportional? Is side length and perimeter proportional? Identify Proportional Relationships 7.2.7 Morgan 8 – Comparing Relationships with Equations Lesson 8 7.2.8 Morgan 9 – Solving Problems about Proportional Relationships Lesson 9 Solving Proportionality Constants Mid-unit Practice Quiz 7.2.9 Morgan 10 – Introducing Graphs of Proportional Relationships Lesson 10 Identifying Proportional Relationship from Graphs Identify Proportional Relationships from Graphs 7.2.10 Morgan 11 – Interpreting Graphs of Proportional Relationships Lesson 11 Constant of Proportionality from Graphs Interpreting Graphs of Proportional Relationships Constant of Proportionality from Graphs Interpreting Graphs of Proportional Relationships 7.2.11 Morgan 12 – Using Graphs to Compare Relationships Lesson 12 Comparing Constants of Proportionality Comparing Constants of Proportionality 7.2.12 Morgan 13 – Two Graphs for Each Relationship Lesson 13 Interpret Constant of Proportionality in Graphs Quiz Number 2 7.2.13 Morgan 14 – Four Representations Lesson 14 7.2.14 Morgan 15 – Using Water Efficiently Lesson 15 Unit 2 Review Khan Academy Unit 2 Test | HuggingFaceTB/finemath |
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