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https://forums.bignerdranch.com/t/purpose-of-finding-distance/2660
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# Purpose of finding distance
#1
What exactly is the purpose of this:
```// How far apart are the 'color' and 'c'? float dist; dist = pow(red - r, 2) + pow(green - g, 2) + pow(blue - b, 2);```
I don’t understand what is going on there at all. I see that we check next to see if it’s the closest distance, but closest distance to what?
#2
That code finds the distance between the elements of NSColor *color and NSColor *c, the format of which are apparently stored in a CGFLoat (which is a typedef for float).
I don’t understand why the pow() function is used here though… why not just subtract the two, like so:
``dist = (red - r) + (green - g) + (blue - b);``
#3
The pow () saves you from getting zero values for distance. Imagine for example the case:
(red - r) == - ((green - g) + (blue - b))
#4
I disagree with ibex10. I believe that the reason why the sample code uses pow() is to calculate the distance between two points on a 2D space. This tells how far “apart” the two colors are.
For a graphical illustration of this, in the color picker, select Color Wheel.
(Of course, I could be wrong, since I am a newbie myself; but this is what came to me.)
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# Can you live off dividends in retirement?
Contents
## How much money do you need to live off dividends?
Using the standard 4% dividend yield, most people need roughly 1 million dollars invested in dividend stocks to be able to live off of the passive income.
## How do I live off dividends forever?
5 Steps to Invest for Passive Income with the Goal of Living Off Dividends
1. Contribute \$200 per month to your dividend portfolio your first year. …
2. Increase your monthly contributions by 25% per year. …
3. Any dividend income you receive should be reinvested into your dividend growth portfolio.
## What happens to dividends in a retirement account?
When you set up your 401(k) allocation, you do not have a choice of what happens with fund dividends; all fund distributions are reinvested. You do not pay any taxes on the dividends earned, since your 401(k) account grows tax-deferred as long as the money remains in the account.
## How much do I need to invest to make \$1000 a month?
To make \$1000 a month in dividends you need to invest between \$342,857 and \$480,000, with an average portfolio of \$400,000. The exact amount of money you will need to invest to create a \$1000 per month dividend income depends on the dividend yield of the stocks. What is dividend yield?
IT IS INTERESTING: What percentage of stock returns come from dividends?
## How much do I need to invest to make \$500 a month in dividends?
In order to make \$500 a month in dividends, you’ll need to invest approximately \$200,000 in dividend stocks. The exact amount will depend on the dividend yields for the stocks you buy for your portfolio. Take a closer look at your budget and decide how much money you can set aside each month to grow your portfolio.
## How much money do I need to invest to make \$3 000 a month?
By this calculation, to get \$3,000 a month, you would need to invest around \$108,000 in a revenue-generating online business. Here’s how the math works: A business generating \$3,000 a month is generating \$36,000 a year (\$3,000 x 12 months).
## What stocks pay dividends monthly?
The following seven monthly dividend stocks all yield 6% or more.
• AGNC Investment Corp. ( ticker: AGNC) …
• Horizon Technology Finance Corp. ( HRZN) …
• LTC Properties Inc. ( LTC) …
• Main Street Capital Corp. ( MAIN) …
• PennantPark Floating Rate Capital Ltd. ( PFLT) …
• Pembina Pipeline Corp. ( PBA)
## Does Warren Buffett reinvest dividends?
While Berkshire Hathaway itself does not pay a dividend because it prefers to reinvest all of its earnings for growth, Warren Buffett has certainly not been shy about owning shares of dividend-paying stocks. Over half of Berkshire’s holdings pay a dividend, and several of them have yields near 4% or higher.
## Do I pay taxes if I reinvest dividends?
Are reinvested dividends taxable? Generally, dividends earned on stocks or mutual funds are taxable for the year in which the dividend is paid to you, even if you reinvest your earnings.
IT IS INTERESTING: Can I still invest in Amazon?
## Do I pay taxes on dividends?
In short, yes. The IRS considers dividends to be income, so you usually need to pay tax on them. Even if you reinvest all of your dividends directly back into the same company or fund that paid you the dividends, you will pay taxes. … Qualified dividends are subject to the lower, capital gains rates.
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## FANDOM
10,821 Pages
In this blog post, I'll be extending the mighty arrow notation 'infinite'ly! (not really)
## Arrows
Basically, there is no last extension.
When extending and generalizing arrow notation, you'll find out that you can still extend it even more. You can even create a function for the nth generalization and still be able to extend.
## Up up up
When using arrow notation, you'll find out that you can use arrow notation to define the number of arrows and you can also use the large number generated using that large number of arrows to define the number of arrows!
$3=3$
$3\uparrow^33=\text{tritri}$
$3\uparrow^{3\uparrow^33}3=...$
$3\uparrow^{3\uparrow^{3\uparrow^33}3}3$
At this point, increasing the number of layers is getting impractical.
Let's denote the symbol for the generalization of the sequence above as a $\rightarrow$.
$3\rightarrow1=3$
$3\rightarrow3=3\uparrow^{3\uparrow^33}3$
We can do some things here:
$3\rightarrow^23=3\rightarrow\rightarrow3=3\rightarrow(3\rightarrow3)$
$3\rightarrow^33$
$3\rightarrow^{3\rightarrow^33}3$
Again:
$3\downarrow3=3\rightarrow^{3\rightarrow^33}3$
And so on...
## In general
At this point, the arrows are colliding with other notations so now let's denote $\uparrow_n$ as the nth generalization shown above. Then,
$3\uparrow_13=3\uparrow3$
$3\uparrow_23=3\rightarrow3$
$3\uparrow_33=3\downarrow3$
Now, how are we going to extend this?
Easy:
$3\leftarrow1=3$
$3\leftarrow2=3\uparrow_33$
$3\leftarrow3=3\uparrow_{3\uparrow_33}3$ (Remember how we generalized the operators)
Although we have extended it, the arrows are very confusing.
## The Operator Space
Ok. Let's calm down.
Here are the operators in the first line: (Yes, first 1D space)
$\uparrow, \rightarrow, \downarrow,...$
Generalizing the up-arrow creates the right-arrow and generalizing the right-arrow creates the down-arrow and so on...
By generalizing I mean generalizing the sequence of terms with increasing layers (like the stuff above).
Then, generalizing these operators, we get the left-arrow.
We could then generalize that to get a nice northeast-arrow and so on...
Here is the second line:
$\leftarrow, \nearrow, \nwarrow,...$
And, of course, we could generalize that line to get another beautiful operator: the southeast-arrow.
And then, we could generalize that and so on...
Before you know it (or you probably already do), we have created our first 2D space. :D
How do we get the second 2D space? We generalize the first terms in each line.
The sequence of operators that are the first in their own lines are:
$\uparrow, \leftarrow, \searrow,...$
By generalizing these, we get the first term in our second 2D space!
We could generalize that to get the first line in the second 2D space and we could then generalize that line...
By generalizing the operators that are first in their own lines in the second 2D space, we get the first operator in the third 2D space!
We can then create our first 3D space and continue on by generalizing the operators that are first in their own plane to get the first term in the first 4D space and so on...
Let's create a function: $Gen()$.
$Gen(n, a, b, c, ...)=n O n$ where $O$ is the ath operator in the bth 1D space in the cth 2D space and so on...
## Beyond The Operator Space
Look at the edit summary.
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# 255483 (number)
255,483 (two hundred fifty-five thousand four hundred eighty-three) is an odd six-digits composite number following 255482 and preceding 255484. In scientific notation, it is written as 2.55483 × 105. The sum of its digits is 27. It has a total of 3 prime factors and 6 positive divisors. There are 170,316 positive integers (up to 255483) that are relatively prime to 255483.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 27
• Digital Root 9
## Name
Short name 255 thousand 483 two hundred fifty-five thousand four hundred eighty-three
## Notation
Scientific notation 2.55483 × 105 255.483 × 103
## Prime Factorization of 255483
Prime Factorization 32 × 28387
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 85161 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,483 is 32 × 28387. Since it has a total of 3 prime factors, 255,483 is a composite number.
## Divisors of 255483
6 divisors
Even divisors 0 6 3 3
Total Divisors Sum of Divisors Aliquot Sum τ(n) 6 Total number of the positive divisors of n σ(n) 369044 Sum of all the positive divisors of n s(n) 113561 Sum of the proper positive divisors of n A(n) 61507.3 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 505.453 Returns the nth root of the product of n divisors H(n) 4.1537 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,483 can be divided by 6 positive divisors (out of which 0 are even, and 6 are odd). The sum of these divisors (counting 255,483) is 369,044, the average is 615,07.,333.
## Other Arithmetic Functions (n = 255483)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 170316 Total number of positive integers not greater than n that are coprime to n λ(n) 28386 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 22418 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 170,316 positive integers (less than 255,483) that are coprime with 255,483. And there are approximately 22,418 prime numbers less than or equal to 255,483.
## Divisibility of 255483
m n mod m 2 3 4 5 6 7 8 9 1 0 3 3 3 4 3 0
The number 255,483 is divisible by 3 and 9.
## Classification of 255483
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
## Base conversion (255483)
Base System Value
2 Binary 111110010111111011
3 Ternary 110222110100
4 Quaternary 332113323
5 Quinary 31133413
6 Senary 5250443
8 Octal 762773
10 Decimal 255483
12 Duodecimal 103a23
20 Vigesimal 1bie3
36 Base36 5h4r
## Basic calculations (n = 255483)
### Multiplication
n×y
n×2 510966 766449 1021932 1277415
### Division
n÷y
n÷2 127742 85161 63870.8 51096.6
### Exponentiation
ny
n2 65271563289 16675774803763587 4260376974189932497521 1088453890496966524264157643
### Nth Root
y√n
2√n 505.453 63.4533 22.4823 12.0635
## 255483 as geometric shapes
### Circle
Diameter 510966 1.60525e+06 2.05057e+11
### Sphere
Volume 6.98513e+16 8.20227e+11 1.60525e+06
### Square
Length = n
Perimeter 1.02193e+06 6.52716e+10 361308
### Cube
Length = n
Surface area 3.91629e+11 1.66758e+16 442510
### Equilateral Triangle
Length = n
Perimeter 766449 2.82634e+10 221255
### Triangular Pyramid
Length = n
Surface area 1.13054e+11 1.96526e+15 208601
## Cryptographic Hash Functions
md5 c8758a3341ce803f15dbede6b458da63 3a6c3d9c5ebc04f35cef718665c78f0db7142590 61c7e94cb11a03f39c7b8921004c3f4e9f0e97e4def589b059a12a7080892427 51471c024d97384749174d75812701c04e15f2a13c24ea8fda43bb9447cae6d0020467d174a2d1b196730ae4054174005d8e1b761515164a5de0d8ee95eaadaa 6ef2df9c197854599a553a72d461621ee699fd5e
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# Types: Int
The integer type `int` is signed and uses twos-complement representation for negative values. At least 64 bits are used, so the range of values that can be stored is at least [-9223372036854775808, 9223372036854775807].
Namespace HH\Lib\Math contains the following integer-related constants: `INT64_MAX`, `INT64_MIN`, `INT32_MAX`, `INT32_MIN`, `INT16_MAX`, `INT16_MIN`, and `UINT32_MAX`.
Refer to your compiler's documentation to find the behavior when the largest `int` value is incremented, the smallest value is decremented, and the unary minus is applied to the smallest value.
Consider the following example:
``````<?hh // strict
namespace Hack\UserDocumentation\Types\Int\Examples\LeapYearTest;
function is_leap_year(int \$yy): bool {
return (((\$yy & 3) === 0) && ((\$yy % 100) !== 0)) || ((\$yy % 400) === 0);
}
<<__EntryPoint>>
function main(): void {
\$year = 2001;
\$result = is_leap_year(\$year);
echo "\$year is " . ((\$result === true) ? "" : "not ") . "a leap year\n";
}
``````
Output
``````2001 is not a leap year
``````
When called, function `is_leap_year` takes one argument, of type `int`, and returns a value of type `bool`. (A year is a leap year if it is a multiple of 4 but not a multiple of 100—for example, 1700, 1800, and 1900 were not leap years—or it's a multiple of 400. Some redundant grouping parentheses have been added to aid readability.)
The bitwise AND operator, `&`, and the remainder operator, `%`, require operands of type `int`.
Like `3`, `0`, `100`, and `400`, `2001` is an `int` literal, so the local variable `\$year` is inferred as having type `int`. (Unlike function parameters such as `\$yy`, or a function return, a local variable cannot have an explicit type.) Then when `\$year` is passed to `is_leap_year`, the compiler sees that an `int` was passed and an `int` was expected, so the call is well-formed.
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## how to calculate ubrrl
8 posts / 0 new
Author
Message
hi everybody,
i have to calculate ubrrl, when ubrr is above 256. their was a special formula for it. im searching on many forums for 1 hour now. i found nothing. and what was the max. for ubrrh 256 or 255?
greetings,
Greetings and welcome to AVR Freaks!
This is a standard-issue "fundamentals" question.
For UBRR values from 0 to 255, UBRRH = 0 (or 0x00, same thing) and the entire value is held in URBBL. That is because 255 = 0xff which just fits into 8 bits.
For larger values, some "overflows" into UBRRH.
The common wisdom is just forget about high and low registers if you are using C and write UBRR = value; the compiler will automatically put the necessary parts in the proper places.
Same goes for 16 bit timer registers and the value generated by the ADC.
By the way, do you have this in the correct forum? Is this really about xMega?
Jim
Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net
Last Edited: Sat. Dec 21, 2019 - 11:53 PM
To use the UBRR formula, we need to know your AVR device type, your AVR system clock speed, and your desired baud rate.
thank you for your responds, its xmega32 and i need timer 1 and timer 0. and forgive my typo i meant ubrrh. i need to know how to calculate ubrrh the baud and systemclockspeed may can change but lets say 14400 Baud and 10MHz speed.
Last Edited: Tue. Dec 24, 2019 - 05:43 PM
jewlord wrote:
... its xmega32 ...
Atmel AVR XMEGA E Manual
[page 290]
21.10 Fractional Baud Rate Generation
due to
AVR1307: Using the XMEGA USART
[page 4]
2.1.3 Baud rate selection
"Dare to be naïve." - Buckminster Fuller
thank you for your respond. i know ubrr for bsacle=0 for u2x=0 is systemclockspeed/(16*baud) -1 and for u2x=1 it is systemclockspeed/(8*baud) -1. and if the value is over 255 (if i remember correctly) you had to subtract the value with 255 and the you had to use a formula. but i cant find the formula anywhere. i may have to calculate it during my exam, otherwise the excel table is good too. i apologize for all the circumstances :D
The formula SHOULD be in the two references that gchapman provided. See Eqn 2.1 of the second reference. It tells you all you need.
What appears to be the problem, here, is the inability to grasp using two registers to hold a 16 bit value (well, 12 bits in this case). Methinks you need to get those fundamentals under your control, first.
Jim
Jim Wagner Oregon Research Electronics, Consulting Div. Tangent, OR, USA http://www.orelectronics.net
Last Edited: Tue. Dec 24, 2019 - 09:56 PM
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# Posts from 2019
## The Broken Physics of “The Umbrella Academy” Finale
Published 3 weeks, 2 days ago
Not long ago, Kat and I got around to watching The Umbrella Academy’s first season on Netflix. I thought it was pretty good! It was a decent mix of good decisions and bad decisions by people in the story, I liked most of the characters and their portrayals, and I thought the narrative arcs came to good places. Not perfect, but good.
Except. I have to talk about the finale, people. I have to get into why the ending, the very last few minutes of season one, just didn’t work for me. And in order to do that, I’m going to deploy, for the first time ever, a WordPress Spoiler Cut™ on this here blog o’ mine, because this post is spoilerrific. Ready? Here we go.
Massive, massive spoilers and a fair amount of science ahead!
## Color Easing Isn’t Always Easy
Published 4 weeks, 1 day ago
A fairly new addition to CSS is the ability to define midpoints between two color stops in a gradient. You can do this for both linear and radial gradients, but I’m going to stick with linear gradients in this piece, since they’re easier to show and visualize, at least for me.
The way they work is that you can define a spot on the gradient where the color that’s a halfway blend between the two color stops is located. Take the mix of `#00F` (blue) with `#FFF` (white), for example. The color midway through that blend is `#8080FF`, a pale-ish blue. By default, that will land halfway between the two color stops. So given `linear-gradient(90deg, blue 0px, white 200px)`, you get `#8080FF` at 100 pixels. If you use a more generic `90deg, blue, white 100%`, then you get `#8080FF` at the 50% mark.
`linear-gradient(90deg, blue, white 100%)`
If you set a midpoint, though, the placement of `#8080FF` is set, and the rest of the gradient is altered to create a smooth progression. `linear-gradient(blue 0px, 150px, white 200px)` places the midway color `#8080FF` at 150 pixels. From 0 to 150 pixels is a gradient from `#F00` to `#8080FF`, and from 150 pixels to 200 pixels is a gradient from `#8080FF` to `#FFF`. In the following case, `#8080FF` is placed at the 80% mark; if the gradient is 200 pixels wide, that’s at 160 pixels. For a 40-em gradient, that midpoint color is placed at 32em.
`linear-gradient(90deg, blue, 80%, white 100%)`
You might think that’s essentially two linear gradients next to each other, and that’s an understandable assumption. For one, that’s what used to be the case. For another, without setting midpoints, you do get linear transitions. Take a look at the following example. If you hover over the second gradient, it’ll switch direction from `270deg` to `90deg`. Visually, there’s no difference, other than the label change.
`linear-gradient(<angle>, blue, white, blue)`
That works out because the easing from color stop to color stop is, in this case, linear. That’s the case here because the easing midpoints are halfway between the color stops—if you leave them out, then they default to 50%. In other words, `linear-gradient(0deg, blue, white, blue)` and `linear-gradient(0deg, blue, 50%, white, 50%, blue)` have the same effect. This is because the midpoint easing algorithm is based on logarithms, and is designed to yield linear easing for a 50% midpoint.
Still, in the general case, it’s a logarithm algorithm (which I love to say out loud). If the midpoint is anywhere other than exactly halfway between color stops, there will be non-linear easing. More to the point, there will be non-linear, asymmetrical easing. Hover over the second gradient in the following example, where there are midpoints set at `10%` and `90%`, to switch it from `270deg` to `90deg`, and you’ll see that it’s only a match when the direction is the same.
`linear-gradient(<angle>, blue, 10%, white, 90%, blue)`
This logarithmic easing is used because that’s what Photoshop does. (Not Mosaic, for once!) Adobe proposed adding non-linear midpoint easing to gradients, and they had an equation on hand that gave linear results in the default case. It was also what developers would likely need to match if they got handed a Photoshop file with eased gradients in it. So the Working Group, rather sensibly, went with it.
The downside is that under this easing regime, it’s really hard to create symmetric non-linear line gradients. It might even be mathematically impossible, though I’m no mathematician. Regardless, its very nature means you can’t get perfect symmetry. This stands in contrast to cubic Bézier easing, where it’s easy to make symmetric easings as long as you know which values to swap. And there are already defined keywords that are symmetric to each other, like `ease-in` and `ease-out`.
If you’re up for the work it takes, it’s possible to get some close visual matches to cubic Bézier easing using the logarithmic easing we have now. With a massive assist from Tab Atkins, who wrote the JavaScript I put to use, I created a couple of CodePens to demonstrate this. In the first, you can see that `linear-gradient(90deg, blue, 66.6%, white)` is pretty close to `linear-gradient(90deg, blue, ease-in, white)`. There’s a divergence around the 20-30% area, but it’s fairly minor. Setting an interim color stop would probably bring it more in line. That’s partly how I got a close match to `linear-gradient(90deg, blue, ease-out, white)`, which came out to be ` linear-gradient(90deg, blue, 23%, #AFAFFF 50%, 68%, white 93%)`.
Those examples are all one-way, however—not symmetrical. So I set up a second CodePen where I explored recreations of a few symmetrical non-linear gradients. The simplest example matches `linear-gradient(90deg, blue, ease-in, white, ease-out, blue)` with `linear-gradient(90deg, blue, 33.3%, white 50%, 61.5%, #5050FF 75%, 84%, blue 93%)`, and they only get more complex from there.
I should note that I make no claim I’ve found the best possible matches in my experiments. There are probably more accurate reproductions possible, and there are probably algorithms to work out what they would be. Instead, I did what most authors would do, were they motivated to do this at all: I set some stops and manually tweaked midpoints until I got a close match. My basic goal was to minimize the number of stops and midpoints, because doing so meant less work for me.
So, okay, we can recreate cubic Bézier easing with logarithmic midpoints. Still, wouldn’t it be cool to just allow color easing using cubic Béziers? That’s what Issue #1332 in the CSS Working Group’s Editor Drafts repository requests. From the initial request, the idea has been debated and refined so that most of the participants seem happy with a syntax like `linear-gradient(red, ease-in-out, blue)`.
The thing is, it’s generally not enough to have an accepted syntax—the Working Group, and more specifically browser implementors, want to see use cases. When resources are finite, requests get prioritized. Solving actual problems that authors face always wins over doing an arguably cool thing nobody really needs. Which is this? I don’t know, and neither does the Working Group.
So: if you have use cases where cubic Bézier easing for gradient color stops would make your life easier, whether it’s for drop shadows or image overlays or something I could never think of because I haven’t faced it, please add them to the GitHub issue!
## Color Me FACE1E55
Published 1 month, 3 weeks ago
There’s a long history in computer programming of using hexadecimal strings that look like English words to flag errors. These are referred to, amusingly, as “magic debug values”, and yes, Wikipedia has the lowdown. One of the most (in)famous is `DEADBEEF`, which was used “on IBM systems such as the RS/6000, also used in the classic Mac OS operating systems, OPENSTEP Enterprise, and the Commodore Amiga”, among others. It’s also become the name of a Gnu/Linux music player, and apparently does not have anything to do with Cult of the Dead Cow, at least not so far as I could determine. Maybe someone with more knowledge can drop a comment.
Anyway, one of the things about these magic debug values is they’re usually eight characters long. Not always, as in the case of `BADC0FFEE0DDF00D` (from RS/6000, again), but usually. Nintendo used `0D15EA5E` in the GameCube and Wii to indicate a normal boot (!), iOS logs `DEAD10CC` when an application terminates in a specific yet incorrect manner, and `FEEDFACE` shows up in PowerPC Mach-O binaries , as well as the VLC Player application. Just to pick a few examples.
The eight-character nature of these magic codes has meant that, for a long time, you couldn’t also use them on the sly to define colors in CSS, because it was limited to the `#RRGGBB` format. Well, those days are over. Long over. Eight-digit hex color values are here, have been here a while, and are widely supported. Here are a few swatches laid over a (fully opaque) white-to-black gradient.
If you’re using Internet Explorer or Edge, those aren’t going to work for you. At least, not until Edge switches over to Blink; then, they should work just fine.
Thanks to the way they were constructed, by only using the letters A-F, most of the colors above are mostly opaque. The last two digits in `#RRGGBBAA` set the alpha channel level of the color, just like the last part of the `rgba()` syntax. Thus, the `EF` at the end of `DEADBEEF` sets the alpha value to `0.937`; `EF` is equivalent to decimal 239, and 239 ÷ 255 = 0.937 (approximately). In other words, `#DEADBEEF` is essentially equivalent to `rgba(222,173,190,0.937)`.
That’s why, of the six swatches, only the sheepish `#baaaaaaa` and the homophonic `#feedbacc` let the background gradient show through more than very slightly; their alpha channels are 0.666 and 0.8, respectively. The rest are 0.929 and up.
Being stuck in the A-F range is fairly constraining, but that’s where hexadecimal and English overlap, so that’s how it goes. However, if you’re willing to turn to leetspeak syntax—that is, allowing yourself to use `0` as a substitute for `O`, `1` for `L` and occasionally `I`, `5` for `S`, `7` for `T`, and so on—then a lot more possibilities open up. In addition to some of the classic error codes like `fee1dead` (Linux), I had fun devising other eight-character color words like `acc0lade` and `face1e55`, not to mention the very nautical `ccccccc5`. (Think about it.) Behold!
There are still more l33t-compliant number substitutions available, like `6` for `G`, but I felt like I was already pushing it with the examples I have. One could also use calculator spelling, where `9` is a stand-in for `g`, and even mix together l33t and calculator syntaxes in the same value. So many possibilities!
You may have noticed one value which creates no color: `#DABBAD00`, which has `00` for its alpha, so it’s fully, completely transparent. It’s fully transparent `#DABBAD`, I suppose, but there’s really no difference between one transparent color and another, as far as I’m concerned. I mean, if a color falls transparent, then there’s nobody to see it, so is it really a color at all? I say thee nay.
If you’re familiar with the way `#RRGGBB` hex values can be represented with the shortened `#RGB` syntax, then it will probably come as little surprise that `#RRGGBBAA` has a shortened `#RGBA` syntax, where each digit is duplicated. This opens the world of four-letter words to us! Here are a few:
Here, we finally have a fully opaque word-color: `#BEEF` expands out to `#BBEEEEFF`, making the alpha value `FF`, which decimal-translates to `255`, which is fully opaque. So we get a nice opaque powdery blue out of `BEEF`, which is counterintuitive in the best possible way. Also, every time I see `BBEEEEFF`, either in print or in my head, I hear Mrs. Which ordering dinner.
And okay, yes, `#F8ED` isn’t a four-letter word, it’s a four-symbol license-plate word. So it’s even cooler.
If you’re thinking about using these in your CSS, you might be concerned about backwards compatibility, since any browser that doesn’t understand four- or eight-digit hexadecimal color values will just drop them on the floor. That might be okay for text coloring, since the text will likely have some color, even if it’s browser-default, which is usually black. For backgrounds, having colors ignored probably less okay, particularly if you set foreground colors that depend on the background colors.
There are a couple of possibilities here. One is to use the cascade and CSS error handling to your advantage, in the time-honored pattern of doing the simpler version first and the more sophisticated version second.
``````#example {
color: #DEA;
}``````
That works in simple scenarios, but for more complicated situations—say, ones where you have foreground and background depending on each other—feature queries are an option to consider, if for no other reason than cleaner organization and legibility.
``````#example {
color: red;
background: #EEE;
}
@supports (color: #ABCD) {
#example {
color: #f00d;
background: #feed;
}
}
``````
Naturally and as usual, you’ll have to figure out what makes the most sense for your situation. Maybe the right answer will be to avoid using these sorts of values at all, although I don’t know where the fun is in that.
At any rate, I hope you’ve enjoyed this little tour of magic debug values, l33tspeak, and color words. As always, `#feedbacc` is more than welcome in the comments!
• Published
• Categorized under CSS
• No responses so far
## CSS4 Color Keyword Distribution Visualization
Published 1 month, 3 weeks ago
Long, long ago—not quite seven years ago, in fact—I built a canvas-based visualization of the distribution of CSS3/SVG color keywords and released it. And there it’s sat, static and inert (despite being drawn with a whooooole lotta JS) ever since.
I’ve always meant to get back to it and make it more interactive. So over the past several evenings, I’ve rebuilt it as an SVG-based visualization. The main point of doing this was so that when you hover the mouse pointer over one of the little color boxes, it will fill the center of the color wheel with the hovered color and tell you its name and HSL values. Which it does, now. It even tries to guess whether the text should be white or black, in order to contrast with the underlying color. Current success rate on that is about 90%, I think. Calculating perceived visual brightness turns out to be pretty hard!
Other things I either discovered, or want to do better in the future:
• Very nearly half the CSS4 (and also CSS3/SVG) color keywords are in the first 90 degrees of hue. More than half are in the first 120 degrees.
• There are a lot of light/medium/dark variant names in the green and blue areas of the color space.
• I wish I could make the color swatches bigger, but when I do that the adjacent swatches overlap each other and one of them gets obscured.
• Therefore, being able to zoom in on parts of the visualization is high on my priority list. All I need is a bit of event monitoring and some viewbox manipulation. Well, that and a bit more time. Done, at least for mouse scroll wheels.
• I’d like to add a feature at some point where you type text, and a list is dynamically filtered to show keywords containing what you typed. And each such keyword has a line connecting it to the actual color swatch in the visualization. I have some ideas for how to make that work.
• I’d love to create a visualization that placed the color swatches in a 3D cylindrical space summarizing hue, lightness. and saturation. Not this week, though.
• I’m almost certain it needs accessibility work, which is also high on my priority list.
• SVG needs conic gradients. Or the ability to wrap a linear gradient along/inside/around a shape like a circle, that would work too. Having to build a conic gradient out of 360 individual `<path>`s is faintly ridiculous, even if you can automate it with JS.
• And also `z-index` awareness. C’mon, SVG, get it together.
Anyway, here it is: CSS4 Color Keyword Distribution. I hope you like it!
## “Stacked ‘Borders’” Published at CSS-Tricks
Published 2 months, 5 days ago
I toyed with the idea of nesting elements with borders and some negative margins to pull one border on top of another, or nesting a border inside an outline and then using negative margins to keep from throwing off the layout. But none of that felt satisfying.
It turns out there are a number of tricks to create the effect of stacking one border atop another by combining a border with some other CSS effects, or even without actually requiring the use of any borders at all. Let’s explore, shall we?
That’s from the introduction to my article “Stacked ‘Borders’”, which marks the first time I’ve ever been published at the venerable upstart CSS-Tricks. (I’m old, so I can call things both venerable and an upstart. You kids today!) In it, I explore ways to simulate the effect of stacking multiple element borders atop on another, including combining box shadows and outlines, borders and backgrounds, and even using border images, which have a much wider support base than you might have realized.
And yes, as per my usual, the images in the piece are all double-dpi :screenshot captures directly from Firefox.
Many thanks to Chris Coyier for accepting the piece, and Geoff Graham for his editorial assistance. I hope you’ll find at least some part of it useful, or better still, interesting. Share and enjoy!
## Legend of the Stalwart Mouse: Return of the MX518
Published 2 months, 6 days ago
I’ve relied on a mouse for about a decade and a half. I don’t mean “relied on a mouse” in the generic sense, but rather in the sense that I’ve relied on one very specific and venerable mouse: a Logitech MX500.
I’ve had it for so long, I’d forgotten how long I’ve had it. I searched for information about its production dates and wouldn’t you know it, Wikipedia has an article devoted solely to Logitech products throughout history, because of course it does, and it lists (among other things) their dates of release. The MX500 was released in 2002, and superseded by the MX510 in 2004. I then remembered a photo I took of my eldest child when she was an infant, trying to chew on a computer mouse. I dug it out of my iPhoto library and yep, it’s my MX500. The picture is dated June 2004.
So I have photographic evidence that I’ve used this specific mouse for 15 years or more. The logo plate on top of the mouse has been worn half-smooth and half-paintless by the palm of my hand, much like the shiny-smooth areas worn into the subtle matte surface texture where the thumb and pinky finger grip the sides. The model and technical information printed on the underside has similarly worn away. It started out with four little oval glide nubs on the underside that held the bottom away from the desk surface; only one remains. Even though, as an optical mouse, it can be used on any surface, I eventually went back to soft mousepads, so as to limit further undercarriage damage.
Why have I been so devoted to this mouse? Well, it’s incredibly well engineered, for one—it’s put up with 15 years of daily use. It’s exactly the right shape for my hand, and it has multiple configurable inputs right where I expect them. There are arrow buttons just above my thumb which I use as forward/backward in browsers, buttons above and below the scroll wheel that I map to Page Up/Page Down, an extra button at almost the apex of the mouse’s back mapped to ⌥⇥ (Option-Tab), and the usual right/left mouse click buttons. Plus the scroll wheel is itself a push-down-to-click button.
Most of these features can be found on one mouse or another, but it’s rare to find them all in one mouse—and next to impossible to find them in a shape and size that feels comfortable to me. I’d occasionally looked at the secondary market, but even used, the MX500 can command three figures. I checked Amazon as I wrote this, and an unused MX500 was listing for two hundred fifty dollars. Unused copies of its successor, the MX510, were selling for even more.
Now, if you were into gaming in the first decade of the 2000s, you may have heard of or used the MX510’s successor, the MX518. Released in 2005, it was basically an MX500/MX510, but branded for gaming, with some optical-sensor upgrades for more tracking precision. The MX518 lasted until 2011, when it was superseded by a different model, which itself was superseded, which et cetera, et cereta, et cetera.
Which brings me to the point of all this. A few weeks ago, after several weeks of sporadic glitches, the scroll wheel on my MX500 almost completely stopped responding to being scrolled. Which maybe doesn’t sound like a big deal, but try going without your scroll wheel for a while. I was surprised to discover how much I relied on it. So, glumly, knowing the model was long out of production and incredibly expensive to buy, I went searching for equivalents.
And that’s when I discovered that Logitech had literally announced less than a week earlier that they were releasing an updated MX518, available for pre-order.
Friends, I have never pre-ordered anything so fast.
This past Thursday afternoon, it arrived. I got it set up and have been working with it since. And I have some impressions.
Physically, the MX518 Legendary (as Logitech has branded it) is 95% a match for my old MX500. It’s ever so slightly smaller, just enough that I can tell but not quite enough to be annoying, odd as that may seem. Otherwise, everything feels like it should. The buttons are crisp and clicky, and right where I expect them. And the scroll wheel… well, it works.
The coloration is different—the surface and buttons are all black, as opposed to the MX500’s black-and-silver two-tone styling. While I miss the two-tone a bit, there’s an upgrade: the smooth black top surface has subtle little sparkles embedded in the paint. Shiny!
On the other hand, configuring the mouse was a bit of an odyssey. First off, let me make clear that I have a weird setup, even for a grumpy old Mac user. I plug a circa-2000 Macally original iKey 104-key keyboard into my 2013 MacBook Pro. (Yes, you have sensed a trend here: when I find hardware I really like, I hang onto it like a rabid weasel. Ditto software.) The “extra” keys on the Macally like Page Up, Home, and so on don’t get recognized by a lot of current software. Even the Finder can’t read the keyboard’s function keys properly. I’ve restored their functionality with the entirely excellent BetterTouchTool, but it remains that the keyboard is just odd in its ancientness.
Anyway, I first opened System Preferences and then the Logitech Control Center pane. It couldn’t find the MX518 Legendary at all. So next I opened the (separate) Logitech Options pane, which drives the wireless mouse I use when I travel. It too was unable to find the MX518.
Some Bing-ing led me to a download for Logitech Gaming Software (hereafter LGS), which I installed. That could see the MX518 just fine. Once I stumbled my way into an understanding of LGS’s UI, I set about trying to configure the MX518’s buttons to do what I wanted.
And could not. In the list of predefined mouse actions that could be assigned to the buttons, precisely none of my desires were listed. No ⌘-arrow combos, no page up or down, not even ⌥⇥ to switch apps. I mean, I guess that’s to be expected: it’s sold as a gaming mouse. LGS has plenty of support for on-the-fly-dee-pee-eye switching and copy-paste and all that. Not so much for document editing and code browsing.
There is a way to assign keyboard combos to buttons, but again, the software could understand precisely none of the combos I wanted to record when I typed them on my Macally. So I went to the MacBook Pro’s built-in keyboard, where I was able to register ⌥⇥, ⌘→, and ⌘←. I could not, however much I tried, register Page Up or Page Down. I pressed Fn, which showed “Fn” in the LGS software, and then pressed the down arrow for Page Down, and as long as I held down both keys, it showed “Page Down”. But as soon as I let go of the down arrow, “Fn” was registered again. No Page Down for me.
Now, recall, this was happening on the laptop’s built-in keyboard. I can’t really blame this one on age of the external Macally. I really think this one might fall on LGS itself; while a 2013 MacBook is old, it’s not that old.
I thought I might be stuck, but I intuited a workaround: I opened the Keyboard Viewer app built into the Finder. With that, I could just click the virtual Page Up and Page Down keys, and LGS registered them without a hiccup. While I was in there, I used it to set the scroll wheel’s middle-button click to trigger Mission Control (F3).
The following key-repeat problem has been fixed and was not the fault of the MX518; see my comment for details on how I resolved it. The one letdown I have is that the buttons don’t appear to repeat keystrokes. So if I hold the button I’ve assigned to Page Down for example, I get exactly one page-down, and that’s it until I release and click the button again. On the MX500, holding down the button assigned to Page Down would just constantly page down until I let go. This was sometimes preferable to scrolling with the scroll wheel, especially for long documents I wanted to very quickly scan for a certain figure or other piece of the page. The same was true for all the buttons: hold it down, and the thing it was configured to do happened repeatedly until you let go.
The MX518 Legendary isn’t doing that. I don’t know if this is an inherent limitation of the mouse, its software, my configuration of it, the interaction of software and operating system, or something else entirely. It’s not an issue forty-nine times out of fifty, but that fiftieth time is annoying.
The other annoyance is one of possibly missed potential. The mouse software has, in keeping with its gaming focus, the ability to set up multiple profiles; that way, you can assign unique actions to the buttons on a per-application basis. I set up a couple of profiles to test it out, but LGS is completely opaque about how to make profiles switch automatically when you switch to an app. I’ll look for an answer online, but it’s annoying that the software promises per-app profiles, and then apparently fails to deliver on that promise.
So after all that, am I happy? Yes. It’s essentially my old mouse, except brand new. My heartfelt thanks to Logitech for bringing this workhorse out of retirement. I look forward to a decade or more with it.
## Breaking Silence
Published 2 months, 1 week ago
So it’s been (checks watch) half a year since I last blogged, yeah, okay, been a while. I took a break, not that you would’ve been able to tell from the sporadic nature of updates before I did so, but a break I took nonetheless. Well, break’s over.
One of things I plan to do is fill in a post I missed writing at the beginning of December: the 25th anniversary of my working with the web. I’ll tell the story in that post, but suffice to say it involves a laptop, a printout of the HTML specification, Microsoft Word 5.1a, a snagged Usenet post, and Mystery Science Theater 3000. Keep circulating the tags!
Before that happens, I’ll be posting a review of the return of a very old, very faithful assistant. I also have an article coming on a site where I’ve never been published before, so that’s exciting—look for an announcement here as soon as it’s public. Stay tuned!
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## Comma inside Round function in measure
Hi all
I have this measure which allows me to put in a card the total amount without all numbers:
```Measure = IF(CALCULATE(SUM(Table[Amount]),ALLSELECTED(Table[Concept]))>999999,
CONCATENATE("\$",CONCATENATE(ROUND(CALCULATE(SUM(Table[Amount])/1000000,ALLSELECTED(Table[Field])),0)," millions")),
IF(ISBLANK(CALCULATE(SUM(Table[Amount]),ALLSELECTED(Table[Field]))),"Zero",
CONCATENATE("\$",CONCATENATE(ROUND(CALCULATE(SUM(Table[Amount])/1000,ALLSELECTED(Table[FIeld])),0)," k"))))```
It gaves me values like this: \$1084 millions
What I want is to put a comma like this when first condition is met: \$1,084 millions | \$10,000 millions
How can i achieve it inside the measure?
Regards,
Julián
1 ACCEPTED SOLUTION
Accepted Solutions
Community Support Team
## Re: Comma inside Round function in measure
As tested, here is a workaround.
1.Create measures as below to replace the formula you use in your [measure].
ROUND(CALCULATE(SUM(Table[Amount])/1000000,ALLSELECTED(Table[Field])),0)->Measure2
ROUND(CALCULATE(SUM(Table[Amount])/1000,ALLSELECTED(Table[FIeld])),0)->Measure4
```Measure2 = ROUND(CALCULATE(SUM('Table'[Amount])/1000000,ALLSELECTED('Table'[Field])),0)
Measure4 = ROUND( CALCULATE(SUM('Table'[Amount])/1000,ALLSELECTED('Table'[Field])),0)```
2. displays the number with your currency locale formatting by measures below
```Measure3 = FORMAT([Measure2], "Currency")
Measure5 = FORMAT([Measure4],"Currency")```
3 concatenate the currency and "millions" or"k".
Measure
```final output = IF(CALCULATE(SUM('Table'[Amount]),ALLSELECTED('Table'[Concept]))>999999,
CONCATENATE([Measure3]," millions"),
IF(ISBLANK(CALCULATE(SUM('Table'[Amount]),ALLSELECTED('Table'[Field]))),"Zero",
CONCATENATE([Measure5]," k")))```
Or you could nest measures in one measure which works as lists of measures above .
```Measure =
VAR Measure2 =
ROUND (
CALCULATE ( SUM ( 'Table'[Amount] ) / 1000000, ALLSELECTED ( 'Table'[Field] ) ),
0
)
VAR Measure4 =
ROUND (
CALCULATE ( SUM ( 'Table'[Amount] ) / 1000, ALLSELECTED ( 'Table'[Field] ) ),
0
)
VAR Measure3 =
FORMAT ( Measure2, "Currency" )
VAR Measure5 =
FORMAT ( Measure4, "Currency" )
RETURN
IF (
CALCULATE ( SUM ( 'Table'[Amount] ), ALLSELECTED ( 'Table'[Concept] ) ) > 999999,
CONCATENATE ( Measure3, " millions" ),
IF (
ISBLANK (
CALCULATE ( SUM ( 'Table'[Amount] ), ALLSELECTED ( 'Table'[Field] ) )
),
"Zero",
CONCATENATE ( Measure5, " k" )
)
)```
Best Regards
Maggie
3 REPLIES 3
Community Support Team
## Re: Comma inside Round function in measure
As tested, we could get the comma in nuber type value by "Modeling"->"Format"->comma, but it can be put into measure in text type.
Best Regards
Maggie
Community Support Team
## Re: Comma inside Round function in measure
As tested, here is a workaround.
1.Create measures as below to replace the formula you use in your [measure].
ROUND(CALCULATE(SUM(Table[Amount])/1000000,ALLSELECTED(Table[Field])),0)->Measure2
ROUND(CALCULATE(SUM(Table[Amount])/1000,ALLSELECTED(Table[FIeld])),0)->Measure4
```Measure2 = ROUND(CALCULATE(SUM('Table'[Amount])/1000000,ALLSELECTED('Table'[Field])),0)
Measure4 = ROUND( CALCULATE(SUM('Table'[Amount])/1000,ALLSELECTED('Table'[Field])),0)```
2. displays the number with your currency locale formatting by measures below
```Measure3 = FORMAT([Measure2], "Currency")
Measure5 = FORMAT([Measure4],"Currency")```
3 concatenate the currency and "millions" or"k".
Measure
```final output = IF(CALCULATE(SUM('Table'[Amount]),ALLSELECTED('Table'[Concept]))>999999,
CONCATENATE([Measure3]," millions"),
IF(ISBLANK(CALCULATE(SUM('Table'[Amount]),ALLSELECTED('Table'[Field]))),"Zero",
CONCATENATE([Measure5]," k")))```
Or you could nest measures in one measure which works as lists of measures above .
```Measure =
VAR Measure2 =
ROUND (
CALCULATE ( SUM ( 'Table'[Amount] ) / 1000000, ALLSELECTED ( 'Table'[Field] ) ),
0
)
VAR Measure4 =
ROUND (
CALCULATE ( SUM ( 'Table'[Amount] ) / 1000, ALLSELECTED ( 'Table'[Field] ) ),
0
)
VAR Measure3 =
FORMAT ( Measure2, "Currency" )
VAR Measure5 =
FORMAT ( Measure4, "Currency" )
RETURN
IF (
CALCULATE ( SUM ( 'Table'[Amount] ), ALLSELECTED ( 'Table'[Concept] ) ) > 999999,
CONCATENATE ( Measure3, " millions" ),
IF (
ISBLANK (
CALCULATE ( SUM ( 'Table'[Amount] ), ALLSELECTED ( 'Table'[Field] ) )
),
"Zero",
CONCATENATE ( Measure5, " k" )
)
)```
Best Regards
Maggie
Highlighted
Member
## Re: Comma inside Round function in measure
Regards,
Julián
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https://www.numbersaplenty.com/14653
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Search a number
14653 is a prime number
BaseRepresentation
bin11100100111101
3202002201
43210331
5432103
6151501
760502
oct34475
922081
1014653
1110011
128591
136892
1454a9
15451d
hex393d
14653 has 2 divisors, whose sum is σ = 14654. Its totient is φ = 14652.
The previous prime is 14639. The next prime is 14657. The reversal of 14653 is 35641.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 13924 + 729 = 118^2 + 27^2 .
It is a cyclic number.
It is not a de Polignac number, because 14653 - 25 = 14621 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 14653.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (14657) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7326 + 7327.
It is an arithmetic number, because the mean of its divisors is an integer number (7327).
214653 is an apocalyptic number.
It is an amenable number.
14653 is a deficient number, since it is larger than the sum of its proper divisors (1).
14653 is an equidigital number, since it uses as much as digits as its factorization.
14653 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 360, while the sum is 19.
The square root of 14653 is about 121.0495766205. The cubic root of 14653 is about 24.4704628077.
The spelling of 14653 in words is "fourteen thousand, six hundred fifty-three".
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https://oeis.org/A264893
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A264893 First differences of A155043. 3
1, 0, 1, 0, 1, -1, 2, -1, 0, 0, 1, -1, 2, -1, 1, 0, 1, -2, 3, -2, 2, -2, 3, -2, 0, 0, 3, -3, 4, -4, 5, -4, 4, -4, 5, -2, 3, -5, 5, -5, 6, -6, 7, -6, 5, -5, 6, -6, 1, 0, 6, -6, 7, -7, 7, -7, 8, -7, 8, -9, 10, -8, 7, 0, 2, -9, 10, -9, 9, -9, 10, -12, 13, -10 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,7 COMMENTS a(n) = A155043(n+1) - A155043(n); a(A264898(n)) = 0; it is not true that in general a(even) >= 0 and a(odd) <= 0: even indices with negative values: 224,226,1088,1090,1368,1520,1860, ... odd indices with positive values: 225,441,445,675,1023,1035,1089,1093, ... . LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 PROG (Haskell) a264893 n = a264893_list !! n a264893_list = zipWith (-) (tail a155043_list) a155043_list CROSSREFS Cf. A155043, A264898. Sequence in context: A269518 A219840 A343220 * A340653 A334744 A136567 Adjacent sequences: A264890 A264891 A264892 * A264894 A264895 A264896 KEYWORD sign,look AUTHOR Reinhard Zumkeller, Nov 27 2015 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 16 14:07 EDT 2021. Contains 345057 sequences. (Running on oeis4.)
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https://courseinfo.canterbury.ac.nz/GetCourseDetails.aspx?course=MATH220&occurrence=23S1(C)&year=2023
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MATH220-23S1 (C) Semester One 2023
# Discrete Mathematics and Cryptography
15 points
Details:
Start Date: Monday, 20 February 2023 End Date: Sunday, 25 June 2023
Withdrawal Dates
Last Day to withdraw from this course:
• Without financial penalty (full fee refund): Sunday, 5 March 2023
• Without academic penalty (including no fee refund): Sunday, 14 May 2023
## Description
Discrete mathematics underpins many areas of modern-day science. This course is an introduction to graph theory and cryptography, two central topics in discrete mathematics.
Discrete mathematics underpins many areas of modern-day science. This course is an introduction to graph theory and cryptography, two central topics in discrete mathematics, each having fundamental links to many branches of science. Graph theory underlies the solution to many problems in a variety of disciplines including operations research and computational biology. Cryptography has applications to all communications security, from state security to online banking and mobile phone conversations. This course is designed for mathematics and computer science students.
Topics covered:
Cryptography: Basic ideas and terminology of cryptography. Shift and affine ciphers. One-time pads. Basic properties of the integers. Euclid’s algorithm. Modular arithmetic. Public key ciphers. The RSA, Rabin and ElGamal ciphers. Diffie-Hellman key exchange. Arithmetic of polynomials over finite fields. Constructing finite fields. Linear and non-linear shift registers.
Graph theory: Concepts and terminology of graphs. Eulerian and Hamiltonian graphs. Complexity, polynomial-time and exponential-time algorithms. Chromatic polynomials. Matchings and Hall’s Marriage Theorem. The Greedy Algorithm. Directed graphs. Network flows.
## Learning Outcomes
• At the end of the course, students will:
• Be familiar with some of the old and modern cryptographic schemes and have developed the necessary mathematics to understand and analyse them.
• Have an understanding of some topics in graph theory with an emphasis on graph algorithms and proof techniques.
This course will provide students with an opportunity to develop the Graduate Attributes specified below:
Critically competent in a core academic discipline of their award Students know and can critically evaluate and, where applicable, apply this knowledge to topics/issues within their majoring subject. Employable, innovative and enterprising Students will develop key skills and attributes sought by employers that can be used in a range of applications. Globally aware Students will comprehend the influence of global conditions on their discipline and will be competent in engaging with global and multi-cultural contexts.
MATH221, MATH231
## Assessment
Assessment Due Date Percentage
Assignments 25%
Test 25%
Final Examination 50%
To obtain a passing grade in this course you must pass the course as a whole (which requires an overall mark of 50% or more) and score at least 40% in the final exam.
## Textbooks / Resources
Buchmann, Johannes; Introduction to cryptography ; 2nd ed; Springer, 2004.
Clark, John. , Holton, Derek Allan; A first look at graph theory ; World Scientific, 1991.
Copies of these books will be on reserve in the Engineering and Physical Sciences Library. Also, there are a number of other good books on cryptography and graph theory in the library.
## Indicative Fees
Domestic fee \$824.00
International fee \$4,750.00
* All fees are inclusive of NZ GST or any equivalent overseas tax, and do not include any programme level discount or additional course-related expenses.
For further information see Mathematics and Statistics .
## All MATH220 Occurrences
• MATH220-23S1 (C) Semester One 2023
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OEIS "Top N" lists?
Leroy Quet qq-quet at mindspring.com
Fri Dec 5 03:32:22 CET 2003
```Or perhaps sequences could be judged upon the integers themselves.
It might be interesting to rank sequences by, say, the sequences with the
fewest number of terms in common with other sequences (compared among
terms appearing in the database, of course).
..Or by the most uncommon absolute-differences among terms, the sequences
whose terms average the highest, the 'most chotic' sequences (ie. least
able to fit to a low-order polynomial, or some other definition of 'most
chaotic'), the sequences which contain the most terms of other sequences
among the terms listed (but in some non-respective order as to avoid
trivial cases), the sequences which contain the longest runs of
consecutive integers but which miss the occasional integer (ranked based
upon some formula involving the lengths of the integer-runs and the
number of ommitted integers among the consecutive integers), etc etc...
Perhaps rankings with more serious motives can be created, such as those
which inspire research into connections between sequences, connections
that would not have been obvious at first to the casual OEIS user without
the computer-aided analysis.
thanks,
Leroy Quet
>In looking at the comments on A002620 just now I'm reminded of why the OEIS
>is such an exciting repository of cultural knowledge. There are about a
>dozen very different interesting views on the sequence, contributed by as
>many authors. It's really something to be proud of.
>
>This got me to wondering if someone with the appropriate software might
>want to find the "Top N" (for N=5 or 10?) sequences in various categories:
> Most references
> Most cited reference (ie bibliography entry)
> Most contributors (by counting the (AT)s)
> Longest formulas (or most formulas? tricky to count!)
> Largest explicitly given integers (ie as 99999999 versus 10^10-1)
> etc
>(with possible sub-categories based on keywords, such as tabl etc)
>
>Perhaps along the way it could collect stats of various kinds (eg averages
>etc).
>
>It would be fun to rerun this census at different times, (eg 1/1/2004,
>A100000, etc), to see how these things might change.
```
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https://us.metamath.org/ilegif/taupi.html
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Mathbox for Jim Kingdon < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > Mathboxes > taupi Unicode version
Theorem taupi 13295
Description: Relationship between and . This can be seen as connecting the ratio of a circle's circumference to its radius and the ratio of a circle's circumference to its diameter. (Contributed by Jim Kingdon, 19-Feb-2019.) (Revised by AV, 1-Oct-2020.)
Assertion
Ref Expression
taupi
Proof of Theorem taupi
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-tau 11488 . 2 inf
2 lttri3 7851 . . . . 5
32adantl 275 . . . 4
4 2re 8797 . . . . . 6
5 pire 12883 . . . . . 6
64, 5remulcli 7787 . . . . 5
76a1i 9 . . . 4
8 2rp 9453 . . . . . . 7
9 pirp 12886 . . . . . . 7
10 rpmulcl 9473 . . . . . . 7
118, 9, 10mp2an 422 . . . . . 6
126recni 7785 . . . . . . 7
13 cos2pi 12901 . . . . . . 7
14 cosf 11418 . . . . . . . . 9
15 ffn 5272 . . . . . . . . 9
1614, 15ax-mp 5 . . . . . . . 8
17 fniniseg 5540 . . . . . . . 8
1816, 17ax-mp 5 . . . . . . 7
1912, 13, 18mpbir2an 926 . . . . . 6
2011, 19elini 3260 . . . . 5
2120a1i 9 . . . 4
22 elinel2 3263 . . . . . . . . . 10
23 fniniseg 5540 . . . . . . . . . . 11
2416, 23ax-mp 5 . . . . . . . . . 10
2522, 24sylib 121 . . . . . . . . 9
2625simprd 113 . . . . . . . 8
2726adantr 274 . . . . . . 7
28 elinel1 3262 . . . . . . . . . . 11
2928rpred 9490 . . . . . . . . . 10
3029adantr 274 . . . . . . . . 9
3128rpgt0d 9493 . . . . . . . . . 10
3231adantr 274 . . . . . . . . 9
33 simpr 109 . . . . . . . . 9
34 0xr 7819 . . . . . . . . . 10
356rexri 7830 . . . . . . . . . 10
36 elioo2 9711 . . . . . . . . . 10
3734, 35, 36mp2an 422 . . . . . . . . 9
3830, 32, 33, 37syl3anbrc 1165 . . . . . . . 8
39 cos02pilt1 12948 . . . . . . . 8
4038, 39syl 14 . . . . . . 7
4127, 40eqbrtrrd 3952 . . . . . 6
42 1red 7788 . . . . . . 7
4342ltnrd 7882 . . . . . 6
4441, 43pm2.65da 650 . . . . 5
4544adantl 275 . . . 4
463, 7, 21, 45infminti 6914 . . 3 inf
4746mptru 1340 . 2 inf
481, 47eqtri 2160 1
Colors of variables: wff set class Syntax hints: wn 3 wa 103 wb 104 w3a 962 wceq 1331 wtru 1332 wcel 1480 cin 3070 csn 3527 class class class wbr 3929 ccnv 4538 cima 4542 wfn 5118 wf 5119 cfv 5123 (class class class)co 5774 infcinf 6870 cc 7625 cr 7626 cc0 7627 c1 7628 cmul 7632 cxr 7806 clt 7807 c2 8778 crp 9448 cioo 9678 ccos 11358 cpi 11360 ctau 11487 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 603 ax-in2 604 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-13 1491 ax-14 1492 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2121 ax-coll 4043 ax-sep 4046 ax-nul 4054 ax-pow 4098 ax-pr 4131 ax-un 4355 ax-setind 4452 ax-iinf 4502 ax-cnex 7718 ax-resscn 7719 ax-1cn 7720 ax-1re 7721 ax-icn 7722 ax-addcl 7723 ax-addrcl 7724 ax-mulcl 7725 ax-mulrcl 7726 ax-addcom 7727 ax-mulcom 7728 ax-addass 7729 ax-mulass 7730 ax-distr 7731 ax-i2m1 7732 ax-0lt1 7733 ax-1rid 7734 ax-0id 7735 ax-rnegex 7736 ax-precex 7737 ax-cnre 7738 ax-pre-ltirr 7739 ax-pre-ltwlin 7740 ax-pre-lttrn 7741 ax-pre-apti 7742 ax-pre-ltadd 7743 ax-pre-mulgt0 7744 ax-pre-mulext 7745 ax-arch 7746 ax-caucvg 7747 ax-pre-suploc 7748 ax-addf 7749 ax-mulf 7750 This theorem depends on definitions: df-bi 116 df-stab 816 df-dc 820 df-3or 963 df-3an 964 df-tru 1334 df-fal 1337 df-nf 1437 df-sb 1736 df-eu 2002 df-mo 2003 df-clab 2126 df-cleq 2132 df-clel 2135 df-nfc 2270 df-ne 2309 df-nel 2404 df-ral 2421 df-rex 2422 df-reu 2423 df-rmo 2424 df-rab 2425 df-v 2688 df-sbc 2910 df-csb 3004 df-dif 3073 df-un 3075 df-in 3077 df-ss 3084 df-nul 3364 df-if 3475 df-pw 3512 df-sn 3533 df-pr 3534 df-op 3536 df-uni 3737 df-int 3772 df-iun 3815 df-disj 3907 df-br 3930 df-opab 3990 df-mpt 3991 df-tr 4027 df-id 4215 df-po 4218 df-iso 4219 df-iord 4288 df-on 4290 df-ilim 4291 df-suc 4293 df-iom 4505 df-xp 4545 df-rel 4546 df-cnv 4547 df-co 4548 df-dm 4549 df-rn 4550 df-res 4551 df-ima 4552 df-iota 5088 df-fun 5125 df-fn 5126 df-f 5127 df-f1 5128 df-fo 5129 df-f1o 5130 df-fv 5131 df-isom 5132 df-riota 5730 df-ov 5777 df-oprab 5778 df-mpo 5779 df-of 5982 df-1st 6038 df-2nd 6039 df-recs 6202 df-irdg 6267 df-frec 6288 df-1o 6313 df-oadd 6317 df-er 6429 df-map 6544 df-pm 6545 df-en 6635 df-dom 6636 df-fin 6637 df-sup 6871 df-inf 6872 df-pnf 7809 df-mnf 7810 df-xr 7811 df-ltxr 7812 df-le 7813 df-sub 7942 df-neg 7943 df-reap 8344 df-ap 8351 df-div 8440 df-inn 8728 df-2 8786 df-3 8787 df-4 8788 df-5 8789 df-6 8790 df-7 8791 df-8 8792 df-9 8793 df-n0 8985 df-z 9062 df-uz 9334 df-q 9419 df-rp 9449 df-xneg 9566 df-xadd 9567 df-ioo 9682 df-ioc 9683 df-ico 9684 df-icc 9685 df-fz 9798 df-fzo 9927 df-seqfrec 10226 df-exp 10300 df-fac 10479 df-bc 10501 df-ihash 10529 df-shft 10594 df-cj 10621 df-re 10622 df-im 10623 df-rsqrt 10777 df-abs 10778 df-clim 11055 df-sumdc 11130 df-ef 11361 df-sin 11363 df-cos 11364 df-pi 11366 df-tau 11488 df-rest 12131 df-topgen 12150 df-psmet 12165 df-xmet 12166 df-met 12167 df-bl 12168 df-mopn 12169 df-top 12174 df-topon 12187 df-bases 12219 df-ntr 12274 df-cn 12366 df-cnp 12367 df-tx 12431 df-cncf 12736 df-limced 12803 df-dvap 12804 This theorem is referenced by: (None)
Copyright terms: Public domain W3C validator
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| 3
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https://stats.stackexchange.com/questions/185911/why-are-bias-nodes-used-in-neural-networks/186284#186284
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# Why are bias nodes used in neural networks?
1. Why are bias nodes used in neural networks?
2. How many you should use?
3. In which layers you should use them: all hidden layers and the output layer?
• Dec 13, 2015 at 18:13
The bias node in a neural network is a node that is always 'on'. That is, its value is set to $1$ without regard for the data in a given pattern. It is analogous to the intercept in a regression model, and serves the same function. If a neural network does not have a bias node in a given layer, it will not be able to produce output in the next layer that differs from $0$ (on the linear scale, or the value that corresponds to the transformation of $0$ when passed through the activation function) when the feature values are $0$.
Consider a simple example: You have a feed forward perceptron with 2 input nodes $x_1$ and $x_2$, and 1 output node $y$. $x_1$ and $x_2$ are binary features and set at their reference level, $x_1=x_2=0$. Multiply those 2 $0$'s by whatever weights you like, $w_1$ and $w_2$, sum the products and pass it through whatever activation function you prefer. Without a bias node, only one output value is possible, which may yield a very poor fit. For instance, using a logistic activation function, $y$ must be $.5$, which would be awful for classifying rare events.
A bias node provides considerable flexibility to a neural network model. In the example given above, the only predicted proportion possible without a bias node was $50\%$, but with a bias node, any proportion in $(0, 1)$ can be fit for the patterns where $x_1=x_2=0$. For each layer, $j$, in which a bias node is added, the bias node will add $N_{j+1}$ additional parameters / weights to be estimated (where $N_{j+1}$ is the number of nodes in layer $j+1$). More parameters to be fitted means it will take proportionately longer for the neural network to be trained. It also increases the chance of overfitting, if you don't have considerably more data than weights to be learned.
1. Bias nodes are added to increase the flexibility of the model to fit the data. Specifically, it allows the network to fit the data when all input features are equal to $0$, and very likely decreases the bias of the fitted values elsewhere in the data space.
2. Typically, a single bias node is added for the input layer and every hidden layer in a feedforward network. You would never add two or more to a given layer, but you might add zero. The total number is thus determined largely by the structure of your network, although other considerations could apply. (I am less clear on how bias nodes are added to neural network structures other than feedforward.)
3. Mostly this has been covered, but to be explicit: you would never add a bias node to the output layer; that wouldn't make any sense.
• Is CNN different in this regard? since when I add bias to my conv layers, the performance (accuracy)degrades! and when I remove them, it actually goes higher! Mar 7, 2016 at 15:21
• @Hossein, not that I know of, but you could ask a new question. I'm not much of an expert there. Mar 7, 2016 at 16:46
• Would I still need bias nodes if my inputs never go to 0? Oct 29, 2018 at 11:39
• @alec_djinn, yes. Almost certainly the model would be biased without them, even if you never have 0 for an input value. By analogy, it may help to read: When is it ok to remove the intercept in a linear regression model? Oct 29, 2018 at 16:57
• @krupeshAnadkat, "The bias node in a neural network is a node that is always 'on'. That is, its value is set to 1 without regard for the data in a given pattern." So you can connect if if you like, just always change the resulting value of the node back to $1$ before you multiply it by the weight, since a bias node is a node whose value is always 1. Feb 3, 2019 at 15:09
In the context of neural networks, Batch Normalization is currently the gold-standard for making smart "bias nodes." Instead of clamping a neuron's bias value, you instead adjust for the covariance of the neuron's input. So in a CNN, you would apply a batch normalization just between the convolutional layer and the next fully connected layer (of say, ReLus). In theory, all fully connected layers could benefit from Batch Normalization but this in practice becomes very expensive to implement since each batch normalization carries its own parameters.
Concerning why, most of the answers already have explained that, in particular, neurons are susceptible to saturated gradients when the input pushes the activation to an extreme. In the case of ReLu's this would be pushed to the left, giving a gradient of 0. In general, when you train a model, you first normalize the inputs to the neural network. Batch Normalization is a way of normalizing the inputs inside the neural network, between layers.
• The bias is just an added number to the next layers activation. One way to visualize it is by having a constant 1 value in the previous layer and one weight (one bias value) for each of the next layers neurons. Feb 3, 2019 at 9:06
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### Category Seymoure3734
manufacturing overhead (MOH) using an actual plant-wide overhead allocation rate using direct labor hours (DLHs). MMI had expected to produce and sell Plantwide rates are the easiest to apply but can cause cost distortion because all overhead resources are treated as though they are equally consumed by all cost 24 Jan 2020 a firm-wide overhead rate or multiple departmental overhead rates. compared to the plant-wide approach because cost allocations are based
Departmental overhead rates are used by many manufacturers instead of using a single, plant-wide overhead rate. The reason for departmental overhead rates 7 Oct 2019 The plantwide overhead rate is a single overhead rate that a company uses to allocate all of its manufacturing overhead costs to products or Using a plant-wide or single overhead rate when a business manufactures or produces a single product or provides a single service is feasible and generally To calculate the plantwide overhead rate, first divide total overhead by the number of direct labor hours used to find the overhead per labor hour. Next, multiply the
## To calculate the plantwide overhead rate, first divide total overhead by the number of direct labor hours used to find the overhead per labor hour. Next, multiply the
A pre-determined overhead rate is normally the term when using a single, plant- wide base to calculate and apply overhead. Overhead is then applied by 1 Explain why a single plantwide overhead rate can distort the cost of a Using single plant wide overhead rate assumes that all the factory overhead cost is plant-wide overhead rateの意味や使い方 工場一括製造間接費配賦率; 総括配賦率 - 約1153万語ある英和辞典・和英辞典。発音・イディオムも分かる英語辞書。 20 Jan 2015 Factory Overhead Subtopic: Departmental Rate and Plantwide Rate Illustration: Job Costing using Plant wide rate and Departmental Rate. (b) The production overhead absorption rates of factories X and Y are calculated lize a single factory-wide recovery rate for all production overheads or a sepa Landen Company uses a single plantwide overhead rate based on direct labor hours. Total budgeted overhead costs are \$200,000. Total budgeted direct manufacturing overhead (MOH) using an actual plant-wide overhead allocation rate using direct labor hours (DLHs). MMI had expected to produce and sell
### A pre-determined overhead rate is normally the term when using a single, plant- wide base to calculate and apply overhead. Overhead is then applied by
Using a plant-wide or single overhead rate when a business manufactures or produces a single product or provides a single service is feasible and generally To calculate the plantwide overhead rate, first divide total overhead by the number of direct labor hours used to find the overhead per labor hour. Next, multiply the Sometimes called the "predetermined overhead rate," your plant-wide figure helps you understand your company profitability. Indirect Costs. Your indirect costs Plant or factory wide (single or blanket) rate is used for the whole factory and is assigned to all cost units irrespective of the departments in which they were
### For example, a company with a simple manufacturing operation that produces similar products could have a plant-wide overhead rate of \$40 per machine hour if it has budgeted \$800,000 of total manufacturing overhead costs and it expects to produce 20,000 machine hours of good output.
Overhead allocation rate = Total overhead / Total direct labor hours = \$100,000 / 4,000 hours = \$25.00 Therefore, for every hour of direct labor needed to make books, Band Book applies \$25 worth of overhead to the product. Divided into the overhead of \$120,000, this comes to \$48 in overhead per labor hour. Product A requires 1.5 hours per unit, so the overhead rate is 1.5 times \$48, or \$72 per unit. For product B, two labor hours are needed per unit, so the overhead per unit equals two times \$48, or \$96. The plantwide overhead rate is a single overhead rate that a company uses to allocate all of its manufacturing overhead costs to products or cost objects . It is most commonly used in smaller entities with simple cost structures . Using a plantwide overhead rate is acceptable in the followin According to a survey 34% of the manufacturing businesses use a single plant wide overhead rate, 44% use multiple predetermined overhead rates and rest of the companies use activity based costing (ABC) system. The company wants to know how much overhead relates to direct labor costs. The company has direct labor expenses totaling \$5 million for the same period. To calculate the overhead rate: Divide \$20 million (indirect costs) by \$5 million (direct labor costs). Overhead rate = \$4 or (\$20/\$5), A pre-determined overhead rate is the rate used to apply manufacturing overhead to work-in-process inventory.The pre-determined overhead rate is calculated before the period begins. The first step is to estimate the amount of the activity base that will be required to support operations in the upcoming period. The second step is to estimate the total manufacturing cost at that level of activity.
## Question: Single Plantwide Factory Overhead Rate Bach Instruments Inc. Makes Three Musical Instruments: Flutes, Clarinets, And Oboes. The Budgeted Factory Overhead Cost Is \$152,800. Overhead Is Allocated To The Three Products On The Basis Of Direct Labor Hours. The Products Have The Following Budgeted Production Volume And Direct Labor Hours Per Unit: Budgeted
To calculate the plantwide overhead rate, first divide total overhead by the number of direct labor hours used to find the overhead per labor hour. Next, multiply the Sometimes called the "predetermined overhead rate," your plant-wide figure helps you understand your company profitability. Indirect Costs. Your indirect costs Plant or factory wide (single or blanket) rate is used for the whole factory and is assigned to all cost units irrespective of the departments in which they were A pre-determined overhead rate is normally the term when using a single, plant- wide base to calculate and apply overhead. Overhead is then applied by
Had the company used a plant-wide rate, the manufacturing overhead rate would have been \$33.33 per MH (\$500,000 divided by 15,000 MH), instead of \$40 for the machining department and \$20 for the finishing department.
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src/HOL/Library/Multiset.thy
author haftmann Fri Aug 27 19:34:23 2010 +0200 (2010-08-27 ago) changeset 38857 97775f3e8722 parent 38287 796302ca3611 child 39198 f967a16dfcdd permissions -rw-r--r--
renamed class/constant eq to equal; tuned some instantiations
``` 1 (* Title: HOL/Library/Multiset.thy
```
``` 2 Author: Tobias Nipkow, Markus Wenzel, Lawrence C Paulson, Norbert Voelker
```
``` 3 *)
```
``` 4
```
``` 5 header {* (Finite) multisets *}
```
``` 6
```
``` 7 theory Multiset
```
``` 8 imports Main
```
``` 9 begin
```
``` 10
```
``` 11 subsection {* The type of multisets *}
```
``` 12
```
``` 13 typedef 'a multiset = "{f :: 'a => nat. finite {x. f x > 0}}"
```
``` 14 morphisms count Abs_multiset
```
``` 15 proof
```
``` 16 show "(\<lambda>x. 0::nat) \<in> ?multiset" by simp
```
``` 17 qed
```
``` 18
```
``` 19 lemmas multiset_typedef = Abs_multiset_inverse count_inverse count
```
``` 20
```
``` 21 abbreviation Melem :: "'a => 'a multiset => bool" ("(_/ :# _)" [50, 51] 50) where
```
``` 22 "a :# M == 0 < count M a"
```
``` 23
```
``` 24 notation (xsymbols)
```
``` 25 Melem (infix "\<in>#" 50)
```
``` 26
```
``` 27 lemma multiset_ext_iff:
```
``` 28 "M = N \<longleftrightarrow> (\<forall>a. count M a = count N a)"
```
``` 29 by (simp only: count_inject [symmetric] expand_fun_eq)
```
``` 30
```
``` 31 lemma multiset_ext:
```
``` 32 "(\<And>x. count A x = count B x) \<Longrightarrow> A = B"
```
``` 33 using multiset_ext_iff by auto
```
``` 34
```
``` 35 text {*
```
``` 36 \medskip Preservation of the representing set @{term multiset}.
```
``` 37 *}
```
``` 38
```
``` 39 lemma const0_in_multiset:
```
``` 40 "(\<lambda>a. 0) \<in> multiset"
```
``` 41 by (simp add: multiset_def)
```
``` 42
```
``` 43 lemma only1_in_multiset:
```
``` 44 "(\<lambda>b. if b = a then n else 0) \<in> multiset"
```
``` 45 by (simp add: multiset_def)
```
``` 46
```
``` 47 lemma union_preserves_multiset:
```
``` 48 "M \<in> multiset \<Longrightarrow> N \<in> multiset \<Longrightarrow> (\<lambda>a. M a + N a) \<in> multiset"
```
``` 49 by (simp add: multiset_def)
```
``` 50
```
``` 51 lemma diff_preserves_multiset:
```
``` 52 assumes "M \<in> multiset"
```
``` 53 shows "(\<lambda>a. M a - N a) \<in> multiset"
```
``` 54 proof -
```
``` 55 have "{x. N x < M x} \<subseteq> {x. 0 < M x}"
```
``` 56 by auto
```
``` 57 with assms show ?thesis
```
``` 58 by (auto simp add: multiset_def intro: finite_subset)
```
``` 59 qed
```
``` 60
```
``` 61 lemma MCollect_preserves_multiset:
```
``` 62 assumes "M \<in> multiset"
```
``` 63 shows "(\<lambda>x. if P x then M x else 0) \<in> multiset"
```
``` 64 proof -
```
``` 65 have "{x. (P x \<longrightarrow> 0 < M x) \<and> P x} \<subseteq> {x. 0 < M x}"
```
``` 66 by auto
```
``` 67 with assms show ?thesis
```
``` 68 by (auto simp add: multiset_def intro: finite_subset)
```
``` 69 qed
```
``` 70
```
``` 71 lemmas in_multiset = const0_in_multiset only1_in_multiset
```
``` 72 union_preserves_multiset diff_preserves_multiset MCollect_preserves_multiset
```
``` 73
```
``` 74
```
``` 75 subsection {* Representing multisets *}
```
``` 76
```
``` 77 text {* Multiset comprehension *}
```
``` 78
```
``` 79 definition MCollect :: "'a multiset => ('a => bool) => 'a multiset" where
```
``` 80 "MCollect M P = Abs_multiset (\<lambda>x. if P x then count M x else 0)"
```
``` 81
```
``` 82 syntax
```
``` 83 "_MCollect" :: "pttrn => 'a multiset => bool => 'a multiset" ("(1{# _ :# _./ _#})")
```
``` 84 translations
```
``` 85 "{#x :# M. P#}" == "CONST MCollect M (\<lambda>x. P)"
```
``` 86
```
``` 87
```
``` 88 text {* Multiset enumeration *}
```
``` 89
```
``` 90 instantiation multiset :: (type) "{zero, plus}"
```
``` 91 begin
```
``` 92
```
``` 93 definition Mempty_def:
```
``` 94 "0 = Abs_multiset (\<lambda>a. 0)"
```
``` 95
```
``` 96 abbreviation Mempty :: "'a multiset" ("{#}") where
```
``` 97 "Mempty \<equiv> 0"
```
``` 98
```
``` 99 definition union_def:
```
``` 100 "M + N = Abs_multiset (\<lambda>a. count M a + count N a)"
```
``` 101
```
``` 102 instance ..
```
``` 103
```
``` 104 end
```
``` 105
```
``` 106 definition single :: "'a => 'a multiset" where
```
``` 107 "single a = Abs_multiset (\<lambda>b. if b = a then 1 else 0)"
```
``` 108
```
``` 109 syntax
```
``` 110 "_multiset" :: "args => 'a multiset" ("{#(_)#}")
```
``` 111 translations
```
``` 112 "{#x, xs#}" == "{#x#} + {#xs#}"
```
``` 113 "{#x#}" == "CONST single x"
```
``` 114
```
``` 115 lemma count_empty [simp]: "count {#} a = 0"
```
``` 116 by (simp add: Mempty_def in_multiset multiset_typedef)
```
``` 117
```
``` 118 lemma count_single [simp]: "count {#b#} a = (if b = a then 1 else 0)"
```
``` 119 by (simp add: single_def in_multiset multiset_typedef)
```
``` 120
```
``` 121
```
``` 122 subsection {* Basic operations *}
```
``` 123
```
``` 124 subsubsection {* Union *}
```
``` 125
```
``` 126 lemma count_union [simp]: "count (M + N) a = count M a + count N a"
```
``` 127 by (simp add: union_def in_multiset multiset_typedef)
```
``` 128
```
``` 129 instance multiset :: (type) cancel_comm_monoid_add proof
```
``` 130 qed (simp_all add: multiset_ext_iff)
```
``` 131
```
``` 132
```
``` 133 subsubsection {* Difference *}
```
``` 134
```
``` 135 instantiation multiset :: (type) minus
```
``` 136 begin
```
``` 137
```
``` 138 definition diff_def:
```
``` 139 "M - N = Abs_multiset (\<lambda>a. count M a - count N a)"
```
``` 140
```
``` 141 instance ..
```
``` 142
```
``` 143 end
```
``` 144
```
``` 145 lemma count_diff [simp]: "count (M - N) a = count M a - count N a"
```
``` 146 by (simp add: diff_def in_multiset multiset_typedef)
```
``` 147
```
``` 148 lemma diff_empty [simp]: "M - {#} = M \<and> {#} - M = {#}"
```
``` 149 by(simp add: multiset_ext_iff)
```
``` 150
```
``` 151 lemma diff_cancel[simp]: "A - A = {#}"
```
``` 152 by (rule multiset_ext) simp
```
``` 153
```
``` 154 lemma diff_union_cancelR [simp]: "M + N - N = (M::'a multiset)"
```
``` 155 by(simp add: multiset_ext_iff)
```
``` 156
```
``` 157 lemma diff_union_cancelL [simp]: "N + M - N = (M::'a multiset)"
```
``` 158 by(simp add: multiset_ext_iff)
```
``` 159
```
``` 160 lemma insert_DiffM:
```
``` 161 "x \<in># M \<Longrightarrow> {#x#} + (M - {#x#}) = M"
```
``` 162 by (clarsimp simp: multiset_ext_iff)
```
``` 163
```
``` 164 lemma insert_DiffM2 [simp]:
```
``` 165 "x \<in># M \<Longrightarrow> M - {#x#} + {#x#} = M"
```
``` 166 by (clarsimp simp: multiset_ext_iff)
```
``` 167
```
``` 168 lemma diff_right_commute:
```
``` 169 "(M::'a multiset) - N - Q = M - Q - N"
```
``` 170 by (auto simp add: multiset_ext_iff)
```
``` 171
```
``` 172 lemma diff_add:
```
``` 173 "(M::'a multiset) - (N + Q) = M - N - Q"
```
``` 174 by (simp add: multiset_ext_iff)
```
``` 175
```
``` 176 lemma diff_union_swap:
```
``` 177 "a \<noteq> b \<Longrightarrow> M - {#a#} + {#b#} = M + {#b#} - {#a#}"
```
``` 178 by (auto simp add: multiset_ext_iff)
```
``` 179
```
``` 180 lemma diff_union_single_conv:
```
``` 181 "a \<in># J \<Longrightarrow> I + J - {#a#} = I + (J - {#a#})"
```
``` 182 by (simp add: multiset_ext_iff)
```
``` 183
```
``` 184
```
``` 185 subsubsection {* Equality of multisets *}
```
``` 186
```
``` 187 lemma single_not_empty [simp]: "{#a#} \<noteq> {#} \<and> {#} \<noteq> {#a#}"
```
``` 188 by (simp add: multiset_ext_iff)
```
``` 189
```
``` 190 lemma single_eq_single [simp]: "{#a#} = {#b#} \<longleftrightarrow> a = b"
```
``` 191 by (auto simp add: multiset_ext_iff)
```
``` 192
```
``` 193 lemma union_eq_empty [iff]: "M + N = {#} \<longleftrightarrow> M = {#} \<and> N = {#}"
```
``` 194 by (auto simp add: multiset_ext_iff)
```
``` 195
```
``` 196 lemma empty_eq_union [iff]: "{#} = M + N \<longleftrightarrow> M = {#} \<and> N = {#}"
```
``` 197 by (auto simp add: multiset_ext_iff)
```
``` 198
```
``` 199 lemma multi_self_add_other_not_self [simp]: "M = M + {#x#} \<longleftrightarrow> False"
```
``` 200 by (auto simp add: multiset_ext_iff)
```
``` 201
```
``` 202 lemma diff_single_trivial:
```
``` 203 "\<not> x \<in># M \<Longrightarrow> M - {#x#} = M"
```
``` 204 by (auto simp add: multiset_ext_iff)
```
``` 205
```
``` 206 lemma diff_single_eq_union:
```
``` 207 "x \<in># M \<Longrightarrow> M - {#x#} = N \<longleftrightarrow> M = N + {#x#}"
```
``` 208 by auto
```
``` 209
```
``` 210 lemma union_single_eq_diff:
```
``` 211 "M + {#x#} = N \<Longrightarrow> M = N - {#x#}"
```
``` 212 by (auto dest: sym)
```
``` 213
```
``` 214 lemma union_single_eq_member:
```
``` 215 "M + {#x#} = N \<Longrightarrow> x \<in># N"
```
``` 216 by auto
```
``` 217
```
``` 218 lemma union_is_single:
```
``` 219 "M + N = {#a#} \<longleftrightarrow> M = {#a#} \<and> N={#} \<or> M = {#} \<and> N = {#a#}" (is "?lhs = ?rhs")proof
```
``` 220 assume ?rhs then show ?lhs by auto
```
``` 221 next
```
``` 222 assume ?lhs thus ?rhs
```
``` 223 by(simp add: multiset_ext_iff split:if_splits) (metis add_is_1)
```
``` 224 qed
```
``` 225
```
``` 226 lemma single_is_union:
```
``` 227 "{#a#} = M + N \<longleftrightarrow> {#a#} = M \<and> N = {#} \<or> M = {#} \<and> {#a#} = N"
```
``` 228 by (auto simp add: eq_commute [of "{#a#}" "M + N"] union_is_single)
```
``` 229
```
``` 230 lemma add_eq_conv_diff:
```
``` 231 "M + {#a#} = N + {#b#} \<longleftrightarrow> M = N \<and> a = b \<or> M = N - {#a#} + {#b#} \<and> N = M - {#b#} + {#a#}" (is "?lhs = ?rhs")
```
``` 232 (* shorter: by (simp add: multiset_ext_iff) fastsimp *)
```
``` 233 proof
```
``` 234 assume ?rhs then show ?lhs
```
``` 235 by (auto simp add: add_assoc add_commute [of "{#b#}"])
```
``` 236 (drule sym, simp add: add_assoc [symmetric])
```
``` 237 next
```
``` 238 assume ?lhs
```
``` 239 show ?rhs
```
``` 240 proof (cases "a = b")
```
``` 241 case True with `?lhs` show ?thesis by simp
```
``` 242 next
```
``` 243 case False
```
``` 244 from `?lhs` have "a \<in># N + {#b#}" by (rule union_single_eq_member)
```
``` 245 with False have "a \<in># N" by auto
```
``` 246 moreover from `?lhs` have "M = N + {#b#} - {#a#}" by (rule union_single_eq_diff)
```
``` 247 moreover note False
```
``` 248 ultimately show ?thesis by (auto simp add: diff_right_commute [of _ "{#a#}"] diff_union_swap)
```
``` 249 qed
```
``` 250 qed
```
``` 251
```
``` 252 lemma insert_noteq_member:
```
``` 253 assumes BC: "B + {#b#} = C + {#c#}"
```
``` 254 and bnotc: "b \<noteq> c"
```
``` 255 shows "c \<in># B"
```
``` 256 proof -
```
``` 257 have "c \<in># C + {#c#}" by simp
```
``` 258 have nc: "\<not> c \<in># {#b#}" using bnotc by simp
```
``` 259 then have "c \<in># B + {#b#}" using BC by simp
```
``` 260 then show "c \<in># B" using nc by simp
```
``` 261 qed
```
``` 262
```
``` 263 lemma add_eq_conv_ex:
```
``` 264 "(M + {#a#} = N + {#b#}) =
```
``` 265 (M = N \<and> a = b \<or> (\<exists>K. M = K + {#b#} \<and> N = K + {#a#}))"
```
``` 266 by (auto simp add: add_eq_conv_diff)
```
``` 267
```
``` 268
```
``` 269 subsubsection {* Pointwise ordering induced by count *}
```
``` 270
```
``` 271 instantiation multiset :: (type) ordered_ab_semigroup_add_imp_le
```
``` 272 begin
```
``` 273
```
``` 274 definition less_eq_multiset :: "'a multiset \<Rightarrow> 'a multiset \<Rightarrow> bool" where
```
``` 275 mset_le_def: "A \<le> B \<longleftrightarrow> (\<forall>a. count A a \<le> count B a)"
```
``` 276
```
``` 277 definition less_multiset :: "'a multiset \<Rightarrow> 'a multiset \<Rightarrow> bool" where
```
``` 278 mset_less_def: "(A::'a multiset) < B \<longleftrightarrow> A \<le> B \<and> A \<noteq> B"
```
``` 279
```
``` 280 instance proof
```
``` 281 qed (auto simp add: mset_le_def mset_less_def multiset_ext_iff intro: order_trans antisym)
```
``` 282
```
``` 283 end
```
``` 284
```
``` 285 lemma mset_less_eqI:
```
``` 286 "(\<And>x. count A x \<le> count B x) \<Longrightarrow> A \<le> B"
```
``` 287 by (simp add: mset_le_def)
```
``` 288
```
``` 289 lemma mset_le_exists_conv:
```
``` 290 "(A::'a multiset) \<le> B \<longleftrightarrow> (\<exists>C. B = A + C)"
```
``` 291 apply (unfold mset_le_def, rule iffI, rule_tac x = "B - A" in exI)
```
``` 292 apply (auto intro: multiset_ext_iff [THEN iffD2])
```
``` 293 done
```
``` 294
```
``` 295 lemma mset_le_mono_add_right_cancel [simp]:
```
``` 296 "(A::'a multiset) + C \<le> B + C \<longleftrightarrow> A \<le> B"
```
``` 297 by (fact add_le_cancel_right)
```
``` 298
```
``` 299 lemma mset_le_mono_add_left_cancel [simp]:
```
``` 300 "C + (A::'a multiset) \<le> C + B \<longleftrightarrow> A \<le> B"
```
``` 301 by (fact add_le_cancel_left)
```
``` 302
```
``` 303 lemma mset_le_mono_add:
```
``` 304 "(A::'a multiset) \<le> B \<Longrightarrow> C \<le> D \<Longrightarrow> A + C \<le> B + D"
```
``` 305 by (fact add_mono)
```
``` 306
```
``` 307 lemma mset_le_add_left [simp]:
```
``` 308 "(A::'a multiset) \<le> A + B"
```
``` 309 unfolding mset_le_def by auto
```
``` 310
```
``` 311 lemma mset_le_add_right [simp]:
```
``` 312 "B \<le> (A::'a multiset) + B"
```
``` 313 unfolding mset_le_def by auto
```
``` 314
```
``` 315 lemma mset_le_single:
```
``` 316 "a :# B \<Longrightarrow> {#a#} \<le> B"
```
``` 317 by (simp add: mset_le_def)
```
``` 318
```
``` 319 lemma multiset_diff_union_assoc:
```
``` 320 "C \<le> B \<Longrightarrow> (A::'a multiset) + B - C = A + (B - C)"
```
``` 321 by (simp add: multiset_ext_iff mset_le_def)
```
``` 322
```
``` 323 lemma mset_le_multiset_union_diff_commute:
```
``` 324 "B \<le> A \<Longrightarrow> (A::'a multiset) - B + C = A + C - B"
```
``` 325 by (simp add: multiset_ext_iff mset_le_def)
```
``` 326
```
``` 327 lemma mset_lessD: "A < B \<Longrightarrow> x \<in># A \<Longrightarrow> x \<in># B"
```
``` 328 apply (clarsimp simp: mset_le_def mset_less_def)
```
``` 329 apply (erule_tac x=x in allE)
```
``` 330 apply auto
```
``` 331 done
```
``` 332
```
``` 333 lemma mset_leD: "A \<le> B \<Longrightarrow> x \<in># A \<Longrightarrow> x \<in># B"
```
``` 334 apply (clarsimp simp: mset_le_def mset_less_def)
```
``` 335 apply (erule_tac x = x in allE)
```
``` 336 apply auto
```
``` 337 done
```
``` 338
```
``` 339 lemma mset_less_insertD: "(A + {#x#} < B) \<Longrightarrow> (x \<in># B \<and> A < B)"
```
``` 340 apply (rule conjI)
```
``` 341 apply (simp add: mset_lessD)
```
``` 342 apply (clarsimp simp: mset_le_def mset_less_def)
```
``` 343 apply safe
```
``` 344 apply (erule_tac x = a in allE)
```
``` 345 apply (auto split: split_if_asm)
```
``` 346 done
```
``` 347
```
``` 348 lemma mset_le_insertD: "(A + {#x#} \<le> B) \<Longrightarrow> (x \<in># B \<and> A \<le> B)"
```
``` 349 apply (rule conjI)
```
``` 350 apply (simp add: mset_leD)
```
``` 351 apply (force simp: mset_le_def mset_less_def split: split_if_asm)
```
``` 352 done
```
``` 353
```
``` 354 lemma mset_less_of_empty[simp]: "A < {#} \<longleftrightarrow> False"
```
``` 355 by (auto simp add: mset_less_def mset_le_def multiset_ext_iff)
```
``` 356
```
``` 357 lemma multi_psub_of_add_self[simp]: "A < A + {#x#}"
```
``` 358 by (auto simp: mset_le_def mset_less_def)
```
``` 359
```
``` 360 lemma multi_psub_self[simp]: "(A::'a multiset) < A = False"
```
``` 361 by simp
```
``` 362
```
``` 363 lemma mset_less_add_bothsides:
```
``` 364 "T + {#x#} < S + {#x#} \<Longrightarrow> T < S"
```
``` 365 by (fact add_less_imp_less_right)
```
``` 366
```
``` 367 lemma mset_less_empty_nonempty:
```
``` 368 "{#} < S \<longleftrightarrow> S \<noteq> {#}"
```
``` 369 by (auto simp: mset_le_def mset_less_def)
```
``` 370
```
``` 371 lemma mset_less_diff_self:
```
``` 372 "c \<in># B \<Longrightarrow> B - {#c#} < B"
```
``` 373 by (auto simp: mset_le_def mset_less_def multiset_ext_iff)
```
``` 374
```
``` 375
```
``` 376 subsubsection {* Intersection *}
```
``` 377
```
``` 378 instantiation multiset :: (type) semilattice_inf
```
``` 379 begin
```
``` 380
```
``` 381 definition inf_multiset :: "'a multiset \<Rightarrow> 'a multiset \<Rightarrow> 'a multiset" where
```
``` 382 multiset_inter_def: "inf_multiset A B = A - (A - B)"
```
``` 383
```
``` 384 instance proof -
```
``` 385 have aux: "\<And>m n q :: nat. m \<le> n \<Longrightarrow> m \<le> q \<Longrightarrow> m \<le> n - (n - q)" by arith
```
``` 386 show "OFCLASS('a multiset, semilattice_inf_class)" proof
```
``` 387 qed (auto simp add: multiset_inter_def mset_le_def aux)
```
``` 388 qed
```
``` 389
```
``` 390 end
```
``` 391
```
``` 392 abbreviation multiset_inter :: "'a multiset \<Rightarrow> 'a multiset \<Rightarrow> 'a multiset" (infixl "#\<inter>" 70) where
```
``` 393 "multiset_inter \<equiv> inf"
```
``` 394
```
``` 395 lemma multiset_inter_count:
```
``` 396 "count (A #\<inter> B) x = min (count A x) (count B x)"
```
``` 397 by (simp add: multiset_inter_def multiset_typedef)
```
``` 398
```
``` 399 lemma multiset_inter_single: "a \<noteq> b \<Longrightarrow> {#a#} #\<inter> {#b#} = {#}"
```
``` 400 by (rule multiset_ext) (auto simp add: multiset_inter_count)
```
``` 401
```
``` 402 lemma multiset_union_diff_commute:
```
``` 403 assumes "B #\<inter> C = {#}"
```
``` 404 shows "A + B - C = A - C + B"
```
``` 405 proof (rule multiset_ext)
```
``` 406 fix x
```
``` 407 from assms have "min (count B x) (count C x) = 0"
```
``` 408 by (auto simp add: multiset_inter_count multiset_ext_iff)
```
``` 409 then have "count B x = 0 \<or> count C x = 0"
```
``` 410 by auto
```
``` 411 then show "count (A + B - C) x = count (A - C + B) x"
```
``` 412 by auto
```
``` 413 qed
```
``` 414
```
``` 415
```
``` 416 subsubsection {* Comprehension (filter) *}
```
``` 417
```
``` 418 lemma count_MCollect [simp]:
```
``` 419 "count {# x:#M. P x #} a = (if P a then count M a else 0)"
```
``` 420 by (simp add: MCollect_def in_multiset multiset_typedef)
```
``` 421
```
``` 422 lemma MCollect_empty [simp]: "MCollect {#} P = {#}"
```
``` 423 by (rule multiset_ext) simp
```
``` 424
```
``` 425 lemma MCollect_single [simp]:
```
``` 426 "MCollect {#x#} P = (if P x then {#x#} else {#})"
```
``` 427 by (rule multiset_ext) simp
```
``` 428
```
``` 429 lemma MCollect_union [simp]:
```
``` 430 "MCollect (M + N) f = MCollect M f + MCollect N f"
```
``` 431 by (rule multiset_ext) simp
```
``` 432
```
``` 433
```
``` 434 subsubsection {* Set of elements *}
```
``` 435
```
``` 436 definition set_of :: "'a multiset => 'a set" where
```
``` 437 "set_of M = {x. x :# M}"
```
``` 438
```
``` 439 lemma set_of_empty [simp]: "set_of {#} = {}"
```
``` 440 by (simp add: set_of_def)
```
``` 441
```
``` 442 lemma set_of_single [simp]: "set_of {#b#} = {b}"
```
``` 443 by (simp add: set_of_def)
```
``` 444
```
``` 445 lemma set_of_union [simp]: "set_of (M + N) = set_of M \<union> set_of N"
```
``` 446 by (auto simp add: set_of_def)
```
``` 447
```
``` 448 lemma set_of_eq_empty_iff [simp]: "(set_of M = {}) = (M = {#})"
```
``` 449 by (auto simp add: set_of_def multiset_ext_iff)
```
``` 450
```
``` 451 lemma mem_set_of_iff [simp]: "(x \<in> set_of M) = (x :# M)"
```
``` 452 by (auto simp add: set_of_def)
```
``` 453
```
``` 454 lemma set_of_MCollect [simp]: "set_of {# x:#M. P x #} = set_of M \<inter> {x. P x}"
```
``` 455 by (auto simp add: set_of_def)
```
``` 456
```
``` 457 lemma finite_set_of [iff]: "finite (set_of M)"
```
``` 458 using count [of M] by (simp add: multiset_def set_of_def)
```
``` 459
```
``` 460
```
``` 461 subsubsection {* Size *}
```
``` 462
```
``` 463 instantiation multiset :: (type) size
```
``` 464 begin
```
``` 465
```
``` 466 definition size_def:
```
``` 467 "size M = setsum (count M) (set_of M)"
```
``` 468
```
``` 469 instance ..
```
``` 470
```
``` 471 end
```
``` 472
```
``` 473 lemma size_empty [simp]: "size {#} = 0"
```
``` 474 by (simp add: size_def)
```
``` 475
```
``` 476 lemma size_single [simp]: "size {#b#} = 1"
```
``` 477 by (simp add: size_def)
```
``` 478
```
``` 479 lemma setsum_count_Int:
```
``` 480 "finite A ==> setsum (count N) (A \<inter> set_of N) = setsum (count N) A"
```
``` 481 apply (induct rule: finite_induct)
```
``` 482 apply simp
```
``` 483 apply (simp add: Int_insert_left set_of_def)
```
``` 484 done
```
``` 485
```
``` 486 lemma size_union [simp]: "size (M + N::'a multiset) = size M + size N"
```
``` 487 apply (unfold size_def)
```
``` 488 apply (subgoal_tac "count (M + N) = (\<lambda>a. count M a + count N a)")
```
``` 489 prefer 2
```
``` 490 apply (rule ext, simp)
```
``` 491 apply (simp (no_asm_simp) add: setsum_Un_nat setsum_addf setsum_count_Int)
```
``` 492 apply (subst Int_commute)
```
``` 493 apply (simp (no_asm_simp) add: setsum_count_Int)
```
``` 494 done
```
``` 495
```
``` 496 lemma size_eq_0_iff_empty [iff]: "(size M = 0) = (M = {#})"
```
``` 497 by (auto simp add: size_def multiset_ext_iff)
```
``` 498
```
``` 499 lemma nonempty_has_size: "(S \<noteq> {#}) = (0 < size S)"
```
``` 500 by (metis gr0I gr_implies_not0 size_empty size_eq_0_iff_empty)
```
``` 501
```
``` 502 lemma size_eq_Suc_imp_elem: "size M = Suc n ==> \<exists>a. a :# M"
```
``` 503 apply (unfold size_def)
```
``` 504 apply (drule setsum_SucD)
```
``` 505 apply auto
```
``` 506 done
```
``` 507
```
``` 508 lemma size_eq_Suc_imp_eq_union:
```
``` 509 assumes "size M = Suc n"
```
``` 510 shows "\<exists>a N. M = N + {#a#}"
```
``` 511 proof -
```
``` 512 from assms obtain a where "a \<in># M"
```
``` 513 by (erule size_eq_Suc_imp_elem [THEN exE])
```
``` 514 then have "M = M - {#a#} + {#a#}" by simp
```
``` 515 then show ?thesis by blast
```
``` 516 qed
```
``` 517
```
``` 518
```
``` 519 subsection {* Induction and case splits *}
```
``` 520
```
``` 521 lemma setsum_decr:
```
``` 522 "finite F ==> (0::nat) < f a ==>
```
``` 523 setsum (f (a := f a - 1)) F = (if a\<in>F then setsum f F - 1 else setsum f F)"
```
``` 524 apply (induct rule: finite_induct)
```
``` 525 apply auto
```
``` 526 apply (drule_tac a = a in mk_disjoint_insert, auto)
```
``` 527 done
```
``` 528
```
``` 529 lemma rep_multiset_induct_aux:
```
``` 530 assumes 1: "P (\<lambda>a. (0::nat))"
```
``` 531 and 2: "!!f b. f \<in> multiset ==> P f ==> P (f (b := f b + 1))"
```
``` 532 shows "\<forall>f. f \<in> multiset --> setsum f {x. f x \<noteq> 0} = n --> P f"
```
``` 533 apply (unfold multiset_def)
```
``` 534 apply (induct_tac n, simp, clarify)
```
``` 535 apply (subgoal_tac "f = (\<lambda>a.0)")
```
``` 536 apply simp
```
``` 537 apply (rule 1)
```
``` 538 apply (rule ext, force, clarify)
```
``` 539 apply (frule setsum_SucD, clarify)
```
``` 540 apply (rename_tac a)
```
``` 541 apply (subgoal_tac "finite {x. (f (a := f a - 1)) x > 0}")
```
``` 542 prefer 2
```
``` 543 apply (rule finite_subset)
```
``` 544 prefer 2
```
``` 545 apply assumption
```
``` 546 apply simp
```
``` 547 apply blast
```
``` 548 apply (subgoal_tac "f = (f (a := f a - 1))(a := (f (a := f a - 1)) a + 1)")
```
``` 549 prefer 2
```
``` 550 apply (rule ext)
```
``` 551 apply (simp (no_asm_simp))
```
``` 552 apply (erule ssubst, rule 2 [unfolded multiset_def], blast)
```
``` 553 apply (erule allE, erule impE, erule_tac [2] mp, blast)
```
``` 554 apply (simp (no_asm_simp) add: setsum_decr del: fun_upd_apply One_nat_def)
```
``` 555 apply (subgoal_tac "{x. x \<noteq> a --> f x \<noteq> 0} = {x. f x \<noteq> 0}")
```
``` 556 prefer 2
```
``` 557 apply blast
```
``` 558 apply (subgoal_tac "{x. x \<noteq> a \<and> f x \<noteq> 0} = {x. f x \<noteq> 0} - {a}")
```
``` 559 prefer 2
```
``` 560 apply blast
```
``` 561 apply (simp add: le_imp_diff_is_add setsum_diff1_nat cong: conj_cong)
```
``` 562 done
```
``` 563
```
``` 564 theorem rep_multiset_induct:
```
``` 565 "f \<in> multiset ==> P (\<lambda>a. 0) ==>
```
``` 566 (!!f b. f \<in> multiset ==> P f ==> P (f (b := f b + 1))) ==> P f"
```
``` 567 using rep_multiset_induct_aux by blast
```
``` 568
```
``` 569 theorem multiset_induct [case_names empty add, induct type: multiset]:
```
``` 570 assumes empty: "P {#}"
```
``` 571 and add: "!!M x. P M ==> P (M + {#x#})"
```
``` 572 shows "P M"
```
``` 573 proof -
```
``` 574 note defns = union_def single_def Mempty_def
```
``` 575 note add' = add [unfolded defns, simplified]
```
``` 576 have aux: "\<And>a::'a. count (Abs_multiset (\<lambda>b. if b = a then 1 else 0)) =
```
``` 577 (\<lambda>b. if b = a then 1 else 0)" by (simp add: Abs_multiset_inverse in_multiset)
```
``` 578 show ?thesis
```
``` 579 apply (rule count_inverse [THEN subst])
```
``` 580 apply (rule count [THEN rep_multiset_induct])
```
``` 581 apply (rule empty [unfolded defns])
```
``` 582 apply (subgoal_tac "f(b := f b + 1) = (\<lambda>a. f a + (if a=b then 1 else 0))")
```
``` 583 prefer 2
```
``` 584 apply (simp add: expand_fun_eq)
```
``` 585 apply (erule ssubst)
```
``` 586 apply (erule Abs_multiset_inverse [THEN subst])
```
``` 587 apply (drule add')
```
``` 588 apply (simp add: aux)
```
``` 589 done
```
``` 590 qed
```
``` 591
```
``` 592 lemma multi_nonempty_split: "M \<noteq> {#} \<Longrightarrow> \<exists>A a. M = A + {#a#}"
```
``` 593 by (induct M) auto
```
``` 594
```
``` 595 lemma multiset_cases [cases type, case_names empty add]:
```
``` 596 assumes em: "M = {#} \<Longrightarrow> P"
```
``` 597 assumes add: "\<And>N x. M = N + {#x#} \<Longrightarrow> P"
```
``` 598 shows "P"
```
``` 599 proof (cases "M = {#}")
```
``` 600 assume "M = {#}" then show ?thesis using em by simp
```
``` 601 next
```
``` 602 assume "M \<noteq> {#}"
```
``` 603 then obtain M' m where "M = M' + {#m#}"
```
``` 604 by (blast dest: multi_nonempty_split)
```
``` 605 then show ?thesis using add by simp
```
``` 606 qed
```
``` 607
```
``` 608 lemma multi_member_split: "x \<in># M \<Longrightarrow> \<exists>A. M = A + {#x#}"
```
``` 609 apply (cases M)
```
``` 610 apply simp
```
``` 611 apply (rule_tac x="M - {#x#}" in exI, simp)
```
``` 612 done
```
``` 613
```
``` 614 lemma multi_drop_mem_not_eq: "c \<in># B \<Longrightarrow> B - {#c#} \<noteq> B"
```
``` 615 by (cases "B = {#}") (auto dest: multi_member_split)
```
``` 616
```
``` 617 lemma multiset_partition: "M = {# x:#M. P x #} + {# x:#M. \<not> P x #}"
```
``` 618 apply (subst multiset_ext_iff)
```
``` 619 apply auto
```
``` 620 done
```
``` 621
```
``` 622 lemma mset_less_size: "(A::'a multiset) < B \<Longrightarrow> size A < size B"
```
``` 623 proof (induct A arbitrary: B)
```
``` 624 case (empty M)
```
``` 625 then have "M \<noteq> {#}" by (simp add: mset_less_empty_nonempty)
```
``` 626 then obtain M' x where "M = M' + {#x#}"
```
``` 627 by (blast dest: multi_nonempty_split)
```
``` 628 then show ?case by simp
```
``` 629 next
```
``` 630 case (add S x T)
```
``` 631 have IH: "\<And>B. S < B \<Longrightarrow> size S < size B" by fact
```
``` 632 have SxsubT: "S + {#x#} < T" by fact
```
``` 633 then have "x \<in># T" and "S < T" by (auto dest: mset_less_insertD)
```
``` 634 then obtain T' where T: "T = T' + {#x#}"
```
``` 635 by (blast dest: multi_member_split)
```
``` 636 then have "S < T'" using SxsubT
```
``` 637 by (blast intro: mset_less_add_bothsides)
```
``` 638 then have "size S < size T'" using IH by simp
```
``` 639 then show ?case using T by simp
```
``` 640 qed
```
``` 641
```
``` 642
```
``` 643 subsubsection {* Strong induction and subset induction for multisets *}
```
``` 644
```
``` 645 text {* Well-foundedness of proper subset operator: *}
```
``` 646
```
``` 647 text {* proper multiset subset *}
```
``` 648
```
``` 649 definition
```
``` 650 mset_less_rel :: "('a multiset * 'a multiset) set" where
```
``` 651 "mset_less_rel = {(A,B). A < B}"
```
``` 652
```
``` 653 lemma multiset_add_sub_el_shuffle:
```
``` 654 assumes "c \<in># B" and "b \<noteq> c"
```
``` 655 shows "B - {#c#} + {#b#} = B + {#b#} - {#c#}"
```
``` 656 proof -
```
``` 657 from `c \<in># B` obtain A where B: "B = A + {#c#}"
```
``` 658 by (blast dest: multi_member_split)
```
``` 659 have "A + {#b#} = A + {#b#} + {#c#} - {#c#}" by simp
```
``` 660 then have "A + {#b#} = A + {#c#} + {#b#} - {#c#}"
```
``` 661 by (simp add: add_ac)
```
``` 662 then show ?thesis using B by simp
```
``` 663 qed
```
``` 664
```
``` 665 lemma wf_mset_less_rel: "wf mset_less_rel"
```
``` 666 apply (unfold mset_less_rel_def)
```
``` 667 apply (rule wf_measure [THEN wf_subset, where f1=size])
```
``` 668 apply (clarsimp simp: measure_def inv_image_def mset_less_size)
```
``` 669 done
```
``` 670
```
``` 671 text {* The induction rules: *}
```
``` 672
```
``` 673 lemma full_multiset_induct [case_names less]:
```
``` 674 assumes ih: "\<And>B. \<forall>(A::'a multiset). A < B \<longrightarrow> P A \<Longrightarrow> P B"
```
``` 675 shows "P B"
```
``` 676 apply (rule wf_mset_less_rel [THEN wf_induct])
```
``` 677 apply (rule ih, auto simp: mset_less_rel_def)
```
``` 678 done
```
``` 679
```
``` 680 lemma multi_subset_induct [consumes 2, case_names empty add]:
```
``` 681 assumes "F \<le> A"
```
``` 682 and empty: "P {#}"
```
``` 683 and insert: "\<And>a F. a \<in># A \<Longrightarrow> P F \<Longrightarrow> P (F + {#a#})"
```
``` 684 shows "P F"
```
``` 685 proof -
```
``` 686 from `F \<le> A`
```
``` 687 show ?thesis
```
``` 688 proof (induct F)
```
``` 689 show "P {#}" by fact
```
``` 690 next
```
``` 691 fix x F
```
``` 692 assume P: "F \<le> A \<Longrightarrow> P F" and i: "F + {#x#} \<le> A"
```
``` 693 show "P (F + {#x#})"
```
``` 694 proof (rule insert)
```
``` 695 from i show "x \<in># A" by (auto dest: mset_le_insertD)
```
``` 696 from i have "F \<le> A" by (auto dest: mset_le_insertD)
```
``` 697 with P show "P F" .
```
``` 698 qed
```
``` 699 qed
```
``` 700 qed
```
``` 701
```
``` 702
```
``` 703 subsection {* Alternative representations *}
```
``` 704
```
``` 705 subsubsection {* Lists *}
```
``` 706
```
``` 707 primrec multiset_of :: "'a list \<Rightarrow> 'a multiset" where
```
``` 708 "multiset_of [] = {#}" |
```
``` 709 "multiset_of (a # x) = multiset_of x + {# a #}"
```
``` 710
```
``` 711 lemma in_multiset_in_set:
```
``` 712 "x \<in># multiset_of xs \<longleftrightarrow> x \<in> set xs"
```
``` 713 by (induct xs) simp_all
```
``` 714
```
``` 715 lemma count_multiset_of:
```
``` 716 "count (multiset_of xs) x = length (filter (\<lambda>y. x = y) xs)"
```
``` 717 by (induct xs) simp_all
```
``` 718
```
``` 719 lemma multiset_of_zero_iff[simp]: "(multiset_of x = {#}) = (x = [])"
```
``` 720 by (induct x) auto
```
``` 721
```
``` 722 lemma multiset_of_zero_iff_right[simp]: "({#} = multiset_of x) = (x = [])"
```
``` 723 by (induct x) auto
```
``` 724
```
``` 725 lemma set_of_multiset_of[simp]: "set_of(multiset_of x) = set x"
```
``` 726 by (induct x) auto
```
``` 727
```
``` 728 lemma mem_set_multiset_eq: "x \<in> set xs = (x :# multiset_of xs)"
```
``` 729 by (induct xs) auto
```
``` 730
```
``` 731 lemma multiset_of_append [simp]:
```
``` 732 "multiset_of (xs @ ys) = multiset_of xs + multiset_of ys"
```
``` 733 by (induct xs arbitrary: ys) (auto simp: add_ac)
```
``` 734
```
``` 735 lemma surj_multiset_of: "surj multiset_of"
```
``` 736 apply (unfold surj_def)
```
``` 737 apply (rule allI)
```
``` 738 apply (rule_tac M = y in multiset_induct)
```
``` 739 apply auto
```
``` 740 apply (rule_tac x = "x # xa" in exI)
```
``` 741 apply auto
```
``` 742 done
```
``` 743
```
``` 744 lemma set_count_greater_0: "set x = {a. count (multiset_of x) a > 0}"
```
``` 745 by (induct x) auto
```
``` 746
```
``` 747 lemma distinct_count_atmost_1:
```
``` 748 "distinct x = (! a. count (multiset_of x) a = (if a \<in> set x then 1 else 0))"
```
``` 749 apply (induct x, simp, rule iffI, simp_all)
```
``` 750 apply (rule conjI)
```
``` 751 apply (simp_all add: set_of_multiset_of [THEN sym] del: set_of_multiset_of)
```
``` 752 apply (erule_tac x = a in allE, simp, clarify)
```
``` 753 apply (erule_tac x = aa in allE, simp)
```
``` 754 done
```
``` 755
```
``` 756 lemma multiset_of_eq_setD:
```
``` 757 "multiset_of xs = multiset_of ys \<Longrightarrow> set xs = set ys"
```
``` 758 by (rule) (auto simp add:multiset_ext_iff set_count_greater_0)
```
``` 759
```
``` 760 lemma set_eq_iff_multiset_of_eq_distinct:
```
``` 761 "distinct x \<Longrightarrow> distinct y \<Longrightarrow>
```
``` 762 (set x = set y) = (multiset_of x = multiset_of y)"
```
``` 763 by (auto simp: multiset_ext_iff distinct_count_atmost_1)
```
``` 764
```
``` 765 lemma set_eq_iff_multiset_of_remdups_eq:
```
``` 766 "(set x = set y) = (multiset_of (remdups x) = multiset_of (remdups y))"
```
``` 767 apply (rule iffI)
```
``` 768 apply (simp add: set_eq_iff_multiset_of_eq_distinct[THEN iffD1])
```
``` 769 apply (drule distinct_remdups [THEN distinct_remdups
```
``` 770 [THEN set_eq_iff_multiset_of_eq_distinct [THEN iffD2]]])
```
``` 771 apply simp
```
``` 772 done
```
``` 773
```
``` 774 lemma multiset_of_compl_union [simp]:
```
``` 775 "multiset_of [x\<leftarrow>xs. P x] + multiset_of [x\<leftarrow>xs. \<not>P x] = multiset_of xs"
```
``` 776 by (induct xs) (auto simp: add_ac)
```
``` 777
```
``` 778 lemma count_filter:
```
``` 779 "count (multiset_of xs) x = length [y \<leftarrow> xs. y = x]"
```
``` 780 by (induct xs) auto
```
``` 781
```
``` 782 lemma nth_mem_multiset_of: "i < length ls \<Longrightarrow> (ls ! i) :# multiset_of ls"
```
``` 783 apply (induct ls arbitrary: i)
```
``` 784 apply simp
```
``` 785 apply (case_tac i)
```
``` 786 apply auto
```
``` 787 done
```
``` 788
```
``` 789 lemma multiset_of_remove1[simp]:
```
``` 790 "multiset_of (remove1 a xs) = multiset_of xs - {#a#}"
```
``` 791 by (induct xs) (auto simp add: multiset_ext_iff)
```
``` 792
```
``` 793 lemma multiset_of_eq_length:
```
``` 794 assumes "multiset_of xs = multiset_of ys"
```
``` 795 shows "length xs = length ys"
```
``` 796 using assms proof (induct xs arbitrary: ys)
```
``` 797 case Nil then show ?case by simp
```
``` 798 next
```
``` 799 case (Cons x xs)
```
``` 800 then have "x \<in># multiset_of ys" by (simp add: union_single_eq_member)
```
``` 801 then have "x \<in> set ys" by (simp add: in_multiset_in_set)
```
``` 802 from Cons.prems [symmetric] have "multiset_of xs = multiset_of (remove1 x ys)"
```
``` 803 by simp
```
``` 804 with Cons.hyps have "length xs = length (remove1 x ys)" .
```
``` 805 with `x \<in> set ys` show ?case
```
``` 806 by (auto simp add: length_remove1 dest: length_pos_if_in_set)
```
``` 807 qed
```
``` 808
```
``` 809 lemma (in linorder) multiset_of_insort [simp]:
```
``` 810 "multiset_of (insort x xs) = {#x#} + multiset_of xs"
```
``` 811 by (induct xs) (simp_all add: ac_simps)
```
``` 812
```
``` 813 lemma (in linorder) multiset_of_sort [simp]:
```
``` 814 "multiset_of (sort xs) = multiset_of xs"
```
``` 815 by (induct xs) (simp_all add: ac_simps)
```
``` 816
```
``` 817 text {*
```
``` 818 This lemma shows which properties suffice to show that a function
```
``` 819 @{text "f"} with @{text "f xs = ys"} behaves like sort.
```
``` 820 *}
```
``` 821
```
``` 822 lemma (in linorder) properties_for_sort:
```
``` 823 "multiset_of ys = multiset_of xs \<Longrightarrow> sorted ys \<Longrightarrow> sort xs = ys"
```
``` 824 proof (induct xs arbitrary: ys)
```
``` 825 case Nil then show ?case by simp
```
``` 826 next
```
``` 827 case (Cons x xs)
```
``` 828 then have "x \<in> set ys"
```
``` 829 by (auto simp add: mem_set_multiset_eq intro!: ccontr)
```
``` 830 with Cons.prems Cons.hyps [of "remove1 x ys"] show ?case
```
``` 831 by (simp add: sorted_remove1 multiset_of_remove1 insort_remove1)
```
``` 832 qed
```
``` 833
```
``` 834 lemma multiset_of_remdups_le: "multiset_of (remdups xs) \<le> multiset_of xs"
```
``` 835 by (induct xs) (auto intro: order_trans)
```
``` 836
```
``` 837 lemma multiset_of_update:
```
``` 838 "i < length ls \<Longrightarrow> multiset_of (ls[i := v]) = multiset_of ls - {#ls ! i#} + {#v#}"
```
``` 839 proof (induct ls arbitrary: i)
```
``` 840 case Nil then show ?case by simp
```
``` 841 next
```
``` 842 case (Cons x xs)
```
``` 843 show ?case
```
``` 844 proof (cases i)
```
``` 845 case 0 then show ?thesis by simp
```
``` 846 next
```
``` 847 case (Suc i')
```
``` 848 with Cons show ?thesis
```
``` 849 apply simp
```
``` 850 apply (subst add_assoc)
```
``` 851 apply (subst add_commute [of "{#v#}" "{#x#}"])
```
``` 852 apply (subst add_assoc [symmetric])
```
``` 853 apply simp
```
``` 854 apply (rule mset_le_multiset_union_diff_commute)
```
``` 855 apply (simp add: mset_le_single nth_mem_multiset_of)
```
``` 856 done
```
``` 857 qed
```
``` 858 qed
```
``` 859
```
``` 860 lemma multiset_of_swap:
```
``` 861 "i < length ls \<Longrightarrow> j < length ls \<Longrightarrow>
```
``` 862 multiset_of (ls[j := ls ! i, i := ls ! j]) = multiset_of ls"
```
``` 863 by (cases "i = j") (simp_all add: multiset_of_update nth_mem_multiset_of)
```
``` 864
```
``` 865
```
``` 866 subsubsection {* Association lists -- including rudimentary code generation *}
```
``` 867
```
``` 868 definition count_of :: "('a \<times> nat) list \<Rightarrow> 'a \<Rightarrow> nat" where
```
``` 869 "count_of xs x = (case map_of xs x of None \<Rightarrow> 0 | Some n \<Rightarrow> n)"
```
``` 870
```
``` 871 lemma count_of_multiset:
```
``` 872 "count_of xs \<in> multiset"
```
``` 873 proof -
```
``` 874 let ?A = "{x::'a. 0 < (case map_of xs x of None \<Rightarrow> 0\<Colon>nat | Some (n\<Colon>nat) \<Rightarrow> n)}"
```
``` 875 have "?A \<subseteq> dom (map_of xs)"
```
``` 876 proof
```
``` 877 fix x
```
``` 878 assume "x \<in> ?A"
```
``` 879 then have "0 < (case map_of xs x of None \<Rightarrow> 0\<Colon>nat | Some (n\<Colon>nat) \<Rightarrow> n)" by simp
```
``` 880 then have "map_of xs x \<noteq> None" by (cases "map_of xs x") auto
```
``` 881 then show "x \<in> dom (map_of xs)" by auto
```
``` 882 qed
```
``` 883 with finite_dom_map_of [of xs] have "finite ?A"
```
``` 884 by (auto intro: finite_subset)
```
``` 885 then show ?thesis
```
``` 886 by (simp add: count_of_def expand_fun_eq multiset_def)
```
``` 887 qed
```
``` 888
```
``` 889 lemma count_simps [simp]:
```
``` 890 "count_of [] = (\<lambda>_. 0)"
```
``` 891 "count_of ((x, n) # xs) = (\<lambda>y. if x = y then n else count_of xs y)"
```
``` 892 by (simp_all add: count_of_def expand_fun_eq)
```
``` 893
```
``` 894 lemma count_of_empty:
```
``` 895 "x \<notin> fst ` set xs \<Longrightarrow> count_of xs x = 0"
```
``` 896 by (induct xs) (simp_all add: count_of_def)
```
``` 897
```
``` 898 lemma count_of_filter:
```
``` 899 "count_of (filter (P \<circ> fst) xs) x = (if P x then count_of xs x else 0)"
```
``` 900 by (induct xs) auto
```
``` 901
```
``` 902 definition Bag :: "('a \<times> nat) list \<Rightarrow> 'a multiset" where
```
``` 903 "Bag xs = Abs_multiset (count_of xs)"
```
``` 904
```
``` 905 code_datatype Bag
```
``` 906
```
``` 907 lemma count_Bag [simp, code]:
```
``` 908 "count (Bag xs) = count_of xs"
```
``` 909 by (simp add: Bag_def count_of_multiset Abs_multiset_inverse)
```
``` 910
```
``` 911 lemma Mempty_Bag [code]:
```
``` 912 "{#} = Bag []"
```
``` 913 by (simp add: multiset_ext_iff)
```
``` 914
```
``` 915 lemma single_Bag [code]:
```
``` 916 "{#x#} = Bag [(x, 1)]"
```
``` 917 by (simp add: multiset_ext_iff)
```
``` 918
```
``` 919 lemma MCollect_Bag [code]:
```
``` 920 "MCollect (Bag xs) P = Bag (filter (P \<circ> fst) xs)"
```
``` 921 by (simp add: multiset_ext_iff count_of_filter)
```
``` 922
```
``` 923 lemma mset_less_eq_Bag [code]:
```
``` 924 "Bag xs \<le> A \<longleftrightarrow> (\<forall>(x, n) \<in> set xs. count_of xs x \<le> count A x)"
```
``` 925 (is "?lhs \<longleftrightarrow> ?rhs")
```
``` 926 proof
```
``` 927 assume ?lhs then show ?rhs
```
``` 928 by (auto simp add: mset_le_def count_Bag)
```
``` 929 next
```
``` 930 assume ?rhs
```
``` 931 show ?lhs
```
``` 932 proof (rule mset_less_eqI)
```
``` 933 fix x
```
``` 934 from `?rhs` have "count_of xs x \<le> count A x"
```
``` 935 by (cases "x \<in> fst ` set xs") (auto simp add: count_of_empty)
```
``` 936 then show "count (Bag xs) x \<le> count A x"
```
``` 937 by (simp add: mset_le_def count_Bag)
```
``` 938 qed
```
``` 939 qed
```
``` 940
```
``` 941 instantiation multiset :: (equal) equal
```
``` 942 begin
```
``` 943
```
``` 944 definition
```
``` 945 "HOL.equal A B \<longleftrightarrow> (A::'a multiset) \<le> B \<and> B \<le> A"
```
``` 946
```
``` 947 instance proof
```
``` 948 qed (simp add: equal_multiset_def eq_iff)
```
``` 949
```
``` 950 end
```
``` 951
```
``` 952 lemma [code nbe]:
```
``` 953 "HOL.equal (A :: 'a::equal multiset) A \<longleftrightarrow> True"
```
``` 954 by (fact equal_refl)
```
``` 955
```
``` 956 definition (in term_syntax)
```
``` 957 bagify :: "('a\<Colon>typerep \<times> nat) list \<times> (unit \<Rightarrow> Code_Evaluation.term)
```
``` 958 \<Rightarrow> 'a multiset \<times> (unit \<Rightarrow> Code_Evaluation.term)" where
```
``` 959 [code_unfold]: "bagify xs = Code_Evaluation.valtermify Bag {\<cdot>} xs"
```
``` 960
```
``` 961 notation fcomp (infixl "\<circ>>" 60)
```
``` 962 notation scomp (infixl "\<circ>\<rightarrow>" 60)
```
``` 963
```
``` 964 instantiation multiset :: (random) random
```
``` 965 begin
```
``` 966
```
``` 967 definition
```
``` 968 "Quickcheck.random i = Quickcheck.random i \<circ>\<rightarrow> (\<lambda>xs. Pair (bagify xs))"
```
``` 969
```
``` 970 instance ..
```
``` 971
```
``` 972 end
```
``` 973
```
``` 974 no_notation fcomp (infixl "\<circ>>" 60)
```
``` 975 no_notation scomp (infixl "\<circ>\<rightarrow>" 60)
```
``` 976
```
``` 977 hide_const (open) bagify
```
``` 978
```
``` 979
```
``` 980 subsection {* The multiset order *}
```
``` 981
```
``` 982 subsubsection {* Well-foundedness *}
```
``` 983
```
``` 984 definition mult1 :: "('a \<times> 'a) set => ('a multiset \<times> 'a multiset) set" where
```
``` 985 "mult1 r = {(N, M). \<exists>a M0 K. M = M0 + {#a#} \<and> N = M0 + K \<and>
```
``` 986 (\<forall>b. b :# K --> (b, a) \<in> r)}"
```
``` 987
```
``` 988 definition mult :: "('a \<times> 'a) set => ('a multiset \<times> 'a multiset) set" where
```
``` 989 "mult r = (mult1 r)\<^sup>+"
```
``` 990
```
``` 991 lemma not_less_empty [iff]: "(M, {#}) \<notin> mult1 r"
```
``` 992 by (simp add: mult1_def)
```
``` 993
```
``` 994 lemma less_add: "(N, M0 + {#a#}) \<in> mult1 r ==>
```
``` 995 (\<exists>M. (M, M0) \<in> mult1 r \<and> N = M + {#a#}) \<or>
```
``` 996 (\<exists>K. (\<forall>b. b :# K --> (b, a) \<in> r) \<and> N = M0 + K)"
```
``` 997 (is "_ \<Longrightarrow> ?case1 (mult1 r) \<or> ?case2")
```
``` 998 proof (unfold mult1_def)
```
``` 999 let ?r = "\<lambda>K a. \<forall>b. b :# K --> (b, a) \<in> r"
```
``` 1000 let ?R = "\<lambda>N M. \<exists>a M0 K. M = M0 + {#a#} \<and> N = M0 + K \<and> ?r K a"
```
``` 1001 let ?case1 = "?case1 {(N, M). ?R N M}"
```
``` 1002
```
``` 1003 assume "(N, M0 + {#a#}) \<in> {(N, M). ?R N M}"
```
``` 1004 then have "\<exists>a' M0' K.
```
``` 1005 M0 + {#a#} = M0' + {#a'#} \<and> N = M0' + K \<and> ?r K a'" by simp
```
``` 1006 then show "?case1 \<or> ?case2"
```
``` 1007 proof (elim exE conjE)
```
``` 1008 fix a' M0' K
```
``` 1009 assume N: "N = M0' + K" and r: "?r K a'"
```
``` 1010 assume "M0 + {#a#} = M0' + {#a'#}"
```
``` 1011 then have "M0 = M0' \<and> a = a' \<or>
```
``` 1012 (\<exists>K'. M0 = K' + {#a'#} \<and> M0' = K' + {#a#})"
```
``` 1013 by (simp only: add_eq_conv_ex)
```
``` 1014 then show ?thesis
```
``` 1015 proof (elim disjE conjE exE)
```
``` 1016 assume "M0 = M0'" "a = a'"
```
``` 1017 with N r have "?r K a \<and> N = M0 + K" by simp
```
``` 1018 then have ?case2 .. then show ?thesis ..
```
``` 1019 next
```
``` 1020 fix K'
```
``` 1021 assume "M0' = K' + {#a#}"
```
``` 1022 with N have n: "N = K' + K + {#a#}" by (simp add: add_ac)
```
``` 1023
```
``` 1024 assume "M0 = K' + {#a'#}"
```
``` 1025 with r have "?R (K' + K) M0" by blast
```
``` 1026 with n have ?case1 by simp then show ?thesis ..
```
``` 1027 qed
```
``` 1028 qed
```
``` 1029 qed
```
``` 1030
```
``` 1031 lemma all_accessible: "wf r ==> \<forall>M. M \<in> acc (mult1 r)"
```
``` 1032 proof
```
``` 1033 let ?R = "mult1 r"
```
``` 1034 let ?W = "acc ?R"
```
``` 1035 {
```
``` 1036 fix M M0 a
```
``` 1037 assume M0: "M0 \<in> ?W"
```
``` 1038 and wf_hyp: "!!b. (b, a) \<in> r ==> (\<forall>M \<in> ?W. M + {#b#} \<in> ?W)"
```
``` 1039 and acc_hyp: "\<forall>M. (M, M0) \<in> ?R --> M + {#a#} \<in> ?W"
```
``` 1040 have "M0 + {#a#} \<in> ?W"
```
``` 1041 proof (rule accI [of "M0 + {#a#}"])
```
``` 1042 fix N
```
``` 1043 assume "(N, M0 + {#a#}) \<in> ?R"
```
``` 1044 then have "((\<exists>M. (M, M0) \<in> ?R \<and> N = M + {#a#}) \<or>
```
``` 1045 (\<exists>K. (\<forall>b. b :# K --> (b, a) \<in> r) \<and> N = M0 + K))"
```
``` 1046 by (rule less_add)
```
``` 1047 then show "N \<in> ?W"
```
``` 1048 proof (elim exE disjE conjE)
```
``` 1049 fix M assume "(M, M0) \<in> ?R" and N: "N = M + {#a#}"
```
``` 1050 from acc_hyp have "(M, M0) \<in> ?R --> M + {#a#} \<in> ?W" ..
```
``` 1051 from this and `(M, M0) \<in> ?R` have "M + {#a#} \<in> ?W" ..
```
``` 1052 then show "N \<in> ?W" by (simp only: N)
```
``` 1053 next
```
``` 1054 fix K
```
``` 1055 assume N: "N = M0 + K"
```
``` 1056 assume "\<forall>b. b :# K --> (b, a) \<in> r"
```
``` 1057 then have "M0 + K \<in> ?W"
```
``` 1058 proof (induct K)
```
``` 1059 case empty
```
``` 1060 from M0 show "M0 + {#} \<in> ?W" by simp
```
``` 1061 next
```
``` 1062 case (add K x)
```
``` 1063 from add.prems have "(x, a) \<in> r" by simp
```
``` 1064 with wf_hyp have "\<forall>M \<in> ?W. M + {#x#} \<in> ?W" by blast
```
``` 1065 moreover from add have "M0 + K \<in> ?W" by simp
```
``` 1066 ultimately have "(M0 + K) + {#x#} \<in> ?W" ..
```
``` 1067 then show "M0 + (K + {#x#}) \<in> ?W" by (simp only: add_assoc)
```
``` 1068 qed
```
``` 1069 then show "N \<in> ?W" by (simp only: N)
```
``` 1070 qed
```
``` 1071 qed
```
``` 1072 } note tedious_reasoning = this
```
``` 1073
```
``` 1074 assume wf: "wf r"
```
``` 1075 fix M
```
``` 1076 show "M \<in> ?W"
```
``` 1077 proof (induct M)
```
``` 1078 show "{#} \<in> ?W"
```
``` 1079 proof (rule accI)
```
``` 1080 fix b assume "(b, {#}) \<in> ?R"
```
``` 1081 with not_less_empty show "b \<in> ?W" by contradiction
```
``` 1082 qed
```
``` 1083
```
``` 1084 fix M a assume "M \<in> ?W"
```
``` 1085 from wf have "\<forall>M \<in> ?W. M + {#a#} \<in> ?W"
```
``` 1086 proof induct
```
``` 1087 fix a
```
``` 1088 assume r: "!!b. (b, a) \<in> r ==> (\<forall>M \<in> ?W. M + {#b#} \<in> ?W)"
```
``` 1089 show "\<forall>M \<in> ?W. M + {#a#} \<in> ?W"
```
``` 1090 proof
```
``` 1091 fix M assume "M \<in> ?W"
```
``` 1092 then show "M + {#a#} \<in> ?W"
```
``` 1093 by (rule acc_induct) (rule tedious_reasoning [OF _ r])
```
``` 1094 qed
```
``` 1095 qed
```
``` 1096 from this and `M \<in> ?W` show "M + {#a#} \<in> ?W" ..
```
``` 1097 qed
```
``` 1098 qed
```
``` 1099
```
``` 1100 theorem wf_mult1: "wf r ==> wf (mult1 r)"
```
``` 1101 by (rule acc_wfI) (rule all_accessible)
```
``` 1102
```
``` 1103 theorem wf_mult: "wf r ==> wf (mult r)"
```
``` 1104 unfolding mult_def by (rule wf_trancl) (rule wf_mult1)
```
``` 1105
```
``` 1106
```
``` 1107 subsubsection {* Closure-free presentation *}
```
``` 1108
```
``` 1109 text {* One direction. *}
```
``` 1110
```
``` 1111 lemma mult_implies_one_step:
```
``` 1112 "trans r ==> (M, N) \<in> mult r ==>
```
``` 1113 \<exists>I J K. N = I + J \<and> M = I + K \<and> J \<noteq> {#} \<and>
```
``` 1114 (\<forall>k \<in> set_of K. \<exists>j \<in> set_of J. (k, j) \<in> r)"
```
``` 1115 apply (unfold mult_def mult1_def set_of_def)
```
``` 1116 apply (erule converse_trancl_induct, clarify)
```
``` 1117 apply (rule_tac x = M0 in exI, simp, clarify)
```
``` 1118 apply (case_tac "a :# K")
```
``` 1119 apply (rule_tac x = I in exI)
```
``` 1120 apply (simp (no_asm))
```
``` 1121 apply (rule_tac x = "(K - {#a#}) + Ka" in exI)
```
``` 1122 apply (simp (no_asm_simp) add: add_assoc [symmetric])
```
``` 1123 apply (drule_tac f = "\<lambda>M. M - {#a#}" in arg_cong)
```
``` 1124 apply (simp add: diff_union_single_conv)
```
``` 1125 apply (simp (no_asm_use) add: trans_def)
```
``` 1126 apply blast
```
``` 1127 apply (subgoal_tac "a :# I")
```
``` 1128 apply (rule_tac x = "I - {#a#}" in exI)
```
``` 1129 apply (rule_tac x = "J + {#a#}" in exI)
```
``` 1130 apply (rule_tac x = "K + Ka" in exI)
```
``` 1131 apply (rule conjI)
```
``` 1132 apply (simp add: multiset_ext_iff split: nat_diff_split)
```
``` 1133 apply (rule conjI)
```
``` 1134 apply (drule_tac f = "\<lambda>M. M - {#a#}" in arg_cong, simp)
```
``` 1135 apply (simp add: multiset_ext_iff split: nat_diff_split)
```
``` 1136 apply (simp (no_asm_use) add: trans_def)
```
``` 1137 apply blast
```
``` 1138 apply (subgoal_tac "a :# (M0 + {#a#})")
```
``` 1139 apply simp
```
``` 1140 apply (simp (no_asm))
```
``` 1141 done
```
``` 1142
```
``` 1143 lemma one_step_implies_mult_aux:
```
``` 1144 "trans r ==>
```
``` 1145 \<forall>I J K. (size J = n \<and> J \<noteq> {#} \<and> (\<forall>k \<in> set_of K. \<exists>j \<in> set_of J. (k, j) \<in> r))
```
``` 1146 --> (I + K, I + J) \<in> mult r"
```
``` 1147 apply (induct_tac n, auto)
```
``` 1148 apply (frule size_eq_Suc_imp_eq_union, clarify)
```
``` 1149 apply (rename_tac "J'", simp)
```
``` 1150 apply (erule notE, auto)
```
``` 1151 apply (case_tac "J' = {#}")
```
``` 1152 apply (simp add: mult_def)
```
``` 1153 apply (rule r_into_trancl)
```
``` 1154 apply (simp add: mult1_def set_of_def, blast)
```
``` 1155 txt {* Now we know @{term "J' \<noteq> {#}"}. *}
```
``` 1156 apply (cut_tac M = K and P = "\<lambda>x. (x, a) \<in> r" in multiset_partition)
```
``` 1157 apply (erule_tac P = "\<forall>k \<in> set_of K. ?P k" in rev_mp)
```
``` 1158 apply (erule ssubst)
```
``` 1159 apply (simp add: Ball_def, auto)
```
``` 1160 apply (subgoal_tac
```
``` 1161 "((I + {# x :# K. (x, a) \<in> r #}) + {# x :# K. (x, a) \<notin> r #},
```
``` 1162 (I + {# x :# K. (x, a) \<in> r #}) + J') \<in> mult r")
```
``` 1163 prefer 2
```
``` 1164 apply force
```
``` 1165 apply (simp (no_asm_use) add: add_assoc [symmetric] mult_def)
```
``` 1166 apply (erule trancl_trans)
```
``` 1167 apply (rule r_into_trancl)
```
``` 1168 apply (simp add: mult1_def set_of_def)
```
``` 1169 apply (rule_tac x = a in exI)
```
``` 1170 apply (rule_tac x = "I + J'" in exI)
```
``` 1171 apply (simp add: add_ac)
```
``` 1172 done
```
``` 1173
```
``` 1174 lemma one_step_implies_mult:
```
``` 1175 "trans r ==> J \<noteq> {#} ==> \<forall>k \<in> set_of K. \<exists>j \<in> set_of J. (k, j) \<in> r
```
``` 1176 ==> (I + K, I + J) \<in> mult r"
```
``` 1177 using one_step_implies_mult_aux by blast
```
``` 1178
```
``` 1179
```
``` 1180 subsubsection {* Partial-order properties *}
```
``` 1181
```
``` 1182 definition less_multiset :: "'a\<Colon>order multiset \<Rightarrow> 'a multiset \<Rightarrow> bool" (infix "<#" 50) where
```
``` 1183 "M' <# M \<longleftrightarrow> (M', M) \<in> mult {(x', x). x' < x}"
```
``` 1184
```
``` 1185 definition le_multiset :: "'a\<Colon>order multiset \<Rightarrow> 'a multiset \<Rightarrow> bool" (infix "<=#" 50) where
```
``` 1186 "M' <=# M \<longleftrightarrow> M' <# M \<or> M' = M"
```
``` 1187
```
``` 1188 notation (xsymbols) less_multiset (infix "\<subset>#" 50)
```
``` 1189 notation (xsymbols) le_multiset (infix "\<subseteq>#" 50)
```
``` 1190
```
``` 1191 interpretation multiset_order: order le_multiset less_multiset
```
``` 1192 proof -
```
``` 1193 have irrefl: "\<And>M :: 'a multiset. \<not> M \<subset># M"
```
``` 1194 proof
```
``` 1195 fix M :: "'a multiset"
```
``` 1196 assume "M \<subset># M"
```
``` 1197 then have MM: "(M, M) \<in> mult {(x, y). x < y}" by (simp add: less_multiset_def)
```
``` 1198 have "trans {(x'::'a, x). x' < x}"
```
``` 1199 by (rule transI) simp
```
``` 1200 moreover note MM
```
``` 1201 ultimately have "\<exists>I J K. M = I + J \<and> M = I + K
```
``` 1202 \<and> J \<noteq> {#} \<and> (\<forall>k\<in>set_of K. \<exists>j\<in>set_of J. (k, j) \<in> {(x, y). x < y})"
```
``` 1203 by (rule mult_implies_one_step)
```
``` 1204 then obtain I J K where "M = I + J" and "M = I + K"
```
``` 1205 and "J \<noteq> {#}" and "(\<forall>k\<in>set_of K. \<exists>j\<in>set_of J. (k, j) \<in> {(x, y). x < y})" by blast
```
``` 1206 then have aux1: "K \<noteq> {#}" and aux2: "\<forall>k\<in>set_of K. \<exists>j\<in>set_of K. k < j" by auto
```
``` 1207 have "finite (set_of K)" by simp
```
``` 1208 moreover note aux2
```
``` 1209 ultimately have "set_of K = {}"
```
``` 1210 by (induct rule: finite_induct) (auto intro: order_less_trans)
```
``` 1211 with aux1 show False by simp
```
``` 1212 qed
```
``` 1213 have trans: "\<And>K M N :: 'a multiset. K \<subset># M \<Longrightarrow> M \<subset># N \<Longrightarrow> K \<subset># N"
```
``` 1214 unfolding less_multiset_def mult_def by (blast intro: trancl_trans)
```
``` 1215 show "class.order (le_multiset :: 'a multiset \<Rightarrow> _) less_multiset" proof
```
``` 1216 qed (auto simp add: le_multiset_def irrefl dest: trans)
```
``` 1217 qed
```
``` 1218
```
``` 1219 lemma mult_less_irrefl [elim!]:
```
``` 1220 "M \<subset># (M::'a::order multiset) ==> R"
```
``` 1221 by (simp add: multiset_order.less_irrefl)
```
``` 1222
```
``` 1223
```
``` 1224 subsubsection {* Monotonicity of multiset union *}
```
``` 1225
```
``` 1226 lemma mult1_union:
```
``` 1227 "(B, D) \<in> mult1 r ==> trans r ==> (C + B, C + D) \<in> mult1 r"
```
``` 1228 apply (unfold mult1_def)
```
``` 1229 apply auto
```
``` 1230 apply (rule_tac x = a in exI)
```
``` 1231 apply (rule_tac x = "C + M0" in exI)
```
``` 1232 apply (simp add: add_assoc)
```
``` 1233 done
```
``` 1234
```
``` 1235 lemma union_less_mono2: "B \<subset># D ==> C + B \<subset># C + (D::'a::order multiset)"
```
``` 1236 apply (unfold less_multiset_def mult_def)
```
``` 1237 apply (erule trancl_induct)
```
``` 1238 apply (blast intro: mult1_union transI order_less_trans r_into_trancl)
```
``` 1239 apply (blast intro: mult1_union transI order_less_trans r_into_trancl trancl_trans)
```
``` 1240 done
```
``` 1241
```
``` 1242 lemma union_less_mono1: "B \<subset># D ==> B + C \<subset># D + (C::'a::order multiset)"
```
``` 1243 apply (subst add_commute [of B C])
```
``` 1244 apply (subst add_commute [of D C])
```
``` 1245 apply (erule union_less_mono2)
```
``` 1246 done
```
``` 1247
```
``` 1248 lemma union_less_mono:
```
``` 1249 "A \<subset># C ==> B \<subset># D ==> A + B \<subset># C + (D::'a::order multiset)"
```
``` 1250 by (blast intro!: union_less_mono1 union_less_mono2 multiset_order.less_trans)
```
``` 1251
```
``` 1252 interpretation multiset_order: ordered_ab_semigroup_add plus le_multiset less_multiset
```
``` 1253 proof
```
``` 1254 qed (auto simp add: le_multiset_def intro: union_less_mono2)
```
``` 1255
```
``` 1256
```
``` 1257 subsection {* The fold combinator *}
```
``` 1258
```
``` 1259 text {*
```
``` 1260 The intended behaviour is
```
``` 1261 @{text "fold_mset f z {#x\<^isub>1, ..., x\<^isub>n#} = f x\<^isub>1 (\<dots> (f x\<^isub>n z)\<dots>)"}
```
``` 1262 if @{text f} is associative-commutative.
```
``` 1263 *}
```
``` 1264
```
``` 1265 text {*
```
``` 1266 The graph of @{text "fold_mset"}, @{text "z"}: the start element,
```
``` 1267 @{text "f"}: folding function, @{text "A"}: the multiset, @{text
```
``` 1268 "y"}: the result.
```
``` 1269 *}
```
``` 1270 inductive
```
``` 1271 fold_msetG :: "('a \<Rightarrow> 'b \<Rightarrow> 'b) \<Rightarrow> 'b \<Rightarrow> 'a multiset \<Rightarrow> 'b \<Rightarrow> bool"
```
``` 1272 for f :: "'a \<Rightarrow> 'b \<Rightarrow> 'b"
```
``` 1273 and z :: 'b
```
``` 1274 where
```
``` 1275 emptyI [intro]: "fold_msetG f z {#} z"
```
``` 1276 | insertI [intro]: "fold_msetG f z A y \<Longrightarrow> fold_msetG f z (A + {#x#}) (f x y)"
```
``` 1277
```
``` 1278 inductive_cases empty_fold_msetGE [elim!]: "fold_msetG f z {#} x"
```
``` 1279 inductive_cases insert_fold_msetGE: "fold_msetG f z (A + {#}) y"
```
``` 1280
```
``` 1281 definition
```
``` 1282 fold_mset :: "('a \<Rightarrow> 'b \<Rightarrow> 'b) \<Rightarrow> 'b \<Rightarrow> 'a multiset \<Rightarrow> 'b" where
```
``` 1283 "fold_mset f z A = (THE x. fold_msetG f z A x)"
```
``` 1284
```
``` 1285 lemma Diff1_fold_msetG:
```
``` 1286 "fold_msetG f z (A - {#x#}) y \<Longrightarrow> x \<in># A \<Longrightarrow> fold_msetG f z A (f x y)"
```
``` 1287 apply (frule_tac x = x in fold_msetG.insertI)
```
``` 1288 apply auto
```
``` 1289 done
```
``` 1290
```
``` 1291 lemma fold_msetG_nonempty: "\<exists>x. fold_msetG f z A x"
```
``` 1292 apply (induct A)
```
``` 1293 apply blast
```
``` 1294 apply clarsimp
```
``` 1295 apply (drule_tac x = x in fold_msetG.insertI)
```
``` 1296 apply auto
```
``` 1297 done
```
``` 1298
```
``` 1299 lemma fold_mset_empty[simp]: "fold_mset f z {#} = z"
```
``` 1300 unfolding fold_mset_def by blast
```
``` 1301
```
``` 1302 context fun_left_comm
```
``` 1303 begin
```
``` 1304
```
``` 1305 lemma fold_msetG_determ:
```
``` 1306 "fold_msetG f z A x \<Longrightarrow> fold_msetG f z A y \<Longrightarrow> y = x"
```
``` 1307 proof (induct arbitrary: x y z rule: full_multiset_induct)
```
``` 1308 case (less M x\<^isub>1 x\<^isub>2 Z)
```
``` 1309 have IH: "\<forall>A. A < M \<longrightarrow>
```
``` 1310 (\<forall>x x' x''. fold_msetG f x'' A x \<longrightarrow> fold_msetG f x'' A x'
```
``` 1311 \<longrightarrow> x' = x)" by fact
```
``` 1312 have Mfoldx\<^isub>1: "fold_msetG f Z M x\<^isub>1" and Mfoldx\<^isub>2: "fold_msetG f Z M x\<^isub>2" by fact+
```
``` 1313 show ?case
```
``` 1314 proof (rule fold_msetG.cases [OF Mfoldx\<^isub>1])
```
``` 1315 assume "M = {#}" and "x\<^isub>1 = Z"
```
``` 1316 then show ?case using Mfoldx\<^isub>2 by auto
```
``` 1317 next
```
``` 1318 fix B b u
```
``` 1319 assume "M = B + {#b#}" and "x\<^isub>1 = f b u" and Bu: "fold_msetG f Z B u"
```
``` 1320 then have MBb: "M = B + {#b#}" and x\<^isub>1: "x\<^isub>1 = f b u" by auto
```
``` 1321 show ?case
```
``` 1322 proof (rule fold_msetG.cases [OF Mfoldx\<^isub>2])
```
``` 1323 assume "M = {#}" "x\<^isub>2 = Z"
```
``` 1324 then show ?case using Mfoldx\<^isub>1 by auto
```
``` 1325 next
```
``` 1326 fix C c v
```
``` 1327 assume "M = C + {#c#}" and "x\<^isub>2 = f c v" and Cv: "fold_msetG f Z C v"
```
``` 1328 then have MCc: "M = C + {#c#}" and x\<^isub>2: "x\<^isub>2 = f c v" by auto
```
``` 1329 then have CsubM: "C < M" by simp
```
``` 1330 from MBb have BsubM: "B < M" by simp
```
``` 1331 show ?case
```
``` 1332 proof cases
```
``` 1333 assume "b=c"
```
``` 1334 then moreover have "B = C" using MBb MCc by auto
```
``` 1335 ultimately show ?thesis using Bu Cv x\<^isub>1 x\<^isub>2 CsubM IH by auto
```
``` 1336 next
```
``` 1337 assume diff: "b \<noteq> c"
```
``` 1338 let ?D = "B - {#c#}"
```
``` 1339 have cinB: "c \<in># B" and binC: "b \<in># C" using MBb MCc diff
```
``` 1340 by (auto intro: insert_noteq_member dest: sym)
```
``` 1341 have "B - {#c#} < B" using cinB by (rule mset_less_diff_self)
```
``` 1342 then have DsubM: "?D < M" using BsubM by (blast intro: order_less_trans)
```
``` 1343 from MBb MCc have "B + {#b#} = C + {#c#}" by blast
```
``` 1344 then have [simp]: "B + {#b#} - {#c#} = C"
```
``` 1345 using MBb MCc binC cinB by auto
```
``` 1346 have B: "B = ?D + {#c#}" and C: "C = ?D + {#b#}"
```
``` 1347 using MBb MCc diff binC cinB
```
``` 1348 by (auto simp: multiset_add_sub_el_shuffle)
```
``` 1349 then obtain d where Dfoldd: "fold_msetG f Z ?D d"
```
``` 1350 using fold_msetG_nonempty by iprover
```
``` 1351 then have "fold_msetG f Z B (f c d)" using cinB
```
``` 1352 by (rule Diff1_fold_msetG)
```
``` 1353 then have "f c d = u" using IH BsubM Bu by blast
```
``` 1354 moreover
```
``` 1355 have "fold_msetG f Z C (f b d)" using binC cinB diff Dfoldd
```
``` 1356 by (auto simp: multiset_add_sub_el_shuffle
```
``` 1357 dest: fold_msetG.insertI [where x=b])
```
``` 1358 then have "f b d = v" using IH CsubM Cv by blast
```
``` 1359 ultimately show ?thesis using x\<^isub>1 x\<^isub>2
```
``` 1360 by (auto simp: fun_left_comm)
```
``` 1361 qed
```
``` 1362 qed
```
``` 1363 qed
```
``` 1364 qed
```
``` 1365
```
``` 1366 lemma fold_mset_insert_aux:
```
``` 1367 "(fold_msetG f z (A + {#x#}) v) =
```
``` 1368 (\<exists>y. fold_msetG f z A y \<and> v = f x y)"
```
``` 1369 apply (rule iffI)
```
``` 1370 prefer 2
```
``` 1371 apply blast
```
``` 1372 apply (rule_tac A=A and f=f in fold_msetG_nonempty [THEN exE, standard])
```
``` 1373 apply (blast intro: fold_msetG_determ)
```
``` 1374 done
```
``` 1375
```
``` 1376 lemma fold_mset_equality: "fold_msetG f z A y \<Longrightarrow> fold_mset f z A = y"
```
``` 1377 unfolding fold_mset_def by (blast intro: fold_msetG_determ)
```
``` 1378
```
``` 1379 lemma fold_mset_insert:
```
``` 1380 "fold_mset f z (A + {#x#}) = f x (fold_mset f z A)"
```
``` 1381 apply (simp add: fold_mset_def fold_mset_insert_aux add_commute)
```
``` 1382 apply (rule the_equality)
```
``` 1383 apply (auto cong add: conj_cong
```
``` 1384 simp add: fold_mset_def [symmetric] fold_mset_equality fold_msetG_nonempty)
```
``` 1385 done
```
``` 1386
```
``` 1387 lemma fold_mset_insert_idem:
```
``` 1388 "fold_mset f z (A + {#a#}) = f a (fold_mset f z A)"
```
``` 1389 apply (simp add: fold_mset_def fold_mset_insert_aux)
```
``` 1390 apply (rule the_equality)
```
``` 1391 apply (auto cong add: conj_cong
```
``` 1392 simp add: fold_mset_def [symmetric] fold_mset_equality fold_msetG_nonempty)
```
``` 1393 done
```
``` 1394
```
``` 1395 lemma fold_mset_commute: "f x (fold_mset f z A) = fold_mset f (f x z) A"
```
``` 1396 by (induct A) (auto simp: fold_mset_insert fun_left_comm [of x])
```
``` 1397
```
``` 1398 lemma fold_mset_single [simp]: "fold_mset f z {#x#} = f x z"
```
``` 1399 using fold_mset_insert [of z "{#}"] by simp
```
``` 1400
```
``` 1401 lemma fold_mset_union [simp]:
```
``` 1402 "fold_mset f z (A+B) = fold_mset f (fold_mset f z A) B"
```
``` 1403 proof (induct A)
```
``` 1404 case empty then show ?case by simp
```
``` 1405 next
```
``` 1406 case (add A x)
```
``` 1407 have "A + {#x#} + B = (A+B) + {#x#}" by (simp add: add_ac)
```
``` 1408 then have "fold_mset f z (A + {#x#} + B) = f x (fold_mset f z (A + B))"
```
``` 1409 by (simp add: fold_mset_insert)
```
``` 1410 also have "\<dots> = fold_mset f (fold_mset f z (A + {#x#})) B"
```
``` 1411 by (simp add: fold_mset_commute[of x,symmetric] add fold_mset_insert)
```
``` 1412 finally show ?case .
```
``` 1413 qed
```
``` 1414
```
``` 1415 lemma fold_mset_fusion:
```
``` 1416 assumes "fun_left_comm g"
```
``` 1417 shows "(\<And>x y. h (g x y) = f x (h y)) \<Longrightarrow> h (fold_mset g w A) = fold_mset f (h w) A" (is "PROP ?P")
```
``` 1418 proof -
```
``` 1419 interpret fun_left_comm g by (fact assms)
```
``` 1420 show "PROP ?P" by (induct A) auto
```
``` 1421 qed
```
``` 1422
```
``` 1423 lemma fold_mset_rec:
```
``` 1424 assumes "a \<in># A"
```
``` 1425 shows "fold_mset f z A = f a (fold_mset f z (A - {#a#}))"
```
``` 1426 proof -
```
``` 1427 from assms obtain A' where "A = A' + {#a#}"
```
``` 1428 by (blast dest: multi_member_split)
```
``` 1429 then show ?thesis by simp
```
``` 1430 qed
```
``` 1431
```
``` 1432 end
```
``` 1433
```
``` 1434 text {*
```
``` 1435 A note on code generation: When defining some function containing a
```
``` 1436 subterm @{term"fold_mset F"}, code generation is not automatic. When
```
``` 1437 interpreting locale @{text left_commutative} with @{text F}, the
```
``` 1438 would be code thms for @{const fold_mset} become thms like
```
``` 1439 @{term"fold_mset F z {#} = z"} where @{text F} is not a pattern but
```
``` 1440 contains defined symbols, i.e.\ is not a code thm. Hence a separate
```
``` 1441 constant with its own code thms needs to be introduced for @{text
```
``` 1442 F}. See the image operator below.
```
``` 1443 *}
```
``` 1444
```
``` 1445
```
``` 1446 subsection {* Image *}
```
``` 1447
```
``` 1448 definition image_mset :: "('a \<Rightarrow> 'b) \<Rightarrow> 'a multiset \<Rightarrow> 'b multiset" where
```
``` 1449 "image_mset f = fold_mset (op + o single o f) {#}"
```
``` 1450
```
``` 1451 interpretation image_left_comm: fun_left_comm "op + o single o f"
```
``` 1452 proof qed (simp add: add_ac)
```
``` 1453
```
``` 1454 lemma image_mset_empty [simp]: "image_mset f {#} = {#}"
```
``` 1455 by (simp add: image_mset_def)
```
``` 1456
```
``` 1457 lemma image_mset_single [simp]: "image_mset f {#x#} = {#f x#}"
```
``` 1458 by (simp add: image_mset_def)
```
``` 1459
```
``` 1460 lemma image_mset_insert:
```
``` 1461 "image_mset f (M + {#a#}) = image_mset f M + {#f a#}"
```
``` 1462 by (simp add: image_mset_def add_ac)
```
``` 1463
```
``` 1464 lemma image_mset_union [simp]:
```
``` 1465 "image_mset f (M+N) = image_mset f M + image_mset f N"
```
``` 1466 apply (induct N)
```
``` 1467 apply simp
```
``` 1468 apply (simp add: add_assoc [symmetric] image_mset_insert)
```
``` 1469 done
```
``` 1470
```
``` 1471 lemma size_image_mset [simp]: "size (image_mset f M) = size M"
```
``` 1472 by (induct M) simp_all
```
``` 1473
```
``` 1474 lemma image_mset_is_empty_iff [simp]: "image_mset f M = {#} \<longleftrightarrow> M = {#}"
```
``` 1475 by (cases M) auto
```
``` 1476
```
``` 1477 syntax
```
``` 1478 "_comprehension1_mset" :: "'a \<Rightarrow> 'b \<Rightarrow> 'b multiset \<Rightarrow> 'a multiset"
```
``` 1479 ("({#_/. _ :# _#})")
```
``` 1480 translations
```
``` 1481 "{#e. x:#M#}" == "CONST image_mset (%x. e) M"
```
``` 1482
```
``` 1483 syntax
```
``` 1484 "_comprehension2_mset" :: "'a \<Rightarrow> 'b \<Rightarrow> 'b multiset \<Rightarrow> bool \<Rightarrow> 'a multiset"
```
``` 1485 ("({#_/ | _ :# _./ _#})")
```
``` 1486 translations
```
``` 1487 "{#e | x:#M. P#}" => "{#e. x :# {# x:#M. P#}#}"
```
``` 1488
```
``` 1489 text {*
```
``` 1490 This allows to write not just filters like @{term "{#x:#M. x<c#}"}
```
``` 1491 but also images like @{term "{#x+x. x:#M #}"} and @{term [source]
```
``` 1492 "{#x+x|x:#M. x<c#}"}, where the latter is currently displayed as
```
``` 1493 @{term "{#x+x|x:#M. x<c#}"}.
```
``` 1494 *}
```
``` 1495
```
``` 1496
```
``` 1497 subsection {* Termination proofs with multiset orders *}
```
``` 1498
```
``` 1499 lemma multi_member_skip: "x \<in># XS \<Longrightarrow> x \<in># {# y #} + XS"
```
``` 1500 and multi_member_this: "x \<in># {# x #} + XS"
```
``` 1501 and multi_member_last: "x \<in># {# x #}"
```
``` 1502 by auto
```
``` 1503
```
``` 1504 definition "ms_strict = mult pair_less"
```
``` 1505 definition "ms_weak = ms_strict \<union> Id"
```
``` 1506
```
``` 1507 lemma ms_reduction_pair: "reduction_pair (ms_strict, ms_weak)"
```
``` 1508 unfolding reduction_pair_def ms_strict_def ms_weak_def pair_less_def
```
``` 1509 by (auto intro: wf_mult1 wf_trancl simp: mult_def)
```
``` 1510
```
``` 1511 lemma smsI:
```
``` 1512 "(set_of A, set_of B) \<in> max_strict \<Longrightarrow> (Z + A, Z + B) \<in> ms_strict"
```
``` 1513 unfolding ms_strict_def
```
``` 1514 by (rule one_step_implies_mult) (auto simp add: max_strict_def pair_less_def elim!:max_ext.cases)
```
``` 1515
```
``` 1516 lemma wmsI:
```
``` 1517 "(set_of A, set_of B) \<in> max_strict \<or> A = {#} \<and> B = {#}
```
``` 1518 \<Longrightarrow> (Z + A, Z + B) \<in> ms_weak"
```
``` 1519 unfolding ms_weak_def ms_strict_def
```
``` 1520 by (auto simp add: pair_less_def max_strict_def elim!:max_ext.cases intro: one_step_implies_mult)
```
``` 1521
```
``` 1522 inductive pw_leq
```
``` 1523 where
```
``` 1524 pw_leq_empty: "pw_leq {#} {#}"
```
``` 1525 | pw_leq_step: "\<lbrakk>(x,y) \<in> pair_leq; pw_leq X Y \<rbrakk> \<Longrightarrow> pw_leq ({#x#} + X) ({#y#} + Y)"
```
``` 1526
```
``` 1527 lemma pw_leq_lstep:
```
``` 1528 "(x, y) \<in> pair_leq \<Longrightarrow> pw_leq {#x#} {#y#}"
```
``` 1529 by (drule pw_leq_step) (rule pw_leq_empty, simp)
```
``` 1530
```
``` 1531 lemma pw_leq_split:
```
``` 1532 assumes "pw_leq X Y"
```
``` 1533 shows "\<exists>A B Z. X = A + Z \<and> Y = B + Z \<and> ((set_of A, set_of B) \<in> max_strict \<or> (B = {#} \<and> A = {#}))"
```
``` 1534 using assms
```
``` 1535 proof (induct)
```
``` 1536 case pw_leq_empty thus ?case by auto
```
``` 1537 next
```
``` 1538 case (pw_leq_step x y X Y)
```
``` 1539 then obtain A B Z where
```
``` 1540 [simp]: "X = A + Z" "Y = B + Z"
```
``` 1541 and 1[simp]: "(set_of A, set_of B) \<in> max_strict \<or> (B = {#} \<and> A = {#})"
```
``` 1542 by auto
```
``` 1543 from pw_leq_step have "x = y \<or> (x, y) \<in> pair_less"
```
``` 1544 unfolding pair_leq_def by auto
```
``` 1545 thus ?case
```
``` 1546 proof
```
``` 1547 assume [simp]: "x = y"
```
``` 1548 have
```
``` 1549 "{#x#} + X = A + ({#y#}+Z)
```
``` 1550 \<and> {#y#} + Y = B + ({#y#}+Z)
```
``` 1551 \<and> ((set_of A, set_of B) \<in> max_strict \<or> (B = {#} \<and> A = {#}))"
```
``` 1552 by (auto simp: add_ac)
```
``` 1553 thus ?case by (intro exI)
```
``` 1554 next
```
``` 1555 assume A: "(x, y) \<in> pair_less"
```
``` 1556 let ?A' = "{#x#} + A" and ?B' = "{#y#} + B"
```
``` 1557 have "{#x#} + X = ?A' + Z"
```
``` 1558 "{#y#} + Y = ?B' + Z"
```
``` 1559 by (auto simp add: add_ac)
```
``` 1560 moreover have
```
``` 1561 "(set_of ?A', set_of ?B') \<in> max_strict"
```
``` 1562 using 1 A unfolding max_strict_def
```
``` 1563 by (auto elim!: max_ext.cases)
```
``` 1564 ultimately show ?thesis by blast
```
``` 1565 qed
```
``` 1566 qed
```
``` 1567
```
``` 1568 lemma
```
``` 1569 assumes pwleq: "pw_leq Z Z'"
```
``` 1570 shows ms_strictI: "(set_of A, set_of B) \<in> max_strict \<Longrightarrow> (Z + A, Z' + B) \<in> ms_strict"
```
``` 1571 and ms_weakI1: "(set_of A, set_of B) \<in> max_strict \<Longrightarrow> (Z + A, Z' + B) \<in> ms_weak"
```
``` 1572 and ms_weakI2: "(Z + {#}, Z' + {#}) \<in> ms_weak"
```
``` 1573 proof -
```
``` 1574 from pw_leq_split[OF pwleq]
```
``` 1575 obtain A' B' Z''
```
``` 1576 where [simp]: "Z = A' + Z''" "Z' = B' + Z''"
```
``` 1577 and mx_or_empty: "(set_of A', set_of B') \<in> max_strict \<or> (A' = {#} \<and> B' = {#})"
```
``` 1578 by blast
```
``` 1579 {
```
``` 1580 assume max: "(set_of A, set_of B) \<in> max_strict"
```
``` 1581 from mx_or_empty
```
``` 1582 have "(Z'' + (A + A'), Z'' + (B + B')) \<in> ms_strict"
```
``` 1583 proof
```
``` 1584 assume max': "(set_of A', set_of B') \<in> max_strict"
```
``` 1585 with max have "(set_of (A + A'), set_of (B + B')) \<in> max_strict"
```
``` 1586 by (auto simp: max_strict_def intro: max_ext_additive)
```
``` 1587 thus ?thesis by (rule smsI)
```
``` 1588 next
```
``` 1589 assume [simp]: "A' = {#} \<and> B' = {#}"
```
``` 1590 show ?thesis by (rule smsI) (auto intro: max)
```
``` 1591 qed
```
``` 1592 thus "(Z + A, Z' + B) \<in> ms_strict" by (simp add:add_ac)
```
``` 1593 thus "(Z + A, Z' + B) \<in> ms_weak" by (simp add: ms_weak_def)
```
``` 1594 }
```
``` 1595 from mx_or_empty
```
``` 1596 have "(Z'' + A', Z'' + B') \<in> ms_weak" by (rule wmsI)
```
``` 1597 thus "(Z + {#}, Z' + {#}) \<in> ms_weak" by (simp add:add_ac)
```
``` 1598 qed
```
``` 1599
```
``` 1600 lemma empty_idemp: "{#} + x = x" "x + {#} = x"
```
``` 1601 and nonempty_plus: "{# x #} + rs \<noteq> {#}"
```
``` 1602 and nonempty_single: "{# x #} \<noteq> {#}"
```
``` 1603 by auto
```
``` 1604
```
``` 1605 setup {*
```
``` 1606 let
```
``` 1607 fun msetT T = Type (@{type_name multiset}, [T]);
```
``` 1608
```
``` 1609 fun mk_mset T [] = Const (@{const_abbrev Mempty}, msetT T)
```
``` 1610 | mk_mset T [x] = Const (@{const_name single}, T --> msetT T) \$ x
```
``` 1611 | mk_mset T (x :: xs) =
```
``` 1612 Const (@{const_name plus}, msetT T --> msetT T --> msetT T) \$
```
``` 1613 mk_mset T [x] \$ mk_mset T xs
```
``` 1614
```
``` 1615 fun mset_member_tac m i =
```
``` 1616 (if m <= 0 then
```
``` 1617 rtac @{thm multi_member_this} i ORELSE rtac @{thm multi_member_last} i
```
``` 1618 else
```
``` 1619 rtac @{thm multi_member_skip} i THEN mset_member_tac (m - 1) i)
```
``` 1620
```
``` 1621 val mset_nonempty_tac =
```
``` 1622 rtac @{thm nonempty_plus} ORELSE' rtac @{thm nonempty_single}
```
``` 1623
```
``` 1624 val regroup_munion_conv =
```
``` 1625 Function_Lib.regroup_conv @{const_abbrev Mempty} @{const_name plus}
```
``` 1626 (map (fn t => t RS eq_reflection) (@{thms add_ac} @ @{thms empty_idemp}))
```
``` 1627
```
``` 1628 fun unfold_pwleq_tac i =
```
``` 1629 (rtac @{thm pw_leq_step} i THEN (fn st => unfold_pwleq_tac (i + 1) st))
```
``` 1630 ORELSE (rtac @{thm pw_leq_lstep} i)
```
``` 1631 ORELSE (rtac @{thm pw_leq_empty} i)
```
``` 1632
```
``` 1633 val set_of_simps = [@{thm set_of_empty}, @{thm set_of_single}, @{thm set_of_union},
```
``` 1634 @{thm Un_insert_left}, @{thm Un_empty_left}]
```
``` 1635 in
```
``` 1636 ScnpReconstruct.multiset_setup (ScnpReconstruct.Multiset
```
``` 1637 {
```
``` 1638 msetT=msetT, mk_mset=mk_mset, mset_regroup_conv=regroup_munion_conv,
```
``` 1639 mset_member_tac=mset_member_tac, mset_nonempty_tac=mset_nonempty_tac,
```
``` 1640 mset_pwleq_tac=unfold_pwleq_tac, set_of_simps=set_of_simps,
```
``` 1641 smsI'= @{thm ms_strictI}, wmsI2''= @{thm ms_weakI2}, wmsI1= @{thm ms_weakI1},
```
``` 1642 reduction_pair= @{thm ms_reduction_pair}
```
``` 1643 })
```
``` 1644 end
```
``` 1645 *}
```
``` 1646
```
``` 1647
```
``` 1648 subsection {* Legacy theorem bindings *}
```
``` 1649
```
``` 1650 lemmas multi_count_eq = multiset_ext_iff [symmetric]
```
``` 1651
```
``` 1652 lemma union_commute: "M + N = N + (M::'a multiset)"
```
``` 1653 by (fact add_commute)
```
``` 1654
```
``` 1655 lemma union_assoc: "(M + N) + K = M + (N + (K::'a multiset))"
```
``` 1656 by (fact add_assoc)
```
``` 1657
```
``` 1658 lemma union_lcomm: "M + (N + K) = N + (M + (K::'a multiset))"
```
``` 1659 by (fact add_left_commute)
```
``` 1660
```
``` 1661 lemmas union_ac = union_assoc union_commute union_lcomm
```
``` 1662
```
``` 1663 lemma union_right_cancel: "M + K = N + K \<longleftrightarrow> M = (N::'a multiset)"
```
``` 1664 by (fact add_right_cancel)
```
``` 1665
```
``` 1666 lemma union_left_cancel: "K + M = K + N \<longleftrightarrow> M = (N::'a multiset)"
```
``` 1667 by (fact add_left_cancel)
```
``` 1668
```
``` 1669 lemma multi_union_self_other_eq: "(A::'a multiset) + X = A + Y \<Longrightarrow> X = Y"
```
``` 1670 by (fact add_imp_eq)
```
``` 1671
```
``` 1672 lemma mset_less_trans: "(M::'a multiset) < K \<Longrightarrow> K < N \<Longrightarrow> M < N"
```
``` 1673 by (fact order_less_trans)
```
``` 1674
```
``` 1675 lemma multiset_inter_commute: "A #\<inter> B = B #\<inter> A"
```
``` 1676 by (fact inf.commute)
```
``` 1677
```
``` 1678 lemma multiset_inter_assoc: "A #\<inter> (B #\<inter> C) = A #\<inter> B #\<inter> C"
```
``` 1679 by (fact inf.assoc [symmetric])
```
``` 1680
```
``` 1681 lemma multiset_inter_left_commute: "A #\<inter> (B #\<inter> C) = B #\<inter> (A #\<inter> C)"
```
``` 1682 by (fact inf.left_commute)
```
``` 1683
```
``` 1684 lemmas multiset_inter_ac =
```
``` 1685 multiset_inter_commute
```
``` 1686 multiset_inter_assoc
```
``` 1687 multiset_inter_left_commute
```
``` 1688
```
``` 1689 lemma mult_less_not_refl:
```
``` 1690 "\<not> M \<subset># (M::'a::order multiset)"
```
``` 1691 by (fact multiset_order.less_irrefl)
```
``` 1692
```
``` 1693 lemma mult_less_trans:
```
``` 1694 "K \<subset># M ==> M \<subset># N ==> K \<subset># (N::'a::order multiset)"
```
``` 1695 by (fact multiset_order.less_trans)
```
``` 1696
```
``` 1697 lemma mult_less_not_sym:
```
``` 1698 "M \<subset># N ==> \<not> N \<subset># (M::'a::order multiset)"
```
``` 1699 by (fact multiset_order.less_not_sym)
```
``` 1700
```
``` 1701 lemma mult_less_asym:
```
``` 1702 "M \<subset># N ==> (\<not> P ==> N \<subset># (M::'a::order multiset)) ==> P"
```
``` 1703 by (fact multiset_order.less_asym)
```
``` 1704
```
``` 1705 ML {*
```
``` 1706 fun multiset_postproc _ maybe_name all_values (T as Type (_, [elem_T]))
```
``` 1707 (Const _ \$ t') =
```
``` 1708 let
```
``` 1709 val (maybe_opt, ps) =
```
``` 1710 Nitpick_Model.dest_plain_fun t' ||> op ~~
```
``` 1711 ||> map (apsnd (snd o HOLogic.dest_number))
```
``` 1712 fun elems_for t =
```
``` 1713 case AList.lookup (op =) ps t of
```
``` 1714 SOME n => replicate n t
```
``` 1715 | NONE => [Const (maybe_name, elem_T --> elem_T) \$ t]
```
``` 1716 in
```
``` 1717 case maps elems_for (all_values elem_T) @
```
``` 1718 (if maybe_opt then [Const (Nitpick_Model.unrep (), elem_T)]
```
``` 1719 else []) of
```
``` 1720 [] => Const (@{const_name zero_class.zero}, T)
```
``` 1721 | ts => foldl1 (fn (t1, t2) =>
```
``` 1722 Const (@{const_name plus_class.plus}, T --> T --> T)
```
``` 1723 \$ t1 \$ t2)
```
``` 1724 (map (curry (op \$) (Const (@{const_name single},
```
``` 1725 elem_T --> T))) ts)
```
``` 1726 end
```
``` 1727 | multiset_postproc _ _ _ _ t = t
```
``` 1728 *}
```
``` 1729
```
``` 1730 declaration {*
```
``` 1731 Nitpick_Model.register_term_postprocessor @{typ "'a multiset"}
```
``` 1732 multiset_postproc
```
``` 1733 *}
```
``` 1734
```
``` 1735 end
```
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# Properties
Label 49.2.c Level $49$ Weight $2$ Character orbit 49.c Rep. character $\chi_{49}(18,\cdot)$ Character field $\Q(\zeta_{3})$ Dimension $2$ Newform subspaces $1$ Sturm bound $9$ Trace bound $0$
# Related objects
## Defining parameters
Level: $$N$$ $$=$$ $$49 = 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 49.c (of order $$3$$ and degree $$2$$) Character conductor: $$\operatorname{cond}(\chi)$$ $$=$$ $$7$$ Character field: $$\Q(\zeta_{3})$$ Newform subspaces: $$1$$ Sturm bound: $$9$$ Trace bound: $$0$$
## Dimensions
The following table gives the dimensions of various subspaces of $$M_{2}(49, [\chi])$$.
Total New Old
Modular forms 18 10 8
Cusp forms 2 2 0
Eisenstein series 16 8 8
## Trace form
$$2 q - q^{2} + q^{4} - 6 q^{8} + 3 q^{9} + O(q^{10})$$ $$2 q - q^{2} + q^{4} - 6 q^{8} + 3 q^{9} - 4 q^{11} + q^{16} + 3 q^{18} + 8 q^{22} - 8 q^{23} + 5 q^{25} + 4 q^{29} - 5 q^{32} + 6 q^{36} + 6 q^{37} - 24 q^{43} + 4 q^{44} - 8 q^{46} - 10 q^{50} + 10 q^{53} - 2 q^{58} + 14 q^{64} - 4 q^{67} + 32 q^{71} - 9 q^{72} + 6 q^{74} - 8 q^{79} - 9 q^{81} + 12 q^{86} + 12 q^{88} - 16 q^{92} - 24 q^{99} + O(q^{100})$$
## Decomposition of $$S_{2}^{\mathrm{new}}(49, [\chi])$$ into newform subspaces
Label Dim $A$ Field CM Traces $q$-expansion
$a_{2}$ $a_{3}$ $a_{5}$ $a_{7}$
49.2.c.a $2$ $0.391$ $$\Q(\sqrt{-3})$$ $$\Q(\sqrt{-7})$$ $$-1$$ $$0$$ $$0$$ $$0$$ $$q-\zeta_{6}q^{2}+(1-\zeta_{6})q^{4}-3q^{8}+3\zeta_{6}q^{9}+\cdots$$
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4 added 120 characters in body edited Jul 6 at 20:17 Aakash Gupta 322 bronze badges I am trying to find coefficient of an exponent term in an expression. For example, consider tt = E^(I*T0*(ω+σ)); Doing Coefficient[tt, E^(I*T0*(ω+σ))] (* 1 *) gives 1. However, Coefficient[tt, E^(I*T0*ω)] (* 0 *) this returns 0. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents. PS: The expression tt is just a dummy expression, my original expression is really big and contains a lot of terms. I am trying to find coefficient of an exponent term in an expression. For example, consider tt = E^(I*T0*(ω+σ)); Doing Coefficient[tt, E^(I*T0*(ω+σ))] (* 1 *) gives 1. However, Coefficient[tt, E^(I*T0*ω)] (* 0 *) this returns 0. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents. I am trying to find coefficient of an exponent term in an expression. For example, consider tt = E^(I*T0*(ω+σ)); Doing Coefficient[tt, E^(I*T0*(ω+σ))] (* 1 *) gives 1. However, Coefficient[tt, E^(I*T0*ω)] (* 0 *) this returns 0. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents. PS: The expression tt is just a dummy expression, my original expression is really big and contains a lot of terms. 3 deleted 51 characters in body edited Jul 6 at 18:39 Roman 15.9k11 gold badge2222 silver badges5555 bronze badges I am trying to find coefficient of an exponent term in an expression. For example, consider in: $$tt = e^{i*T0*(\omega + \sigma)};$$ tt = E^(I*T0*(ω+σ)); Doing in: $$\text{Coefficient}[tt, e^{i*T0*(\omega + \sigma)}]$$ out: 1 Coefficient[tt, E^(I*T0*(ω+σ))] (* 1 *) gives 1. However, in: $$\text{Coefficient}[tt, e^{i*T0*\omega}]$$ out: 0 Coefficient[tt, E^(I*T0*ω)] (* 0 *) this returns $$0$$0. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpandPowerExpand will work, but it also did not. I couldn't find the command to expand exponents . I am trying to find coefficient of an exponent term in an expression. For example, consider in: $$tt = e^{i*T0*(\omega + \sigma)};$$ Doing in: $$\text{Coefficient}[tt, e^{i*T0*(\omega + \sigma)}]$$ out: 1 gives 1. However, in: $$\text{Coefficient}[tt, e^{i*T0*\omega}]$$ out: 0 this returns $$0$$. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents I am trying to find coefficient of an exponent term in an expression. For example, consider tt = E^(I*T0*(ω+σ)); Doing Coefficient[tt, E^(I*T0*(ω+σ))] (* 1 *) gives 1. However, Coefficient[tt, E^(I*T0*ω)] (* 0 *) this returns 0. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents. 2 edited body edited Jul 6 at 17:37 Aakash Gupta 322 bronze badges I am trying to find coefficient of an exponent term in an expression. For example, consider in: $$tt = e^{i*T0*(\omega + \sigma)};$$ Doing in: $$\text{Coefficient}[tt, e^{i*T0*(\omega + \sigma)}]$$ out: 1 gives 1. However, in: $$\text{Coefficient}[tt, e^{i*T0*\omega}]$$ out: 0 this returns $$0$$. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but isit also did not. I couldn't find the command to expand exponents I am trying to find coefficient of an exponent term in an expression. For example, consider in: $$tt = e^{i*T0*(\omega + \sigma)};$$ Doing in: $$\text{Coefficient}[tt, e^{i*T0*(\omega + \sigma)}]$$ out: 1 gives 1. However, in: $$\text{Coefficient}[tt, e^{i*T0*\omega}]$$ out: 0 this returns $$0$$. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but is also did not. I couldn't find the command to expand exponents I am trying to find coefficient of an exponent term in an expression. For example, consider in: $$tt = e^{i*T0*(\omega + \sigma)};$$ Doing in: $$\text{Coefficient}[tt, e^{i*T0*(\omega + \sigma)}]$$ out: 1 gives 1. However, in: $$\text{Coefficient}[tt, e^{i*T0*\omega}]$$ out: 0 this returns $$0$$. How do I obtain $$e^{i*T0*\sigma}$$ as output instead of $$0$$? I thought PowerExpand will work, but it also did not. I couldn't find the command to expand exponents 1 asked Jul 6 at 17:29 Aakash Gupta 322 bronze badges
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# How do you adjust closing price when offering stock dividends?
Contents
The adjusted closing price shows the stock’s value after posting a dividend. For example, if a share with a closing price of \$100 paid a \$5 dividend per share, the adjusted closing price would be \$95 in order to account for the newly reduced value caused by the dividend.
## Does adjusted close account for dividends?
1. Adjusted closing price for dividends
• Adjusted closing price for dividends. A dividend reduces the value of stocks since it is recognised as capital lost from the company. …
• Adjusted closing price for dividends. A dividend reduces the value of stocks since it is recognised as capital lost from the company.
## How do you change the closing price of a dividend?
To calculate the adjustment factor, we subtract the \$2.00 dividend from Monday’s closing price (\$40.00 – \$2.00 = \$38.00). Then, we divide 38.00 by 40.00 to determine the dividend adjustment in percentage terms. The result is 0.95. Lastly, we multiply all historical prices prior to the dividend by the factor of 0.95.
## Why are stock prices adjusted for dividends?
The reason for the adjustment is that the amount paid out in dividends no longer belongs to the company, and this is reflected by a reduction in the company’s market cap. … Historical prices stored on some public websites also adjust the past prices of the stock downward by the dividend amount.
## How do you calculate adjusted closing price?
If a company announces a dividend payment, you’d subtract the amount of the dividend from the share price to calculate the adjusted closing price. Let’s say a company’s closing price is \$100 per share and it distributes a dividend of \$2 per share. You’d subtract the \$2 dividend from the closing price of \$100.
## Should I use closing price or adjusted closing price?
Overall, the adjusted closing price will give you a better idea of the overall value of the stock and help you make informed decisions about buying and selling, while the closing stock price will tell you the exact cash value of a share of stock at the end of the trading day.
## What is the adjusted daily closing stock price?
What Is the Adjusted Closing Price? The adjusted closing price amends a stock’s closing price to reflect that stock’s value after accounting for any corporate actions. It is often used when examining historical returns or doing a detailed analysis of past performance.
Adjusted close is the closing price after adjustments for all applicable splits and dividend distributions. Data is adjusted using appropriate split and dividend multipliers, adhering to Center for Research in Security Prices (CRSP) standards.
## What is split adjustment factor?
Split adjusted refers to how historical stock prices are portrayed in the event that a company has issued a stock split for its shares in the past. When reviewing price data, whether in tables or on charts, split adjusted data will reflect the increase in price as if there had been no split in the shares.
THIS IS FUN: How do you invest in yourself?
## Do dividends go down when stock price goes down?
The final long-winded answer: You will often see companies cut their dividends when there is a severe economic crash, but not in reaction to a market correction. Since dividends are not a function of stock price, market fluctuations and stock price fluctuations on their own do not affect a company’s dividend payments.
## How do you calculate stock price after dividend?
To figure the new average price after a stock dividend, convert the percentage of the stock dividend to a decimal by dividing by 100. Then, add it to 1. Finally, divide the initial stock price by the result to find the new stock price.
## How long do you have to hold stock to get dividend?
In order to receive the preferred 15% tax rate on dividends, you must hold the stock for a minimum number of days. That minimum period is 61 days within the 121-day period surrounding the ex-dividend date. The 121-day period begins 60 days before the ex-dividend date.
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CC-MAIN-2022-05
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latest
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http://mathhelpforum.com/new-users/206798-please-help-derivatives-functions-print.html
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• Nov 5th 2012, 06:37 AM
ZKR
Hello,
I`m a total noob in math and barely remember anything from high school past one variable equations.
If someone can help me with several issues here, I`d be most appreciated.
It is given that the second derivative of a function is g"(x) = 4
and that g`(0) = 0 and g(0) = 4.
What`s the function g(x) ?
Any help is most appreciated.
Thanks
• Nov 5th 2012, 06:46 AM
fkf
We have
g''(x) = 4, g'(0) = 0 and g(0) = 4
Start with the function g''(x) = 4 and find its primitive function
g''(x) = 4 => g'(x) = 4x+C
Since g'(0) = 4 we have
g'(x) = 4x+C => 4*0+C = 0 <=> C = 0
Finding the primitive function of f'(x) gives us
g'(x) = 4x => g(x) = 2x^2+D
Since g(0) = 4 we have
g(x) = 2x^2+D => 4 = 2*0^2+D <=> D = 4
Thus, g(x) = 2x^2+4
• Nov 5th 2012, 06:48 AM
skeeter
\$\displaystyle g''(x) = 4\$
antiderivative ...
\$\displaystyle g'(x) = 4x + C\$
\$\displaystyle g'(0) = 0 \implies C = 0\$
\$\displaystyle g'(x) = 4x\$
antiderivative ...
\$\displaystyle g(x) = 2x^2 + C\$
\$\displaystyle g(0) = 4 \implies C = 4\$
\$\displaystyle g(x) = 2x^2 + 4\$
next time, post calculus questions in the calculus forum ... the lobby is just a place for introductions.
• Nov 5th 2012, 06:50 AM
HallsofIvy
Quote:
Originally Posted by fkf
We have
g''(x) = 4, g'(0) = 4 and g(0) = 4
The problem gave g'(0)= 0.
Quote:
Start with the function g''(x) = 4 and find its primitive function
g''(x) = 4 => g'(x) = 4x+C
Since g'(0) = 4 we have
g'(x) = 4x+C => 4*0+C = 4 <=> C = 4
Finding the primitive function of f'(x) gives us
g'(x) = 4x+4 => g(x) = 2x^2+4x+D
Since g(0) = 4 we have
g(x) = 2x^2+4x+D => 4 = 2*0^2+4*0+D <=> D = 4
Thus, g(x) = 2x^2+4x+4
• Nov 5th 2012, 07:03 AM
fkf
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http://oeis.org/A005881
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This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A005881 Theta series of planar hexagonal lattice (A2) with respect to edge. (Formerly M0187) 3
2, 2, 0, 4, 2, 0, 4, 0, 0, 4, 4, 0, 2, 2, 0, 4, 0, 0, 4, 4, 0, 4, 0, 0, 6, 0, 0, 0, 4, 0, 4, 4, 0, 4, 0, 0, 4, 2, 0, 4, 2, 0, 0, 0, 0, 8, 4, 0, 4, 0, 0, 4, 0, 0, 4, 4, 0, 0, 4, 0, 2, 0, 0, 4, 4, 0, 8, 0, 0, 4, 0, 0, 0, 6, 0, 4, 0, 0, 4, 0, 0, 4, 0, 0, 6, 4, 0, 4, 0, 0, 4, 4, 0, 0, 4, 0, 4, 0, 0, 4, 4, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Also number of ways of writing n as the sum of a triangular number and three times a triangular number. The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice. Given g.f. A(x), then q^(1/2)*A(q) is denoted phi_1(z) where q=exp(Pi*i*z) in Conway and Sloane. Cubic AGM theta functions: a(q) (see A004016), b(q) (A005928), c(q) (A005882). REFERENCES N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Antti Karttunen, Table of n, a(n) for n = 0..10000 J. H. Conway and N. J. A. Sloane, Sphere Packings, Lattices and Groups, Springer-Verlag, p. 103. see Equ. (13). M. D. Hirschhorn, The number of representations of a number by various forms, Discrete Mathematics 298 (2005), 205-211. G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2 N. J. A. Sloane, Theta series and magic numbers for diamond and certain ionic crystal structures, J. Math. Phys. 28 (1987), 1653-1657. N. J. A. Sloane and B. K. Teo, Theta series and magic numbers for close-packed spherical clusters, J. Chem. Phys. 83 (1985) 6520-6534. FORMULA Expansion of q^(-1) * (a(q) - a(q^4)) / 3 in powers of q^2 where a() is a cubic AGM theta function. - Michael Somos, Nov 05 2006 a(n) = 2*A033762(n). MAPLE d:=proc(r, m, n) local i, t1; t1:=0; for i from 1 to n do if n mod i = 0 and i-r mod m = 0 then t1:=t1+1; fi; od: t1; end; [seq(2*(d(1, 3, 2*n+1)-d(2, 3, 2*n+1)), n=0..120)]; MATHEMATICA a[n_] := 2*DivisorSum[2n+1, KroneckerSymbol[-12, #]*Mod[(2n+1)/#, 2]& ]; Table[a[n], {n, 0, 105}] (* Jean-François Alcover, Dec 02 2015, adapted from PARI *) PROG (PARI) {a(n) = if( n<0, 0, n = 2*n + 1; 2 * sumdiv(n, d, kronecker( -12, d) * (n/d%2)))}; /* Michael Somos, Nov 05 2006 */ (PARI) {a(n) = if( n<0, 0, n = 8*n + 4; 2 * sum(j=1, sqrtint(n\3), (j%2) * issquare(n - 3*j^2)))}; /* Michael Somos, Nov 05 2006 */ CROSSREFS Cf. A033762. Sequence in context: A286123 A253243 A201396 * A218875 A218869 A144458 Adjacent sequences: A005878 A005879 A005880 * A005882 A005883 A005884 KEYWORD nonn AUTHOR STATUS approved
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Last modified January 23 02:40 EST 2019. Contains 319365 sequences. (Running on oeis4.)
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| 3.203125
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CC-MAIN-2019-04
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latest
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https://www.jsmount.com/insert-node-in-linked-list/
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Insert node in Linked list | Linked list insertion | insertion in a linked list in data structure | Linked list insertion algorithm | How to insert a node in a linked list | Linked list insertion algorithm | Singly linked list node insertion | insert node in linked list program
### 💡What is a Linked List?
Linked lists are linear data structures that store information in discrete nodes, or individual items. These nodes store the information as well as a pointer to the following node in the list.
A linked list is a dynamic data structure so it can grow and shrink at runtime by allocating and deallocating memory. So there is no need to give the initial size of the linked list.
### 💡Difference between Array and LinkedList?
Array stores the data elements in a contiguous memory zone. Elements are stored in Linked Lists at random, or we could say anyplace in the memory area.
In Array, The elements are not dependent on each other. In the linked list, The data elements are dependent on each other.
In an array, When performing any action, such as insertion or deletion, it takes more time.
When performing any operation, such as insertion or deletion, the linked list requires less time.
``Structure: [1, next] -> [2, next] -> [3, next] -> null``
### 💻 Create a Node Class
```class Node {
value;
next;
constructor(value) {
this.value = value;
this.next = null; // initially next is null
}
}
console.log(head) // Node {value: 1, next: Node}
```
Insertion at beginning
```function insertionAtStart(head, value) {
let newNode = new Node(value);
}```
``````var head = new Node(1); // initial list
console.log(list1)
console.log(list2)``````
``````{
"value": 3,
"next": {
"value": 1,
"next": {
"value": 2,
"next": null
}
}
}``````
Insertion at End
```function insertionAtEnd(head, value) {
while (current.next != null) { // finding last element
current = current.next;
}
current.next = new Node(value); // assign new node to the last element next.
}
Insertion a element at start or end in empty linked list
```function insert(head, value) {
}
}
insert(null, 1);```
Complete program of insertion at end
```// TC = O(n)
} else {
while (current.next != null) { // finding last element
current = current.next;
}
current.next = new Node(value); // assign new node to the last element next.
}
}```
Insert at kth position
```// TC = O(n)
if (k == 0) { // means new element will become Head
let newNode = new Node(value);
} else {
let i = 0; // 0 based indexing
// If I do (i < k) - it will reach to kth index while I have to stop at (k-1)
while (i < k - 1 && curr.next != null) {
curr = curr.next;
i++;
}
let newNode = new Node(value);
newNode.next = curr.next;
curr.next = newNode;
}
}```
``````
insertAtK(head, 'elephant', 3); // insert at 3rd position
{
"value": "apple",
"next": {
"value": "banana",
"next": {
* "value": "cat", -- This is (k - 1) position
"next": {
* "value": "elephant", -- // this is newly added item at kth position
"next": {
"value": "dog",
"next": null
}
}
}
}
} ``````
```// TC = O(n)
// head here represents first item of list.
let items = '';
while (curr.next != null) {
items = items + curr.value + ' ';
curr = curr.next;
}
items = items + curr.data + ' ';
console.log(items)
}
printList(head); // apple banana cat elephant
printList(new Node('pen')) // pen```
#### ❓What is diff in an array and linked list in case of insertion at the kth position while both take O(n) time complexity?
Insertion at the kth position in Array takes O(n)
Insertion at Kth position in the Linked list takes O(n)
However, linked lists are still preferred. Because when we add an item to an array at the kth position, we must move all items following that index to the next available location and copy their data, that process of copying and relocation takes a lot of effort.
While in the Linked list, everything is a Connection, which means two nodes are joined with only a connection. Therefore, if we add a new item, we simply update the connections between the previous and subsequent items. Everything is still as it was, in the same places. That makes a lot of sense.
Insert node in Linked list | Linked list insertion | insertion in a linked list in data structure | Linked list insertion algorithm | How to insert a node in a linked list | Linked list insertion algorithm | Singly linked list node insertion | insert node in linked list program | How to insert a node in a linked list
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http://physexams.com/flashcard/Relation_between_electric_force_and_electric_field/3
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# Blog & News
Electrostatic
## Relation between electric force and electric field
To find the electric field at each point in vicinity of a charged particle $q$, place a small and insignificant positive charge, called test charge, $q_0$ at that point and then measure the force $\vec{F}$acting on it. The electric field $\overrightarrow{E}$ due to that charged point charge $q$ is defined as
$\vec{E}=\frac{\vec{F}}{q_0}$
Electric field is a vector quantity that its magnitude is $E=F/q_0$ and its direction is in the same direction as the force acting on the test charge. In the other words, electric field points in opposite direction of the electric force acting on a negative charged particle.
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https://cs.stackexchange.com/questions/120606/collision-entropy-definition
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# Collision entropy definition
The collision entropy is defined as the Renyi entropy for the case $$\alpha = 2$$. It is given by
$$\mathrm{H}_{2}(X)=-\log \sum_{i=1}^{n} p_{i}^{2} \tag{1}$$
Take two random variables $$X$$ and $$X'$$ which follow the same probability distribution. The probability of a collision is then simply $$P_{\rm coll} = \sum_{i=1}^{n} p_{i}^{2}$$. I would then expect that we say the collision entropy is just $$H(P_{\rm coll})$$ i.e.
$$-\left(\sum_{i=1}^{n} p_{i}^{2}\right)\log\left(\sum_{i=1}^{n} p_{i}^{2}\right) - \left(1 -\sum_{i=1}^{n} p_{i}^{2}\right)\log\left(1-\sum_{i=1}^{n} p_{i}^{2}\right)$$
This is in analogy with the binary entropy but with the probability replaced with the probability of a collision.
What is the motivation behind choosing $$(1)$$ to be the definition of collision entropy?
• When the distribution is uniform, all Renyi entropies are the same. In contrast, your proposal is always between 0 and 1. Feb 12, 2020 at 15:01
When $$X$$ is distributed uniformly over a domain of size $$n$$, then Shannon entropy, collision entropy and min-entropy are all equal to $$\log n$$. In a sense, all of these parameters measure the amount of uncertainty in $$X$$.
In contrast, your proposed definition is always between $$0$$ and $$1$$, tending to zero as $$X$$ gets more unpredictable. This is quite different from other notions of entropy, in which zero stands for predictable.
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https://digitalcommons.cwu.edu/source/2013/oralpresentations/87/
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| 310,998,525
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## Oral Presentations
#### Title
An Estimation of the Proportion of Abundant Numbers
#### Document Type
Oral Presentation
SURC 202
16-5-2013
16-5-2013
#### Abstract
For 2,500 years mathematicians have studied the “Sum of Divisors” function σ(n). To give an example, σ(12) = 28 because the divisors of 12 are 1, 2, 3, 4, 6 and 12, and the sum of the divisors of 12 is given by 1+2+3+4+6+12 = 28. Now depending on the value of σ(n) – n we classify these numbers as either deficient, perfect or abundant. Perfect numbers are numbers for which σ(n) – n = n, deficient numbers are numbers for which σ(n) – n < n and abundant numbers are numbers for which σ(n) – n > n. The question I am asking is: what is the proportion of abundant numbers? The best current bounds for this proportion are .2476171 and .2476475. I have written code in Java and used it to generate data to try to improve this estimation using statistics. This presentation will cover what an abundant number is, how I generated my data, how I used statistics to try to close the gap for the estimation of the proportion of abundant numbers, and what the result were.
Dominic Klyve
Mathematics
#### Share
COinS
May 16th, 1:30 PM May 16th, 1:50 PM
An Estimation of the Proportion of Abundant Numbers
SURC 202
For 2,500 years mathematicians have studied the “Sum of Divisors” function σ(n). To give an example, σ(12) = 28 because the divisors of 12 are 1, 2, 3, 4, 6 and 12, and the sum of the divisors of 12 is given by 1+2+3+4+6+12 = 28. Now depending on the value of σ(n) – n we classify these numbers as either deficient, perfect or abundant. Perfect numbers are numbers for which σ(n) – n = n, deficient numbers are numbers for which σ(n) – n < n and abundant numbers are numbers for which σ(n) – n > n. The question I am asking is: what is the proportion of abundant numbers? The best current bounds for this proportion are .2476171 and .2476475. I have written code in Java and used it to generate data to try to improve this estimation using statistics. This presentation will cover what an abundant number is, how I generated my data, how I used statistics to try to close the gap for the estimation of the proportion of abundant numbers, and what the result were.
| 605
| 2,249
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| 4.1875
| 4
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CC-MAIN-2021-17
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longest
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en
| 0.852064
|
http://www.numbersaplenty.com/12356789
| 1,547,694,864,000,000,000
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text/html
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crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00248.warc.gz
| 363,321,065
| 3,422
|
Search a number
12356789 is a prime number
BaseRepresentation
bin101111001000…
…110010110101
3212020210022212
4233020302311
511130404124
61120503205
7210013424
oct57106265
925223285
1012356789
116a7a925
12417ab05
132738513
1418d92bb
15114140e
hexbc8cb5
12356789 has 2 divisors, whose sum is σ = 12356790. Its totient is φ = 12356788.
The previous prime is 12356777. The next prime is 12356801. The reversal of 12356789 is 98765321.
It is a balanced prime because it is at equal distance from previous prime (12356777) and next prime (12356801).
It can be written as a sum of positive squares in only one way, i.e., 11336689 + 1020100 = 3367^2 + 1010^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-12356789 is a prime.
It is a plaindrome in base 10.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12356749) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6178394 + 6178395.
It is an arithmetic number, because the mean of its divisors is an integer number (6178395).
Almost surely, 212356789 is an apocalyptic number.
It is an amenable number.
12356789 is a deficient number, since it is larger than the sum of its proper divisors (1).
12356789 is an equidigital number, since it uses as much as digits as its factorization.
12356789 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 90720, while the sum is 41.
The square root of 12356789 is about 3515.2224680666. Note that the first 3 decimals coincide. The cubic root of 12356789 is about 231.1897331141.
The spelling of 12356789 in words is "twelve million, three hundred fifty-six thousand, seven hundred eighty-nine".
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Cody
# Problem 44443. Matrix to vector transformation
Solution 1426783
Submitted on 28 Jan 2018 by Jihye Sofia Seo
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3;4 5 6;7 8 9]; y_correct = [1 3 5 7 9 8 6 4 2]' assert(isequal(vectorizeit(x),y_correct))
y_correct = 1 3 5 7 9 8 6 4 2 x = 1 2 3 4 5 6 7 8 9 even = 2 4 6 8 odd = 1 3 5 7 9 even = 8 6 4 2 odd = 1 3 5 7 9
2 Pass
x = [5 8 9 ;7 22 9]; y_correct = [5 7 9 22 8]' assert(isequal(vectorizeit(x),y_correct))
y_correct = 5 7 9 22 8 x = 5 7 8 9 22 even = 8 22 odd = 5 7 9 even = 22 8 odd = 5 7 9
3 Pass
x = [88 99; 0 64]; y_correct = [99 88 64 0]' assert(isequal(vectorizeit(x),y_correct))
y_correct = 99 88 64 0 x = 0 64 88 99 even = 0 64 88 odd = 99 even = 88 64 0 odd = 99
4 Pass
x = [0 0; 1 1]; y_correct = [1 0]' assert(isequal(vectorizeit(x),y_correct))
y_correct = 1 0 x = 0 1 even = 0 odd = 1 even = 0 odd = 1
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Appendix E
# Appendix E - Axia College Material Appendix E Fueling Up...
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Axia College Material Appendix E Fueling Up Motorists often complain about rising gas prices. Some motorists purchase fuel-efficient vehicles and participate in trip reduction plans, such as carpooling and using alternative transportation. Other drivers try to drive only when necessary. Application Practice Answer the following questions. Use Equation Editor to write mathematical expressions and equations. First, save this file to your hard drive by selecting Save As from the File menu. Click the white space below each question to maintain proper formatting. 1. Imagine you are at a gas station filling your tank with gas. The function C ( g ) represents the cost C of filling up the gas tank with g gallons. Given the equation ) ( 03 . 3 ) ( g g C = a. What does the number 3.03 represent? The price per gallon of gas (Exactly 3 dollars and 3 cents in this case) b. Find C (2). \$6.06 c. Find C (9). \$27.27 d. For the average motorist, name one value for g that would be inappropriate for this
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### Page1 / 2
Appendix E - Axia College Material Appendix E Fueling Up...
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All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Perfect Square To Divisibility By 56 (Posted on 2010-04-01)
N is a positive integer such that each of 3*N + 1 and 4*N + 1 is a perfect square.
Is N always divisible by 56?
If so, prove it. Otherwise, give a counterexample.
See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Does this explain why 7 is a factor of n? | Comment 12 of 14 |
1. 56 = 7*8. The part of the solution showing 8 is a factor has already been done.<o:p></o:p>
2. For all m, m^2-1 = (m-1)(m+1)<o:p></o:p>
3. 3n+1 is a perfect square, x ^2. 3n= (x-1)(x+1) = k<o:p></o:p>
4. 4n+1 is a perfect square, y^2. 4n= (y-1)(y+1) = l<o:p></o:p>
5. 7n = k+l<o:p></o:p>
6. n=(k+l)/7<o:p></o:p>
7. n is an integer. Therefore n is divisible by 7.<o:p></o:p>
Posted by broll on 2010-04-04 04:09:42
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# Number 3447
### Properties of number 3447
Cross Sum:
Factorization:
3 * 3 * 383
Divisors:
1, 3, 9, 383, 1149, 3447
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
d77
Base 32:
3bn
sin(3447)
-0.62322440157742
cos(3447)
-0.78204305845552
tan(3447)
0.79691827046997
ln(3447)
8.1452595665169
lg(3447)
3.5374412834079
sqrt(3447)
58.711157372343
Square(3447)
### Number Look Up
Look Up
3447 which is pronounced (three thousand four hundred forty-seven) is a unique figure. The cross sum of 3447 is 18. If you factorisate the figure 3447 you will get these result 3 * 3 * 383. The figure 3447 has 6 divisors ( 1, 3, 9, 383, 1149, 3447 ) whith a sum of 4992. The number 3447 is not a prime number. The figure 3447 is not a fibonacci number. 3447 is not a Bell Number. 3447 is not a Catalan Number. The convertion of 3447 to base 2 (Binary) is 110101110111. The convertion of 3447 to base 3 (Ternary) is 11201200. The convertion of 3447 to base 4 (Quaternary) is 311313. The convertion of 3447 to base 5 (Quintal) is 102242. The convertion of 3447 to base 8 (Octal) is 6567. The convertion of 3447 to base 16 (Hexadecimal) is d77. The convertion of 3447 to base 32 is 3bn. The sine of the figure 3447 is -0.62322440157742. The cosine of the number 3447 is -0.78204305845552. The tangent of 3447 is 0.79691827046997. The root of 3447 is 58.711157372343.
If you square 3447 you will get the following result 11881809. The natural logarithm of 3447 is 8.1452595665169 and the decimal logarithm is 3.5374412834079. that 3447 is special number!
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# Shell Method Vs Washer Method
When it comes to calculus, the Shell Method and Washer Method are two techniques used to calculate the volume of a solid of revolution, a shape created by rotating a 2D figure around an axis. While they have different approaches, both methods are designed to achieve the same results. In this article, we will discuss Shell Method Vs Washer Method, their differences, and applications.
What is the Shell Method?
The Shell Method is a technique used in calculus to calculate the volume of a solid of revolution. It involves finding the volume of a shell formed by slicing the solid into thin cylindrical shells, each having a height of “dx” and a radius “r”. The volume of each shell is then calculated using the formula “2πrh*dx,” where “r” is the radius of the shell and “h” is the height of the shell. The total volume of the solid is found by adding the volumes of all the individual shells together.
Applications
The Shell Method is often used to calculate the volume of solids of revolution that have easy-to-determine functions. For instance, a solid created by revolving the curve y = √x around the x-axis from x = 0 to x = 4 can be easily calculated using the Shell Method.
What is the Washer Method?
The Washer Method is another technique used to calculate the volume of a solid of revolution. Unlike the Shell Method, the Washer Method slices the solid into discs instead of shells. A disc is created by slicing the solid perpendicular to the axis of rotation, generating a “washer” shape, just like the shape that’s formed when the washer is removed from a bolt. Each disc has an outer radius “R” and an inner radius “r,” and its volume is calculated as π(R^2-r^2)dx.
Applications
The Washer Method is often used to calculate the volume of solids of revolution with more challenging functions. For instance, a solid created by revolving the curve y = 9 – x^2 around the x-axis from x = -3 to x = 3 may be more challenging to calculate using the Shell Method, but it can be readily determined using the Washer Method.
Differences Between the Shell Method and Washer Method
The primary difference between the Shell Method and the Washer Method is the shape into which they slice the solid of revolution. The Shell Method slices the solid into thin cylindrical shells, while the Washer Method splits it into thinner discs. While the Shell Method is usually easier to use when the shape of the solid of revolution is easily identifiable, the Washer Method is often preferred when the solid has a more complicated shape.
Another difference between these techniques is the “dx” height. In the Shell Method, this height is parallel to the axis of revolution. In contrast, the “dx” height in the Washer Method is perpendicular to the axis of rotation. As a result, the “dx” values used in both techniques have different meanings.
Conclusion
In summary, the Shell Method and Washer Method are essential techniques used in calculus to calculate the volume of solids of revolution. They are two different approaches that can be employed depending on the shape and complexity of the solid of revolution. The Shell Method splits the solid into cylindrical shells and calculates the volume by adding the volumes of all the individual shells together. The Washer Method, on the other hand, slices the solid into many thin discs and determines the volume by adding the volumes of all the individual slices together. By understanding these techniques, you can choose the best method for the job and solve problems with ease.
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## STP Puzzle Overview
I’ve named a new type of blog post: an STP Puzzle. The big idea is pretty simple. The problem tells you some (but not all) facts about Spanning Tree Protocol (STP) in a small switched network. Your job: determine as many STP facts as you can, things that are for sure true. The more difficult part may be figuring out which facts you simply cannot know based on the limited information given.
Hopefully, these STP puzzles help us all practice applying STP knowledge. Applying networking knowledge to a different scenario may well be the toughest thing to learn by just reading, but practicing with a mentor helps a lot. Hopefully these puzzles can fill that role to some extent.
The rest of this post provide some background and strategies for this type of problem, so take a look at the details below the fold!
## STP Puzzle Pieces
Each STP puzzle acts like a jigsaw puzzle with missing pieces. Sometimes, you can put all the known pieces together, sometimes not, but always leaving a few holes. Sometimes, you can visualize the picture on the missing pieces, sometimes not.
Each STP puzzle problem statement gives you some pieces of an STP topology, for example, a switch’s Bridge ID (BID). Your job is to put the known pieces together. Then, apply STP rules to determine other STP facts.
For example, some of the STP puzzles use a figure like Figure 1. This topology shows a typical campus switch design, with switches S1 and S2 as distribution switches, and S3 plus S4 as access switches:
Figure 1: Typical STP Puzzle Topology
An STP puzzle problem statement might tell you that S3 and S4 are not the root switch, and also tell you the BID of S1 and S2. The problem might not tell you which switch is the root switch. The problem might not tell you enough to decide which switch is root, but maybe just to rule out a few of the switches so you know that they are not the root switch. Regardless, one task in the puzzle is to determine the root switch if possible, then find the various root ports (RPs), designated ports (DPs), and so on.
These puzzles purposefully do not give you all the information you need, just to make you think about STP a little harder. As a result, these problems require a fair amount of abstract thought. You may also jump to conclusions because of similar examples you may have seen in a book you might have read, but part of the point is to give you examples so you can think and apply the concepts to a different example.
The rest of this post walks through some suggestions on how to solve each STP puzzle.
## Step 1: Determine the Root Switch
To put a jigsaw puzzle together, you first find the corners, and then try to find and fit all the edge pieces together. With STP puzzles, to do the equivalent, find the root switch first.
The STP topology revolves around the root switch. If you know the STP Bridge ID (BID) of all the switches, just choose the root as the switch with the numerically-lowest BID.
For these puzzles, however, you may not know all the BIDs, so you need a strategy. Here’s what I suggest: rule out switches as not possibly being the root, until only one is left. If you can’t rule out all but one switch, you don’t have enough information to identify the root. Here are the specific suggested steps:
1. For switches with known BIDs, rule out switches with a worse (numerically-higher) BID compared to others.
2. Rule out switches that have a Root Port (RP). (Only non-root switches have a root port.)
3. From the remaining candidates, pick one, and try to rule it out based on other known information, like known RPs.
For that last step, if you have narrowed down the possibilities, assume one switch is root, and look at information about known Root Ports (RPs) and Designated Ports (DPs), plus the cost to reach that assumed root switch through various paths. Then puzzle out the logic to see if the known RP and DP information works if that switch is root. If not, rule that switch out as well.
Note that this last step requires a lot of thinking, and is where some of the most interesting work happens. Rather than give three pages on it here, look to future STP puzzles for examples of how to work through that logic.
## Determine the Root Port of Each Non-Root Switch
Assume for a moment that you have determined the root switch. The next step would be to determine the root port (RP) for each non-root switch.
By definition, the rules to determine a switch’s RP are:
1. Pick the local port that is part of the least-cost path through the network to reach the root
2. If the cost ties, use these tiebreakers, but only consider ports whose root cost tied:
1. Choose the port connected to the switch with the numerically lowest BID
2. If the previous tiebreaker fails, choose the port connected to the port on the neighboring switch with the lowest STP port priority
3. If that also ties, choose the port connected to the port on the neighboring switch with the lowest internal port number
Those last two tiebreakers only occur when you have multiple connections from the local switch to one neighboring switch. They are frankly hard to visualize without an example; I’ll eventually add a few such examples to one of the future STP puzzles for discussion.
To determine the root cost, start at a non-root switch. Then determine each path from that non-root switch to the root. For each path, add up the port cost for each outgoing port. Note that the STP port cost is per interface, per VLAN.
For instance, in Figure 1, for switch S3, if S1 were root, three paths exist:
• S3 F0/1 to S1
• S3 F0/2 to S2, S2 F0/1 to S1
• S3 F0/2 to S2, S2 F0/4 to S4, S4 F0/1 to S1
All three paths could be S3’s lowest cost path to reach the root, depending on the per-VLAN port cost settings on the interfaces. Add them up, and choose the lowest cost.
As usual, if you know all the information, like all the per-VLAN port costs in this case, finding the root port is relatively simple. However, finding each root port with partial information can be much harder. The point of the puzzle is not to make the problem difficult just to be difficult, but to give us all a chance to think about exactly how a switch makes each STP choice.
## Determine the Designated Port
Each link between switches will have one designated port (DP). In some cases, finding the DP is easy, and in others, it requires that you know each switch’s root cost (RC). First, for review, switches use these rules to choose whether they are a DP on an interface:
1. If the switch is the only switch sending STP Hellos onto that link, for that VLAN, that port is a DP.
2. If the local switch hears Hellos from other switches, the switch advertising the lowest root cost wins, and becomes the DP.
3. If a tie occurs:
1. The switch with the lowest BID wins
2. If a tie still exists, the switch is somehow connected to itself for STP; break the tie by picking the port with the lowest internal port priority, and if a tie, the lowest internal port number.
Those rules are important, but for these puzzles, and then for the exam, you can go a little faster by stepping back for a moment and thinking about a few facts about STP. For instance, the root switch ha no RP port, and the other switches have exactly one RP. The root switch’s ports are all DPs, because they always win the DP election. Finally, on any switch-to-switch link, you end up with either an RP and DP on opposite ends, or a DP and a port in blocking state (because that port is neither RP nor DP). Summarizing what to do, and why:
1. Mark all ports on the root switch as DP. (The root switch’s local ports always win the DP election.)
2. Find any known RPs, and mark the port on the other end of the link as a DP. (If a port is an RP, the port on the other end of the link will always be a DP.)
3. Find any known blocking ports, and mark the other end of the link as a DP. (For any port in a blocking state, it by definition lost the DP election on that link.)
4. Be careful on links with a known DP, because without more information, you don’t know if the other end is an RP or not.
5. Otherwise, use normal STP rules.
## Mark all non-RP and non-DP as Blocking
In an STP topology, RP and DP ports are placed in a forwarding state, and all the rest of the ports are placed into a blocking state. This part is simple: just note the ports that are neither RP nor DP, and you know the ports that block.
## Note What You Cannot Know
Finally, do not miss out on one of the best features of these puzzles: figuring out what you cannot possibly know. If you cannot determine the root switch, at least list the shorter list of possible root switches. If you can determine a switch’s RP, but not the root cost, note that fact. A lot of useful learning exists in uncovering those assumptions you may be making.
## Puzzle Rules
For each puzzle, you will be given a problem statement that includes the following kinds of facts:
• Switch Bridge ID
• Switch priority (a part of the bridge ID)
• Root port
• Per-VLAN port cost
• Per-VLAN port priority
• Notes about a switch using default settings
You can list your answers any way you like. Here in the blog, I will typically give some commentary that matches the sections in this overview post, and summarize the answer in a figure.
Some rules:
1. The puzzles use CCNA-level concepts only.
2. The problem lists partial information, so you may not be able to determine all STP facts. Part of your job is to figure out what you cannot tell from the information given.
3. Unless otherwise stated, assume that the problem relates to the STP topology for VLAN 1
4. Unless otherwise stated, assume all switch-to-switch links are up and working physically, are performing VLAN trunking in the chosen VLAN. That is, the trunking state is not preventing STP from using the link.
5. Do *not* assume that the switches use default configuration.
6. If you have questions or comments, make sure and list you reasoning that leads up to the question or point.
That’s it. Ask questions on the process here if you have them, and engage on the puzzles as I get them posted. Enjoy!
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[…] the puzzle problem statement, then the answers won’t make as much sense. Especially for this one, the new post this week that gives some suggestions on how to attack these STP puzzles can help. The realization that on a switch-to-switch link, the […]
[…] Overview of how to approach these STP puzzles […]
[…] Overview of how to approach STP puzzles […]
[…] Overview of how to approach STP puzzles […]
I hope all is well.
I am slightly confused with something which is mentioned in the CCNA book in regards to STP forward delay timer he said ‘
STP leaves the interface in each interim state for a time equal to the forward delay timer, which defaults to 15 seconds.
As a result, a convergence event that causes an interface to change from blocking to forwarding requires 30 seconds to transition from blocking to forwarding.
The interim states are listening + learning, so how much is each 7.5? Or is learning just the interim state of 15 seconds.
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# Plotting the objective function
11 views (last 30 days)
Vivek on 3 Feb 2023
Answered: Sarthak on 7 Mar 2023
The below shown objective function has to be plotted and while doing so its going in infinite recursion.
The functions m and b are external functions defined in different function files.
function P1=f(x0)
M=5;
x0=[5,8];
% x0=[12,13];
M2=m(M,x0(1));
M3=m(M2,x0(2));
M4=m(M3,(x0(1)+x0(2)));
beta1=b(M,(x0(1)));
beta2=b(M2,(x0(2)));
beta3=b(M3,((x0(1)+x0(2))));
s1=sin(beta1);
s2=sin(beta2);
s3=sin(beta3);
t1n=(13.824)*((M2*M3*M)^2)*((s1*s2*s3)^2);
t1d=(((0.4)*(M3^2)*(s3^2))+2)*(((0.4)*(M2^2)*(s2^2))+2)*(((0.4)*(M^2)*(s1^2))+2);
t1=(t1n/t1d)^(3.5);
t2n=13.824;
t2d=((2.8*(M3^2)*(s3^2))-0.4)*((2.8*(M2^2)*(s2^2))-0.4)*((2.8*(M^2)*(s1^2))-0.4);
t2=(t2n/t2d)^(2.5);
P1=(t1*t2);
P= P1*(-1);
% surf(
% fplot(x0,f)
end
Vivek on 5 Feb 2023
But the objective function is f(x0) and it is scripted in different file and the function has to be returned while plotting.
The below graph is returned by using Z = arrayfun(@(x,y) f([x,y]),X,Y);
Torsten on 5 Feb 2023
Edited: Torsten on 5 Feb 2023
Your objective function accepts a vector with two elements [x y].
x1 = linspace(1,100,55);
x2 = linspace(1,100,55);
[X, Y] = meshgrid(x1, x2);
Z = arrayfun(@(x,y) f([x,y]),X,Y);
This code passes all tuples (x1(i),y1(j)) to your function f, evaluates f at these points and saves the result in Z(i,j).
If the subsequent command
surfc(X, Y, Z)
gives you a surface with constant z value, your function f does not seem to behave properly.
Sarthak on 7 Mar 2023
Hi,
It looks like the function ‘f(x0)’ is not recursively calling itself, but it might be stuck in an infinite loop due to some other reason. One possible reason could be that the external functions ‘m’ and ‘b’ are calling back the ‘f(x0)’ function, which can lead to an infinite recursion. Another reason could be that the values of ‘M’, ‘x0(1)’, and ‘x0(2)’ are not changing during the execution, leading to a loop that never terminates.
To debug this issue, you can try printing the values of the variables ‘M’, ‘x0(1)’, and ‘x0(2)’ at different points in the function to see if they are changing as expected. You can also try commenting out parts of the code to see which part is causing the infinite loop.
Once you have identified the issue, you can modify the code accordingly to fix the problem.
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85% off out of \$31.95 how much do you have to pay ?
Question
Updated 3/31/2014 11:18:01 AM
Edited by sujaysen [3/31/2014 11:17:27 AM], Edited by sujaysen [3/31/2014 11:17:48 AM], Confirmed by sujaysen [3/31/2014 11:18:01 AM]
Original conversation
User: 0.38888888888 as a fraction
User: 85% off out of \$31.95 how much do you have to pay ?
Weegy: 0.15*\$31.95 = \$4.79, so you have to pay \$4.79
Question
Updated 3/31/2014 11:18:01 AM
Edited by sujaysen [3/31/2014 11:17:27 AM], Edited by sujaysen [3/31/2014 11:17:48 AM], Confirmed by sujaysen [3/31/2014 11:18:01 AM]
Rating
0
0.38888888888 as a fraction = 7/18;
Confirmed by sujaysen [3/31/2014 11:18:38 AM]
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# How to find the sum of squares representations of a number? [duplicate]
I have a number, say 1885, I can prove that n can be represented as a sum of squares:
Thm: n is the sum of 2 squares IF AND ONLY IF each prime factor of n that is congruent to 3(mod4) occurs to an even power in the Prime Power Decomposition of n. 1885=5x13x29, all of which are congruent to 1(mod4)
In the question I am given that 1885 can be represented as a sum of squares in 4 distinct ways. How do I go about solving for them?
## marked as duplicate by Guy Fsone, Namaste, Krish, Misha Lavrov, Robert ZNov 26 '17 at 8:13
• Factorise it, and then express each prime factor as a sum of two squares. – Lord Shark the Unknown Nov 25 '17 at 18:15
• Can you provide details for how you proved your theorem? – James Arathoon Nov 25 '17 at 18:33
The Fibonacci identity is
$$(a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2 = (ac-bd)^2 + (ad+bc)^2.$$
So after you find $5=2^2+1^2$ and $13 = 3^2+2^2$ you have
$$65 = (2^2+1^2)(3^2+2^2) = 8^2 + 1^2 = 4^2 + 7^2.$$
Then repeat when you find $29 = 5^2+2^2.$
$1885=5\times 13\times 29$
$5=2^2+1^2$
$13=2^2+3^2$
$29=2^2+5^2$
$\ldots$
$1885=6^2+ 43^2=11^2+ 42^2=21^2+38^2=27^2+34^2$
Hope this can be useful
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# 7.1 Triangle Application Theorems and 7.2 Two Proof-Oriented Theorems - PowerPoint PPT Presentation
1 / 8
7.1 Triangle Application Theorems and 7.2 Two Proof-Oriented Theorems. Objective: To apply theorems about the interior angles, the exterior angles, and the midlines of triangles. Index Card. Definition Exterior angles (page 296) Be sure to include a diagram!!!. Theorems – INDEX CARDS.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
### Download Presentation
7.1 Triangle Application Theorems and 7.2 Two Proof-Oriented Theorems
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## 7.1 Triangle Application Theorems and 7.2 Two Proof-Oriented Theorems
Objective: To apply theorems about the interior angles, the exterior angles, and the midlines of triangles.
### Index Card
Definition
• Exterior angles (page 296)
Be sure to include a diagram!!!
### Theorems – INDEX CARDS
• The sum of the measures of the three angles of a triangles is 180.
• The measure of an exterior angles of a triangles is equal to the sum of the measures of the remote interior angles.
• If a segment joining the midpoints of two sides of a triangle is parallel to the third side, then its length is one-half the length of the third side (Midline Theorem).
• If two angles of one triangle are congruent to two angles of a second triangle, then the third angles are congruent (No-Choice Theorem).
See pages 295, 296, and 302
Don’t forget to draw diagrams for each!!!!!
G
M
J
H
Example 1
In the diagram as marked, if mG = 50, find mM.
Solution
2x + 2y + 50 = 180
2x + 2y = 130
x + y = 65
65 + mM = 180
mM = 115
50
x
y
x
y
Example 2
The vertex angle of an isosceles triangle is twice as large as one of the base angles. Find the measure of the vertex angle.
Let m vertex = 2x
x + x + 2x = 180
4x = 180
x = 45
m vertex = 2(45) = 90
2x
x
x
Example 3
In ΔDEF, the sum of the measures of D and E is 110. The sum of the measures of E and F is 150. Find the sum of the measures of D and F.
Solution
D + E + F = 180
110 + F = 180
F = 70
D + 150 = 180
D = 30
D + F = 30 + 70 = 100
A
D
B
E
C
A
D
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Math and Arithmetic
Units of Measure
Volume
Weight and Mass
# How many kilograms are in two liters?
838485
###### 2010-12-31 13:21:16
• This is a trick question. The kilogram is a unit of mass or weight, while the liter is a unit of liquid volume. In Canada we've used the Metric System since 1977. To compare your question to one using units in the the Imperial System, it would be like asking: How many pounds are there in a gallon? See how inane that is? The answer depends on what substance you are measuring. If it is water, the answer is about two kilograms. If it's two liters of molten lead, well, that weighs far more. (Read the related FAQ question on the right.)
• Actually if density= mass divided by volume, then mass= density times volume!!
• No conversion. You can't convert them. Kilograms are a unit of mass or weight. Liters are a measure of volume, normally associated with liquids.
• Scroll down to related links and look at "Explanation of Litre - Wikipedia".
A correction to this answer. The kilogram is the SI unit for MASS, not weight. Weight is measured in newtons.
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## Related Questions
This depends on the density of the substance liters - Volume kilograms - Weight 30 liters of water = 30 kilograms.
There is no equivalence. Liters measure volume, kilograms measure mass.
Liters can't be converted to kilograms. Liters measure volume, while kilograms measure mass.
Liters and kilograms of what substance? Liters are a measure of volume, kilograms are a measure of mass. You will need the substance's density to find its mass in kilograms.
Since kilograms and liters measure different things, there is no standard conversion - 25 liters may have more kilograms, or less kilograms, depending on the substance involved.
This cannot be sensibly answered. Liters is a measure of volume, kilograms is a measure of weight or mass.
Liters can't be converted to kilograms. Liters measure volume, while kilograms measure mass.
Liters can't be converted to kilograms. Liters measure volume, while kilograms measure mass.
2,000 liters of plain water weighs approximately 2,000 (1,998.7) kilograms.
Kilograms can't be converted to liters. Kilograms measure mass, while liters measure volume.
Kilograms and liters really measure different things (mass, versus volume), so there is no general conversion between the two.
Liters can't be converted to kilograms. Liters measure volume, while kilograms measure mass.
Liters are a measurement for the volume of a liquid; kilograms are a measurement for weight. Please reask your question. How many liters of fresh water equels 1.9 kilograms? How many liters of mercury, or sea water? Sea water from the Pacific Ocean or the Dead Sea? Now really... please be more specific. ;-)
That depends on the density of what you are dealing with:800 liters of air has a mass of 0.96 kilograms 800 liters of water has a mass of 800 kilograms 800 liters of mercury has a mass of 10880 kilograms
There is no general conversion between kilograms and liters. A liter is a unit of volume, while a kilogram is a unit of mass.
Assuming you mean "How many kilos of water in 90 liters?", the answer is 90 kg.
Kilograms can't be converted to liters. Kilograms measure mass, while liters measure volume.
Kilograms can't be converted to liters. Kilograms measure mass, while liters measure volume.
Liters is a measure of volumen. Kilograms is a unit of mass.
A kilogram is a unit of mass. a litre is a unit of capacity. The two units are therefore incompatible.
A kilogram is a unit of mass. A litre is a unit of capacity. The two units are therefore incompatible.
Liters can't be converted to kilograms. Liters measure volume, while kilograms measure mass.
###### Units of MeasureMath and ArithmeticVolumeMedication and DrugsWeight and Mass
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# Advanced Causal Measurements (ACM)
Tags:
Time Limit: 1 s Memory Limit: 128 MB
Submission：0 AC：0 Score：100.00
## Description
Causality is a very important concept in theoretical physics. The basic elements in a discussion of causality are events. An event e is described by its time of occurrence t, and its location, x, and we write e = (t,x). For our concerns, all events happen in the one dimensional geometric space and thus locations are given by a single real number x as a coordinate on x-axis. Usually, theoretical physicists like to define the speed of light to be 1, so that time and space have the same units (actual physical units frighten and confuse theorists).
One event e1 = (t1,x1) is a possible cause for a second event e2 = (t2,x2) if a signal emitted at e1 could arrive at e2. Signals can't travel faster than the speed of light, so this condition can be stated as:
e1 is a possible cause for e2 iff t2 >= t1+|x2-x1|
Thus an event at (-1,1) could cause events at (0,0), (1,2), and (1,3), for example, but could not have caused events at (1,4) or (-2,1). Note that one event can cause several others.
Recently, scientists have observed several unusual events in the geometrically one dimensional universe, and using current theories, they know how many causes were responsible for these observations, but they know nothing about the time and space coordinates of the causes. You asked to write a program to determine the latest time at which the earliest cause could have occurred (i.e. the time such that at least one cause must have occurred on or before this time). Somewhat surprisingly, all the observed events have both space and time coordinates expressed by integer numbers in the range -1000000 ≤ tx ≤ 1000000.
The figure on the right illustrates the first case from input: the earliest single event as a possible cause of all four events.
The first line of input is the number of cases which follow. Each case begins with a line containing the number n of events and the number m of causes, 1 ≤ n, m ≤ 100000. Next follows n lines containing the t and x coordinates for each event.
Output consists of a single line for each case in the format as in the sample output, giving the latest time at which the earliest cause could have occurred, this will be an integer as our time units are not divisible.
## Samples
input
4 4 1 1 -1 1 3 1 4 2 6 4 2 1 -1 1 3 1 4 2 6 4 3 1 -1 1 3 1 4 2 6 4 4 1 -1 1 3 1 4 2 6
output
Case 1: -2 Case 2: 0 Case 3: 0 Case 4: 1
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# Equivalence of Secret IV and Secret Prefix MAC
I read that the secret $IV$ and secret prefix $MAC$ methods are equivalent. Intuitively, this sounds reasonable because the $IV$ and the secret prefix are both the foundation of the inner state at the point of time when the actual message is hashed/compressed.
E.g. in "MDx-MAC and Building Fast MACs from Hash Functions" (link) by Preneel and Oorschot. Section 4.1 The Secret Prefix Method: "If the key consists of a complete block, this corresponds to a hash function with a secret IV."
With secret $IV$ $MAC$ I mean that the key is written into the $IV$ variables / the internal state before the message is hashed.
With secret prefix $MAC$ I consider the function $H(k\ ||\ m)$ where $H$ is a hash function, $k$ is the secret key, $m$ is the message and $||$ denotes concatenation.
I guess the security properties of these two schemes are indeed equivalent but how does this work in practice? For example some attacks against $MD4$ $MAC$s involve the work of offline $MD4$ computations. I can do that on messages with standard $MD4$ in the secret prefix $MAC$ setting (just leaving the unknown key out), but how can one do an offline computation when the secret $IV$ is unknown. How should the hash function be initialized and how (if at all) are the attacks applicable?
Examples of such attacks are:
New Key-Recovery Attacks on HMAC/NMAC-MD4 and NMAC-MD5 (this targets $HMAC$ instead of secret prefix $MAC$, but I guess the attack is adaptable)
Password Recovery Attack on Authentication Protocol MD4(Password || Challenge) (... must leave out the reference here, due to missing repudiation)
• Where did you read this? Because in the current state this question does not make much sense to me. I'm not saying that you are incorrect or wrong in any way, it could just be me not understanding what's going on, hence a request for clarification. Nov 22, 2016 at 18:07
• Just updated the question. I hope it is more clear now. Nov 23, 2016 at 12:10
There are a number of assumptions being made here, and I will make them explicit in case this was not what you had in mind.
The first assumption is that we are using a Merkle-Damgard hash. That is, given a compression function $C(h, m)$, we have (assuming a properly padded message) $$H_{\text{IV}}(m_0 \| m_1\| \dots \| m_n) = C(C(C(C(\text{IV}, m_0), m_1), \dots), m_n).$$
The secret-prefix MAC is simply $H_{\text{IV}}(k \| m)$. I also assume that secret-prefix MAC uses the entire first block, that is, the secret key $k$ is the same size as $m_0$ above.
The secret-IV MAC sets the $\text{IV}$ to be the key: $H_{k}(m)$. But notice we can easily convert a secret prefix into a secret IV by setting $k' = C(\text{IV}, k)$. Therefore if you have a key-recovery attack on a secret-IV MAC, the process is exactly the same with secret-prefix, except the output you obtain is $C(\text{IV}, k)$ instead of $k$. Likewise, an attack on secret-prefix MAC is highly likely to target $C(\text{IV}, k)$ and not $k$ directly. To define the IV, you cannot use the standard APIs for hashing (e.g., md4_init(&ctx)), but will likely have to write your own code.
Recovering this secret IV, in either case, will allow you to forge any message and completely destroy the security of the MAC. In the secret-prefix case, however, recovering the IV does not immediately translate to recovering the original key.
• Thanks for your answer, it explains parts of the relationship between the secret-prefix and the secret-IV. In fact, you pointed out how one could use an attack against secret-IV MAC to attack a secret-prefix MAC (guess this still requires inverting the compression function). But originally, my question targeted the other way around. Most attacks seem to aim against secret-prefix MAC and I was wondering how to use this to attack secret-IV MAC. Nov 23, 2016 at 14:23
• What attacks are you thinking of? Note that the MD4 password recovery attack you linked to violates our assumptions: the key (password) and the message share the same block, and this gives more power to the attacker. The NMAC attack you link to looks like a secret-IV recovery to me, but I only skimmed it. Nov 23, 2016 at 14:36
• Ok, you are right. It seems a concrete answer depends on a concrete attack. I had the hope there is a more general relation. Maybe I will post a further comment or a new more specific question when I have something more specific in mind. Nov 23, 2016 at 15:30
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# Earl can stuff advertising circulars into envelopes at the rate of 45
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Earl can stuff advertising circulars into envelopes at the rate of 45 [#permalink]
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05 Feb 2018, 05:54
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Earl can stuff advertising circulars into envelopes at the rate of 45 envelopes per minute and Ellen requires a minute and a half to stuff the same number of envelopes. Working together, how long will it take Earl and Ellen to stuff 300 envelopes?
(A) 15 minutes
(B) 4 minutes
(C) 3 minutes 30 seconds
(D) 3 minutes 20 seconds
(E) 2 minutes
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Re: Earl can stuff advertising circulars into envelopes at the rate of 45 [#permalink]
### Show Tags
05 Feb 2018, 06:12
Ellen rate = 2/3 * Earl rate = 2/3 * 45 = 30
combined rate = 45+30=75 per minute
300/75=4 min
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Earl can stuff advertising circulars into envelopes at the rate of 45 [#permalink]
### Show Tags
05 Feb 2018, 06:19
Bunuel wrote:
Earl can stuff advertising circulars into envelopes at the rate of 45 envelopes per minute and Ellen requires a minute and a half to stuff the same number of envelopes. Working together, how long will it take Earl and Ellen to stuff 300 envelopes?
(A) 15 minutes
(B) 4 minutes
(C) 3 minutes 30 seconds
(D) 3 minutes 20 seconds
(E) 2 minutes
Earl's rate - 45 envelopes per minute (or) 3 envelopes in 4 seconds
Ellen's rate - 45 envelopes per minute and a half (or) 2 envelopes in 4 seconds.
Working together they can stuff 3+2 = 5 envelopes in 4 seconds.
Therefore, the time taken by both of them to fill 300 envelopes is $$300*\frac{4}{5}$$ = 240 seconds = 4 minutes(Option B)
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Re: Earl can stuff advertising circulars into envelopes at the rate of 45 [#permalink]
### Show Tags
05 Feb 2018, 19:13
Earl's rate= 45 Envelopes/min
Ellen's rate= 45* 2/3= 30 Envelopes/min
so Both together can stuff 75 envelope/min.
75 Envelope ---> 1 min
300 Envelope ---> 300/75= 4 min
Ans B= 4min
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Re: Earl can stuff advertising circulars into envelopes at the rate of 45 [#permalink]
### Show Tags
06 Feb 2018, 16:55
Bunuel wrote:
Earl can stuff advertising circulars into envelopes at the rate of 45 envelopes per minute and Ellen requires a minute and a half to stuff the same number of envelopes. Working together, how long will it take Earl and Ellen to stuff 300 envelopes?
(A) 15 minutes
(B) 4 minutes
(C) 3 minutes 30 seconds
(D) 3 minutes 20 seconds
(E) 2 minutes
We see that Earl’s rate is 45 envelopes/minute and Ellen’s rate is 45/1.5 = 30 envelopes/minute. Thus, their combined rate is 45 + 30 = 75 envelopes/minute.
Thus, it will take them 300/75 = 4 minutes to stuff 300 envelopes.
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Re: Earl can stuff advertising circulars into envelopes at the rate of 45 &nbs [#permalink] 06 Feb 2018, 16:55
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(Analysis by Avichal Goel)
Let's say the plates have labels $L_1, L_2, \ldots, L_n$.
Bessie's first move is to take the top plate from the dirty stack and make a new soapy stack. If $L_2 < L_1$, then it makes sense for Bessie to place it on top of the first plate. On the other hand, if $L_2 > L_1$, then Bessie should instead make a new stack to the right of the first one. (Take a moment to draw it out on paper to convince yourself that this would result in Elsie receiving the plates in the correct order.)
From here, we can make some generalizations. Specifically, we should place a given $L_i$ on the stack with the smallest $L_j$ such that $L_j > L_i$, which ensures that all plates on stacks to the left have a smaller label.
|1|
|2| |*| |7|
|4| |6| |9| |10| |15|
For example, if $L_i = 5$, it should be placed at the location marked by the asterisk in the example above, allowing Elsie to process the plates in the order $1 \rightarrow 2 \rightarrow 4 \rightarrow 5 \rightarrow 6 \rightarrow 7 \rightarrow 9 \rightarrow 10 \rightarrow 15$ as desired.
So far so good, but there might be a plate with a smaller label that is already on our stack. Consider the following modification of the above example, where we are still trying to place $L_i = 5$.
|?|
|1| |4| |7|
|2| |6| |9| |10| |15|
In this case, we cannot simply add $5$ to the stack with $6$, because it would get picked up by Elsie before $4$, which we don't want. In fact, there's no way to place $5$ in the current configuration while still maintaining the correct order. So, from here, the only way to continue is by letting Elsie clear out $1, 2, \text{ and } 4$, and then placing $5$ on top of $6$.
These observations give us our final algorithm: place each plate on the leftmost stack that would preserve the correct order, but only after removing smaller labels already on that stack. We should keep track of the maximum label that we've removed so far because any future plates with smaller labels cannot be processed.
Check out my code below, which runs in $\mathcal{O}(n)$.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, base[N];
vector<int> items[N];
int main() {
cin >> n;
int placed = 0, ans = n;
for (int i = 0; i < n; i++) {
int x; cin >> x;
// impossible to add this plate
if (x < placed) {
ans = i;
break;
}
// plates that go on this stack
for (int j = x; j > 0 && !base[j]; j--) {
base[j] = x;
}
// remove plates with smaller labels
while (!items[base[x]].empty() && items[base[x]].back() < x) {
placed = items[base[x]].back();
items[base[x]].pop_back();
}
// add this plate to the stack
items[base[x]].push_back(x);
}
cout << ans << endl;
}
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http://ghcguide.haskell.jp/8.2.2/libraries/containers-0.5.10.2/Data-Tree.html
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containers-0.5.10.2: Assorted concrete container types
Copyright (c) The University of Glasgow 2002 BSD-style (see the file libraries/base/LICENSE) [email protected] portable Trustworthy Haskell98
Data.Tree
Description
Multi-way trees (aka rose trees) and forests.
Synopsis
# Documentation
data Tree a Source #
Multi-way trees, also known as rose trees.
Constructors
Node FieldsrootLabel :: alabel valuesubForest :: Forest azero or more child trees
Instances
type Forest a = [Tree a] Source #
# Two-dimensional drawing
Neat 2-dimensional drawing of a tree.
Neat 2-dimensional drawing of a forest.
# Extraction
flatten :: Tree a -> [a] Source #
The elements of a tree in pre-order.
levels :: Tree a -> [[a]] Source #
Lists of nodes at each level of the tree.
foldTree :: (a -> [b] -> b) -> Tree a -> b Source #
Catamorphism on trees.
# Building trees
unfoldTree :: (b -> (a, [b])) -> b -> Tree a Source #
Build a tree from a seed value
unfoldForest :: (b -> (a, [b])) -> [b] -> Forest a Source #
Build a forest from a list of seed values
unfoldTreeM :: Monad m => (b -> m (a, [b])) -> b -> m (Tree a) Source #
Monadic tree builder, in depth-first order
unfoldForestM :: Monad m => (b -> m (a, [b])) -> [b] -> m (Forest a) Source #
Monadic forest builder, in depth-first order
unfoldTreeM_BF :: Monad m => (b -> m (a, [b])) -> b -> m (Tree a) Source #
Monadic tree builder, in breadth-first order, using an algorithm adapted from Breadth-First Numbering: Lessons from a Small Exercise in Algorithm Design, by Chris Okasaki, ICFP'00.
unfoldForestM_BF :: Monad m => (b -> m (a, [b])) -> [b] -> m (Forest a) Source #
Monadic forest builder, in breadth-first order, using an algorithm adapted from Breadth-First Numbering: Lessons from a Small Exercise in Algorithm Design, by Chris Okasaki, ICFP'00.
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For CE transistor amplifier, the audio signal voltage across the collector resistance of 2k $\Omega$ is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is I k$\Omega$, then the input signal voltage is
(a) 10 mV (b) 20 mV
(c) 30 mV (d) 15 mV
Concept Questions :-
Applications of Transistor
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The given circuit has two ideal diodes connected as shown in the figure below. The current flowing through the resistance ${R}_{1}$ will be
(a) 2.5 A (b) 10.0 A
(c) 1.43 A (d) 3. 13 A
Concept Questions :-
Applications of PN junction
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
What is the output Y in the following circuit, when all the three inputs A,B,C are first 0 and then 1 ?
1. 0,1 2. 0,0
3. 1,0 4. 1,1
Concept Questions :-
Logic gates
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
To get output 1 for the following circuit, the correct choice for the input is
(a) (b)
(c) (d)
Concept Questions :-
Logic gates
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
A n-p-n transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuits is 192 Ω the voltage gain and the power gain of the amplifier will respectively be
(a)3.69,3.84 (b)4,4 (c)4,3.69 (d)4,3.84
Concept Questions :-
Transistor
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
If in a p-n junction, a square input signal of 10V is applied as shown,
then the output across RL will be
(a) (b) (c) (d)
Concept Questions :-
Logic gates
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
In the given figure, a diode D is connected to an external resistance R = 100 Ω and an e.m.f of 3.5 V. If the barrier potential developed across the diode is 0.5 V, the current in the circuit will be
1. 30mA
2. 40mA
3. 20mA
4. 35mA
Concept Questions :-
PN junction
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The input signal given to a CE amplifier having a voltage gain of 150 is Vi=2cos(15t+π/3). The corresponding output signal will be
(a)300cos(15t+π/3)
(b)75cos(15t+2π/3)
(c)2cos(15t+5π/3)
(d)300cos(15t+4π/3)
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The given graph represents V-I characteristic for a semiconductor device. Which of the following statement is correct?
(a) It is V-I characteristic for solar cell where point A represents open circuit voltage and point B short circuit current
(b) It is for a solar cell and points A and B represent open circuit voltage and current, respectively
(c) It is for a photodiode and points A and B represent open circuit voltage and current, respectively
(d) It is for a LED and points A and B represents open circuit voltage and short circuit current respectively
Concept Questions :-
Applications of PN junction
High Yielding Test Series + Question Bank - NEET 2020
Difficulty Level:
The barrier potential of a p-n junction depends on
(i)type of semiconductor material
(ii)amount of doping
(iii)temperature
Which one of the following is correct
1. (i) and (ii)only
2. (ii) only
3. (ii) and (iii)only
4. (i),(ii) and (iii)
Concept Questions :-
PN junction
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https://it.mathworks.com/matlabcentral/cody/problems/2555-do-operation-as-per-given-string/solutions/2884055
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Cody
# Problem 2555. Do operation as per given string
Solution 2884055
Submitted on 27 Aug 2020 by Rafael S.T. Vieira
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = '3+5'; y_correct = 8; assert(isequal(str_ari(x),y_correct))
2 Pass
x = '13*5'; y_correct = 65; assert(isequal(str_ari(x),y_correct))
3 Pass
x = '145'; y_correct = 145; assert(isequal(str_ari(x),y_correct))
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https://oalevelsolutions.com/past-papers-solutions/cambridge-international-examinations/as-a-level-mathematics-9709/pure-mathematics-p1-9709-01/year-2011-october-november-p1-9709-12/cie_11_on_9709_12_q_5/
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# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2011 | Oct-Nov | (P1-9709/12) | Q#5
Hits: 612
Question
i. Sketch, on the same diagram, the graphs of and for .
ii. Verify that is a root of the equation , and state the other root of this equation for which .
iii. Hence state the set of values of , for , for which
Solution
i.
We are required to sketch and for .
First we sketch for .
We can sketch the graph of for as follows.
We can find the points of the graph as follows.
Now we sketch for .
We can sketch the graph of for as follows.
We can find the points of the graph as follows.
Red curve is for while orange curve is for .
ii.
We have sketched both sides of the equation in (i) and we can see that both curves intersect at points. Therefore, there are two roots of the equation.
If we substitute in the given equation;
Using calculator we can find that;
Hence, both sides of the equation are equal for , therefore, is the root of the equation.
We utilize the symmetry property of to find another solution (root) of :
Symmetry Property or
Therefore;
If we substitute in the above equation;
Using calculator we can see that;
Hence other root for is .
We can also check this root for .
Using calculator we can see that;
Hence, roots of equation for are and .
iii.
We do not need to do tedious calculations for this part, we are required to “state”.
By closely observing the sketch graphed in (i) we can see that two points of intersection of the two curves are roots of the equation as found and discussed in (ii). Values of at these intersection points are are and .
By mere observation we can tell that curve remains below the curve of when and .
Therefore range of values of for the given condition is;
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How resonance is produced?
15 views
How resonance is produced?
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What is the formula for resonance?
What is the effect of resonance?
What is resonance in a Sonometer?
What is the use of resonance?
What is resonance in electricity?
What is meant by resonance?
What is resonance in power system?
What is current resonance?
Why noise of motor is produced? How it can be reduced?
How eddy current is produced in a transformer?
How much electricity is produced by a power plant?
Why a parallel resonant circuit is called Rejector circuit?
What is the natural resonant frequency?
What is the definition of resonant frequency?
Static electricity is produced by _____.
If a 220 V heater is used on 110 V supply, the heat produced by it will be nearly (a) one half (b) twice (c) one-fourth (d) four times
Which gas is produced during charging of a lead-acid battery
Harmonic distortion of the signal is produced in an RC coupled transistor amplifier. The probable component responsible for this distortion is (A) The transistor itself (B) the power supply Vcc (C) the coupling capacitor Cc (D) the biasing resistors
In a double squirrel cage induction motor, good starting torque is produced as (A) Large current flows through the outer high resistance winding. (B) Large current flows through the inner high resistance winding. (C) Small current flows through the outer high resistance winding. (D) Large current flows through the inner low resistance winding.
The torque produced in wattmeter is proportional to
For controlling the vibration of the disc of ac energy meter, damping torque is produced by
The maximum power in a synchronous motor (for a given excitation) is produced when the power angle is equal to: (A) 0° (B) 45° (C) 60° (D) 90°
In DC generators, armature reaction is produced by the: (A) field current (B) armature conductors (C) field pole winding (D) load current in armature
The torque produced in a 4-pole machine is 100 Nm. If machine is re-wound with 6 ploes, other things remaining the unchanged, then the torque produced would be (1) 66.67 Nm (2) 100 Nm (3) 150 Nm (4) 133.33 Nm
In electric discharge lamps, light is produced by (1) cathode ray emission (2) ionization in a gas or vapours (3) heating effect of current (4) magnetic effect of current
During the resistance welding heat produced at the joint is proportional to (a) I2R (b) kVA (c) Current (d) Voltage
In a d.c. motor, unidirectional torque is produced with the help of
Heat in a conductor is produced on the passage of electric current due to?
The noise produced by a transformer is termed as?
Which of the following statements is incorrect? A) During parallel resonance impedance at resonance is minimum B) During parallel resonance current is magnified C) During series resonance current at resonance is maximum D) During series as well as parallel resonance, the power factor is unity
Which of the following is true at resonance in LC circuits? A) The series impedance of the two elements is at a maximum and the parallel impedance is at minimum B) The series impedance of the two elements is at a maximum and the parallel impedance is also at maximum C) The series impedance of the two elements is at a minimum and the parallel impedance is also at minimum D) The series impedance of the two elements is at a minimum and the parallel impedance is at maximum
A network is said to be under resonance when the voltage and the current at the network input terminals are (A) in phase (B) out of phase (C) in phase quadrature (D) in phase, and have equal magnitudes
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# Tagged Questions
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# Truss Analysis
In this article, you can study truss analysis techniques with examples. Analysis of truss by the techniques of joints and by the means of section are explained in brief in the article.
We have studied about the basics of the equilibrium of bodies; now we will debate about the trusses that are basically utilized in creation of stable load-bearing structures. The instances of these are the sides of the bridges or tall tower structures or those towers that carry electricity wires. Schematic diagram of a structure on the side of a bridge is drawn and illustrated in the figure below.
The structure as per shown above is fundamentally structure with two dimensions. It is also called plane truss. Similarly, on the other hand, a tower of microwave oven or telephones are a three-dimensional structure. Consequently, there are basically 2 types of trusses – Plane trusses just like on the sides of a bridge and another one being space trusses like the TV towers. Here, we will be studying about plane trusses in which the basis elements are joined together in the plane.
To stimulate the structure of a plane truss, we will now take a slender rod (12) amongst points 1 and 2 and join it to a attached pin joint at 1 (see figure 2).
Presently I put a pin (pin2) at point 2 at the upper end and suspend a weight W on it. The question is whether there arises the need to hold the weight by then, what other least backings would it be advisable for us to give? For rods we are to make just pin joints (We accept everything is in this plane and the structures does not topple side ways). Since bar (12) tends to turn clockwise, we stop the rightward development of point 2 by associating a bar (23) on it and afterward prevent direct 3 from moving toward the privilege by interfacing it to point 1 by another pole (13). Every one of the joints in this structure are stick rods.. Notwithstanding, in spite of this the whole structure still tends to swing to turn clockwise on the grounds that there is a torque on it because of W. To counter this, we connect a wheel on point 3 and put it on the ground. This is the absolute minimum that we require to hold the weight is place. The triangle made by poles shapes the premise of a plane truss.
Note
At this point, the question may arise as to explain need the parallel rod (13). It is due to the fact that the point 3 will then keep shifting to the right making the entire construction unhinged. Rod (13) will two forces functioning on it: one perpendicular force because of the wheel and the other one will be at the end 2. Nevertheless, these two forces will not be collinear so exclusive of the rod (13) the system shall not be in the position of the equilibrium. Commonly, in a truss all of the joint has to be associated to minimum three or at least two rods and additionally one external support as well.
Now we will analyze forces in the construction structure that were just created. For the purpose of straightforwardness, now we will choose lengths of all rods to be equivalent. To obtain the forces I, now we will look at all the forces on every one of the pin and bargain circumstances under which the pins will be in the equilibrium position. The initial thing we mark down that every one of the rod in equilibrium when they are exposed to the influence of two forces which is exerted by the pins at their conclusions. As we had studied in the previous chapter, in circumstances like these the forces will have to be collinear and consequently along the rods only. Therefore each one of the rod is under a tensile or compaction force. As a result, rods (12), (23) and (13) experience forces as illustrated in figure 3.
We have to pay attention to the fact that we have taken all the forces to be compaction. If the actual forces are ductile in nature, the result will lead out to be negative. We will now look at pin 2. The singular forces acting on pin 2 are F12 because rod (12) and F23 because rod (23). Additional, it is pulled downwards due to the weight W. Therefore, forces acting on pin 2 look like illustrated in the figure 4.
Applying equilibrium condition to pin (2) results to
We will now examine at pin 3 (see figure 4). It is in equilibrium condition under forces F23, normal reaction N and a parallel force F13.
Applying equilibrium condition yields
Because the direction of F13 is coming out to be negative, the direction should be opposite to that assumed. Balance of forces in the vertical direction provides
Consequently, we see that the weight is retained with these three rods. The construction structure is determinate and it retains the weight in place as well.
Even if we substitute the pin joints by a minor plate (also called the gusset plate) with at least two or three pins in these, the study relics pretty much the same as previous owing to the fact that the pins are so near organized that they barely generate any moment about the joints. Even in those case when if the rods are fused together at the joints, to a higher extent of accurateness majority of the force is undertaken longitudinally on the rods, even though some very minor (insignificant) moment is formed by the joints and might be feasible bending of the rods.
From this point, we are all ready to construct a truss and examine it as well. Now, we are going to construct it by augmenting higher and more of triangles fused together. Now, consequently when the triangles are added, the member of joints j and additionally the number of members (rods) m are related shown below:
m = 2j – 3
After this, it will make a truss statically determinate. This is easily comprehendible as follows. First of all, we will consider the whole truss as one system. If those case where it has to be statically determinate, there has to be just three unidentified forces on it due to the fact for forces in a plane there are at least three equilibrium conditions. Attaching one of its ends a pin joint and keeping the other one on the roller does that (roller also provides the augmented benefit that it can aid in regulating any alteration in the interval of a member because of different causes like deformations). If we want to compute these outside forces and the force in every member of the truss, the complete number of unknowns converts m + 3. We resolve for these unknowns by writing equality circumstances for each pin; there has to be 2j such equations. For the system to be completely determinate we need to have m + 3 = 2j , which is the condition as state above. If the case arise where we need to augment any more members, these are dismissed. On the other side, lesser number of members would create the truss unbalanced and it will breakdown when loaded. This will occur since the truss will not be capable to deliver the prerequisite number of forces for all equilibrium circumstances to be fulfilled. Statically determinate trusses are also called simple trusses.
Exercise 1: As per illustrated in the figure 5 are three usually utilized trusses on the sides of bridges. Demonstrate that all three of them are simple trusses.
The questions may arise as why we keep the trusses on the bridges.. As our advanced examination will display they allocate the load over all elements and thus building the bridge stronger and more robust.
We will now work towards the process to get the forces made in numerous arms of a truss when it is laden outwardly. This is carried out under the given expectations:
1. In those cases, where the middle line of the members of a truss meet at a given point, that point will be assumed as a pin joint. It is a very critical assumption since we have seen previously while presenting a truss (triangle with pin joint), the load is transported on to another member of the trusses so that forces will remain fundamentally collinear with the member.
2. All of the external loads are exerted on pin connections.
3. All members’ weight is correspondingly distributed on connecting pins.
There are basically two approaches of shaping forces in the members of a truss – Technique of joints and system of sections. We initiate with the technique of joints:
Truss Analysis- Method of Joints
In this technique of joints, we shall analyze the equilibrium of the pin at the joints. Because of the facts that the forces are synchronized at the pin, it is exclusive of moment equation and just two equations for equilibrium viz. . Consequently, we will begin our analysis at a point where one identified load and at maximum two unidentified forces are there. The mass of all the members is allocated into two halves and that is reinforced by each pin. To a degree, we have by this time referred to this technique while presenting trusses. We will now illustrate it with help of examples.
Example 1: we will take truss ABCDEF as illustrated in figure 6 and load it at the specific given point E by 5000N. The stretch of small members of the truss is 4m and the length of the diagonal members is m. we will now ascertain the forces in all of the members of this truss taking after the assumption that they are weightless.
Thereafter, now we will take each point to be a pin joint and begin corresponding forces on all of the existing pins. Due to the fact that the pin E has an exterior load of 5000N one might need to commence from there. Though, E point consists of more than 2 unidentified forces so we will not be able to begin at E. We thus first give the truss as a entire and catch reactions of ground at the given points A and D owing to the fact that then at points A and D their will persist just two unidentified forces. The horizontal reaction Nx at point A is nil due to the fact that there is no exterior horizontal force on the system. To ascertain N2 I take moment about A to get
Which through equation results to
In the techniques of the joints, we will now commence at pin A and balance the numerous forces. We by now expect the course and demonstration their almost at A (figure 7). All the approaches that the diagonals make are 45° .
The only equations we now have to be concerned about are the force balance equations.
We have to make sure that the force on the member AB and AF working to be contrary to the forces on the pin ( Newton ‘s IIIrd law). Consequently, force on member AB is compaction (pushes pin A away) however that on AF is ductile (pulls A towards the aforementioned).
Now we will undertake the joint F where force AF is recognized and two forces BF and FE are unidentified. For pint F
Now we will analyze the point B because now there are only two unidentified forces there. At point B
Negative sign indicates that while we have revealed FBE to be compressive, it is really tensile.
At this point, we will now contemplate point C and equilibrium the forces there. I have previously projected the course of the forces and presented FCE to be ductile whereas FCD to be compressive
From there, we will go to pin D where the normal reaction is N and balance forces there.
Therefore, forces in numerous adherents of the truss have been ascertained. They are
Now the question that arises is the process of how we got all the forces exclusive of utilization of equations at all joints. Remember that is how we had acquired the statical determinacy condition. We did not have to utilize all joints since previously we had preserved the structure as an unabridged and had acquired two equations from there. So one joint – in this case E – has got no need to be examined. Though, given the fact that the truss is statically determinate, all these forces have to balance at point E, where the load shall be exerted also. How the situation would vary if each member of the truss had weight. Let’s assume that every members weighs approximately 500N, then we will assume that the load is separated correspondingly among two pins supporting the member the loading of the truss would seem as set in figure 8 (loading because of the weight as shown in red). Except at points A and D the loading because the weight is 750N; at the A and D points it is 500N.
Now the exterior reaction at each end will be:
The additional 2000N could be computed from the moment equation or they can also be computed straightaway by understanding that the new augmented weight is faultlessly symmetric about the midpoint of the truss and consequently will be correspondingly separated among the two supports. For the process of balancing forces at other pins, we trail the same process as above, ensuring though that all of the pin has an external loading because of the weight of each member. We’ll solve for forces in some member of the truss. Looking at pin A, we obtain:
Now, we move to point F and see that the forces are
Now we can likewise solve for other pins in the truss we will complete that as an exercise for you.
After the demonstration of the method of joints, we will now shift on to look at the technique of sections that straight provides the force on a preferred member of the truss.
Truss Analysis- Method of sections
Just like the name suggests, in this technique of sections we make segments through a truss and then compute the force in the members of the truss though which the cut is created. For instance, if we take the trick we just solved in the technique of joints and create a section S1, S2 (see figure 9), we can then easily ascertain the forces in members BC, BE and FE by bearing in mind the equilibrium of the segment to the left or the right of the unit.
We will now show this. Similar to the method of joints, we initiate by first of all shaping the reactions at the exterior support of the truss by considering it as a completely inflexible body. In the current case scenario, it provides N at D and N at A. Now we will assume the section of the truss on the left (see figure 10).
Because the complete section is in equilibrium, . Note down that we are now utilizing all three equations for equilibrium because the forces in all of the members are not simultaneous. The course of force in each member, we can now ascertain it by inspection. Consequently, the force in the section of members BE must be pointing down due to the fact that there is no other member that can provide a descending force to counterbalance N reaction at A. This obviously demonstrates that F BE is tensile. Correspondingly, to counter the torque about B created by N force at A, the force on FE will also has to be from F to E. As a result, this force is also ductile. If we next contemplate the balance of torque about A, N and FFE do not provide any torque about A. So to counter torque created by FBE , the force on BC must act towards B, thus then making the force compressive.
We will now calculate individual forces. FFE is the simplest to compute. For it we take the moment about B. This provides
4 × = 4 × F FE
F FE = N
Now, we calculate FBE . For this, we use the equation . It gives
Lastly, for calculation of FBC , we can utilize either the equation about A or
Finally we have ascertained the forces in these three members straight without computing forces going from one joint to next joint and have preserved ample amount of time and effort during the whole process. The forces on the right segment shall be contrary to those on the left segments at points from side to side which the segment is cut. This could be utilized to check the result and it will be an exercise for you.
After this diagram we will show the different steps that are undertaken in order to solve for forces in members of a truss by the technique of sections:
1. first of all, create a cut to divide the truss into segment, passing the cut through members where the force is desirable.
2. Create the cut through three member of a truss due to the fact that with three equilibrium equations viz. we can resolve for a maximum of three forces.
3. Exert equilibrium conditions and resolve for the preferred forces.
In utilization of the method of sections, ingenuity lies in creating a proper. The technique after a way of directly computing required force circumventing the hard work included in utilizing the method of joints where we will have to solve for each joint.
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# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data
Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 18 - Mean, Median and Mode of Ungrouped Data.
All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.
Exercise/Page
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise MCQ
Solution 1
Solution 2
Solution 3
Solution 4
Correct option: (b)
If each observation of the data is decreased by 8 then their mean is also decreased by 8.
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18A
Solution 1(v)
Prime numbers between 50 and 80 are as follows:
53, 59, 61, 67, 71, 73, 79
Total prime numbers between 50 and 80 = 7
Solution 1(iv)
Factors of 20 are: 1,2,4,5,10,20
Solution 1(iii)
First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35
Therefore, Mean =20
Solution 1(ii)
First ten odd numbers are:
1,3,5,7,9,11,13,15, 17, and 19
Solution 1(i)
first eight natural numbers are:
1,2,3,4,5,6,7and 8
Solution 2
Solution 3
Sol.3
Solution 4
Solution 5
Total numbers of observations = 5
Thus, last three observations are (9 + 4), (9 + 6) and (9 + 8),
i.e. 13, 15 and 17
Solution 6
Mean weight of the boys =48 kg
Sum of the weight of6 boys =(48x6)kg =288kg
Sum of the weights of 5 boys=(51+45+49+46+44)kg=235kg
Weight of the sixth boy=(sum of the weights of 6 boys ) - (sum of the weights of 5 boys)
=(288-235)=53kg.
Solution 7
Calculated mean marks of 50 students =39
calculated sum of these marks=(39x 50)=1950
Corrected sum of these marks
=[1950-(wrong number)+(correct number)]
=(1950-23+43) =1970
correct mean =
Solution 8
Let the given numbers be x1,X2......X24
Solution 9
Let the given numbers be x1, x2.....x20
Then , the mean of these numbers =
Solution 10
Let the given numbers be x1, x2.......x15
Then the mean of these numbers=27
Solution 11
Solution 12
Solution 13
Mean of 6 numbers = 23
Sum of 6 numbers =(23x6 )=138
Again , mean of 5 numbers =20
Sum of 5 numbers=(20x 5 ) =100
The excluded number= (sum of 6 numbers )-(sum of 5 numbers)
=(138-100) =38
The excluded number=38.
Solution 14
Mean height of 30 boys = 150 cm
Total height of 30 boys = 150 × 30 = 4500 cm
Correct sum = 4500 - incorrect value + correct value
= 4500 - 135 + 165
= 4530
Solution 15
Mean weight of 34 students = 46.5 kg
Total weight of 34 students =(34x46.5)kg =1581 kg
Mean weight of 34 students and the teacher =(46.5+0.5)kg=47kg
Total weight of 34 students and the teacher
=(47x35)kg =1645kg
Weight of the teacher =(1645-1581)kg= 64kg
Solution 16
Mean weight of 36 students = 41 kg
Total weight of 36 students = 41x 36 kg = 1476kg
One student leaves the class mean is decreased by 200 g.
New mean =(41-0.2)kg = 40.8 kg
Total weight of 35 students = 40.8x35 kg = 1428 kg.
the weight of the student who left =(1476-1428)kg =48 kg.
Solution 17
Mean weight of 39 students =40 kg
Total weight of 39 students = 40x 39) = 1560 kg
One student joins the class mean is decreased by 200 g.
New mean =(40-0.2)kg = 39.8 kg
Total weight of 40 students =(39.8x40)kg=1592 kg.
the weight of new student
= Total weight of 40 students - Total weight of 39 students
= 1592-1560 = 32 kg
Solution 18
The increase in the average of 10 oarsmen = 1.5 kg
Total weight increased =(1.5x10) kg=15 kg
Since the man weighing 58 kg has been replaced,
Weight of the new man =(58+15)kg =73kg.
Solution 19
Mean of 8 numbers=35
Total sum of 8 numbers = 35x8 = 280
Since One number is excluded, New mean = 35 - 3 = 32
Total sum of 7 numbers = 32x7 = 224
the excluded number = Sum of 8 numbers - Sum of 7 numbers
= 280 - 224 = 56
Solution 20
Mean of 150 items = 60
Total Sum of 150 items = 150x60 = 9000
Correct sum of items =[(sum of 150 items)-(sum of wrong items)+(sum of right items)]
= [9000 - (52 + 8) + (152 + 88)]
= [9000-(52+8)+(152+88)]
= 9180
Correct mean =
Solution 21
Mean of 31 results=60
Total sum of 31 results = 31x60 = 1860
Mean of the first 16 results =16x58=928
Total sum of the first 16 results=16x58=928
Mean of the last 16 results=62
Total sum of the last 16 results=16x62=992
The 16th result = 928 + 992 - 1860
= 1920 - 1860 = 60
The 16th result = 60.
Solution 22
Mean of 11 numbers = 42
Total sum of 11 numbers = 42x11 = 462
Mean of the first 6 numbers = 37
Total sum of first 6 numbers = 37x6 = 222
Mean of the last 6 numbers = 46
Total sum of last 6 numbers = 6x46 = 276
The 6th number= 276 + 222 - 462
= 498 - 462 = 36
The 6th number = 36
Solution 23
Mean weight of 25 students = 52kg
Total weight of 25 students = 52x25 kg=1300 kg
Mean of the first 13 students = 48 kg
Total weight of the first 13 students = 48x13 kg = 624kg
Mean of the last 13 students = 55 kg
Total weight of the last 13 students = 55x13 kg = 715 kg
The weight of 13th student
= Total weight of the first 13 students + Total weight of the last 13 students - Total weight of 25 students
= 624+715-1300 kg
= 39 kg.
Therefore, the weight of 13th student is 39 kg.
Solution 24
Mean score of 25 observations = 80
Total score of 25 observations = 80x25 = 2000
Mean score of 55 observations = 60
Total score of 55 observations = 60x55 =3300
Total no. of observations = 25+55 =80 observations
Total score = 2000+3300 = 5300
Mean score =
Solution 25
Average marks of 4 subjects = 50
Total marks of 4 subjects = 50x4 = 200
36 + 44 + 75 + x = 200
155 + x = 200
x = 200 - 155 = 45
The value of x = 45
Solution 26
Let the distance of mark from the staring point be x km.
Then , time taken by the ship reaching the marks=
Time taken by the ship reaching the starting point from the marks =
Total time taken =
Total distance covered =x+x=2x km.
Solution 27
Total number of students = 50
Total number of girls = 50-40 = 10
Average weight of the class = 44 kg
Total weight of 50 students= 44x 50 kg = 2200kg
Average weight of 10 girls = 40 kg
Total weight of 10 girls = 40x10 kg = 400 kg
Total weight of 40 boys = 2200-400 kg =1800 kg
the average weight of the boys =
Solution 28
Total earnings of the year
= Rs. (3 × 18720 + 4 × 20340 + 5 ×21708 + 35340)
= Rs. (56160 + 81360 + 108540 + 35340)
= Rs. 281400
Number of months = 12
Solution 29
Average weekly payment of 75 workers = Rs. 5680
Total weekly payment of 75 workers = Rs. (75 × 5680) = Rs. 426000
Mean weekly payment of 25 workers = Rs. 5400
Total weekly payment of 25 workers = Rs. (25 × 5400) = Rs. 135000
Mean weekly payment of 30 workers = Rs. 5700
Total weekly payment of 30 workers = Rs. (30 × 5700) = Rs. 171000
Number of remaining workers = 75 - 25 - 30 = 20
Therefore, Total weekly payment of remaining 20 workers
= Rs. (426000 - 135000 - 171000)
= Rs. 120000
Solution 30
Let the ratio of number of boys to the number of girls be x : 1.
Then,
Sum of marks of boys = 70x
Sum of marks of girls = 73 × 1 = 73
And, sum of marks of boys and girls = 71 × (x + 1)
70x + 73 = 71(x + 1)
70x + 73 = 71x + 71
x = 2
Hence, the ratio of number of boys to the number of girls is 2 : 1.
Solution 31
Mean monthly salary of 20 workers = Rs. 45900
Total monthly salary of 20 workers = Rs. (20 × 45900) = Rs. 918000
Mean monthly salary of 20 workers + manager = Rs. 49200
Total monthly salary of 20 workers + manager = Rs. (21 × 49200) = Rs. 1033200
Therefore, manager's monthly salary = Rs. (1033200 - 918000) = Rs. 115200
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18B
Solution 1
Solution 2
For calculating the mean , we prepare the following frequency table :
Weight (in kg) (Xi) No of workers (fi) fiXi 60 63 66 69 72 4 3 2 2 1 240 189 132 138 72 771
Solution 3
Solution 4
For calculating the mean , we prepare the following frequency table :
Age (in years) (Xi) Frequency (fi) fiXi 15 16 17 18 19 20 3 8 9 11 6 3 45 128 153 198 114 60 698
Solution 5
For calculating the mean , we prepare the following frequency table :
Variable (Xi) Frequency (fi) fiXi 10 30 50 70 89 7 8 10 15 10 70 240 500 1050 890
Solution 6
Solution 7
Solution 8
We prepare the following frequency table :
(Xi) (fi) fiXi 3 5 7 9 11 13 6 8 15 P 8 4 18 40 105 9P 88 52
303 + 9p = 8(41+p)
303 + 9p= 328 + 8p
9p - 8p = 328 -303
P=25
the value of P=25
Solution 9
We prepare the following frequency distribution table:
(Xi) (fi) fiXi 15 20 25 30 35 40 8 7 P 14 15 6 120 140 25p 420 525 240
1445 + 25p = (28.25)(50+p)
1445 + 25p = 1412.50 + 28.25p
-28.25p + 25p = -1445 + 1412.50
-3.25p = -32.5
the value of p=10
Solution 10
We prepare the following frequency distribution table:
(Xi) (fi) fiXi 8 12 15 P 20 25 30 12 16 20 24 16 8 4 96 192 300 24p 320 200 120
1228 + 24p = 1660
24p = 1660-1228
24p = 432
the value of p =18
Solution 11
Solution 12
Let f1 and f2 be the missing frequencies.
We prepare the following frequency distribution table.
(Xi) (fi) fixi 10 30 50 70 90 17 f1 32 f2 19 170 30f1 1600 70f2 1710 Total 120 3480 + 30f1 + 70f2
Here,
Thus, .......(1)
Also,
Substituting the value of f1 in equation 1, we have,
f2=52 - 28 = 24
Thus, the missing frequencies are f1 =28 and f2=24 respectively.
Solution 13
Solution 14
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18C
Solution 1(i)
Arranging the data in accending order, we have
2,2,3, 5, 7, 9, 9, 10, 11
Here n = 9, which is odd
Solution 1(ii)
Arranging the data in ascending order , we have
6, 8, 9, 15, 16, 18, 21, 22, 25
Here n = 9, which is odd
Solution 1(iii)
Arranging data in ascending order:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here n = 11 odd
Solution 1(iv)
Arranging the data in ascending order , we have
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here n = 13, which is odd
Solution 2(iii)
Arranging the data in ascending order , we have
3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81 Here n = 12, which is even
Solution 2(ii)
Arranging the data in ascending order , we have
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here n = 10, which is even
Solution 2(i)
Arranging the data in ascending order , we have
9, 10, 17, 19, 21, 22, 32, 35
Here n = 8, which is even
Solution 3
Arranging the data in ascending order , we have
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here n = 15, which is odd
Thus, the median score is 23.
Solution 4
Total number of students = n = 9 (odd)
Arranging heights (in cm) in ascending order, we have
144, 145, 147, 148, 149, 150, 152, 155, 160
Solution 5
Arranging the weights of 8 children in ascending order, we have
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here , n= 8 , which is even
Solution 6
Arranging the ages of teachers in ascending order , we have
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n =10, which is even
Solution 7
The ten observations in ascending order:
10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41
Here, n =10, which is even
Solution 8
Total number of observations = n = 10 (even)
Median = 65
Solution 9
Total number of observations = n = 8 (even)
Median = 25
Solution 10
Total number of observations = n = 11 (odd)
Arranging data in ascending order, we have
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Now, 41 and 55 are replaced by 61 and 75 respectively.
Arranging new data in ascending order, we have
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 18 - Mean, Median and Mode of Ungrouped Data Page/Excercise 18D
Solution 1
Arrange the given data in ascending order we have
0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6
Let us prepare the following table:
Observations(x) 0 1 2 3 4 5 6 Frequency 2 1 1 1 1 2 4
As 6 ocurs the maximum number of times i.e. 4, mode = 6
Solution 2
Arranging the given data in ascending order , we have:
15, 20, 22, 23, 25, 25, 25, 27, 40
The frequency table of the data is :
Observations(x) 15 20 22 23 25 27 40 Frequency 1 1 1 1 3 1 1
As 25 ocurs the maximum number of times i.e. 3, mode = 25
Solution 3
Arranging the given data in ascending order , we have:
1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9,
The frequency table of the data is :
Observations(x) 1 2 3 4 5 6 7 8 9 Frequency 2 1 2 1 2 2 1 1 5
As 9, occurs the maximum number of times i.e. 5, mode = 9
Solution 4
Arranging the given data in ascending order , we have:
9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60
The frequency table of the data is :
Observations(x) 9 19 27 28 30 32 35 50 60 Frequency 1 1 1 1 1 1 1 4 1
As 50, ocurs the maximum number of times i.e. 4, mode = 50
Thus, the modal score of the cricket player is 50.
Solution 5
Total number of observations = n = 7
Mean = 18
Thus, data is as follows:
3, 21, 25, 17, 24, 19, 17
The most occurring value is 17.
Hence, the mode of the data is 17.
Solution 6
Number of values = n = 9 (odd)
Numbers in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55, 56, 57
Thus, we have
52, 53, 54, 54, 55, 55, 55, 56, 57
The most occurring number is 55.
Hence, the mode of the data is 55.
Solution 7
Mode of the data = 25
So, we should have the value 25 occurring maximum number of times in the given data.
That means, x + 3 = 25
x = 22
Thus, we have 24, 15, 40, 23, 27, 26, 22, 25, 20, 25.
Arranging data in ascending order, we have
15, 20, 22, 23, 24, 25, 25, 26, 27, 40
Number of observations = 10 (even)
Solution 8
Total number of observations = n = 9 (odd)
Median = 45
Thus, we have
42, 43, 44, 44, 45, 45, 45, 46, 47
The most occurring value is 45.
Hence, the mode of the data is 45.
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Reasoning Quiz On Puzzle Day 17 Bag | | GovernmentAdda
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# Reasoning Quiz On Puzzle Day 17 Bag
130+ SSC CGL Previous Papers With Solution Free >>
## Reasoning Quiz On Puzzle Day 17 Bag
I. Study the following information carefully to answer the given questions
A, B, C, D, E, F and G are seven people live on seven different floors of a building but not necessarily in the same order. The lowermost floor of the building is numbered 1, the one above that is numbered 2 and so on till the topmost floor is numbered 7. Each one of them spends different amount per month. i.e. 20000, 15000, 10000, 25000, 30000, 35000 and 40000(But not necessarily in the same order). Each one of them likes different fruit i.e Lemon, Apple, Grapes, Mango, Pomegranate, Orange and Banana(But not necessarily in the same order).
The person who likes Mango lived immediately below the one who likes Banana. The one who spends Rs.20000 lives immediately above the one who spends Rs.40000. Only one person lives between B and E. The one who likes Grapes lives on even numbered floor while the one who likes Apple fruit lives on odd numbered floor and not on the third floor. B lives on one of the floors above E. Neither C nor A spends Rs.25000. The one who likes orange live one of the floors above Banana. E does not spend Rs. 10000. Person A lives on an odd numbered floor but not on the floor numbered three. The one who spends Rs.30000 lives immediately above A. Only two people live between A and the one who spends Rs.10000. The one who spends Rs.15000 lives on one of the odd numbered floors above D. The person who likes orange does not live in the floor numbered 5 while the person who likes Mango does not live in the floor numbered 2. Only three people live between C and the one who spends 15000. The one who spends Rs.10000 lives immediately above C. F spends 10000. The one who likes Lemon lives immediately above the one who likes Orange. The person who likes Grapes lives in middle floor exactly.
Explanation
Floor Persons Expenditure Fruit 7 G 25000 Lemon 6 B 30000 Orange 5 A 15000 Apple 4 E 20000 Grapes 3 D 40000 Pomegranate 2 F 10000 Banana 1 C 35000 Mango
1. Which of the following is the Expenditure of B?
A. Rs. 20000
B. Rs. 40000
C. Rs. 35000
D. Rs. 30000
E. None of these
D. Rs. 30000
2. Which of the following combination is true as per the given arrangement?
A. A – 15000 – Apple
B. C – 30000 – Banana
C. B – 35000 – Pomegranate
D. F – 20000 – Grapes
E. None of these.
A. A – 15000-Apple
3. Who among the following lives in floor no 4?
A. A
B. C
C. D
D. F
E. E
E. E
4. Four among the following form a group in a certain way.Which of the following does not belong to Group?
A. G – 15000
B. A – 40000
C. B – 20000
D. F – 40000
E. E – 10000
D. F – 40000
5. Who among the following likes Mango?
A. A
B. C
C. D
D. F
E. None of these.
B. C
II. Study the following information carefully to answer the given questions
Seven students namely viz A, B, C, D, E, F and G of seven different colleges have seminar on seven different days, namely viz Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday of the same week but necessarily in the same order. Each student stays in a hostel in different floor. The lower most floor of the building is numbered 1, the one above that is numbered 2 and so on till the topmost floor is numbered 7.Each one of them likes different Games i.e Temple Run, Fruit Ninja, Clash of clans, Pokemon Go, Candy crush, Angry Bird and Crossy Road.(But not necessarily in the same order).
The one who likes Angry Bird lives immediately below the one who likes Candy crush. A does not like Fruit Ninja. G stayed in the second floor and has a seminar on Wednesday. The one who stayed in the first floor has a seminar on Saturday. The person who likes Temple Run lives on the floor above the floor numbered 4. More than four persons live between the person who likes Temple Run and the one Who likes Crossy Road. B has a seminar immediately before E. B does not have seminar on any of the days after G. The one who stayed in the seventh floor does not have a seminar on any of the days on or before Friday. The one who stayed in the third floor has a seminar immediately after C. E does not stay in the fifth floor. More than two persons live between the one who likes Candy crush and the one who likes Pokemon Go. The one who stays in sixth floor does not have a seminar immediately before or after G. F does not have any seminar on Sunday and does not live in third floor. D does not have seminar on any one of the days before A.
Explanation
Floor Person Day Games 7 D Sunday Temple Run 6 B Monday Candy crush 5 C Thursday Angry Bird 4 E Tuesday Fruit Ninja 3 A Friday Clash of clans 2 G Wednesday Pokémon Go 1 F Saturday Crossy Road
1. F stays in which of the following floor?
A. 1
B. 2
C. 3
D. 4
E. None of these.
A. 1
2. Who among the following likes Fruit Ninja?
A. A
B. C
C. B
D. F
E. E
E. E
3. Four among the following form a group in a certain way. Which of the following does not belong to Group ?
A. D – Monday
B. C – Tuesday
C. A – Wednesday
D. G – Saturday
E. E – Sunday
E. E – Sunday
4. Which of the following is correctly matched?
A. D – Monday – Temple Run
B. C – Tuesday – Candy crush
C. A – Friday – Clash of clans
D. G – Saturday – Fruit Ninja
E. E – Sunday – Pokémon Go
C. A – Friday – Clash of clans
5. Who among the following have seminar on Monday?
A. A
B. C
C. B
D. D
E. E
C. B
I. Study the following information carefully to answer the given questions
Six People – A, C, D, E, F, and G live in eight different floors of building (but not necessarily in the same order). Two of the floors in the building are vacant. The lowermost floor of the building is numbered one, the one above that is numbered two, and so on till the topmost floor is numbered eight. Each one of them likes a different brands of perfumes, namely Adidas, Axe, Fogg, Gillette, Nivea and NIKE (but not necessarily in the same order). All have a different number of chocolates 2, 3, 4, 5, 6 and 8 but not in same order.
The difference of chocolates of C and G is equal to the number of chocolates hold by F. The number of floors above F is same as the number of floors between F and D. F lives an odd numbered floor above the floor numbered four. There are three people live between the two vacant floors. Only three floors between D and the one who likes Gillette. The one who likes Nivea lives immediately above G. The difference of chocolates of D and A is equal to the number of chocolates hold by E. Only three floors between G and A. The one who likes NIKE lives immediately above the one who likes Adidas. C has more chocolates than G. C lives immediately above the one who likes Fogg. D has more chocolates than A. The number of floors between F and the one who likes AXE is only one. The one who likes AXE perfume immediately below one of the vacant floors. The floor number of the vacant floors are even – number. Only two floors are there between the one who likes NIKE and Gillette. C lived immediately below one of the vacant floors and not on the ground floor. C neither lived on floor number 5 nor floor number 3. A has more than 3 chocolates. The one who likes Axe lives on one of the floors below the floor number 4. G does not have 4 chocolates. A has neither 2 nor 6 chocolates. The number of chocolates hold by C, G and F together is “4” less than the number of chocolates hold by E, A and D together.
Explanation
Floor No Person Perfume Number of Chocolates 8 _ _ _ 7 C Nivea 6 6 G Fogg 2 5 F Gillette 4 4 _ _ _ 3 E AXE 3 2 A NIKE 5 1 D Adidas 8
1. Which of the following Statements is true with respect to the given information?
A. G has 2 chocolates
B. E lives immediately above C
C. Only three people live between F and the one who has 5 chocolates.
D. D likes Fogg.
E. All the given statements are true.
Answer – G has 2 chocolates
2. Who amongst the following lives exactly between the fourth floor and the one who has 8 chocolates?
A. B, C
B. G, D
C. F, G
D. A, B
E. E, A
3. Which of the following pair represents vacant floors?
A. 4, 6
B. 2, 6
C. 4, 8
D. 2, 8
E. 6, 8
4. Four of the following five are alike in a certain way and so form a group. Which one of the following does not belong to the group?
A. G – Nivea
B. C – Gillette
C. F – Nivea
E. D – AXE
Answer – A. G – Nivea
5. The number of chocolates hold by F is?
A. Four
B. Three
C. None
D. Five
E. Two
II. Study the following information carefully to answer the given questions
Ten students namely viz P, Q, R, S, T, U, V, W, X and Y of ten different colleges but not necessarily in the same order have an exam on five different days starting from Monday to Friday of the same week. Each student have exam at two different time slots, i.e 08.00 AM or 11.00 A.M. Each one of them likes different brands of pens, namelyParker, Cello, Camlin, Papermate, Natraj, Sheaffer, Lemy, Shanghai Hero, Aurora and Watermen Pens (but not necessarily in the same order).
Only two people have exam between U and Y. Neither T nor V does not have the exam on Friday. X has the exam on Tuesday at 08.00 A.M. W does not have the exam at 11.00 AM. The number of people who have exam between V and S is same as the number of people who have exam between R and W. S does not have exam on any one of the days after T. U does not have exam on any of the days after W. Q has exam immediately before X. X does not have exam on any of the days before V. The one who has exam at 08.00 A.M. immediately before Y. S has exam immediately after the day of one who has the exam on Monday. U does not have the exam at 11.00 A.M. Only three people have exam between V and T. Four people have exam between the one who likes Parker pen and the one who likes Watermen Pens. The one who likes Watermen Pens has the exam on Wednesday. Y does not like Parker. Four people have exam between the one who likes Camlin and the one who likes Aurora. The one who likes Aurora has the exam on Thursday. X does not like Camlin. The one who likes Cello has exam immediately after the one who likes Natraj and immediately before the one who likes Papermate. The one who likes Shanghai Hero has exam immediately after the one who likes Sheaffer. The one who likes Sheaffer has the exam on Thursday.
Explanation
Person Day Time Slot Pen V Mon 08.00 AM Parker Q Mon 11.00 AM Camlin X Tue 08.00 AM Natraj S Tue 11.00 AM Cello T Wed 08.00 AM Papermate R Wed 11.00 AM Watermen Pens U Thu 08.00 AM Aurora P Thu 11.00 AM Sheaffer W Fri 08.00 AM Shanghai Hero Y Fri 11.00 AM Lemy
1. How many persons have the exam at 11’0 clock between X and W?
A. 5
B. 6
C. 3
D. 4
E. None of these.
2. Who among the following person has the exam at 8 A.M?
A. Y
B. W
C. P
D. R
E. S
3. Four among the following form a group in a certain way. Which of the following does not belong to Group?
A. Q – Tuesday
B. S – Wednesday
C. V – Tuesday
D. P – Friday
E. W – Friday
Answer – E. W – Friday
4. Which of the following is correctly matched?
A. X – Monday – Parker
B. S – Tuesday – Cello
C. Q – Friday – Lemy
D. V – Tuesday – Papermate
E. X – Wednesday – Aurora
Answer – B. S – Tuesday – Cello
5. Who among the following like Shanghai Hero and Lemy?
A. P, Q
B. R, S
C. T, V
D. W, Y
E. V, X
I. Study the following information carefully to answer the given questions.
Seven people, namely P,Q,R,S,T,U and V like seven different e-commerce websites namely Amazon, Flipkart, Snapdeal, E-bay, Jabong, Myntra and Paytm but not necessarily in the same order. Each people also works in the same office but at a different department on the basis of experience namely Administration (ADMIN), Marketing & Sales, (M&S), Accounts (ACC), Production (PO), Quality Management (QM), Human Resources (HR), and Public Relations (PR), but not necessarily in the same order. Each person also like different cars namely viz – Audi, BMW, Ford, Fiat, Hyundai, Chevrolet and Ferrari.
Note: Each person has been allocated to a department as per increasing order of experience with the one in ADMIN being the least experienced whilst the one in PR Being the most experienced.
T neither has the least experience than the one who likes Snapdeal. T neither has the least experience nor he works in QM. Q does not work in QM. The one who likes Flipkart does not work in PO. The person who likes Myntra has more experience than the one who likes Fiat. The one in Quality Management likes Chevrolet. The person who likes Jabong also likes the Hyundai car. Persons who have the least experience and most experience like BMW and Ferrari car respectively. Only one person has less experience than U. V likes Paytm and has more experience than the one who likes Amazon. S has less experience than the one in PO, but more experience than the one who likes Snapdeal. The one who has less experience than U likes E-bay. Only one person has more experience than P. P does not like Audi. The one in Marketing and Sales like Ford. The one in HR likes Jabong. Only two people have more experience than the one who likes Amazon
Explanation
Person Job E-Commerce Car V PR Paytm Ferrari P HR Jabong Hyundai R QM Amazon Chevrolet T PO Myntra Audi S ACC Flipkart Fiat U M&S Snapdeal Ford Q ADMIN E-bay BMW
1. As per the given arrangement, ADMIN is related to Ferrari and PR is related to Hyundai in a certain way. To which of the following is ACC related to the same way?
A. Ferrari
B. BMW
C. Ford
D. Fiat
E. Chevrolet
2. Which of the following pairs of people who have more experience than P less experience than S?
A. V, P
B. V, U
C. R, V
D. T, Q
E. R, P
3. Which combination represents the department that T works in and the movie he likes?
A. QM – Amazon
B. PO – Snapdeal
C. PO – Myntra
D. ACC – E-bay
Answer – C. PO – Myntra
4. Who amongst the following likes Fiat?
A. S
B. R
C. P
D. Q
E. Other than those given as options
5. Which of the following e-commerce websites does Q like?
A. Snapdeal
B. Myntra
C. Amazon
D. E-bay
E. Flipkart
II. Study the following information carefully to answer the given questions.
Seven persons – A, B, C, D, E, F and G – went to tour in the months of February, March, April May, July, October and December but not necessarily in the same order. Each one of them likes different brand of cycle viz., Firefox, Hercules, Atlas, BSA, Hero, Montra and Kross but not necessarily in the same order. Each person also like seven different brand of bikes namely viz – Honda, Yamaha, Suzuki, Harley Davidson, TVS, Royal Enfield and Vespa.
There are two persons went to tour between the one who likes Honda and the one who likes Vespa. E does not like Atlas. The person who likes Montra went to tour in the month having less than 31 days. The person who likes Honda went to tour on one of the months after March which has less than 31 days. The one who likes Hero went to tour in the month having less than 31 days. There is only one person between A and the person who likes Hero. The person who likes Vespa went to tour immediately before the one who likes Suzuki. G went to tour in that month which has less than 31 days. F went to tour immediately after G. Only one person went to tour between A and the who likes BSA. F does not like Harley Davidson. A does not like Montra. The one who likes Firefox went to tour immediately before the one who likes Kross. The person who likes Yamaha went to tour immediately before the one who likes Royal Enfield and immediately after the one who likes Honda. The one who likes Atlas went to tour immediately before A. C went to tour immediately after A. Only two persons went to tour between C and B.
Explanation
Month Person Cycles Bike February G Montra Harley Davidson March F Hercules Kit-Kat April B Hero Honda May D Atlas Yamaha July A Firefox Royal Enfield October C Kross Vespa December E BSA Suzuki
1. Which of the following brand of cycles is liked by C?
A. Firefox
B. BSA
C. Montra
D. Hercules
E. Kross
2. Which of the following combinations of Month-Person-Cycle-Bike is correct?
A. March – G – Firefox – Harley Davidson
B. July – A – Firefox – Royal Enfield
C. October – E – Montra – Yamaha
D. May – C – Atlas – TVS
E. April – F – Hero – Vespa
Answer – B. July – A – Firefox – Royal Enfield
3. Which of the following statements is true with respect to the given arrangement?
A. C went to tour in October
B. A likes Kross
C. D went to tour immediately before E.
D. E went to tour in July
E. None of the given statements is true
Answer – A. C went to tour in October
4. Who among the following went to tour in May?
A. F
B. A
C. C
D. D
E. B
5. Who among the following likes Royal Enfield?
A. E
B. C
C. A
D. G
E. B
I. Study the following information carefully to answer the given questions.
Seven Persons – A, B, C, D, E, F, and G – live on separate floors of a seven storey-ed building, but not in the same order. The ground floor of the building is numbered 1, the floor above it 2 and so on until the topmost floor is numbered 7. Each person likes different vegetables – Cabbage, Potato, Tomato, Onion, Carrot, Radish and Bean, but not necessarily in the same order. Each person has 7 different weight of their favorite vegetables starting from 1kg to 10 kg. The weight of Onion is more than 2 kg. The total weight of Carrot and Radish is 10 kg. The person who likes Onion lives on floor numbered four. A does not live on the lowermost floor. A lives on any odd numbered floor below the one who likes Onion. Only two persons live between A and the person who likes Bean. Only one person lives between B and F. The total weight of Cabbage is square of the total weight of Carrot while The total weight of Bean is square of the total weight of Onion. F lives on an even numbered floor and does not like Onion. Only three persons live between the persons who like Cabbage and Tomato respectively. The person who likes Cabbage live on any floor above the B’s floor. The person who likes Cabbage does not live on the topmost floor. G lives on an even numbered floor but neither immediately above nor immediately below the floor of A. C does not like Cabbage or Tomato. Only two persons live between D and the one who likes Onion. The person who likes Carrot lives on the floor immediately above the floor of the person who likes Raddish. The difference between the weight of the Tomato and Radish is 2 kg. The floor number and the weight of favorite vegetable is same for the person C.
Explanation
Floor No Person Vegetables weight 7 C Potato 7 kg 6 G Bean 9 kg 5 E Cabbage 4 kg 4 B Onion 3 kg 3 A Carrot 2 kg 2 F Raddish 8 kg 1 D Tomato 6 kg
1. D has how many kg of favourite vegetable?
A. 2 kg
B. 8 kg
C. 6 kg
D. 4 kg
E. No one
2. Which of the following statements is/are true according to the given information?
A. E lives on floor numbered 5 and he does not like Onion
B. A likes Carrot and he does not live on floor numbered 4
C. C likes Potato and he does not have 6 kg.
D. Only two persons live between the floors of E and F
E. All the statements are true.
Answer – E. All the statements are true.
3. Who among the following lives on the floor immediately above the floor of A?
A. B
B. F
C. G
D. C
E. No one
4. Who among the following lives exactly between the floors on which B and F live?
A. F
B. A
C. D
D. C
E. No one
5. Who among the following does like Carrot?
A. A
B. D
C. B
D. C
E. No one
II. Study the following information carefully to answer the given questions
P, Q, R, S, T, U, V and W live on eight different floors of a building but not necessarily in the same order. The ground floor is numbered one and the floor above it is numbered two and so on. The top most floor is numbered eight. Each of them likes a different news paper viz The Times of India, The Economic Times, Business Standard, THE HINDU, The Indian Express, The Financial Express, Deccan Chronicle and THE TRIBUNE but not necessarily in the same order.
P lives immediately above U, who lives on an odd numbered floor. The one who likes The Indian Express does not live on first floor. The one who likes The Economic Times lives on an even numbered floor but not on floor number 8. Only two persons live between U and the one who likes The Economic Times. Only one person lives between U and the one who likes The Indian Express. Neither T nor R lives on first floor. Only one person lives between R and S, who likes Business Standard. Only two persons live between T and P. Q lives on an even numbered floor and immediately above R. The one who likes THE HINDU lives on an even numbered floor and live immediately above the person who likes THE TRIBUNE. R does not like The Indian Express or THE TRIBUNE. Only two persons live between W and the one who likes The Times of India. Q lives on floor number four. The one who likes The Financial Express does not live on odd numbered floor. W does not like Deccan Chronicle.
Explanation
Floor Person News paper 8 P THE HINDU 7 U THE TRIBUNE 6 W The Financial Express 5 T The Indian Express 4 Q The Economic Times 3 R The Times of India 2 V Deccan Chronicle 1 S Business Standard
1. Who among the following lives on floor 5?
A. S
B. R
C. V
D. T
E. None of these
2. Which of the following News papers does P like?
A. THE HINDU
B. THE TRIBUNE
C. The Economic Times
D. Deccan Chronicle
E. None of these
3. Who among the following likes The Financial Express News paper?
A. Q
B. R
C. W
D. S
E.None of these
4. Who among the following lives between P and W?
A. P
B. R
C. S
D. U
E.None of these
5. Which of the following is true as per the arrangement?
A. 4-R-The Times of India
B. 7-W-THE HINDU
C. 5-T-THE TRIBUNE
D. 3-Q-The Economic Times
E. None of these
Answer – E. None of these
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# Homework Help: Division question
1. Oct 24, 2004
### roger
HI
please can someone help me with this quick question on division :
If I have to do 3 divided by 4
____
4 |3
4 doesn't go into 3 so I put a zero on top .
When it gets to 4 into 2(from 30-4x7), it doesn't go, so why don't I put a zero on top but instead put a zero beside the 2 to make it 20 ?
thankyou
roger
2. Oct 24, 2004
### dav2008
I have no idea what you're doing.
If you're doing 3 divided by 4 using long division:
http://home.comcast.net/~iberiaforums/division.GIF [Broken]
Last edited by a moderator: May 1, 2017
3. Oct 24, 2004
### arildno
roger:
Do you want to perform 3:4 or 4:3?
4. Oct 24, 2004
### roger
DEAR ARILDNO,
I AM TRYING TO DO 3 DIVIDED BY 4 LONG DIVISION
BUT WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?
ROGER
5. Oct 24, 2004
### dav2008
You never put a 0 on top.
What you do in the first step is put a decimal point on top.
6. Oct 24, 2004
### roger
when you do 7x4 =28
and then 30 minus 28 = 2.
from here
4 doesnt go into 2 so why don't I put a zero on top ?
This is what I want to know ....
thanx
roger
7. Oct 25, 2004
### arildno
Now, since to write down the long-division algoritm is a bit difficult (and besides, we do it differently in Norway from the US), I will show you the RATIONALE behind the long division technique instead; long division is simply a condensed version of what I'll present.
1. What is meant by "division" in this case?
Ordinarily, "division" is meant to be that process that rewrites a number, given as a FRACTION, into the equivalent DECIMAL REPRESENTATION of that number (which, as it happens, is a particular TYPE of fractional representation).
2. Let's look at the number given by fraction 3/4
We want to write 3/4 in its decimal representation, that is to find digits $$a_{i}$$ between 0 and 9, so that we have:
$$\frac{3}{4}=0.a_{1}a_{2}a_{3}....$$
Where the notation $$0.a_{1}a_{2}a_{3}....$$ MEANS:
$$0.a_{1}a_{2}a_{3}....\equiv\frac{a_{1}}{10}+\frac{a_{2}}{100}+\frac{a_{3}}{1000}++$$
3. Let's start!
a) We note that 3<4, so our first step is to multiply 3/4 with an appropriate representation of the number "1":
$$\frac{3}{4}=1*\frac{3}{4}=\frac{10}{10}*\frac{3}{4}=\frac{1}{10}*(\frac{30}{4})$$
Henceforth, we will work with the expression included in the parenthesis.
b)
We note that 30>4, and the greatest multiple of 4 which is less than 30, is 4*7=28.
Hence, we write 30=4*7+2
We therefore have the equality:
$$\frac{30}{4}=\frac{4*7+2}{4}$$
c)We now use the fact that we can split up a sum in the numerator into a sum of fractions:
$$\frac{4*7+2}{4}=\frac{4*7}{4}+\frac{2}{4}=7+\frac{2}{4}$$
The last step follows since 4 is a common factor in both the numerator and denominator in the first fraction.
d) Hence we have shown:
$$\frac{3}{4}=\frac{1}{10}*(7+\frac{2}{4})$$
This can be rewritten as:
$$\frac{3}{4}=\frac{7}{10}+\frac{1}{10}(\frac{2}{4})=0.7+\frac{1}{10}(\frac{2}{4})$$
e) We will now work with the parenthesized 2/4.
Since 2<4, we multiply 2/4 by an appropriate version of 1:
$$\frac{2}{4}=1*\frac{2}{4}=\frac{10}{10}*\frac{2}{4}=\frac{1}{10}*(\frac{20}{4})$$
f) We note that 20=5*4, so we have:
$$\frac{2}{4}=\frac{1}{10}*(\frac{5*4}{4})=\frac{1}{10}*(5)=\frac{5}{10}$$
g) We now look back at the equation in d):
$$\frac{3}{4}=0.7+\frac{1}{10}(\frac{2}{4})$$
With the result from f), we have:
$$\frac{3}{4}=0.7+\frac{1}{10}(\frac{5}{10})=0.7+\frac{5}{100}=0.7+0.05=0.75$$
And that's our result..
Last edited: Oct 25, 2004
8. Oct 25, 2004
### HallsofIvy
I'm confused about your question "WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?" because I certainly WOULD put a 0 on top!
4 goes into 3 0 times so I would put a 0 before the decimal point and continue:
__0.
4) 3.0
Now, 4 divides into 30 (ignore the decimal point now- we've already taken care of it) 7 times: 4*7= 28
__0.7
4) 3.00
_28_
20
and now 4 divides into 20 5 times, evenly
__0.75_
4) 3.000
_2 8__
20
20
Of course, you don't HAVE to put the "0 on top" because 0, after all, means NOTHING! 0.75 is the same as .75 although I think you will find that "0.75" is preferred in "formal" writing.
9. Oct 25, 2004
### roger
I'm confused about your question "WHY DONT I PUT A ZERO ON TOP FOR 4 INTO 2 ?" because I certainly WOULD put a 0 on top!
4 goes into 3 0 times so I would put a 0 before the decimal point and continue:
__0.
4) 3.0
Now, 4 divides into 30 (ignore the decimal point now- we've already taken care of it) 7 times: 4*7= 28
__0.7
4) 3.00
_28_
20
and now 4 divides into 20 5 times, evenly
this is where my concern was...
I thought it was 4 divides into 2 NOT 20.
So is that to say because the decimal point has been put down, it is really the calculation of 30 divided by 4 ?
Because if it had been 2 instead of 20, my question was if 4 does not go into 2, then why do I not put a zero on top to indicate this .And then I would have brought down a zero to make it 20 and proceed from there......
If I ever have to bring down a zero, where does it come from ?
__0.75_
4) 3.000
_2 8__
20
20
Of course, you don't HAVE to put the "0 on top" because 0, after all, means NOTHING! 0.75 is the same as .75 although I think you will find that "0.75" is preferred in "formal" writing
10. Oct 26, 2004
### HallsofIvy
"Because if it had been 2 instead of 20, my question was if 4 does not go into 2, then why do I not put a zero on top to indicate this "
And my answer was you certainly can "put a zero on top to indicate this."
You don't HAVE to since ".75" is commonly understood as "0.75" although, in my opinion, 0.75 is better.
11. Oct 26, 2004
### roger
It still doesn't answer my query because if I had done it my way, I would have got an answer of 0.705
This is wrong it should be 0.75 but I'm referring to the zero after the 7 not before the decimal point ?
Roger
12. Oct 26, 2004
### faust9
Ok, I don't understand why you WANT to put the zero on the top?
Lets walk through an easy example the way you want to do it and the correct way:
116/4
___2
4)116
2*4=8
11-8=3
Now 3/4=0
___20
4)116
bring down the 6
36/4=9
___209
4)116
Does that look like the correct answer? No. The reason you don't put the 0 on the top is that you don't redivide the remainder from one step to the next.
Good luck.
13. Oct 27, 2004
### dav2008
Your question is like asking "why don't you add a zero when you divide 30/5 and get 6, not 60"
You just don't. That's how it's done.
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# 2006 AMC 10B Problems/Problem 15
## Problem
Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$. What is the area of rhombus $BFDE$?
$[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); label("F", F, dir(point--F)); [/asy]$ $\mathrm{(A) \ } 6\qquad \mathrm{(B) \ } 4\sqrt{3}\qquad \mathrm{(C) \ } 8\qquad \mathrm{(D) \ } 9\qquad \mathrm{(E) \ } 6\sqrt{3}$
## Solution
Using properties of a rhombus, $\angle DAB = \angle DCB = 60 ^\circ$ and $\angle ADC = \angle ABC = 120 ^\circ$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$.
The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD = s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3}$
The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}$. Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $\left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3}$
Let $x$ be the area of rhombus $BFDE$. Then $\frac{x}{24} = \frac{1}{3}$, so $x = 8 \Longrightarrow \boxed{\mathrm{(C)}}$.
## Solution 2
Triangle DAB is equilateral so triangles $DEA$, $AEB$, $BED$, $BFD$, $BFC$ and $CFD$ are all congruent with angles $30^\circ$, $30^\circ$ and $120^\circ$ from which it follows that rhombus $BFDE$ has one third the area of rhombus $ABCD$ i.e. $8 \Longrightarrow \boxed{\mathrm{(C)}}$.
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A chordwise offset of the wing-pitch axis enhances rotational aerodynamic forces on insect wings: a numerical study
Abstract
Most flying animals produce aerodynamic forces by flapping their wings back and forth with a complex wingbeat pattern. The fluid dynamics that underlies this motion has been divided into separate aerodynamic mechanisms of which rotational lift, that results from fast wing pitch rotations, is particularly important for flight control and manoeuvrability. In our study we focus on the generation of these rotational forces during the wing-reversal and come to a new and remarkably simpel rotational lift model.
Type
Background
Insects fly by flapping there wings back and forth, contrary to birds which often move there wings up and down. To understand this unique wingbeat pattern scientist have tried to link the motion of the wing to the forces the insect generates Ellington1984. In the end this will enable us to gain insight in how an insect flies, but also how the morphology of the insect influences the force generation. Essentially this helps us understand how an insects stays in the air.
Before we can say anything about the force generation we need to know how an insect moves. Unfortunately a flying insect moves really fast, but luckily for us high-speed camera’s can be used to slow down this motion. Lets look at a great movie on how a fruit-fly flies captured by Muijres2014
It might be difficult to see but the insect moves the wings to the front of its body (horizontal to the ground) under an angle of around $$45^o$$, when it can no longer move forward it rotates its wing around the length of the wing and then move back again.
This rotation is the focus of our manuscript. It has been discovered some time ago by Dickinson1999 that pitching a wing around its longitudinal axis that is moving forward with a certain stroke velocity generates a force. These forces are known as rotational forces. However, as mentioned by Bomphrey2017 it appears that mosquitoes also generates rotational forces at wing-reversal, when the wing is not moving forward or backward, so what gives?
The idea from Nakata2015 is that there is an additional rotational lift force, called “rotational drag”, which is linked to the rotational velocity about the longitudinal axis of the wing. This means that an insect can generate rotational forces at wing-reversal.
Our approach
Based on the this notion from all the previous mentioned authors we set out to do a parametric study, varying the forward velocity, wing shape, and longitudinal rotational velocity systematically to understand how these additional rotational forces are generated.
For this study we used computational fluid mechanics, which is a method that can simulate the movement of the air if the motion of the animal is known. The solver which we used is open-source and can be found here: IBAMR Bhalla2013.
From our results we noticed that indeed Nakata2015 was correct, that insects can generate additional rotational forces at wing-reversal. We also found that the geometry influences these forces in a very systematic way. The additional rotational forces are influenced by the asymmetry in the shape of the wing along the longitudinal axis. This means that for a symmetrical wing no additional forces are generated!
From our parametric study we also where able to create a very simple rotational lift model
$$F_{\mathrm{rotational}} = C_{F, \mathrm{rotational}} \rho \left(\sqrt{S_{xx}S_{yy}} \omega_{\mathrm{stroke}} \omega_{\mathrm{pitch}} + S_{x|x|} \omega_{\mathrm{pitch}}^2 \right)$$
where the coefficient $$C_{F,\mathrm{rotational}} \approx \frac{2}{3}\pi$$, $$\rho$$ the density of the air, $$\sqrt{S_{xx}S_{yy}}$$ the symmetric second moment of area, $$\omega_{\mathrm{stroke}}$$ the stroke velocity, $$\omega_{\mathrm{pitch}}$$ the pitch velocity and $$S_{x|x|}$$ the asymmetric second moment of area and finally $$F_{\mathrm{rotational}}$$ the rotational forces, which are defined perpendicular to the wing surface.
Conclusion
This asymmetric second moment of area $$S_{x|x}$$ is essentially a measure of how much wing surface is above or below the pitch axis. This means that if we move the pitch axis more towards the leading edge (introducing the chord-wise offset) the wing-asymmetry increases and so do the rotational forces. This also holds true the other way around, if we decrease the pitch axis, moving it closer to the symmetry axis of the wing the rotational forces decrease.
These rotational forces are of importance for the insect because it is been shown by Muijres2014 that the kinematics is adapted at wing-reversal when a maneuver is performed.
Bibliography
[Ellington1984] Ellington, The aerodynamics of hovering insect flight. IV. aerodynamic mechanisms, Philosophical transaction royal society London, (1984).
[Muijres2014] Muijres, Elzinga, Melis & Dickinson, Flies evade looming targets by executing rapid visually directed banked turns, Science, 344, 172-177 (2014).
[Dickinson1999] Dickinson, Lehmann & Sane, Wing roation and the aerodynamic basis of insect flight, Science, 284, 1954-1960 (1999).
[Bomphrey2017] Bomphrey, Nakata, Phillips & Walker, Smart wing rotation and trailing-edge vortices enable high frequency mosquito flight, Nature, 544, 92-96 (2017).
[Nakata2015] Nakata, Liu Hao & Bomphrey, A CFD-informed quasi-steady model of flapping-wing aerodynamics, Journal of Fluid Mechanics, 783, 323-343 (2015).
[Bhalla2013] Bhalla, Bale, Griffith & Patankar, A unified mathematical framework and an adaptive numerical method for fluid–structure interaction with rigid, deforming, and elastic bodies, Journal of computational physics, 250, 446-476 (2013).
Wouter G. van Veen
Aspiring developer & scientist
I use computational fluid mechanics to research the fundaments of insect flight.
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# Capacitor between emitter and collector in fm modulator
1. ## Capacitor between emitter and collector in fm modulator
what does Capacitor between emitter and collector do in this fm modulator(5pf)
http://www.circuitstoday.com/wp-cont...er-circuit.jpg
dc and ac analysis?
2. ## Re: Capacitor between emitter and collector in fm modulator
The circuit is a FM transmitter and that capacitor is used for Oscillator Feedback..
•
3. ## Re: Capacitor between emitter and collector in fm modulator
could you tell me how it works?
if we use ac analys it we know in small signal it should be shorted.how does it work?
4. ## Re: Capacitor between emitter and collector in fm modulator
Originally Posted by baby_1
could you tell me how it works?
if we use ac analys it we know in small signal it should be shorted.how does it work?
You can not do AC analysis with this circuit because it's an oscillator that works autonom.
You can make a transient analysis by providing start-up conditions.
Note:AC analysis is done in "steady state occured" circuits.
1 members found this post helpful.
•
•
6. ## Re: Capacitor between emitter and collector in fm modulator
The base of Q2 is grounded to RF. If a little bit of noise gets into Q2's emitter such as to go positive, this reduces the forward bias on its base-emitter junction, so its collector current falls and the collector voltage rises. So the voltages at Q2's emitter and collector are in phase. The tuned circuit L1 & C6 only provide gain at their resonant frequency, so as the gain is peaked at this frequency the circuit oscillates at this frequency. C5 is to provide the positive feedback and stops the low input impedance of the emitter damping the Q of the tuned circuit by having a high impedance.
Frank
1 members found this post helpful.
7. ## Re: Capacitor between emitter and collector in fm modulator
Thanks a lot dear chuckey .
could you tell me how can we calculate the C5 feedback capacitor value accroding the circuit & frequecny?
Thanks
8. ## Re: Capacitor between emitter and collector in fm modulator
The input to the "amplifier" is the emitter impedance of the transistor. I can't remember the formula, buts something like Hfe/Ic(in mA), anyway its low, say 25 ohms. So the voltage across R6, sends some current down through R6 to earth and some into the emitter (the importent bit!). Because of the action of the transistor this current (or at least 99% of it), flows through the tuned circuit, causing a voltage to appear across it, there is also the output resistance of the transistor shunting the tuned circuit (high resistance 20K?). Its is this voltage that sends the current through C5 into the emitter and causes the volt drop across R6.
Frank
1 members found this post helpful.
9. ## Re: Capacitor between emitter and collector in fm modulator
Hi Baby,
some value must be wrong in the schematic.
As is, Q1 goes in saturation.
C7 should be 22 pf or 0.022 nF.
Regards
Z
10. ## Re: Capacitor between emitter and collector in fm modulator
Well spotted Zorro! . If Q1 takes 1mA, then Ve = 1V, Vb= 1.6V, so Vcc -1.6V = 7.4V, so R2 : R3 = ~5:1, so I would make R2= 100K. As for C7, ever tried buying a .022 Pf capacitor?, I reckon 22Pf is about right.
Frank
11. ## Re: Capacitor between emitter and collector in fm modulator
Thanks dear frank.
didn't remeber where you read this anaylise or book that you have been read before?
•
12. ## Re: Capacitor between emitter and collector in fm modulator
I can't remember, I first came across transistors in about 1962. Wow!, that was almost fifty years ago. I have been learning about them ever since. Perhaps one day I'll understand them.
Regards
Frank
1 members found this post helpful.
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INCHES VICE INCH RATIONALE PLURAL VERSUS SINGULAR MEASUREMENTS RATIONALE Wow! I actually remembered to write this. The wonders will never cease. The main thesis in this treatise will seek to explain my rationale for using the plural relative to measuring distance, diameter, weight, and volume. When I worked in the engineering field for 29 years, there was always a controversial area germane to measuring various things. I continually clashed with engineers, mathematicians and other fellow travelers. Like most of my views, I have always stood alone in most cases. I shall now state my contention. When one measures something, one must use the plural when the measurement is more than 1 unit. Let me use an example. The following controversial examples went on for 29 years. As part of my job, I often had to measure diameters of various pipes such as culverts. If, for example, the diameter measured 48 inches i.e. 4 feet, I would always say it is a 4 feet pipe. The other engineers would say, "No, it is a 4 foot pipe or 48 inch pipe." I would insist on feet or inches and would both orally and written refer to the plural. Now, follow my rationale. Look at your feet right now. You will see 2 FEET. If you cover one of your feet, what do you now see? One FOOT. Am I not correct? Of course I am. Now, relative to the "foot" measure, our system of measure is based someone's foot size back in antiquity. Therefore, as I just pointed out, the plural of "foot" is "feet" which is more than one "foot". I think we call can agree on that, I trust. Now, if something is 4 feet, that is the plural of "foot". If something is exactly 1 foot, than it is one foot in length. Can we all agree on that? All right, if a culvert is more than 1 foot in diameter, it is referred to as "feet" correct. Now let us apply this same rationale for other measurements. If some thing weighs 1 ton, that is the singular, if something weighs more than 1 ton, that is plural, correct? So then, if something weighs 4 tons, then it is 4 tons! Not 4 "ton". It always aggravates me when someone refers to more than 1 ton as "ton", i.e., that culvert weighs 4 "ton". UGGGHHH Now, I will take an example of inconsistency of the people who ague with me about the plural. If something contains 1 ounce, it is singular. If some thing contains more than 1 ounce, then my position states it has to be the plural, i.e. "ounces". These same "singular people" always refer to it as ounces. For example, they say that glass has 3 ounces of liquid in it. Ah Huhhh! I then point out their inconsistency, which proves my contention! Get it? Whenever, I pointed out this blatant inconsistency, all the other engineers would ignore it and insist on using their incorrect references. In conclusion, I believe their insistence in continuing their errors is based on a lifetime of incorrect usage of others. I have always stood contrary to the "herd mentality". I have always been my own man and have never followed the crowd especially when I know I'm right. Oh Uhhh! It just occurred to me that I forgot to take my morning pills. I have to take 6 PILLS every morning, NOT 6 PILL. I REST MY CASE...
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# Convert 30,000 Milliliters to US Barrels (Dry)
30,000 Milliliters (ml)
1 ml = 8.6e-06 bbl (US)
=
0.259455 US Barrels (Dry) (bbl (US))
1 bbl (US) = 115,627.12 ml
## Data Volume converter
### Q: How many Milliliters in a US Barrel (Dry)?
The answer is 115,627.12 US Barrel (Dry)
### Q: How do you convert 30000 Milliliter (ml) to US Barrel (Dry) (bbl (US))?
30000 Milliliter is equal to 0.259455 US Barrel (Dry). Formula to convert 30000 ml to bbl (US) is 30000 / 115627.1236
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View the step-by-step solution to:
Question
# <pre class="ql-syntax">I need <span class="hljs-keyword">help</span> writing
a scenario topic <span class="hljs-keyword">using</span> statics <span class="hljs-keyword">including</span> four variables </pre>
University of Maryland University College
STAT200 - Assignment #1: Descriptive Statistics Data Analysis Plan Identifying Information
Student (Full Name): Yesenia Vasquez Martinez
Class: Statistics 200 –Assignment #1: Descriptive statistics data analysis plan
Instructor:tdr
Date:
Scenario: Please write a few lines describing your scenario and the four variables (in addition to income)
you have selected.
Use Table 1 to report the variables selected for this assignment. Note: The information for the required
variable, “Income,” has already been completed and can be used as a guide for completing information
on the remaining variables.
Table 1. Variables Selected for the Analysis
Variable Name in the Data
Set Variable 1: “Income” Description
(See the data dictionary for describing the
variables.)
Annual household income in USD. Variable 2: Type of Variable
(Qualitative or
Quantitative)
Quantitative Quantitative
Total Amount of Annual Expenditure on Food Variable 3:
Variable 4:
Marital Status of Head of Household
Variable 5: Qualitative
Quantitative Reason(s) for Selecting the Variables and Expected Outcome(s):
1. Variable 1: “Income” -Income is the upper limit of the budget to be determined, therefore, it is an important factor in helping the household establish the budget accordingly.
The higher the income, the higher the planned budget will be, and the lower income will
restrict the budget. STAT200: Assignment #1 - Descriptive Statistics Analysis Plan - Template
Page 1 of 3 2. Variable 2: “Food “- Food is an essential expense and a budget cannot be completed without including the expense related to this factor. Therefore, food becomes an essential part
of budget planning for each household
3. Variable 3: “ “ - 4. Variable 4: “Marital status “ -The marital status of the head of the family allows us to know the routine expenses. A married head has higher expenses compared to a single person.
Therefore, it is one of the prerequisite factors that affect family budget planning.
5. Variable 5: “Family size “-The purpose of the data is to plan the budget for the home. More family members would be more expenses, therefore it is important to know the number
of family members. Data Set Description:
Proposed Data Analysis:
Measures of Central Tendency and Dispersion
Complete Table 2. Numerical Summaries of the Selected Variables and briefly explain why you choose
those measurements. Note: The information for the required variable, “Income,” has already been
completed and can be used as a guide for completing information on the remaining variables. Table 2. Numerical Summaries of the Selected Variables
Variable
Name
Variable 1:
“Income” Measures of Central
Tendency and Dispersion
● Number of
Observations
Median
Sample Standard
Deviation Rationale for Why Appropriate I am using median for two reasons:
1. If there are any outliers or the data is not normally
distributed, the median is the best measure of
central tendency.
2. The variable is quantitative.
I am using sample standard deviation for three reasons:
1. The data is a sample from a larger data set.
2. It is the most commonly used measure of
dispersion.
3. The variable is quantitative. STAT200: Assignment #1 - Descriptive Statistics Analysis Plan - Template
Page 2 of 3 Variable 2:
Marital Status of Head of
Household
Variable 3:
Variable 4:
Variable 5: Graphs and/or Tables
Complete Table 3. Type of Graphs and/or Table for Selected Variables and briefly explain why you
choose those graphs and/or tables. Note: The information for the required variable, “Income,” has
already been completed and can be used as a guide for completing information on the remaining
variables.
Table 3. Type of Graphs and/or Tables for Selected Variables
Variable
Name Graph and/or Table Rationale for why Appropriate? Variable 1:
“Income” Graph: I will use the histogram to
show the normal distribution of data. Histogram is one of the best plot to show the
normal distribution of quantitative level data . Variable 2:
Variable 3:
Variable 4:
Variable 5: STAT200: Assignment #1 - Descriptive Statistics Analysis Plan - Template
Page 3 of 3
STAT200 Introduction to Statistics
Dataset for Written Assignments
Description of Dataset:
The data is a random sample from the US Department of Labor’s 2016 Consumer Expenditure Surveys (CE) and provides information about the
composition of households and their annual expenditures (https://www.bls.gov/cex/). It contains information from 30 households, where a survey
responder provided the requested information; it is all self-reported information. This dataset contains four socioeconomic variables (whose names start
with SE) and four expenditure variables (whose names start with USD).
Description of Variables/Data Dictionary:
The following table is a data dictionary that describes the variables and their locations in this dataset (Note: Dataset is on second page of this document):
Variable Name
Location in Dataset
Variable Description
Coding
UniqueID# First Column Unique number used to identify each survey
responder Each responder has a unique
number from 1-30 SE-MaritalStatus
SE-Income
SE-FamilySize Second Column
Third Column
Fourth Column
Fifth Column Not Married/Married
Amount in US Dollars
Age in Years
Number of People in Family USD-Food
USD-Meat
USD-Bakery
USD-Fruits Sixth Column
Seventh Column
Eighth Column
Ninth Column Marital Status of Head of Household
Total Annual Household Income
Age of the Head of Household
Total Number of People in Family (Both Adults
and Children)
Total Amount of Annual Expenditures on Food
Total Amount of Annual Expenditure on Meat
Total Amount of Annual Expenditure on Bakery
Total Amount of Annual Expenditure on Fruit Amount in US Dollars
Amount in US Dollars
Amount in US Dollars
Amount in US Dollars How to read the data set: Each row contains information from one household. For instance, the first row of the dataset starting on the next page shows
us that: the head of household is not married and is 39 years old, has an annual household income of \$96,727, a family size of 2, annual food
expenditures of \$7,051, and spends \$904 on meat, \$345 on bakery items, and \$759 on fruit. UniqueID# SE-MaritalStatus SE-Income SE-AgeHeadHousehold SE-FamilySize USD-Food USD-Meat USD-Bakery USD-Fruits 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30 Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Not Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married
Married 96727
95366
95432
96886
97469
95744
98717
94929
97912
96244
96621
97681
96697
96522
96664
95208
106622
95801
97611
97835
107235
101890
107511
95385
106627
107795
107338
105601
96362
99610 39
48
51
44
35
52
40
59
49
56
54
53
49
43
53
52
49
54
44
30
38
48
56
50
56
51
67
19
37
36 2
2
1
2
4
4
3
2
1
4
2
4
2
4
3
4
4
3
5
5
6
2
3
4
3
3
2
4
2
2 7051
7130
7089
6982
6900
7040
7036
6948
6937
7073
7000
7097
6971
6991
7051
8970
10865
9395
9037
8671
10856
11089
10682
9101
10363
11278
11710
10330
8789
9513 904
904
900
917
915
906
889
899
913
918
911
921
898
922
906
1116
1554
1211
1147
1062
1322
1481
1428
1179
1561
1408
1533
1377
983
721 345
344
350
359
335
353
348
345
353
338
344
341
357
349
346
452
534
449
449
390
549
541
564
450
585
544
541
568
355
367 759
760
765
752
773
753
768
771
770
773
768
767
779
758
772
979
1240
1018
994
1005
1156
1157
1169
1001
1178
1231
1324
1098
1146
1025
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# Logarithmic Function – Graph, Definition, Domain and Range
Here you will learn what is logarithmic function graph, formula, domain and range.
Let’s begin –
## Logarithmic Function
Definition : If a > 0 and a $$\ne$$ 1, then the function defined by f(x) = $$log_a x$$, x > 0 is called the logarithmic function.
We know that logarithmic function and the exponential function are inverse of each other.
i.e $$log_a x$$ = y $$\iff$$ x = $$a^y$$
## Logarithmic Function Graph
Case 1 : When a > 1,
In this case, we have
y = $$log_a x$$ = $$\begin{cases} < 0, & \text{for}\ 0 < x < 1 \\ = 0, & \text{for}\ x = 1 \\ > 0, & \text{for}\ x > 1 \end{cases}$$.
Also, the value of y increase with increase in x.
Thus, the graph f(x) = $$log_a x$$ for a > 1 is :
Case 2 : When 0 < a < 1
In this case, we have
y = $$log_a x$$ = $$\begin{cases} > 0, & \text{for}\ 0 < x < 1 \\ = 0, & \text{for}\ x = 1 \\ < 0, & \text{for}\ x > 1 \end{cases}$$.
Also, the value of y decrease with the increase in x.
Thus, the graph y = $$log_a x$$ for 0 < a <1 is :
## Domain and Range
We observe that the domain of the logarithmic function is the set of all non-negative real numbers i.e. $$(0, \infty)$$ and the range is the set R of all real numbers.
Domain : $$(0, \infty)$$
Range : R
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Cody
# Problem 12. Fibonacci sequence
Solution 185192
Submitted on 3 Jan 2013 by Philipp
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; f = 1; assert(isequal(fib(n),f))
2 Pass
%% n = 6; f = 8; assert(isequal(fib(n),f))
f = 1 1 2 f = 1 1 2 3 f = 1 1 2 3 5 f = 1 1 2 3 5 8
3 Pass
%% n = 10; f = 55; assert(isequal(fib(n),f))
f = 1 1 2 f = 1 1 2 3 f = 1 1 2 3 5 f = 1 1 2 3 5 8 f = 1 1 2 3 5 8 13 f = 1 1 2 3 5 8 13 21 f = 1 1 2 3 5 8 13 21 34 f = 1 1 2 3 5 8 13 21 34 55
4 Pass
%% n = 20; f = 6765; assert(isequal(fib(n),f))
f = 1 1 2 f = 1 1 2 3 f = 1 1 2 3 5 f = 1 1 2 3 5 8 f = 1 1 2 3 5 8 13 f = 1 1 2 3 5 8 13 21 f = 1 1 2 3 5 8 13 21 34 f = 1 1 2 3 5 8 13 21 34 55 f = 1 1 2 3 5 8 13 21 34 55 89 f = 1 1 2 3 5 8 13 21 34 55 89 144 f = 1 1 2 3 5 8 13 21 34 55 89 144 233 f = 1 1 2 3 5 8 13 21 34 55 89 144 233 377 f = 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 f = 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 f = Columns 1 through 8 1 1 2 3 5 8 13 21 Columns 9 through 16 34 55 89 144 233 377 610 987 Column 17 1597 f = Columns 1 through 8 1 1 2 3 5 8 13 21 Columns 9 through 16 34 55 89 144 233 377 610 987 Columns 17 through 18 1597 2584 f = Columns 1 through 8 1 1 2 3 5 8 13 21 Columns 9 through 16 34 55 89 144 233 377 610 987 Columns 17 through 19 1597 2584 4181 f = Columns 1 through 8 1 1 2 3 5 8 13 21 Columns 9 through 16 34 55 89 144 233 377 610 987 Columns 17 through 20 1597 2584 4181 6765
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# 17. Exercises. (ii)
2.1) Consider the initial value problem
Find a two-term perturbation approximation u=u(t) for
2.2)
Verify the following order relations
:
a)
b)
c)
d)
2.3) Use Poincaré-Lindstedt’s method to get a two-term perturbation approximation y=y(t) to the problem
2.4) Consider the initial value problem
Use regular perturbation methods to achieve a three-term approximate solution y=y(t) for t>0.
2.5) Show that regular perturbation fails for the boundary value problem
Find the exact solution y=y(t). If
show that
is large. If
show that
Find the inner and outer approxiamtions from the exact solution.
2.6) Suppose that
and use singular perturbation to find an approximate solution y=y(t) to the problem
2.7) Show why singular perturbation fails for the boundary value problem
by comparing with the exact solution y=y(t).
2.8)
Try singular perturbation on the boundary value problem
Discuss the result.
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# UW AMATH 352 - Lecture 5: Floating Point Arithmetic
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Lecture 5: Floating Point ArithmeticAMath 352Wed., Apr. 71 / 15Binary Representation and Base 2 ArithmeticMost computers today use binary or base 2 arithmetic. This isnatural since on/off gates can represent a 1 (on) or a 0 (off), andthese are the only two digits in base 2. In base 10, a naturalnumber is represented by a sequence of digits from 0 to 9, with theright-most digit representing 1’s (or 100’s), the next representing10’s (or 101’s), the next representing 100’s (or 102’s), etc. In base2, the digits are 0 and 1, and the right-most digit represents 1’s (or20’s), the next represents 2’s (or 21’s), the next 4’s (or 22’s), etc.2 / 15Binary Representation of Natural NumbersConsider the decimal number 27. To find its binary representation,first find the highest power of 2 that is less than or equal to 27;this is 24, so a 1 goes in the fifth position from the right of thenumber: 1 .Then subtract 24from 27 to find that the remainder is 11. Since23is less than 11, a 1 goes in the next position to the right: 11 .Subtracting 23from 11 leaves 3, which is less than 22, so a 0 goesin the next position: 110 . Since 21is less than 3, a 1 goes in thenext position, and since 3 − 21= 1, another 1 goes in theright-most position to give 27 = 110112.3 / 15Binary Representation of Natural NumbersConsider the decimal number 27. To find its binary representation,first find the highest power of 2 that is less than or equal to 27;this is 24, so a 1 goes in the fifth position from the right of thenumber: 1 .Then subtract 24from 27 to find that the remainder is 11. Since23is less than 11, a 1 goes in the next position to the right: 11 .Subtracting 23from 11 leaves 3, which is less than 22, so a 0 goesin the next position: 110 . Since 21is less than 3, a 1 goes in thenext position, and since 3 − 21= 1, another 1 goes in theright-most position to give 27 = 110112.3 / 15Binary Representation of Natural NumbersConsider the decimal number 27. To find its binary representation,first find the highest power of 2 that is less than or equal to 27;this is 24, so a 1 goes in the fifth position from the right of thenumber: 1 .Then subtract 24from 27 to find that the remainder is 11. Since23is less than 11, a 1 goes in the next position to the right: 11 .Subtracting 23from 11 leaves 3, which is less than 22, so a 0 goesin the next position: 110 . Since 21is less than 3, a 1 goes in thenext position, and since 3 − 21= 1, another 1 goes in theright-most position to give 27 = 110112.3 / 15Binary Arithmetic with Natural NumbersBinary arithmetic is carried out in a similar way to decimalarithmetic, except that when adding binary numbers one mustremember that 1 + 1 is 102. To add the two numbers 10 = 10102and 27 = 110112, we align their binary digits and do the additionas below:1 0 1 0+ 1 1 0 1 1− − − − − −1 0 0 1 0 1You can check that 1001012is equal to 37. Subtraction is similar,with borrowing from the next column being necessary whensubtracting 1 from 0. Multiplication and division follow similarpatterns.4 / 15Rational Numbers in Base 2Just as we represent rational numbers using decimal expansions,we can also represent them using binary expansions. The digits tothe right of the decimal point in base 10 represent 10−1’s (tenths),10−2’s (hundredths), etc., while those to the right of the binarypoint in base 2 represent 2−1’s (halves) 2−2’s (fourths), etc. Forexample, the fraction 11/2 is 5.5 in base 10, while it is 101.12inbase 2: one 22, one 20, and one 2−1.Not all rational numbers can be represented with finite decimalexpansions. The number 1/3, for example, is .333, with the barover the 3 meaning that this digit is repeated infinitely manytimes. The same is true for binary expansions, although thenumbers that require an infinite binary expansion may be differentfrom the ones that require an infinite decimal expansion.5 / 15Rational Numbers in Base 2Just as we represent rational numbers using decimal expansions,we can also represent them using binary expansions. The digits tothe right of the decimal point in base 10 represent 10−1’s (tenths),10−2’s (hundredths), etc., while those to the right of the binarypoint in base 2 represent 2−1’s (halves) 2−2’s (fourths), etc. Forexample, the fraction 11/2 is 5.5 in base 10, while it is 101.12inbase 2: one 22, one 20, and one 2−1.Not all rational numbers can be represented with finite decimalexpansions. The number 1/3, for example, is .333, with the barover the 3 meaning that this digit is repeated infinitely manytimes. The same is true for binary expansions, although thenumbers that require an infinite binary expansion may be differentfrom the ones that require an infinite decimal expansion.5 / 151/10 in Base 2For example, the number 1/10 = 0.1 in base 10 has the repeatingbinary expansion: 0.00011002. To see this, one can do binary longdivision in a similar way to base 10 long division:.0 0 0 1 1 0 0− − − − − − − − −1 0 1 0 / 1. 0 0 0 0 0 0 0 01 0 1 0− − − −1 1 0 01 0 1 0− − − −1 0 0 0 06 / 15Fixed Point RepresentationA computer word consists of a certain number of bits, which canbe either on (to represent 1) or off (to represent 0). Some earlycomputers used fixed point representation, where one bit is used todenote the sign of a number, a certain number of the remainingbits are used to store the part of the binary number to the left ofthe binary point, and the remaining bits are used to store the partto the right of the binary point. The difficulty with this system isthat it can store numbers only in a very limited range. If, say, 16bits are used to store the part of the number to the left of thebinary point, then the left-most bit represents 215, and numbersgreater than or equal to 216cannot be stored. Similarly, if, say, 15bits are used to store the part of the number to the right of thebinary point, then the right-most bit represents 2−15and nopositive number smaller than 2−15can be stored.7 / 15Floating Point RepresentationA more flexible system is floating point representation, which isbased on scientific notation. Here a number is written in the form±m × 2E, where 1 ≤ m < 2. Thus, the number 10 = 10102wouldbe written as 1.0102× 23, while110= 0.00011002would be writtenas 1.10011002× 2−4.The computer word consists of three fields: one for the sign, onefor the exponent E , and one for the significand m. A singleprecision word consists of 32 bits: 1
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# Trapezoid 3d
Click a category below to view related FAQs. Download Trapezoid stock photos. The Computer-Aided Design ("CAD") files and all associated content posted to this website are created, uploaded, managed and owned by third party users. The Trapezoid Area Calculator will calculate the area of a trapezoid if you enter in the height, the length of the top, and the length of the top of the trapezoid. lateral sides, angle at the base and other base 3. A quadrilateral has 4 straight sides. 1) I create a horizontal rectangle shape. 5 cm high - you can triple those measurements for a nicer effect! Next, take one of your dusty CD cases, break off the holders and the sides, and cut out four trapezoids. A trapezoidal prism is a three dimensional geometric shape that consists of a trapezoid or trapezium shape on one cross section, and a rectangle on the other cross sections. The triangle creeps into this page because we need to know its area in order to compute the area of a trapezoid. You may be assuming that a 3 dimensional square is called a cube and that by analogy there must be a 3D trapezoid but this assumption is incorrect. Next your white trapezoid with the longest side on the bottom. w k aAulQlc arDiwg[hwt\sJ erXeEswemrkvoeydL. However, since I have an 8:12 roof/wall pitch and a 4:12 ceiling pitch I am selecting slope to ceiling. by KLucky13 May 1, 2014. The perimeter of a trapezoid is the sum of all its sides. The formula for the area of a sector is (angle / 360) x height x π x radius 2. The louvers can open and close to control solar energy, light and glare. A wedge is a subclass of the prismatoids with the base and opposite ridge in two parallel planes. A trapezoid is defined as a quadrilateral with two parallel sides. Engineering ToolBox - SketchUp Extension - Online 3D modeling! Add standard and customized parametric components. Diecast Aluminum (R100-R111, R190 & R191, R196, R270-273 and R280 series). The Perimeter is the distance around the edges. A wide variety of 3d trapezoid options are available to you, such as roof. Put together two identical shapes to make a new shape. Learn to create 3D Cone in 4 easy steps. An area is the size of a two-dimensional surface. If it were me, I wouldn't try to build each trapezoid from separate parts. by cypax Jul 25, 2014. UHF Quad-Trapezoid-Loop + 11RR 3D View: enlarge 278KB, 1017x1024 17 UHF Quad-Trapezoid-Loop + 11RR Front View [1 large square = 2. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. The architect designs the shape of the building, chooses the materials it will be built with, and it is architect’s responsibility to prove that the building will be durable, saf. A quadrilateral with only two of its sides parallel is called a trapezium, or trapezoid. Draw a 3D trapezoid with help from an artist in this free video clip. The formula for the area of a sector is (angle / 360) x height x π x radius 2. kjmarkey May 7, 2017, 12:48pm #1. In the video above, we look at 3D tools including the Extrude, Revolve, Sweep and Loft features, within Fusion 360. Drawing Types. trapz performs numerical integration via the trapezoidal method. K-12 students may refer the below formulas to know what are all the parameters are being. She has a Bachelor in English Education and her MBA. An ovoid is a 3d oval, so why isn't a trapezoid a 3d trapeze(-ium), i. com where you can measure up, measure down or measure all around! This page includes Measurement worksheets for length, area, angles, volume, capacity, mass, time and temperature in Metric, U. Opposite faces are parallel. 7 yd 11 yd 3) 48 in 6. A quadrilaterally-based pyramid is a wedge in which one of the edges between two trapezoid faces has collapsed into. The trapezoid is a plane figure which has surface Area, but no volume but if there was a 3d figure your equation would be. Let's get started!. Find the word "Workplane" on your screen and use the 3D modeler's salute to make sure you know which axis is which. In the aforementioned post, I described a shape, namely, an irregular triangular prism, that was analogous to a trapezoid in terms of its volume formula (the proof is located here). sh3f library of 3D models contained in the 3DModels-Scopia-1. The illusion also uses perspective angles. An optical grade trapezoid stereoscopic mirror is great for photography, telescopes, 3D printers, film transfer devices and so much more! skip to Main Content (419) 787-4526 [email protected] Browse synchronous timing belt width in stock and ready for shipping right now on the internet. This trapezoid shape will also be reused throughout the arch. When the problem has a solution, the outputs are: the angles A, B, C and D, the height h, the area and the lengths of the diagonals AC and. Instructions. A trapezoid can be drawn on 2d surfaces like board,page etc. Some regular solids have been named since antiquity, and some are still being named using greek forms. Of course the simple solution to THIS example is to profile the trapezoid and extrude IT -- BUT since my. A trapezoid is a quadrilateral (four sided figure) with (at least) one pair of parallel sides. a flat shape with four…. Rotating 2D shapes to make 3D Shapes. Next your white trapezoid with the longest side on the bottom. Click a category below to view related FAQs. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. Contact Author. In another words, Centroid of a Trapezoid is geometrically lies on the median. You can print them in high resolution and play with toddlers by coloring simple designs or with older children with more difficult patterns. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. REGENTS EXAM REVIEW CLASSES. Use the trapezoid calculator to find the area of the end of the gold bar with bases 4 cm and 2 cm and height 3 cm. I would like to split it up into N number of layers, lets say when we have N=5 the layers are S,H,S,H,S where volume of all S = Volume of all H. There are equilateral, isosceles, and right triangles. Load Saved JSON Load Saved Table. In this article, you will learn to create the diagram in a step by step way using basic 3D tools in PowerPoint. STEP 3: To find the lengths of AB and CD, we need to find x. 2 L 3BY3S 3D q —0 22 q 7. QuickShip available in: Sugar Maple (QS4600BKSSM) Grey Nebula (QS4600B. In OpenSCAD Extrusion is always performed on the projection (shadow) of the 2d object xy plane and along the Z axis; so if you rotate or apply other transformations to the 2d object before extrusion, it's shadow shape is what is extruded. Trapezoid Amandazine. and Trapezoids For use with pages 537–544 Solve. The discussion led to a neat generalization of the trapezoid to higher dimensions different from the one that envisions the 3D trapezoid as a pyramidal frustum. Use the trapezoid calculator to find the area of the end of the gold bar with bases 4 cm and 2 cm and height 3 cm. 7 3D Trapezoid models available for download. This tool calculates the moment of inertia I (second moment of area) of a trapezoid. The Isosceles Trapezoid can sometimes cause quite a bit of confusion. I guess what I ultimately want is to take the dimensions of the hallway as represented by the 2d trapezoid plane and calculate the dimensions of the hallway without the perspective distortion. 3D Printer Accessories. Stick them together. How To: Classify a Triangle as an Isosceles Triangle. The second section features shapes that must be measured by the student first. Lamp Problems - Lamp not lighting, replaced lamp still not working, and more. This is an excellent review for the FCAT math test. Please enter angles in degrees, here you can convert angle units. The median of the trapezoid m connects the midpoints of the two non-parallel sides c and d where m = (a+b)÷2. It would be nice if a formula to find the lengths of the diagonal sides of trapezoid was included [3] 2018/12/03 10:50 Female / Under 20 years old / Elementary school/ Junior high-school student / Very /. Three common open channel cross sections, the rectangle, trapezoid, and triangle, are covered in this article. first you'd figure out the dimensions of a cone if you tended the sides of the "trapezoidal cylinder" until. With the CSS transform property you can use the following 3D transformation methods: The rotateX () method rotates an element. Then place the red trapezoid through next longest side on the bottom. It is very beneficial when precise light control is important in venues such as galleries or. Find the word "Workplane" on your screen and use the 3D modeler's salute to make sure you know which axis is which. Trapezoid (area) (a + b) x h 2 Don’t forget, that you’ll also need to know how to calculate the perimeter of an object. Find the perimeter of the trapezoid. Multi-step Trapezoid Method. Grab this free printable set of foldable 3D shapes here from Math Geek Mama. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. The formula for the area of a trapezoid is A = ½(b 1 +b 2)h, where b 1 and b 2 are the lengths of the bases and h is the height. A facile method has been developed to detect pathogenic bacteria using magnetic nanoparticle clusters (MNCs) and a 3D-printed helical microchannel. Ellipse Area Formula and Ellipse Perimeter Formula. Back to Contents. Relationship to 3D-Shapes. e the shape at the top of this post? We need to look at it in a different way. The MSI Carrara White Trapezoid Mosaic Tile lends a distinct pattern to your kitchen, bathroom or any other area in your home or commercial space. Contact Author. And here you have some additional examples for inspiration. Note that the area of each trapezoid is the sum of a rectangle and a triangle. Then draw a 90 degree angle at one end of the base, using a protractor. Is there some way I can take a rectangular selection and reduce the height on just one side of it to create a trapezoid? (The idea is to make it look as though the object is being viewed from an angle. HAVING SAID THIS, it is a good idea for you as a parent/tutor to have your child draw or deconstruct 3d-shapes from a trapezoid that they do have to deal with, such as the Cuboid and the triangular prism. The Best 3d Wooden Star Plans Free Download PDF And Video. A prism with six congruent faces. Find the best 1 free fonts in the Trapezoid style. 3d rendered medically accurate illustration of the trapezoid bone;. Program Provide a mixed-use building with ground floor commercial space of 7,000 SF, with eight upper story apartment units. Rotate Triangle. Trapezoid is a two dimensional plane or geometric shape represents quadrilateral with either no sides parallel to each other or with only two sides are parallel to each other in SI & US customary units measurements. The area of a trapezoid is basically the average width times the altitude, or as a formula: b1, b2 are the lengths of each base. A prism with six congruent faces. A freely available dataset of iconic gesture performances refering to 20 different 3D objects. Moreover, forward and. It's not a shape that has been rotated into the 3rd dimension (like an oval is to become an ovoid), so it's more like a type of prism, i. Find the area of just the trapezoidal face A = 1 2h(b1+b2), then multiply it by the length of the entire prism. Various shapes and units of measurement are used. Trapezoid Calculator. Measure 3 inches in from the opposite corner of the same side and make another mark. The height of a. In case of arbitrary spaced samples, the two functions trapz and simps are available. It doubles as wall or floor tile and has a detailed pattern of arrangement with tile pieces sized on a 12 in. QuickShip available in: Sugar Maple (QS4600BKSSM) Grey Nebula (QS4600B. There are certain properties of trapezoids that identify them as trapezoids-The base angles and the diagonals of an isosceles trapezoid are equal. Three-dimensional figures, also referred to as 3D objects, are figures in space that have length, width and depth. Therefore, for a trapezoid with sides a, b, c and d, the formula of the perimeter can be written as-Perimeter= a+b+c+d. Download this FREE Revit Family / model of a Trapezoid office desk including castors in 3D view. Comparisons: A wedge is a parallelepiped where a face has collapsed into a line. 3DPX-002128 Santiago Ramón y Cajal - Calyx of Held Discover 3D Prints. The top countries of suppliers are India, China, from which the percentage of 3d trapezoid supply is 1%, 99% respectively. 4" opposite side, 0. Since a trapezoid. The part starts out looks looking like this: Length: 12. When inserting a trapezoid window into a gable end wall with a cathedral ceiling I can't seem to get the desired results. 7 3D Trapezoid models available for download. However by combining the 2D mediums you can generate a 3D object. In theory, using this setting on a meter will allow you to scale it, to rotate it, to flip it, to skew it in any way you choose. Find the word "Workplane" on your screen and use the 3D modeler's salute to make sure you know which axis is which. Has control for Top and Bottom stretch. For smooth operation, layers should be as low as possible. 3D design 'This is a Freaking Trapezoid!' created by Cashew Man with Tinkercad. To create more complicated geometry, such as polygons, polylines, or stars you use another class from this package, GeneralPath. Guided notes and practice problems combined – see link below: trapezoid and kite notes. Calculations at a right trapezoid (or right trapezium). It has three dimensions: length, width, and height. Find the average of the sum of the two bases and multiply by the height to find the area of the trapezoids. Moreover, forward and. Volume of a Trapezoid Calculator. UNI Trapezoid Motorcycle Air Filters for Honda, Trapezoid Plastic Motorcycle Air Filters for Suzuki, Backgammon Doubling Cubes, Trapezoid Motorcycle Air Filters for Kawasaki Ninja ZX6R, Trapezoid Plastic Motorcycle Air Filters for Yamaha YZF R1, Cube YuXin Brain Teasers & Cube/Twist Puzzles, Ice Cube Ice Cube Bags, Shining 3D 3D Scanners,. 00001255 6 64 0. The word "workplane" will be parallel with the X axis. In coordinate geometry, if you know the coordinates of the four vertices, you can. pair of parallel sides a right trapezoid has two right angles Free download right wallpaper, here we provide some of wallpaper on right, as well asright image and right Picture. Our Tech Support team tackles tough technical questions every day. And here you have some additional examples for inspiration. In geometry, a trapezoid is a quadrilateral (four-sided figure) in which only one pair of opposite sides are parallel. Enclosure Depth (Less Lid) Small End. The bone resembles a four-sided table in the geometrical shape of a trapezoid, with its dorsal side (on the side of the back of the hand) being two times wider than its palmar side. Lines AC (or q) and BD (or p) are called diagonals The line perpendicular to lines AD & BC is called the height or altitude. Therefore, for a trapezoid with sides a, b, c and d, the formula of the perimeter can be written as-Perimeter= a+b+c+d. The given data consists of: As the actual height is not given, we have to use equation no. This is because people often refer to it as simply a Trapezoid, but as you will see from this page, even though it is a trapezoid, it has some extra special features making it an isosceles shape. 3D Shapes. Clip Art of 2D and 3D Geometric Shapes. The center of gravity will be in the intersection between the middle line CD and the line between the triangles centers of gravity. Tabbing to the second worksheet, you’ll find the 3D cube, which starts with a cube (defined with one center point and 16 corner points), three. To construct trapezoids: Type Trapezoid at the command prompt. 5" Width: 1. 3DPX-002128 Santiago Ramón y Cajal - Calyx of Held Discover 3D Prints. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. How many trapezoids can she make? Harrison drew a trapezoid. Specialty Shapes Specialty Shape windows utilize advanced engineering and the highest quality construction to create open, light-filled spaces that blur the line between indoors and out. As a guide, threads larger than M12 or 1/2″ can be successfully printed with 0. Moving forward, we can extrude our 2D sketch into a 3D object. 0 Likes | 114 Downloads | 422 Views. where h is the height, the 127 dimension in your drawings, B 1 and B 2 are the areas of the two endcap trapezoids, and M is the area of a midsection cut so B 1 = (16/2)*(15 + 7. Then put glue on the emblem and place it in the center. Trapezoid 3D models ready to view, buy, and download for free. Buy trapezoid CG textures & 3D model from $6. K8200 / 3Drag double z-axis spindle TR16x4. 7 yd 11 yd 3) 48 in 6. Guided notes and practice problems combined – see link below: trapezoid and kite notes. What is the best way to create a 3-D trapezoid 9. Integration tool performs numerical integration on the active data plot using the trapezoidal rule. Position the nails so that they will drive directly through the angle face of the adjoining boards without shearing out through the side. Relationship to 3D-Shapes. The area of a trapezoid is h(b1+b2)/2, where b1 and b2 are the lengths of the two bases and h is the height of the trapezoid. The second section features shapes that must be measured by the student first. Find the area of just the trapezoidal face A = 1 2h(b1+b2), then multiply it by the length of the entire prism. A shape is the outer form of an object or figure such as a circle, triangle, square, rectangle, parallelogram, trapezoid, rhombus, octagon, pentagon, and hexagon. Round to the nearest tenth. Locations in the cube are defined by x-, y- and z- coordinates. To get to it, I'll go up to the Edit menu in the Menu Bar and I'll choose Transform. In this article, you will learn to create the diagram in a step by step way using basic 3D tools in PowerPoint. The lateral area is the surface area of a 3D figure, but excluding the area of any bases. The acute trapezoid has two acute angles located on each side of the long base. Again only put glue on the narrowest edge. Is there some way I can take a rectangular selection and reduce the height on just one side of it to create a trapezoid? (The idea is to make it look as though the object is being viewed from an angle. Slicing a rectangular pyramid. Geometry Library - Students write and illustrate books to make a class library of math term books. CLASSIFY SHAPE RECTANGLE (opposite side congruent/90 degree angles) QUADRILATERAL (4 sided polygon) POLYGON (closed figure made up of line segments). Mfr's Part #: 48-22-1901. This trapezoid shape will also be reused throughout the arch. Again only put glue on the narrowest edge. These illustrations allow to review the basis of the anatomy of the shoulder and to locate them more easily on an MRI by using cross-references. When inserting a trapezoid window into a gable end wall with a cathedral ceiling I can't seem to get the desired results. How to use the trapezoid calculator Enter the 4 sides a, b, c and d of the trapezoid in the order as positive real numbers and press "calculate" with b being the short base and d being the long base (d > b). Use a craft knife to cut the lined area in the isosceles trapezoid shapes in teach wing. If you don't want to perform the calculation by yourself, just enter the data in the fields and press the button to see the results below. Each lower base angle is supplementary to […]. 3D Printing Threads and Screw. The area of a trapezoid is the space contained within its 4 sides. A quadrilateral having two and only two sides parallel is called a trapezoid. A trapezoid is a 2 dimensional figure. A trapezoidal prism is a three-dimensional figure that consists of two trapezoids on opposite faces connected by four rectangles. basic geometry shapes basic geometric shapes vector basic geometric shapes outline of shapes quatrefoil shape basic geometric geometric character vector geometric shapes 2d simple oval diamond shape outline. Octagon & Trapezoid Diecast Aluminum Enclosures 3D Solid Model. All orders are custom made and most ship worldwide within 24 hours. Explanation:. first uses simple geometry: Use the formula for the volume of a cone: 1/3 * pi * r^2 * h. 3D crystal trapezoid awards make wonderful corporate gifts recognition awards promotional pieces and keepsakes. Note that the area of each trapezoid is the sum of a rectangle and a triangle. Let's get started!. The shadow of this shape is called Trapezium. Faye has 3triangles and parallelograms. Although a static 2D trapezoid is usually perceived to 3D de pth, once a trapezoid is se t into rotary motion, the d epth perception becomes stronger and typically is bistable–only one of two. If not load a new reel. So the output would be the dimensions of a 2d rectangular plane with dimensions relative to the original trapezoid. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. Clock Constrain Easing Keyboard Mouse 1D Mouse 2D Mouse Press Mouse Functions Mouse Signals Storing Input Rollover. Since a trapezoid. Use MathJax to format equations. Load Saved JSON Load Saved Table. Trapezoid 3D models. 3d rendering. The parallel sides a and b are called bases, and h is the altitude or distance between the bases. A shape known as a Trapezoid in North America or a Trapezium in the rest of the English speaking world, can come in many shapes, but has to have at least one pair of parallel running sides. Printable Trapezoid Shape to Cut Out for Projects. The trapezoid cut diamonds are used mainly for side stones and are cut in brilliant-cut (shown above left) or step-cut (shown above right). 2 for solving this problem. Area: Rectangles, Triangles, Parallelograms, Trapezoids Name_____ ©Z h2[0D1J6o fKvuvtcaK `SioEfctWwqaDrGeS gLZL_Ct. How to use square in a sentence. This feature promotes mathematical understanding of 3D graphs and helps in learning solid figures. Explanation:. Geometry Library - Students write and illustrate books to make a class library of math term books. The Trapezoid Table is a great alternative to individual desks. 5) Make a trapezoid whose bases are sloping upward. mesh sheet, making installation simple. The Computer-Aided Design ("CAD") files and all associated content posted to this website are created, uploaded, managed and owned by third party users. trapezium definition: 1. A Trapezoidal Prism contains six faces. By using this website, you agree to our Cookie Policy. Buy Trapezoid Hsampc Scott Sort Order Best Match / Completing / Lowest Prices / Price High To Low / New Browse By Pricing$58 / $170 /$239 / $407 /$551 / $863 /$1205 / $1930 /$2712 / $4059. Affordable and search from millions of royalty free images, photos and vectors. Area of Trapezoids involving Unit Conversion. Lateral side (leg) and angle at the base 3. First, we need a little terminology/notation out of the way. One of the coolest, but undoubtedly most confusing additions to Rainmeter is the TransformationMatrix setting. Popular Trapezoid 3D models View all. A trapezoidal prism has six faces, eight vertices and 12 edges. All faces meet at 90-degree angles. Trapezium (shape) synonyms, Trapezium (shape) pronunciation, Trapezium (shape) translation, English dictionary definition of Trapezium (shape). a flat shape with four sides, where two of the sides are parallel 2. Clock Constrain Easing Keyboard Mouse 1D Mouse 2D Mouse Press Mouse Functions Mouse Signals Storing Input Rollover. The various existing versions of MS word change how I would direct you to menus, however, generally one method is to simply draw the shape you want with the 'draw line' tool (where ever it is found on your version). A trapezoid midsegment is parallel to the set of parallel lines in a trapezoid and is equal to the average of the lengths of the bases. In mathematics, and more specifically in numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral. What is the triangle’s area? 2. Trapezoid-shaped floor tiles allow for unique and interesting compositions. Medically accurate illustration of the trapezoid ligament The trapezoid ligament. are important when setting up a table saw to make an object involving the shape of a regular polygon. Directions: Use pattern blocks to help you solve the problems. With the picture cut into a trapezoid, adjust the angle of the trapezoid so that you can see the entirety of the shape that you want to have pop-out of your picture (creating the 3D effect). First, draw the long base. The area is in whatever designation square units you have used for the entries. The three-dimensional figure sides are its flat surfaces. Trapezoid Box Description. How to use square in a sentence. A trapezoid is defined as a quadrilateral with two parallel sides. Now that we've seen several types of quadrilaterals that are parallelograms, let's learn about figures that do not have the properties of parallelograms. …I'll drag a pyramid onto my work plane, and then…orbit and zoom in. A trapezoid midsegment is related to a triangle midsegment given that both of their lengths are proportional to the bases. If PQ = 34 and NP = 16, find MP. I printed a large round piece for a buddy of mine the other day and the print came out great, like always. More information is needed. Although a static 2D trapezoid is usually perceived to 3D de pth, once a trapezoid is se t into rotary motion, the d epth perception becomes stronger and typically is bistable–only one of two. Stainless Steel Backsplash offers +35 unique and affordable metal and non-metal tile patterns in stainless steel, copper & aluminum, including glass and stone mixed designs. Every font is free to download and preview for your projects. 613 Trapezoid clip art images on GoGraph. Find the side of a right trapezoid if given 1. Now for some reason, this printer cannot print a circle. With inequalities, you can add colored shading to your Desmos graph. 1590TRPB. The louvers can open and close to control solar energy, light and glare. Among this six faces, four faces are rectangular and remaining two faces are trapezoidal. Grab this free printable set of foldable 3D shapes here from Math Geek Mama. Stick them together. Mostly the bases are parallel. @3dmedieval: You are not mapping the trapezoid into rectangle, you are mapping it into a trapezoid region in the texture. The area, the angles and the diagonals of a Trapezoid are calculated given its 4 sides. Moment of Inertia. The quadrilateral is a. 1) Using templates. Title: Grade 3 geometry worksheet: Naming quadrilaterals Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter. Watch this video to learn how to draw a right trapezoid using given specifications for lengths of sides. To understand how to calculate square footage we must first begin with the definition of area. Bluprint - Woodworking Get 3d Wooden Star Plans: Learn techniques & deepen your practice with classes from pros. triangle ( x1, y1, x2, y2, x3, y3) float: x-coordinate of the first point. To find the height and base of the trapezoid that is used in the area of a trapezoid formula, we need to use the following formulas. Tocheri a,b,*, A. This fact and the properties of quadrilaterals can be used to calculate angles. Our 3D laser etched crystals are produced using our digital laser etching systems that engrave your 3D images inside optical crystal. We’re the ideal introduction to Autodesk, the leader in 3D design, engineering and entertainment software. To find the height and base of the trapezoid that is used in the area of a trapezoid formula, we need to use the following formulas. More Geometry Gifs. Outside the US, "trapezium" = US::"trapezoid", and US::"trapezium" is just "irregular". 3D models created by Kator Legaz. The formula is 1/2 height * sum of parallel. Slicing a rectangular pyramid. - [Instructor] Next up is rotating 3D shapes. Perfect for a new product launch, new slogan launch, or a company logo!. Three common open channel cross sections, the rectangle, trapezoid, and triangle, are covered in this article. A trapezoid (or trapezium) is a tetragon with two parellel sides. This last equation is the one that tells us how the 3D-spheres (projections of the 4D-sphere in 3D-space) that we want to draw depend on time: the left-hand side of the last equation is the 3D-radius of those spheres, while the right-hand side is a function of t that employs two constants: the predetermined radius d of our 4D-sphere, and the. By using this website, you agree to our Cookie Policy. Thus the area of the trapezoid is A 1 2 a b a b 1 2 a b 2. Rotate Rectangle. Khan Academy is a 501(c)(3. Diecast Aluminum (R100-R111, R190 & R191, R196, R270-273 and R280 series). Revolved Trapezoid Calculate the volume, side lengths, angle, etc. Ok, obviously a newbie question. There are four bones in the distal row: It's easy to confuse the location of the trapezoid with the adjacent and similarly named trapezium bone. If not, explain why not! 1) Make a trapezoid that has one right angle. An acute trapezoid has two adjacent acute angles on its longer base edge, while an obtuse trapezoid has one acute and one obtuse angle on each base. 3D Trapezoid models are ready for animation, games and VR / AR projects. 2-mm layers, while smaller threads should be printed with thinner layers. Stick them together. stp files (zipped) Width. So I need to find a way to change my copied rectangle into a trapezoid that will make it look natural on the new image. mesh sheet, making installation simple. This tool calculates the moment of inertia I (second moment of area) of a trapezoid. Rotate Triangle. The machine has been identified as the first of its kind and it can generate a powerful and. ,so it is a 2d shape. By using 3D and Edit Shapes you can easily change the look and feel for your PowerPoint 3D diagrams, for example, we have edited the shape from the 3D pyramid to use a trapezoid in PowerPoint and this is the result. I would like to split it up into N number of layers, lets say when we have N=5 the layers are S,H,S,H,S where volume of all S = Volume of all H. You can also choose from colored steel, glazed steel 3d trapezoid There are 344 suppliers who sells 3d trapezoid on Alibaba. Tabbing to the second worksheet, you’ll find the 3D cube, which starts with a cube (defined with one center point and 16 corner points), three. Connect your entire product development process in a single cloud-based platform with tools for 3D CAD, CAM, and CAE. * Tile thickness could vary based on size. trapezium definition: 1. Step 2: The side-lengths of the trapezoid are 3 cm, 3 cm, 6 cm and 11 cm. Sails We known heights 220, 165 and 132 of sail. She has been mightily, saved by grace and is grateful for God’s sovereignty throughout her life’s journey. See the complete profile on LinkedIn and discover akinlolu’s connections and jobs at similar companies. So the volume of a 3D trapezoid prism is (adding depth d): d * h * (a + b) / 2. I am trying to draw a shelf that is a isoscoles trapazoid. A three-dimensional trapezoid is also known as a trapezoidal prism. Solution: Given base lengths are b 1 = 6 cm, b 2 = 11 cm, and h = 5 cm. It’s easy to draw 3D graphs using templates. I like to tell this story as part of the lesson when I'm teaching classifying quadrilaterals. Now, we are going to make hole. Explanation:. Again only put glue on the narrowest edge. Google will ask you to confirm Google Drive access. Experiment with reflections across any line, revolving around any line (which yields a 3-D image), rotations about any point, and translations in any direction. The integration of light and shadow effects create dimensional vertical surfaces using simple shapes that shifts and transform. This 1,448 square foot condo features 3 bedrooms and 2. Trapezoid stock photos and images (1,103) Best Match Fresh. Find the best 1 free fonts in the Trapezoid style. Here you will find our list of different Geometric shapes. 2 Name Tru-Size™ Angled Trapezoid Diamond Knife 1. 3DCasement windows can show open in 3D if a Hinge Side is specified. 5 cm high - you can triple those measurements for a nicer effect! Next, take one of your dusty CD cases, break off the holders and the sides, and cut out four trapezoids. Trapezoid is a quadrilateral with two parallel sides and centroid of a trapezoid lies between two bases. Sci Rep 5. The triangle creeps into this page because we need to know its area in order to compute the area of a trapezoid. Bluprint - Woodworking Get 3d Wooden Star Plans: Learn techniques & deepen your practice with classes from pros. Medically accurate illustration of the trapezoid ligament The trapezoid ligament. Buy Aluminum Alloy Trapezoid Ruler now and monitor Aluminum Alloy Trapezoid Ruler deals remotely using browser, xml, email, to save on Aluminum Alloy Trapezoid Ruler. 3D Geometry; Euclidean Geometry Area Warmup Area From Rectangles to Triangles Area From Rectangles to Parallelograms and Trapezoids If the rectangle and. Pattern is on swatches panel The trapezoid. 22 cm Correct Answer: A. K8200/3Drag Z-motor mount. The area of a trapezoid is the space contained within its 4 sides. If you have problems using this site, or have other questions, please feel free to contact us. I need some help please. The parallel sides of a trapezoid are called the bases. 3) Make a right isosceles trapezoid (an isosceles trapezoid with a right angle) 4) Make a trapezoid whose bases are VERTICAL. a solid having bases or ends that are parallel. Special transform node that modifies the input in a perspective/trapezoid warp manner. The acute trapezoid has two acute angles located on each side of the long base. Layer height is an important parameter when printing threads. E-series windows added (Eagle Windows removed). Two sides of the quadrilateral can cross each other --a complex quadrilateral -- which makes your quadrilateral look like two adjoining triangles. The following figure shows a trapezoid to the left, and an isosceles trapezoid on the right. Comparisons: A wedge is a parallelepiped where a face has collapsed into a line. An optical grade trapezoid stereoscopic mirror is great for photography, telescopes, 3D printers, film transfer devices and so much more! skip to Main Content (419) 787-4526 [email protected] You can print them in high resolution and play with toddlers by coloring simple designs or with older children with more difficult patterns. Downloadable Item(s) Can be received immediately. Look at the filament reel and see if there’s any filament left. This fact and the properties of quadrilaterals can be used to calculate angles. Square definition is - an instrument having at least one right angle and two straight edges used especially to lay out or test right angles. Enter the lengths of the two parallel sides a and c and either base b or slant side d. Area of trapezoids, derive area formula of trapezoids, solve problems using area of trapezoids, examples, worksheets, examples and step by step solutions, How to find the height of a trapezoid given the area, How to use the area of a trapezoid to find a missing base length. I found volume of a cross section using Isosceles Trapezoids (three equilateral triangles together), using the equation x 1/2 * cos(pi * x) + 3, x from [0,7. Free worksheets for classifying quadrilaterals With this worksheet generator, you can make worksheets for classifying (identifying, naming) quadrilaterals, in PDF or html formats. * Tile thickness could vary based on size. Area of trapezoids, derive area formula of trapezoids, solve problems using area of trapezoids, examples, worksheets, examples and step by step solutions, How to find the height of a trapezoid given the area, How to use the area of a trapezoid to find a missing base length. trapezoid n. …Then pull a little bit further. One of the coolest, but undoubtedly most confusing additions to Rainmeter is the TransformationMatrix setting. Multiply this by the height (distance between the parallel sides. It’s not a shape that has been rotated into the 3rd dimension (like an oval is to become an ovoid), so it’s more like a type of prism, i. If you know the 3D-view, you can three-dimensionally look at the following two cube pairs. Popular Trapezoid 3D models View all. Lines AB and DC are the non-parallel sides and are called legs. This pom nut use for Tr8 lead screws, Prusa i3 MK3 MK2 MK2 42 Stepper Motor 3D Printer Z Axis Trapezoid Motor Screw Bolt Spacing: 19mm, Lead Screw Diameter: 8mm, Pitch: 2mm, Lead of thread: 8mm Material: POM Color: Black After-sales guarantee: each defective item can be replacement for free or refund. Rotate Circle. 2mm Product Number E0130 1. A hexagon is a polygon with six sides. Solution for 8) In isosceles trapezoid MNPQ with MN || QP, diagonal MPI MQ. of a trapezoid shape revolved around an external axis. In B&B and the handout from Jacobs you got the Exclusive Definition. The nonparallel sides are called legs. Choose the number of decimal places and click Calculate. Engineering ToolBox - SketchUp Extension - Online 3D modeling! Add standard and customized parametric components. The angles on the same side of a leg are called adjacent angles such as. Degree Radian. Base of Triangle and Trapezoid. 3D Objects (Prisms, Cylinders. 38 KB File Count 1 Create Date February 5, 2020 Last Updated February 5, 2020 Geo Tile - Trapezoid - 3D - CAD. 613 Trapezoid clip art images on GoGraph. Well what do I mean by that?. A Trapezoidal Prism contains six faces. Solution for 8) In isosceles trapezoid MNPQ with MN || QP, diagonal MPI MQ. 25 lbs; The Collier Trapezoid Cube can be displayed two ways, so your imprint options are endless! Lay this award on the long flat surface or stand it tall. The Surface Area of a trapezoid = ½(b1+b2) x h X Height of figure. The Best 3d Wooden Star Plans Free Download PDF And Video. What is the best way to create a 3-D trapezoid 9. This picture features a trapezoid (trapezium). Formula for Volume of a Trapezoidal Prism. Circular Trapezoid. 1 Trapezoid Trapezoid Macromedia Fontographer 4. Of course there are always options to use the elements to creat many different cards. Surface Areas. See also geometric figures, 1D, 2D. Knives - Reusable / Diamond Knives / Phaco Knives / MICS™ / Tru-Size™ Angled Trapezoid Diamond Knife 1. A quadrilaterally-based pyramid is a wedge in which one of the edges between two trapezoid faces has collapsed into. lateral sides, angle at the base and other base 3. DTT-TZ3060E-13H 3d (DWG 320 KB. The altitude (or height) of a trapezoid is the perpendicular distance between the two bases. In this problem the lengths for each of the bases and the height of the isosceles trapezoid is provided in the question prompt. The given data consists of: As the actual height is not given, we have to use equation no. Buy Trapezoid Hsampc Scott Sort Order Best Match / Completing / Lowest Prices / Price High To Low / New Browse By Pricing$58 / $170 /$239 / $407 /$551 / $863 /$1205 / $1930 /$2712 / \$4059. As we see the perimeter of the parallelogram is equal the perimeter of the rectangle. L = al + cl + bl + dl Substitute each piece into the equation: S = 2*h/2 (a + b) + al. Include UNITS in your answers. When you open the spreadsheet, you’ll see the 2D rotation, then the 2D translation and finally 2D scaling. A trapezoid is a quadrilateral with exactly one pair of parallel sides (the parallel sides are called bases). The trapezoid has 2 sides that are parallel lines. Trapezium Tables in stock at the best prices. She has been mightily, saved by grace and is grateful for God’s sovereignty throughout her life’s journey. by KLucky13 May 1, 2014. With the picture cut into a trapezoid, adjust the angle of the trapezoid so that you can see the entirety of the shape that you want to have pop-out of your picture (creating the 3D effect). Missing values are ignored. sh3f library of 3D models contained in the 3DModels-Scopia-1. The hydraulic radius for open channel flow is defined as the cross sectional area of flow divided by the wetted perimeter. JMAP resources include Regents Exams in various formats, Regents Books sorting exam questions by State Standard: Topic, Date, Type and at Random, Regents Worksheets sorting exam questions by State Standard: Topic, Type and at Random, an Algebra I Study Guide, and Algebra I Lesson Plans. Now that we've seen several types of quadrilaterals that are parallelograms, let's learn about figures that do not have the properties of parallelograms. stp files (zipped) Width. Has control for Top and Bottom stretch. Perimeter of a Trapezoid = a + b1 + b2 + c. 6) Make a trapezoid whose longer base is on the top of the shape. First, we need a little terminology/notation out of the way. Trapezoid pyramid problem 3: A base length of the trapezoid pyramid is 6 cm, 11 cm and its height of the pyramid is 5 cm. The standard formula for the area of a trapezoid is where is the height of the trapezoid and and are the lengths of the trapezoid's two bases. Use strict inequalities (< and >) for dotted lines and non-strict inequalities (<= or >=) for a solid line. …Then pull a little bit further. Rotate Circle. Each lower base angle is supplementary to […]. Engineering ToolBox - SketchUp Extension - Online 3D modeling! Add standard and customized parametric components. 5 lbs; The Collier Trapezoid Cube can be displayed two ways, so your imprint options are endless! Lay this award on the long flat surface or stand it tall. Ques: Laura drew a trapezoid with dimensions as shown in the figure. Manuel calculates the areas of triangles using the formula Area 1_ 2 bh, where b is the base of the triangle and h is the corresponding height. Learn all about Aluminum Alloy Trapezoid Ruler here!. It’s easy to draw 3D graphs using templates. This is an excellent review for the FCAT math test. When the problem has a solution, the outputs are: the angles A, B, C and D, the height h, the area and the lengths of the diagonals AC and BD of the trapezoid. Perfect for a new product launch, new slogan launch, or a company logo!. Solved Example on Trapezoid. This packet is full of fun printables that will help your students learn about a variety of 2D shapes and 3D shapes! The shapes included in this packet are: circle, square, triangle, rectangle, oval, rhombus, trapezoid, pentagon, hexagon, sphere, cube, cone and cylinder. Trapezoid Picture. Various shapes and units of measurement are used. Rectangular Prism. ) Any help on this would be greatly appreciated. Right trapezoids are used in the trapezoidal rule for estimating areas under a curve. Imagine a soup can. The three-dimensional figure sides are its flat surfaces. ©K-5MathTeachingResources. Download 39,490 trapezoid free vectors. They also detained one suspect during two days of fighting. Available in any file format including FBX, OBJ, MAX, 3DS, C4D. If it were me, I wouldn't try to build each trapezoid from separate parts. Draw and write about the new shape you made from the two shapes. Formula for Volume of a Trapezoidal Prism. pair of parallel sides a right trapezoid has two right angles Free download right wallpaper, here we provide some of wallpaper on right, as well asright image and right Picture. Polygons (trapezoids) from lines I want to make polygon (trapezoid) from each line segment that has information about trapezoid width at start and at the end. (I'm assuming you want a trapezoid that is symmetrical - there isn't really a 3D shape for what you describe if it isn't). To learn some basics about trapezoids, scroll to the bottom of the page, or dive into our resources to begin mastering these geometric shapes. The sum of interior angles in a quadrilateral is 360°. Under LAYERS - ROTATE / ZOOM, I found some tools that seem to work. 3D design 'This is a Freaking Trapezoid!' created by Cashew Man with Tinkercad. A quadrilateral having two and only two sides parallel is called a trapezoid. The area of a trapezoid is the space contained within its 4 sides. To calculate the volume of this shape you need to enter the lengths of these two sides, the distance between these lengths and the depth of the slab. 3D Transmographer: Build your own polygon and transform it in the Cartesian coordinate system. area of kites and traps page 1 of 2. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. Welcome to The Volume and Surface Area of Trapezoid Prisms Math Worksheet from the Measurement Worksheets Page at Math-Drills. A cone is a three-dimensional shape that tapers smoothly from its typically circular base to a common point called the apex (or vertex). A trapezoid is a four-sided shape with at least one set of parallel sides. Return from Basic Shapes Game to Geometry Math Games or to Elementary Math Games. Look it up now!. Before discussing the formula for the area of a trapezoid, ask the students what they did when they decomposed the trapezoid into triangles. I like to tell this story as part of the lesson when I'm teaching classifying quadrilaterals. For more intricate graphs, you can also use inequalities with restrictions to shade selected parts of the graph. This Measurement Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. Trapezoid 3D models ready to view, buy, and download for free. Terminals for reliable power- and signal transmission. 2) Make an isosceles trapezoid. Perimeter is the total of all the sides of a polygon. Medically accurate illustration of the trapezoid ligament The trapezoid ligament. Mouse over the elements below to see the difference between a 2D and a 3D transformation: The numbers in the table specify the first browser version that fully supports the property. CylinderGeometry which base sides are always the same. Draw a 3D trapezoid with help from an artist. Since I wrote the post entitled 3D Analogue to the Trapezoid, I have learned that people already have something in mind when they type in terms to this effect. cube = 6 a 2. 5 in the box labeled ' The area of the trapezoid is' The area of the trapezoid is 37. Khan Academy is a 501(c)(3. Back Page of 15 style Stock Image by kongvector 0 / 0 Clown trapezoid mascot cartoon style Picture by kongvector 0 / 0 Green speech bubble Trapezoid shape 3D Stock Images by djmilic 0 / 0 Elf trapezoid character cartoon style Stock Images by kongvector 0 / 0 Geometric. Of course there are always options to use the elements to creat many different cards. The simple formula for a 4-sided polygon is P= S 1 + S 2 + S 3 + S 4 As a way to get started, complete the following pages on the area of various shapes. #N#GeoGebra Classic. Related Surface Area Calculator | Volume Calculator. One typical use for this model is packing pralines. lateral sides, angle at the base and other base 3. …Then pull a little bit further. The properties of the trapezoid are as follows: The bases are parallel by definition.
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The registrar's office at State University would like to determine a 95% confidence interval
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# Question : The registrar's office at State University would like to determine a 95% confidence interval : 2150491
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Solve the problem.
6) What is zα/2 when α = 0.01?
A) 2.575
B) 1.96
C) 2.33
D) 1.645
7) What is the confidence level of the following confidence interval for μ?
overbar(x) ± 0.99((σ/√(n)))
A) 99%
B) 90%
C) 80%
D) 67%
8) The registrar's office at State University would like to determine a 95% confidence interval for the mean commute time of its students. A member of the staff randomly chooses a parking lot and surveys the first 200 students who park in the chosen lot on a given day. The confidence interval is
A) meaningful because the sample size exceeds 30 and the Central Limit Theorem ensures normality of the sampling distribution of the sample mean.
B) meaningful because the sample is representative of the population.
C) not meaningful because of the lack of random sampling.
D) not meaningful because the sampling distribution of the sample mean is not normal.
9) A 90% confidence interval for the mean percentage of airline reservations being canceled on the day of the flight is (3.9%, 7.3%). What is the point estimator of the mean percentage of reservations that are canceled on the day of the flight?
A) 3.65%
B) 3.4%
C) 5.60%
D) 1.70%
10) A 95% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was (\$130,771, \$146,241). To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width?
A) Increase the sample size and increase the confidence level.
B) Decrease the sample size and increase the confidence level.
C) Increase the sample size and decrease the confidence level.
D) Decrease the sample size and decrease the confidence level.
11) Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 375 of its members at random and monitor the working time of each of them for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. Which of the following should be used to estimate the parameter of interest for this problem?
A) A large sample confidence interval for p.
B) A small sample confidence interval for μ.
C) A large sample confidence interval for μ.
D) A small sample confidence interval for p.
12) Explain what the phrase 95% confident means when we interpret a 95% confidence interval for μ.
A) 95% of the observations in the population fall within the bounds of the calculated interval.
B) The probability that the sample mean falls in the calculated interval is 0.95.
C) In repeated sampling, 95% of similarly constructed intervals contain the value of the population mean.
D) 95% of similarly constructed intervals would contain the value of the sampled mean.
14) A retired statistician was interested in determining the average cost of a \$200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers) and constructed the following 95 percent confidence interval for the mean cost of the term life insurance: (\$850.00, \$1050.00). What value of alpha was used to create this confidence interval?
A) 0.05
B) 0.10
C) 0.01
D) 0.025
SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
15) Suppose (1,000, 2,100) is a 95% confidence interval for μ. To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Explain why an increase in sample size will lead to a narrower interval of the estimate of μ.
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Answer the question True or False.
23) Find zα/2 for the given value of α.
α = 0.05
A) 2.81
B) 0.33
C) 1.96
D) 1.645
24) Determine the confidence level for the given confidence interval for μ.
overbar(x) ± 1.34((σ/√(n)))
A) 95.5%
B) 91%
C) 41%
D) 82%
25) A random sample of n measurements was selected from a population with unknown mean μ and known standard deviation σ. Calculate a 95% confidence interval for μ for the given situation. Round to the nearest hundredth when necessary.
n = 160, overbar(x) = 68, σ = 15
A) 68 ± 2.32
B) 68 ± 0.18
C) 68 ± 29.4
D) 68 ± 1.95
26) A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was (\$124,443, \$140,235). Give a practical interpretation of the interval.
A) We are 90% confident that the mean salary of the sampled CEOs falls in the interval \$124,443 to \$140,235.
B) 90% of all CEOs in the electronics industry have salaries that fall between \$124,443 to \$140,235.
C) We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval \$124,443 to \$140,235.
D) 90% of the sampled CEOs have salaries that fell in the interval \$124,443 to \$140,235.
27) A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours. Estimate the mean credit hours taken by a student each quarter using a 95% confidence interval. Round to the nearest thousandth.
A) 15.2 ± .018
B) 15.2 ± .188
C) 15.2 ± .012
D) 15.2 ± .285
28) A random sample of 250 students at a university finds that these students take a mean of 15 credit hours per quarter with a standard deviation of 2.5 credit hours. The 99% confidence interval for the mean is 15 ± 0.407. Interpret the interval.
A) The probability that a student takes 14.593 to 15.407 credit hours in a quarter is 0.99.
B) 99% of the students take between 14.593 to 15.407 credit hours per quarter.
C) We are 99% confident that the average number of credit hours per quarter of the sampled students falls in the interval 14.593 to 15.407 hours.
D) We are 99% confident that the average number of credit hours per quarter of students at the university falls in the interval 14.593 to 15.407 hours.
29) The director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 81 different 24-hour periods and determines the number of admissions for each. For this sample, overbar(x) = 15.6 and s2 = 16. Estimate the mean number of admissions per 24-hour period with a 99% confidence interval.
A) 15.6 ± .440
B) 15.6 ± .127
C) 15.6 ± 1.144
D) 15.6 ± 4.578
30) Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 301 of its members at random and monitor the working time of each of them for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. If the mean and standard deviation of the sample are overbar(x) = 9.3 hours and s = 2.3 hours, find a 95% confidence interval for the true mean number of hours a union member is absent per month. Round to the nearest thousandth.
A) 9.3 ± .126
B) 9.3 ± .015
C) 9.3 ± .171
D) 9.3 ± .260
31) Parking at a large university can be extremely difficult at times. One particular university is trying to determine the location of a new parking garage. As part of their research, officials are interested in estimating the average parking time of students from within the various colleges on campus. A survey of 338 College of Business (COBA) students yields the following descriptive information regarding the length of time (in minutes) it took them to find a parking spot. Note that the "Lo 95%" and "Up 95%" refer to the endpoints of the desired confidence interval.
Variable N Lo 95% CI Mean Up 95% CI SD
Parking Time 338 9.1944 10.466 11.738 11.885
Give a practical interpretation for the 95% confidence interval given above.
A) 95% of the COBA students had parking times that fell between 9.19 and 11.74 minutes.
B) We are 95% confident that the average parking time of the 338 COBA students surveyed falls between 9.19 and 11.74 minutes.
C) 95% of the COBA students had parking times of 10.466 minutes.
D) We are 95% confident that the average parking time of all COBA students falls between 9.19 and 11.74 minutes.
32) Parking at a large university can be extremely difficult at times. One particular university is trying to determine the location of a new parking garage. As part of their research, officials are interested in estimating the average parking time of students from within the various colleges on campus. A survey of 338 College of Business (COBA) students yields the following descriptive information regarding the length of time (in minutes) it took them to find a parking spot. Note that the "Lo 95%" and "Up 95%" refer to the endpoints of the desired confidence interval.
Variable N Lo 95% CI Mean Up 95% CI SD
Parking Time 338 9.1944 10.466 11.738 11.885
Explain what the phrase "95% confident" means when working with a 95% confidence interval.
A) In repeated sampling, 95% of the population means will fall within the interval created.
B) 95% of the observations in the population will fall within the endpoints of the interval.
C) In repeated sampling, 95% of the intervals created will contain the population mean.
D) In repeated sampling, 95% of the sample means will fall within the interval created.
35) How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show that the sample mean was \$70.00 and prior sampling indicated that the population standard deviation was \$17.50. Use this information to create a 95 percent confidence interval for the population mean.
A) 70 ± 1.645((17.50/√(60)))
B) 70 ± 1.671((17.50/√(60)))
C) 70 ± 1.833((17.50/√(60)))
D) 70 ± 1.960((17.50/√(60)))
## Solution 5 (1 Ratings )
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## Welcome Everyone!
Hello everybody, my name is Sarah Qarizadha and I will be teaching you a little about the grade 10 curriculum. Today i'll be teaching you about Quadratic Relations. I will be teaching you how to answer Vertex Form, Factored Form and Standard Form questions. All of this is apart of the grade 10 curriculum, so if your brain is looking for a good lesson or just a refresh, then this is a great website for you!
## Learning Goals
1. To be able to solve all vertex form equations
2. Finding the value of 'a'
3. To use quadratic equations to solve word problems in vertex form
## Definitions Of Unit #1
Parabola : This is a curve, shaped as a arch. Its distance from a fixed point is equal to its distance from a fixed line
Vertex Form : The vertex are the points (h,k) on a parabola. The vertex is when both the points meet. The x-intercept is 'k' and the y-intercept is 'h'.
AOS (Axis Of Symmetry) : The AOS is a straight vertical line that goes right down the middle of the parabola
## Vertex Form
y = a (x-h)^2 + k
## Introduction To Vertex form
1. Vertex is (h,k)
2. 'a' tells us if its stretched or compressed, and the direction of opening
3. 'h' tells us if its going to be left or right on the graph/horizontal translation
4. 'k' tells is if its a vertical translation
5. Use the step pattern
6. To find the y-intercept, set x=0 and then solve for the y-intercept
7. Then to solve, set y=0 and solve for x or expand and simplify to get the standard form
8. Then use the Quadratic Formula
## Equation Question Examples
Example #1 - Determine an equation when given the vertex and determining the 'a' value:
Vertex : (-3,5)
One of the points are (3,7)
Step 1 : Plug in the vertex into the equation
y=a(x-h)^2+k
y=a(x+3)^2+5
Step 2 : Find the 'a' value. Plug in (3,7) in the equation
y=a(x+3)^2+5
7=a(3+3)^2+5
7=a(36)+5
7=36a+5
7-5=36a
2/36 = 36a/36
18=a
Step 3 : Fill in final equation
y=18(x+3)^2+5
Example #2 - Determining an equation when given the vertex and when given (x,y) :
Example #3 - Isolating For 'x' :
*We are finding the x-intercept when we are isolating for 'x'
Graphing Vertex Form
We can easily use a step pattern to graph y = 2(x-4)^2 - 6
In the step pattern, your supposed to multiply the 'a' value by 1, 3 and 5
Vertex : (4,-6)
AOS : -4
Optimal value : -6
'a' : 2
Direction Of Opening : Up, because the 'a' value is positive
By using the step pattern, we can easily figure out what parabola will be :
• Step 1 : 1x2 = 2
• Step 2 : 3x2 = 6
• Step 3 : 5x2 =10
## Using Vertex Form
At a baseball game, a fan throws a baseball from the stadium back onto the field. The height in meters of a ball t seconds after being thrown is modeled by the function h = -4.9 (t-2)^2 + 45
a) What is the maximum height of the ball?
The Vertex : (2,45)
Maximum height of the ball is 45 meters
b) When did the maximum height occur?
The maximum height is 2 seconds
c) What is the height of the ball after 1 second?
h = -4.9 (1-2)^2 + 45
h = 40.1 meters
Therefore, the height of the ball is 40.1 meters after 1 second
d) What is the initial height of the ball?
When t=0 Solve for h
h = -4.9 (0-2)^2 + 45
=25.4
Therefore, the initial height was 25.4 meters
## Mapping notation
(x,y) - (x+h, ay+k)
Example - y=(x+7)^2 ~ (x-7, y)
## Graphing Using Transformations
Quick Way of Graphing a Quadratic Function in Vertex Form
## Learning Goals
1. To be able to graph a quadratic equation in factored form
2. Turning Factored Form, Quadratic Equations into Standard form by using FOIL and simplifying
## Factored Form Equation
y = a (x-r) (x-s)
Factors And Zeros
A zero of a parabola is another name for the x- intercepts and in order to find the x- intercepts you must set y=0
Key points
• When the (a) value changes the zeroes do not change
• When the (a) value changes the axis of symmetry do not change
• When the (a) value changes the optimal value does change
## Introduction To Factored Form
The value of a gives you the shape and direction of opening
The value of r and s give you the x-intercepts
axis of symmetry, AOS: = (r+s / 2)
1. Sub this x value into equation to find
2. the optimal value
3. to find the y-intercept, set x=0 and solve for y
4. Solve using the factors
Types of Factoring:
• Greatest Common Factor
• Simple factoring (a=1)
• Complex factoring
• Special case - Difference of squares
• Special case – Perfect square
## Factoring Simple Trinomials
1. Identify a,b, and c in the trinomial ax^2 + bx + c
2. Write down all factor pairs of c
3. Identify which factor pair from the previous step sums up to b
4. Substitute factor pairs into two binomials
## Factoring by grouping
Step 1: Decide if the four terms have a GCF. If so, factor out the GCF. Do not forget to include the GCF as part of your final answer.
Step 2: Group first two terms together and the last two terms together.
Step 3: Factor out the GCF from each of the two groups.
Step 4: The one thing that the two groups have in common should be what is in parenthesis, write whats outside the brackets as a parenthesis
Step 5: Determine if the remaining factors can be factored any further.
## Common Factoring
Step 1: Determine the GCF of the given terms. The greatest common factor or GCF is the largest factor that all terms have in common.
Step 2: Factor out or divide out the GCF from each term. You could check your answer at the point by distributing the GCF to see if you get the original question. Factoring out the GCF is the first step in many factoring problems.
## Word Problem Using Factored Form
A toy rocket is shot into the air. It's path is approximated by the function h = -5t ^2 + 25t where h(t) is the height in meters and t is the time in seconds. When will the rocket hit the ground?
h = -5t ^2 + 25t
h = -5t (t-5)
t=0 t=5
Therefore, the rocket hit the ground at 5 seconds
## Difference Of Squares
Equation
a^2 - b^2 = (a - b)(a + b)
or
a^2 - b^2 = (a + b)(a - b)
Step 1: Decide if theirs a GCF. If so, factor out the GCF. Do not forget to include the GCF as part of your final answer.
Step 2: So, all you need to do to factor these types of problems is to determine what numbers squares will produce the desired results.
Step 3: Determine if the remaining factors can be factored any further.
## Perfect Squares
Equation
(a + b)^2 = a^2 + 2ab + b^2
(a - b)^2 = a^2 - 2ab + b^2
Step 1. Verify that the first term and the third term are both perfect squares. (This means that the coefficients are perfect squares
Step 2. Verify that the middle term is twice the product of the square roots of the first and third term.
Step. 3. Use the standard form above to write the factored form.
20160412 184558
## learning goals
1. Find the number of zeros that a quadratic relationship has by calculating the discriminant.
2. Solve using the quadratic formula or by factoring or by completing the square to get vertex form
## Standard form equation
y = ax^2 + bx + c
## Introduction to standard form
The value of a gives you the shape and direction of opening
The value of c is the y-intercept
Solve using the quadratic formula, to get the x-intercepts
MAX or MIN? Complete the square to get vertex form
Quadratic equation : The quadratic equation is a formula used to get the x-intercepts. I recommend using this, if the equation isn't easily factor-able.
The Discriminant : A discriminant is to know how many x-intercepts there will be. Using this formula: b^2 - 4ac
• There can be two solutions, one solutions or no solution.
• There are two solutions when the discriminant is a positive.
• There is one solution when the discriminant equals to 0
• There is no solution, when the discriminant is negative
Turning Into Vertex Form And Completing The Square :
Question: y=3x^2 +6x -4
• Step One: Focus on the 3x^2 +6x, and factor out the GCF (which is 3 in this case) and leave the -4 out.
y=3(x^2 +2x) -4
• Step Two: Divide the 'b' value by 2 and then square the result.
(2/2)^2=1
• Step Three: Add the result in the bracket and also subtract it. So the equation looks like this: y= 3(x^2 +2x +__ -__) -4
y=3(x^2 +2x +1 -1) -4
• Step Four: Take the negative one outside of the brackets, therefore it will be multiply with 3.
y=3(x^2 +2x +1) -3 -4
• Step Five: factor the trinomial in the bracket.
y=3(x+1)^2 -7
• *Remember Vertex Form Equation* y=a(x-h)^2+k
• Therefore, vertex is (-1, -7)
## word problem using quadratic formula
1) A toy rocket is launched upward at an initial velocity of 51 m/s, from a height of 1.3 , above the ground, the height of the toy rocket, in meters, after t seconds is modeled by the equation h= -4.9t^2 + 51t +1.3
a) How long does it take the rocket to fall to the ground, rounded to the nearest hundredth of a second?
b) Find the times when the toy rocket is at a height of 95.7 m above the ground. Round your answers to the nearest tenth.
c) What is the maximum height of the toy rocket? At what times does it teach this height? Round you answer to the nearest tenth.
## Video explaining standard form - Quadratics
Quadratics was a new unit for me where I got to learn many interesting things about parabolas. I first struggled with parabolas, but later on, I understood it more and more. I feel like if I redid my three unit tests for quadratics, then I would do fairly well. Making this website has made me understand the concepts of quadratics much more. Below is an image of my Quadratics Standard Form Unit Test. I chose to show this test, because i feel that this unit was my favorite. It was an easy unit to understand, and i enjoyed answering the questions for the different quizzes and tests. I chose to not show any of the tips assignments because I did not get the marks that i thought i would get. I feel like if got to redo any assignment or test, then it would be the various Tips assignments. Overall, i know that if i were to ask more questions in class, finish my homework and practice with extra worksheets, then i would be passing math with over an 85% for sure.
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Extended Static Checking for Haskell (ESC/Haskell). Dana N. Xu University of Cambridge advised by Simon Peyton Jones Microsoft Research, Cambridge. Program Errors Give Headache!. Module UserPgm where f :: [Int]->Int f xs = head xs `max` 0 : … f [] …. Module Prelude where - PowerPoint PPT Presentation
### Transcript of Extended Static Checking for Haskell (ESC/Haskell)
• Extended Static Checking for Haskell(ESC/Haskell)Dana N. XuUniversity of Cambridge
advised by Simon Peyton Jones Microsoft Research, Cambridge
• Module UserPgm where
f :: [Int]->Intf xs = head xs `max` 0
: f [] Program Errors Give Headache!Glasgow Haskell Compiler (GHC) gives
Exception: Prelude.head: empty listModule Prelude where
head :: [a] -> ahead (x:xs) = xhead [] = error empty list
• Preconditionshead xs @ requires { not (null xs) }head (x:xs) = x
f xs = head xs `max` 0
Warning: f [] calls head which may fail heads precondition!
f_ok xs = if null xs then 0 else head xs `max` 0A precondition (ordinary Haskell)not :: Bool -> Boolnot True = Falsenot False = Truenull :: [a] -> Boolnull [] = Truenull (x:xs) = False
• Postconditionsrev xs @ ensures { null \$res ==> null xs }rev [] = []rev (x:xs) = rev xs ++ [x]
f xs = case (rev xs) of [] -> head xs (x:xs) ->
Warning: f [] calls head which may fail heads precondition!A postcondition (ordinary Haskell)(==>) :: Bool -> Bool -> Bool(==>) True x = x(==>) False x = True
• Functions without Annotationsdata T = T1 Bool | T2 Int | T3 T T
noT1 :: T -> BoolnoT1 (T1 _) = FalsenoT1 (T2 _) = TruenoT1 (T3 t1 t2) = noT1 t1 && noT1 t2
(&&) True x = x(&&) False x = FalseNo abstraction is more compact than the function definition itself!
• Expressiveness of the Specification LanguagesumT :: T -> IntsumT x @ requires { noT1 x }sumT (T2 a) = asumT (T3 t1 t2) = sumT t1 + sumT t2
rmT1 :: T -> TrmT1 x @ ensures { noT1 \$res }rmT1 (T1 a) = if a then T2 1 else T2 0rmT1 (T2 a) = T2 armT1 (T3 t1 t2) = T3 (rmT1 t1) (rmT1 t2)
For all crash-free t::T, sumT (rmT1 t) will not crash.
• Higher Order Functionsall :: (a -> Bool) -> [a] -> Boolall f [] = Trueall f (x:xs) = f x && all f xs
filter f xs @ ensures { all f \$res }filter f [] = []filter f (x:xs) = case (f x) of True -> x : filter f xs False -> filter f xs
• Various Exampleszip xs ys @ requires { length xs == length ys }zip xs ys @ ensures { length \$res == length xs }
f91 n @ requires { n n 10
ones @ ensures { even \$res == True } ones @ ensures { even \$res == False } ones = 1 : ones
even [] = Trueeven [x] = Falseeven (_:_:xs) = even xs
• Sortingsorted [] = Truesorted (x:[]) = Truesorted (x:y:xs) = x sorted \$res }insertsort xs @ ensures { sorted \$res }
merge xs ys @ ensures { sorted xs & sorted ys ==> sorted \$res }mergesort xs @ ensures { sorted \$res }
bubbleHelper :: [Int] -> ([Int], Bool)bubbleHelper xs @ ensures { not (snd \$res) ==> sorted (fst \$res) } bubblesort xs @ ensures { sorted \$res }
• What we cant dog1 x = case (prime x > square x) of True -> x False -> error urk
g2 xs ys = case (rev (xs ++ ys) == rev ys ++ rev xs) of True -> xs False -> error urk
Hence, three possible outcomes: (1) Definitely Safe (no crash, but may loop)(2) Definite Bug (definitely crashes)(3) Possible BugCrash!Crash!
• LanguageSyntaxFollowing HaskellsLazy semantics
• Preprocessing1. Filling in missing pattern matchings.2. Type checking the pre/postconditions.head xs @ requires { xs /= [] }head :: [a] -> ahead (x:xs) = xhead :: Eq a => [a] -> a
• Symbolic Pre/Post CheckingA the definition of each function f,assuming the given precondition holds,we checkNo pattern matching failurePrecondition of all calls in the body of f holdsPostcondition holds for f itself.
• Given f x = e, f.pre and f.postTheorem: if so, then given precondition of f holds:1. No pattern matching failure2. Precondition of all calls in the body of f holds3. Postcondition holds for f itselfGoal: show fchk is crash-free!
• The Representative FunctionAll crashes in f are exposed in f#No need to look inside OK calls
• Simplifier
• Expressive specification does not increase the complication of checkingfilter f xs @ ensures { all f \$res }
filterchk f xs =case xs of [] -> True (x:xs) -> case (all f (filter f xs)) of True -> all f (filter f xs)
• Arithmetic via External Theorem Proverfoo :: Int -> Int -> Intfoo i j @ requires { i > j }foo# i j = case i > j of False -> BAD foo True ->
goo i = foo (i+8) i
goochk i = case (i+8 > i) ofFalse -> BAD fooTrue -> >>ThmProveri+8>I>>Valid!case i > j of True -> case j < 0 ofFalse -> case i > 0 of False -> BAD f>>ThmProverpush(i> j)push(j0)>>Valid!
• Counter-Example Guided UnrollingsumT :: T -> IntsumT x @ requires { noT1 x }sumT (T2 a) = asumT (T3 t1 t2) = sumT t1 + sumT t2After simplifying sumTchk, we may have:case ((OK noT1) x) ofTrue -> case x of T1 a -> BAD sumT T2 a -> a T3 t1 t2 -> case ((OK noT1) t1) of False -> BAD sumT True -> case ((OK noT1) t2) of False -> BAD sumT True -> (OK sumT) t1 + (OK sumT) t2
• Step 1:Program Slicing Focus on the BAD Pathscase ((OK noT1) x) ofTrue -> case x of T1 a -> BAD sumT T3 t1 t2 -> case ((OK noT1) t1) of False -> BAD sumT True -> case ((OK noT1) t2) of False -> BAD sumT
• Step 2: Unrollingcase ((\x -> case x of T1 a -> False T2 a -> True T3 t1 t2 -> (OK noT1) t1 && (OK noT1) t2) x) ofTrue -> case x of T1 a -> BAD sumT T3 t1 t2 -> case ((OK noT1) t1) of False -> BAD sumT True -> case ((OK noT1) t2) of False -> BAD sumT
• Step 2: Unrollingcase ((\x -> case x of T1 a -> False T2 a -> True T3 t1 t2 -> (OK noT1) t1 && (OK noT1) t2) x) ofTrue -> case x of T1 a -> BAD sumT T3 t1 t2 -> case ((OK noT1) t1) of False -> BAD sumT True -> case ((OK noT1) t2) of False -> BAD sumT (OK noT1) t1(OK noT1) t2((OK noT1) t1)((OK noT1) t1)
• Keeping Known Informationcase (case (NoInline ((OK noT1) x)) of True -> (\x -> case x of T1 a -> False T2 a -> True T3 t1 t2 -> (OK noT1) t1 && (OK noT1) t2) x) ofTrue -> case x of T1 a -> BAD sumT T3 t1 t2 -> case ((OK noT1) t1) of False -> BAD sumT True -> case ((OK noT1) t2) of False -> BAD sumT (OK noT1) t1(OK noT1) t2((OK noT1) t1)((OK noT1) t1)
• Counter-Example Guided Unrolling The Algorithm
• Tracing
• Counter-Example Generationf3chk xs z = case xs of [] -> 0 (x:y) -> case x > z of True -> Inside f2 (Inside f1 (BAD f1)) False ->
Warning : f3 (x:y) z where x>z calls f2 which calls f1 which may fail f1s precondition!
• ContributionsChecks each program in a modular fashion on a per function basis.Pre/postcondition annotations are in Haskell.Allow recursive function and higher-order function Unlike VC generation, we treat pre/postcondition as boolean-valued functions and use symbolic simplification.Handle user-defined data typesBetter control of the verification processFirst time that Counter-Example Guided approach is applied to unrolling.Produce a trace of calls that may lead to crash at compile-time.Our prototype works on small but significant examples Sorting: insertion-sort, merge-sort, bubble-sortNested recursion
• Future WorkAllowing pre/post declaration for data types.data A where A1 :: Int -> A A2 :: [Int] -> [Bool] -> A A2 x y @ requires { length x == length y}Allowing pre/post declaration for parameters in higher-order function.map f xs @ requires {all f.pre xs}map f [] = []map f (x:xs) = f x : map f xsAllowing polymorphism and support full Haskell.
**************************
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# Bohr's postulate: Electron not losing energy
1. Mar 15, 2013
### SweatingBear
Hey forum.
One of the major issues physicists had at the brink of the entrance of modern physics was that an electron simply could not orbit around the nucleus since it would successively lose energy and consequently spiral into the nucleaus, collapsing the whole atom (and model as well) altogether.
But why was this an issue? From what I have understood, no work is done unto a particle subjected to centripetal accelerationen i.e. no energy is expended. Therefore, the electron shouldn't be emitting radiation and losing energy, right? Or is the electron really emitting radiation in its orbit according to the model? I do not see how.
(PS: High-school level)
2. Mar 15, 2013
### Staff: Mentor
It's nothing to do with the work done (or not done) on the orbiting electron. The problem comes from classical electrodynamics, where an accelerating charge (such as an electron moving in a circle) would be expected to give off electromagnetic waves and thus lose energy.
3. Mar 15, 2013
### Staff: Mentor
4. Mar 15, 2013
### SweatingBear
Ah, of course, I remember that! "Accelerating charges emit electromagnetc waves"! But is it also valid if the accelerationen is perpendicular to the velocity? Centripetal movement doesn't always imply change in the particles speed.
5. Mar 15, 2013
### Staff: Mentor
Yes.
True. But it's still accelerating, which is the important thing.
6. Mar 16, 2013
### Jano L.
It was an issue because people thought that if the system radiates EM waves, it has to lose energy. Decrease of energy of the atom leads to collapse and this was unwanted result.
It is not the force of the nucleus that was supposed to oppose the motion, but the additional "force of radiation reaction", which is basically the sum of forces which parts of charged body (electron) exert on themselves. In relativity, if the body is extended, this sum need not be zero (the law of action-reaction does not hold). If the electron is charged extended body, this force can be calculated approximately and it turns out to be proportional to $\dot{\mathbf a}$, which for circular motion points against the velocity. So this "self-force" would do negative work on the charged extended body and bring it down to the nucleus.
7. Mar 16, 2013
### SweatingBear
Hm, alright. I was comparing the circular motion with e.g. a satellite orbiting the earth, and from what I have understood no energy per se is required to keep it in its orbit, hence me wondering why the electron ought to collapse.
8. Mar 16, 2013
### SweatingBear
Besides, even if the electron were to plummet into the nucleus: Wouldn't the electrostatic force repel it away from the protons?
9. Mar 16, 2013
### Staff: Mentor
No, the positively charged protons attract the electrons.
10. Mar 16, 2013
### SweatingBear
Oh lord, of course, how silly of me!
Thankful to everybody for the replies.
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Code source de code_beatrix.algorithm.tsp
```# -*- coding: utf-8 -*-
"""
Function solving the TSP problem
"""
import random
[docs]def distance_point(p1, p2):
"""
Returns the Euclidian distance between two points.
Retourne la distance euclidienne entre deux points.
:param p1: point 1
:param p2: point 2
:return: distance
"""
d = 0
for a, b in zip(p1, p2):
d += (a - b) ** 2
return d ** 0.5
[docs]def distance_circuit(points):
"""
Computes the distance of this circuit.
Calcule la longueur d'un circuit.
:param points: list of points, the circuit assumes they are giving in that order
:return: distance
"""
d = 0
for i in range(1, len(points)):
d += distance_point(points[i - 1], points[i])
return d + distance_point(points[0], points[-1])
[docs]def permutation(points, i, j):
"""
Switches two points and returns a new path.
Echange deux points et retourne le nouveau circuit.
:param points: circuit
:param i: first index
:param j: second index (< len(points))
:return: new circuit
"""
points = points.copy()
points[i], points[j] = points[j], points[i]
return points
[docs]def reverse(points, i, j):
"""
Reverses a sub part of circuit.
Retourne une partie du circuit.
:param points: circuit
:param i: first index
:param j: second index (<= len(points))
:return: new circuit
"""
points = points.copy()
if i > j:
i, j = j, i
c = points[i:j]
c.reverse()
points[i:j] = c
return points
[docs]def voyageur_commerce_simple(points):
"""
Solves the TSP using basic permutations,
points are 2D coordinates.
Résoud le problème du voyageur de commerce.
:param points: list of points
"""
d0 = distance_circuit(points)
dnew = d0
n = len(points) - 1
first = True
while dnew < d0 or first:
first = False
d0 = dnew
# first pass : random
for i in range(len(points)):
h1 = random.randint(0, n)
h2 = random.randint(0, n)
p = permutation(points, h1, h2)
d = distance_circuit(p)
if d < dnew:
dnew = d
points = p
h1 = random.randint(0, n)
h2 = random.randint(h1 + 1, n + 1)
p = reverse(points, h1, h2)
d = distance_circuit(p)
if d < dnew:
dnew = d
points = p
# second pass : no reverse
for i in range(len(points)):
for j in range(i + 1, len(points) + 1):
p = reverse(points, i, j)
d = distance_circuit(p)
if d < dnew:
dnew = d
points = p
return points
[docs]def plot_circuit(points, ax=None, **kwargs):
"""
Plots the circuit on a graph.
Dessine la solution du voyageur de commerce.
:param points: points
:param ax: axe
:param kwargs: sent to ``plt.subplots``
:return: ax
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# Quantum cryptography
33 %
67 %
Information about Quantum cryptography
Education
Published on March 8, 2014
Author: HimanshuShekhar13
Source: slideshare.net
## Description
find about quantum cryptography working principle
QUANTUM CRYPTOGRAPHY BY: HIMANSHU SHEKHAR PRASOON SRIVASTAVA
TRADITIONAL CRYPTOGRAPHY • The problem with public-key cryptology is that it's based on the staggering size of the numbers created by the combination of the key and the algorithm used to encode the message. • The problem with Private key algorithm is that there's almost always a place for an unwanted third party to listen in and gain information the users don't want that person to have.
ONE OF THE GREAT CHALLENGES OF CRYPTOLOGY To Keep Unwanted Parties Or Eavesdroppers From Learning Of Sensitive Information. And the solution is “QUANTUM CRYPTOLOGY”
QUANTUM CRYPTOLOGY Quantum physics is a branch of science that deals with discrete, indivisible units of energy called quanta as described by the Quantum Theory. There are five main ideas represented in Quantum Theory: 1. Energy is not continuous, but comes in small but discrete units. 2. The elementary particles behave both like particles and like waves. 3. The movement of these particles is inherently random. 4. It is physically impossible to know both the position and the momentum of a particle at the same time. The more precisely one is known, the less precise the measurement of the other is. 5. The atomic world is nothing like the world we live in.
QUANTUM CRYPTOLOGY BASICS 1.Quantum cryptography uses photons to transmit a key 2. Each type of a photon's spin represents one piece of information -usually a 1 or a 0, for binary code 3.Polarize them through either the X or the + filters, so that each polarized photon has one of four possible states: (|), (--), (/) or ( ).
QUANTUM CRYPTOLOGY WORKING • Alice sends Bob her photons using an LED, so that each polarized photon has one of four possible states: (|), (--), (/) or ( ). • As Bob receives these photons, he decides whether to measure each with either his + or X filter -- he can't use both filters together. • . After the entire transmission, Bob and Alice have a non-encrypted discussion about the transmission. • Bob: Plus Alice: Correct • Bob: Plus Alice: Incorrect • Bob: X Alice: Correct • A third party listening in on their conversation can't determine what the actual photon sequence is.
REALIZATATION OF QUANTUM CRYPTOLOGY USING MATLAB 1.Here we will realize our “X” and “+” filter using following table. “+”table “X” table 11 10 01 00 00 01 10 11 2. Filter selection: a) By LFSR b) Simple random sequence which will be retained by the sender.
Our message: “lovely professional university” Its symbol output: Transmitted Data “Few binary sequence” Binary data Sequence Symbol Transmitted Limiter Used 01 10 11 00 01 10 11 11 01 11 2 0 3 1 0 2 1 1 1 1 + + X X X X + + X + Received Data Sequence: Received Sequence 2 0 3 1 0 2 1 1 1 1 Predicted Limiter X + + + X X + X + + Received Symbol 10 10 00 11 01 10 11 00 10 11
Table representing the asking process: Bob Replies for his choice of filter X + + + X X + X + + Response W R W W R R R W W R W- Wrong R- Right Hence, Bob Gets information about the choice of filter and corrects himself for the wrong one (either discards it and ask for retransmission, or makes a guess). Knowledge of changed sequence If the choice of filter used is correct but the data matched for parity is found wrong, then the receiver gets to know that the data sequence has been tampered. Hence, he can stop the transmission further knowing that the communication channel is being tapped.
COMPANIES IN THIS FIELD
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Water Tank Reinforcement Learning Environment Model
This example shows how to create a water tank reinforcement learning Simulink® environment that contains an RL Agent block in the place of a controller for the water level in a tank. To simulate this environment, you must create an agent and specify that agent in the RL Agent block. For an example that trains an agent using this environment, see Create Simulink Environment and Train Agent.
```mdl = "rlwatertank"; open_system(mdl)```
This model already contains an RL Agent block, which connects to the following signals:
• Scalar action output signal
• Vector of observation input signals
• Scalar reward input signal
• Logical input signal for stopping the simulation
Actions and Observations
A reinforcement learning environment receives action signals from the agent and generates observation signals in response to these actions. To create and train an agent, you must create action and observation specification objects.
The action signal for this environment is the flow rate control signal that is sent to the plant. To create a specification object for an action channel carrying a continuous signal, use `rlNumericSpec`.
```actionInfo = rlNumericSpec([1 1]); actionInfo.Name = "flow";```
If the action signal takes one of a discrete set of possible values, create the specification using the `rlFiniteSetSpec` function.
For this environment, there are three observation signals sent to the agent, specified as a vector signal. The observation vector is ${\left[\begin{array}{ccc}\int \mathit{e}\text{\hspace{0.17em}}\mathrm{dt}& \mathit{e}& \mathit{h}\end{array}\right]}^{\mathit{T}\text{\hspace{0.17em}}}$, where:
• $\mathit{h}$ is the height of the water in the tank.
• $\mathit{e}=\mathit{r}-\mathit{h}$, where $\mathit{r}$ is the reference value for the water height.
Compute the observation signals in the generate observations subsystem.
`open_system(mdl + "/generate observations")`
Create a three-element vector of observation specifications. Specify a lower bound of 0 for the water height, leaving the other observation signals unbounded.
```observationInfo = rlNumericSpec([3 1],... LowerLimit=[-inf -inf 0 ]',... UpperLimit=[ inf inf inf]'); observationInfo.Name = "observations"; observationInfo.Description = "integrated error, error, and measured height";```
If the actions or observations are represented by bus signals, create specifications using the `bus2RLSpec` function.
Reward Signal
Construct a scalar reward signal. For this example, specify the following reward.
`$\mathrm{reward}=10\left(|\mathit{e}|<0.1\right)-1\left(|\mathit{e}|\ge 0.1\right)-100\left(\mathit{h}\le 0||\mathit{h}\ge 20\right)$`
The reward is positive when the error is below `0.1` and negative otherwise. Also, there is a large reward penalty when the water height is outside the `0` to `20` range.
Construct this reward in the calculate reward subsystem.
`open_system(mdl + "/calculate reward")`
Stop Signal
To terminate training episodes and simulations, specify a logical signal to the `isdone` input port of the block. For this example, terminate the episode if $\mathit{h}\le 0$ or $\mathit{h}\ge 20$.
Compute this signal in the stop simulation subsystem.
`open_system(mdl + "/stop simulation")`
Create Environment Object
Create an environment object for the Simulink model.
`env = rlSimulinkEnv(mdl,mdl + "/RL Agent",observationInfo,actionInfo);`
Reset Function
You can also create a custom reset function that randomizes parameters, variables, or states of the model. In this example, the reset function randomizes the reference signal and the initial water height and sets the corresponding block parameters.
`env.ResetFcn = @(in)localResetFcn(in);`
Local Function
```function in = localResetFcn(in) % Randomize reference signal h = 3*randn + 10; while h <= 0 || h >= 20 h = 3*randn + 10; end in = setBlockParameter(in, ... "rlwatertank/Desired \nWater Level", ... Value=num2str(h)); % Randomize initial height h = 3*randn + 10; while h <= 0 || h >= 20 h = 3*randn + 10; end in = setBlockParameter(in, ... "rlwatertank/Water-Tank System/H", ... InitialCondition=num2str(h)); end```
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### 3.1903 $$\int \frac{(d+e x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^4} \, dx$$
Optimal. Leaf size=139 $-\frac{e^2}{\left (c d^2-a e^2\right )^3 (a e+c d x)}+\frac{e}{2 \left (c d^2-a e^2\right )^2 (a e+c d x)^2}-\frac{1}{3 \left (c d^2-a e^2\right ) (a e+c d x)^3}-\frac{e^3 \log (a e+c d x)}{\left (c d^2-a e^2\right )^4}+\frac{e^3 \log (d+e x)}{\left (c d^2-a e^2\right )^4}$
[Out]
-1/(3*(c*d^2 - a*e^2)*(a*e + c*d*x)^3) + e/(2*(c*d^2 - a*e^2)^2*(a*e + c*d*x)^2) - e^2/((c*d^2 - a*e^2)^3*(a*e
+ c*d*x)) - (e^3*Log[a*e + c*d*x])/(c*d^2 - a*e^2)^4 + (e^3*Log[d + e*x])/(c*d^2 - a*e^2)^4
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Rubi [A] time = 0.0984611, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.057, Rules used = {626, 44} $-\frac{e^2}{\left (c d^2-a e^2\right )^3 (a e+c d x)}+\frac{e}{2 \left (c d^2-a e^2\right )^2 (a e+c d x)^2}-\frac{1}{3 \left (c d^2-a e^2\right ) (a e+c d x)^3}-\frac{e^3 \log (a e+c d x)}{\left (c d^2-a e^2\right )^4}+\frac{e^3 \log (d+e x)}{\left (c d^2-a e^2\right )^4}$
Antiderivative was successfully verified.
[In]
Int[(d + e*x)^3/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^4,x]
[Out]
-1/(3*(c*d^2 - a*e^2)*(a*e + c*d*x)^3) + e/(2*(c*d^2 - a*e^2)^2*(a*e + c*d*x)^2) - e^2/((c*d^2 - a*e^2)^3*(a*e
+ c*d*x)) - (e^3*Log[a*e + c*d*x])/(c*d^2 - a*e^2)^4 + (e^3*Log[d + e*x])/(c*d^2 - a*e^2)^4
Rule 626
Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c*x)/e)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]
Rule 44
Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])
Rubi steps
\begin{align*} \int \frac{(d+e x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^4} \, dx &=\int \frac{1}{(a e+c d x)^4 (d+e x)} \, dx\\ &=\int \left (\frac{c d}{\left (c d^2-a e^2\right ) (a e+c d x)^4}-\frac{c d e}{\left (c d^2-a e^2\right )^2 (a e+c d x)^3}+\frac{c d e^2}{\left (c d^2-a e^2\right )^3 (a e+c d x)^2}-\frac{c d e^3}{\left (c d^2-a e^2\right )^4 (a e+c d x)}+\frac{e^4}{\left (c d^2-a e^2\right )^4 (d+e x)}\right ) \, dx\\ &=-\frac{1}{3 \left (c d^2-a e^2\right ) (a e+c d x)^3}+\frac{e}{2 \left (c d^2-a e^2\right )^2 (a e+c d x)^2}-\frac{e^2}{\left (c d^2-a e^2\right )^3 (a e+c d x)}-\frac{e^3 \log (a e+c d x)}{\left (c d^2-a e^2\right )^4}+\frac{e^3 \log (d+e x)}{\left (c d^2-a e^2\right )^4}\\ \end{align*}
Mathematica [A] time = 0.100535, size = 117, normalized size = 0.84 $-\frac{\frac{\left (c d^2-a e^2\right ) \left (11 a^2 e^4+a c d e^2 (15 e x-7 d)+c^2 d^2 \left (2 d^2-3 d e x+6 e^2 x^2\right )\right )}{(a e+c d x)^3}+6 e^3 \log (a e+c d x)-6 e^3 \log (d+e x)}{6 \left (c d^2-a e^2\right )^4}$
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x)^3/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^4,x]
[Out]
-(((c*d^2 - a*e^2)*(11*a^2*e^4 + a*c*d*e^2*(-7*d + 15*e*x) + c^2*d^2*(2*d^2 - 3*d*e*x + 6*e^2*x^2)))/(a*e + c*
d*x)^3 + 6*e^3*Log[a*e + c*d*x] - 6*e^3*Log[d + e*x])/(6*(c*d^2 - a*e^2)^4)
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Maple [A] time = 0.052, size = 135, normalized size = 1. \begin{align*}{\frac{{e}^{3}\ln \left ( ex+d \right ) }{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}}}+{\frac{1}{ \left ( 3\,a{e}^{2}-3\,c{d}^{2} \right ) \left ( cdx+ae \right ) ^{3}}}+{\frac{e}{2\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{2} \left ( cdx+ae \right ) ^{2}}}+{\frac{{e}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{3} \left ( cdx+ae \right ) }}-{\frac{{e}^{3}\ln \left ( cdx+ae \right ) }{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x)
[Out]
e^3/(a*e^2-c*d^2)^4*ln(e*x+d)+1/3/(a*e^2-c*d^2)/(c*d*x+a*e)^3+1/2*e/(a*e^2-c*d^2)^2/(c*d*x+a*e)^2+e^2/(a*e^2-c
*d^2)^3/(c*d*x+a*e)-e^3/(a*e^2-c*d^2)^4*ln(c*d*x+a*e)
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Maxima [B] time = 1.07751, size = 556, normalized size = 4. \begin{align*} -\frac{e^{3} \log \left (c d x + a e\right )}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}} + \frac{e^{3} \log \left (e x + d\right )}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}} - \frac{6 \, c^{2} d^{2} e^{2} x^{2} + 2 \, c^{2} d^{4} - 7 \, a c d^{2} e^{2} + 11 \, a^{2} e^{4} - 3 \,{\left (c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x}{6 \,{\left (a^{3} c^{3} d^{6} e^{3} - 3 \, a^{4} c^{2} d^{4} e^{5} + 3 \, a^{5} c d^{2} e^{7} - a^{6} e^{9} +{\left (c^{6} d^{9} - 3 \, a c^{5} d^{7} e^{2} + 3 \, a^{2} c^{4} d^{5} e^{4} - a^{3} c^{3} d^{3} e^{6}\right )} x^{3} + 3 \,{\left (a c^{5} d^{8} e - 3 \, a^{2} c^{4} d^{6} e^{3} + 3 \, a^{3} c^{3} d^{4} e^{5} - a^{4} c^{2} d^{2} e^{7}\right )} x^{2} + 3 \,{\left (a^{2} c^{4} d^{7} e^{2} - 3 \, a^{3} c^{3} d^{5} e^{4} + 3 \, a^{4} c^{2} d^{3} e^{6} - a^{5} c d e^{8}\right )} x\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="maxima")
[Out]
-e^3*log(c*d*x + a*e)/(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8) + e^3*log(e*
x + d)/(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8) - 1/6*(6*c^2*d^2*e^2*x^2 +
2*c^2*d^4 - 7*a*c*d^2*e^2 + 11*a^2*e^4 - 3*(c^2*d^3*e - 5*a*c*d*e^3)*x)/(a^3*c^3*d^6*e^3 - 3*a^4*c^2*d^4*e^5 +
3*a^5*c*d^2*e^7 - a^6*e^9 + (c^6*d^9 - 3*a*c^5*d^7*e^2 + 3*a^2*c^4*d^5*e^4 - a^3*c^3*d^3*e^6)*x^3 + 3*(a*c^5*
d^8*e - 3*a^2*c^4*d^6*e^3 + 3*a^3*c^3*d^4*e^5 - a^4*c^2*d^2*e^7)*x^2 + 3*(a^2*c^4*d^7*e^2 - 3*a^3*c^3*d^5*e^4
+ 3*a^4*c^2*d^3*e^6 - a^5*c*d*e^8)*x)
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Fricas [B] time = 1.99822, size = 961, normalized size = 6.91 \begin{align*} -\frac{2 \, c^{3} d^{6} - 9 \, a c^{2} d^{4} e^{2} + 18 \, a^{2} c d^{2} e^{4} - 11 \, a^{3} e^{6} + 6 \,{\left (c^{3} d^{4} e^{2} - a c^{2} d^{2} e^{4}\right )} x^{2} - 3 \,{\left (c^{3} d^{5} e - 6 \, a c^{2} d^{3} e^{3} + 5 \, a^{2} c d e^{5}\right )} x + 6 \,{\left (c^{3} d^{3} e^{3} x^{3} + 3 \, a c^{2} d^{2} e^{4} x^{2} + 3 \, a^{2} c d e^{5} x + a^{3} e^{6}\right )} \log \left (c d x + a e\right ) - 6 \,{\left (c^{3} d^{3} e^{3} x^{3} + 3 \, a c^{2} d^{2} e^{4} x^{2} + 3 \, a^{2} c d e^{5} x + a^{3} e^{6}\right )} \log \left (e x + d\right )}{6 \,{\left (a^{3} c^{4} d^{8} e^{3} - 4 \, a^{4} c^{3} d^{6} e^{5} + 6 \, a^{5} c^{2} d^{4} e^{7} - 4 \, a^{6} c d^{2} e^{9} + a^{7} e^{11} +{\left (c^{7} d^{11} - 4 \, a c^{6} d^{9} e^{2} + 6 \, a^{2} c^{5} d^{7} e^{4} - 4 \, a^{3} c^{4} d^{5} e^{6} + a^{4} c^{3} d^{3} e^{8}\right )} x^{3} + 3 \,{\left (a c^{6} d^{10} e - 4 \, a^{2} c^{5} d^{8} e^{3} + 6 \, a^{3} c^{4} d^{6} e^{5} - 4 \, a^{4} c^{3} d^{4} e^{7} + a^{5} c^{2} d^{2} e^{9}\right )} x^{2} + 3 \,{\left (a^{2} c^{5} d^{9} e^{2} - 4 \, a^{3} c^{4} d^{7} e^{4} + 6 \, a^{4} c^{3} d^{5} e^{6} - 4 \, a^{5} c^{2} d^{3} e^{8} + a^{6} c d e^{10}\right )} x\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="fricas")
[Out]
-1/6*(2*c^3*d^6 - 9*a*c^2*d^4*e^2 + 18*a^2*c*d^2*e^4 - 11*a^3*e^6 + 6*(c^3*d^4*e^2 - a*c^2*d^2*e^4)*x^2 - 3*(c
^3*d^5*e - 6*a*c^2*d^3*e^3 + 5*a^2*c*d*e^5)*x + 6*(c^3*d^3*e^3*x^3 + 3*a*c^2*d^2*e^4*x^2 + 3*a^2*c*d*e^5*x + a
^3*e^6)*log(c*d*x + a*e) - 6*(c^3*d^3*e^3*x^3 + 3*a*c^2*d^2*e^4*x^2 + 3*a^2*c*d*e^5*x + a^3*e^6)*log(e*x + d))
/(a^3*c^4*d^8*e^3 - 4*a^4*c^3*d^6*e^5 + 6*a^5*c^2*d^4*e^7 - 4*a^6*c*d^2*e^9 + a^7*e^11 + (c^7*d^11 - 4*a*c^6*d
^9*e^2 + 6*a^2*c^5*d^7*e^4 - 4*a^3*c^4*d^5*e^6 + a^4*c^3*d^3*e^8)*x^3 + 3*(a*c^6*d^10*e - 4*a^2*c^5*d^8*e^3 +
6*a^3*c^4*d^6*e^5 - 4*a^4*c^3*d^4*e^7 + a^5*c^2*d^2*e^9)*x^2 + 3*(a^2*c^5*d^9*e^2 - 4*a^3*c^4*d^7*e^4 + 6*a^4*
c^3*d^5*e^6 - 4*a^5*c^2*d^3*e^8 + a^6*c*d*e^10)*x)
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Sympy [B] time = 2.6889, size = 668, normalized size = 4.81 \begin{align*} \frac{e^{3} \log{\left (x + \frac{- \frac{a^{5} e^{13}}{\left (a e^{2} - c d^{2}\right )^{4}} + \frac{5 a^{4} c d^{2} e^{11}}{\left (a e^{2} - c d^{2}\right )^{4}} - \frac{10 a^{3} c^{2} d^{4} e^{9}}{\left (a e^{2} - c d^{2}\right )^{4}} + \frac{10 a^{2} c^{3} d^{6} e^{7}}{\left (a e^{2} - c d^{2}\right )^{4}} - \frac{5 a c^{4} d^{8} e^{5}}{\left (a e^{2} - c d^{2}\right )^{4}} + a e^{5} + \frac{c^{5} d^{10} e^{3}}{\left (a e^{2} - c d^{2}\right )^{4}} + c d^{2} e^{3}}{2 c d e^{4}} \right )}}{\left (a e^{2} - c d^{2}\right )^{4}} - \frac{e^{3} \log{\left (x + \frac{\frac{a^{5} e^{13}}{\left (a e^{2} - c d^{2}\right )^{4}} - \frac{5 a^{4} c d^{2} e^{11}}{\left (a e^{2} - c d^{2}\right )^{4}} + \frac{10 a^{3} c^{2} d^{4} e^{9}}{\left (a e^{2} - c d^{2}\right )^{4}} - \frac{10 a^{2} c^{3} d^{6} e^{7}}{\left (a e^{2} - c d^{2}\right )^{4}} + \frac{5 a c^{4} d^{8} e^{5}}{\left (a e^{2} - c d^{2}\right )^{4}} + a e^{5} - \frac{c^{5} d^{10} e^{3}}{\left (a e^{2} - c d^{2}\right )^{4}} + c d^{2} e^{3}}{2 c d e^{4}} \right )}}{\left (a e^{2} - c d^{2}\right )^{4}} + \frac{11 a^{2} e^{4} - 7 a c d^{2} e^{2} + 2 c^{2} d^{4} + 6 c^{2} d^{2} e^{2} x^{2} + x \left (15 a c d e^{3} - 3 c^{2} d^{3} e\right )}{6 a^{6} e^{9} - 18 a^{5} c d^{2} e^{7} + 18 a^{4} c^{2} d^{4} e^{5} - 6 a^{3} c^{3} d^{6} e^{3} + x^{3} \left (6 a^{3} c^{3} d^{3} e^{6} - 18 a^{2} c^{4} d^{5} e^{4} + 18 a c^{5} d^{7} e^{2} - 6 c^{6} d^{9}\right ) + x^{2} \left (18 a^{4} c^{2} d^{2} e^{7} - 54 a^{3} c^{3} d^{4} e^{5} + 54 a^{2} c^{4} d^{6} e^{3} - 18 a c^{5} d^{8} e\right ) + x \left (18 a^{5} c d e^{8} - 54 a^{4} c^{2} d^{3} e^{6} + 54 a^{3} c^{3} d^{5} e^{4} - 18 a^{2} c^{4} d^{7} e^{2}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**3/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**4,x)
[Out]
e**3*log(x + (-a**5*e**13/(a*e**2 - c*d**2)**4 + 5*a**4*c*d**2*e**11/(a*e**2 - c*d**2)**4 - 10*a**3*c**2*d**4*
e**9/(a*e**2 - c*d**2)**4 + 10*a**2*c**3*d**6*e**7/(a*e**2 - c*d**2)**4 - 5*a*c**4*d**8*e**5/(a*e**2 - c*d**2)
**4 + a*e**5 + c**5*d**10*e**3/(a*e**2 - c*d**2)**4 + c*d**2*e**3)/(2*c*d*e**4))/(a*e**2 - c*d**2)**4 - e**3*l
og(x + (a**5*e**13/(a*e**2 - c*d**2)**4 - 5*a**4*c*d**2*e**11/(a*e**2 - c*d**2)**4 + 10*a**3*c**2*d**4*e**9/(a
*e**2 - c*d**2)**4 - 10*a**2*c**3*d**6*e**7/(a*e**2 - c*d**2)**4 + 5*a*c**4*d**8*e**5/(a*e**2 - c*d**2)**4 + a
*e**5 - c**5*d**10*e**3/(a*e**2 - c*d**2)**4 + c*d**2*e**3)/(2*c*d*e**4))/(a*e**2 - c*d**2)**4 + (11*a**2*e**4
- 7*a*c*d**2*e**2 + 2*c**2*d**4 + 6*c**2*d**2*e**2*x**2 + x*(15*a*c*d*e**3 - 3*c**2*d**3*e))/(6*a**6*e**9 - 1
8*a**5*c*d**2*e**7 + 18*a**4*c**2*d**4*e**5 - 6*a**3*c**3*d**6*e**3 + x**3*(6*a**3*c**3*d**3*e**6 - 18*a**2*c*
*4*d**5*e**4 + 18*a*c**5*d**7*e**2 - 6*c**6*d**9) + x**2*(18*a**4*c**2*d**2*e**7 - 54*a**3*c**3*d**4*e**5 + 54
*a**2*c**4*d**6*e**3 - 18*a*c**5*d**8*e) + x*(18*a**5*c*d*e**8 - 54*a**4*c**2*d**3*e**6 + 54*a**3*c**3*d**5*e*
*4 - 18*a**2*c**4*d**7*e**2))
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Giac [B] time = 1.34534, size = 973, normalized size = 7. \begin{align*} \frac{2 \,{\left (c^{3} d^{6} e^{3} - 3 \, a c^{2} d^{4} e^{5} + 3 \, a^{2} c d^{2} e^{7} - a^{3} e^{9}\right )} \arctan \left (-\frac{2 \, c d x e + c d^{2} + a e^{2}}{\sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}}\right )}{{\left (c^{6} d^{12} - 6 \, a c^{5} d^{10} e^{2} + 15 \, a^{2} c^{4} d^{8} e^{4} - 20 \, a^{3} c^{3} d^{6} e^{6} + 15 \, a^{4} c^{2} d^{4} e^{8} - 6 \, a^{5} c d^{2} e^{10} + a^{6} e^{12}\right )} \sqrt{-c^{2} d^{4} + 2 \, a c d^{2} e^{2} - a^{2} e^{4}}} - \frac{6 \, c^{5} d^{8} x^{5} e^{5} + 15 \, c^{5} d^{9} x^{4} e^{4} + 11 \, c^{5} d^{10} x^{3} e^{3} + 3 \, c^{5} d^{11} x^{2} e^{2} + 3 \, c^{5} d^{12} x e + 2 \, c^{5} d^{13} - 18 \, a c^{4} d^{6} x^{5} e^{7} - 30 \, a c^{4} d^{7} x^{4} e^{6} + 5 \, a c^{4} d^{8} x^{3} e^{5} + 15 \, a c^{4} d^{9} x^{2} e^{4} - 15 \, a c^{4} d^{10} x e^{3} - 13 \, a c^{4} d^{11} e^{2} + 18 \, a^{2} c^{3} d^{4} x^{5} e^{9} - 70 \, a^{2} c^{3} d^{6} x^{3} e^{7} - 30 \, a^{2} c^{3} d^{7} x^{2} e^{6} + 60 \, a^{2} c^{3} d^{8} x e^{5} + 38 \, a^{2} c^{3} d^{9} e^{4} - 6 \, a^{3} c^{2} d^{2} x^{5} e^{11} + 30 \, a^{3} c^{2} d^{3} x^{4} e^{10} + 70 \, a^{3} c^{2} d^{4} x^{3} e^{9} - 30 \, a^{3} c^{2} d^{5} x^{2} e^{8} - 120 \, a^{3} c^{2} d^{6} x e^{7} - 56 \, a^{3} c^{2} d^{7} e^{6} - 15 \, a^{4} c d x^{4} e^{12} - 5 \, a^{4} c d^{2} x^{3} e^{11} + 75 \, a^{4} c d^{3} x^{2} e^{10} + 105 \, a^{4} c d^{4} x e^{9} + 40 \, a^{4} c d^{5} e^{8} - 11 \, a^{5} x^{3} e^{13} - 33 \, a^{5} d x^{2} e^{12} - 33 \, a^{5} d^{2} x e^{11} - 11 \, a^{5} d^{3} e^{10}}{6 \,{\left (c^{6} d^{12} - 6 \, a c^{5} d^{10} e^{2} + 15 \, a^{2} c^{4} d^{8} e^{4} - 20 \, a^{3} c^{3} d^{6} e^{6} + 15 \, a^{4} c^{2} d^{4} e^{8} - 6 \, a^{5} c d^{2} e^{10} + a^{6} e^{12}\right )}{\left (c d x^{2} e + c d^{2} x + a x e^{2} + a d e\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^4,x, algorithm="giac")
[Out]
2*(c^3*d^6*e^3 - 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*e^7 - a^3*e^9)*arctan(-(2*c*d*x*e + c*d^2 + a*e^2)/sqrt(-c^2*d^
4 + 2*a*c*d^2*e^2 - a^2*e^4))/((c^6*d^12 - 6*a*c^5*d^10*e^2 + 15*a^2*c^4*d^8*e^4 - 20*a^3*c^3*d^6*e^6 + 15*a^4
*c^2*d^4*e^8 - 6*a^5*c*d^2*e^10 + a^6*e^12)*sqrt(-c^2*d^4 + 2*a*c*d^2*e^2 - a^2*e^4)) - 1/6*(6*c^5*d^8*x^5*e^5
+ 15*c^5*d^9*x^4*e^4 + 11*c^5*d^10*x^3*e^3 + 3*c^5*d^11*x^2*e^2 + 3*c^5*d^12*x*e + 2*c^5*d^13 - 18*a*c^4*d^6*
x^5*e^7 - 30*a*c^4*d^7*x^4*e^6 + 5*a*c^4*d^8*x^3*e^5 + 15*a*c^4*d^9*x^2*e^4 - 15*a*c^4*d^10*x*e^3 - 13*a*c^4*d
^11*e^2 + 18*a^2*c^3*d^4*x^5*e^9 - 70*a^2*c^3*d^6*x^3*e^7 - 30*a^2*c^3*d^7*x^2*e^6 + 60*a^2*c^3*d^8*x*e^5 + 38
*a^2*c^3*d^9*e^4 - 6*a^3*c^2*d^2*x^5*e^11 + 30*a^3*c^2*d^3*x^4*e^10 + 70*a^3*c^2*d^4*x^3*e^9 - 30*a^3*c^2*d^5*
x^2*e^8 - 120*a^3*c^2*d^6*x*e^7 - 56*a^3*c^2*d^7*e^6 - 15*a^4*c*d*x^4*e^12 - 5*a^4*c*d^2*x^3*e^11 + 75*a^4*c*d
^3*x^2*e^10 + 105*a^4*c*d^4*x*e^9 + 40*a^4*c*d^5*e^8 - 11*a^5*x^3*e^13 - 33*a^5*d*x^2*e^12 - 33*a^5*d^2*x*e^11
- 11*a^5*d^3*e^10)/((c^6*d^12 - 6*a*c^5*d^10*e^2 + 15*a^2*c^4*d^8*e^4 - 20*a^3*c^3*d^6*e^6 + 15*a^4*c^2*d^4*e
^8 - 6*a^5*c*d^2*e^10 + a^6*e^12)*(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)^3)
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+0
help plz
0
87
1
I need help with a counting problem.
In how many ways can three points be selected from the grid below so that they form a right isosceles triangle?
Mar 13, 2023
#1
+195
0
To form a right isosceles triangle, we need to select three points such that two of them lie on one of the grid's horizontal lines and the third lies on a line that is perpendicular to that line and passes through the midpoint of the first two points.
There are 5 horizontal lines and 5 vertical lines we can choose from. Let's consider the cases where the two points are on the first horizontal line:
- If we choose the two points that are closest to each other, they are one unit apart, and there are two points on the second horizontal line that would form a right isosceles triangle with them.
- If we choose two points that are two units apart, there are no points on the second horizontal line that would form a right isosceles triangle with them.
- If we choose two points that are three units apart, there is only one point on the second horizontal line that would form a right isosceles triangle with them.
Therefore, for each horizontal line, we have a total of 2 + 0 + 1 = 3 ways to select two points that are one, two, or three units apart. For each of these pairs of points, there is exactly one point on a vertical line that would form a right isosceles triangle with them. Therefore, the total number of ways to select three points that form a right isosceles triangle is:
5 horizontal lines x 3 ways to choose the pair of points on each line x 1 way to choose the third point on a perpendicular line = 15.
Therefore, there are 15 ways to select three points from the given grid that form a right isosceles triangle.
Mar 13, 2023
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# Fit with Gaussian
Hello,
I am trying to fit a gaussian to a data file I have. When the I view the histogram, it has a mean value around close to the peak at which I want it to fit around. How would I comprise a code that would fit about the ‘mean’ of a histogram, and then output (cout) the ‘chi^2/ndf’ of the fit?
Also, how would I separate the ‘mean’ of the histogram and the ‘mean’ of the fit that I obtain? (The reason why is because my ultimate goal is to iterate the fit to histogram until the ‘chi^2/ndf’ is sufficiently low, any help with this also would be appreciated to but help with just the initial question is very much appreciated.)
Thank You,
Edwin
Hello,
Maybe this would make my question more clear, I tried the following code and it shows up with subsequent erros:
``````root [41] TF1 *f1 = new TF1("f1","gaus",h1.GetMean(),-.5,.5)
Error in <TF1::TF1>: can not find any function at the address 0x3dacb9c. This function requested for f1
root [42] h1.Fit("f1","H")
Unknown function: f1
(class TFitResultPtr)64677040
root [43] ``````
Basically I’m trying to define a new function with the predefined gaussian function, and set the parameters so that the range to fit is set around the Mean (i.e. Mean-0.5 to the Mean+0.5). Any help would be much appreciated.
Thank you,
Edwin
Hi,
you can define your function as following:
``````TF1 *f1 = new TF1("f1","gaus",-.5,.5);
// set initial parameters (not really needed for gaus)
f1->SetParameters(h1.GetMaximum(), h1.GetMean(), h1.GetRMS() );
h1.Fit("f1")``````
Note that the initial parameters are not needed when fitting a pre-defined gaussian function, since in this case they are computed automatically.
Also the option H is not valid when fitting an histogram. It is for a linear fitting of a TGraph.
Best Regards
Lorenzo
Thank You so much,
I just have another question if you wouldn’t mind:
After I fit the historgram, I’d like to be able to call and output the Mean value of the fit. So I try the following but get the subsequent errors:
``````root [18] gfit.GetMean()
Error: Can't call TF1::GetMean() in current scope (tmpfile):1:
Possible candidates are...
(in TF1)
(in TFormula)
*** Interpreter error recovered ***
root [19] ``````
I’m not sure what this means or how to resolve it. I thought it would be as easy as I stated because I can call the Chisquare and NDF via:
``````root [5] double chisq=gfit->GetChisquare();
root [6] double ndf=gfit->GetNDF();
root [7] double chisqdf=chisq/ndf;
root [8] cout << "Chisquare: " << chisq << "/" << ndf << " : " << chisqdf << endl;
Chisquare: 19.0439/10 : 1.90439``````
So, I am left a bit confused. Any help would be great. Thank you.
Edwin
Hi,
for getting the function parameters like the mean of your gaussian do:
``double mean = gfit->GetParameter("Mean");``
or
``double mean = gfit->GetParameter(1);``
Lorenzo
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# How far is Natuna Ranai from Banjarmasin?
The distance between Banjarmasin (Syamsudin Noor International Airport) and Natuna Ranai (Ranai Airport) is 670 miles / 1079 kilometers / 582 nautical miles.
The driving distance from Banjarmasin (BDJ) to Natuna Ranai (NTX) is 1081 miles / 1739 kilometers, and travel time by car is about 113 hours 40 minutes.
670
Miles
1079
Kilometers
582
Nautical miles
1 h 46 min
121 kg
## Distance from Banjarmasin to Natuna Ranai
There are several ways to calculate the distance from Banjarmasin to Natuna Ranai. Here are two standard methods:
Vincenty's formula (applied above)
• 670.290 miles
• 1078.727 kilometers
• 582.466 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 672.100 miles
• 1081.640 kilometers
• 584.039 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Banjarmasin to Natuna Ranai?
The estimated flight time from Syamsudin Noor International Airport to Ranai Airport is 1 hour and 46 minutes.
## Flight carbon footprint between Syamsudin Noor International Airport (BDJ) and Ranai Airport (NTX)
On average, flying from Banjarmasin to Natuna Ranai generates about 121 kg of CO2 per passenger, and 121 kilograms equals 267 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Banjarmasin to Natuna Ranai
See the map of the shortest flight path between Syamsudin Noor International Airport (BDJ) and Ranai Airport (NTX).
## Airport information
Origin Syamsudin Noor International Airport
City: Banjarmasin
Country: Indonesia
IATA Code: BDJ
ICAO Code: WAOO
Coordinates: 3°26′32″S, 114°45′46″E
Destination Ranai Airport
City: Natuna Ranai
Country: Indonesia
IATA Code: NTX
ICAO Code: WION
Coordinates: 3°54′31″N, 108°23′16″E
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# Potential and torque of electric dipole in E field
1. Sep 10, 2011
### nthnschager
1. The problem statement, all variables and given/known data
Q: A polar molecule has a dipole moment of magnitude 20 e*pm that makes an angle of 20 degrees with a uniform electric field of magnitude 3.0*10^3 N/C. FInd the magnitude of the torque on the dipole, and the potential energy of the system.
2. Relevant equations
torque = pE sin(theta)
where p is dipole moment and E is electric field mag.
3. The attempt at a solution
in the explanation in my book, (20 e*pm)(3*10^3)(sin20)
is reduced to (.02)(1.6*10^-19 C)(10^-9 m)(3*10^3 N/C)(sin20)
This might be a stupid question but why is 20 reduced to .02? I could not figure out what happened there.
Also, how do you minimize the potential energy when a dipole is is in an electric field? when U = -PEcos(theta), where U is potential energy, P is dipole moment, E is electric field, and theta is angle between direction of dipole moment and electric field. is pot. energy minimized when theta = zero or when theta = ninety?
In other words, I do not understand the concept of negetive potential energy. Would it be considered "minimized" when the potential energy is at zero or negetive?
Last edited: Sep 10, 2011
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Trending ▼ ResFinder
# ICSE Class X Prelims 2021 : Geography (Hume McHenry Memorial High School of S. D. A., Pune)
3 pages, 53 questions, 3 questions with responses, 3 total responses, 2 0
Saisha Chavan Hume McHenry Memorial High School of S. D. A., Pune
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Hume McHenry Memorial High School Salisbury Park Pune 411037 Academic Year 2020 2021 Semester Examination - 2 Subject Geography Date:- 15/12/2020 Class: - X Marks : 80 PART :- I - 30 MARKS ( all questions are compulsory) Question 1: Study the extract Survey of India Map No. G43 S/10 and answer the following questions: Appendix (a). What is the scale of the map in R.F form ? (b). How do you describe the feature associated with the point 574256 ? (c). Find the direction of village Goreli from the village Makawal. (d). Why do you find few white patches in the map area ? (e). Give two map evidences to show that the area is economically backward. (f). Describe the drainage pattern found in the grid square 5827 and 5524. (g). Give four figure grid references of the following. (1). Village Butri (2). Dry tank to the south of Settlement Mitan. (h). In what ways does the pattern of settlement in grid 6029 differ from that in Grid Square 5531 ? (i). On which bank of the nadi is the Village Nimbora situated ? Give reasons For your answer. (j). What is the direct distance from village Solatra to Village Rampura Khera ? (k). What is the meaning of 3r in grid 6029 ? Question 2: In the outline map of India.... (a). Mark and label Karakoram Range (b). Mark and label Satpura (c). Shade and label Gangetic Plains (d). Label the river Narmada (e). Label the river Chambal (f). Label the river Cauveri (g). Shade and name Gulf of kutch (h). Mark and name Karakoram Pass (i). Mark with a dot and name Chennai (j). Mark with a dot and name Allahabad (1 marks) (2 marks) (2 marks) (2 marks) (2 marks) (2 marks) (2 marks) (2 marks) (2 marks) (2 marks) (1 mark) (1) (1) (1) (1) (1) (1) (1) (1) (1) (1) PART:- II - 50 MARKS (attempt any 5 questions ) Question 3: 1. (i).Name the type of climate prevailing over India. (ii). Mention any two factors responsible for it. 2. What is the direction of the summer monsoon ? Why? 3. Give geographical reasons for the following : (i). Even in summer Shimla is cooler than Delhi. (2) (2) (3)
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# Determining wave speed
1. Jun 29, 2015
### DiracPool
I had a thought experiment related to wave speed and frequency.
Lets say we had a wave emitter that emitted transverse plane waves at a regular frequency. Of course we could imagine this as a radio tower, but I wanted it the thought experiment to be more general. Say the emitter is sending out the waves at a frequency of 100 cycles per second (cps). We'll set the wavelength to one meter and we'll set the medium of propagation to one whereby the velocity of the wave in this scenario is 100 m/s.
So what we have is velocity=wavelength x frequency which equals 100=(1)(100)
Now we have an observer Bob watch these waves come in at 100 cps and he gets comfortable seeing this steady, predictable display. After a while, though, we switch the medium between the emitter and Bob so that the speed of the wave is cut in half, down to 50 m/s. Subsequently, Bob now sees the waves coming in at 50 cps.
My question is, is there any way Bob can tell that the wave has slowed down rather than the emitter instead just slowed down it's rate of emission? In other words, can Bob tell that speed of the wave has halved rather than the wavelength of the wave has doubled? Sure, he could ask the emitter tower to stop transmitting, wait a while, and then send a pulse and measure the speed of the pulse, but this thought experiment assumes that this is not possible, all there is a continuously oscillating wave moving through Bob's vicinity.
Part B of the question is the extrapolation to light. I understand that light travels at the constant speed of c in all instances, and that this speed has been measured directly without regard to doppler shift. My question, though, is that, without those direct measurements and Maxwell's equations, etc., if we just had to rely on directly measuring a continuously emitting light source, is there any property inherent in that signal that would allow us to distinguish the velocity from the wavelength/frequency?
2. Jun 29, 2015
### harrylin
It's different from what you think. At constant distance, the frequency in steady state is constant; changing the medium changes the wavelength and only has a temporary effect on the received frequency (depending on the speed of switching) as the number of cycles in transit increases. After that, Bob receives the waves again at 100 cps (it's impossible to loose cycles!) and the wavelength is halved.
Without other means of investigation I think that the emitter tower can perfectly simulate the effect of a medium change for Bob. They only have to gradually reduce and then again increase the frequency to the original value. Note that the wavelength in your illustration has halved because the speed has halved.
According to GR this is only locally the case; non-locally determined the speed of light is reduced in a gravitational field.
- https://en.wikisource.org/wiki/The_...Perihelion-motion_of_the_paths_of_the_Planets.
In steady state the frequency does not depend on the velocity.
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# Difference between low, high and low_limit, high_limit?
Why do we need to redefine something we’ve already defined?
We’ve defined low_limit and high_limit with equations and then when we tell the program get_boundaries(100, 20) this should tell what each key word in the equation means. Perhaps I take the ease with which computer do things normal for granted but having to introduce low and high seems to be an unnecessary step logically speaking.
The variables that are defined inside the function are local only to that function. They cannot be seen. We need to return their values to see those.
``````def foo(m, n):
return m - n, m + n
a, b = foo(100, 20)
``````
What we are doing is unpacking a sequence. The return value contains multiple values, so it is packaged as a sequence (a tuple, probably).
1 Like
I would say it’s more to keep the code clear and simple to understand as if you use the same variable from the function like
Blockquote
print(low_limit)#would give you error to define the variable since you did define it as low
print(high_limit)
Blockquote
more or less keeping thing clear and consistent to understand when you review code later.
When naming variables we try to include as much information as can given.
``````low
high
``````
Those are very general, with not a lot of interpretive information.
``````low_limit
high_limit
``````
Those are less general, and imply lower and upper bounds. More specific, and easier to rationalize for the reader.
2 Likes
I was a little confused at first by over-complicating something simple. You must define the new variables before they can be used:
``````def get_boundaries(target, margin):
low_limit = target - margin
high_limit = margin + target
return low_limit, high_limit
low = low_limit
high = high_limit
low, high = get_boundaries(100, 20)
``````
-lif
Hello, @tera7835798474.
Welcome to the forums!
I formatted your code, so we can see how you had originally indented it. Please review this post: How do I format code in my posts?
Regarding your code, these lines will never be executed:
Consider what `return` does. There is also no need for those lines to be executed. Values assigned to variable inside of a function have local scope (are only accessible inside the function). The values that are `return`ed by the function are ‘unpacked’ and assigned to `low` and `high` in this line:
``````low, high = get_boundaries(100, 20)
``````
1 Like
You are correct the variables inside a function are local to that function unless shenanigans are engaged, but that is outside the scope of this question.
If you’re curious there’s a good discussion of it over on stack overflow:
It has several commented code examples to help get the finer points across.
While a function can have multiple return statements. It can only return once.
So in this case it returns the first calculation, then exits.
This is why the example saves the calculations as variables and returns them both in a single return.
3 Likes
There are two schools of thought on the subject of `return`. That much is clear.
One school accepts multiple return branches, the other insists upon one, only. Can we find legitimate arguments for each scenario?
See if we can fill out this question with more information, enough evidence for us to draw our own conclusion. Be sure to link to any sources quoted.
This is a very good explanation, thank you for clearing this up!
I understand how it makes sense but it was very confusing as we have not set parameters in this way so far in the course.
I found it much more intuitive to remain consistent with the variable naming scheme. I also added a print() line to check my work which seemed to vindicate my approach. Let me know if this doesn’t make sense to anyone else as I am pretty new to this stuff.
Hello everyone. I was also confused as to why low_limit and high_limit were abbreviated with low and high when we printed the equation.
I decided to change the abbreviation to something else. Bob for low_limit and Mary for high_limit. I printed and got the same output of 120 and 80.
This was my syntax:
def get_boundaries(target, margin):
low_limit = target - margin
high_limit = margin + target
return low_limit, high_limit
bob, mary = get_boundaries(100, 20)
print(bob)
print(mary)
This made me conclude that the abbreviation (or variable of the variable) is defined by the position within the commas of:
bob (1st position), mary (second position) = get_boundaries(100, 20)
in comparison to the position of the variables low_limit, high_limit within the return line:
return low_limit (1st position), high_limit (2nd position)
Hope this helps!
2 Likes
``````def get_boundaries(target, margin):
low_limit = target - margin
high_limit = margin + target
return low_limit, high_limit
# Two vairables areassigned to the same function call thereby returning two different values/results.
low, high = get_boundaries(100, 20)
# A print statement of the above removes the confusion
print(low, high)
``````
Thanks, this helped me, I was clueless as to how I got my answer when the variable names do not match up, but it’s not about the variable names. You are calling the return values in order and the names do not have to match up, which confused me at first but seemed like the only explainable way for my programs results.
Why is my code wrong ?
def get_boundaries(target, margin):
low_limit = target - margin
high_limit = target + margin
return low_limit, high_limit
low, high = get_boundaries(100, 20)
return low, high
We cannot see how your blocks are indented, but the last line does not belong. The line above it should not be indented (not inside the block).
you’re meant to save the return value, like this
low = print(low_limit)
high = print(high_limit)
@mtf’s response about checking indentation and a moving a particular line outside the function block would be correct. In your code you’re using print as a function, which will print to the console, but it will return `None` to the names `low` and `high`.
Maybe I need more coffee, and I’m sure the answer is simple, but, I still have this question. Why save the values globally(?) when you can plug in any two numbers in the `print()` function?
``````def get_boundaries(target, margin):
low_limit = target - margin
high_limit = margin + target
return(low_limit, high_limit)
#low, high = get_boundaries(100,20)
print(get_boundaries(100, 20))
print(get_boundaries(200, 40))
print(get_boundaries(50, 95)
``````
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Pre-calculus involves graphing, dealing with angles and geometric shapes such as circles and triangles, and finding absolute values. In solving pre-calculus problems, a number of formulas come in handy. Cheat Sheet Pre-Calculus. Much more than a workbook, this study aid provides pre-calculus problems ranked from easy to advanced, with detailed explanations and step-by-step solutions for each one. How to Find Local Extrema with the First Derivative Test, How to Work with 45-45-90-Degree Triangles, A Quick Guide to the 30-60-90 Degree Triangle, Pre-Calculus Workbook For Dummies Cheat Sheet. This cheat sheet provides the most frequently used formulas, with brief descriptions of what the letters and symbols represent. The following table shows the basic values for each. 599 x 939 png 70kB. Exponents, of course, indicate the operation of raising a number to a power, or in other words, of multiplying a number by itself. You discover new ways to record solutions with interval notation and you plug trig identities into your equations. © 2005 Paul Dawkins You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. Calculus requires knowledge of other math disciplines. Summary. Even though you’re involved with pre-calculus, you remember your old love, algebra, and that fact that absolute values then usually had two possible solutions. Pre-calculus involves graphing, dealing with angles and geometric shapes such as circles and triangles, and finding absolute values. Pre-Calculus - Algebra Review for PreCalculus. Oct 29, 2018 - Pre-Calculus bridges Algebra II and Calculus. The always-true, never-changing trig identities are grouped by subject in the following lists: Mary Jane Sterling aught algebra, business calculus, geometry, and finite mathematics at Bradley University in Peoria, Illinois for more than 30 years. 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Although most of the questions on the exam are multiple-choice, there are some questions that require students to enter a numerical answer. .. To make studying and working out problems in calculus easier, make sure you know basic formulas for geometry, trigonometry, integral calculus, and differential calculus. Apr 16, 2019 - Precalculus for dummies cheat sheet by bertha on Indulgy.com Pre-Calculus FINAL EXAM Formula Sheet - Docsity. Pre-calculus involves graphing, dealing with angles and geometric shapes such as circles and triangles, and finding absolute values. By Mark Ryan . Pre-Calculus For Dummies Cheat Sheet - dummies Pre-Calculus bridges Algebra II and Calculus. You will need graph paper. It includes formulas, the laws of logarithmic functions, trigonometric values of basic angles, conic section equations, and interval notation. opposite sin hypotenuse q= hypotenuse csc opposite q= adjacent cos hypotenuse q= hypotenuse sec adjacent q= opposite tan adjacent q= adjacent cot opposite q= Unit circle definition For this definition q is any angle. But you can take some of the fear of studying Calculus away by understanding its basic principles, such as derivatives and antiderivatives, integration, and solving compound functions. The following list explains how the equations for each are unique: Among the trickier aspects of pre-calculus for beginners to remember are the rules for exponents and logarithms. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. She is the author of several For Dummies books, including Algebra Workbook For Dummies, Algebra II For Dummies, and Algebra II Workbook For Dummies. Pre-Calculus For Dummies Cheat Sheet - dummies. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. You discover new ways to record solutions with interval notation, and you plug trig identities into your equations. Calculus Cheat Sheet Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. Discover (and save!) Title. She is the author of Trigonometry For Dummies and Finite Math For Dummies. Pre-Calculus For Dummies Cheat Sheet - dummies. | Algebra cheat ... 768 x 1024 jpeg 74kB. Pre-Calculus bridges Algebra II and Calculus. The following formulas show how to format solution sets in interval notation. 1275 x 1650 png 47kB. 582 x 804 gif 131kB. Calculus Cheat Sheet Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. It has a radius of 1, hence the unit. Feb 26, 2015 - This Pin was discovered by Gemini Wolf. 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An Introduction to Game Theory in Poker
A quick look up on the internet, gives us the definition of Game Theory as:
The branch of mathematics concerned with the analysis of strategies for dealing with competitive situations where the outcome of a participant's choice of action depends critically on the actions of other participants.
Having a game theory optimal (GTO) strategy means that we find the best solution for our situation irrespective of what our opponent does. A standard example is the Prisoner's Dilemma. Here's a link to the Wikipedia entry and a YouTube video if you didn't know about this already.
What does GTO mean in poker?
Being a good poker player means making the best decisions based on the incomplete information you have. To bridge this very chasm between incomplete information and the best decision, we take the help of game theory. Just like in the Prisoner's Dilemma, we do not know what the opponent's strategy is. So we make the most optimal game theory decision in a vacuum. Then, as we gain more information about an opponent, we adjust your strategy based on this new information.
Two Examples:
1. Bet Size
On the flop, your opponent bets 1/3rd pot. What do you do with middle pair? Most likely, you would call. What if he makes it 1/2 pot? Or 3/4 pot? Or 2x pot? 4x pot? At what point do you decide that you won't call this bet?
As your opponent's bet size increases, the odds that you get become worse. In order to not get exploited, we have to start letting go of some hands. So we start folding our weak pairs, then middle pairs good kickers, then middle pairs strong kickers, and sometimes even top pairs with weak-medium kickers.
2. Bluff catching
Similar logic goes into bluff catching. For example, say you have JTo and board runs out Jack high - you have top pair medium kicker. Your opponent, whom you don't know anything about, bets big all three streets. What do you do? If you always call down with just a top pair, you will be often up against QQ+ or better top pairs. If you always fold, you open yourself up to getting bluffed. To find a balance, we use a GTO approach.
As we move from flop to turn and to the river, we start folding the bottom of our range. This means that out of all the hands that we could have in this spot, we start folding the worst ones. If we find ourselves here with top pairs often, we start folding the ones with the worst kickers. If we call with every Jack on the flop, we fold some like J9, JT on the turn. On the river, since we arrive with QJ+, we have to fold some of these and call with some. So we can fold the worst Jacks - QJ, KJ and call down with AJ.
In the first example, we did not know what our opponent bet-sizes meant. In the second, we do not know when our opponent is bluffing or when he is value-betting. Hence we chose a strategy that allows us to make the best decision no matter what he does. Does it mean that we will win every pot? No. But it does mean that in the long run, after playing thousands of hands, and facing the same situations many times, the summation of all our decisions will come out to be a profitable number.
How does GTO compare with Exploitative strategy?
Exploitative strategy means that when you have absolute reads on your opponents, you deviate from the GTO approach to make the best decision in that particular instance. For example, if you know that the villain never bluffs when he bets big all three streets, but you are at the top of your range, it maybe a fine GTO call, but you can make an exploitative fold knowing that you are always beat.
No single strategy is better than the other. Usually a combination of both is the way to go - because even though GTO style is profitable, mixing it with exploitative might be even more profitable. This is especially true when you are playing against weaker opponents. They have certain tendencies and leaks that are better suited for an exploitative style of play. But against stronger opponents, who have fewer obvious leaks, a GTO style is much more favorable.
As the poker software and poker AI has become better, more and more players are employing a GTO approach, especially when the stakes are high and the competition is good. But since no player is playing a perfectly GTO strategy, adding an exploitative style to your game is a good idea.
Hopefully, this primer was useful and may it help you make more profitable decisions in the future. And next time try employing this strategy in any one aspect of your game - value-betting, check-raising, bluffing etc.
Cheers,
Mayank
This article was written by Mayank Jain. Mayank is a writer who plays poker for a living. He writes at http://mayankja.in/ on mindfulness, travel and art, among other things.
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# AGED Sudoku
Version 1.4.4
Install +5 K From 44 Rates 4.3 Category Puzzle Size 24 MB Last Update 2023 June 2
# AGED Sudoku
Aged Studio Limited
Version 1.4.4
Install +5 K From 44 Rates 4.3 Category Puzzle Size 24 MB Last Update 2023 June 2
View the antivirus scan results
## Introduction
Classic sudoku game now can be played on your phone and tablet. Sudoku is a logic number-placement puzzle game. To win a sudoku game, you need to fill the 9×9 grid with numbers so that there will be no repetitive number in each row, each column and each 3×3 sub-grid, namely numbers 1-9 will only appear once in each row, column or sub-grid. Also, it is a relaxing causal game, but it requires your focus to figure out the solution. Therefore, sudoku will help you exercise your brain and memory because you need to think and utilize your logic when you are playing sudoku.
Our sudoku is committed to improve your game experience. It has 4 levels of difficulty for players with different capabilities of sudoku, from beginners to masters on sudoku. Also, it has various functions to help you win the game, such as Hint, Undo/Redo, Take Notes, etc. Moreover, you can practice your skills and challenge yourself by playing Daily Challenge and My Challenge. We are sure you can have great fun and better game experience when you are playing our sudoku.
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User Reviews - 44 Rates
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Home / Power Conversion / Convert Joule/hour to Terawatt
# Convert Joule/hour to Terawatt
Please provide values below to convert joule/hour [J/h] to terawatt [TW], or vice versa.
From: joule/hour To: terawatt
### Joule/hour to Terawatt Conversion Table
Joule/hour [J/h]Terawatt [TW]
0.01 J/h2.7777777777778E-18 TW
0.1 J/h2.7777777777778E-17 TW
1 J/h2.7777777777778E-16 TW
2 J/h5.5555555555556E-16 TW
3 J/h8.3333333333333E-16 TW
5 J/h1.3888888888889E-15 TW
10 J/h2.7777777777778E-15 TW
20 J/h5.5555555555556E-15 TW
50 J/h1.3888888888889E-14 TW
100 J/h2.7777777777778E-14 TW
1000 J/h2.7777777777778E-13 TW
### How to Convert Joule/hour to Terawatt
1 J/h = 2.7777777777778E-16 TW
1 TW = 3.6E+15 J/h
Example: convert 15 J/h to TW:
15 J/h = 15 × 2.7777777777778E-16 TW = 4.1666666666667E-15 TW
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A008677 Expansion of 1/((1-x^3)*(1-x^5)*(1-x^7)). 2
1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 3, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 14, 13, 14, 15, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,11 COMMENTS Number of partitions of n into parts 3, 5, and 7. - Joerg Arndt, Aug 17 2013 Number of different total numbers of kicks, tries and converted tries which lead to a score of n in a rugby (union) match. - Matthew Scroggs, Jul 09 2015 REFERENCES L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 114, [6t]. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 230 M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7. Index entries for linear recurrences with constant coefficients, signature (0,0,1,0,1,0,1,-1,0,-1,0,-1,0,0,1). FORMULA a(n) = a(n-3)+a(n-5)+a(n-7)-a(n-8)-a(n-10)-a(n-12)+a(n-15) for n>=15. - David Neil McGrath, Sep 03 2014 G.f.: 1 / ((1 - x^3) * (1 - x^5) * (1 - x^7)). Euler transform of length 7 sequence [ 0, 0, 1, 0, 1, 0, 1]. - Michael Somos, Sep 30 2014 a(n) = a(-15-n) for all n in Z. - Michael Somos, Sep 30 2014 0 = a(n) - a(n+3) - a(n+5) + a(n+8) - [mod(n, 7) == 6] for all n in Z. - Michael Somos, Sep 30 2014 a(n) = round(n^2/210 + n/14 + 5/21) + r(n) where r(n) = 1 if n == 0, 3, 10, 15, 45, 75, 80, 87, or 90 mod 105, r(n) = -1 if n == 4, 11, 16, 44, 46, 74, 79 or 86 mod 105, r(n) = 0 otherwise. - Robert Israel, Jul 09 2015 EXAMPLE G.f. = 1 + x^3 + x^5 + x^6 + x^7 + x^8 + x^9 + 2*x^10 + x^11 + 2*x^12 + ... MAPLE S:= series(1 / ((1 - x^3) * (1 - x^5) * (1 - x^7)), x, 101): seq(coeff(S, x, j), j=0..100); # Robert Israel, Jul 09 2015 MATHEMATICA CoefficientList[Series[1 / ((1 - x^3) (1 - x^5) (1 - x^7)), {x, 0, 100}], x] (* Vincenzo Librandi, Jun 23 2013 *) PROG (PARI) a(n)=[1, 0, -2, 2, -2, 0, 1][n%7+1]/7+[2, -1, 0, 0, -1][n%5+1]/5+[2, -1, -1][n%3+1]/9+(3*n^2+45*n+148)/630; \\ Tani Akinari, Aug 17 2013 (PARI) a(n)=floor((n^2+15*n+86)/210+(n%3<1)/3+3*(n%5<1)/5) \\ Tani Akinari, Sep 30 2014 CROSSREFS Sequence in context: A218469 A230502 A280253 * A036497 A211976 A035460 Adjacent sequences: A008674 A008675 A008676 * A008678 A008679 A008680 KEYWORD nonn,easy AUTHOR EXTENSIONS Typo in name fixed by Vincenzo Librandi, Jun 23 2013 STATUS approved
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# Roman Numerals
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## Leah Kuhn
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#### Transcript of Roman Numerals
Roman Numerals
The Subtractive Principle
The Subtractive Principle
Reflection
What the Letters Mean
What the Letters Mean
Here is a chart that has all the Roman Numerals, and what they stand for. The system is based on seven different symbols in total. These symbols can be used to write any Roman Numeral!
Roman Numeral Value
I 1
V 5
X 10
L 50
C 100
D 500
M 1,000
Once you have familiarized yourself with these, let's move on to the Additive Principle, where we'll begin to learn how to construct other numbers using these.
Practice
Check out http://www.abcya.com/roman_numerals.htm for a Roman Numeral game that has increasing levels of difficulty.
Practice
Create at least two problems for the following operations: addition, subtraction, multiplication, and division. Write your problems in Roman Numerals and make an answer key.
Finding Roman Numerals
Roman numerals were used by the Ancient Romans as their numbering system. We still use them sometimes today. You see them in the Super Bowl's numbering system, after king's names (King Henry IV), in outlines, and other places. Roman numerals are base 10 or decimal, like the numbers we use today. They are not entirely positional, however, and there is no number zero.
The system for writing Roman Numerals is an additive system. This means to calculate the value of a Roman Numeral, you simply have to add up all the digits in the Roman Numeral!
Examples:
VIII = 5 + 1 + 1 + 1 = 8
XXXIII = 10 + 10 + 10 + 1 + 1 +1 = 33
CLXXVIII = 100 + 50 + 10 + 10 + 5 + 1 + 1 + 1 = 178
MMX = 1,000 + 1,000 + 10 = 2,010
There are a couple of other rules to follow:
Roman Numerals are always written with the largest numbers on the left, down to the smallest on the right. For example, you wouldn't write 8 above as IIIV, it's written as VIII.
Roman Numerals are always written with the shortest number of numerals as possible. For example, you wouldn't write 5 as IIIII, you'd always write it as V.
Video Lesson
To watch Roman Numeral Instruction go to:
http://www.roman-numerals.org/videos.html
So, remember how you are not to put smaller Roman Numerals to the left of larger ones? There's one exception where you're allowed to do this. For numbers where you'd have to write four numerals in a row, there is a shortcut to save precious writing space. You can write a smaller numeral to the left of a larger one to subtract from it! The table below has all the allowed subtractive combinations:
Roman
Numeral Value
IV 4
IX 9
XL 40
XC 90
CD 400
CM 900
As you can see above, there are only certain cases where subtraction is allowed. You can remember this because the numeral being subtracted is always 1 or 2 numerals away from the numeral being subtracted from. So you can subtract I from V or X, but not L, and X from L or C, but not D.
You might think it would be easy to write 99 as IC, but only the combinations above are allowed. You can have multiple subtractions in the same Roman Numeral, which we need to write 99 correctly, as XCIX = XC + IX = 90 + 9 = 99.
1. What was the most interesting thing you learned about Roman Numerals?
2. What was the most challenging thing to remember about Roman Numerals?
3. Which do you think is a more effective number system: ours or Roman numerals? Why?
Full transcript
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# Simple Interest- Aptitude Questions and Answers - in Hindi
At what rate percent per annum will a sum of money doubles in 8 years?
प्रतिवर्ष किस ब्याज की दर पर 8 साल में राशि दोगुना होगा?
A)
12 %
B)
12.5 %
C)
13 %
D)
15 %
12.5 %
Explanation :
Let Principal = P, Rate = R% per annum, Time = n years.
Simple Interest (SI) = (P*R*T )/100
Amount =P + (P*R*T )/100
Thus, 2P =P + (P*R*8 )/100
⇒ P= P*R*8 / 100
⇒ R = 100 /8
So, R=12.5%
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# How to Build Rubble Stone Walls
## Overview
Rubble stone is the by-product of either natural geological occurrences or work in a quarry. While it is usually used as fill for mortared walls, it may also be used on its own to create beautiful, rustic rubble stone walls that will enhance any landscape. It is also an economical resource, and building with it is almost like solving a puzzle as you work to fit the pieces together.
## Gathering Stone
### Step 1
Lay out a rough outline in string of the size and shape of the rubble stone walls you want to build. Determine the height, length and width of each.
### Step 2
Calculate the amount of rubble stone needed for each wall by multiplying the three measurements to get the total cubic feet. Fifteen cubic feet of wall requires one ton of stone. Add extra to ensure you have enough and to account for the irregular shape of rubble stone and the varying density of the rocks.
### Step 3
Collect or purchase the required amount of rubble stones. Shale, slate, schist and limestone are good choices for building rubble stone walls as they are easily split if necessary.
### Step 4
Sort the stones into piles of similar sizes and shapes to make the process of fitting the rubble stones together easier and quicker.
## Building the Wall
### Step 1
Dig a trench the length of the wall. It should be at least a foot wide for a rubble stone wall up to three feet high, wider if the rubble stones are wide, and at least six inches deep.
### Step 2
Stack the largest rubble stones in the trench to form the wall's footing. This will help prevent the wall from sliding forward. Make sure the stones are level and complete the whole length of the wall before continuing.
### Step 3
Use a tamper to pack the rubble stones in place, adding dirt as needed. Fill in space between the larger rocks with smaller rocks, using the hammer and chisel to break them up if necessary.
### Step 4
Place the second layer of rubble stones so that they fill any openings created by the first layer. Add dirt and smaller rocks to fill in the spaces. Water carefully poured on the fill will help it settle. Try to maintain the rule of two stones over one, one stone over two for maximum strength and stability. Build each layer in this manner until the desired height is reached.
### Step 5
Split some of the larger rubble stones that are left so that they are flat and place on the very top of the wall to give it a more finished appearance.
## Things You'll Need
• Measuring Tape
• Wooden stakes
• String
• Shovel
• Work gloves
• Rubble stone
• Dirt
• Tamper
• Hammer and stone chisel
• Safety goggles
• Carpenters level
• Water
## References
• Building a Dry-Stacked Stone Wall
• How to Build a Dry Stone Wall
Keywords: rubble stone, stone walls, building rubble stone walls
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##### Question Info
This question is public and is used in 157 tests or worksheets.
Type: Multiple-Choice
Category: Functions and Relations
Standards: HSF-IF.A.1
Author: janawilliams01
View all questions by janawilliams01.
# Functions and Relations Question
View this question.
Add this question to a group or test by clicking the appropriate button below.
## Grade 9 Functions and Relations CCSS: HSF-IF.A.1
Which relation represents a function?
1. {(2,6), (-3,6), (4,9), (2,10)}
2. {(-4,4), (-3, 3), (-2,2), (-1,1), (-4, 0)}
3. {(3,5), (4, 8), (2, 9)}
4. {(8, 0), (5,2), (8,-1), (4,-2)}
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Cody
# Problem 44421. Portfolio diversification: choose your stocks !
Solution 2903735
Submitted on 2 Sep 2020
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
stock1=[0.1 0.3 0.22 -.15 ] ; stock2=[0.3 0.4 -0.13 -0.22 ]; stock3=[0.6 -0.3 0.44 0.05]; y_correct ={'stock2' 'stock3'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock2'} {'stock3'} ans = 1×2 cell array {'stock1'} {'stock2'}
Assertion failed.
2 Fail
stock1=[-0.1 0.4 -0.14 -.32 ]; stock2=[0.35 -0.10 0.66 0.18 ]; stock3=[0.62 -0.3 0.44 0.05]; y_correct = {'stock1' 'stock2'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock1'} {'stock2'} ans = 1×2 cell array {'stock2'} {'stock3'}
Assertion failed.
3 Fail
stock1=[0.3 0.4 -0.8 0.5 ]; stock2=[0.62 0.2 -0.3 0.05]; stock3=[0.35 -0.10 0.66 0.18 ]; y_correct = {'stock1' 'stock3'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock1'} {'stock3'} ans = 1×2 cell array {'stock1'} {'stock2'}
Assertion failed.
4 Fail
stock1=[-0.5 -0.2 0.35 0.02 ]; stock2=[0.2 0.15 -0.3 0.13]; stock3=[0.45 -0.04 0.66 0.2 ]; y_correct = {'stock1' 'stock2'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock1'} {'stock2'} ans = 1×2 cell array {'stock1'} {'stock3'}
Assertion failed.
5 Fail
stock1=[0.5 0.2 0.35 0.4 ]; stock2=[-0.2 -0.15 -0.3 0.13]; stock3=[-0.45 -0.04 0.33 0.15 ]; y_correct = {'stock1' 'stock3'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock1'} {'stock3'} ans = 1×2 cell array {'stock2'} {'stock3'}
Assertion failed.
6 Fail
stock1=[-0.32 -0.2 0.35 -0.4 ]; stock2=[0.2 0.15 0.3 0.13]; stock3=[-0.45 -0.04 -0.33 0.15 ]; y_correct = {'stock2' 'stock3'} assert(isequal(your_fcn_name(stock1,stock2,stock3),y_correct))
y_correct = 1×2 cell array {'stock2'} {'stock3'} ans = 1×2 cell array {'stock1'} {'stock2'}
Assertion failed.
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# Mathematicians Solve Sum-of-Three-Cubes Problem for the Number 42
It was pretty cool earlier this year when Andrew Booker, a professor at the U.K.'s University of Bristol, revealed a solution to expressing the number 33 as the sum of three cubes.
But now, Booker and a Massachusetts Institute of Technology professor Andrew Sutherland have found a way to solve the Diophantine Equation (x^3+y^3+z^3=k) for the lone remaining integer under 100 that hadn't been conquered: the number 42.
As this University of Bristol press release details, Booker and Sutherland did the massive amount of calculations required with the help of Charity Engine. The website recruits volunteers across the world to install an app on their PCs (over 400,000 PCs were involved), which allows their unused computational power to be aggregated to form the equivalent of a supercomputer. The software used to solve for 42 is the same code that was used for the number 33, previously discovered by Booker and published in the journal "Research in Number Theory." Charity Engine rents out its services to raise funds for worthy causes, such as CARE, Oxfam and others.
The answer, which took over a million hours of calculating to find, is:
X = -80538738812075974 Y = 80435758145817515 Z = 12602123297335631
In this YouTube segment from Numberphile, Booker explains how the problem was solved:
As Booker explains in the video, though, they aren't through finding the sum of three cubes for numbers. 114 is next on the agenda. But he also wants to see if there is a third solution for the number 3, a question first posed by mathematician Louis Mordell back in the early 1950s.
But solving for 42 has a special significance. Fans of Douglas Adams' 1979 novel "The Hitchhiker's Guide to the Galaxy," will recall it as the answer that the giant computer Deep Thought offers to the meaning of life, the universe, and everything. And in Lewis Carroll's "Alice's Adventures in Wonderland," there's the famous Rule Forty-two cited by the King of Hearts, which requires all persons more than a mile high to leave the court. It's also the jersey number worn by the Brooklyn Dodgers' Jackie Robinson, who integrated Major League Baseball in 1947. Fifty years later, then-acting baseball commissioner Bud Selig permanently retired number 42, so no other player can wear it.
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Hosted by The Math Forum
Weighted Quadrilaterals
A 3x4 rectangle has sides 3,4,3,4 and diagonals 5,5. Its "weight" is said to be 3+4+3+4+5+5 = 24. Find a quadrilateral with all sides and diagonals having integer length and whose weight is smaller than 24.
Bonus goes to the solution(s) with the smallest weights.
Source: Math Gazette March 1997, p. 18. Article by B. E. Peterson and J. H. Jordan. Result due to H. Harborth and A. Kemnitz.
© Copyright 1997 Stan Wagon. Reproduced with permission.
The Math Forum
2 October 1998
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# Trigonometric Functions of Sum and Difference of Two Angles
• Difficulty Level : Hard
• Last Updated : 20 Dec, 2021
Trigonometry is a branch of mathematics, which deal with the angles, lengths, and heights of triangles and their relationships. It had played an important role to calculate complex functions or large distances which were not possible to calculate without trigonometry. While solving problems with trigonometry, we came across many situations where we have to calculate the trigonometric solutions for the sum of angles or differences of angles. E.g.
Here,
Which is a tangent trigonometric ratio, with an angle opposite to BC.
tan(θ+Φ) =
If θ = 30° and Φ = 45°. We know the trigonometric angles of 45° and 30°, but we don’t know the trigonometric angle of (45° + 30° = 75°). So, to simplify these types of problems. We will get to learn trigonometric formulae or identities of sum and differences of two angles which will make things easier.
Before moving further first we will see the signs of the trigonometric functions in the four quadrants. These signs play an important role in trigonometry.
### Trigonometric identities
Now we are going to find the trigonometric identities. As we know that
sin(-x) = – sin x
cos(-x) = cos x
Because only cos and sec are positive in the fourth quadrant. So, now we prove some results regarding sum and difference of angles:
Let’s consider a unit circle (having radius as 1) with centre at the origin. Let x be the ∠DOA and y be the ∠AOB. Then (x + y) is the ∠DOB. Also let (– y) be the ∠DOC.
Therefore, the coordinates of A, B, C and D are
A = (cos x, sin x)
B = [cos (x + y), sin (x + y)]
C = [cos (– y), sin (– y)]
D = (1, 0).
As, ∠AOB = ∠COD
Adding, ∠BOC both side, we get
∠AOB + ∠BOC = ∠COD + ∠BOC
∠AOC = ∠BOD
In â–³ AOC and â–³ BOD
OA = OB (radius of circle)
∠AOC = ∠BOD (Proved earlier)
OC = OD (radius of circle)
△ AOC ≅ △ BOD by SAS congruency.
By using distance formula, for
AC2 = [cos x – cos (– y)]2 + [sin x – sin(–y]2
AC2 = 2 – 2 (cos x cos y – sin x sin y) …………….(i)
And, now
Similarly, using distance formula, we get
BD2 = [1 – cos (x + y)]2 + [0 – sin (x + y)]2
BD2 = 2 – 2 cos (x + y) …………….(ii)
As, △ AOC ≅ △ BOD
AC = BD, So AC2 = BD2
From eq(i) and eq(ii), we get
2 – 2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y)
So,
cos (x + y) = cos x cos y – sin x sin y
Take y = -y, we get
cos (x + (-y)) = cos x cos (-y) – sin x sin (-y)
cos (x – y) = cos x cos y + sin x sin y
Now, taking
cos (-(x + y)) = cos ((-x) – y) (cos (-θ) = sin θ)
sin (x – y) = sin x cos y – cos x sin y
take y = -y, we get
sin (x – (-y)) = sin x cos (-y) – cos x sin (-y)
sin (x + y) = sin x cos y + cos x sin y
The derived formulae for trigonometric ratios of compound angles are as follows:
sin (A + B) = sin A cos B + cos A sin B ………………..(1)
sin (A – B) = sin A cos B – cos A sin B ………………..(2)
cos (A + B) = cos A cos B – sin A sin B ..………………(3)
cos (A – B) = cos A cos B + sin A sin B ………………..(4)
By using these formulae, we can obtain some important and mostly used form:
(1) Take, A =
In eq(1) and (3), we get
sin (+B) = cos B
cos (+B) = – sin A
(2) Take, A = π
In eq(1), (2), (3) and (4) we get
sin (Ï€ + B) = – sin B
sin (Ï€ – B) = sin B
cos (Ï€ ± B) = – cos B
(3) Take, A = 2Ï€
In eq(2) and (4) we get
sin (2Ï€ – B) = – sin B
cos (2Ï€ – B) = cos B
Similarly for cot A, tan A, sec A, and cosec A
(4)
Here, A, B, and (A + B) is not an odd multiple of π/2, so, cosA, cosB and cos(A + B) are non-zero
tan(A + B) = sin(A + B)/cos(A + B)
From eq(1) and (3), we get
tan(A + B) = sin A cos B + cos A sin B/cos A cos B – sin A sin B
Now divide the numerator and denominator by cos A cos B we get
tan(A + B) =
(5)
As we know that
So, on putting B = -B, we get
(6)
Here, A, B, and (A + B) is not a multiple of π, so, sinA, sinB and sin(A + B) are non-zero
cot(A + B) = cos(A + B)/sin(A + B)
From eq(1) and (3), we get
cot(A + B) = cos A cos B – sin A sin B/sin A cos B + cos A sin B
Now divide the numerator and denominator by sin A sin B we get
cot(A + B) =
(7)
As we know that
So, on putting B = -B, we get
Here, we will establish two sets of transformation formulae: Factorization and Defactorization formulae.
### Defactorization Formulae
In trigonometry, defactorisation means converting a product into a sum or difference. The defactorization formulae are:
(1) 2 sin A cos B = sin (A + B) + sin (A – B)
Proof:
As we know that
sin (A + B) = sin A cos B + cos A sin B ………………………(1)
sin (A – B) = sin A cos B – cos A sin B ………………………(2)
By adding eq(1) and (2), we get
2 sin A cos B = sin (A + B) + sin (A – B)
(2) 2 cos A sin B = sin (A + B) – sin (A – B)
Proof:
As we know that
sin (A + B) = sin A cos B + cos A sin B ………………………(1)
sin (A – B) = sin A cos B – cos A sin B ………………………(2)
By subtracting eq(2) from (1), we get
2 cos A sin B = sin (A + B) – sin (A – B)
(3) 2 cos A cos B = cos (A + B) + cos (A – B)
Proof:
As we know that
cos (A + B) = cos A cos B – sin A sin B ………………………(1)
cos (A – B) = cos A cos B + sin A sin B ………………………(2)
By adding eq(1) and (2), we get
2 cos A cos B = cos (A + B) + cos (A – B)
(4) 2 sin A sin B = cos (A – B) – cos (A + B)
Proof:
cos (A + B) = cos A cos B – sin A sin B ………………………(1)
cos (A – B) = cos A cos B + sin A sin B ………………………(2)
By subtracting eq(3) from (4), we get
2 sin A sin B = cos (A – B) – cos (A + B)
Example 1. Convert each of the following products into the sum or difference.
(i) 2 sin 40° cos 30°
(ii) 2 sin 75° sin 15°
(iii) cos 75° cos 15°
Solution:
(i) Given: A = 40° and B = 30°
Now put all these values in the formula,
2 sin A cos B = sin (A + B) + sin (A – B)
We get
2 sin 40° cos 30° = sin (40 + 30) + sin (40 – 30)
= sin (70°) + sin (10°)
(ii) Given: A = 75° and B = 15°
Now put all these values in the formula,
2 sin A sin B = cos (A – B) – cos (A + B)
We get
2 sin 75° sin 15° = cos (75-15) – cos (75+15)
= cos (60°) – cos (90°)
(iii) Given: A = 75° and B = 15°
Now put all these values in the formula,
2 cos A cos B = cos (A + B) + cos (A – B)
We get
cos 75° cos 15° = 1/2(cos (75+15) + cos (75-15))
= 1/2 (cos (90°) + cos (60°))
Example 2. Solve for
Solution:
Using the formula
2 cos A cos B = cos (A + B) + cos (A – B)
Hence,
= 0
### Factorisation Formulae
In trigonometry, factorisation means converting sum or difference into the product. The factorisation formulae are:
(1) sin (C) + sin (D) = 2 sin cos
Proof:
We have
2 sin A cos B = sin (A + B) + sin (A – B) ………………………(1)
So now, we are taking
A + B = C and A – B = D
Then, A = and B =
Now put all these values in eq(1), we get
2 sin () cos () = sin (C) + sin (D)
Or
sin (C) + sin (D) = 2 sin () cos ()
(2) sin (C) – sin (D) = 2 cos sin
Proof:
We have
2 cos A sin B = sin (A + B) – sin (A – B) ………………………(1)
So now, we are taking
A + B = C and A – B = D
Then, A = and B =
Now put all these values in eq(1), we get
2 cos () sin () = sin (C) – sin (D)
Or
sin (C) – sin (D) = 2 cos () sin ()
(3) cos (C) + cos (D) = 2 cos cos
Proof:
We have
2 cos A cos B = cos (A + B) + cos (A – B) ………………………(1)
So now, we are taking
A + B = C and A – B = D
Then, A = and B =
Now put all these values in eq(1), we get
2 cos () cos () = cos (C) + cos (D)
Or
cos (C) + cos (D) = 2 cos () cos ()
(4) cos (C) – cos (D) = 2 sin sin
Proof:
We have
2 sin A sin B = cos (A – B) – cos (A + B) ………………………(1)
So now, we are taking
A + B = C and A – B = D
Then, A = and B =
Now put all these values in eq(1), we get
2 sin () sin () = cos (C) – cos (D)
Or
cos (C) – cos (D) = 2 sin () sin ()
Explain 1. Express each of the following as a product
(i) sin 40° + sin 20°
(ii) sin 60° – sin 20°
(iii) cos 40° + cos 80°
Solution:
(i) Given: C = 40° and D = 20°
Now put all these values in the formula,
sin (C) + sin (D) = 2 sin cos
We get
sin 40° + sin 20° = 2 sin cos
= 2 sin cos
= 2 sin 30° cos 10°
(ii) Given: C = 60° and D = 20°
Now put all these values in the formula,
sin (C) – sin (D) = 2 cos sin
We get
sin 60° – sin 20° = 2 cos sin
= 2 cos sin
= 2 cos 40° sin 20°
(iii) Given: C = 80° and D = 40°
Now put all these values in the formula,
cos (C) + cos (D) = 2 cos cos
We get
cos 40° + cos 80° = 2 cos cos
= 2 cos cos
= 2 cos 60° cos 20°
Example 2. Prove that: 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x
Solution:
Lets take LHS
1 + cos 2x + cos 4x + cos 6x
Here, cos 0x = 1
So,
(cos 0x + cos 2x) + (cos 4x + cos 6x)
Using formula
cos (C) + cos (D) = 2 cos cos
We get
(2 cos cos ) + (2 cos cos )
(2 cos x cos x) + (2 cos 5x cos x)
Taking 2 cos x common, we have
2 cos x (cos x + cos 5x)
Again using the formula
cos (C) + cos (D) = 2 cos cos
We get
2 cos x (2 cos cos )
2 cos x (2 cos 3x cos 2x)
4 cos x cos 2x cos 3x
LHS = RHS
Hence proved
### Trigonometric ratios of multiple angles (2A) in terms of angle A
The trigonometric ratios of an angle in a right triangle define the relationship between the angle and the length of its sides. sin 2x or cos 2x, etc. are also, one such trigonometrical formula, also known as double angle formula, as it has a double angle in it.
(1) sin 2A = 2 sin A cos A
Proof:
As we know that
sin (A + B) = sin A cos B + cos A sin B ………………..(1)
Now taking B = A, in eq(1), we get
sin (A + A) = sin A cos A + cos A sin A
sin 2A = 2 sin A cos A
(2) cos 2A = cos2 A – sin2 A
Proof:
As we know that
cos (A + B) = cos A cos B – sin A sin B ………………..(1)
Now taking B = A, in eq(1), we get
cos (A + A) = cos A cos A + sin A sin A
cos 2A = cos2 A – sin2 A
(3) cos 2A = 2cos2 A – 1
Proof:
As we know that
cos 2A = cos2 A – sin2 A ………………..(1)
We also know that
sin2 A + cos2 A = 1
So, sin2 A = 1 – cos2 A
Now put the value of sin2 A in eq(1), we get
cos 2A = cos2 A – (1 – cos2 A)
cos 2A = cos2 A – 1 + cos2 A
cos 2A = 2cos2 A – 1
(4) cos 2A = 1 – 2sin2 A
Proof:
As we know that
cos 2A = 2cos2 A – 1 ………………..(1)
We also know that
sin2 A + cos2 A = 1
So, cos2 A = 1 – sin2 A
Now put the value of sin2 A in eq(1), we get
cos 2A = 2(1 – sin2 A) – 1
cos 2A = 2 – 2sin2 A) – 1
cos 2A = 1 – 2sin2
(5) cos 2A =
Proof:
As we know that
cos 2A = cos2 A – sin2
So, now dividing, by sin2 A + cos2 A = 1, we get
cos 2A =
Again dividing the numerator and denominator by cos2 A, we get
cos 2A =
cos 2A =
(6) sin 2A =
Proof:
As we know that
sin (A + B) = sin A cos B + cos A sin B ………………..(1)
Now taking B = A, in eq(1), we get
sin (A + A) = sin A cos A + cos A sin A
sin 2A = 2 sin A cos A
As we also know that sin2 A + cos2 A = 1
So, now dividing, by sin2 A + cos2 A = 1, we get
sin 2A =
Now, on dividing the numerator and denominator by cos2 A, we get
sin 2A =
(7) tan 2A =
Proof:
As we know that
………………..(1)
Now taking B = A, in eq(1), we get
tan(A + A) =
tan 2A =
Example: Prove that
(i) = tan θ
(ii) = cot θ
(iii) cos 4x = 1 – 8 sin2x cos2x
Solution:
(i) sin 2θ = 2 sin θ cos θ ………..(from identity 1)
and, 1 + cos 2θ = 2cos2θ ………..(from identity 3)
= tan θ
Hence Proved
(ii) sin 2θ = 2 sin θ cos θ ………..(from identity 1)
and, 1 – cos 2θ = 2sin2θ ………..(from identity 4)
= cot θ
Hence Proved
(iii) cos 4x = cos 2(2x)
= 1 – 2sin2(2x) (using 16)
= 1 – 2(sin(2x))2
= 1 – 2(2 sin x cos x) (using identity 1)
= 1 – 2(4 sin2 x cos2 x)
cos 4x = 1 – 8 sin2 x cos2 x
Hence Proved
### Trigonometric ratios of multiple angles (3A) in terms of angle A
The trigonometric ratios of an angle in a right triangle define the relationship between the angle and the length of its sides. sin 3x or cos 3x, etc. are also, one such trigonometrical formula, also known as triple angle formula, as it has a triple angle in it.
(1) sin 3A = 3sin A – 4 sin3A
Proof:
Let’s take LHS
sin 3A = sin(2A + A)
Using identity
sin (A + B) = sin A cos B + cos A sin B
We get
sin 3A = sin 2A cos A + cos 2A sin A
= 2sin A cos A cos A + (1 – 2 sin2A)sin A
= 2sin A(1 – sin2A) + sin A – 2 sin3A
= 2sin A – 2sin3A + sin A – 2 sin3A
sin 3A = 3sin A – 4 sin3A
(2) cos 3A = 4 cos3A – 3cos A
Proof:
Let’s take LHS
sin 3A = sin(2A + A)
Using identity
cos (A + B) = cos A cos B – sin A sin B
We get
cos 3A = cos 2A cos A – sin 2A sin A
= (2cos2A – 1)cos A – 2sin A cos A sin A
= (2cos2A – 1)cos A – 2cos A(1 – cos2A)
= 2cos3A – cos A – 2cos A + 2cos3A)
cos 3A = 4 cos3A – 3cos A
(3) tan 3A =
Proof:
Let’s take LHS
tan 3A = tan(2A + A)
Using identity
We get
tan 3A =
Example 1. Solve 2sin3xsinx.
Solution:
We have 2sin3xsinx
We also write as y = y1 . y2 ….(1)
Here, y1 = 2sin3x
y2 = sinx
So let’s solve y1 = 2sin3x
Using identity
sin 3A = 3sin A – 4 sin3A
We get
y1 = 2(sin x – 4 sin3x)
= 2sin x – 8 sin3x
Now put these values in eq(1), we get
y = (2sin x – 8 sin3x)(sinx)
= 2sin2 x – 8 sin4x
Example 2. Solve 2tan3xtanx.
Solution:
We have 2tan3xtanx
We also write as y = y1 . y2 ….(1)
Here, y1 = 2tan3x
y2 = tanx
So let’s solve y1 = 2tan3x
Using identity
tan 3A =
We get
y1 = 2()
Now put these values in eq(1), we get
y = ()(tanx)
My Personal Notes arrow_drop_up
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# Functions with Inverses Functions with Inverses
## Presentation on theme: "Functions with Inverses Functions with Inverses"— Presentation transcript:
Functions with Inverses Functions with Inverses
One-to-One Functions Functions with Inverses One-to-One Functions We here take up the discussion of one-to-one functions: what they are, how they work, and conditions necessary for their existence. One-to-One Functions 8/10/2013
Functions with Inverses
One-to-One Functions Functions with Inverses One-to-One Functions Definition A function f is a one-to-one function if no two ordered pairs of f have the same second component Note: One-to-one is often written as 1-1 One-to-One Functions We lay the groundwork for the study of inverse functions by considering one-to-one functions. The definition is framed as simply as possible in the same terms as was the definition of function itself – that is, in terms of a set of ordered pairs. While function is defined as a set of ordered pairs no two of which have the same first component, a one-to-one (1-1) function is first of all a function, but one in which no two ordered pairs have the same second component. Thus, a 1-1 function is a set of ordered pairs no two of which have either the same first or second components. The examples show functions which are 1-1, as well as some that are not Some are shown as sets of ordered pairs (with no formula) and some defined by algebraic expressions (or formulas). One example shows a set of ordered pairs which is not a function, since it contains the distinct pairs (5, 3) and (5, 6) having the same first component. This would say that the functional value of 5 is 3, but also that the functional value of 5 is 6. If the function is called f, then f(5) = 3 and f(5) = 6. Since f(5) = 3 ≠ 6 = f(5) we then have the awkward condition that f(5) ≠ f(5). 8/10/2013 One-to-One Functions One-to-One Functions 8/10/2013
Functions with Inverses
One-to-One Functions Functions with Inverses One-to-One Functions 1-1 Examples: 1. f = { (1, 3), (2, 5), (3, 2), (7, 1) } 2. g = { (1, 3), (2, 5), (3, 6), (7, 3) } 3. h = { (5, 3), (2, 9), (5, 6), (8, 7) } 4. f(x) = (x – 4)2 + 7 5. g(x) = x + 1 6. f(x) = |x + 1| NOT 1-1 NOT a function NOT 1-1 WHY ? One-to-One Functions We lay the groundwork for the study of inverse functions by considering one-to-one functions. The definition is framed as simply as possible in the same terms as was the definition of function itself – that is, in terms of a set of ordered pairs. While function is defined as a set of ordered pairs no two of which have the same first component, a one-to-one (1-1) function is first of all a function, but one in which no two ordered pairs have the same second component. Thus, a 1-1 function is a set of ordered pairs no two of which have either the same first or second components. The examples show functions which are 1-1, as well as some that are not Some are shown as sets of ordered pairs (with no formula) and some defined by algebraic expressions (or formulas). One example shows a set of ordered pairs which is not a function, since it contains the distinct pairs (5, 3) and (5, 6) having the same first component. This would say that the functional value of 5 is 3, but also that the functional value of 5 is 6. If the function is called f, then f(5) = 3 and f(5) = 6. Since f(5) = 3 ≠ 6 = f(5) we then have the awkward condition that f(5) ≠ f(5). NOT 1-1 WHY ? 8/10/2013 One-to-One Functions One-to-One Functions 8/10/2013
Functions with Inverses
One-to-One Functions Functions with Inverses Horizontal Line Test No horizontal line intersects the graph of a 1-1 function more than once Examples x y(x) x y(x) 1-1 function Not 1-1 Horizontal Line Test Graphical interpretation of the 1-1 property can be done with the horizontal line test, analogous to the vertical line test for the graph of a function. The examples start with the graph of a 1-1 function, followed by the graphs of two functions that are not 1-1. The graph in the lower center passes the horizontal line test, but NOT the vertical line test – that is, the graph does not even represent a function and hence cannot be a 1-1 function. The last graph is clearly the graph of function, but is not 1-1 as shown by the horizontal line test. 8/10/2013 One-to-One Functions One-to-One Functions 8/10/2013
Functions with Inverses
One-to-One Functions Functions with Inverses Horizontal Line Test No horizontal line intersects the graph of a 1-1 function more than once More Examples x y(x) x y(x) x y(x) Horizontal Line Test Graphical interpretation of the 1-1 property can be done with the horizontal line test, analogous to the vertical line test for the graph of a function. The examples start with the graph of a 1-1 function, followed by the graphs of two functions that are not 1-1. The graph in the lower center passes the horizontal line test, but NOT the vertical line test – that is, the graph does not even represent a function and hence cannot be a 1-1 function. The last graph is clearly the graph of function, but is not 1-1 as shown by the horizontal line test. Not 1-1 Not 1-1 NOT a function ! 8/10/2013 One-to-One Functions One-to-One Functions 8/10/2013
Functions with Inverses
Think about it ! 8/10/2013 One-to-One Functions One-to-One Functions 8/10/2013
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# Mathematics / Year 6 / Number and Algebra / Fractions and decimals
Curriculum content descriptions
Compare fractions with related denominators and locate and represent them on a number line (ACMNA125)
Elaborations
• demonstrating equivalence between fractions using drawings and models
General capabilities
• Numeracy Numeracy
ScOT terms
Number lines, Related denominators
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A bag conta
Given:
Bag I contains 4 white and 5 black balls.
Bag II contains 3 white and 4 black balls.
A ball is transferred from bag I to bag II and then a ball is drawn from bag II.
There are two mutually exclusive ways to draw a white ball from bag II –
a. A white ball is transferred from bag I to bag II, and then, a white ball is drawn from bag II
b. A black ball is transferred from bag I to bag II, and then, a white ball is drawn from bag II
Let E1 be the event that white ball is drawn from bag I and E2 be the event that black ball is drawn from bag I.
Now, we have
We also have
Let E3 denote the event that white ball is drawn from bag II.
Hence, we have
We also have
Using the theorem of total probability, we get
P(E3) = P(E1)P(E3|E1) + P(E2)P(E3|E2)
Thus, the probability of the drawn ball being white is.
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This site is supported by donations to The OEIS Foundation.
The October issue of the Notices of the Amer. Math. Soc. has an article about the OEIS.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A030681 Smallest nontrivial extension of n-th palindromic prime which is a square. 2
25, 36, 529, 729, 1156, 101124, 131044, 15129, 181476, 191844, 3136, 35344, 373321, 383161, 727609, 7573504, 7873636, 797449, 919681, 929296, 1030153216, 1050148836, 10601536, 1131111424, 1141155961, 124211025 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Chai Wah Wu, Table of n, a(n) for n = 1..10000 MATHEMATICA d[n_]:=IntegerDigits[n]; Table[i=1; While[!IntegerQ[Sqrt[x=FromDigits[Flatten[{d[n], d[i]}]]]], i++]; x, {n, Select[Prime[Range[1500]], Reverse[x=d[#]]==x &]}] (* Jayanta Basu, May 24 2013 *) CROSSREFS Cf. A030682. Sequence in context: A077437 A077676 A030671 * A167363 A035383 A179803 Adjacent sequences: A030678 A030679 A030680 * A030682 A030683 A030684 KEYWORD nonn,base AUTHOR STATUS approved
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Last modified September 25 20:53 EDT 2018. Contains 315425 sequences. (Running on oeis4.)
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https://www.programmingmag.com/product/cs2134-homework-assignment-3b-solved/
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Sale!
# CS 2134 Homework Assignment 3B solved
Original price was: \$35.00.Current price is: \$30.00.
Category:
## Description
Programming Part:
Enter data from the file MTA_train_stop_data.txt. The data from this assignment is from
the information in this file for this assignment.1
)
In the batch phase you will read all the data from the file called MTA_train_stop_data.txt into a
container of type vector.
Your program will define the class trainStopData. It has the following private member variables :
string stop_id; // id of train stop (1st token)
string stop_name; // name of station (4th token)
double stop_lat; // latitude of train stop location
double stop_lon; // longitude of train stop location
Your class should also have a constructor and the following public member functions:
string get_id( ) const
string get_stop_name( ) const
double get_latitude( ) const
double get_longitude( ) const
1
Written Part:
The C++ STL has many functions and functors. Here is your chance to try some of them. In a
program when you use an STL algorithm add #include, and when you use an STL
For many of these problems you will need to look up information online. Here are some sources:
http://en.cppreference.com/w/cpp
http://www.cplusplus.com/reference/algorithm/
http://www.cplusplus.com/reference/std/functional/
http://www.sgi.com/tech/stl/
Fill in the correct code where you see a ****
vector A {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
vector B {1, 2, 1, 2, 1, 2};
vector C{1, 2, 3, 1, 2, 3};
vector D(6);
vector E(10);
1. Copy the first 6 items of vector A into vector D
copy(****, ****, ****);
// D now equals {1, 2, 3, 4, 5, 6}
2. Count the number of ones in vector B
cout << count(****, ****, 1);
//prints out the number of times a one appears in B
3. In C++, there is a way to construct a unary functor from a binary functor. To do this you
use an adapter, a function called bind1st or bind2nd. We use bind1st in this example to
convert the STL binary predicate functor not_equal_to into a unary predicate by setting its
first value to 1.
Count the number of items that are not equal to one in B
cout << count_if(B.begin( ), ****, bind1st(not_equal_to( ), 1));
/*prints out the number of times a one doesn’t appear in B.*/
4. Test to determine if the first 3 items of A are the same as C.
bool same;
same = equal(A.begin( ), ****, ****);
2
5. Find the first item in vector A which equals 5
vector::iterator vecItr;
vecItr = find(****, ****, 5);
// returns an iterator to 5 in vector A
if (vecItr != A.end( ))
cout << ****;
// prints out the value pointed to by vecItr
6. Find the first item in C that is greater than 2
vecItr = find_if(****, ****, bind2nd(greater(), 2));
// returns an iterator to 3 in C
if (vecItr != C.end( ))
cout << ****;
// prints out the value pointed to by vecItr
3
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https://www.schooltube.com/tag/tagid/how%20to%20divide%20mixed%20numbers
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Search for tag: "how to divide mixed numbers"
How to Divide a Mixed Number by a Fraction | Math with Mr. J
Welcome to Dividing Mixed Numbers by Fractions with Mr. J! Need help with how to divide a mixed number by a fraction? You're in the right place! Whether you're just starting out, or need…
From MathMrJ _ on May 22nd, 2021 0 likes 4 plays
How to Divide a Fraction by a Mixed Number | Math with Mr. J
Welcome to Dividing Fractions by Mixed Numbers with Mr. J! Need help with how to divide a fraction by a mixed number? You're in the right place! Whether you're just starting out, or need…
From MathMrJ _ on May 22nd, 2021 0 likes 3 plays
Dividing Mixed Numbers by Whole Numbers | Math with Mr. J
Welcome to Dividing Mixed Numbers by Whole Numbers with Mr. J! Need help with how to divide mixed numbers by whole numbers? You're in the right place! Whether you're just starting out, or…
From MathMrJ _ on May 22nd, 2021 0 likes 2 plays
Dividing Mixed Numbers and Whole Numbers | Math With Mr. J
Welcome to Dividing Mixed Numbers and Whole Numbers (mixed numbers divided by whole numbers and whole numbers divided by mixed numbers) with Mr. J! Need help with how to divide mixed numbers and…
From MathMrJ _ on May 22nd, 2021 0 likes 2 plays
Dividing Mixed Numbers and Fractions | Math with Mr. J
Welcome to Dividing Mixed Numbers and Fractions (dividing mixed numbers by fractions and dividing fractions by mixed numbers) with Mr. J! Need help with how to divide mixed numbers and fractions?…
From MathMrJ _ on May 22nd, 2021 0 likes 4 plays
Dividing Mixed Numbers | Math with Mr. J
Welcome to Dividing Mixed Numbers with Mr. J! Need help with how to divide mixed numbers? You're in the right place! Whether you're just starting out, or need a quick refresher, this is…
From MathMrJ _ on May 16th, 2021 0 likes 4 plays
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http://etc.usf.edu/clipart/keyword/radius/2
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### Ellipse Fourth Method Case 1
Draftsman's fourth method for drawing an ellipse, case 1
### Ellipse Third Method
Draftsman's third method for drawing an ellipse
### Conjugate Diameters of an Ellipse
Illustration showing that if one diameter is conjugate to a second, the second is conjugate to the first.
### Construction of an Ellipse
Illustration of half of an ellipse and its auxiliary circle used to construct an ellipse by points,…
### Focal Radii of an Ellipse
Illustration of half of an ellipse. "If d denotes the abscissa of a point of an ellipse, r and r' its…
### Line Bisecting Angle Between Focal Radii on Ellipse
Illustration of half of an ellipse. "If through a point P of an ellipse a line is drawn bisecting the…
### Ordinate and Major Axis of Ellipse
Illustration of half of an ellipse. The square of the ordinate of a point in an ellipse is to the product…
### Parallel Tangents to an Ellipse
Illustration showing that tangents drawn at the ends of any diameter are parallel to each other.
### Point Distances to Foci on Ellipse
Illustration of half of an ellipse. "The sum of the distances of any point from the foci of an ellipse…
### Tangent From External Point to an Ellipse
Illustration of how to draw a tangent to an ellipse from an external point.
### Tangent to an Ellipse
Diagram an ellipse with a tangent line that illustrates "A line through a point on the ellipse and bisecting…
### Tangents to an Ellipse
Illustration showing the tangents drawn at two corresponding points of an ellipse and its auxiliary…
### Construction Of An Equilateral Triangle Inscribed In A Circle
An illustration showing how to construct an equilateral triangle inscribed in a circle. "With the radius…
### Forearm Bones
This diagram shows the bones of the right fore-arm. H, the humerus; R, the radius; and U, the ulna.
### Forearm Bones
Transverse section through the bones of the forearm (radius and ulna), taken at about the middle of…
### Back View of the Bones of the Forearm
Front view of the bones of the forearm
### Bones and Ligaments of Forearm
Bones of the right forearm,with their ligaments; volar aspect.
### Cross Section One Inch above the Styloid Process of the Forearm
Section one inch above the styloid process of the right radius.
### Cross Section Three Inches above the Styloid Process of the Forearm
Section three inches above the styloid process of the right radius.
### Cross Section Through the Styloid Process of the Forearm
Section through the styloid process of the right radius.
### Cross Section Two Inches above the Styloid Process of the Forearm
Section two inches above the styloid process of the right radius.
### Dorsal View of the Bones of the Forearm
The bones of the right forearm, dorsal view.
### Front View of the Bones of the Forearm
Front view of the bones of the forearm
### Muscles and Nerves of the Forearm
The muscles and nerves on the front of the forearm and hand. The pronator radii teres, flexor carpi…
### Section Across Forearm
Section across the forearm in the middle third. Labels: A, pronator radii teres; B, flexor carpi radialis;…
### Forearm, Section of
A section across the forearm a short distance below the elbow-joint. R and U, its two supporting bones,…
### Ventral View of the Bones of the Forearm
The bones of the right forearm, ventral view.
### Geometry
Two students drawing geometrical shapes on the chalkboard. A student is also using a scale while students…
### Guiding Needle Point of Compass to Little Finger
The little finger on the left hand can guide the needle point of the compass to the center.
### Hand Ligaments
A anterior view of the ligaments of the wrist and hand.
### Hand Ligaments
A posterior view of the ligaments of the wrist and hand.
### Construction Of A Hexagon In A Circle
An illustration showing how to construct a hexagon in a given circle. "The radius of the circle is equal…
### Tangent to a Hyperbola
Diagram part of a hyperbola with a tangent line that illustrates "A line through a point on the hyperbola…
### Elbow Joint
"Showing how the Ends of the Bones are shaped to form the Elbow Joint. The cut ends of a few ligaments…
### Left Scaphoid
A view of the left scaphoid seen from behind.
### Left Scaphoid
A view of the left scaphoid seen from in front.
### Left Semilunar
A view of the left semilunar from the external surface.
### Noctule Bat
"Skeleton and volar Membranes of the Noctule Bat. c, clavicle; h, humerus; r, radius; u, ulna; d1, first…
### Construction Of A Pentagon Inscribed In A Circle
An illustration showing how to construct a pentagon inscribed in a circle. "Draw the diameter AB, and…
### Construction Of A Pentagon On A Line
An illustration showing how to construct a pentagon on a given line. "From B erect BC perpendicular…
### Regular Pentagon With Circle Inscribed
Illustration showing a circle inscribed in a regular pentagon. Or, a regular pentagon circumscribed…
### Circle With 2 Perpendicular Diameters
Illustration of a circle with center O and diameters AB and CD perpendicular to each other.
### Construction of Radii of a Circle
Illustration used to draw lines which are radii of a circle where the center is inaccessible.
### Radius
The plan of development of the radius for three centuries.
### Radius
The radial bone in the human arm.
### Radius
Anterior view of radius (bone of the arm) of the right side. Labels: 1, cylindrical head; 2, surface…
### Radius
The radius, a bone of the arm.
### Radius and Ulna
An anterior view of the radius and ulna.
### Radius and Ulna
A posterior view of the radius and ulna.
### Radius from Before
Right radius from before.
### Radius from Behind
Right radius from behind.
### Broken Radius
"When a bone is broken, blood trickles out between the injured parts, and afterwards gives place to…
### Carpal Articular Surface of the Radius
Carpal articular surface of the radius, and triangular fibro cartilage of the wrist.
### Colle's Fraction of the Radius
Showing the situation of Colle's fracture of the radius, with fracture of the styloid of the ulna. The…
### Fractured Radius
Fracture of the shaft of the radius.
### Fractured Radius
Fracture of the lower end of the radius.
### Inner Aspect of Radius
Right radius, inner aspect.
### Orbicular Ligament of the Radius
Orbicular ligament of the radius.
### Ring Made of Concentric Circles
Illustration of ring (small circle in larger concentric circle).
### Ring With Piece Cut Out
Illustration of ring (small circle in larger concentric circle) with piece cut out.
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https://community.khronos.org/t/equation/26076
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# equation???
Hi,
Can someone explain to me the following equation means to what. Expecially the “%” symbol refer to what.
move = (move - 5) % 360;
glutPostRedisplay();
Thanks
% is modulus, this just decrements move my 5 every time it is called. Presumably its intended to keep move between 0 and 360 though it should be move = (move+5) % 360 for that to work properly
Originally posted by chowe6685:
% is modulus, this just decrements move my 5 every time it is called. Presumably its intended to keep move between 0 and 360 though it should be move = (move+5) % 360 for that to work properly
Yup, i got it. Thanks
Maybe I misunderstood something, but modulus doesn’t decrement.
If c=a % b then c is the remainder of the integer division a/b. So 10%6 is 4 for example. Whether the result of a modulus is allowed to be negative varies, although I think it’s allowed in C, but I’m not sure.
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http://learnbonds.com/bond-yield-calculator/
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# Bond Yield Calculator – Converting APR into APY
The bond yield calculator converts APR into APY, which allows you to better compare two different bonds. Information about interest earned on investments is sometimes given in one of two ways. The first is in terms of the annual percentage interest rate, or APR. The second is the annual percentage yield – the APY. The APY includes an effect of compounding interest that the APR does not, and for that reason the APY is always higher than the APR. It’s important therefore to be able to compare like with like, and this is where the Yield calculator can help.
### How do you use the Bond Yield Calculator?
You’ll need to put in the following information so that the bond yield calculator can then display the figure for the APY:
Annual Interest Rate: This is for entering the APR. For example, if the APR is 3.4%, then enter 3.4
Number Of Compounding Periods: The assumption is that the total period concerned is one year. So if you enter 365, this means that interest is compounded daily; 12 would mean monthly, 4 would mean quarterly, and so on. By changing the number of periods you’ll also see that the more frequently the interest is compounded (more compounding periods in the year), the more the APY (the result given by the bond yield calculator) increases as well.
Annual Effective Interest Rate: When you’ve entered in the data above into the bond yield calculator and clicked on “Calculate”, this is the result. You’ll see the APY that corresponds to the APR and the frequency at which the interest is compounded.
### What are the limitations of the Yield Calculator?
The bond yield calculator only works in one direction, from APR to APY. What happens if you want to calculate in the opposite direction meaning you want to find out what the APR is by first starting with the APY? You’ll have to make an initial estimate of the APR, see how close it gets you to the APY that you already have and then repeat the process until you home in on the right APR figure.
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https://www.nag.com/numeric/nl/nagdoc_26.2/nagdoc_fl26.2/html/f06/f06ztf.html
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# NAG Library Routine Document
## 1Purpose
f06ztf (zsymm) performs one of the matrix-matrix operations
$C←αAB + βC or C←αBA + βC ,$
where $A$ is a complex symmetric matrix, $B$ and $C$ are $m$ by $n$ complex matrices, and $\alpha$ and $\beta$ are complex scalars.
## 2Specification
Fortran Interface
Subroutine f06ztf ( side, uplo, m, n, a, lda, b, ldb, beta, c, ldc)
Integer, Intent (In) :: m, n, lda, ldb, ldc Complex (Kind=nag_wp), Intent (In) :: alpha, a(lda,*), b(ldb,*), beta Complex (Kind=nag_wp), Intent (Inout) :: c(ldc,*) Character (1), Intent (In) :: side, uplo
#include <nagmk26.h>
void f06ztf_ (const char *side, const char *uplo, const Integer *m, const Integer *n, const Complex *alpha, const Complex a[], const Integer *lda, const Complex b[], const Integer *ldb, const Complex *beta, Complex c[], const Integer *ldc, const Charlen length_side, const Charlen length_uplo)
The routine may be called by its BLAS name zsymm.
None.
None.
## 5Arguments
1: $\mathbf{side}$ – Character(1)Input
On entry: specifies whether $B$ is operated on from the left or the right.
${\mathbf{side}}=\text{'L'}$
$B$ is pre-multiplied from the left.
${\mathbf{side}}=\text{'R'}$
$B$ is post-multiplied from the right.
Constraint: ${\mathbf{side}}=\text{'L'}$ or $\text{'R'}$.
2: $\mathbf{uplo}$ – Character(1)Input
On entry: specifies whether the upper or lower triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'U'}$
The upper triangular part of $A$ is stored.
${\mathbf{uplo}}=\text{'L'}$
The lower triangular part of $A$ is stored.
Constraint: ${\mathbf{uplo}}=\text{'U'}$ or $\text{'L'}$.
3: $\mathbf{m}$ – IntegerInput
On entry: $m$, the number of rows of the matrices $B$ and $C$; the order of $A$ if ${\mathbf{side}}=\text{'L'}$.
Constraint: ${\mathbf{m}}\ge 0$.
4: $\mathbf{n}$ – IntegerInput
On entry: $n$, the number of columns of the matrices $B$ and $C$; the order of $A$ if ${\mathbf{side}}=\text{'R'}$.
Constraint: ${\mathbf{n}}\ge 0$.
5: $\mathbf{alpha}$ – Complex (Kind=nag_wp)Input
On entry: the scalar $\alpha$.
6: $\mathbf{a}\left({\mathbf{lda}},*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the second dimension of the array a must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$ if ${\mathbf{side}}=\text{'L'}$ and at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$ if ${\mathbf{side}}=\text{'R'}$.
On entry: the symmetric matrix $A$; $A$ is $m$ by $m$ if ${\mathbf{side}}=\text{'L'}$, or $n$ by $n$ if ${\mathbf{side}}=\text{'R'}$.
• If ${\mathbf{uplo}}=\text{'U'}$, the upper triangular part of $A$ must be stored and the elements of the array below the diagonal are not referenced.
• If ${\mathbf{uplo}}=\text{'L'}$, the lower triangular part of $A$ must be stored and the elements of the array above the diagonal are not referenced.
7: $\mathbf{lda}$ – IntegerInput
On entry: the first dimension of the array a as declared in the (sub)program from which f06ztf (zsymm) is called.
Constraints:
• if ${\mathbf{side}}=\text{'L'}$, ${\mathbf{lda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$;
• if ${\mathbf{side}}=\text{'R'}$, ${\mathbf{lda}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
8: $\mathbf{b}\left({\mathbf{ldb}},*\right)$ – Complex (Kind=nag_wp) arrayInput
Note: the second dimension of the array b must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: the $m$ by $n$ matrix $B$.
9: $\mathbf{ldb}$ – IntegerInput
On entry: the first dimension of the array b as declared in the (sub)program from which f06ztf (zsymm) is called.
Constraint: ${\mathbf{ldb}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
10: $\mathbf{beta}$ – Complex (Kind=nag_wp)Input
On entry: the scalar $\beta$.
11: $\mathbf{c}\left({\mathbf{ldc}},*\right)$ – Complex (Kind=nag_wp) arrayInput/Output
Note: the second dimension of the array c must be at least $\mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{n}}\right)$.
On entry: the $m$ by $n$ matrix $C$.
If ${\mathbf{beta}}=0$, c need not be set.
On exit: the updated matrix $C$.
12: $\mathbf{ldc}$ – IntegerInput
On entry: the first dimension of the array c as declared in the (sub)program from which f06ztf (zsymm) is called.
Constraint: ${\mathbf{ldc}}\ge \mathrm{max}\phantom{\rule{0.125em}{0ex}}\left(1,{\mathbf{m}}\right)$.
None.
Not applicable.
## 8Parallelism and Performance
f06ztf (zsymm) is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this routine. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
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