url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://rdrr.io/rforge/spatialwarnings/src/R/coarse_graining.R | 1,575,853,419,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540515344.59/warc/CC-MAIN-20191208230118-20191209014118-00354.warc.gz | 512,454,066 | 12,733 | # R/coarse_graining.R In spatialwarnings: Spatial Early Warning Signals of Ecosystem Degradation
#### Documented in coarse_grain
```#
#' @title Matrix coarse-graining
#'
#' @description This function averages the spatial data locally. It divides
#' the input matrix into submatrices of dimension \code{subsize} and
#' averages the spatial data in these submatrices. By doing this, the
#' dimension of resultant matrix is reduced by a factor of
#' \code{subsize}.
#'
#' @details If the data is classified into discrete units, the calculation of
#' variance and skewness can give spurious results irrelevant to the proximity
#' to transition. Therefore, discrete data should be 'coarse-grained' before
#' calculating the spatial early warning signals. However, this can also be
#' applied to continuous state data.
#'
#' @references
#'
#' Sankaran, S., Majumder, S., Kefi, S. and Guttal, V. (2017). Implications
#' of being discrete and spatial for detecting early warning signals of
#' regime shifts. Ecological Indicators.
#'
#' @param mat A matrix
#'
#' @param subsize Dimension of the submatrix. This has to be a positive
#' integer smaller than the dimension of input matrix.
#'
#' @return A matrix of reduced dimension.
#'
#'
#' @examples
#' rmat <- matrix(runif(20*10) > .5,
#' ncol = 20, nrow = 10)
#' rmat.cg <- coarse_grain(rmat, subsize = 2)
#'
#' par(mfrow = c(1, 2))
#' image(rmat)
#' title('Raw matrix')
#' image(rmat.cg)
#' title('Coarse-grained matrix')
#'
#'@export
coarse_grain <- function(mat, subsize) {
if ( subsize < 1 ) {
warning('Cannot coarse-grain a matrix with a subsize argument under 1, ',
'returning the matrix unchanged')
return(mat)
}
coarse_grain_cpp(mat, subsize)
}
```
## Try the spatialwarnings package in your browser
Any scripts or data that you put into this service are public.
spatialwarnings documentation built on May 2, 2019, 5:16 p.m. | 513 | 1,915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-51 | latest | en | 0.647416 |
https://www.physicsforums.com/threads/subspaces-in-r4.113358/ | 1,623,803,162,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621699.22/warc/CC-MAIN-20210616001810-20210616031810-00248.warc.gz | 855,344,817 | 17,139 | # Subspaces in R4
Gold Member
I'm so lost!
1. W is the set of all vectors in R4 such that x1 + x3 = x2 + x4. Is W a subspace of R4 and Why?
How do i get started here? i'm thoroughly confused on this whole idea of vector spaces and such.
## Answers and Replies
matt grime
Homework Helper
What are the axioms you need to check, and what can you verify and how?
HallsofIvy
Homework Helper
I find looking at the definitions helpful in such a case.
Gold Member
So far we're looking at vector addition, scalar multiplication and the zero vector. Plugging in 0,0,0,0 produces a vector that is in the set. I'm not sure how to show vector addition however. Do i need to turn that equation into a function x1= x2 + x4 - x3?
Hurkyl
Staff Emeritus
Gold Member
What you need to do is to treat
"(a, b, c, d) is in W"
as being synonymous with
"a + c = b + d"
Gold Member
Can i set a new vector equal to like (y1 + y3 = y2 + y4) and add the equations?
Hurkyl
Staff Emeritus
Gold Member
No. You need to treat
"(a, b, c, d) is in W"
as being synonymous with
"a + c = b + d"
So far we're looking at vector addition, scalar multiplication and the zero vector.
So what, abstractly speaking, are you trying to show?
matt grime
Homework Helper
Take two elements that satisfy the condition, add them, does the result satisfy the condition?
suppose i wanted to show that the subset V of R^2 defined by (x,y) in V if x=0, then let (x,y) and (s,t) be in V, the sum is (x+s,y+t), now is x+s zero when x and s are zero?
Gold Member
Hurkyl said:
No. You need to treat
"(a, b, c, d) is in W"
as being synonymous with
"a + c = b + d"
Ok so how can I show that they add up using two different vectors? I know for example, (1,3,4,2) are in W but how do i go about showing that all such vectors satisify vector addition and multiplication?
Hurkyl said:
So what, abstractly speaking, are you trying to show?
I guess i'm trying to show that every vector that follows a + c = b + d is a vector in W. Is that right?
Gold Member
matt grime said:
Take two elements that satisfy the condition, add them, does the result satisfy the condition?
suppose i wanted to show that the subset V of R^2 defined by (x,y) in V if x=0, then let (x,y) and (s,t) be in V, the sum is (x+s,y+t), now is x+s zero when x and s are zero?
If I pick v = (1, 3, 4, 2) and u = ( 3, 6, 5, 2), they add up to be (4, 9, 9, 4) which is also in the set. I can also do cv, c being say, -3, to get cv = ( -3, -9, -12, -6) which is also in the set. How can i show that it works for all numbers?
benorin
Homework Helper
Gold Member
Formally, we have W is defined by
$$W=\left\{ (x_1,x_2,x_3,x_4)\in\mathbb{R}^4 : x_1 + x_3 = x_2 + x_4 \right\}$$
The requirement $x_1 + x_3 = x_2 + x_4$ may be written as $x_4 = x_1 - x_2 + x_3$ and notice how once $x_1, x_2, x_3$ are chosen, $x_4$ is determined, and so it would seem we only need three parameters to express points in W...
If you need more, post on that wise.
don't use numbers...use variables
you are given a condition x1+x3=x2+x4, someone above
said look at (a,b,c,d) and you yourself posted can i do x1=...
NOW combine those 2 concepts to get a general vector satisfying your given condition. Note you should ask yourself how many unknowns should remain in your general vector
matt grime
Homework Helper
i am going to state categorically, and i'm sorry to do so, that you should ignore benorin.
it's very easy, so stop making it difficult.
let's do my example, since I refuse to answer your homework,
Define W in R^2 by (x,y) is in W if and only if x=0.
let's show it is a subspace.
1. (0,0) is in W, clearly since the first coordinate is zero
2. if (x,y) and (u,v) are in W the (x+u,y+v) is in W since x+u=0 because x=0 and u=0
3. if (x,y) is in W then so is t(x,y)=(tx,ty) since x=0 implies tx=0
and we are done.
so how about it, now?
You see how to do it?
benorin
Homework Helper
Gold Member
If $$\forall\alpha , \beta\in\mathbb{R},x,y\in W\Rightarrow (\alpha x+\beta y) \in W ,$$ then W is a subspace. That's it. Simple.
Parts (1), (2), and (3) of your test correspond to the cases $$\alpha=\beta=0, \alpha=\beta=1, \mbox{ and }\alpha=t,\beta=0,$$ respectfully.
matt grime
Homework Helper
benorin said:
The requirement $x_1 + x_3 = x_2 + x_4$ may be written as $x_4 = x_1 - x_2 + x_3$ and notice how once $x_1, x_2, x_3$ are chosen, $x_4$ is determined, and so it would seem we only need three parameters to express points in W...
i meant it was this comment that should be ignored. i don't see what choosing and determing things has to do with proving its a subspace
benorin
Homework Helper
Gold Member
I was attempting to lead to the representation of vectors in W as vectors of the form (x1,x2,x3,x1-x2+x3) is all. | 1,497 | 4,751 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-25 | latest | en | 0.966084 |
https://cybersphere.me/numbers/ | 1,709,306,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475311.93/warc/CC-MAIN-20240301125520-20240301155520-00512.warc.gz | 187,857,380 | 29,958 | # Everything About Numbers In A Sheet Worker
In a recent forum post, I described how to make sure your numbers have a fixed number of decimal places. Often when you use money, you have 2 decimals, like \$3.30, and when using weights you might have three deminals like 123.650Kg. Handling such situations can be very simple, but is also sometimes very complicated. I thought that would make a great topic for a post, then realised there was a lot more to say about how JavaScript handles numbers.
I’ve briefly covered numbers before (see a brief introduction to data types, how to do arithmetic, and an introduction to strings), and I’ll revisit some of that here while covering the topic more comprehensively. I’ll go into way too much detail explaining why this is sometimes complicated – this is a massive post!
## The Basics
### Strings, Numbers, and other Data types
A string is basically any bunch of letters and numbers, like “word” or “this is a string”, or even “area 51”.
A number either has decimal places or it doesn’t. We call the latter an integer, but in javascript, both types are numbers and are treated the same. You can add an integer like 7 and a floating point number like 7.12, and JS won’t complain – you’ll get a new floating point number 14.12. If you really want that final number to be an integer (14), you’ll have to do something to that number- JS won’t do it automatically.
There are other data types, like arrays and objects, but we are mainly concerned with numbers and strings here.
### Creating Attributes
When you create attributes in HTML, it always starts out as a string.
```.wp-block-code {
border: 0;
}
.wp-block-code > span {
display: block;
overflow: auto;
}
.shcb-language {
border: 0;
clip: rect(1px, 1px, 1px, 1px);
-webkit-clip-path: inset(50%);
clip-path: inset(50%);
height: 1px;
margin: -1px;
overflow: hidden;
position: absolute;
width: 1px;
word-wrap: normal;
word-break: normal;
}
.hljs {
box-sizing: border-box;
}
.hljs.shcb-code-table {
display: table;
width: 100%;
}
.hljs.shcb-code-table > .shcb-loc {
color: inherit;
display: table-row;
width: 100%;
}
.hljs.shcb-code-table .shcb-loc > span {
display: table-cell;
}
.wp-block-code code.hljs:not(.shcb-wrap-lines) {
white-space: pre;
}
.wp-block-code code.hljs.shcb-wrap-lines {
white-space: pre-wrap;
}
.hljs.shcb-line-numbers {
border-spacing: 0;
counter-reset: line;
}
.hljs.shcb-line-numbers > .shcb-loc {
counter-increment: line;
}
.hljs.shcb-line-numbers .shcb-loc > span {
}
.hljs.shcb-line-numbers .shcb-loc::before {
border-right: 1px solid #ddd;
content: counter(line);
display: table-cell;
text-align: right;
-webkit-user-select: none;
-moz-user-select: none;
-ms-user-select: none;
user-select: none;
white-space: nowrap;
width: 1%;
}
```<input type="number" name="attr_example_number" value="10">
<input type="text" name="attr_example_string" value="18">
```Code language: HTML, XML (xml)```
In this case, example_number has a value of “10” and example_string is “18” (note the quotes). While one is defined as a number, both are strings by default. You can confirm this in a sheet worker:
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = values.example_number; // equals "10"
let s = values.example_string; // equals "18"
console.info({n, s});
});```Code language: JavaScript (javascript)```
Here we use `console.info` to output the named variables to the console, and interrogate their values. These start with the default values for those attributes, but they are both strings despite one being given a type of number. All attributes on Roll20 are strings (except when they aren’t – more on that below).
### Logical Comparisons
Since they are strings, there are some things we can’t do, no matter how much we want to.
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = values.example_number; // equals "10"
let s = values.example_string; // equals "18"
let bool = false;
console.info({n, s, bool});
if(n < s) {
bool = true;
}
console.info({n, s, bool});
});```Code language: JavaScript (javascript)```
Here we want to test if the variable `n` is below `s`. There are many types of logical tests you might want to do. But they are both strings, so this doesnt work. That `bool` value will always be false. 10 is below 18, but “10” is not below “18” (it’s not greater either). These are strings, and logical comparisons don’t work (here, in JavaScript).
For this to work, we need a way to convert those strings into numbers. We’ll come back to that in a surprising way, but first…
### What happens when we try to add strings together?
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = values.example_number; // equals "10"
let s = values.example_string; // equals "18"
console.info({n, s});
let total = n + s;
console.info({total}); // total = "1018", why is that?
});```Code language: JavaScript (javascript)```
You might expect the result to be “28”, but these two variables are strings, so JS “adds” the values as if they were strings (which they are) – thus, it just plonks one on the end of the other, producing “1018”.
So + a tricky operator to use. What about other arithmetic operators, like -, /, and *?
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = values.example_number; // equals "10"
let s = values.example_string; // equals "18"
console.info({n, s});
let total = s - n;
console.info({total}); // total = 8
});```Code language: JavaScript (javascript)```
Javascript always tries to treat strings as numbers, if it can, so the * / and – operators work properly. Notice the total here is not a string (no enclosing quotes: 8 and not “8”).
### Arithmetic in Javascript
You can take advantage of this when initially creating the variables.
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = 1 * values.example_number; // equals 10 (this is a number)
let s = 1 * values.example_string; // equals 18 (not a string)
console.info({n, s});
let total = n + s;
console.info({total}); // total = 28, why is that?
});```Code language: JavaScript (javascript)```
Here, by multiplying the initial objects by 1, JS has converted them to a number. This only works because the value is a number, but it is saved as a string. If the value was something like “word”, which is not a number (1 * word makes no sense), you’ll see an error.
Later in the worker, JS sees you are trying to add two numbers together using +, and since they are actually numbers (not strings containing a number), it adds them properly.
The rule is: when you are adding things together, if they are all numbers, JS adds them properly. If even one of the items is a string, every element is treated as a string, which may produce surprising results.
Here’s a mixed example.
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let n = 1 * values.example_number; // equals 10 (this is a number)
let s = values.example_string; // equals "18" (this is a string)
console.info({n, s});
let total = n + s;
console.info({total}); // total = "1018"
});```Code language: JavaScript (javascript)```
Notice in this example, we converted a string to a number – “10” become 10. This was done by multiplying by 1. If you perform any arithmetic operation, the result is converted to a number. we could have done any of these:
``````on('change:example_number', () => {
getAttrs(['example_number'], values => {
let m = 1 * values.example_number; // equals 10 (this is a number)
let d = (values.example_number) / 1; // this is also a number: 10
let s = values.example_number - 0; // here is another a number: 10
let a = values.example_number + 0; // oh oh, this is not a number: "100"
console.info({m, d, s, a});
});```Code language: JavaScript (javascript)```
In that last one, the + operator does double duty in JavaScript. It can be used as addition if every element is a number, but also is used as the concatenation operator (add one string to the end of another). Here, one of the elements is a string, so concatenation is performed.
As an aside, if you’re creating a HTML macro, this is one of the biggest source of errors. Make sure each number is enclosed in parentheses. But that’s another topic.
## The Big Topic: Coercing Strings Into Numbers
So, numbers might be stored in strings. In Roll20, that’s common. When we really want to treat those attributes as numbers, it’s a good idea to make sure they are actually numbers and not strings. It’s not always necessary – but it is safer (especially if addition or logical tests are involved).
When you convert data from one type to another, it’s called coercing – we want to coerce strings into numbers. Javascript provides many, many ways to do this. We have seen that we can use arithmetic operators, but there are a few functions created specifically for this purpose. Lets look at them,
### Parsing vs Conversion
Here are the four most common functions for this: parseInt, parseFloat, Number, and + (yes, the dreaded + again).
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let pi = parseInt(values.example_number); // equals 10
let pf = parseFloat(values.example_string); // equals 18
let n = Number(values.example_number); // equals 10
let p = +values.example_string; // equals 18
console.info({pi, pf, n, p});
});```Code language: JavaScript (javascript)```
The first two operate in a similar way, the last two act a little differently.
parseInt and parseFloat look at the string, and if it starts with a number, it truncates everything after the number ends. parseInt looks for integers and parseFloat looks for floating point numbers.
So, if your string is, say “14.7 kg”, parseInt will give a result of 14, and parseFloat will give 14.7. Everything after the number required is snipped off. Notice the number isn’t rounded. If you want the number to be rounded, you’ll have to grab it as a floating point, then round it and save the new value.
``````on('change:example_weight', () => {
getAttrs(['example_weight'], values => {
let pf = parseFloat(values.example_weight); // equals 14.7
let rounded = Math.round(pf); // equals 15
console.info({pf, rounded});
});```Code language: JavaScript (javascript)```
parseInt always gives you the number without any decimal places, and parseFloat keeps any decimal places – if there are any. It doesn’t add extra decimal places, so if the number is already an integer that’s what you get.
### Number and +
Number() works just like parseFloat(), except it looks at the whole value, and if there is anything it that can’t be treated as a pure number, it’ll return an error.
So if you try Number(“14.7 kg”) you’ll get an error. With that space and “kg”, it is not a number, and so this fails. A human can see it’s a number, but to a computer it’s just a string of numbers and letters – it is a string. But if you then try Number(“14.7”) you’ll get the expected 14.7.
The + operator is just a shorthand for Number() – it is exactly the same operator, but is very quick to type.
### Errors
So there are two types of function here, and it looks like the first two are more robust. But they can create errors too. If the attribute starts with a number, they’ll extract that number. If there are multiple numbers, only the first is grabbed, and if the string does not start with a number, you’ll get an error.
``````let mult = parseInt("7/11"); // equals 7
let ending = parseInt("Area 51"); // equals NaN - an error
console.info({mult, ending});```Code language: JavaScript (javascript)```
That first one fails because of the slash – it’s read as the letter “/” not a divisor, because it’s part of a string. So there are two numbers 7 and 11, and the function grabs only the first.
The second fails because the string does not start with a number.
Number would fail on both of these two.
### Default Values
If you have an error here, the sheet worker probably fails – especially when you try to do something with those variables later. A common solution is to create default values. That looks like this:
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let pi = parseInt(values.example_number) || 0; // equals 10
let p = +values.example_string || 0; // equals 18
console.info({pi, p});
});```Code language: JavaScript (javascript)```
The || 0 ending says “OR 0”. If the number calculation is an error, the second value, 0, will be used. You might set a different default value, as appropriate. If the variable is going to be a divisor later, setting it to zero is a bad idea- you don’t want to divide by zero. For example:
``````on('change:example_number change:example_string', () => {
getAttrs(['example_number', 'example_string'], values => {
let pi = parseInt(values.example_number) || 1; // equals 10
let p = +values.example_string || 100000; // equals 18
console.info({pi, p});
});```Code language: JavaScript (javascript)```
### A Thing About Default Values
Here’s a thing people often do:
``let pi = parseInt(values.example1) || 0 + parseInt(values.example2) || 0 ;`Code language: JavaScript (javascript)`
There’s a very serious but also very subtle problem here. It’s not related to the method used to coerce values. It happens with parseFloat, Number, +, or any other method used to coerce values.
Let’s say the first parseInt is an error, what happens?
JS sees the first value contains an error, then jumps to the next value, which is:
``let pi = 0 + parseInt(values.example2) || 0 ;`Code language: JavaScript (javascript)`
That is probably (almost certainly) not what you want. It also means if the first parseInt is NOT an error, you get this:
``let pi = parseInt(values.example1);`Code language: JavaScript (javascript)`
The bit after the || symbol only happens if the first one generates an error, so that second parseInt is completely ignored.
Each parseInt should be isolated, like this:
``let pi = (parseInt(values.example1) || 0) + (parseInt(values.example2) || 0);`Code language: JavaScript (javascript)`
In this way, each branch is properly isolated and looked at separately. All those extra brackets can make calculations look complicated, which is why I’m fond of functions like these:
``````const int = (score, fallback = 0) => parseInt(score) || fallback;
const float = (score, fallback = 0) => parseFloat(score) || fallback;```Code language: JavaScript (javascript)```
These one line functions can be put at the start of your script block, and then when you are tempted to write parseInt in a worker, you can instead type:
``````let example1 = int(values.example1);
let example_total = int(values.example2) + int(values.example3);```Code language: JavaScript (javascript)```
It properly isolates each value, and the typing is shorter. Win-win. You can see more examples of this kind of function over here.
### Other Methods
Any method of manipulating numbers can be used to coerce strings into numbers, just as long as the string is actually a number. The methods above (parseInt, parseFloat, Number, +) are designed for this and are good methods to use. But here are some other methods that also work:
• Arithmetic: like x1, -0, /1. But avoid +.
• Rounding: You can use Math.round, Math.ceil, Math.floor, and Math.trunc (truncate) to round strings to numbers. Math.round(“7.7”) will properly give a result of 8, and that result is a number. Math.floor and Math.trunc both round down, but they work diffferent on negative numbers. Math.floor(“-1.1”) gives -2, and Math.trunc(“-1.1”) gives -1. trunc simply truncates to the first decimal point, while floor rounds down – and that works differently if the number is negative.
• Math functions: there are other Math.something functions, and they’ll all turn a string into a number. Basically any attempt to treat a string as a number produces a number.
Remember to include default values, because if the string is not a number these methods generate an error.
### Which To Use?
There’s no real reason to use Number() when + is available. The choice between + and the parse functions is entirely one of personal preferences. Remember that JS doesn’t consider Integers to be special. If the original data is an integer, the value you get will always be an integer, whichever you use.
So there’s an argument to always use +. Personally, I use parseInt whenever the number should be an integer and the player can change the value, and + otherwise. You can’t rely on an input or textarea to contain the data you expect, if players can alter that data. (Some players deliberately try to break things, while others simply make typos. It’s better to be safe.)
Use whichever serves you best. (But always create a default value.)
## Data Is Always Stored As A String (except when it’s not)
We are often told that Roll20 saves all attributes as strings, and this is true. Except when it’s not.
### A Brief Revision of Data Types
JavaScript supports several data types, and the most common we’ll deal with are strings, numbers, arrays, and objects. You can create them in code, like this:
``````let a_string = "a string";
let a_number = 13;
let an_array = [12, "a_string", string];
let an_object = {
a_number: 13,
a_string: a_string
a_defined_string: "this is a string",
};
// you can create empty variables using the delimters, but number is different
let empty_string = "";
let empty_number = 0;
let empty_array = [];
let empty_object = {};
let undefined_variable;```Code language: JavaScript (javascript)```
That last variable created there doesn’t have a date type, but will gain one when you give it data. in JS, data type changes when needed. Any variable can have its data type changed.
JS tries to be smart, and converts data from one type to another as needed. It tries to be smart – it doesn’t always succeed. So it’s a good idea to make sure variables are of the right type when you do something with them.
Variables can be arrays and objects, and even functions (which can be mind minding). But for this essay we are mostly concerned with strings and numbers.
### A Few Things About Data Types
``````let a_string = "a string";
a_string = 13; // no longer a string - now has a value of 13 and is a number.```Code language: JavaScript (javascript)```
If you define variables with const instead of let, they can’t be changed. They are constant.
``````const a_string = "a string";
a_string = 13; // returns an error, because consts cant be changed.```Code language: JavaScript (javascript)```
One thing that might trip you up: when creating empty objects or arrays, the data type can’t be changed, but the contents can.
``````const empty_object = {};
empty_object.string = "testing" // this doesn't return an error
// now the object is {string: "testing"}```Code language: JavaScript (javascript)```
### Attributes as Numbers
When you grab a variable from an attribute it is, by default, a string. But when you modify an attribute in code, it is often coerced into a number. Then when you save it, it might be saved as a number. Here’s an extremely complex sheet worker to demonstrate this. I’ll explain it afterwards.
``````on('change:example_string', () => {
getAttrs(['example_number'], values => {
let example = parseInt(values.example_number) || 0; // a number equalling 10
setAttrs({
example_number: example
}, () => {
getAttrs(['example_number'], values => {
let example2 = values.example_number; // a <strong>number</strong> equalling 10
console.info({example2});
});
});
});
});```Code language: JavaScript (javascript)```
First we have a normal sheet worker that watches the example_string attribute for changes. But the sheet worker doesn’t touch that, to make it easy to avoid infinite loops. (Try to imagine what would happen if we watched example_number for changes, then in the worker, changed it).
Then the worker grabs the value of example_number and coerces it into a number, and with setAttrs saves that modified attribute.
Now the second part of the setAttrs function that starts () => is another function that runs after the first one has completed. It only runs after the setAttrs has completed. You can use this technique to run functions after a change.
Our last worker gets the attribute again after its change, and now we see it is NOT a string – it is a number.
### What This Tells Us
This is a fairly convoluted way of showing that when you save a number, it might remain a number data type. It is NOT a string.
This isn’t really useful, because you can’t be sure if an attribute contains a string or a number, so you have to treat it as if it contains a string. But this means that if you don’t realise the attribute can be of either type and just treat it as a number you can encounter subtle errors.
On the other hand, if you treat is always being a string, you won’t have any errors. Using parseInt or Number() on a value that is already a number won’t cause any problems.
So always treat variables from character sheets as if they were strings. They might not be, but it doesn’t matter – the code will work properly.
## Getting a Fixed Number of Decimals
Now on to the original reason for writing this post, which is almost comically simple.
### Types of Number
In JavaScript, the number data type is always a floating point number. You can force it to be an integer, but in javascript, an integer is just a Number with no decimal places.
While this may seem obvious, other programming languages often treat integers and floating point numbers as fundamentally different, and mean adding one to the other fails unless you coerce them into the same data type. You don’t need to worry about that.
So parseInt and parseFloat both create a number data type, but parseInt removes any decimal places. They are still the same data type.
This means you don’t technically have to use parseInt. If use Number or + on a number that is already an integer it stays an integer. New decimal places aren’t created.
### Creating A Fixed Number of Decimal Places
The problem is, you sometimes want to show a number of decimal places. For example, money is often shown as \$3.30 (dollars as integers, and cents as decimals), or weight in thousandths (like 120.650 Kg).
JavaScript will also truncate those to the minimum number of places needed, like 3.3 when you wanted 3.30, or 120.65 when you wanted 120.650.
So can force a number of decimal places to be shown with the toFixed function, like this:
``````let money = 3.3;
let weight = 120.65
let dollars = money.toFixed(2);
let kg = weight.toFixed(3);```Code language: JavaScript (javascript)```
### Rounding
So, put the number of decimals you want in brackets, and leave empty for 0. By default, the toFixed function rounds to the nearest.
``````let money = 3.3;
let weight = 120.65
let dollars = money.toFixed(); // a number, 3
let kg = weight.toFixed(); // a number, 121```Code language: JavaScript (javascript)```
This will show the proper number of integers even if an integer. So 3 can become 3.00.
### Bare Numbers
You might find it necessary to apply toFixed to a bare number, like this:
``let money = 3.toFixed(2);`Code language: JavaScript (javascript)`
I can’t remember a time I’ve had to do this, but if you are ever tempted to do this, know that it will fail. But if you surround the number with parenthesis it works. Like this:
``let money = (3).toFixed(2);`Code language: JavaScript (javascript)`
There are many situations where JS functions fail, but if you put parenthesis around the entity, it suddenly works. This is a handy way to trick JS into thinking there is something there that it can work on, so it does. I’ll probably come back to this technique in a later article – it is sometimes useful.
## A Decimal Places Function
The previous section described how to convert a number to one with a specific number of decimal places. You can create a function to make this easier.
``````// a function to convert a number to specifical decimals
const places = (score, decimals = 0) => +score.toFixed(decimals);```Code language: JavaScript (javascript)```
Place this function at the start of your script block, then when you want to show convert a number to a specific number of decimals, you can do this:
``````// create a value purely for this example
let value = 43.765432;
// use the places function to convert it into a number with two decimal places.
let money = places(value, 2);
// result = 43.77```Code language: JavaScript (javascript)```
The places function takes 1 or 2 parameters. You can pass it a value to be acted on, and, optionally, a number of decimal places. For example:
``````// create a number with no decimal places (an integer)
let integer = places(value);
// create a number with three decimal places
let weight = places(value, 3);
// create a number with a variable number of decimal places
let decimals = 1;
let final_total = places(value, decimals);```Code language: JavaScript (javascript)```
If you’re having to create a lot of numbers with specific numbers of decimal places, a function like this could be very handy.
## Conclusion
So there we have it. How to convert strings to numbers, why we do it the way we do, and how to create numbers with specific numbers of decimal places. Is there anything you want to know? Are there any questions about this article?
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 6,178 | 25,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | longest | en | 0.85087 |
https://blog.csdn.net/Yasola/article/details/78149321 | 1,544,965,492,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827727.65/warc/CC-MAIN-20181216121406-20181216143406-00559.warc.gz | 559,673,027 | 34,239 | # Codeforces 864F Cities Excursions（离线处理+Tarjan）
F. Cities Excursions
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
There are n cities in Berland. Some pairs of them are connected with m directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (x, y) there is at most one road from x to y.
A path from city s to city t is a sequence of cities p1, p2, … , pk, where p1 = s, pk = t, and there is a road from city pi to city pi + 1 for each i from 1 to k - 1. The path can pass multiple times through each city except t. It can’t pass through t more than once.
A path p from s to t is ideal if it is the lexicographically minimal such path. In other words, p is ideal path from s to t if for any other path q from s to t pi < qi, where i is the minimum integer such that pi ≠ qi.
There is a tourist agency in the country that offers q unusual excursions: the j-th excursion starts at city sj and ends in city tj.
For each pair sj, tj help the agency to study the ideal path from sj to tj. Note that it is possible that there is no ideal path from sj to tj. This is possible due to two reasons:
there is no path from sj to tj;
there are paths from sj to tj, but for every such path p there is another path q from sj to tj, such that pi > qi, where i is the minimum integer for which pi ≠ qi.
The agency would like to know for the ideal path from sj to tj the kj-th city in that path (on the way from sj to tj).
For each triple sj, tj, kj (1 ≤ j ≤ q) find if there is an ideal path from sj to tj and print the kj-th city in that path, if there is any.
Input
The first line contains three integers n, m and q (2 ≤ n ≤ 3000,0 ≤ m ≤ 3000, 1 ≤ q ≤ 4·105) — the number of cities, the number of roads and the number of excursions.
Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), denoting that the i-th road goes from city xi to city yi. All roads are one-directional. There can’t be more than one road in each direction between two cities.
Each of the next q lines contains three integers sj, tj and kj (1 ≤ sj, tj ≤ n, sj ≠ tj, 1 ≤ kj ≤ 3000).
Output
In the j-th line print the city that is the kj-th in the ideal path from sj to tj. If there is no ideal path from sj to tj, or the integer kj is greater than the length of this path, print the string ‘-1’ (without quotes) in the j-th line.
Example
input
7 7 5
1 2
2 3
1 3
3 4
4 5
5 3
4 6
1 4 2
2 6 1
1 7 3
1 3 2
1 3 5
output
2
-1
-1
2
-1
### 题目大意
有一张V(V3000)$V(V \leq 3000)$的有向图，有Q(Q4×105)$Q(Q \leq 4 \times 10^5)$组询问，没次询问从一个点到另一个点字典序最小的路径的第k$k$个点。
### 解题思路
由于只用3000$3000$个点，我们可以考虑对查询离线处理，把所有起点相同的查询放到一组，以这个起点进行dfs，当出现环的时候，后面的点答案都是-1，用Tarjan判定一下有向环即可。总时间复杂度O(V2)$O(V^2)$
### AC代码
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <cstdlib>
#include <string>
#include <map>
#include <bitset>
using namespace std;
#define INF 0x3f3f3f3f
#define LL long long
#define ULL unsigned long long
#define fi first
#define se second
#define mem(a,b) memset((a),(b),sizeof(a))
#define sqr(x) ((x)*(x))
const int MAXV=3000+3;
const int MAXQ=400000+3;
struct Query
{
int u, v, k, id;
bool operator < (const Query &other)const
{
return u<other.u;
}
}query[MAXQ];
int V, E, Q;
vector<int> G[MAXV];
vector<Query> save_q[MAXV];
int ans[MAXQ];
int dfn[MAXV], low[MAXV], tmpdfn, path[MAXV], vis[MAXV];
void dfs(int u, int deep, bool cant)
{
path[deep]=u;
dfn[u]=low[u]=tmpdfn++;
vis[u]=1;
if(!cant)
for(int i=0;i<save_q[u].size();++i)
if(save_q[u][i].k<=deep)
ans[save_q[u][i].id]=path[save_q[u][i].k];
for(int i=0;i<G[u].size();++i)
{
int v=G[u][i];
if(!vis[v])
{
dfs(v, deep+1, cant|(low[u]<dfn[u]));
low[u]=min(low[u], low[v]);
cant|=low[v]<dfn[v];
}
else if(vis[v]==1)
low[u]=min(low[u], dfn[v]);
}
vis[u]=2;
}
int main()
{
scanf("%d%d%d", &V, &E, &Q);
for(int i=0;i<E;++i)
{
int u, v;
scanf("%d%d", &u, &v);
G[u].push_back(v);
}
for(int i=1;i<=V;++i)
sort(G[i].begin(), G[i].end());
for(int i=0;i<Q;++i)
{
scanf("%d%d%d", &query[i].u, &query[i].v, &query[i].k);
--query[i].k;
query[i].id=i;
}
sort(query, query+Q);
mem(ans, -1);
for(int i=0;i<Q;++i)
{
save_q[query[i].v].push_back(query[i]);
if(i==Q-1 || query[i].u!=query[i+1].u)
{
tmpdfn=0;
mem(vis, 0);
dfs(query[i].u, 0, 0);
for(int j=1;j<=V;++j)
save_q[j].clear();
}
}
for(int i=0;i<Q;++i)
printf("%d\n", ans[i]);
return 0;
} | 1,592 | 4,528 | {"found_math": true, "script_math_tex": 5, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-51 | longest | en | 0.919248 |
https://www.in2013dollars.com/1920-CAD-in-1918?amount=1 | 1,590,922,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00137.warc.gz | 762,477,456 | 21,305 | # \$1 in 1918 is worth \$0.84 in 1918
\$
## Value of \$1 from 1920 to 1918
According to Statistics Canada consumer price index, prices in 1918 are 16.07% lower than average prices since 1920. The Canadian dollar experienced an average inflation rate of 9.16% per year during this period, causing the real value of a dollar to decrease.
In other words, \$1 in 1920 is equivalent in purchasing power to about \$0.84 in 1918, a difference of \$-0.16 over 2 years.
The 1918 inflation rate was 13.25%. The inflation rate in 1920 was 9.80%. The 1920 inflation rate is higher compared to the average inflation rate of 2.53% per year between 1920 and 2020.
Cumulative price change -16.07% Average inflation rate 9.16% Converted amount (\$1 base) \$0.84 Price difference (\$1 base) \$-0.16 CPI in 1920 11.200 CPI in 1918 9.400 Inflation in 1918 13.25% Inflation in 1920 9.80%
## Buying power of \$1 in 1918
This chart shows a calculation of buying power equivalence for \$1 in 1918 (price index tracking began in 1914).
For example, if you started with \$1, you would need to end with \$0.84 in order to "adjust" for inflation (sometimes refered to as "beating inflation").
When \$1 is equivalent to \$0.84 over time, that means that the "real value" of a single Canadian dollar decreases over time. In other words, a dollar will pay for fewer items at the store.
This effect explains how inflation erodes the value of a dollar over time. By calculating the value in 1918 dollars, the chart below shows how \$1 is worth less over 2 years.
According to Statistics Canada, each of these CAD amounts below is equal in terms of what it could buy at the time:
Dollar inflation: 1918-1920
Year Dollar Value Inflation Rate
1918 \$1.00 13.25%
1919 \$1.09 8.51%
1920 \$1.19 9.80%
1921 \$1.00 -16.07%
1922 \$0.98 -2.13%
1923 \$0.98 0.00%
1924 \$0.97 -1.09%
1925 \$0.99 2.20%
1926 \$0.98 -1.08%
1927 \$0.97 -1.09%
1928 \$0.98 1.10%
1929 \$1.00 2.17%
1930 \$0.94 -6.38%
1931 \$0.84 -10.23%
1932 \$0.78 -7.59%
1933 \$0.76 -2.74%
1934 \$0.77 1.41%
1935 \$0.79 2.78%
1936 \$0.80 1.35%
1937 \$0.83 4.00%
1938 \$0.81 -2.56%
1939 \$0.83 2.63%
1940 \$0.87 5.13%
1941 \$0.93 6.10%
1942 \$0.96 3.45%
1943 \$0.98 2.22%
1944 \$0.96 -2.17%
1945 \$0.98 2.22%
1946 \$1.03 5.43%
1947 \$1.18 14.43%
1948 \$1.29 9.01%
1949 \$1.30 0.83%
1950 \$1.37 5.74%
1951 \$1.52 10.85%
1952 \$1.50 -1.40%
1953 \$1.50 0.00%
1954 \$1.50 0.00%
1955 \$1.51 0.71%
1956 \$1.55 2.82%
1957 \$1.59 2.05%
1958 \$1.63 2.68%
1959 \$1.65 1.31%
1960 \$1.67 1.29%
1961 \$1.67 0.00%
1962 \$1.70 1.91%
1963 \$1.73 1.88%
1964 \$1.77 1.84%
1965 \$1.82 3.01%
1966 \$1.88 3.51%
1967 \$1.96 3.95%
1968 \$2.04 4.35%
1969 \$2.14 4.69%
1970 \$2.16 1.00%
1971 \$2.27 4.93%
1972 \$2.38 5.16%
1973 \$2.61 9.38%
1974 \$2.94 12.65%
1975 \$3.21 9.42%
1976 \$3.39 5.63%
1977 \$3.71 9.40%
1978 \$4.03 8.60%
1979 \$4.43 9.76%
1980 \$4.91 11.06%
1981 \$5.51 12.12%
1982 \$6.02 9.27%
1983 \$6.30 4.59%
1984 \$6.53 3.72%
1985 \$6.82 4.40%
1986 \$7.11 4.21%
1987 \$7.40 4.19%
1988 \$7.69 3.88%
1989 \$8.10 5.26%
1990 \$8.50 4.99%
1991 \$8.82 3.75%
1992 \$9.01 2.17%
1993 \$9.16 1.65%
1994 \$9.18 0.23%
1995 \$9.34 1.74%
1996 \$9.54 2.16%
1997 \$9.62 0.78%
1998 \$9.71 1.00%
1999 \$9.97 2.63%
2000 \$10.29 3.20%
2001 \$10.36 0.72%
2002 \$10.76 3.80%
2003 \$10.98 2.08%
2004 \$11.21 2.13%
2005 \$11.45 2.09%
2006 \$11.64 1.67%
2007 \$11.91 2.38%
2008 \$12.05 1.16%
2009 \$12.21 1.32%
2010 \$12.50 2.35%
2011 \$12.79 2.30%
2012 \$12.89 0.83%
2013 \$13.05 1.24%
2014 \$13.24 1.47%
2015 \$13.46 1.61%
2016 \$13.66 1.50%
2017 \$13.87 1.56%
2018 \$13.91 0.31%
2019 \$14.19 1.99%
2020 \$14.51 2.25%*
* Compared to previous annual rate. Not final. See inflation summary for latest 12-month trailing value.
This conversion table shows various other 1918 amounts in 1920 dollars, based on the -16.07% change in prices:
Conversion Table: Value of a dollar in 1920
Initial value Equivalent value
\$1 dollar in 1918 \$1.19 dollars in 1920
\$5 dollars in 1918 \$5.96 dollars in 1920
\$10 dollars in 1918 \$11.91 dollars in 1920
\$50 dollars in 1918 \$59.57 dollars in 1920
\$100 dollars in 1918 \$119.15 dollars in 1920
\$500 dollars in 1918 \$595.74 dollars in 1920
\$1,000 dollars in 1918 \$1,191.49 dollars in 1920
\$5,000 dollars in 1918 \$5,957.45 dollars in 1920
\$10,000 dollars in 1918 \$11,914.89 dollars in 1920
\$50,000 dollars in 1918 \$59,574.47 dollars in 1920
\$100,000 dollars in 1918 \$119,148.94 dollars in 1920
\$500,000 dollars in 1918 \$595,744.68 dollars in 1920
\$1,000,000 dollars in 1918 \$1,191,489.36 dollars in 1920
## How to Calculate Inflation Rate for \$1, 1918 to 1920
Our calculations use the following inflation rate formula to calculate the change in value between 1918 and 1920:
CPI in 1918 CPI in 1920
×
=
Then plug in historical CPI values. The Canadian CPI was 11.2 in the year 1920 and 9.4 in 1918:
9.411.2
×
\$1
=
\$0.84
\$1 in 1920 has the same "purchasing power" or "buying power" as \$0.84 in 1918.
To get the total inflation rate for the 2 years between 1918 and 1920, we use the following formula:
CPI in 1918 - CPI in 1920CPI in 1920
×
100
=
Cumulative inflation rate (2 years)
Plugging in the values to this equation, we get:
9.4 - 11.211.2
×
100
=
-16%
Politics and news often influence economic performance. Here's what was happening at the time:
• The League of Nations' first assembly takes place in Paris.
• The British Parliament passes the Fourth Home Rule Act, dividing Ireland into two parts and leading to a yearlong Anglo-Irish war in the southern counties.
• The Catholic Church canonizes Joan of Arc, formally making her a saint.
• The 19th Amendment to the constitution is ratified, granting U.S. woman suffrage for the first time.
## Data Source & Citation
Raw data for these calculations comes from the government of Canada's annual Consumer Price Index (CPI), established in 1914 and computed by Statistics Canada (StatCan).
You may use the following MLA citation for this page: “\$1 in 1920 → 1918 | Canada Inflation Calculator.” Official Inflation Data, Alioth Finance, 31 May. 2020, https://www.officialdata.org/1920-CAD-in-1918?amount=1. | 2,356 | 6,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-24 | latest | en | 0.89112 |
https://teacherblog.co/?p=4008 | 1,597,203,857,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738864.9/warc/CC-MAIN-20200812024530-20200812054530-00575.warc.gz | 511,200,199 | 11,481 | Great Math Products!
Base Ten Number Line
Multiplication Tricks
Doubles
Telling Time Misconceptions
Equivalent Fractions
Simplifying Fractions
Clock Fractions
Math Fact Motivation
Math Night 2012
Bulletin Board Ideas
Classroom Management
Lines and Angles
Freebies
# Try These Fun Christmas Stations!
I wanted my students to enjoy their last class with me before the Christmas break, so I planned several stations for them to work on. The students really enjoyed the engagement of every station.
Station 1: Students picked out a pine cone and were instructed to find seeds inside the pinecone. I showed them a picture of what pine cone seeds looked like. I gave them magnifying glasses so they could look closely and deeply into the pine cones. I instructed them to draw a picture of their pine cone and to draw a picture of the seed. In preparation for the next station, I instructed them to find two more pine cones that were really “pretty” and “tree shaped”. I really wanted them to find two that were opened up really well so they could see the difference in station 2.
Station 2: I had students place two nice flared out pinecones in water. Students made predictions before they placed one pinecone in cold water and one pinecone in warm water. I have no sink in my room, so I had a gallon jug set up for the cold water and a crock pot set up with warm water for them to dip from with a measuring cup. Make sure to have paper towels on hand for any water that is spilled. Now, students can watch the pinecones for a while, but it will take about 30-45 minutes before students will see any changes. I make students look and notice what happened. I don’t tell them. Now the last group that does this station probably won’t notice what happens. They won’t have enough time in an hour long class period to notice the pinecones change.
(So what does happen? Pinecones close up when wet. Why does this happen? Pinecones don’t want to release their seeds when weather is wet because they want to allow their seeds to fly as far away as possible during drier weather so that they won’t compete with the parent tree.)
After students have noticed the pinecones closing up, make sure you leave time at the end of the stations to ask the students why they think the pine cones are closing up in the water.
In case you are wondering, yes the pinecones will open back up when dried out.
Station 3: What happens when you place a peppermint under water? After students have watched, then what happens when you place M & M’s under water? This experiment wasn’t that impressive in my opinion, but the kids loved it! The experiment works best in a white bowl. The plates were way too messy when kids tried to throw out the water. Water, of course, spilled everywhere.
Students predicted first what happened with the peppermint. Most guessed it would turn white, but few guessed it would streak the way it did. You must keep the dish still and not shake the desk to get the full effect of this.
After students watch the peppermint for about five minutes, have them predict what would happen when they place out M & M’s around the bowl. I bought the small ones so there would be more color. If students place them around the edge of the bowl, it makes a dramatic rainbow like this. Again keep the bowl as still as possible.
Station 4: Light up a Christmas light with alligator wires and batteries. This station by far had the biggest WOW factor for the kids. I didn’t give kids many instructions with this station. I wanted them to do experimentation to figure this out. I made the kids predict which type of batteries would work to light up the battery. They lit up a small LED bulb. (this wasn’t actually a Christmas light bulb). To get these bulbs I bought a \$1 LED flashlight at War-Mart and banged the flashlight with a hammer until I broke it enough to get the lightbulbs out. It’s cheaper to buy bulbs that way :0. To light up the bulb students need at least three batteries placed end to end with at least two alligator clip wires–one each touched to the positive and negative ends of the battery. I used D batteries because I had a bunch in a science kit, but this will work with smaller batteries. I hovered around this station because I knew students might struggle here. I made it a point to talk about how scientists struggle until they figure things out. We talk about how Edison didn’t give up the first time his improvements of the light bulb didn’t work and they shouldn’t either. Now in reality, it would have been a lot cooler if I actually did have Christmas light bulbs, but I already had LED bulbs on hand so we used them.
ChristmasScience Here is an editable document I used when students visited science stations. This is nothing fancy, but you are welcome to use it for your classroom and make it your own. | 1,135 | 4,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-34 | latest | en | 0.963101 |
http://magoosh.com/gmat/2016/gmat-tuesday-og-reading-comprehension-1/ | 1,477,384,059,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720000.45/warc/CC-MAIN-20161020183840-00063-ip-10-171-6-4.ec2.internal.warc.gz | 161,915,992 | 18,874 | offers hundreds of practice questions and video explanations. Go there now.
# GMAT Tuesday: OG Reading Comprehension #1
Welcome! We are looking at the first question in the reading comprehension section of the Official Guide to the GMAT. This is the second video in a series of 5. In this video, I cover how to approach detail questions, the common wrong answers in detail questions, and of course, how to arrive at the right answer.
## Video Transcript
Hello, welcome to GMAT Tuesdays. My name is Kevin and I’m here from Magoosh. I’m sure you guys all know about Pokemon Go, all the kids are playing it. I was like, man, I can play Pokemon Go. Look at this Mewtwo Pokemon I caught. They’re super rare, isn’t that awesome?
Let’s dive into reading comprehension. In the official guide to the GMAT, we’re looking at question number one, the very first one. We’re in this book. There are newer official guides, but you’ll find this question in there as well. The previous video we did was about the passage. So now, we’re diving into question number one.
### Passage Review: GMAT Official Guide Reading Comprehension Question #1
So, one thing to review before we dive into the question is what we discussed in the previous video. We went over the main idea, which is there’s two theories about fish schooling. We talked about the structure, first paragraph’s the intro, second paragraph talks about idea or theory number A, not number A, idea A.
And then paragraph number three is the critics of theory A and introduces theory B and then the fourth paragraph is all about theory B, the confusion effect. And then we talk about the tone, it is balanced-ish. In that the author is, basically, just presenting two theories of schooling.
### Official Guide Reading Comp Question #1
So, question number one. I’m gonna just read straight from the passage, cause that’s easy. The question says, according to the passage, theory B states that which of the following is a factor that enables a schooling fish to escape predators? Great. So, this question is a detail question. I know that, because of the phrase “according to the passage.”
And when I know I’m dealing with a detail type question, I know there are certain things I need to do. This is my plan. The first one, attack the passage, which we all ready did. Second one, we’re gonna rephrase the question, which we can do right now. The question is basically just asking, what does theory B say about schooling and escaping predators?
Then I’m gonna do some research in the passage, I’m going to anticipate what an answer might sound like. So, sort of predict what the answer might be. And then finally, step five, go into the answer choices. Eliminate wrong answers until I arrive at the right answer. One other thing to remember is that they like to have common wrong answers with specific question types.
So with detail questions, we have three typical wrong answers. One, an inference, if an answer choice is making an inference, it’s wrong, cause we’re looking for a detail in the passage, which means it’ll be written in the passage somewhere.
[Two] A distortion of the passage where they take something in the passage and twist it, make it wrong just a little bit.
And then finally [three], when they introduce new information. Sometimes it’s related to what the topic is, but it isn’t in the passage at all. So, keep an eye out for the new information. Let’s dive into the answer choices.
So answer choice A mentions compact groups, as a reason and this is a distortion of the passage. Compact groups that’s just what schooling is, but it doesn’t, that’s not what theory B is talking about. Theory B is explaining why compact groups would help the fish to avoid a predator. So this is a good example of a distortion, we can eliminate it.
Answer choice B is awesome. It is the right answer, cause they mentioned movement of the fish and I know I’m looking for theory B. So, I’m looking in the fourth paragraph for an answer. And if you look at lines 35 to 40, that’s where you find the description about the movement of the fish allows them to avoid being chomped on by predators.
Answer choice C, inability to detect. This is a distortion of the passage, because this is what theory A says. So, this is a perfect example of how they’re distorting the passage. We have a question that’s asking about the second theory and they provide an answer choice that is about the first theory. So they wanna make sure you’re paying attention and reading closely and know what they’re actually asking about, cause you would find them in the passage, but it’s not related to theory B, it’s related to theory A.
And answer choice D says that the fish can hide behind each other, which is an interesting concept, but not mentioned in the passage. We have no details and no information about fish hiding behind each other.
And then finally, answer choice E, that the fish can move with great speed. Maybe that might be a reason, but it’s not brought up in the passage. There’s nothing in the fourth or third paragraph where we learn about theory A or theory B, excuse me, where the fish move really fast and that helps them do avoid the predator. So, this is new information not included in the passage. So, easy to eliminate. Alright, all you GMAT studiers.
I hope this was helpful. If you have any questions, leave comments down below. And if you need any more help, head over to Magoosh where there’s a lot of friendly and fun people who can help you with their GMAT studies and help you capture Pokemon.
This is how you play Pokemon Go right? You just capture, capture.
By the way, sign up for our 1 Week Free Trial to try out Magoosh GMAT Prep! | 1,252 | 5,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2016-44 | latest | en | 0.942005 |
https://stats.stackexchange.com/questions/168118/transferring-percentage-scores-to-raw-scores | 1,579,376,978,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00293.warc.gz | 676,377,603 | 30,638 | # Transferring percentage scores to raw scores
I have a question ranging from 1 to 5. I have 10 subjects, 6 persons select 4 and 4 persons select 5, So the average question is 4.4. If I calculate percentage for response modes ( 1 to 5), I get the following result
1 2 3 4 5
0 0 0 60 40
How I can get the values of 4.4 based on these percentages. Is that possible? Thank you for your help
• I'm not sure what you're asking for..Are you asking "how do I compute an average for values in a table? Is this an exercise for some class? – Glen_b -Reinstate Monica Aug 21 '15 at 1:08
• @Glen- Based on the above table, how I can get 4.4 ( the average question). – User20100 Aug 21 '15 at 18:15
• How does that calculation relate to your title? – Glen_b -Reinstate Monica Aug 22 '15 at 7:41
For tabulated values, where you have outcomes $x_i$ with proportions $p_i$:
x 0 1 2 3 4 5
then the sample mean is $\sum_i p_i x_i=0+0+0+0+2.4+2.0=4.4$ | 324 | 961 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.88224 |
http://www.markedbyteachers.com/gcse/science/osmosis-is-the-passage-of-water-molecules-from-a-weaker-solution-into-a-stronger-solution-through-a-partially-permeable-membrane.html | 1,529,887,565,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867304.92/warc/CC-MAIN-20180624234721-20180625014721-00380.warc.gz | 441,178,028 | 20,093 | • Join over 1.2 million students every month
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# Osmosis is the passage of water molecules from a weaker solution into a stronger solution through a partially permeable membrane.
Extracts from this document...
Introduction
Introduction Osmosis is the passage of water molecules from a weaker solution into a stronger solution through a partially permeable membrane. In this case, Osmosis applies to our practical because we are trying to see if a sugared solution (sucrose) has an effect on potatoes. I aim to investigate the effect of varying concentration of a certain sugar solution on the amount of osmotic activity between the solution and a potato chip of a given size. I will also endeavour to find the exact (or as near as possible) concentration of sucrose solution when no osmosis occurs and the concentrations are equal. Preliminary Work In a sense, preliminary work is as important as the actual experiment. It allows you to make hypotheses before performing actual tests. For my preliminary work, I used a 50 mm piece of potato. It was easy to cut 50 mm (using a Vernier calliper) and to weigh them (using electronic scales to two decimal places). We used 10 mls of sugar solution/distilled water- this was the amount needed to cover one piece of potato. This was put in a test tube along with a piece of the potato and these were then put in a test tube rack. The first thing we needed to observe was how different concentrations affected the weight and length of the potato. We therefore needed to observe each end of the spectrum. We tried 0.0 molar concentration (the lowest) and 0.8 molar (the highest.) We then left each test until the next lesson, and then took the measurement of the potato chips once more. This is my results table: Concentration Initial length Final Length Percentage difference Initial mass Final mass Percentage difference 0 molar (water) 5 cm 5.15 cm 3 % 1.69 g 1.76 g 4.14 % 0 molar (water) 5 cm 5.2 cm 4 % 1.63 g 1.70 g 4.29 % 0 molar (water) ...read more.
Middle
If done with potato cells the cells would increase in length volume and mass because of the extra water. If these potato cells were placed in a solution with a low water concentration, then the opposite would happen. Water would move out of the cell into the solution. In extreme cases the cell membrane breaks away from the cell wall and the cell is referred to as plasmolysed. The potato cells will have decreased in length, volume and mass. The greater the concentration of water in the external solution the greater the amount of water that enters the cell by osmosis. The smaller the concentration of water in the external solution the greater the amount of water that leaves the cell. However, there will be a point where the concentrations of water inside and outside the potato cells are equal (isotonic). At this point there will be no change in the length, volume and mass of the potato, as the net movement of water will be zero, no osmosis has occurred. I predict that the higher the sugar concentration, the more water will transfer from the potato into the sucrose solution making the potato decrease in mass and length, and vice versa for a lower sugar concentrations. I believe this will happen because the solution is diluted with water; the water molecules will transfer over to the stronger solution: whether this is the solution or the potato. Thus occurring osmosis. I also predict that if a sample floats in the solution then it will gain weight and osmosise and if it sinks then the potato will ex-osmosise. Results Table Concentration Initial mass/ g Final mass/ g Percentage change in mass/ % Average percentage change in mass/ % 0.0 1.04 1.20 +15.38 0.0 1.03 1.17 +13.60 0.0 1.01 1.18 +16.83 +15.27 0.2 1.03 1.08 +4.85 0.2 1.03 1.04 +0.97 0.2 1.01 1.03 +1.98 +2.6 0.3 1.00 0.98 -2.00 0.3 1.01 0.98 -2.97 0.3 1.04 1.00 -3.85 -2.94 0.4 1.03 0.92 -10.68 0.4 1.06 1.01 -4.75 ...read more.
Conclusion
There are no averages that are too far out indicating that the results are fairly accurate: there is not too much variability. The results in my results table are fairly similar within the concentrations with the maximum difference only being 5 percent between runs in each concentration. I would have preferred if it was less but there was not time to repeat the experiment over any more times. I am confident about my values, trend and therefore my conclusions to the extent that I believe the negative correlation trend to be accurate. I am not completely positive that a straight line is truly the line of best fit and I am also not sure as to whether my line is accurate as it is difficult to plot a straight line of best fit. If my results were slightly different I may have decided to draw a curve but then this would not fit in with the logic that mass percentage loss and gain are directly proportional to the concentration of sugar solution. I also had to round off my average percentage change in mass to make it fit onto graph paper and also be easier to draw: two decimal places were rounded off to one which has made my graph and therefore possibly my conclusions less accurate. To continue the experiment further I would like to have been able to do more concentrations and to actually test out the conclusion that there would be 0 percentage change with a solution of 0.24 moles. It would also improve accuracy to have repeated the experiment with perhaps different lengths and diameters of potato chips: as we are measuring percentage change then the size would not matter if the whole experiment was re-run with the same size and then the results compared to the experiment which has been done. Secondary sources include: Biology lessons Internet (various biology and coursework sites) Encyclopaedias Textbooks (various but mainly The Living World by Michael Roberts) ...read more.
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# Related GCSE Life Processes & Cells essays
1. ## Investigating the effect of Sucrose Concentration on the Rate of Osmosis in Potato Chips.
5 star(s)
change in mass for varied concentrations of sucrose concentration in the 1st and 2nd sets of experiments, they show a trend. The results show that when the sucrose concentration is low and the water concentration is higher outside the potato cells, the mass of the potato chip, after an hour increases.
2. ## Aim: To find out the effect of altering the concentration of sugar solution on ...
making use of the measuring cylinder so the sugar solution is accurately measured out and poured into the test tube for each experiment. * The concentration of the potato chips will need to be kept constant - this will be done by using similar types of potato chips.
1. ## Osmosis is defined as 'the movement of water molecules from an area of high ...
Evaluation The accuracy of the investigation was adequate, however the concentrations were measured using a measuring cylinder and could be made more accurate using titration. Also when the potato was dried to remove surface liquid it was not necessarily done the same on each potato, a more accurate and uniform way of drying would improve the accuracy further.
2. ## Use scientific knowledge and practical skills to determine the effect of a range of ...
The potato cells will have decreased in length, volume and mass. The greater the concentration of water in the external solution the greater the amount of water that enters the cell by osmosis. The smaller the concentration of water in the external solution the greater the amount of water that leaves the cell.
1. ## Osmosis is the diffusion of water through a semi-permeable membrane, moving from a dilute ...
A plant cell in this state is known as being FLACCID. Prediction By using my background information on the subject, I will be making an educated prediction of the final outcome of the experiment that I will later be carrying out.
2. ## Osmosis, Aim: Investigate the movement of osmosis through a selectively permeable membrane, in ...
osmosis lose some of its water, and the water will diffuse into the solution of glucose, causing the potato chip to lose water, thus decreasing in weight and length. However, if a potato chip is placed into a solution of 0 molar concentration, it should gain weight, width and length.
1. ## Osmosis, Aim: To investigate the movement of osmosis through a selectively permeable membrane, ...
This will give me a very varied set of results from which I hope to make a decent conclusion. If any of the non-variables below are not kept constant it would mean it would not be a fair test: - I will be carrying out this experiment at room temperature.
2. ## Osmosis, Aim: To investigate the effect of varying concentration of a certain sugar solution ...
1st attempt 1.76 1.6 -9.090909091 2nd attempt 1.71 1.56 -8.771929825 1.0m: 1st attempt 1.74 1.59 -8.620689655 2nd attempt 1.4 1.25 -10.71428571 Percentage change in accordance to the varying solutions Concentration Average % change 0.00m 4.94128412 0.25m -2.11022273 0.5m -5.16089356 0.75m -8.93141946 1.0m -9.66748768 These preliminary results give me an overall
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• Ideas and feedback to | 2,341 | 9,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-26 | latest | en | 0.931606 |
https://everipedia.org/wiki/lang_en/Darboux%2527s_theorem | 1,719,209,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865074.62/warc/CC-MAIN-20240624052615-20240624082615-00233.warc.gz | 211,198,711 | 14,246 | Everipedia is now IQ.wiki - Join the IQ Brainlist and our Discord for early access to editing on the new platform and to participate in the beta testing.
# Darboux's theorem
Darboux's theorem is a theorem in the mathematical field of differential geometry and more specifically differential forms, partially generalizing the Frobenius integration theorem. It is a foundational result in several fields, the chief among them being symplectic geometry. The theorem is named after Jean Gaston Darboux[1] who established it as the solution of the Pfaff problem.[2]
One of the many consequences of the theorem is that any two symplectic manifolds of the same dimension are locally symplectomorphic to one another. That is, every 2n-dimensional symplectic manifold can be made to look locally like the linear symplectic space Cn with its canonical symplectic form. There is also an analogous consequence of the theorem as applied to contact geometry.
## Statement and first consequences
The precise statement is as follows.[3] Suppose thatis a differential 1-form on an n dimensional manifold, such thathas constantrankp. If
everywhere,
then there is a local system of coordinatesin which
.
If, on the other hand,
everywhere,
then there is a local system of coordinates 'in which
.
Note that ifeverywhere andthenis acontact form.
In particular, suppose thatis a symplectic 2-form on an n=2m dimensional manifold M. In a neighborhood of each point p of M, by thePoincaré lemma, there is a 1-formwith. Moreover,satisfies the first set of hypotheses in Darboux's theorem, and so locally there is acoordinate chartU near p in which
.
Taking an exterior derivative now shows
The chart U is said to be a Darboux chart around p.[4] The manifold M can be covered by such charts.
To state this differently, identifywithby letting. Ifis a Darboux chart, thenis thepullbackof the standard symplectic formon:
## Comparison with Riemannian geometry
This result implies that there are no local invariants in symplectic geometry: a Darboux basis can always be taken, valid near any given point. This is in marked contrast to the situation in Riemannian geometry where the curvature is a local invariant, an obstruction to the metric being locally a sum of squares of coordinate differentials.
The difference is that Darboux's theorem states that ω can be made to take the standard form in an entire neighborhood around p. In Riemannian geometry, the metric can always be made to take the standard form at any given point, but not always in a neighborhood around that point.
• Carathéodory-Jacobi-Lie theorem, a generalization of this theorem.
• Symplectic basis
## References
[1]
Sep 20, 2019, 6:31 PM
[2]
Sep 20, 2019, 6:31 PM
[3]
Sep 20, 2019, 6:31 PM
[4]
Citation Linkopenlibrary.orgCf. with McDuff and Salamon (1998) p. 96.
Sep 20, 2019, 6:31 PM
[5]
Citation Linkgallica.bnf.fr"Sur le problème de Pfaff"
Sep 20, 2019, 6:31 PM
[6]
Sep 20, 2019, 6:31 PM
[7]
Citation Linkneo-classical-physics.infoG. Darboux, "On the Pfaff Problem," transl. by D. H. Delphenich
Sep 20, 2019, 6:31 PM
[8]
Citation Linkneo-classical-physics.infoG. Darboux, "On the Pfaff Problem (cont.)," transl. by D. H. Delphenich
Sep 20, 2019, 6:31 PM
[9]
Citation Linkgallica.bnf.fr"Sur le problème de Pfaff"
Sep 20, 2019, 6:31 PM
[10] | 887 | 3,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.932116 |
https://biology.stackexchange.com/questions/80390/how-can-i-approach-this-data-that-has-a-different-total-number-of-results-betwee | 1,561,405,447,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00007.warc.gz | 379,986,571 | 35,960 | # How can I approach this data that has a different total number of results between samples?
I was not sure whether I should post this in Biology Stack Exchange or Math Stack Exchange so I posted it in both.
I was planning on doing a small research for my math class (a final project for the year) on population dynamics. I am going to use yeast cells and I will approximate the population of cells per medium (there will be several containers with yeast cells) using a colorimeter and a microscope. I will then graph the results to find a logarithmic regression for the graph that best represents the data, then I'd find the growth rate of the population of yeast cells. Up to here I have no issues.
I also wanted to calculate a distribution curve for the size of the yeast cells at three different points of the growing phase, t1, t2 & t3 (I would take a sample of the mediums with yeast cells, and using a microscope and a grid I would approximate the size of the yeast cells and separate them by size). The problem is that each sample will most likely have a different total number of cells (also varying depending on the medium). This would mean that the samples I take at, for example, t1 , will have a different total number of cells in them, thus making it impossible for me to average the number of cells with a size between two intervals. I do not want to average the total number of cells mainly because a sample might have 50 cells in it meanwhile another might have 200 cells in it.
Additionally, I wanted to make a binomial distribution too, by classifying the cells as either dead or alive (at t1, t2 & t3). I would calculate the probability of a cell being dead or alive by calculating the ratio between alive cells and total number of cells (I am not sure if this is also correct). Nonetheless, I face the same problem here: there will be different values regarding the total amount of cells per sample.
My question is: should I average the total number of cells per sample? Or is there some other way I can approach this problem?
• I don't understand your paragraph 2. Are you trying to compare the mean size of cells at different times? Then, why can't you just compute the mean (whether the sample are of different size or not). Are you thinking about making some stat test? What test do you have in mind? – Remi.b Jan 9 at 3:18
• Similarly to the second paragraph, I don't understand what is the problem you are phasing. Maybe because I misunderstand what you are trying to do. Having a different sample size will affect your standard deviation for each sample, but you can still compute unbiased mean size and unbiased fraction of dead cells. A priori, the fraction of interest will be number of dead cells over total number of cells (unlike what you suggest), – Remi.b Jan 9 at 3:20
• In all case, this question is not a good fit for Biology.SE (nor is it for math.SE). If I understand you correctly, your problem is a problem of data analysis and the question should rather be asked on stats.SE. I am voting to close as off-topic. – Remi.b Jan 9 at 3:22
• The OP is right that it may not be valid to directly compare correlated samples of unequal size. One way to approach the problem is to use mixed effects models. For the size measurements you could use a linear mixed model, whereas for the probability of survival a generalised linear mixed model with the binomial distribution (logistic regression model). Ideally, you could combine both models into one multivariate model that takes everything into account. – vkehayas Jan 9 at 20:34
• The correct stack for this question is neither Biology nor Math, but rather stats.stackexchange.com. – vkehayas Jan 9 at 20:35 | 812 | 3,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-26 | longest | en | 0.957561 |
http://www.physicstutorials.org/pt/43-Conservation_Of_Momentum | 1,544,415,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823303.28/warc/CC-MAIN-20181210034333-20181210055833-00484.warc.gz | 468,445,438 | 3,892 | ## Conservation Of Momentum
CONSERVATION OF MOMENTUM
As I said before to give something acceleration we must apply an external force. If there is no force then object continue its motion. For momentum change we must apply impulse, in other words there must be external applied force to change momentum of the object. If there is no force applied then momentum of the system is conserved in magnitude and direction.
Pinitial=Pfinal
To understand conservation of momentum we will examine a collision of two objects. Look at the given picture, two ball having masses m1 and m2 and velocities V1 and V2 collide. If there is no external force acting on the system; momentum of the system is conserved.
During the collision balls exert force to each other. From the Newton’s third law these forces are equal in magnitude and opposite in direction. We can say action and reaction to these forces. Picture given below shows these forces at the instance of collision.
Ball having m2 mass exerts force on m1 and ball having m2 mass exerts force on m1. Since collision occurs from the interaction of the bodies then time of collision or interaction must be equal.
F1=-F2 and t1=t2
Impulses of the balls are;
Impulse1=F1. t1 and Impulse2= -F2. t2
F1. t1 = -F2. t2
Impulses of the balls are equal in magnitude. As we said before impulse is equal to the change of momentum. Thus, we can say that momentum changes of the balls are also equal in magnitude and opposite in direction.
m1 (V1final-V1initial) =-m2 (V2final-V2initial)
Conservation of momentum law says that one object loses momentum and other one gains it. Total momentum of the system is conserved.
Example: Bullet shown in the picture collides to a fixed block.0, 2 s is the interaction time of bullet with block. If the velocity of the bullet is 250m/s after the collision, find the resistance of the block to the bullet.
Resistance of the block to the bullet is -250N. We use conservation of momentum to find the change in momentum and using the impulse momentum equation we find force that block apply to bullet.
Example: Two cars are stationary at the beginning. If the car having 10kg mass starts to move to the east with a velocity of 5m/s, find thevelocity of the car having mass 4kg with respect to the ground.
V1 is 12,5m/s to the west. If you want to find the velocity of the small car with respect to big car you should benefit from relative motion and do your calculations according to the vector properties, in other words you must consider the directions of the velocities.
Example: Two blocks move with the given velocities. When they collide they stick and move together.Find the velocity of the blocks after the collision.
Example: Two bullets having velocities 550m/s and 200m/s moves towards to a block that having mass 12kg. If the bullets stick to the block and they move together find the velocity of the final system.
We find momentum of the bullets first then; we calculate total initial momentum by adding them using their vector directions. Finally using conservation of momentum law we find the velocity of the system.
Author:
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https://devskill.com/Course/Detail/data-structure-algorithm-fundamentals | 1,674,960,527,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00783.warc.gz | 208,707,424 | 12,068 | Registration Closed
### Data Structure & Algorithm Fundamentals
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This course solely focuses on data structure and algorithm fundamentals. There are a lot of people both from CS and Non-CS background who do not have adequate understandings on data structures and algorithms. Data structures and algorithms are the most important topics to learn in order to become a successful software engineer or ace technical interviews. If you know basic programming but want to learn DS&A, this course is for you! Course highlight: Explanation of the core concepts of basic data structure and algorithms with practical examples. Detailed explanation on Complexity analysis with working examples (both time and memory). Introduction to Graph Theory, greedy & searching techniques, recursions, and lot more!
### Who Can Join This Course?
Basic knowledge (i.e. loops, arrays, I/O etc) of at least one programming language.
### Course Outline
#### Complexity Analysis
1. Time Complexity Analysis
i) Big-O Notation.
ii) How to calculate time complexity of a function.
2. Memory Complexity Analysis
i) Capacity of each data type variable and it's calculation.
ii) How to calculate memory complexity of a piece of code.
iii) Concepts of Heap Memory & Stack Memory.
#### Data Structures
i) Introduction, why we need it, how it works.
ii) Implementation in C++
2. Stack, Queue
i) Introduction, why we need it, how it works.
ii) Implementation in C++ with static array, dynamic array and linked list.
3. Single and Multidimensional arrays
i) Introduction.
ii) Conversion between various dimensional arrays into 1D arrays.
iii) Application and necessity of arrays
4. Introduction to Graph Theory
i) Graph modeling
#### Algorithms
1. Sorting Algorithms (Bubble, Insertion etc)
i) Introduction
ii) Implementation in C++
2. Searching Techniques (Binary Search)
i) Introduction
ii) Implementation in C++
3. Greedy Techniques
i) Introduction
ii) Discussion on a few related problems.
4. Recursion
i) Intro to recursion and backtracking
ii) Relevant basic problem solving i.e. 8-queen, all permutations, fibonacci etc
5. Shortest Path
i) Various graph traversing algorithms i.e. DFS, BFS
ii) How to model a graph based on real life scenario and finding shortest path
#### Technical Assignments
There will be a set of technical assignments after completion of each module.
No Upcoming Batch
#### Instructors
I am Md Sadman Sakib, an engineering graduate from University of Dhaka, specialized in Computer Science & Engineering. Currently working as a Software Engineer at Google. I love to learn,code,eat and travel! My CF Handle: reverse_macro
### What Students Said
Its a highly recommended course for learning Basic Algorithm and Data Structure. Thorough discussion on each topic and answering topic related questions made it simple to understand.
Data Structure & Algorithm Fundamentals Batch: 1
This course has been helping me a lot to improve my Data Structure and Algorithm knowledge. Md Sadman Sakib vai is an amazing instructor.
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This course is meant to be a starting point of problem solving world by learing entry level knowledge about data structure and algorithm. So this course helped me to make my footprint in problem solving. Though this course is not a C++ learing course and our teacher explicitely told us no need to worry about C++(even we present psudo code before him, students are welcomed), at least one class(C++ basics) would have been good for us who didn't code in C++. Overall I learned a lot from here and I have no complain. If I could not get benefit from this couse, I'm responsible for that by not practicing that much. I feel really grateful to course instructor Md Sadman Sakib bhai(many time I asked questions out of our syllabus and he answered very politely which is rere in Bangladesh).
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ডাটা স্ট্রাকচার এবং এলগোরিদমের প্রথম কোর্সটি অত্যন্ত সুগঠিত এবং যথার্থ ছিল। কোর্সের ইন্সটাক্টর সাকিব ভাই আমাদেরকে প্রতিটি টপিক মার্জিত এবং সুনিপুণভাবে বুঝিয়েছিলেন যা আমাদেরকে কনসেপ্টগুলো পরিষ্কারভাবে বুঝতে সহায়তা করেছিল। আমরা কোর্স থেকে শুধু কোর্সের টপিক সংক্রান্ত লেখাপড়া করিনি বরং ভাই আমাদেরকে ডাটা স্ট্রাকচার এবং এলগোরিদমের প্রয়োগ নিয়ে সফ্টওয়্যার ইন্ডাস্ট্রিতে অনেক 'রিয়েল ফ্যাক্টগুলো' তুলে ধরেছিলেন এবং আমাদেরকে সেই অনুযায়ী প্রস্তুত হতে পরামর্শ দিয়েছিলেন যা অত্যন্ত উপকারী ছিল। আমি এই কোর্স থেকে অনেক উপকৃত হয়েছি এবং বিশেষত যারা সি এস ফিল্ডে বিগিনার আছেন, আমরা অনেকসময় কোর্সের উপর্যুক্ত টপিকগুলো নিয়ে হেলা-ফেলা করে থাকি। জ্ঞানের পরিধি অনুযায়ী, যদি নিজেকে আমরা এই সকল বিষয়ে যাচাই করতে চাই তাহলে আমার মতে এই কোর্সটি করে দেখতে পারেন এতে আপনাদের অনেক প্রশ্নের উত্তর নিহিত আছে। সর্বশেষে, ডেভস্কিল টিম এবং সাকিব ভাইকে ধন্যবাদ কোর্সটি সুসজ্জিতভাবে পরিচালনা করার জন্য।
Md. Yousuf
Data Structure & Algorithm Fundamentals Batch: 1
আলহামদুলিল্লাহ। অত্যন্ত সুন্দর একটি কোর্স। পড়ালেখা, এসাইনমেন্ট, কন্টেস্ট দিয়ে ভরপুর। শেখার সাথে সাথে প্রয়োগ করা যায়, চর্চা করা যায়। শিক্ষক অত্যন্ত বন্ধুত্বপূর্ণ এবং ক্লাসে, ক্লাসের বাইরে যেকোনো সমস্যা সমাধান করতে চেষ্টা করেন। ইচ্ছামতো প্ৰশ্ন জিজ্ঞাসা করা যায়। আর ক্লাসে শুধু গৎবাঁধা পড়ালেখা না এই পড়ালেখার সাথে কর্মজীবন এর সম্পৃক্ততা এবং প্রয়োজনীয়তা ইত্যাদি সম্পর্কেও ধারণা অর্জন করা যায়। সবকিছু মিলিয়ে খুব সুন্দর একটি অভিজ্ঞতা। শুধু একটি জিনিস। ঢিলেমি করে এই কোর্স পার করতে পারবেন না। আর কাজের কাজ সময়ে না করলে পিছিয়ে পড়াটা খুবই স্বাভাবিক। তাই কোর্স করতে চাইলে যথাযথ পরিশ্রম করার নিয়ত করে শুরু করবেন। | 2,645 | 5,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-06 | longest | en | 0.796525 |
https://converter.ninja/length/millimeters-to-kilometers/839-mm-to-km/ | 1,601,250,117,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00139.warc.gz | 334,443,720 | 5,244 | # 839 millimeters in kilometers
## Conversion
839 millimeters is equivalent to 0.000839 kilometers.[1]
## Conversion formula How to convert 839 millimeters to kilometers?
We know (by definition) that: $1\mathrm{mm}=1e-06\mathrm{km}$
We can set up a proportion to solve for the number of kilometers.
$1 mm 839 mm = 1e-06 km x km$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{km}=\frac{839\mathrm{mm}}{1\mathrm{mm}}*1e-06\mathrm{km}\to x\mathrm{km}=0.000839\mathrm{km}$
Conclusion: $839 mm = 0.000839 km$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 kilometer is equal to 1191.89511323004 times 839 millimeters.
It can also be expressed as: 839 millimeters is equal to $\frac{1}{\mathrm{1191.89511323004}}$ kilometers.
## Approximation
An approximate numerical result would be: eight hundred and thirty-nine millimeters is about zero kilometers, or alternatively, a kilometer is about one thousand, one hundred and ninety-one point nine zero times eight hundred and thirty-nine millimeters.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 343 | 1,264 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-40 | latest | en | 0.697725 |
https://www.careerstoday.in/ncert-solutions/ncert-solutions-for-exercise-1-4-class-12-maths-chapter-1-relations-and-functions | 1,720,898,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00555.warc.gz | 623,186,498 | 11,977 | # NCERT Solutions for Exercise 1.4 Class 12 Maths Chapter 1 - Relations and Functions
NCERT Solutions for class 12 maths chapter 1 exercise 1.4 talks about commutative, associative, binary operations etc. Exercise 1.4 Class 12 Maths has questions which includes finding whether a relation follows binary operation or not. After analysis from previous year questions it is clear that NCERT Solutions for class 12 maths chapter 1 exercise 1.4 plays an important role to score well in CBSE class 12 board exam. Also it has a significant contribution in competitive exams like JEE main. Hence it is recommended for students to solve all the questions of Class 12th maths chapter 1 exercise 1.4 to score well in their examination. The NCERT chapter Relations and Functions also consists the following exercise for practice.
Relations and Functions Exercise 1.1
Relations and Functions Exercise 1.2
Relations and Functions Exercise 1.3
Relations and Functions Miscellaneous Exercise
## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions: Exercise 1.4
Question:1(i) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(i) On $Z^+$ , define ∗ by $a * b = a - b$
(i) On $Z^+$ , define ∗ by $a * b = a - b$
It is not a binary operation as the image of $(1,2)$ under * is $1\ast 2=1-2$ $=-1 \notin Z^{+}$ .
Question:1(ii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(ii) On $Z^+$ , define ∗ by $a * b = ab$
(ii) On $Z^+$ , define ∗ by $a * b = ab$
We can observe that for $a,b \in Z^+$ ,there is a unique element ab in $Z^+$ .
This means * carries each pair $(a,b)$ to a unique element $a * b = ab$ in $Z^+$ .
Therefore,* is a binary operation.
Question:1(iii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(iii) On $R$ , define ∗ by $a * b = ab^2$
(iii) On $R$ , define ∗ by $a * b = ab^2$
We can observe that for $a,b \in R$ ,there is a unique element $ab^{2}$ in $R$ .
This means * carries each pair $(a,b)$ to a unique element $a * b = ab^{2}$ in $R$ .
Therefore,* is a binary operation.
Question:1(iv) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.
(iv) On $Z^+$ , define ∗ by $a * b = | a - b |$
(iv) On $Z^+$ , define ∗ by $a * b = | a - b |$
We can observe that for $a,b \in Z^+$ ,there is a unique element $| a - b |$ in $Z^+$ .
This means * carries each pair $(a,b)$ to a unique element $a * b = | a - b |$ in $Z^+$ .
Therefore,* is a binary operation.
Question:1(v) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this
(v) On $Z^+$ , define ∗ by $a * b = a$
(v) On $Z^+$ , define ∗ by $a * b = a$
* carries each pair $(a,b)$ to a unique element $a * b = a$ in $Z^+$ .
Therefore,* is a binary operation.
Question:2(i) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(i)On $Z$ , define $a * b = a-b$
a*b=a-b
b*a=b-a
$a*b\neq b*a$
so * is not commutative
(a*b)*c=(a-b)-c
a*(b*c)=a-(b-c)=a-b+c
(a*b)*c not equal to a*(b*c), so * is not associative
Question:2(ii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(ii) On $Q$ , define $a * b = ab + 1$
(ii) On $Q$ , define $a * b = ab + 1$
ab = ba for all $a,b \in Q$
ab+1 = ba + 1 for all $a,b \in Q$
$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Q$
$(1*2)*3 = (1\times 2+1) * 3 = 3 * 3 = 3\times 3+1 = 10$
$1*(2*3) = 1 * (2\times 3+1) = 1 * 7 = 1\times 7+1 = 8$
$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in Q$
$\therefore$ operation * is not associative.
Question:2(iii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(iii) On $Q$ , define $a * b = \frac{ab}{2}$
(iii) On $Q$ , define $a * b = \frac{ab}{2}$
ab = ba for all $a,b \in Q$
$\frac{ab}{2}=\frac{ba}{2}$ for all $a,b \in Q$
$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Q$
$\therefore$ operation * is commutative.
$(a*b)*c = \frac{ab}{2}*c = \frac{(\frac{ab}{2})c}{2} = \frac{abc}{4}$
$a*(b*c) = a*\frac{bc}{2} = \frac{a(\frac{bc}{2})}{2} = \frac{abc}{4}$
$\therefore$ $(a*b)*c=a*(b*c)$ $;$
$\therefore$ operation * is associative.
Question:2(iv) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(iv) On $Z^+$ , define $a * b = 2^{ab}$
(iv) On $Z^+$ , define $a * b = 2^{ab}$
ab = ba for all $a,b \in Z^{+}$
2ab = 2ba for all $a,b \in Z^{+}$
$\Rightarrow$ $a\ast b=b\ast a$ for $a,b \in Z^{+}$
$\therefore$ the operation is commutative.
$(1*2)*3 = 2^{1\times 2} * 3 = 4 * 3 = 2^{4\times 3} = 2^{12}$
$1*(2*3) = 1 * 2^{2\times 3} = 1 * 64 = 2^{1\times 64}=2^{64}$
$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in Z^{+}$
$\therefore$ operation * is not associative.
Question:2(v) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(v) On $Z^+$ , define $a * b = a^b$
(v) On $Z^+$ , define $a * b = a^b$
$1\ast 2 = 1^{2}= 1$ and $2\ast 1 = 2^{1}= 2$
$\Rightarrow$ $1\ast 2\neq 2\ast 1$ for $1,2 \in Z^{+}$
$\therefore$ the operation is not commutative.
$(2\ast 3)\ast 4 = 2^{3} \ast 4 = 8 \ast 4 = 8^{ 4}=2^{12}$
$2\ast (3\ast 4) = 2 \ast 3^{ 4} = 2 \ast 81 = 2^{81}$
$\therefore$ $(2\ast 3)\ast 4\neq 2\ast (3\ast 4)$ $;$ where $2,3,4 \in Z^{+}$
$\therefore$ operation * is not associative.
Question:2(vi) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.
(vi) On $R - \{-1 \}$ , define $a * b = \frac{a}{b +1}$
(iv) On $R - \{-1 \}$ , define $a * b = \frac{a}{b +1}$
$1\ast 2 = \frac{1}{2+1}=\frac{1}{3}$ and $2\ast 1 = \frac{2}{2+1}= \frac{2}{3}$
$\Rightarrow$ $1\ast 2\neq 2\ast 1$ for $1,2 \in R - \{-1 \}$
$\therefore$ the operation is not commutative.
$(1\ast 2)\ast 3 = (\frac{1}{2+1}) \ast 3 = \frac{1}{3} \ast 3 = \frac{\frac{1}{3}}{3+1}= \frac{1}{12}$
$1\ast (2\ast 3) = 1 \ast (\frac{2}{3+1}) = 1 \ast \frac{2}{4} = 1 \ast \frac{1}{2} = \frac{1}{\frac{1}{2}+1} = \frac{2}{3}$
$\therefore$ $(1\ast 2)\ast 3\neq 1\ast (2\ast 3)$ $;$ where $1,2,3 \in R - \{-1 \}$
$\therefore$ operation * is not associative.
Question:3 Consider the binary operation $\wedge$ on the set $\{1, 2, 3, 4, 5\}$ defined by $a \wedge b = min \{a, b\}$ . Write the operation table of the operation $\wedge$ .
$\{1, 2, 3, 4, 5\}$
$a \wedge b = min \{a, b\}$ for $a,b \in \{1, 2, 3, 4, 5\}$
The operation table of the operation $\wedge$ is given by :
$\wedge$ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5
Question:4(i) Consider a binary operation ∗ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).
(i) Compute $(2 * 3) * 4$ and $2 * (3 * 4)$
(Hint: use the following table)
(i)
$(2 * 3) * 4 = 1*4 =1$
$2 * (3 * 4) = 2*1=1$
Question:4(ii) Consider a binary operation ∗ on the set $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).
(ii) Is ∗ commutative?
(Hint: use the following table)
(ii)
For every $a,b \in\{1, 2, 3, 4, 5\}$ , we have $a*b = b*a$ . Hence it is commutative.
Question:4(iii) Consider a binary operation ∗ on the set { $\{1, 2, 3, 4, 5\}$ given by the following multiplication table (Table 1.2).
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
(Hint: use the following table)
(iii) (2 ∗ 3) ∗ (4 ∗ 5).
from the above table
$(2*3)*(4*5)=1*1=1$
Question:5 Let ∗′ be the binary operation on the set $\{1, 2, 3, 4, 5\}$ defined by $a *' b = H.C.F. \;of\;a\;and\;b$ . Is the operation ∗′ same as the operation ∗ defined
$a *' b = H.C.F. \;of\;a\;and\;b$ for $a,b \in \{1, 2, 3, 4, 5\}$
The operation table is as shown below:
$\ast$ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5
The operation ∗′ same as the operation ∗ defined in Exercise 4 above.
Question:6(i) let ∗ be the binary operation on N given by . Find
(i) 5 ∗ 7, 20 ∗ 16
a*b=LCM of a and b
(i) 5 ∗ 7, 20 ∗ 16
$5*7 = L.C.M \, of\, 5\, \, and \, 7=35$
$20*16 = L.C.M \, of\, 20\, \, and \, 16 =80$
Question:6(ii) Let ∗ be the binary operation on N given by $a * b = L.C.M. \;of \;a\; and \;b$ . Find
(ii) Is ∗ commutative?
$a * b = L.C.M. \;of \;a\; and \;b$
(ii) $L.C.M. \;of \;a\; and \;b = L.C.M. \;of \;b\; and \;a$ for all $a,b \in N$
$\therefore \, \, \, \, \, \, \, a*b = b*a$
Hence, it is commutative.
Question:6(iii) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
(iii) Is ∗ associative?
a $*$ b = L.C.M. of a and b
(iii) $a,b,c \in N$
$(a*b)*c = (L.C.M \, of\, a\, and \, b)*c= L.C.M\, of \, a,b\, and\, c$
$a*(b*c) = a*(L.C.M \, of\, b\, and \, c)= L.C.M\, of \, a,b\, and\, c$
$\therefore \, \, \, \, \, \, \, \, \, (a*b)*c=a*(b*c)$
Hence, the operation is associative.
Question:6(iv) Let ∗ be the binary operation on N given by $a* b = L.C.M.\; of \;a\; and \;b$ . Find
(iv) the identity of ∗ in N
$a* b = L.C.M.\; of \;a\; and \;b$
(iv) the identity of ∗ in N
We know that $L.C.M.\; of \;a\; and \;1 = a = L.C.M.\; of 1\, \, and \;a\;$
$\therefore$ $a*1=a=1*a$ for $a \in N$
Hence, 1 is the identity of ∗ in N.
Question 6(v) Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find
(v) Which elements of N are invertible for the operation ∗?
$a* b = L.C.M.\; of \;a\; and \;b$
An element a is invertible in N
if $a*b=e=b*a$
Here a is inverse of b.
a*b=1=b*a
a*b=L.C.M. od a and b
a=b=1
So 1 is the only invertible element of N
Question:7 Is ∗ defined on the set $\{1, 2, 3, 4, 5\}$ by $a * b = L.C.M. \;of \;a\; and \;b$ a binary operation? Justify your answer.
$a * b = L.C.M. \;of \;a\; and \;b$
A = $\{1, 2, 3, 4, 5\}$
Operation table is as shown below:
$*$ 1 2 3 4 5 1 1 2 3 4 5 2 2 2 6 4 10 3 3 6 3 12 15 4 4 4 12 4 20 5 5 10 15 20 5
From the table, we can observe that
$2*3=3*2=6 \notin A$
$2*5=5*2=10 \notin A$
$3*4=4*3=12 \notin A$
$3*5=5*3=15 \notin A$
$4*5=5*4=20 \notin A$
Hence, the operation is not a binary operation.
Question:8 Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary
operation on N?
a ∗ b = H.C.F. of a and b for all $a,b \in A$
H.C.F. of a and b = H.C.F of b and a for all $a,b \in A$
$\therefore \, \, \, \, a*b=b*a$
Hence, operation ∗ is commutative.
For $a,b,c \in N$ ,
$(a*b)*c = (H.C.F \, of\, a\, and\, b)*c= H.C.F\, of \, a,b,c.$
$a*(b*c )= a*(H.C.F \, of\, b\, and\, c)= H.C.F\, of \, a,b,c.$
$\therefore$ $(a*b)*c=a*(b*c)$
Hence, ∗ is associative.
An element $c \in N$ will be identity for operation * if $a*c=a= c*a$ for $a \in N$ .
Hence, the operation * does not have any identity in N.
Question:9(i) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(i) $a * b = a - b$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defined as $a * b = a - b$ .It is observed that:
$\frac{1}{2}*\frac{1}{3}= \frac{1}{2}-\frac{1}{3}=\frac{1}{6}$
$\frac{1}{3}*\frac{1}{2}= \frac{1}{3}-\frac{1}{2}=\frac{-1}{6}$
$\therefore$ $\frac{1}{2}*\frac{1}{3}\neq \frac{1}{3}*\frac{1}{2}$ here $\frac{1}{2},\frac{1}{3} \in Q$
Hence, the * operation is not commutative.
It can be observed that
$(\frac{1}{2}*\frac{1}{3})*\frac{1}{4} = \left ( \frac{1}{2}-\frac{1}{3}\right )*\frac{1}{4}=\frac{1}{6}*\frac{1}{4}=\frac{1}{6}-\frac{1}{4}=\frac{-1}{12}$
$\frac{1}{2}*(\frac{1}{3}*\frac{1}{4})= \frac{1}{2}*\left ( \frac{1}{3} - \frac{1}{4}\right ) = \frac{1}{2}*\frac{1}{12} = \left ( \frac{1}{2} - \frac{1}{12} \right ) = \frac{5}{12}$
$\left ( \frac{1}{2}*\frac{1}{3} \right )*\frac{1}{4}\neq \frac{1}{2}*(\frac{1}{3}*\frac{1}{4})$ for all $\frac{1}{2},\frac{1}{3}, \frac{1}{4} \in Q$
The operation * is not associative.
Question:9(ii) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(ii) $a*b = a^2 + b^2$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defines as $a*b = a^2 + b^2$ .It is observed that:
For $a,b \in Q$
$a*b=a^{2}+b^{2}= b^{2}+a^{2}=b*a$
$\therefore$ $a*b=b*a$
Hence, the * operation is commutative.
It can be observed that
$(1*2)*3 =(1^{2}+2^{2})*3 = 5*3 = 5^{2}+3^{2} = 25+9 =34$
$1*(2*3) =1*(2^{2}+3^{2}) = 1*13 = 1^{2}+13^{2} = 1+169 =170$
$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$
The operation * is not associative.
Question:9(iii) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(iii) $a * b = a + ab$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defines as $a * b = a + ab$ .It is observed that:
For $a,b \in Q$
$1 * 2 = 1+1\times 2 =1 + 2 = 3$
$2 * 1= 2+2\times 1 =2 + 2 = 4$
$\therefore$ $1*2\neq 2*1$ for $1,2 \in Q$
Hence, the * operation is not commutative.
It can be observed that
$1*(2*3) =1*(2+3\times 2) = 1*8 = 1+1\times 8 = 1+8 =9$
$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$
The operation * is not associative.
Question:9(iv) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(iv) $a * b = (a-b)^2$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defined as $a * b = (a-b)^2$ .It is observed that:
For $a,b \in Q$
$a * b = (a-b)^2$
$b* a = (b-a)^2 = \left [ -\left ( a-b \right ) \right ]^{2} = (a-b)^{2}$
$\therefore$ $a*b = b* a$ for $a,b \in Q$
Hence, the * operation is commutative.
It can be observed that
$(1*2)*3 =(1-2)^{2}*3 = 1*3 =(1-3)^{2}= (-2)^{2} =4$
$1*(2*3) =1*(2-3)^{2} = 1*1 =(1-1)^{2} = 0^{2} =0$
$(1*2)*3 \neq 1*(2*3)$ for all $1,2,3 \in Q$
The operation * is not associative.
Question:9(v) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(v) $a * b = \frac{ab}{4}$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defines as $a * b = \frac{ab}{4}$ .It is observed that:
For $a,b \in Q$
$a * b = \frac{ab}{4}$
$b* a = \frac{ba}{4}$
$\therefore$ $a*b = b* a$ for $a,b \in Q$
Hence, the * operation is commutative.
It can be observed that
$(a*b)*c =(\frac{ab}{4})*c = \frac{\frac{ab}{4}c}{4}=\frac{abc}{16}$
$a*(b*c) =a*(\frac{bc}{4}) = \frac{\frac{bc}{4}a}{4}=\frac{abc}{16}$
$(a*b)*c = a*(b*c)$ for all $a,b,c \in Q$
The operation * is associative.
Question:9(vi) Let ∗ be a binary operation on the set Q of rational numbers as follows:
(vi) $a* b = ab^2$ Find which of the binary operations are commutative and which are associative.
On the set Q ,the operation * is defines as $a* b = ab^2$ .It is observed that:
For $a,b \in Q$
$1* 2 = 1\times 2^2=1\times 4=4$
$2* 1 = 2\times 1^2=2\times 1=2$
$\therefore$ $1*2\neq 2*1$ for $1,2 \in Q$
Hence, the * operation is not commutative.
It can be observed that
$(1*2)*3 = (1\times 2^{2})*3 = 4*3 = 4\times 3^{2}=4\times 9=36$
$1*(2*3) = 1*(2\times 3^{2}) = 1*18 = 1\times 18^{2}=1\times 324=324$
$(1*2)*3\neq 1*(2*3)$ for all $1,2,3 \in Q$
The operation * is not associative.
Question:10 Find which of the operations given above has identity.
An element $p \in Q$ will be identity element for operation *
if $a*p = a = p*a$ for all $a \in Q$
$(v) a * b = \frac{ab}{4}$
$a * p = \frac{ap}{4}$
$p * a = \frac{pa}{4}$
$a*p = a = p*a$ when $p=4$ .
Hence, $(v) a * b = \frac{ab}{4}$ has identity as 4.
However, there is no such element $p \in Q$ which satisfies above condition for all rest five operations.
Hence, only (v) operations have identity.
Question:11 Let $A = N \times N$ and ∗ be the binary operation on A defined by $(a, b) * (c, d) = (a + c, b + d)$ Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.
$A = N \times N$ and ∗ be the binary operation on A defined by
$(a, b) * (c, d) = (a + c, b + d)$
Let $(a,b),(c,d) \in A$
Then, $a,b,c,d \in N$
We have
$(a, b) * (c, d) = (a + c, b + d)$
$(c,d)*(a,b) = (c+a,d+b)= (a+c,b+d)$
$\therefore \, \, \, \, \, (a,b)*(c,d)=(c,d)*(a,b)$
Thus it is commutative.
Let $(a,b),(c,d),(e,f) \in A$
Then, $a,b,c,d,e,f \in N$
$[(a, b) * (c, d)]*(e,f)= [(a + c, b + d)]*(e,f) =[(a+c+e),(b+d+f)]$
$(a, b) * [(c, d)*(e,f)]= (a,b)*[(c + e, d + f)] =[(a+c+e),(b+d+f)]$
$\therefore \, \, \, \, \, [(a,b)*(c,d)]*(e,f)=(a, b) * [(c, d)*(e,f)]$
Thus, it is associative.
Let $e= (e1,e2) \in A$ will be a element for operation * if $(a*e)=a=(e*a)$ for all $a= (a1,a2) \in A$ .
i.e. $(a1+e1,a2+e2)= (a1,a2)= (e1+a1,e2+a2)$
This is not possible for any element in A .
Hence, it does not have any identity.
Question:12(i) State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation ∗ on a set N, $a*a = a,\; \forall a\in N$
(i) For an arbitrary binary operation ∗ on a set N, $a*a = a,\; \forall a\in N$
An operation * on a set N as $a*b=a+b\, \, \, \, \forall\, \, a,b \in N$
Then , for b=a=2
$2*2= 2+2 = 4\neq 2$
Hence, statement (i) is false.
Question:12(ii) State whether the following statements are true or false. Justify.
(ii) If ∗ is a commutative binary operation on N, then $a *(b*c) =( c*b)*a$
(ii) If ∗ is a commutative binary operation on N, then $a *(b*c) =( c*b)*a$
R.H.S $=(c*b)*a$
$=(b*c)*a$ (* is commutative)
$= a*(b*c)$ ( as * is commutative)
= L.H.S
$\therefore$ $a *(b*c) =( c*b)*a$
Hence, statement (ii) is true.
Question:13 Consider a binary operation ∗ on N defined as $a * b = a^3 + b^3$ . Choose the correct answer.
(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?
A binary operation ∗ on N defined as $a * b = a^3 + b^3$ .
For $a,b \in N$
$a * b = a^3 + b^3 = b^{3}+a^{3}=b*a$
Thus, it is commutative.
$(1*2)*3 = (1^{3}+2^{3})*3=9*3 =9^{3}+3^{3}=729+27=756$
$1*(2*3) = 1*(2^{3}+3^{3})=1*35 =1^{3}+35^{3}=1+42875=42876$
$\therefore \, \, \, \, \, (1*2)*3 \neq 1*(2*3)$ where $1,2,3 \in N$
Hence, it is not associative.
Hence, B is the correct option.
More About NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.4
The NCERT Class 12 Maths chapter Relations and Functions has a total of 5 exercises including miscellaneous exercise. Exercise 1.4 Class 12 Maths covers solutions to 13 main questions and their sub-questions. Most of the questions are related to concepts of binary operations which includes commutative, associative operations etc. Hence NCERT Solutions for Class 12 Maths chapter 1 exercise 1.4 is recommended for learning these concepts to score well in the exam.
Also Read| NCERT Notes For Class 12 Mathematics Chapter 1
Benefits of NCERT Solutions for Class 12 Maths Chapter 1 Exercise 1.4
• The Class 12 Maths NCERT syllabus chapter 1.4 exercise is provided above in a comprehensive manner which is solved by subject matter experts .
• Students are recommended to practice Exercise 1.4 Class 12 Maths to prepare for topics like binary operations which includes commutative, associative operations etc.
• These Class 12 Maths chapter 1 exercise 1.4 solutions can be referred by students to revise just before the exam.
• NCERT syllabus for Class 12 Maths chapter 1 exercise 1.4 can be used to prepare similar topics of physics also.
### Also see-
• NCERT exemplar solutions class 12 maths chapter 1
• NCERT solutions for class 12 maths chapter 1
NCERT Solutions Subject Wise
• NCERT solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology
• NCERT solutions for class 12 mathematics
Subject wise NCERT Exemplar solutions
• NCERT Exemplar Class 12th Maths
• NCERT Exemplar Class 12th Physics
• NCERT Exemplar Class 12th Chemistry
• NCERT Exemplar Class 12th Biology
Happy learning!!!
Class 10 NSTSE 5-Year Analysis: Which Chapters Get More Weightage And Why 9 min read Apr 11, 2022 Read More Ask Doubts, Gain Confidence 4 min read Mar 05, 2022 Read More | 7,888 | 20,438 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 310, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2024-30 | latest | en | 0.891552 |
https://fliphtml5.com/xjqi/stod | 1,685,621,324,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647810.28/warc/CC-MAIN-20230601110845-20230601140845-00588.warc.gz | 307,408,485 | 14,808 | Chapter 8: Coefficient of Determination Every sample has some variation in it ... Predict the number of muggings if there are 9 police officers on duty. 5 Read More
Chapter 8: Coefficient of Determination - Keller ISD Schools
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Chapter 8: Coefficient of DeterminationEvery sample has some variation in it (unless all the values are identical, and that's unlikely to happen).The total variation is made up of two parts, the part that can be explained by the regression equation andthe part that can't be explained by the regression equation. The coefficient of determination, r 2, is useful because it gives the proportion of the variance (fluctuation) of one variable that is predictable from the other variable. It is a measure that allows us to determine how certain one can be in making predictions from a certain model/graph. The coefficient of determination is the ratio of the explained variation to the total variation. The coefficient of determination is such that 0 < r 2 < 1, and denotes the strength of the linear association between x and y. The coefficient of determination represents the percent of the data that is the closest to the line of best fit. For example, if r = 0.922, then r 2 = 0.850, which means that 85% of the total variation in y can be explained by the linear relationship between x and y (as described by the regression equation). The other 15% of the total variation in y remains unexplained. The coefficient of determination is a measure of how well the regression line represents the data. If the regression line passes exactly through every point on the scatter plot, it would be able to explain all of the variation. The further the line is away from the points, the less it is able to explain. regression lineobserved valuey total unexplained deviation explained by regression 1
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2. The scatterplot shows the advertised prices (in thousands of dollars) plotted against ages (in years) for a random sample of Plymouth Voyagers on several dealers‟ lots.A computer printout showing the results of a straight Plymouth Voyagers Scatter Plotline to the data by the method of least squares gives: 14 Price = 12.37 – 1.13 Age 12 R-sq = 75.5% Price_100010a) Find the correlation coefficient for the relationship between price and age of Voyagers based on these 8 data. 6b) What is the slope of the regression line? Interpret it in the context of these data. 4c) How will the size of the correlation coefficient 2 change if the 10-year-old Voyager is removed from the data set? Explain. 2 4 6 8 10 Age_in_yearsd) How will the slope of the LSRL change if the 10- year-old Voyager is removed from the data? 3
P:04
3. In one of the Boston city parks there has been a problem with muggings in the summer months. A police cadet took a random sample of 10 days (out of the 90-day summer) and compiled the following data. For each day, x represents the number of police officers on duty in the park and y represents the number of reported muggings on that day. x 10 15 16 1 4 6 18 12 14 7 y 5 2 1 978 1 5 3 6 a) Sketch a scatter plot. b) What is the correlation coefficient? Interpret in terms of the problem. c) Interpret the slope in terms of the problem. d) Find the coefficient of determination and interpret in terms of the problem. e) Predict the number of muggings if there are 9 police officers on duty. 4
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4. The average prices (in dollars) per ounce of gold and silver for the years 1986 through 1994 are given below. Year 1986 1987 1988 1989 1990 1991 1992 1993 1994 Gold 368 478 438 383 385 363 345 361 389 Silver 5.47 7.01 6.53 5.50 4.82 4.04 3.94 4.30 5.30 a. What is the explanatory variable? Explain. b. Find the LSRL for gold predicting silver. c. Interpret the slope and y-intercept. d. What is the correlation coefficient? Interpret. e. Find the LSRL for silver predicting gold. f. Interpret the slope and y-intercept. g. What is the correlation coefficient? Interpret. Compare your answer to part „d‟. h. What is the coefficient of determination? Interpret. 5
P:06
5. How quickly can athletes return to their sport following injuries requiring surgery? The paper “Arthroscopic Distal Clavicle Resection for Isolated Atraumatic Osteolysis in Weight Lifters” (American Journal of Sports Medicine, 1998) discovered their was a moderate positive (r = .55) linear relationship between a lifters age and the number of days after arthroscopic shoulder survgery before being able to return to their sport between 10 weight lifters. The average age of the weight lifters 30.4 with standard deviation of 2.875 years. The average number of days before being able to return to their sport was 3.2 days with a standard deviation of 1.398 days. a. Determine the LSRL to predict the number of days based on the age of the weight lifter. b. Determine the coefficient of determination and interpret in terms of the problem. c. Given the spread of the lifters was from 26 to 34 years old, predict the number of days for a 28 year old lifter. Do you feel this prediction is accurate? Explain. 6 | 1,264 | 5,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-23 | latest | en | 0.932885 |
http://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-10-section-10-3-simplifying-radical-expressions-exercise-set-page-703/57 | 1,524,326,106,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945232.48/warc/CC-MAIN-20180421145218-20180421165218-00426.warc.gz | 429,423,241 | 14,415 | ## Algebra: A Combined Approach (4th Edition)
$5r^3s^4$
$\sqrt[3] {125r^9s^{12}}=5r^3s^4$ because $(5)^3=125$ $(r^3)^3=r^9$ $(s^4)^3=r^{12}$ | 72 | 141 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-17 | latest | en | 0.547824 |
https://fluxml.ai/FastAI.jl/dev/Flux@0.13.17/ref/Flux.Dropout.html | 1,722,781,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00243.warc.gz | 199,240,718 | 39,841 | # `Dropout`
`struct` defined in module `Flux`
``````
Dropout(p; [dims, rng, active])
``````
Layer implementing with the given probability. This is used as a regularisation, i.e. to reduce overfitting.
While training, it sets each input to ``` 0``` (with probability ``` p```) or else scales it by ``` 1 / (1 - p)```, using the function. While testing, it has no effect.
By default the mode will switch automatically, but it can also be controlled manually via ``` Flux.testmode!```, or by passing keyword ``` active=true``` for training mode.
By default every input is treated independently. With the ``` dims``` keyword, instead it takes a random choice only along that dimension. For example ``` Dropout(p; dims = 3)``` will randomly zero out entire channels on WHCN input (also called 2D dropout).
Keyword ``` rng``` lets you specify a custom random number generator. (Only supported on the CPU.)
# Examples
``````
julia> m = Chain(Dense(ones(3,2)), Dropout(0.4))
Chain(
Dense(2 => 3), # 9 parameters
Dropout(0.4),
)
julia> m(ones(2, 7)) # test mode, no effect
3×7 Matrix{Float64}:
2.0 2.0 2.0 2.0 2.0 2.0 2.0
2.0 2.0 2.0 2.0 2.0 2.0 2.0
2.0 2.0 2.0 2.0 2.0 2.0 2.0
julia> Flux.trainmode!(m) # equivalent to use within gradient
Chain(
Dense(2 => 3), # 9 parameters
Dropout(0.4, active=true),
)
julia> m(ones(2, 7))
3×7 Matrix{Float64}:
0.0 0.0 3.33333 0.0 0.0 0.0 0.0
3.33333 0.0 3.33333 0.0 3.33333 0.0 3.33333
3.33333 3.33333 0.0 3.33333 0.0 0.0 3.33333
julia> y = m(ones(2, 10_000));
julia> using Statistics
julia> mean(y) # is about 2.0, same as in test mode
1.9989999999999961
julia> mean(iszero, y) # is about 0.4
0.4003``````
Methods
There are 3 methods for `Flux.Dropout`: | 664 | 1,819 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-33 | latest | en | 0.692132 |
https://worksheetgenius.com/unit-converter/convert-897-w-to-gw/ | 1,638,737,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363216.90/warc/CC-MAIN-20211205191620-20211205221620-00517.warc.gz | 681,752,102 | 5,178 | # Convert 897 w to gw
In this article I will show you how to convert 897 watts into gigawatts. Throughout the explanation below I might also call it 897 w to gw. They are the same thing!
## How to Convert Watts to Gigawatts
A watt is smaller than a gigawatt. I know that a w is smaller than a gw because of something called conversion factors.
Put very simply, a conversion factor is a number that can be used to change one set of units to another, by multiplying or dividing it. So when we need to convert 897 watts into gigawatts, we use a conversion factor to get the answer.
The conversion factor for w to gw is:
1 w = 1.0E-9 gw
Now that we know what the conversion factor is, we can easily calculate the conversion of 897 w to gw by multiplying `1.0E-9` by the number of watts we have, which is 897.
897 x 1.0E-9 = 8.97E-7 gw
So, the answer to the question "what is 897 watts in gigawatts?" is 8.97E-7 gw.
## Watts to Gigawatts Conversion Table
Below is a sample conversion table for w to gw:
Watts (w) Gigawatts (gw)
0.010
0.10
10
20
30
50
100
200
500
1000
10000
## Best Conversion Unit for 897 w
Sometimes when you work with conversions from one unit to another, the numbers can get a little confusing. Especially when dealing with really large numbers.
I've also calculated what the best unit of measurement is for 897 w.
To determine which unit is best, I decided to define that as being the unit of measurement which is as low as possible, without going below 1. Smaller numbers are more easily understood and can make it easier for you to understand the measurement.
The best unit of measurement I have found for 897 w is watts and the amount is 897 w. | 435 | 1,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-49 | latest | en | 0.94058 |
https://fr.mathworks.com/matlabcentral/cody/problems/44486-vector-push/solutions/2098667 | 1,607,075,429,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141735395.99/warc/CC-MAIN-20201204071014-20201204101014-00599.warc.gz | 301,005,988 | 17,270 | Cody
# Problem 44486. Vector push
Solution 2098667
Submitted on 19 Jan 2020 by OSHER GORTENSTEIN
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
filetext = fileread('push.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
v = [1, 2, 7]; x = []; [v, n] = push(v, x); v_correct = [1, 2, 7]; n_correct = 3; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
v = 1 2 7
3 Pass
v = []; x = [5, 6, 8]; [v, n] = push(v, x); v_correct = [5, 6, 8]; n_correct = 3; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
V = 5 6 8 v = 5 6 8
4 Pass
v = []; x = [7; 3; 9; 4]; [v, n] = push(v, x); v_correct = [7; 3; 9; 4]; n_correct = 4; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
V = 7 3 9 4 v = 7 3 9 4
5 Pass
v1 = randi(10, 1, 5); x = 5; [v, n] = push(v1, x); v_correct = [v1, 5]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
v = 9 1 6 3 5 5
6 Pass
v = [ones(4, 1)]; x = [3, 5]; [v, n] = push(v, x); v_correct = [1; 1; 1; 1; 3; 5]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
x = 3 5 V = 1 1 1 1 3 5 v = 1 1 1 1 3 5
7 Pass
v = [zeros(1, 4)]; x = [3; 0]; [v, n] = push(v, x); v_correct = [0, 0, 0, 0, 3, 0]; n_correct = 6; assert(isequal(v, v_correct)); assert(isequal(n, n_correct)); v = [8, 2]; x = []; [v, n] = push(v, x); v_correct = [8, 2]; n_correct = 2; assert(isequal(v, v_correct)); assert(isequal(n, n_correct));
v = 0 0 0 0 3 0 v = 8 2
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After adding my movement script in godot. It seems the move speed is a lot faster when moving diagonally. (IE. up and to the right or down and to the right(left)). Just wondering what is causing this as the velocity doesn’t show an increase other than it showing both the x and y vector.
thank you,
1 Like
I haven’t done this course yet, but could it be that it sees the move as one move unit?
Diagonally would be a slightly longer distance than any of the cardinal directions, so if it sees the movement as one move unit, it would traverse that slightly longer distance in the same time span as a cardinal direction move?
If it makes sense?
1 Like
I did end up finding out why it does it on another forum. It seems it has something to do with the way vectors are calculated and using the two numbers on a diagonal ends up adding about 40% more move speed. I have to normalize something to make it work properly. I don’t fully understand it all yet so I’m continuing on until I have completed this section.
2 Likes
Hi Slothmonger,
You are actually pretty correct its about 41% increase.
When you finally calculate the movement you need to normalise the velocity and then multiply by speed.
direction = velocity.normalize()
move_and_slide(direction * speed)
This would limit the velocity to a value between 0 and 1
I found this thread on reddit that may help the understanding a little better | 313 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.957745 |
https://brainly.in/question/123777 | 1,485,029,056,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00575-ip-10-171-10-70.ec2.internal.warc.gz | 791,613,289 | 10,686 | # A man throws a ball weighing 500 gm vertically upwards with the speed of 10 m/s. Find: (a) What will be its initial momentum? (b) What would its momentum be at the highest point?
2
by PopGirl
2015-06-05T20:05:53+05:30
### This Is a Certified Answer
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(a) 5N/sec[p=mv]
(b) Zero. Because velocity of the ball will be zero at the higest point
That I know, but my question is that, here it is given that speed = 10 m/s. But p = mv , v = velocity. Then how can u place 10 m/s in place of v?
cause there is no change in direction.
Ok, and how come the momentum at the highest point is zero?
cause the velocity will be zero at the highest point due to negative acceleration
THANK U! :)
2015-06-06T04:09:05+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Initial velocity (u) = 10 m/s
Final velocity (v) = 0 m/s (at the highest point, speed becomes zero)
Mass (m) = 500 gm = 500/1000 = 0.5 kg
(a) Initial momentum = Mass × Initial velocity
= 0.5 × 10
= 5 kg.m/s
(b) Momentum at the highest point = Final momentum
= Mass × Final velocity
= 0.5 × 0
= 0 kg.m/s | 432 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-04 | latest | en | 0.922511 |
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mesk
Topic Author
Posts: 2
Joined: January 12th, 2018, 1:44 pm
### Calculation of fed hike/cut probabilities
How do I calculate the probability of Fed cutting the rate using both Fed Fund Futures and OIS rates?
I have attached the Fed hike probabilities quoted on Bloomberg on the 24th of January 2018. The second column shows hike probability, while the third and fourth shows cut probability and probability of no change, respectively.
[attachment=0]FED ssh'er 2018-01-24.PNG[/attachment]
I have figured out how to calculate the hike probability (by following http://www.cmegroup.com/education/fed-f ... lator.html), and since the cut probabilities are 0%, I can also calculate the no change probability by following
P(no change) = 100% - P(hike).
But if the cut wasn't 0%, the formula would be
P(cut) + P(hike) + P(no change) = 100%
Does anyone know how to calculate P(cut)?
Attachments
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Last edited by mesk on February 9th, 2018, 12:08 pm, edited 1 time in total.
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https://essaypantry.com/2-study-the-histogram-which-should-show-a-distinctly-bi-modal/ | 1,701,451,764,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00587.warc.gz | 279,105,638 | 18,877 | # 2. Study the histogram, which should show a distinctly bi-modal…
2. Study the histogram, which should show a distinctly bi-modal distribution (i.e., 2 humps).
a) [5 pts] What kind of crust does each of the peaks or modes represent? Print out and then annotate the histogram with a pencil, indicating which parts of the histogram correspond to which kind of crust. Before proceeding with this, refresh your memory about isostasy, which explains the connection between crustal thickness, density, and elevation.
b) [5 pts] Why is one hump broad and the other narrow? Your answer should consider how the crust is created and/or modified by a range of geologic processes.
3. a) What is the average horizontal spatial resolution (in km) of our map at the equator? In other words, how much area do the pixels represent (you can think of a pixel as one square in this grid)? Look in the Workspace window, where you can see the number of rows and columns in ZI, our elevation grid that is used to make the map. It helps to know the latitude and longitude ranges of the data — you can see the minimum and maximum values in the Workspace. Give the spatial dimensions of each pixel at the equator in units of km NS and km EW. It helps to know that 1° of latitude is about 111 km (the equatorial circumference is 40,000 km; the circumference at the poles is 0).
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# CBSE Class 9 Answered
Shyam throws a heavy stoneout of his boat as a result the boat moves in the opposte direction , why?
Asked by kishor.kp14 | 25 Jul, 2019, 03:48: PM
Coordintates of Centre of mass with respect to some reference point do not change if external force is not acting on it.
when the stone is in boat , the point of center of mass also in boat.
This is because from a refernce point, stone and boat are at same distance.
When stone is thrown out, the distance between stone and reference point is getting changed.
Hence centre of mass will shift slightly in the direction of motion of stone.
In order to make the coordinates of centre of mass constant, the boat moves in opposite direction and
thereby Center of mass remains at same position
Answered by Thiyagarajan K | 25 Jul, 2019, 04:18: PM
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Asked by ayushyadav70303 | 12 Jul, 2022, 07:32: AM | 404 | 1,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.917879 |
https://de.mathworks.com/matlabcentral/cody/problems/470-scoring-for-oriented-dominoes/solutions/1179539 | 1,579,335,995,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592394.9/warc/CC-MAIN-20200118081234-20200118105234-00287.warc.gz | 409,034,324 | 15,749 | Cody
# Problem 470. Scoring for oriented dominoes
Solution 1179539
Submitted on 3 May 2017 by arvid Martens
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
list = [1 2; 5 4; 2 4]; order = [1 3 2]; y_correct = 1; assert(isequal(scoreOrderedDominoes(list, order),y_correct))
list = 1 2 2 4 5 4 score = 1
2 Pass
list = [1 2; 5 4; 2 4]; order = [2 1 3]; y_correct = 3; assert(isequal(scoreOrderedDominoes(list, order),y_correct))
list = 5 4 1 2 2 4 score = 3
3 Pass
list = [1 2; 5 4; 2 4; 1 6; 7 5; 4 5; 4 8]; order = [2 1 3 7 4 6 5]; y_correct = 14; assert(isequal(scoreOrderedDominoes(list, order),y_correct))
list = 5 4 1 2 2 4 4 8 1 6 4 5 7 5 score = 14
4 Pass
list = [1 2; 6 6; 2 4]; order = [2 1 3]; y_correct = 5; assert(isequal(scoreOrderedDominoes(list, order),y_correct))
list = 6 6 1 2 2 4 score = 5
5 Pass
list = [1 2; 5 4; 2 4]; order = [1 2 3]; y_correct = 5; assert(isequal(scoreOrderedDominoes(list, order),y_correct))
list = 1 2 5 4 2 4 score = 5 | 469 | 1,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2020-05 | latest | en | 0.485717 |
https://answersherald.com/how-many-ounces-in-a-pint-of-blood/ | 1,679,740,087,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945323.37/warc/CC-MAIN-20230325095252-20230325125252-00043.warc.gz | 124,214,451 | 17,278 | Answers To All Questions In The World
# How Many Ounces in a Pint?
#### ByJohn Amelia
Dec 6, 2021
If you are wondering how many ounces in a pinted beverage, you’ve come to the right place. There are many ways to convert the measurement of liquids, and knowing how to change it can help you make the best possible decision. How Many Ounces in a Pint? The most common unit of volume is the pint. Its symbol is pt. It is equal to an eighth of a gallon, and is used in both the United States and British systems. A British liquid-only pint is approximately 20% larger than its American counterpart.
### Measurements For Liquids:
The US and British systems use different measurements for liquids. A pint contains 16 fluid ounces. To convert imperial pints to US fluid ounces, multiply the US volume by 16. A half-pint is equal to 8 fluid ounces (or 0.2 liters). A pint of lager, with a dash of lemonade, is the same as one cup of water.
### Ounces In A Pint In Different Countries:
In the UK, a pint has 20 fluid ounces, while the US liquid pint is 16 ounces. A half-pint contains eight fluid oz. (one cup). A half-pint is equal to 16 oz., or 0.2 liters. Both units are used for liquids and solids. As a bartender, you should know how to measure liquids in both systems.
In the US, a pint is 16 ounces. In the UK, a punnet is a smaller version of a pint. A US pint is also called a punnet, but that term only applies to liquids. In the UK, a pint is equal to a large basket of fruit or tomatoes, and an Imperial pint is equivalent to a small bowl of nuts.
In the UK, a pint of beer weighs over a pound, which is about half of a gallon. In the US, a pint is equal to 16 ounces, while a US pint contains a pound and a half. A cup contains four ounces of liquid. For comparison, a pint of ice cream is equal to a quart.
### A Pint Of Blueberries:
A pint of blueberries, for example, is approximately equivalent to two cups, or 12 ounces. However, the units of measurement for other fruit vary. For example, a pint of blueberries is significantly smaller than a pint of strawberries, which are bigger and juicier. In addition, a pint of strawberries will contain a smaller number of ounces than a pound.
If you’re wondering how many ounces are in a pint, you should remember that a single can of blueberries contains approximately two cups. When you’re preparing recipes, make sure to measure in ounces, rather than pints. If you’re using imperial measurements, a pint is equivalent to 20 fluid oz. If you’re converting the measurements between a US cup and a British pint, it’s best to convert them.
### How To Calculate The Amount Of Liquid In A Pint?
A pint is a container used to measure volume. To calculate the amount of liquid in a pint, you must multiply the pint’s volume by sixteen. If you’re using the US system, then the pint is equal to eighteen fluid oz. Alternatively, you can use a fluid oz. instead of a pint. A US standard oz. is the same as a UK one.
### Examples:
If you’re looking for a pint of Ben & Jerry’s, you should know that it contains forty grams of saturated fat. A pint of Ben & Jerry’s can spike your blood’s fat levels. It’s important to note that a pint contains about one-tenth of an adult’s blood. For this reason, the volume of an American pint is equal to about a quarter of a gallon.
If you’re looking for a drink that contains more liquid than you need, you should know the volume of the beverage. A pint is a common beverage, but it can also be a little confusing. The smallest liquid, like a beer, will be eight ounces. A full pint of sour cream will contain 16 oz. It’s worth noting that a bottle of sour cream is about a quarter of a litre.
#### John Amelia
Hey, John here, a content writer. Writing has always been one of the things that I’m passionate about. Whenever I have something on my mind, I would jot it down or type it in my notes. No matter how small or pathetic it seems, You will really enjoy my writing.
#### Girly Tattoo Ideas
error: Can not copy! | 952 | 4,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-14 | latest | en | 0.932642 |
http://www.dreamincode.net/forums/topic/283307-understanding-selection-sort/page__pid__1647756__st__0 | 1,441,248,379,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645298065.68/warc/CC-MAIN-20150827031458-00218-ip-10-171-96-226.ec2.internal.warc.gz | 406,147,907 | 20,409 | # Understanding selection sort.
Page 1 of 1
## 1 Replies - 1030 Views - Last Post: 20 June 2012 - 12:44 PM
### #1 Harro1
Reputation: 0
• Posts: 1
• Joined: 19-June 12
# Understanding selection sort.
Posted 19 June 2012 - 08:04 PM
Hi I am having not sure if I am on the right track and could use some assistance before submitting my assisgnment.
I have an array of numbers that I need to create an algorithm for the selection sort, show it as pseudocode OR a flowchart, and desk check it before submitting. (23, 19, 35, 12, 30)is there something more I should fix or add?
```For K= 0 to n – 1
MAX = 0
For j = 0 to n – k - 1
If item[j] > item[MAX]
MAX = item[j]
end if
end for
swap item[n - k - 1] with item[MAX]
end for
end SelectionSort
```
Is This A Good Question/Topic? 0
## Replies To: Understanding selection sort.
### #2 macosxnerd101
• Games, Graphs, and Auctions
Reputation: 11343
• Posts: 42,767
• Joined: 27-December 08
## Re: Understanding selection sort.
Posted 20 June 2012 - 12:44 PM
*Moved to Software Development*
You want to assign j to MAX, not item[j]. Remember that j is the index, and item[j] is the element. So if j is 10 and item[j] is 2, swapping item[MAX] is the same as swapping item[item[j]]. | 389 | 1,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2015-35 | longest | en | 0.821113 |
https://community.powerbi.com/t5/Desktop/Calendar-and-Weekly-exchange-rate/td-p/1320829 | 1,611,003,098,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703515235.25/warc/CC-MAIN-20210118185230-20210118215230-00075.warc.gz | 271,966,994 | 101,230 | cancel
Showing results for
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Helper I
## Calendar and Weekly exchange rate
Hello,
I need help with this issue.
I have a weekly exchange rate report, it is downloaded every saturday..
Then in the report if I need a exchange rate of Monday or Tuesday by example is necesary use the last valid exchange rate downloaded.
By Example:
The exchange rate used for the day 07/13/2020 must be 1.131850 (Downloaded on 07/11/2020). that is the last vlid exchange rate
Somebody know how do it?
This is the table i have:
Date Currency From Currency To Date IN JDE Exchange Rate 7/10/2020 EUR USD 7/11/2020 EUR USD 7/11/2020 1.131850 7/12/2020 EUR USD 7/13/2020 EUR USD 7/14/2020 EUR USD 7/15/2020 EUR USD 7/16/2020 EUR USD 7/17/2020 EUR USD 7/18/2020 EUR USD 7/18/2020 1.142250 7/19/2020 EUR USD 7/20/2020 EUR USD 7/21/2020 EUR USD 7/22/2020 EUR USD 7/23/2020 EUR USD 7/24/2020 EUR USD 7/25/2020 EUR USD 7/25/2020 1.162550
1 ACCEPTED SOLUTION
Super User IV
@gomezc73 , locate the attachment with the solution.
Proud to be a Super User!
7 REPLIES 7
Super User IV
@gomezc73 , locate the attachment with the solution.
Proud to be a Super User!
Helper I
Hello, I tried using
New exchange Rate = maxx(filter('Table', [Date]<=EARLIERER('Table'[Date])),LASTNONBLANKVALUE('Table'[Date IN JDE],max('Table'[ Exchange Rate ]))))
And it works only with a "Currency From", By Example if the tables is only From EUR to USD.
But when i added other Currencys in the tables the formula doesn't work. It bring the MAX Exchange rates From other Currencys..
How The formula can be improve to consider the "Currency From"?
` `
Helper I
`Perfect!!!. That's just what i needed.. You are a genious!!`
Super User III
Hi,
Regards,
Ashish Mathur
http://www.ashishmathur.com
Helper I
Hello
Yes, the first column is the calendar date, and the fourth column is the date on which the Exchange Rate was obtained.
when I search for
• the calendar date,
• from and
• CURRENCY UP,
it only gets me the exchange rate of the Saturdays and the rest of days is blank, so I need to repeat that exchange rate for the next few days, until I get the exchange rate for the next Saturday and so on.
Calendar Date Currency From Currency To Download Date Exchange Rate 1/6/2018 AED USD 1/6/2018 0.272300 1/7/2018 AED USD 0.272300 1/8/2018 AED USD 0.272300 1/9/2018 AED USD 0.272300 1/10/2018 AED USD 0.272300 1/11/2018 AED USD 0.272300 1/12/2018 AED USD 0.272300 1/13/2018 AED USD 1/13/2018 0.272300 1/14/2018 AED USD 0.272300 1/15/2018 AED USD 0.272300 1/16/2018 AED USD 0.272300 1/17/2018 AED USD 0.272300 1/18/2018 AED USD 0.272300 1/19/2018 AED USD 0.272300 1/20/2018 AED USD 1/20/2018 0.272300
Super User IV
@gomezc73 - Can you use Fill Down in Power Query? Otherwise, if you are doing a column, you could do something like:
``````Final Exchange Rate =
IF(
[Exchange Rate] <> "",
[Exchange Rate],
VAR __Date = MAXX(FILTER('Table',[Calendar Date] < EARLIER([Calendar Date]) && [Currency From] = EARLIER([Currency From]) && [Currency To] = EARLIER([Currency To]) && [Exchange Rate] <> ""),[Calendar Date])
RETURN MAXX(FILTER('Table',[Calendar Date] = __Date),[Exchange Rate])
``````
Somehow though, not sure we are all starting from the same place you are.
---------------------------------------
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Super User IV
@gomezc73 - You are basically looking at Lookup Min/Max kind of - https://community.powerbi.com/t5/Quick-Measures-Gallery/Lookup-Min-Max/m-p/985814#M434
``````Measure =
VAR __Date = MAX('Table'[Date])
VAR __RateDate = MAXX(FILTER('ExchangeRates',[Date]<=__Date && [Exchange Rate] <> ""),[Date])
RETURN
MAXX(FILTER('ExchangeRates',[Date] = __RateDate),[Exchange Rate])``````
---------------------------------------
##### I have a NEW book! DAX Cookbook from Packt
Over 120 DAX Recipes!
Proud to be a Super User!
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#### Happy New Year from Power BI
This is a must watch for a message from Power BI! | 1,254 | 4,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.805142 |
https://stats.stackexchange.com/questions/8375/survival-model-for-predicting-churn-time-varying-predictors | 1,653,743,451,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00226.warc.gz | 601,848,896 | 69,879 | # Survival Model for Predicting Churn - Time-varying predictors?
I am looking to build a predictive model for predicting churn and looking to use a discrete time survival model fitted to a person-period training dataset (one row for each customer and discrete period they were at risk, with an indicator for event – equaling 1 if the churn happened in that period, else 0).
• I am fitting the model using ordinary logistic regression using the technique from Singer and Willet.
• The churn of a customer can happen anywhere during a month, but it is only at the end of the month that we know about it (i.e. sometime during that month they left). 24 months is being used for training.
• The time variable being used is the origin time of the sample - all customers active as of 12/31/2008 - they all receive t=0 as of Jan 2009 (not the classical way to do it, but I believe the way when building a predictive model versus a traditional statistical one). A covariate used is the tenure of the customer at that point in time.
• There are a series of covariates that were constructed – some that do not change across the rows of the dataset (for a given customer) and some that do.
• These time variant covariates are the issue and what is causing to me question a survival model for churn prediction (compared to a regular classifier that predicts churn in the next x months based on current snapshot data). The time-invariant ones describe activity the month prior and are expected to be important triggers.
The implementation of this predictive model, at least based on my current thinking, is to score the customer base at the end of each month, calculating the probability / risk of churn sometime during the next month. Then again for the next 1,2 or 3 months. Then for the next 1,2,3,4,5,6 months. For the 3 and 6 month churn probability, I would be using the estimated survival curve.
The problem:
When it comes to thinking about scoring, how can I incorporate time-varying predictors? It seems like I can only score with time-invariant predictors or to include those that are time invariant, you have to make them time invariant – set to the value “right now”.
Does anyone have experience or thoughts on this use of a survival model?
Update based on @JVM comment:
The issue is not with estimating the model, interpreting coefficients, plotting the hazard/survival plots of interesting covariate values using the training data etc. The issue is in using the model to forecast risk for a given customer. Say at the end of this month, I want to score everyone who is still an active customer with this model. I want to forecast that risk estimate out x periods (risk of closing the account at the end of next month. risk of closing the account at the end of two months from now, etc.). If there are time varying covariates, their values are unknown out any future periods, so how to utilize the model?
Final Update:
A person period data set will have an entry for each person and each time period they are at risk. Say there are J time periods (maybe J =1...24 for 24 months) Lets say I construct a discrete time survival model, where for simplicity we just treat time T as linear and have two covariates X and Z where X is time-invariant, meaning it is constant in every period for the ith person and Z is time varying, meaning that each record for the ith person can take on a different value. For example, X may be the customers gender and Z might be how much they were worth to the company in the prior month. The model for the logit of the hazard for the ith person in the jth time period is :
$logit(h(t_{ij}))=\alpha_{0}+\alpha_{1}T_{j}+\beta_{1}X_{i}+\beta_{2}Z_{ij}$
So the issue is, when using time varying covariates, and forecasting (into the yet unseen future) with new data, the $Z_{j}$ are unknown.
The only solutions I can think are:
• Don't use time varying covariates like Z. This would greatly weaken the model to predict the event of churning though since, for example, seeing a decrease in Z would tell us the customer is disengaging and perhaps preparing to leave.
• Use time varying covariates but lag them (like Z was above) which allows us to forecast out however many periods we have lagged the variable (again, thinking of the model scoring new current data).
• Use time varying covariates but keep them as constants in the forecast (so the model was fitted for varying data but for prediction we leave them constant and simulate how changes in these values, if later actually observed, will impact risk of churning.
• Use time varying covariates but impute their future values based on a forecast from known data. E.g. Forecast the $Z_{j}$ for each customer.
• A couple questions: First, are you wedded to using a discrete time model? What you want to do may be simpler in a parametric survival model. Second, would you be able to provide an example of what you mean by churn; ow maybe just an example of what your data look like? Mar 16, 2011 at 22:33
• @JWM. Churn means the customer canceled their account. Since I know only the month the customer canceled, I think discrete time is appropriate. I believe the same issue with time varying predictors exists if the model is discrete or continuous time (Cox or AFT) No? Mar 17, 2011 at 1:22
• I apologize for still not quite grasping your concern. You should be able to estimate your model well enough. Though, because you will likely have measurement error in your TVC and dependent variable, you should probably use a lag of the TVC in the model. Otherwise, you risk using a covariate value which is measured after the occurrence of the event as a predictor for that event. You can see the time-consistency issue. While I know your goal is prediction, if you want reasonable confidence intervals, you will want to estimate robust standard errors (at least). Mar 23, 2011 at 4:04
• @JVM apparently I have failed to explain myself well. I added a response in the question. Mar 23, 2011 at 17:38
• @B_Miner regarding your four proposed solutions. Isn't there a problem about your t=0? I suppose your are standing at time $s>0$ and want to know $P(T_i>s+x|F_s)$. In other words you are looking for the conditional survival probabilities and not the initial survival curve. Am I right or am I missing something? I just proposed a question on this here: stats.stackexchange.com/questions/346338/… May 15, 2018 at 14:13
## 2 Answers
Thank you for the clarification, B_Miner. I don't do a lot of forecasting myself, so take what follows with a pinch of salt. Here is what I would do as at least a first cut at the data.
• First, formulate and estimate a model that explains your TVCs. Do all of the cross-validation, error checking, etc., to make sure you have a decent model for the data.
• Second, formulate and estimate a survival model (of whatever flavor). Do all of the cross-validation, error checking, to make sure this model is reasonable as well.
• Third, settle on a method of using the forecasts from the TVCs model as the basis of forecasting risks of churn and whatever else you want. Once again, verify that the predictions are reasonable using your sample.
Once you have a model that you think is reasonable, I would suggest bootstrapping the data as a way to incorporate the error in the first TVC model into the second model. Basically, apply steps 1-3 N times, each time taking a bootstrap sample from the data and producing a set of forecasts. When you have a reasonable number of forecasts, summarize them in any way you think is appropriate for your task; e.g., provide mean risk of churn for each individual or covariate profile of interest as well as 95% confidence intervals.
• @JVM. Do I read you correctly that your suggestion is my last solution option above (in the final update). That you develop models to forecast each Z (TVC) to use in the actual survival model? Mar 27, 2011 at 16:13
As I see it there are two survival analysis paradigms that could be used. The Cox regression framework allows time varying covariates and would produce an estimate for the risk of cancellation conditioned on any particular set of covariates relative to the mean level of cancellation. The glm framework with Poisson errors is also a proportional hazards model and is particularly suited to discrete intervals. JVM has pointed out that there is potential error in using incomplete data in the current month, but the sense I get is that you want an estimate that is conditional on the latest value of a co-variate or set of covariates. Better description of the data situation could yield better worked examples.... | 1,935 | 8,627 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | longest | en | 0.949873 |
http://stackoverflow.com/questions/10350115/what-is-difference-between-array-and-arraylist-to-get-prime-numbers/10350163 | 1,455,433,959,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701169261.41/warc/CC-MAIN-20160205193929-00049-ip-10-236-182-209.ec2.internal.warc.gz | 213,908,017 | 19,641 | # what is difference between Array and ArrayList to get Prime numbers?
i was solving problem about the prime numbers, i found solution using Arrays, but i wanted to use the array list for some issues, one of them is to understand the Arraylist well. but i found that the results are not the same when i use the arraylist, here are the 2 Codes:
//using Arrays
import java.util.Arrays;
public class Prime {
public static void main(String[] args) {
}
public static void primeNumber(int end) {
for (int i = 0; i < end; i++) {
}
for (int i = 2; i < primeNumbers.length; i++) {
for (int j = i+i; j < primeNumbers.length; j += i) {
}
}
}
for (int j = 2; j < primeNumbers.length; j++) {
System.out.println(j);
}
}
}
}
//Using ArrayList
import java.util.ArrayList;
public class Prime {
public static void main(String[] args) {
}
public static void primeNumber(int end) {
for (int i = 0; i < end; i++) {
}
for (int i = 2; i < primeNumbers.size(); i++) {
for (int j = i+i; j < primeNumbers.size(); j += i) {
}
}
}
for (int j = 2; j < primeNumbers.size(); j++) {
System.out.println(j);
}
}
}
}
Can i know what is the wrong i did with using the Arraylist, and if i wanted to use the arraylist for the same result what should i do.
Great thanks for the help.
-
You don't need import java.util.Arrays; – Rosdi Kasim Apr 27 '12 at 12:05
you should use
since add() adds a new element at the specified position (meaning the ArrayList grows by one element), but set() only sets the value of the element at specified position.
-
Great thanks, really fast and simple solution. – Ahmed Nassar Apr 27 '12 at 11:59 | 432 | 1,605 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2016-07 | longest | en | 0.690265 |
https://us.metamath.org/nfeuni/f1ococnv1.html | 1,716,687,996,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058861.60/warc/CC-MAIN-20240526013241-20240526043241-00603.warc.gz | 520,688,614 | 4,830 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > f1ococnv1 GIF version
Theorem f1ococnv1 5310
Description: The composition of a one-to-one onto function's converse and itself equals the identity relation restricted to the function's domain. (Contributed by set.mm contributors, 13-Dec-2003.)
Assertion
Ref Expression
f1ococnv1 (F:A1-1-ontoB → (F F) = ( I A))
Proof of Theorem f1ococnv1
StepHypRef Expression
1 cnvcnv 5062 . . 3 F = F
21coeq2i 4877 . 2 (F F) = (F F)
3 f1ocnv 5299 . . 3 (F:A1-1-ontoBF:B1-1-ontoA)
4 f1ococnv2 5309 . . 3 (F:B1-1-ontoA → (F F) = ( I A))
53, 4syl 15 . 2 (F:A1-1-ontoB → (F F) = ( I A))
62, 5syl5eqr 2399 1 (F:A1-1-ontoB → (F F) = ( I A))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1642 ∘ ccom 4721 I cid 4763 ◡ccnv 4771 ↾ cres 4774 –1-1-onto→wf1o 4780 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-13 1712 ax-14 1714 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 ax-nin 4078 ax-xp 4079 ax-cnv 4080 ax-1c 4081 ax-sset 4082 ax-si 4083 ax-ins2 4084 ax-ins3 4085 ax-typlower 4086 ax-sn 4087 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-nan 1288 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-eu 2208 df-mo 2209 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-ne 2518 df-ral 2619 df-rex 2620 df-reu 2621 df-rmo 2622 df-rab 2623 df-v 2861 df-sbc 3047 df-nin 3211 df-compl 3212 df-in 3213 df-un 3214 df-dif 3215 df-symdif 3216 df-ss 3259 df-pss 3261 df-nul 3551 df-if 3663 df-pw 3724 df-sn 3741 df-pr 3742 df-uni 3892 df-int 3927 df-opk 4058 df-1c 4136 df-pw1 4137 df-uni1 4138 df-xpk 4185 df-cnvk 4186 df-ins2k 4187 df-ins3k 4188 df-imak 4189 df-cok 4190 df-p6 4191 df-sik 4192 df-ssetk 4193 df-imagek 4194 df-idk 4195 df-iota 4339 df-0c 4377 df-addc 4378 df-nnc 4379 df-fin 4380 df-lefin 4440 df-ltfin 4441 df-ncfin 4442 df-tfin 4443 df-evenfin 4444 df-oddfin 4445 df-sfin 4446 df-spfin 4447 df-phi 4565 df-op 4566 df-proj1 4567 df-proj2 4568 df-opab 4623 df-br 4640 df-co 4726 df-ima 4727 df-id 4767 df-xp 4784 df-cnv 4785 df-rn 4786 df-dm 4787 df-res 4788 df-fun 4789 df-fn 4790 df-f 4791 df-f1 4792 df-fo 4793 df-f1o 4794 This theorem is referenced by: f1cocnv1 5312 f1ocnvfv1 5476 enmap2lem3 6065 enmap1lem3 6071
Copyright terms: Public domain W3C validator | 1,351 | 2,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-22 | latest | en | 0.118619 |
https://www.numbersaplenty.com/147987 | 1,701,471,083,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00412.warc.gz | 1,039,750,022 | 3,294 | Search a number
147987 = 36729
BaseRepresentation
bin100100001000010011
321112000000
4210020103
514213422
63101043
71154310
oct441023
9245000
10147987
11a1204
1271783
1352488
143bd07
152dcac
hex24213
147987 has 28 divisors (see below), whose sum is σ = 262320. Its totient is φ = 81648.
The previous prime is 147977. The next prime is 147997. The reversal of 147987 is 789741.
It is an interprime number because it is at equal distance from previous prime (147977) and next prime (147997).
It is not a de Polignac number, because 147987 - 27 = 147859 is a prime.
It is a Smith number, since the sum of its digits (36) coincides with the sum of the digits of its prime factors.
It is an inconsummate number, since it does not exist a number n which divided by its sum of digits gives 147987.
It is not an unprimeable number, because it can be changed into a prime (147937) by changing a digit.
It is a polite number, since it can be written in 27 ways as a sum of consecutive naturals, for example, 5089 + ... + 5117.
2147987 is an apocalyptic number.
147987 is a deficient number, since it is larger than the sum of its proper divisors (114333).
147987 is an frugal number, since it uses more digits than its factorization.
147987 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 54 (or 39 counting only the distinct ones).
The product of its digits is 14112, while the sum is 36.
The square root of 147987 is about 384.6907849169. The cubic root of 147987 is about 52.8941759329.
It can be divided in two parts, 14 and 7987, that added together give a triangular number (8001 = T126).
The spelling of 147987 in words is "one hundred forty-seven thousand, nine hundred eighty-seven". | 495 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-50 | latest | en | 0.894575 |
https://www.physicsforums.com/threads/i-really-need-help-fast-i-dont-know-what-im-doing-please-help.24217/ | 1,519,468,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815544.79/warc/CC-MAIN-20180224092906-20180224112906-00145.warc.gz | 920,492,374 | 14,407 | 1. May 5, 2004
### BananaRed
Problem 1. A 5.00-kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed 0.100m from equilibrium and released. The speed of the block is 1.20m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is 0.300. Determine the speed of the block at the eqilibrium position of the spring.
The second problem I have is: The reel has radius R and moment of inertia I. One end is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest. Find the angular speed of the reel when the spring is again unstretched.
If someone could be help me I would really appreciate it
2. May 5, 2004
### gnome
Don't post the same question several times. (It confuses us.) | 276 | 1,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-09 | longest | en | 0.934973 |
https://www.reddit.com/r/HomeworkHelp/comments/14knze/how_do_you_put_y_tan2xpi2_in_a_ti_83_plus/ | 1,519,519,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816068.93/warc/CC-MAIN-20180224231522-20180225011522-00610.warc.gz | 909,897,154 | 22,357 | ×
[–] 0 points1 point (0 children)
tan(2(x+(pi/2)))
Remember that whenever you write a trigonometric function, there's always supposed to be parenthesis, but sometimes it's implied.
tanx
is really
tan(x)
therefore:
tan2(x+(pi/2))
is really
tan(2(x+(pi/2)))
[–] 0 points1 point (0 children)
Are you trying for tan(2x) or tan2 (x)?
[–] 0 points1 point (0 children)
I think you mean tan<squared>(x+pi/2).
Instead of that extra 2 try this:
(tan(x+(pi/2)))*2 (where * is the exponent key by the clear button) | 171 | 525 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2018-09 | latest | en | 0.884692 |
http://projecteuler.net/problem=197 | 1,409,648,971,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535921872.11/warc/CC-MAIN-20140909033921-00174-ip-10-180-136-8.ec2.internal.warc.gz | 443,781,553 | 2,152 | ## Investigating the behaviour of a recursively defined sequence
### Problem 197
Published on Friday, 6th June 2008, 10:00 pm; Solved by 3080
Given is the function f(x) = ⌊230.403243784-x2⌋ × 10-9 ( ⌊ ⌋ is the floor-function),
the sequence un is defined by u0 = -1 and un+1 = f(un).
Find un + un+1 for n = 1012. | 112 | 315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2014-35 | latest | en | 0.863539 |
http://gmatclub.com/forum/a-five-digit-number-divisible-by-3-is-to-be-formed-using-136900.html?fl=similar | 1,485,210,993,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560283008.19/warc/CC-MAIN-20170116095123-00256-ip-10-171-10-70.ec2.internal.warc.gz | 113,555,103 | 61,026 | A five-digit number divisible by 3 is to be formed using : GMAT Problem Solving (PS)
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# A five-digit number divisible by 3 is to be formed using
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A five-digit number divisible by 3 is to be formed using [#permalink]
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06 Aug 2012, 08:23
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A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:
A. 122
B. 210
C. 216
D. 217
E. 225
[Reveal] Spoiler: OA
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]
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06 Aug 2012, 08:30
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miteshsholay wrote:
A five-digit number divisible by 3 is to be formed using numerical 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways this can be done is:
A. 122
B. 210
C. 216
D. 217
E. 225
First step:
We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
We have six digits: 0, 1, 2, 3, 4, 5. Their sum=15.
For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
Second step:
We have two set of numbers:
1, 2, 3, 4, 5 and 0, 1, 2, 4, 5. How many 5 digit numbers can be formed using these two sets:
1, 2, 3, 4, 5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
0, 1, 2, 4, 5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, desired # would be total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96
120+96=216
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]
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Re: A five-digit number divisible by 3 is to be formed using [#permalink]
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29 Feb 2016, 18:59
we can eliminate the incorrect answer choices by applying logic.
we definitely know that for number to be divisible, we must have the sum of the digits divisible by 3.
so we can use all the digits except for 0.
we can arrange these in 5! ways, or 120 ways.
now...
5+4+1+2+0 = 12, divisible by 3.
we can see that we can arrange these numbers in more than 2 ways. thus, A is eliminated.
B = 210-120 = 90. hold
C = 216-120 = 96 - hold
D - 217-120 = 97 - looks more like a prime number..not divisible by 2, by 3, by 5, by 7.
E - 225-120=105. 105=3*35=3*3*7 - we can't have 7, as we have max 6 digits in total, so definitely out.
between B and C:
90=3*3*2*5
96=4*4*3*2
well, we definitely can't use in the combination for 0,2,1,4,5 - 0 for tens of thousands, thus, we'll have max 4 possible ways to arrange for this digit.
B is out, and C remains.
this all takes less than 1 min to figure out.
Re: A five-digit number divisible by 3 is to be formed using [#permalink] 29 Feb 2016, 18:59
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Display posts from previous: Sort by | 1,956 | 6,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-04 | latest | en | 0.855919 |
http://mathhelpforum.com/advanced-algebra/140687-inverse-matrix.html | 1,513,596,029,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00019.warc.gz | 178,064,110 | 11,378 | # Thread: Inverse of a matrix?
1. ## Inverse of a matrix?
Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.
$2x_1 + 3x_2 = 1\ and\ x_1 + 7x_2 = 6$. I have put this in the form Ax = b as asked by the question.
$\left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right]$.
I know that if I multiply both sides by the inverse I get $x = A^{-1}b$. To find the inverse from here $A^{-1} = \frac{1}{det(A)}\cdot A_{cof}T$ (where the T is for transposed).
So $A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$
I calculated the det(A) as being 11 (14 -3) so $A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$
If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there
Thanks
2. Originally Posted by Beard
Hi, I'm having a problem finding the inverse of a matrix in that I have no idea where I am going wrong. I am given two simultaneous equations and told to find the inverse of A and therefore determine the value of x.
$2x_1 + 3x_2 = 1\ and\ x_1 + 7x_2 = 6$. I have put this in the form Ax = b as asked by the question.
$\left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right] = \left[\begin{array}{c}1\\6\end{array}\right]$.
I know that if I multiply both sides by the inverse I get $x = A^{-1}b$. To find the inverse from here $A^{-1} = \frac{1}{det(A)}\cdot A_{cof}T$ (where the T is for transposed).
So $A = \left[\begin{array}{cc}2&3\\1&7\end{array}\right] \ \ A_{cof} = \left[\begin{array}{cc}2&-1\\-3&7\end{array}\right] \ \ A_{cof}T = \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$
I calculated the det(A) as being 11 (14 -3) so $A^{-1} = \frac{1}{11} \cdot \left[\begin{array}{cc}2&-3\\-1&7\end{array}\right]$
If something hasn't already gone wrong then here is where it doesn't make sense from me. Just to make sure that it was correct I decided to multiply A by the inverse I just got and to see if I got the identity matrix. I didn't. Can anyone help me in finding where I have gone wrong. I can do the rest from there
Thanks
Your error is in finding $A_{cof}T$, which should be::
$\left[\begin{array}{cc}7&-3\\-1&2\end{array}\right]$
3. Hello, Beard!
Your inverse matrix is wrong . . .
given: . $\begin{array}{ccc} 2x + 3y &=& 1 \\ x + 7y &=& 6 \end{array}$
I have put this in the form Ax = b as asked by the question.
. . $\left[\begin{array}{cc}2&3\\1&7\end{array}\right] \left[\begin{array}{c}x \\ y\end{array}\right] \:=\: \left[\begin{array}{c}1\\6\end{array}\right]$
So: . $A^{-1} \:=\: \frac{1}{11} \cdot \left[\begin{array}{cc}2&\text{-}3\\\text{-}1&7\end{array}\right]$ . . . . no
There is a "formula" for the inverse of a 2-by-2 matrix.
We have: . $\left[\begin{array}{cc} a&b \\ c&d \end{array}\right]$
[1] Switch the elements on the main diagonal: . $\left[\begin{array}{cc}{\color{red}d}&b\\c&{\color{red}a }\end{array}\right]$
[2] Change the signs on the other diagonal: . $\left[\begin{array}{cc}d & {\color{red}\text{- }}\!b \\ {\color{red}\text{- }}\!c & a\end{array}\right]$
[3] Divide by the determinant: . ${\color{red}\frac{1}{ad-bc}}\,\left[\begin{array}{cc} d & \text{- }\!b \\ \text{- }\!c & a\end{array}\right]$
Therefore: . $\left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1} \;=\;\frac{1}{ad-bc}\,\left[\begin{array}{cc}d&\text{- }\!b \\ \text{- }\!c & a \end{array}\right]$
4. I know can see how I went wrong, the examples I was trying to follow had ones for a and c looking at sorobans method so I was unable to distinguish.
Thanks | 1,470 | 4,127 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-51 | longest | en | 0.896224 |
https://gmatclub.com/forum/jane-started-baby-sitting-when-she-was-18-years-old-whenever-she-baby-118718.html?kudos=1 | 1,723,048,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00518.warc.gz | 223,769,507 | 129,776 | Last visit was: 07 Aug 2024, 09:31 It is currently 07 Aug 2024, 09:31
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Jane started baby-sitting when she was 18 years old. Whenever she baby
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Re: Jane started baby-sitting when she was 18 years old. Whenever she baby [#permalink]
Jane 18 22 32
Babies <=9
oldest baby = 9 year old baby that she could have baby sat when she was 18.
oldest baby's current age = 9+14 (as its been 14 years since she sat that baby) = 23
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Re: Jane started baby-sitting when she was 18 years old. Whenever she baby [#permalink]
ruturaj wrote:
Jane started baby-sitting when she was 18 years old. Whenever she baby-sat for a child, that child was no more than half her age at the time. Jane is currently 32 years old, and she stopped baby-sitting 10 years ago. What is the current age of the oldest person for whom Jane could have baby-sat?
A. 21
B. 22
C. 23
D. 24
E. 25
Check two extreme cases:
Jane = 18, child = 9, years ago = 32-18 = 14 --> child's age now = 9+14 = 23;
Jane = 22, child = 11, years ago = 32-22 = 10 --> child's age now = 11+10 = 21.
Open discussion of this question is here: https://gmatclub.com/forum/jane-started ... 14749.html
Re: Jane started baby-sitting when she was 18 years old. Whenever she baby [#permalink]
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# Effets sur la médiane et la moyenne : supprimer une valeur extrême
## Transcription de la vidéo
and I played five rounds of golf and her lowest score was an 80 the firt the scores of the first four rounds and the lowest round are shown in the following dot plot and we see it right over here the lowest round she scores an 8000 scores a 90 once a 92 once and 94 once and a 96 once it was discovered that Ana broke some rules when she scored 80 so that scores cheating didn't help her so that score will be removed from the data set so they removed that 80 right over there were just left from the with the scores from the other four rounds how will the removal of the lowest round affect the mean and the median so let's actually think about the median first so the median is the middle number so over here when you had five data points the middle data point is going to be the one that has two to the left and two to the right so the median up here is going to be ninety two the median up he there is ninety two and what's the median once you remove this now you only have four data points when you're trying to find the median of an even number of numbers you look at the middle two numbers so that's a 92 and a 94 and then you take the average of them you go halfway between them to figure out the median so the median here is going to be let me do that a little bit clearer the median over here is going to be halfway between 92 and 94 which is 93 so the median the median is 93 median is 93 so removing the lowest data point in this case increase the median so the median let me write down here so the median increased by a little bit the median increases now what's going to happen to the mean what's going to happen to the mean well one way to think about it without even doing any calculations is if you remove a number that is lower than the mean lower than the existing mean and I haven't calculated what the existing mean is but if you remove that the mean is going to go up the mean is going to go up so hopefully that gives you some intuition if you removed a number that's larger than the mean your mean is your mean is going to go down because you don't have that large number anymore if you have remove a number that's lower than the mean well you take that out you don't have that that small number bringing the average down and so the mean will go up but let's verify it mathematically so let's calculate the mean over here so we're going to add 80 plus 90 plus 90 to plus 90 4 plus 96 those are our data points and that gets us 2 + 4 6 plus 6 is 12 and then we have 1 plus 8 is 9 and we essentially this is so these are 9 and you have another 9 another night another night another 9 you essentially have is this five nines right over here so this is going to be 450 450 - so that's the sum of the scores of these five rounds and then you divide it by the number of rounds you have so it'd be 450 2 divided by 5 so 450 2 divided by 5 is going to give us five goes into doesn't go into 4 it goes into 45 9 times 9 times 5 is 45 you subtract you get 0 bring down the 2 5 goes into 2 0 times 0 times 5 is 0 2 times 5 is 0 subtract you have 2 left over so you can say that the mean here the mean here is 90 and two-fifths maybe not 9 into fifths 90 and two-fifths so the mean is right around here so that's the mean of these data points right over there and if you remove it what is the mean going to be so here we're just going to take our 90 plus our 92 plus our 94 plus our 96 add them together so let's see 2 plus 4 plus 6 is 12 and then you add these together going to get 37 372 divided by 4 because I have four data points now not five four goes into three let me do this in a place where you can see it so 4 goes into 372 goes into thirty seven nine times 9 times 4 is 36 subtract you know one bring down the two goes exactly three times three times four is 12 you have no remainder so the median and the mean here are both so this is also the mean the mean here is also 93 so you see that the median the median went from 92 to 93 it increased the mean went from 90 and two-fifths to 93 so the mean increased by more than the median they both increased but the mean increased by more and it makes sense because you this number was way way below all of these over here so you can imagine if you take this out the median should increase by a good amount but let's see which of these choices are what we just described but the mean and the median will decrease nope both the mean and the median will decrease nope both the mean and the median will increase but the mean will increase by more than the median that's exactly that's exactly what happened the mean went from 90 and 2/5 or 90.4 went from ninety point four or 90 and two-fifths to 93 and then the median only increased by one so this is the right answer | 1,214 | 5,081 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2021-21 | latest | en | 0.926695 |
https://muangsingtravel.com/southeast-asian-countries/how-do-you-calculate-yearly-salary-in-the-philippines.html | 1,653,175,121,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00570.warc.gz | 469,736,631 | 18,070 | # How do you calculate yearly salary in the Philippines?
Contents
## How do you calculate annual income in the Philippines?
To convert to annual income:
1. Hourly: Multiply by 2,000.
2. Daily: Multiply by 200.
3. Weekly: Multiply by 50.
4. Monthly: Multiply by 12.
## How do you calculate your yearly salary?
To determine your weekly salary, multiply your hourly salary by the number of hours you worked in each week. Since there are 52 weeks per year, multiply that number by 52. This will result in your annual salary. If you worked 40 hours in one week, at an hourly rate of \$10 per hour, you would be making \$20,800 annually.
## How do you calculate your salary step by step?
How to Calculate your Take-Home Salary?
1. Step 1- Calculate Gross Salary. Gross Salary= Basic Salary + HRA + Other Allowances. …
2. Step 2- Calculate Taxable Income. Taxable Income = Income (Gross Salary + other income) – Deductions. …
3. Step 3- Calculate Income Tax. …
4. Step 4- Calculating take-home salary.
## How do you calculate yearly salary from monthly salary?
Your total yearly take-home salary = gross salary – total deductions = ₹9.50 lakhs – ₹48,700 = ₹9,01,300. Now, your monthly take-home salary = annual salary/12 = ₹9,01,300/12 = ₹75,108. To do away with the tedious calculations, most people prefer the take-home salary calculator in India.
ЭТО ИНТЕРЕСНО: What extent English law is applicable in Malaysia?
## How do you calculate monthly salary in the Philippines 2020?
For monthly-paid employees, here is how it is computed:
1. (Basic Monthly Salary x 12) / (Total Working Days in a Year) = DAILY RATE.
2. Important: The total working days in a year (TWD) may vary from one employee to another. …
3. Basic Salary: ₱ 23,000.
4. Total Working Days in a Year: 261. …
5. (23,000 x 12) / (261) = ₱ 1,057.47.
## How is payslip calculated?
Rules for calculating payroll taxes
1. Income Tax formula for FY (2019 – 2020) = (Basic + Allowances – Deductions) * 12 – (IT Declarations + Standard deduction)
2. Deductions are the sum of PF, ESI, and PT etc.
3. TDS is calculated on (Basic + Allowances – Deductions) * 12 – (IT Declarations + Standard deduction)
## How is monthly salary calculated formula?
If an organization uses 26 as the fixed number of base days each month, an employee who joins on September 21 and whose monthly salary is Rs 26,000, will get paid Rs 10,000 for the 10 days in September; the per-day pay is calculated as Rs 26,000/26 = Rs 1,000.
## What is 5000 a month yearly?
\$5,000 a month is how much per year? If you make \$5,000 per month, your Yearly salary would be \$60,003. This result is obtained by multiplying your base salary by the amount of hours, week, and months you work in a year, assuming you work 40 hours a week. | 731 | 2,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-21 | latest | en | 0.88152 |
https://studylib.net/doc/11901895/12.11-application-of-integration | 1,606,268,105,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00325.warc.gz | 498,960,352 | 13,189 | # 12.11 Application of Integration
```Course: Accelerated Engineering Calculus I
12.11
Instructor: Michael Medvinsky
Application of Integration
12.11.1
Area between curves
Consider that we interesting to compute an area between two curves. How can we do
that? The idea is similar to the one used to find the area of
a
1
1
1
a+b
trapeze. For a example ha − hc = h ( b + c ) − hc = hb + hc = h
h
2
2
2
2
b
c
Let’s rewrite this example so the area will be between 2 curves
⎧x −b
h x≥b
⎪
, thus
y1 = h and y2 = ⎨ a − b
⎪⎩0
else
x−b
h ( x − b)
a
∫0 y1 − y2 dx = ∫0 h dx − ∫b a − b h dx = hx 0 − 2 a − b
a
a
a
2 a
= ha −
b
a−b
a+b
h=h
2
2
Thm: The area of region bounded by continuous curves y=f(x) and y=g(x) is given by
and the lines x=a and x=b, where f(x)>g(x) for all x in [a,b] is given by
b
A = ∫ f ( x ) − g ( x ) dx
a
If f(x) and g(x) are velocities of moving objects and f(a)=g(a) the area is interpreted as
distance between the objects at time x=b.
Ex 1.
To find the area enclosed between parabolas 5 − x 2 and x 2 − 3 we first
need to find the points of intersection, which will be the limits of the integral.
Thus: 5 − x 2 = x 2 − 3 ⇒ 2x 2 = 5 + 3 = 8 ⇒ x 2 = 4 ⇒ x = ±2 and so
2
A=
2
2
2
2
∫ ( 5 − x ) − ( x − 3) dx = 2 ∫ 4 − x dx =
−2
−2
2
⎛
x ⎞
8⎞ ⎛
8⎞⎞
16 ⎞ 64
⎛⎛
⎛
2 ⎜ 4x − ⎟ = 2 ⎜ ⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎟ = 2 ⎜ 16 − ⎟ =
⎝
⎠
⎝
⎠
⎝
⎝
3 ⎠ −2
3
3 ⎠
3⎠ 3
⎝
3
Ex 2.
Draw curves y 2 = 5 − x and y 2 = 3 + x . How to find the areas enclosed
between them? It is possible to split it into 2 subdomains and sum their areas.
The other way is to see that this is the same graph as in previous example
(except that x and y were switched) and to integrate with respect to y.
Course: Accelerated Engineering Calculus I
Ex 3.
Consider we compute the area under a parametric curve given
by x = r cost,
y = r sint from x=-r to x=r:
r
∫ y ( x ) dx
−r
r2
=−
2
12.11.1.1
Instructor: Michael Medvinsky
0
0
0
∫ y (t ) x '(t ) dt = ∫ r sint ( −r sint ) dt = −r ∫ sin
=
2
x=r cost
dx=x '( t )dt=− r sintdt π
π
2
t dt =
π
1 ⎛ sin 2t ⎞
r 2π
1−
cos
2t
dt
=
−r
t
−
=
(
)
⎜
⎟
∫π
2⎝
2 ⎠π
2
0
0
2
Area in Polar coordinates (appendix H) Polar coordinates are given by x = r cosθ ,
y = r sin θ ⇒ r 2 = x 2 + y 2
tan θ =
y
.A
x
straight line (a ray) that passes through origin and making angle θ 0 with polar axis is
given by formula θ = θ 0 .
The rest of Appendix H.1 is for self-reading.
The area enclosed between θ1 and θ 2 and constant radius r is given
1
2
by A = r 2 (θ 2 − θ1 ) . The area enclosed between θ1 and θ 2 and a function r = f (θ ) in
θ2
θ
( )
2
n
1
1
1
polar coordinates is given by A = ∫ r 2 dθ = ∫ f 2 (θ ) dθ = ∑ f 2 θ *j Δθ
2
2
j=1 2
θ1
θ1
Ex 4. Find area enclosed by one loop of the four leaved rose r = cos2 2θ
π /4
π /4
π /4
1
1
1
1
A = ∫ r 2 dθ = ∫ cos 2 2θ dθ = ∫ (1+ cos 4θ ) dθ =
2
2 − π /4
2 − π /4 2
− π /4
=
1
2
π /4
∫
0
π /4
1
1
π
(1+ cos 4θ ) dθ = ⎛⎜⎝ θ + sin 4θ ⎞⎟⎠ =
2
4
8
0
``` | 1,343 | 2,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-50 | latest | en | 0.832524 |
https://www.oreilly.com/library/view/core-php-programming/0130893986/0130893986_ch10lev1sec1.html | 1,563,535,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526210.32/warc/CC-MAIN-20190719095313-20190719121313-00477.warc.gz | 780,983,655 | 6,922 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
### Common Math
The functions in this section offer most of the common mathematical operations that are part of arithmetic, geometry, and trigonometry. Most of these functions work on either doubles or integers. The return type will be the same as the argument. Unless a specific type is called for, I've written "number" to indicate that either an integer or a double is expected.
#### number abs(number value)
The abs function returns the absolute value of a number. This is the number itself if it's positive, or the number multiplied by negative one (−1) if negative.
```<?
//prints 13
print(abs(-13));
?>
```
#### double acos(double value)
The acos function returns the arc cosine of the value argument. Trying to find the arc cosine of a value greater than one or less ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 241 | 1,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2019-30 | latest | en | 0.829281 |
https://oeis.org/A057956 | 1,720,953,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514564.41/warc/CC-MAIN-20240714094004-20240714124004-00764.warc.gz | 378,278,258 | 4,059 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A057956 Number of prime factors of 5^n - 1 (counted with multiplicity). 17
2, 4, 3, 6, 4, 7, 3, 8, 5, 7, 3, 10, 3, 7, 7, 11, 4, 11, 5, 11, 6, 8, 4, 13, 8, 7, 9, 10, 5, 14, 4, 14, 6, 8, 9, 16, 5, 10, 6, 15, 4, 16, 4, 12, 12, 8, 3, 17, 4, 13, 8, 12, 5, 19, 10, 13, 7, 9, 4, 21, 5, 9, 11, 18, 8, 15, 7, 14, 9, 16, 4, 22, 5, 10, 16, 14, 7, 14, 5, 20, 11, 10, 5, 22, 9, 10 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Max Alekseyev, Table of n, a(n) for n = 1..502 (first 448 terms from Amiram Eldar) S. S. Wagstaff, Jr., The Cunningham Project FORMULA Mobius transform of A085030. - T. D. Noe, Jun 19 2003 a(n) = A001222(A024049(n)). - Amiram Eldar, Feb 01 2020 MATHEMATICA PrimeOmega[5^Range[90]-1] (* Harvey P. Dale, Dec 16 2017 *) CROSSREFS bigomega(b^n-1): A057951 (b=10), A057952 (b=9), A057953 (b=8), A057954 (b=7), A057955 (b=6), this sequence (b=5), A057957 (b=4), A057958 (b=3), A046051 (b=2). Cf. A001222, A024049, A074479, A085030. Sequence in context: A341844 A372713 A015049 * A151734 A175027 A269837 Adjacent sequences: A057953 A057954 A057955 * A057957 A057958 A057959 KEYWORD nonn AUTHOR Patrick De Geest, Nov 15 2000 STATUS approved
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Last modified July 14 05:06 EDT 2024. Contains 374291 sequences. (Running on oeis4.) | 708 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-30 | latest | en | 0.692141 |
https://stat.ethz.ch/pipermail/r-sig-mixed-models/2011q1/012443.html | 1,642,398,698,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00647.warc.gz | 590,330,439 | 4,215 | # [R-sig-ME] lme4 convergence problem
Murray Jorgensen maj at stats.waikato.ac.nz
Thu Jan 13 22:39:29 CET 2011
```Dear List & Maintainers,
I am currently running some scripts for some of the data sets in the
book by Andrew Gelman and Jennifer Hill. (2006). (Data Analysis Using
Regression and Multilevel/Hierarchical Models. Cambridge University
Press.) "arm"
However I have encountered a convergence difference between two versions
of R and its packages which seems to be a purely lme4 issue.
I have been mostly using a Dell laptop running Windows XP. Previously I
had been using R 2.11.1 but with packages at 2.10.1. Now I have brought
R and the packages both up to 2.12.1.
Here is my script: it uses the attached file which has itself been
constructed using a a script from the "arm" website but is self-contained.
library(lme4)
heights.clean =
attach(heights.clean)
y <- log(earn)
x <- height
n <- length(y)
n.age <- 3
n.eth <- 4
age <- age.category
# regression of log (earnings) on height, age, and ethnicity
M1 <- lmer (y ~ x + (1 + x | eth))
summary (M1)
M1 <- lmer (y ~ x + (1 + x | eth), verbose = TRUE)
Verbose output from R 2.12.1 and lme4_0.999375-37:
> M1 <- lmer (y ~ x + (1 + x | eth), verbose = TRUE)
0: 3113.7366: 0.0947958 0.00141417 0.00000
1: 3113.5186: 0.00000 0.00000 -0.991536
2: 3113.1254: 0.00106590 0.00111325 -0.991536
3: 3112.9224: 0.000649400 0.000651990 -0.991536
4: 3112.9133: 0.000831104 0.000668572 -0.990942
5: 3112.9121: 0.000715470 0.000730468 -0.991549
6: 3112.9117: 0.000904225 0.000536358 -0.990990
7: 3112.9114: 0.000874136 0.000514266 -0.990990
8: 3112.9112: 0.000889956 0.000518593 -0.990956
9: 3112.9111: 0.000908394 0.000472281 -0.990901
10: 3112.9110: 0.000923320 0.000460105 -0.990753
11: 3112.9106: 0.000995159 0.000309059 -0.989729
12: 3112.9103: 0.00103245 0.00000 -0.988338
13: 3112.9103: 0.00104480 1.00631e-09 -0.988338
14: 3112.9103: 0.00103952 1.72994e-08 -0.988327
15: 3112.9103: 0.00103952 2.02562e-08 -0.988327
16: 3112.9103: 0.00103951 2.01450e-08 -0.988327
17: 3112.9103: 0.00103951 2.01450e-08 -0.988327
18: 3112.9103: 0.00103951 2.01450e-08 -0.988327
Warning message:
In mer_finalize(ans) : false convergence (8)
Verbose output from R 2.11.1 and lme4_0.999375-32:
> M1 <- lmer (y ~ x + (1 + x | eth), verbose = TRUE)
0: 3113.7366: 0.0947958 0.00141417 0.00000
1: 3113.5186: 0.00000 0.00000 -0.991536
2: 3113.1254: 0.00106590 0.00111325 -0.991536
3: 3112.9224: 0.000649400 0.000651990 -0.991536
4: 3112.9133: 0.000831104 0.000668572 -0.990942
5: 3112.9121: 0.000715470 0.000730468 -0.991549
6: 3112.9117: 0.000904407 0.000536168 -0.990990
7: 3112.9114: 0.000874337 0.000514111 -0.990990
8: 3112.9112: 0.000890081 0.000518391 -0.990956
9: 3112.9111: 0.000908420 0.000472254 -0.990901
10: 3112.9110: 0.000923309 0.000460131 -0.990753
11: 3112.9106: 0.000994079 0.000311231 -0.989738
12: 3112.9103: 0.00103382 0.00000 -0.988215
13: 3112.9102: 0.00103963 3.55942e-10 -0.988215
14: 3112.9099: 0.00114005 0.000142779 -0.893828
15: 3112.9099: 0.00113930 9.68963e-05 -0.893828
16: 3112.9099: 0.00114352 9.70670e-05 -0.893805
17: 3112.9099: 0.00114631 9.68698e-05 -0.893782
18: 3112.9099: 0.00114593 8.96231e-05 -0.893737
19: 3112.9099: 0.00114697 8.94093e-05 -0.893635
20: 3112.9099: 0.00114840 8.74568e-05 -0.893534
21: 3112.9099: 0.00114767 8.69236e-05 -0.893331
22: 3112.9099: 0.00114960 8.45012e-05 -0.892925
23: 3112.9097: 0.00119713 0.000103484 -0.854718
24: 3112.9096: 0.00123584 0.000118803 -0.823861
25: 3112.9094: 0.00128530 0.000135095 -0.793003
26: 3112.9092: 0.00134505 0.000150925 -0.755646
27: 3112.9088: 0.00148850 0.000172432 -0.680932
28: 3112.9087: 0.00158189 0.000173348 -0.654113
29: 3112.9080: 0.00172167 0.000172246 -0.591365
30: 3112.9078: 0.00178884 0.000169040 -0.566266
31: 3112.9071: 0.00202232 2.33096e-05 -0.506894
32: 3112.9067: 0.00207199 0.00000 -0.501077
33: 3112.9063: 0.00219029 0.00000 -0.475171
34: 3112.9059: 0.00228481 0.00000 -0.456053
35: 3112.9052: 0.00250518 6.54281e-05 -0.417817
36: 3112.9048: 0.00260386 6.61956e-05 -0.402523
37: 3112.9041: 0.00281388 6.37610e-05 -0.371934
38: 3112.9037: 0.00294734 5.86692e-05 -0.358875
39: 3112.9029: 0.00315988 1.53970e-05 -0.332756
40: 3112.9023: 0.00333904 5.76185e-05 -0.315483
41: 3112.9022: 0.00337566 5.68599e-05 -0.315484
42: 3112.9022: 0.00337228 2.04001e-05 -0.315482
43: 3112.9022: 0.00339045 1.50348e-05 -0.315412
44: 3112.9022: 0.00338041 1.25433e-05 -0.315376
45: 3112.9022: 0.00338454 4.34985e-07 -0.315304
46: 3112.9007: 0.00380902 4.53446e-07 -0.277795
47: 3112.9001: 0.00401754 0.00000 -0.262792
48: 3112.8984: 0.00456608 0.00000 -0.231839
49: 3112.8956: 0.00545144 1.95070e-06 -0.195416
50: 3112.8928: 0.00652406 1.16379e-05 -0.175790
51: 3112.8916: 0.00660908 0.00000 -0.163760
52: 3112.8890: 0.00773482 4.99796e-06 -0.139726
53: 3112.8806: 0.00972168 0.00000 -0.121025
54: 3112.8786: 0.0114761 5.99051e-06 -0.0970285
55: 3112.8760: 0.0109215 6.10952e-06 -0.107449
56: 3112.8703: 0.0130487 0.00000 -0.0891341
57: 3112.8672: 0.0146986 0.00000 -0.0791535
58: 3112.8538: 0.0174766 0.00000 -0.0712682
59: 3112.8464: 0.0196425 9.24134e-08 -0.0648869
60: 3112.8229: 0.0261589 1.12351e-07 -0.0530892
61: 3112.8218: 0.0301145 0.00000 -0.0522032
62: 3112.8064: 0.0310301 2.94121e-08 -0.0482541
63: 3112.7853: 0.0370813 0.00000 -0.0428581
64: 3112.7768: 0.0392902 5.51626e-07 -0.0404835
65: 3112.7574: 0.0448577 0.00000 -0.0371558
66: 3112.7034: 0.0688045 9.08273e-06 -0.0271719
67: 3112.5859: 0.0947196 0.000137210 -0.0259377
68: 3112.5520: 0.120260 0.00000 -0.0214016
69: 3112.4267: 0.138009 0.00000 -0.0220440
70: 3112.2524: 0.241689 0.00000 -0.0175340
71: 3112.0653: 0.252030 5.74985e-06 -0.0195906
72: 3111.6754: 0.350688 0.00000 -0.0176928
73: 3111.5729: 0.396312 0.00000 -0.0169346
74: 3111.4491: 0.441941 0.00000 -0.0166439
75: 3110.9290: 0.624460 0.00000 -0.0161161
76: 3110.0355: 1.03912 0.00000 -0.0160134
77: 3109.9542: 1.12017 0.00000 -0.0157653
78: 3109.8078: 1.29685 0.00000 -0.0157324
79: 3109.7519: 1.45344 0.00000 -0.0157254
80: 3109.7469: 1.52828 0.00000 -0.0156867
81: 3109.7461: 1.53677 0.00000 -0.0157060
82: 3109.7460: 1.53350 0.00000 -0.0157010
83: 3109.7460: 1.53352 0.00000 -0.0157010
The two iterations start off at the same values but start to diverge
around step 12.
Murray
--
Dr Murray Jorgensen http://www.stats.waikato.ac.nz/Staff/maj.html
Department of Statistics, University of Waikato, Hamilton, New Zealand
Email: maj at waikato.ac.nz Fax 7 838 4155
Phone +64 7 838 4773 wk Home +64 7 825 0441 Mobile 021 0200 8350
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``` | 3,581 | 7,242 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-05 | latest | en | 0.69176 |
https://socratic.org/questions/how-do-you-find-the-domain-and-range-of-log-x-9 | 1,719,069,661,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00568.warc.gz | 462,773,699 | 5,850 | # How do you find the domain and range of log(x-9)?
Domain: $x > 9 \mathmr{and} \left(9 , \infty\right)$
The range of $y$ is all of $\mathbb{R} \mathmr{and} \left(\infty , \infty\right)$
$y = \log \left(x - 9\right)$. Domain: $x - 9 > 0 \mathmr{and} x > 9 \mathmr{and} \left(9 , \infty\right)$
Range : As $x$ approaches closer and closer to 9 from right, $y$ gets more and more negative. So Range is all real numbers i.e $\left(\infty , \infty\right)$ graph{log(x-9) [-40, 40, -20, 20]} [Ans] | 195 | 493 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.618976 |
http://darknessorlight.glaser.cz/documentation.collision.circle-line-old/ | 1,606,436,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189030.27/warc/CC-MAIN-20201126230216-20201127020216-00635.warc.gz | 26,489,777 | 7,746 | # Circle-Line collision (old solution)
This describes correct solution for circle-line collision with vector.
However, it involves lot of calculations, and this solution is not suitable, as because of the infinitesimal numbers problem it was extremely difficult to aproximate the solutions and keep the game collision functioning correctly.
This is sub-layout for documentation pages
Top
If collision between circle and line segment (which is line limited on interval $<a,b>$) is resolved, then, using line segment, other geometrical shapes can be created (such as triangle or polygon), and thus, the collision between them and circle is already solved.
#### 1 Definitions
• Term line reffers to infinite line, not line segment
• Let $C$ be circle with centre $C_s = (c_x, c_y)$ that we want to move
• Let $L_s$ be Line Segment defined by 2 points, $A, B$
• Let $L$ be Line defined also by 2 points, $A, B$
• Let $v=(v_x, v_y)$ be vector, defining the movement of the circle $C$
• Let $v'$ be the final movement vector (the result)
• $L_s$ touches $C \Leftrightarrow C$ and $L_s$ have exactly $1$ intersection point
• $L_s$ collides $C \Leftrightarrow C$ and $L_s$ have exactly $2$ intersection points
#### 2 Objective
Imagine, that being given movement vector $v$, the circle $C$ would collide $L_s$ during movement, if we move $C$ by $v$.
The objective is to determine vector $v'$ such as, when moving $C$ by vector $v'$, then $C$ will collide $L_s$ in exactly one point.
The situation is illustrated by following image
#### 3 Mathematical solution
The problem is divided to 6 cases, all of them will be discussed below.
• If $C$ does not touch nor collides line $L_s$ before movement begins
• Case1: $C$ will collide $L_s$ during movement at one of points $A$ or $B$, where $A$, $B$ are points defining the line $L_s$
• Case2: $C$ will collide $L_s$ during movement at point $P$. Let $M$ be, set of all points on line $L_s$. Then, for $P$ applies: $P \in \{M\} / \{A, B\}$
• Case3: $C$ will not collide $L_s$ during movement at all
• If $C$ touches $L_s$ before movement begins
• Case4: $C$ will collide $L_s$ during movement
• Case5: $C$ will not collide $L_s$ during movement
• Case6: $C$ collides $L_s$ before movement begins
##### 3.1 Circle - Line segment intersection relationship
To solve all the cases, we need to first check for relation between $C$ and $L$ (do they collide each other? Or touch?)
##### 3.1.1 Circle touches Line segment
For this, following statements apply:
• Let $P$ be set of all intersection points between $C$ and $L$. Applies $|P| \in \{0,1,2\}$
• If $|P|=0$, then $C$ does not touch $L_s$
• If $|P|=1 \wedge P \in L$, then $C$ touches $L_s$
• If $|P|=2 \wedge P_x,P_y \in P: P_x = B \wedge P_y \notin L_s \wedge |A C_s| \geq r$, then $C$ touches $L_s$
• If $|P|=2 \wedge P_x,P_y \in P: P_x = A \wedge P_y \notin L_s \wedge |B C_s| \geq r$, then $C$ touches $L_s$
To explain the math above in plain words:
• If $C$ and $L$ have 0 intersection points, then they do not touch.
• If $C$ and $L$ have 1 intersection point, then they touch.
• If $C$ and $L$ have 2 intersection points:
• If one of the intersection points is $A$, and the other intersection point does not lie on the line segment $L_s$ and the B point lies outside the circle, then they touch
• If one of the intersection points is $B$, and the other intersection point does not lie on the line segment $L_s$ and the A point lies outside the circle, then they touch
##### 3.1.2 Circle collides Line segment
For this, following statements apply:
• Let $P$ be set of all intersection points between $C$ and $L$. Applies $|P| \in \{0,1,2\}$
• If $|P| \in \{0,1\}$, then $C$ does not collide $L_s$
• If $|P|=2 \wedge P_x,P_y \in P: P_x \in L_s \wedge P_y \in L_s$, then $C$ collides $L_s$
• If $|P|=2 \wedge P_x,P_y \in P: P_x = A \wedge |B C_s| \leq r$, then $C$ collides $L_s$
• If $|P|=2 \wedge P_x,P_y \in P: P_x = B \wedge |A C_s| \leq r$, then $C$ collides $L_s$
To explain the math above in plain words:
• If $C$ and $L$ have 0 r 1 intersection points, then they do not collide.
• If $C$ and $L$ have 2 intersection points:
• If both intersection points lie on $L_s$, then they collide
• If one of the intersection points is $A$, and the other intersection point does not lie on the line segment $L_s$ and the $B$ point lies inside or on the circle, then they collide
• If one of the intersection points is $B$, and the other intersection point does not lie on the line segment $L_s$ and the $A$ point lies inside or on the circle, then they collide
##### Case 1: $C$ will collide $L_s$ during movement at one of points $A$ or $B$, where $A$, $B$ are points defining the line $L_s$
To solve this, similar mathematical approach is used as in case2.
It is strongly recommended to read that first, and then come back here for better understanding...
By taking circle equation and line equation and combining them together, the point A (later also B) is plugged into the equations. If there is solution, and the final movement vector $v$ matches criteria (smaller than original movement vector, etc...for more details, read through case2), then the circle is considered to be colliding line segment at point A (or B, respectivelly).
##### Case 2: $C$ will collide $L_s$ during movement at point $P$. Let $M$ be, set of all points on line $L_s$. Then, for $P$ applies: $P \in \{M\} / \{A, B\}$
Let's reiterate, that $L$ is line, not a line segment. For simplicity, I will explain the solution on LINE, understanding that narrowing the solution down for line segment involves limitation on interval and adding few other edge cases to entire solution.
Let's start by having the equation of circle $C$:
Now, we perform translation of the centre by vector $v$:
However, we do not know how far to move the circle to get exactly one intersection point with $L$. Therefore, we use parameter $t$:
The equation above describes all circles, that are moved on the line defined by movement vector $v'$
Now, we will solve system of 2 equations:
Where the second equation is the equation of $L$ in general form
We express $x$ from the line equation (alternatively, $y$ must be expressed in some cases to avoid diving by 0):
That value is plugged into the circle equation to get:
Now, considering $t$ as the polynom variable, if discriminant equals 0, then there is exactly one solution (circle collides line in one point only)
The discriminant is then:
Considering some input values for circle centre, movement vector, radius, etc., this would be more readable result:
The thing left is to compute the $t$, which is the movement vector multiplier.
Final solution vectors (if the solutions exist) will be:
$v'_1 = (v_x * t_1, v_y * t_1)$
$v'_2 = (v_x * t_2, v_y * t_2)$
The situation is illustrated by following image
Of course, the final vectors $v'_1, v'_2$ might not face in same direction as $v'$.
In that case, the solution is discarted.
Moreover, if $|t_1| > 1 \lor |t_2| > 1)$, then $t_1$, respectivelly $t_2$ is discarted.
That is because we require final vector to have same or smaller length then the original move vector
##### Case 3: $C$ will not collide $L_s$ during movement at all
If none of Case1 or Case2 holds and yields a solution, then the circle will not collide line segment at all during movement.
##### Case 4 and 5: $C$ touches $L_s$ before movement begins, and $C$ will (or not) collide $L_s$ during movement
For this one, imagine following figure:
• Let $t$ be tangent to $C$ in point $P$, which is touch point of $C$ and $L_s$ (in this example, $P=A$)
• Let $u=|AE|$, $v=|AD|$ be two movement vectors, where we position them to begin in touch point $P=A$
• $u=|AE|$ represents movement that would be allowed, whereas $v=|AD|$ represents movement that would NOT be allowed
In general case, imagining vector $u=|AD|$, then movement of $C$ using vector $u$ is valid $\iff D$ is on same side of line $t$ as centre of circle $C$ or $D \in t$
##### Case 6: $C$ collides $L_s$ before movement begins
Trivial case, when is done check for whether $C$ collides $L_s$, and if so, then no movement is done... | 2,327 | 8,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2020-50 | latest | en | 0.889452 |
https://www.physicsforums.com/threads/dropping-a-mass-directly-over-the-north-pole.516812/ | 1,527,473,027,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00049.warc.gz | 807,206,915 | 21,318 | # Dropping a mass directly over the north pole
1. Jul 25, 2011
### starz_above
Hi all,
I'd like to understand what would happen to a mass which is dropped at the north/south pole.
Let's say you raised a mass to a certain height over a certain point, which we will call A. Point A is at the center of the north pole. The height at which the mass is raised is such that it will take over 10 mins to hit the ground, let's say it's 6km high above point A.
When the mass is dropped would it land on point A? Let's assume that there no external forces to alter the descent in anyway. We can assume that the mass falls in a vacuum.
2. Jul 25, 2011
### mikeph
Why wouldn't you expect it to?
3. Jul 25, 2011
### nasu
If you neglect air resistance, 6km is way too low to take 10 min.
4. Jul 25, 2011
### starz_above
I was thinking whether the earth's orbit around the sun would have any effect.
If the mass falls directly onto point A after 10 mins, what causes it to stay above point A as it falls given that the earth is travelling at a tangent below it?
5. Jul 25, 2011
### starz_above
You're right. Let's say it takes about 35 seconds to land on point A.
6. Jul 25, 2011
### fluidistic
The biggest effect that you'd have to take into account would be Corriolis effect. But since you are over a pole, I don't think it applies. It would have been quite different over the equator I think.
7. Jul 25, 2011
### Delta Kilo
Acceleration due to Sun/Moon gravity will be pretty much the same for the mass and for the whole Earth as long as the distance between the centres of mass of the body and the Earth is small compared to the distance to the Sun and/or Moon. Assuming that the mass is at rest relative to the Earth at the moment of release, the mass and the Earth will be in the same orbit.
If the mass is moved significant distance from the Earth, gravitational influences due to other bodies will no longer cancel out and you'd have to calculate the motion using orbital mechanics.
8. Jul 25, 2011
### Shoku Z
The object initially was on the surface of earth and so the velocities of Earth and the object are same.We carry it up and then drop it.observing from earth,the trajectory will be a straight line.This is because the velocity of the object is same as Earth.The only other force acting upon them is that gravitational force of sun.Hence,the trajectory appears to be a straight line from Earth.
9. Jul 25, 2011
### Shoku Z
In short,Delta is right.I didnt see his post!
10. Jul 25, 2011
### starz_above
I understand that the mass and the earth will be in the same orbit and hence have the same velocity. However if the mass lands on point A, then doesn't that mean the mass must have a higher velocity than that of the earth in order to land on point A since it takes x amount of time descend. If the mass has the same velocity as that of the earth then it should land "behind" point A in the path of the earth's orbit around the sun.
11. Jul 25, 2011
### fluidistic
Delta and Shoku z, you guys are missing that the Earth isn't an inertial frame. There is a Coriolis effect and I can assure you that if you drop an object say from a skyscrapper and if you remove all air (so that there's no air resistance), the object trajectory WON'T be a straight line, despite what you said. However I think that the Coriolis effect is inexistant in both poles of the Earth. I have no idea about the centrifugal force though. I'll wait for some expert.
I'm assignated a problem in classical mechanics where I have to throw upward an object and calculate at what distance it will fall from me. I'll have to take into account Coriolis effect.
12. Jul 25, 2011
### Delta Kilo
The Earth is influences by the gravitation of Sun, Moon and other bodies. As a result, the Earth moves with acceleration let's say $\vec{a}$ (pointing roughly towards the Sun). The mass will receive the same acceleration $\vec{a}$ from Sun, Moon etc, plus acceleration $\vec{g}$ towards the centre of the Earth. So the acceleration of the mass relative to the Earth will be just $\vec{g}$ towards the centre. Since initial relative velocity is 0, the resulting relative velocity and hence the relative trajectory will be directed toward the centre of the Earth. When I say relative, I mean non-inertial reference frame with the origin at the centre of the earth and axis pointing in fixed direction in space (moving with acceleration $\vec{a}$ but not rotating).
13. Jul 25, 2011
### fluidistic
Look, here is a problem I see in the assignments of the classical mechanics from my university:
I'm unsure of the expression of the distance. It's written $4 \Omega V_0 ^3 \cos L/3g^2$.
So that I'm almost 100% sure that the Coriolis effect is the most important effect that can effect the object. However as I said twice, right over the poles I think it's worth 0 (maybe L is worth pi/2 and the right expression is $\frac{4 \Omega V_0 ^3 \cos (L)}{3g^2}$?) So that your object would indeed fall over the north pole. The answer depends where you are on Earth and the 2 poles are the special cases where there's no appreciable displacement of objects thrown upward.
14. Jul 25, 2011
### DaveC426913
starsz_above has a point. If we set aside the pull of the Earth for a moment, once the object is released, it is essentially in orbit. But it is not quite in the same orbit as the Earth. Its orbital plane is tilted to cross Earth's.
This effect will be enhanced or reduced, depending on what season it is. If it is summer, then the object's initial position is slightly inside the Earth's orbit (because the axial tilt is causing the north pole to tilt inward); if in the winter, the object's initial position is outside the Earth's orbit.
These are small effects, but they're there.
15. Jul 26, 2011
### mikeph
If you're ignoring the sun, then the problem can be reduced to a single dimension (object ---- Earth's COG) since the object is placed on the axis of rotation of the Earth.
If you're including the sun, it's a circular restricted three body problem (chaotic), and it's not too clear from the question what the initial conditions are (no external forces = ??, if you let the particle go, does that mean giving it zero velocity? Then it won't even land on the Earth which is moving at 30 km/s in orbit.
16. Jul 26, 2011
### starz_above
The mass has inital zero velocity wrt the earth. To clarify, the mass is placed at a certain height directly above the north pole, then it is released to fall without any external forces being added and falls just due to the earth's gravity.
So initially both the mass and the earth are travelling at 30 km/s in orbit.
17. Jul 26, 2011
### mikeph
So the Earth is subject to the sun's gravity but the mass isn't? Aside from breaking physical laws, it wouldn't land at the same point because that point is following a circular trajectory (or some kind of 3D spirograph pattern if you account for the seasons) and the mass isn't.
18. Jul 26, 2011
### FireStorm000
Let us make the following assumptions:
Gravity from all sources apply
Zero influence due to air resistance
Zero influence due to earth's magnetic field
That leaves us with the following forces which have a component perpendicular to the acceleration due to earth's gravity:
-tidal stresses:
--Acceleration due to the moon; the moon is a different distance from earth's CoM than the object, because the earth's axis of rotation isn't perpendicular to the moon's orbital plane.
--Acceleration due to the sun; again, the sun is a different distance from earth's CoM than the object, because the earth's axis of rotation isn't perpendicular to the sun's orbital plane.
Assuming the object was placed exactly distance X above the north or south pole, defining above as being normal to the surface of a perfect sphere, and the velocity of the object is the same as the instantaneous rate of change of position of a point positioned x distance from the surface along the axis, then those are the only other forces applying
19. Jul 26, 2011
### starz_above
The mass is also subject to the sun's gravity. When I said no external forces I meant to say no air resistance and it isn't pushed in any direction when it's released. Apologies for not being clear.
As i understand it, the trajectory of the mass would also be some kind of 3D spirograph pattern as you suggested.
What is puzzling me is that if the earth was not orbiting the sun then you would get the same result ie. the mass landing directly on top of the north pole. Initial thoughts tell me that if the earth is orbiting the sun then the mass should not fall directly on top of the north pole because in order to do so the mass must increase it's horizontal velocity component in order to "catch up" with the north pole as it is falling.
20. Jul 26, 2011
### starz_above
Your assumptions are correct. Thanks for putting it clearly.
21. Jul 26, 2011
### mikeph
That is not the same result. If there is no sun the problem is one-dimensional and it will land on the pole. If there is a sun it's a three body restricted problem with an entirely different (chaotic) trajectory and will not land on the pole. You could calculate it with arbitrary accuracy if you have a powerful enough computer and some decent numerical equations.
22. Jul 26, 2011
### olivermsun
You could also posit the calculation to be arbitrarily accurate by calling the mass a "small test mass." :) That still leaves the problem interesting in the usual sense of the non-inertial frame.
But even if you allowed the small mass to be (slightly) massive, would you really expect any interesting chaotic trajectory between the mass starting a small distance above the pole and striking the earth (therefore ending the trajectory)?
23. Jul 26, 2011
### mikeph
Even with a small test mass it's chaotic (restricted three body problem) because the sun also moves around the barycentre of the Earth-sun system. If you consider that the body has a non-negligible mass (general three body system) it's even worse. In both cases the trajectories are, for general initial conditions, chaotic.
24. Jul 26, 2011
### olivermsun
But my point is, are these "general" initial conditions? The test mass does not seem to continue along its path for very long in the problem as stated.
25. Jul 26, 2011
### mikeph
The starting coordinates are not the Lagrange points (the only five analytical solutions to the problem), so yeah it's a general initial condition. You're right, it will probably meet the Earth's surface soon after it's released, but where exactly on the surface I believe is impossible to compute. | 2,514 | 10,657 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-22 | longest | en | 0.953772 |
https://number.academy/601456 | 1,723,479,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00626.warc.gz | 348,382,212 | 11,238 | # Number 601456 facts
The even number 601,456 is spelled 🔊, and written in words: six hundred and one thousand, four hundred and fifty-six. The ordinal number 601456th is said 🔊 and written as: six hundred and one thousand, four hundred and fifty-sixth. Color #601456. The meaning of the number 601456 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 601456. What is 601456 in computer science, numerology, codes and images, writing and naming in other languages
## What is 601,456 in other units
The decimal (Arabic) number 601456 converted to a Roman number is (D)(C)MCDLVI. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
601456 seconds equals to 6 days, 23 hours, 4 minutes, 16 seconds
601456 minutes equals to 1 year, 2 months, 3 weeks, 4 days, 16 hours, 16 minutes
### Codes and images of the number 601456
Number 601456 morse code: -.... ----- .---- ....- ..... -....
Sign language for number 601456:
Number 601456 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
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### Advanced math operations
#### Is Prime?
The number 601456 is not a prime number. The closest prime numbers are 601451, 601457.
#### Factorization and factors (dividers)
The prime factors of 601456 are 2 * 2 * 2 * 2 * 37591
The factors of 601456 are 1, 2, 4, 8, 16, 37591, 75182, 150364, 300728, 601456.
Total factors 10.
Sum of factors 1165352 (563896).
#### Powers
The second power of 6014562 is 361.749.319.936.
The third power of 6014563 is 217.576.298.971.426.816.
#### Roots
The square root √601456 is 775,535944.
The cube root of 3601456 is 84,411436.
#### Logarithms
The natural logarithm of No. ln 601456 = loge 601456 = 13,307109.
The logarithm to base 10 of No. log10 601456 = 5,779204.
The Napierian logarithm of No. log1/e 601456 = -13,307109.
### Trigonometric functions
The cosine of 601456 is -0,336063.
The sine of 601456 is -0,94184.
The tangent of 601456 is 2,80257.
## Number 601456 in Computer Science
Code typeCode value
PIN 601456 It's recommended that you use 601456 as your password or PIN.
601456 Number of bytes587.4KB
CSS Color
#601456 hexadecimal to red, green and blue (RGB) (96, 20, 86)
Unix timeUnix time 601456 is equal to Wednesday Jan. 7, 1970, 11:04:16 p.m. GMT
IPv4, IPv6Number 601456 internet address in dotted format v4 0.9.45.112, v6 ::9:2d70
601456 Decimal = 10010010110101110000 Binary
601456 Decimal = 1010120001011 Ternary
601456 Decimal = 2226560 Octal
601456 Decimal = 92D70 Hexadecimal (0x92d70 hex)
601456 BASE64NjAxNDU2
601456 MD50fdec7c490a1e5e36bff2b2ddf518352
601456 SHA1b19a4fc45f98014de34abe19e5aa0c34174fa10c
601456 SHA2241e2c70f3d13f8859a4b6a9341424b181d8d80c4904f0ca014c152053
601456 SHA2565d336a8b56b49566ac20e974de8ecf91e9c6a2510e3594176be1680da1caedf6
601456 SHA384edf9f359b32a93fa171468c6e10c73bc56a7137f7dc8264cbe915fe57d671893a2d69e849ffa7d546eebba46c741ec29
More SHA codes related to the number 601456 ...
If you know something interesting about the 601456 number that you did not find on this page, do not hesitate to write us here.
## Numerology 601456
### Character frequency in the number 601456
Character (importance) frequency for numerology.
Character: Frequency: 6 2 0 1 1 1 4 1 5 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 601456, the numbers 6+0+1+4+5+6 = 2+2 = 4 are added and the meaning of the number 4 is sought.
## № 601,456 in other languages
How to say or write the number six hundred and one thousand, four hundred and fifty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 601.456) seiscientos uno mil cuatrocientos cincuenta y seis German: 🔊 (Nummer 601.456) sechshunderteinstausendvierhundertsechsundfünfzig French: 🔊 (nombre 601 456) six cent un mille quatre cent cinquante-six Portuguese: 🔊 (número 601 456) seiscentos e um mil, quatrocentos e cinquenta e seis Hindi: 🔊 (संख्या 601 456) छः लाख, एक हज़ार, चार सौ, छप्पन Chinese: 🔊 (数 601 456) 六十万一千四百五十六 Arabian: 🔊 (عدد 601,456) ستمائة و ألف ألف و أربعمائة و ستة و خمسون Czech: 🔊 (číslo 601 456) šestset jedna tisíc čtyřista padesát šest Korean: 🔊 (번호 601,456) 육십만 천사백오십육 Danish: 🔊 (nummer 601 456) sekshundrede og ettusindfirehundrede og seksoghalvtreds Hebrew: (מספר 601,456) שש מאות ואחד אלף ארבע מאות חמישים ושש Dutch: 🔊 (nummer 601 456) zeshonderdéénduizendvierhonderdzesenvijftig Japanese: 🔊 (数 601,456) 六十万千四百五十六 Indonesian: 🔊 (jumlah 601.456) enam ratus satu ribu empat ratus lima puluh enam Italian: 🔊 (numero 601 456) seicentounomilaquattrocentocinquantasei Norwegian: 🔊 (nummer 601 456) seks hundre og en tusen fire hundre og femtiseks Polish: 🔊 (liczba 601 456) sześćset jeden tysięcy czterysta pięćdziesiąt sześć Russian: 🔊 (номер 601 456) шестьсот одна тысяча четыреста пятьдесят шесть Turkish: 🔊 (numara 601,456) altıyüzbindörtyüzellialtı Thai: 🔊 (จำนวน 601 456) หกแสนหนึ่งพันสี่ร้อยห้าสิบหก Ukrainian: 🔊 (номер 601 456) шістсот одна тисяча чотириста п'ятдесят шість Vietnamese: 🔊 (con số 601.456) sáu trăm lẻ một nghìn bốn trăm năm mươi sáu Other languages ...
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## Frequently asked questions about the number 601456
• ### How do you write the number 601456 in words?
601456 can be written as "six hundred and one thousand, four hundred and fifty-six". | 2,062 | 6,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.738099 |
https://www.coursehero.com/tutors-problems/Trigonometry/16591755-Math-30-1Trigonometry-I-need-hell-with-these-questions-i-am-having-pr/ | 1,561,581,932,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000545.97/warc/CC-MAIN-20190626194744-20190626220744-00383.warc.gz | 707,304,869 | 24,670 | View the step-by-step solution to:
# I SVX3LE Name Math 30-1 6.1 and 6.2 Trigonometric Identities Quiz 1. Determine the value of each expression. a) log2 cos(57 b) log. (tan ( 5x c) sin
Math 30-1:Trigonometry
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5. Verify sec0-cos 0 = sin 0 tan 0 numerically for the value 0=135
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b) cos(7+ A)-cos(7 - A)
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9. If sin A =-, and ZA is in quadrant IV, determine the exact value of sin 2A
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Browse Documents | 750 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2019-26 | latest | en | 0.670724 |
https://pvillage.org/archives/31837 | 1,675,075,982,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00493.warc.gz | 456,562,471 | 13,056 | Pvillage.org
# What is the delta s value?
## What is the delta s value?
A negative delta S corresponds to a spontaneous process when the magnitude of T * delta S is less than delta H (which must be negative). delta G = delta H – (T * delta S). A negative delta S would mean that the products have a lower entropy than the reactants, which is not spontaneous by itself.
## What does T Delta S mean in chemistry?
change in entropy
∆S is the change in entropy (disorder) from reactants to products. R is the gas constant (always positive) T is the absolute temperature (Kelvin, always positive) What it means: If ∆H is negative, this means that the reaction gives off heat from reactants to products.
What is s in chemistry entropy?
In chemistry, the standard molar entropy is the entropy content of one mole of pure substance at a standard state of pressure and any temperature of interest. is usually given the symbol S°, and has units of joules per mole kelvin (J⋅mol−1⋅K−1). Unlike standard enthalpies of formation, the value of S° is absolute.
Is dS the same as Delta S?
S is the entropy of the system. A pop-up calculator is available to calculate the enthalpy and Gibb’s free energy changes in reactions….Free Energy.
Enthalpy Change Entropy Change Spontaneous Reaction?
Exothermic (dH < 0) Decrease (dS < 0) Only at low temps, if |T dS| < |dH|
G
The symbol for free energy is G, in honor of American scientist Josiah Gibbs (1839-1903), who made many contributions to thermodynamics. The change in Gibbs free energy is equal to the change in enthalpy minus the mathematical product of the change in entropy, multiplied by the Kelvin temperature.
### What is entropy in chemistry for dummies?
Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Instead, they are two very different ways of looking at a system.
Why is entropy J K?
In classical thermodynamics, the second law of thermodynamics states that the entropy of an isolated system always increases or remains constant. Thermodynamic entropy has the dimension of energy divided by temperature, which has a unit of joules per kelvin (J/K) in the International System of Units.
Which is more likely to be spontaneous Delta H or delta G?
Let us first look at enthalpy: an exothermic reaction is more likely to be spontaneous; if Delta H is negative, this makes it more likely that Delta G will also be negative and, therefore, spontaneous.
#### Where can I find the NBS table of chemical thermodynamic properties?
*Taken from “The NBS Tables of Chemical Thermodynamic Properties” (1982) and “CRC Handbook of Chemistry and Physics”, 1st Student Edition (1988)
#### How to calculate the standard entropy change in chemistry?
(20.2.3) Δ S o = Σ n S o (products) − Σ n S o (reactants) The standard entropy change is equal to the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants. The symbol ” n ” signifies that each entropy must first be multiplied by its coefficient in the balanced equation. | 738 | 3,211 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | latest | en | 0.906991 |
https://electrical-engineering-portal.com/dimensioning-of-power-distribution-systems-2 | 1,670,179,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00094.warc.gz | 261,174,739 | 35,428 | Save 50% on all EEP Academy courses with Enterprise Membership Plan and study specialized LV/MV/HV technical articles & guides.
# Dimensioning of Power Distribution Systems (2)
Home / Technical Articles / Dimensioning of Power Distribution Systems (2)
Continued from the first part: Dimensioning of Power Distribution Systems (1)
## Circuit types
### Distribution circuit
Dimensioning of cable routes and devices follows the maximum load currents to be expected at this distribution level.
As a rule:
Ib max = ∑ installed capacity x simultaneity factor
Switching/protective device and connecting line are to be matched with regard to overload and short-circuit protection.
In order to ensure overload protection, the standardized conventional (non-)tripping currents referring to the devices in application have to be observed. A verification based merely on the rated device current or the setting value Ir would be insufficient.
## Basic rules for ensuring overload protection:
### Rated current rule
#### Non-adjustable protective equipment
Ib ≤ In ≤ Iz
The rated current In of the selected device must be between the calculated maximum load current Ib and the maximum permissible load current Iz of the selected transmission medium (cable or busbar).
#### Adjustable protective equipment
Ib ≤ Ir ≤ Iz
The rated current Ir of the overload release must be between the calculated maximum load current Ib and the maximum permissible load current Iz of the selected transmission medium (cable or busbar).
#### Tripping current rule
I2 ≤ 1.45 x Iz
The maximum permissible load current Iz of the selected transmission medium (cable or busbar) must be above the conventional tripping current I2-/1.45 of the selected device.
The test value I2 is standardized and varies according to the type and characteristics of the protective equipment applied.
## Basic rules for ensuring short-circuit protection:
### Short-circuit energy
K2S2 ≥ I2t
(K = Material coefficient; S = Cross-section)
The amount of energy that is set free when a short-circuit occurs – and up to the moment it is cleared automatically – must be less than the energy that the transmission medium can carry as a maximum or there will be irreparable damage. As a standard, this basic rule applies in the time range up to max. 5 s.
Below 100 ms of short-circuit breaking time, the let-through energy of the protective device (according to the equipment manufacturer’s specification) must be taken into account.
When devices with a tripping unit are used, observance of this rule across the entire characteristic device curve must be verified. A mere verification in the range of the maximum short-circuit current applied (Ik max) is not always sufficient, in particular when time-delayed releases are used.
### Short-circuit time
ta (Ik min) ≤ 5 s
The resulting current-breaking time of the selected protective equipment must ensure that the calculated minimum short-circuit current Ik min at the end of the transmission line or protected line is automatically cleared within 5 s at the most.
Overload and short-circuit protection need not necessarily be provided by one and the same device. If required, these two protection targets may be realized by a device combination. The use of separate switching protective devices could also be considered, i.e., at the start and end of a cable route. As a rule, devices applied at the end of a cable route can ensure overload protection for that line only.
## Final circuits
The method for coordinating overload and short-circuit protection is practically identical for distribution and final circuits. Besides overload and short-circuit protection, the protection of human life is also important for all circuits.
### Protection against electric shock
ta (Ik1 min) ≤ ta perm
If a 1-phase fault to earth (Ik1 min) occurs, the resulting current breaking time ta for the selected protective equipment must be shorter than the maximum permissible breaking time ta perm that is required for this circuit according to IEC 60364-4-41/DIN VDE 0100-410 to ensure the protection of persons.
Because the required maximum current breaking time varies according to the rated system voltage and the type of load connected (stationary and non-stationary loads), protection requirements regarding minimum breaking times ta perm may be transferred from one load circuit to other circuits. Alternatively, this protection target may also be achieved by observing a maximum touch voltage.
Because final circuits are often characterized by long supply lines, their dimensioning is often affected by the maximum permissible voltage drop.
As far as the choice of switching protective devices is concerned, it is important to bear in mind that long connecting lines are characterized by high impedances, and thus strong attenuation of the calculated short-circuit currents.
Depending on the system operating mode (coupling open, coupling closed) and the medium of supply (transformer or generator), the protective equipment and its settings must be configured for the worst-case scenario for short-circuit currents.
In contrast to supply or distribution circuits, where the choice of a high-quality tripping unit is considered very important, there are no special requirements on the protective equipment of final circuits regarding the degree of selectivity to be achieved.
The use of a tripping unit with LI characteristics is normally sufficient.
## Summary
Basically, the dimensioning process itself is easy to understand and can be performed using simple means. Its complexity lies in the procurement of the technical data on products and systems required. This data can be found in various technical standards and regulations as well as in numerous product catalogs.
An important aspect in this context is the cross-circuit manipulation of dimensioned components owing to their technical data. One such aspect is the above mentioned inheritance of minimum current breaking times of the non-stationary load circuit to other stationary load or distribution circuits.
Another aspect is the mutual impact of dimensioning and network calculation (short-circuit), e.g., for the use of short-circuit current-limiting devices.
In addition, the complexity of the matter increases, when different national standards or installation practices are to be taken into account for dimensioning.
Reference: Siemens – Power Engineering Guide Edition 7.0 (chap. Switchgear and Substations) | 1,257 | 6,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-49 | longest | en | 0.850432 |
https://popflock.com/learn?s=Pressure | 1,652,925,317,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00502.warc.gz | 530,446,823 | 38,089 | Pressure
Get Pressure essential facts below. View Videos or join the Pressure discussion. Add Pressure to your PopFlock.com topic list for future reference or share this resource on social media.
Pressure
Pressure
Common symbols
p, P
SI unitpascal [Pa]
In SI base unitsN/m2, 1 kg/(m·s2), or 1 J/m3
Derivations from
other quantities
p = F / A
DimensionM L-1 T-2
Pressure as exerted by particle collisions inside a closed container
Pressure (symbol: p or P) is the force applied perpendicular to the surface of an object per unit area over which that force is distributed.[1] Gauge pressure (also spelled gage pressure)[a] is the pressure relative to the ambient pressure.
Various units are used to express pressure. Some of these derive from a unit of force divided by a unit of area; the SI unit of pressure, the pascal (Pa), for example, is one newton per square metre (N/m2); similarly, the pound-force per square inch (psi) is the traditional unit of pressure in the imperial and U.S. customary systems. Pressure may also be expressed in terms of standard atmospheric pressure; the atmosphere (atm) is equal to this pressure, and the torr is defined as 1760 of this. Manometric units such as the centimetre of water, millimetre of mercury, and inch of mercury are used to express pressures in terms of the height of column of a particular fluid in a manometer.
## Definition
Pressure is the amount of force applied perpendicular to the surface of an object per unit area. The symbol for it is "p" or P.[2] The IUPAC recommendation for pressure is a lower-case p.[3] However, upper-case P is widely used. The usage of P vs p depends upon the field in which one is working, on the nearby presence of other symbols for quantities such as power and momentum, and on writing style.
### Formula
Mathematically:
${\displaystyle p={\frac {F}{A}},}$[4]
where:
${\displaystyle p}$ is the pressure,
${\displaystyle F}$ is the magnitude of the normal force,
${\displaystyle A}$ is the area of the surface on contact.
Pressure is a scalar quantity. It relates the vector area element (a vector normal to the surface) with the normal force acting on it. The pressure is the scalar proportionality constant that relates the two normal vectors:
${\displaystyle d\mathbf {F} _{n}=-p\,d\mathbf {A} =-p\,\mathbf {n} \,dA.}$
The minus sign comes from the fact that the force is considered towards the surface element, while the normal vector points outward. The equation has meaning in that, for any surface S in contact with the fluid, the total force exerted by the fluid on that surface is the surface integral over S of the right-hand side of the above equation.
It is incorrect (although rather usual) to say "the pressure is directed in such or such direction". The pressure, as a scalar, has no direction. The force given by the previous relationship to the quantity has a direction, but the pressure does not. If we change the orientation of the surface element, the direction of the normal force changes accordingly, but the pressure remains the same.[]
Pressure is distributed to solid boundaries or across arbitrary sections of fluid normal to these boundaries or sections at every point. It is a fundamental parameter in thermodynamics, and it is conjugate to volume.[5]
### Units
Mercury column
The SI unit for pressure is the pascal (Pa), equal to one newton per square metre (N/m2, or kg·m-1·s-2). This name for the unit was added in 1971;[6] before that, pressure in SI was expressed simply in newtons per square metre.
Other units of pressure, such as pounds per square inch (lbf/in2) and bar, are also in common use. The CGS unit of pressure is the barye (Ba), equal to 1 dyn·cm-2, or 0.1 Pa. Pressure is sometimes expressed in grams-force or kilograms-force per square centimetre (g/cm2 or kg/cm2) and the like without properly identifying the force units. But using the names kilogram, gram, kilogram-force, or gram-force (or their symbols) as units of force is expressly forbidden in SI. The technical atmosphere (symbol: at) is 1 kgf/cm2 (98.0665 kPa, or 14.223 psi).
Since a system under pressure has the potential to perform work on its surroundings, pressure is a measure of potential energy stored per unit volume. It is therefore related to energy density and may be expressed in units such as joules per cubic metre (J/m3, which is equal to Pa). Mathematically:
${\displaystyle p={\frac {F\cdot {\text{distance}}}{A\cdot {\text{distance}}}}={\frac {\text{Work}}{\text{Volume}}}={\frac {\text{Energy (J)}}{{\text{Volume }}({\text{m}}^{3})}}.}$
Some meteorologists prefer the hectopascal (hPa) for atmospheric air pressure, which is equivalent to the older unit millibar (mbar). Similar pressures are given in kilopascals (kPa) in most other fields, except aviation where the hecto- prefix is commonly used. The inch of mercury is still used in the United States. Oceanographers usually measure underwater pressure in decibars (dbar) because pressure in the ocean increases by approximately one decibar per metre depth.
The standard atmosphere (atm) is an established constant. It is approximately equal to typical air pressure at Earth mean sea level and is defined as .
Because pressure is commonly measured by its ability to displace a column of liquid in a manometer, pressures are often expressed as a depth of a particular fluid (e.g., centimetres of water, millimetres of mercury or inches of mercury). The most common choices are mercury (Hg) and water; water is nontoxic and readily available, while mercury's high density allows a shorter column (and so a smaller manometer) to be used to measure a given pressure. The pressure exerted by a column of liquid of height h and density ? is given by the hydrostatic pressure equation , where g is the gravitational acceleration. Fluid density and local gravity can vary from one reading to another depending on local factors, so the height of a fluid column does not define pressure precisely. When millimetres of mercury (or inches of mercury) are quoted today, these units are not based on a physical column of mercury; rather, they have been given precise definitions that can be expressed in terms of SI units.[7] One millimetre of mercury is approximately equal to one torr. The water-based units still depend on the density of water, a measured, rather than defined, quantity. These manometric units are still encountered in many fields. Blood pressure is measured in millimetres of mercury in most of the world, and lung pressures in centimetres of water are still common.[]
Underwater divers use the metre sea water (msw or MSW) and foot sea water (fsw or FSW) units of pressure, and these are the standard units for pressure gauges used to measure pressure exposure in diving chambers and personal decompression computers. A msw is defined as 0.1 bar (= 100000 Pa = 10000 Pa), is not the same as a linear metre of depth. 33.066 fsw = 1 atm[8] (1 atm = 101325 Pa / 33.066 = 3064.326 Pa). Note that the pressure conversion from msw to fsw is different from the length conversion: 10 msw = 32.6336 fsw, while 10 m = 32.8083 ft.[8]
Gauge pressure is often given in units with "g" appended, e.g. "kPag", "barg" or "psig", and units for measurements of absolute pressure are sometimes given a suffix of "a", to avoid confusion, for example "kPaa", "psia". However, the US National Institute of Standards and Technology recommends that, to avoid confusion, any modifiers be instead applied to the quantity being measured rather than the unit of measure.[9] For example, rather than .
Differential pressure is expressed in units with "d" appended; this type of measurement is useful when considering sealing performance or whether a valve will open or close.
Presently or formerly popular pressure units include the following:
• atmosphere (atm)
• manometric units:
• centimetre, inch, millimetre (torr) and micrometre (mTorr, micron) of mercury,
• height of equivalent column of water, including millimetre (mm ), centimetre (cm ), metre, inch, and foot of water;
• imperial and customary units:
• non-SI metric units:
• bar, decibar, millibar,
• msw (metres sea water), used in underwater diving, particularly in connection with diving pressure exposure and decompression,
• kilogram-force, or kilopond, per square centimetre (technical atmosphere),
• gram-force and tonne-force (metric ton-force) per square centimetre,
• barye (dyne per square centimetre),
• kilogram-force and tonne-force per square metre,
• sthene per square metre (pieze).
### Examples
The effects of an external pressure of 700 bar on an aluminum cylinder with 5 mm (0.197 in) wall thickness
As an example of varying pressures, a finger can be pressed against a wall without making any lasting impression; however, the same finger pushing a thumbtack can easily damage the wall. Although the force applied to the surface is the same, the thumbtack applies more pressure because the point concentrates that force into a smaller area. Pressure is transmitted to solid boundaries or across arbitrary sections of fluid normal to these boundaries or sections at every point. Unlike stress, pressure is defined as a scalar quantity. The negative gradient of pressure is called the force density.[]
Another example is a knife. If we try to cut with the flat edge, force is distributed over a larger surface area resulting in less pressure, and it will not cut. Whereas using the sharp edge, which has less surface area, results in greater pressure, and so the knife cuts smoothly. This is one example of a practical application of pressure.[]
For gases, pressure is sometimes measured not as an absolute pressure, but relative to atmospheric pressure; such measurements are called gauge pressure. An example of this is the air pressure in an automobile tire, which might be said to be "220 kPa (32 psi)", but is actually 220 kPa (32 psi) above atmospheric pressure. Since atmospheric pressure at sea level is about 100 kPa (14.7 psi), the absolute pressure in the tire is therefore about 320 kPa (46 psi). In technical work, this is written "a gauge pressure of 220 kPa (32 psi)". Where space is limited, such as on pressure gauges, name plates, graph labels, and table headings, the use of a modifier in parentheses, such as "kPa (gauge)" or "kPa (absolute)", is permitted. In non-SI technical work, a gauge pressure of 32 psi (220 kPa) is sometimes written as "32 psig", and an absolute pressure as "32 psia", though the other methods explained above that avoid attaching characters to the unit of pressure are preferred.[9]
Gauge pressure is the relevant measure of pressure wherever one is interested in the stress on storage vessels and the plumbing components of fluidics systems. However, whenever equation-of-state properties, such as densities or changes in densities, must be calculated, pressures must be expressed in terms of their absolute values. For instance, if the atmospheric pressure is 100 kPa (15 psi), a gas (such as helium) at 200 kPa (29 psi) (gauge) (300 kPa or 44 psi [absolute]) is 50% denser than the same gas at 100 kPa (15 psi) (gauge) (200 kPa or 29 psi [absolute]). Focusing on gauge values, one might erroneously conclude the first sample had twice the density of the second one.[]
### Scalar nature
In a static gas, the gas as a whole does not appear to move. The individual molecules of the gas, however, are in constant random motion. Because we are dealing with an extremely large number of molecules and because the motion of the individual molecules is random in every direction, we do not detect any motion. If we enclose the gas within a container, we detect a pressure in the gas from the molecules colliding with the walls of our container. We can put the walls of our container anywhere inside the gas, and the force per unit area (the pressure) is the same. We can shrink the size of our "container" down to a very small point (becoming less true as we approach the atomic scale), and the pressure will still have a single value at that point. Therefore, pressure is a scalar quantity, not a vector quantity. It has magnitude but no direction sense associated with it. Pressure force acts in all directions at a point inside a gas. At the surface of a gas, the pressure force acts perpendicular (at right angle) to the surface.[]
A closely related quantity is the stress tensor ?, which relates the vector force ${\displaystyle \mathbf {F} }$ to the vector area ${\displaystyle \mathbf {A} }$ via the linear relation ${\displaystyle \mathbf {F} =\sigma \mathbf {A} }$.
This tensor may be expressed as the sum of the viscous stress tensor minus the hydrostatic pressure. The negative of the stress tensor is sometimes called the pressure tensor, but in the following, the term "pressure" will refer only to the scalar pressure.[]
According to the theory of general relativity, pressure increases the strength of a gravitational field (see stress-energy tensor) and so adds to the mass-energy cause of gravity. This effect is unnoticeable at everyday pressures but is significant in neutron stars, although it has not been experimentally tested.[10]
## Types
### Fluid pressure
Fluid pressure is most often the compressive stress at some point within a fluid. (The term fluid refers to both liquids and gases - for more information specifically about liquid pressure, see section below.)
Water escapes at high speed from a damaged hydrant that contains water at high pressure
Fluid pressure occurs in one of two situations:
1. An open condition, called "open channel flow", e.g. the ocean, a swimming pool, or the atmosphere.
2. A closed condition, called "closed conduit", e.g. a water line or gas line.
Pressure in open conditions usually can be approximated as the pressure in "static" or non-moving conditions (even in the ocean where there are waves and currents), because the motions create only negligible changes in the pressure. Such conditions conform with principles of fluid statics. The pressure at any given point of a non-moving (static) fluid is called the hydrostatic pressure.
Closed bodies of fluid are either "static", when the fluid is not moving, or "dynamic", when the fluid can move as in either a pipe or by compressing an air gap in a closed container. The pressure in closed conditions conforms with the principles of fluid dynamics.
The concepts of fluid pressure are predominantly attributed to the discoveries of Blaise Pascal and Daniel Bernoulli. Bernoulli's equation can be used in almost any situation to determine the pressure at any point in a fluid. The equation makes some assumptions about the fluid, such as the fluid being ideal[11] and incompressible.[11] An ideal fluid is a fluid in which there is no friction, it is inviscid[11] (zero viscosity).[11] The equation for all points of a system filled with a constant-density fluid is[12]
${\displaystyle {\frac {p}{\gamma }}+{\frac {v^{2}}{2g}}+z=\mathrm {const} ,}$
where:
p, pressure of the fluid,
${\displaystyle {\gamma }}$ = ?g, density × acceleration of gravity is the (volume-) specific weight of the fluid,[11]
v, velocity of the fluid,
g, acceleration of gravity,
z, elevation,
${\displaystyle {\frac {p}{\gamma }}}$, pressure head,
${\displaystyle {\frac {v^{2}}{2g}}}$, velocity head.
### Explosion or deflagration pressures
Explosion or deflagration pressures are the result of the ignition of explosive gases, mists, dust/air suspensions, in unconfined and confined spaces.
### Negative pressures
Low-pressure chamber in Bundesleistungszentrum Kienbaum, Germany
While pressures are, in general, positive, there are several situations in which negative pressures may be encountered:
• When dealing in relative (gauge) pressures. For instance, an absolute pressure of 80 kPa may be described as a gauge pressure of -21 kPa (i.e., 21 kPa below an atmospheric pressure of 101 kPa). For example, abdominal decompression is an obstetric procedure during which negative gauge pressure is applied intermittently to a pregnant woman's abdomen.
• Negative absolute pressures are possible. They are effectively tension, and both bulk solids and bulk liquids can be put under negative absolute pressure by pulling on them.[13] Microscopically, the molecules in solids and liquids have attractive interactions that overpower the thermal kinetic energy, so some tension can be sustained. Thermodynamically, however, a bulk material under negative pressure is in a metastable state, and it is especially fragile in the case of liquids where the negative pressure state is similar to superheating and is easily susceptible to cavitation.[14] In certain situations, the cavitation can be avoided and negative pressures sustained indefinitely,[14] for example, liquid mercury has been observed to sustain up to in clean glass containers.[15] Negative liquid pressures are thought to be involved in the ascent of sap in plants taller than 10 m (the atmospheric pressure head of water).[16]
• The Casimir effect can create a small attractive force due to interactions with vacuum energy; this force is sometimes termed "vacuum pressure" (not to be confused with the negative gauge pressure of a vacuum).
• For non-isotropic stresses in rigid bodies, depending on how the orientation of a surface is chosen, the same distribution of forces may have a component of positive pressure along one surface normal, with a component of negative pressure acting along another surface normal.
• The stresses in an electromagnetic field are generally non-isotropic, with the pressure normal to one surface element (the normal stress) being negative, and positive for surface elements perpendicular to this.
• In cosmology, dark energy creates a very small yet cosmically significant amount of negative pressure, which accelerates the expansion of the universe.
### Stagnation pressure
Stagnation pressure is the pressure a fluid exerts when it is forced to stop moving. Consequently, although a fluid moving at higher speed will have a lower static pressure, it may have a higher stagnation pressure when forced to a standstill. Static pressure and stagnation pressure are related by:
${\displaystyle p_{0}={\frac {1}{2}}\rho v^{2}+p}$
where
${\displaystyle p_{0}}$ is the stagnation pressure,
${\displaystyle \rho }$ is the density,
${\displaystyle v}$ is the flow velocity,
${\displaystyle p}$ is the static pressure.
The pressure of a moving fluid can be measured using a Pitot tube, or one of its variations such as a Kiel probe or Cobra probe, connected to a manometer. Depending on where the inlet holes are located on the probe, it can measure static pressures or stagnation pressures.
### Surface pressure and surface tension
There is a two-dimensional analog of pressure - the lateral force per unit length applied on a line perpendicular to the force.
Surface pressure is denoted by ?:
${\displaystyle \pi ={\frac {F}{l}}}$
and shares many similar properties with three-dimensional pressure. Properties of surface chemicals can be investigated by measuring pressure/area isotherms, as the two-dimensional analog of Boyle's law, , at constant temperature.
Surface tension is another example of surface pressure, but with a reversed sign, because "tension" is the opposite to "pressure".
### Pressure of an ideal gas
In an ideal gas, molecules have no volume and do not interact. According to the ideal gas law, pressure varies linearly with temperature and quantity, and inversely with volume:
${\displaystyle p={\frac {nRT}{V}},}$
where:
p is the absolute pressure of the gas,
n is the amount of substance,
T is the absolute temperature,
V is the volume,
R is the ideal gas constant.
Real gases exhibit a more complex dependence on the variables of state.[17]
### Vapour pressure
Vapour pressure is the pressure of a vapour in thermodynamic equilibrium with its condensed phases in a closed system. All liquids and solids have a tendency to evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid or solid form.
The atmospheric pressure boiling point of a liquid (also known as the normal boiling point) is the temperature at which the vapor pressure equals the ambient atmospheric pressure. With any incremental increase in that temperature, the vapor pressure becomes sufficient to overcome atmospheric pressure and lift the liquid to form vapour bubbles inside the bulk of the substance. Bubble formation deeper in the liquid requires a higher pressure, and therefore higher temperature, because the fluid pressure increases above the atmospheric pressure as the depth increases.
The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial vapor pressure.
### Liquid pressure
When a person swims under the water, water pressure is felt acting on the person's eardrums. The deeper that person swims, the greater the pressure. The pressure felt is due to the weight of the water above the person. As someone swims deeper, there is more water above the person and therefore greater pressure. The pressure a liquid exerts depends on its depth.
Liquid pressure also depends on the density of the liquid. If someone was submerged in a liquid more dense than water, the pressure would be correspondingly greater. Thus, we can say that the depth, density and liquid pressure are directly proportionate. The pressure due to a liquid in liquid columns of constant density or at a depth within a substance is represented by the following formula:
${\displaystyle p=\rho gh,}$
where:
p is liquid pressure,
g is gravity at the surface of overlaying material,
? is density of liquid,
h is height of liquid column or depth within a substance.
Another way of saying the same formula is the following:
${\displaystyle p={\text{weight density}}\times {\text{depth}}.}$
The pressure a liquid exerts against the sides and bottom of a container depends on the density and the depth of the liquid. If atmospheric pressure is neglected, liquid pressure against the bottom is twice as great at twice the depth; at three times the depth, the liquid pressure is threefold; etc. Or, if the liquid is two or three times as dense, the liquid pressure is correspondingly two or three times as great for any given depth. Liquids are practically incompressible - that is, their volume can hardly be changed by pressure (water volume decreases by only 50 millionths of its original volume for each atmospheric increase in pressure). Thus, except for small changes produced by temperature, the density of a particular liquid is practically the same at all depths.
Atmospheric pressure pressing on the surface of a liquid must be taken into account when trying to discover the total pressure acting on a liquid. The total pressure of a liquid, then, is ?gh plus the pressure of the atmosphere. When this distinction is important, the term total pressure is used. Otherwise, discussions of liquid pressure refer to pressure without regard to the normally ever-present atmospheric pressure.
The pressure does not depend on the amount of liquid present. Volume is not the important factor - depth is. The average water pressure acting against a dam depends on the average depth of the water and not on the volume of water held back. For example, a wide but shallow lake with a depth of 3 m (10 ft) exerts only half the average pressure that a small 6 m (20 ft) deep pond does. (The total force applied to the longer dam will be greater, due to the greater total surface area for the pressure to act upon. But for a given 5-foot (1.5 m)-wide section of each dam, the 10 ft (3.0 m) deep water will apply one quarter the force of 20 ft (6.1 m) deep water). A person will feel the same pressure whether their head is dunked a metre beneath the surface of the water in a small pool or to the same depth in the middle of a large lake. If four vases contain different amounts of water but are all filled to equal depths, then a fish with its head dunked a few centimetres under the surface will be acted on by water pressure that is the same in any of the vases. If the fish swims a few centimetres deeper, the pressure on the fish will increase with depth and be the same no matter which vase the fish is in. If the fish swims to the bottom, the pressure will be greater, but it makes no difference what vase it is in. All vases are filled to equal depths, so the water pressure is the same at the bottom of each vase, regardless of its shape or volume. If water pressure at the bottom of a vase were greater than water pressure at the bottom of a neighboring vase, the greater pressure would force water sideways and then up the narrower vase to a higher level until the pressures at the bottom were equalized. Pressure is depth dependent, not volume dependent, so there is a reason that water seeks its own level.
Restating this as energy equation, the energy per unit volume in an ideal, incompressible liquid is constant throughout its vessel. At the surface, gravitational potential energy is large but liquid pressure energy is low. At the bottom of the vessel, all the gravitational potential energy is converted to pressure energy. The sum of pressure energy and gravitational potential energy per unit volume is constant throughout the volume of the fluid and the two energy components change linearly with the depth.[18] Mathematically, it is described by Bernoulli's equation, where velocity head is zero and comparisons per unit volume in the vessel are
${\displaystyle {\frac {p}{\gamma }}+z=\mathrm {const} .}$
Terms have the same meaning as in section Fluid pressure.
### Direction of liquid pressure
An experimentally determined fact about liquid pressure is that it is exerted equally in all directions.[19] If someone is submerged in water, no matter which way that person tilts their head, the person will feel the same amount of water pressure on their ears. Because a liquid can flow, this pressure isn't only downward. Pressure is seen acting sideways when water spurts sideways from a leak in the side of an upright can. Pressure also acts upward, as demonstrated when someone tries to push a beach ball beneath the surface of the water. The bottom of a boat is pushed upward by water pressure (buoyancy).
When a liquid presses against a surface, there is a net force that is perpendicular to the surface. Although pressure doesn't have a specific direction, force does. A submerged triangular block has water forced against each point from many directions, but components of the force that are not perpendicular to the surface cancel each other out, leaving only a net perpendicular point.[19] This is why water spurting from a hole in a bucket initially exits the bucket in a direction at right angles to the surface of the bucket in which the hole is located. Then it curves downward due to gravity. If there are three holes in a bucket (top, bottom, and middle), then the force vectors perpendicular to the inner container surface will increase with increasing depth - that is, a greater pressure at the bottom makes it so that the bottom hole will shoot water out the farthest. The force exerted by a fluid on a smooth surface is always at right angles to the surface. The speed of liquid out of the hole is ${\displaystyle \scriptstyle {\sqrt {2gh}}}$, where h is the depth below the free surface.[19] This is the same speed the water (or anything else) would have if freely falling the same vertical distance h.
### Kinematic pressure
${\displaystyle P=p/\rho _{0}}$
is the kinematic pressure, where ${\displaystyle p}$ is the pressure and ${\displaystyle \rho _{0}}$ constant mass density. The SI unit of P is m2/s2. Kinematic pressure is used in the same manner as kinematic viscosity ${\displaystyle \nu }$ in order to compute the Navier-Stokes equation without explicitly showing the density ${\displaystyle \rho _{0}}$.
Navier-Stokes equation with kinematic quantities
${\displaystyle {\frac {\partial u}{\partial t}}+(u\nabla )u=-\nabla P+\nu \nabla ^{2}u.}$
## Notes
1. ^ The preferred spelling varies by country and even by industry. Further, both spellings are often used within a particular industry or country. Industries in British English-speaking countries typically use the "gauge" spelling.
## References
1. ^ Knight, PhD, Randall D. (2007). "Fluid Mechanics". Physics for Scientists and Engineers: A Strategic Approach (google books) (2nd ed.). San Francisco: Pearson Addison Wesley. p. 1183. ISBN 978-0-321-51671-8. Retrieved 2020. Pressure itself is not a Force, even though we sometimes talk "informally" about the "force exerted by the pressure. The correct statement is that the Fluid exerts a force on a surface. In addition, Pressure is a scalar, not a vector.
2. ^ Giancoli, Douglas G. (2004). Physics: principles with applications. Upper Saddle River, N.J.: Pearson Education. ISBN 978-0-13-060620-4.
3. ^ McNaught, A. D.; Wilkinson, A.; Nic, M.; Jirat, J.; Kosata, B.; Jenkins, A. (2014). IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). 2.3.3. Oxford: Blackwell Scientific Publications. doi:10.1351/goldbook.P04819. ISBN 978-0-9678550-9-7. Archived from the original on 2016-03-04.
4. ^ R Nave. "Pressure". Hyperphysics. Georgia State University, Dept. of Physics and Astronomy. Retrieved .
5. ^ Alberty, Robert A. (2001). "USE OF LEGENDRE TRANSFORMS IN CHEMICAL THERMODYNAMICS (IUPAC Technical Report)" (PDF). Pure Appl. Chem. 73 (8): 1349-1380. doi:10.1351/pac200173081349. S2CID 98264934. Retrieved 2021. See Table 1 Conjugate pairs of variables ... (p.1357)
6. ^ "14th Conference of the International Bureau of Weights and Measures". Bipm.fr. Archived from the original on 2007-06-30. Retrieved .
7. ^ International Bureau of Weights and Measures (2006), The International System of Units (SI) (PDF) (8th ed.), p. 127, ISBN 92-822-2213-6, archived (PDF) from the original on 2021-06-04, retrieved
8. ^ a b "U.S. Navy Diving Manual (Chapter 2:Underwater Physics)" (PDF). pp. 2-32. Archived (PDF) from the original on 2017-02-02.
9. ^ a b "Rules and Style Conventions for Expressing Values of Quantities". NIST. 2 July 2009. Archived from the original on 2009-07-10. Retrieved .
10. ^ Vishwakarma, Ram Gopal (2009). "Einstein's gravity under pressure". Astrophysics and Space Science. 321 (2): 151-156. arXiv:0705.0825. Bibcode:2009Ap&SS.321..151V. doi:10.1007/s10509-009-0016-8. S2CID 218673952.
11. Finnemore, John, E. and Joseph B. Franzini (2002). Fluid Mechanics: With Engineering Applications. New York: McGraw Hill, Inc. pp. 14-29. ISBN 978-0-07-243202-2.{{cite book}}: CS1 maint: multiple names: authors list (link)
12. ^ NCEES (2011). Fundamentals of Engineering: Supplied Reference Handbook. Clemson, South Carolina: NCEES. p. 64. ISBN 978-1-932613-59-9.
13. ^ Imre, A. R. (2007). "How to generate and measure negative pressure in liquids?". Soft Matter under Exogenic Impacts. NATO Science Series II: Mathematics, Physics and Chemistry. Vol. 242. pp. 379-388. doi:10.1007/978-1-4020-5872-1_24. ISBN 978-1-4020-5871-4. ISSN 1568-2609.
14. ^ a b Imre, A. R; Maris, H. J; Williams, P. R, eds. (2002). Liquids Under Negative Pressure (Nato Science Series II). Springer. doi:10.1007/978-94-010-0498-5. ISBN 978-1-4020-0895-5.
15. ^ Briggs, Lyman J. (1953). "The Limiting Negative Pressure of Mercury in Pyrex Glass". Journal of Applied Physics. 24 (4): 488-490. Bibcode:1953JAP....24..488B. doi:10.1063/1.1721307. ISSN 0021-8979.
16. ^ Karen Wright (March 2003). "The Physics of Negative Pressure". Discover. Archived from the original on 8 January 2015. Retrieved 2015.
17. ^ P. Atkins, J. de Paula Elements of Physical Chemistry, 4th Ed, W. H. Freeman, 2006. ISBN 0-7167-7329-5.
18. ^ Streeter, V. L., Fluid Mechanics, Example 3.5, McGraw–Hill Inc. (1966), New York.
19. ^ a b c Hewitt 251 (2006)[full ] | 7,500 | 31,915 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.874717 |
http://math.stackexchange.com/questions/20178/given-a-commutative-ring-r-and-an-epimorphism-rm-to-rn-is-then-m-geq-n | 1,469,536,451,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824853.47/warc/CC-MAIN-20160723071024-00319-ip-10-185-27-174.ec2.internal.warc.gz | 157,943,458 | 19,856 | # Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?
If $\varphi:R^{m}\to R^{n}$ is an epimorphism of free modules over a commutative ring, does it follow that $m \geq n$?
This is obviously true for vector spaces over a field, but how would one show this over just a commutative ring?
-----Edit
Is there any way to use the following?
If $\varphi : M \to M'$ is an epimorphism of left $S$-modules and $N$ is any right $S$-module then $id_N \otimes \varphi$ is an epimorphism.
-
@Chandru1: what's wrong with the question? It seems perfectly clear to me. (In fact it is a standard question: one of the exercises in Atiyah-Macdonald.) Hint: tensor with $R/\mathfrak{m}$, where $\mathfrak{m}$ is a maximal ideal of $R$. – Pete L. Clark Feb 3 '11 at 2:57
@Chandru1: The question is: If $\varphi: R^{m} \to R^{n}$ is a epimorphism of free modules over a commutative ring, does it follow that $m \geq n$? As Pete pointed out, this follows from the first sentence in the question by tensoring with $R / \mathfrak{m}$. – t.b. Feb 3 '11 at 3:05
@Pete: You could have put that as an answer. – anonymous Feb 3 '11 at 3:09
Google "invariant basis number" (IBN) for more than you wanted to know. – Bill Dubuque Feb 3 '11 at 3:50
Commenting so that this will be visible at the top: This question is Problem 1.4 on the homework for Math 620, at the University of Buffalo. As user "Student" points out, every question user6560 has asked is a homework question from that course. math.buffalo.edu/~badzioch/MTH620/Homework_files/hw1.pdf – David Speyer Feb 13 '11 at 19:28
As mentioned in the comments to the question, it’s the first part of Exercise 2.11 in Atiyah-MacDonald, and I refer to the comments for an answer.
The second part of Exercise 2.11 (which is perhaps more interesting) has been the subject of this MO question.
I especially like Balazs Strenner’s answer.
EDIT. Here is Exercise 2.11 of Atiyah-MacDonald. Let $A$ be a nonzero commutative ring and $\phi:A^m\to A^n$ an $A$-linear map. Then
(a) $m\ge n$ if $\phi$ is surjective,
(b) $m\le n$ if $\phi$ is injective,
As pointed out in the comments to the question, there is an obvious proof of (a) [tensor with $A/\mathfrak m$, $\mathfrak m$ maximal]. My favorite proof of (b) is Strenner's one mentioned above. A natural question is: Can one use Strenner's argument to prove (a) and (b) at one go? I'll try to do that below.
Lemma. Let $B$ be a commutative ring, $A$ a subring, $b$ a nonzero element of $B$ which is integral over $A$ and which is not a zero divisor. Then there is a nonzero $a$ in $A$ and a monic $f$ in $A[X]$ such that $a=bf(b)$.
Proof. Let $g\in A[X]$ be a least degree monic polynomial annihilating $b$. Such exists because $b$ is integral over $A$. The constant term $a$ of $g$ is nonzero because $b$ is nonzero and not a zero divisor. QED
Assume (a) [resp. (b)] is false. Then there is an $n$ and a surjective [resp. injective] endomorphism $b$ of $A^n$ satisfying $b(e_n)=0$ [resp. $b(A^n)\subseteq A^{n-1}$], where $e_n$ is the last vector of the canonical basis and $A^{n-1}$ is the span of all the other vectors of this basis. Then $b$ is integral over $A$ by Cayley-Hamilton, and we get a contradiction by using the lemma (with $B:=A[b]$) and applying $a=bf(b)$ to $e_n$.
-
@PierreYvesGaillard Is there a reason why $b(e_n)$ is in $A^{n-1}$ and not $A^n$? – user38268 Apr 12 '12 at 13:55
Dear @BenjaminLim: We assume that (b) is false, which means (swapping $m$ and $n$) that there is an injective $\phi:A^n\to A^m$ with $n > m$. Putting $b:=i\circ\phi$, where $i$ is the natural injection $A^m\hookrightarrow A^n$, we get $be_j\in A^m\subset A^{n-1}$ for all $j$. – Pierre-Yves Gaillard Apr 12 '12 at 14:34
@RobertCardona - Which isomorphism are you referring to? – Pierre-Yves Gaillard Nov 6 '15 at 11:53
@RobertCardona - The assumption of Atiyah-MacDonald' exercise is that $A$ is a nonzero commutative ring and $\phi:A^m\to A^n$ is an $A$-linear map. (There is no $R$.) Sorry, I don't understand your answer. – Pierre-Yves Gaillard Nov 6 '15 at 15:42 | 1,317 | 4,076 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-30 | latest | en | 0.905159 |
http://www.numbersaplenty.com/2396 | 1,590,393,567,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388012.14/warc/CC-MAIN-20200525063708-20200525093708-00561.warc.gz | 196,864,035 | 3,387 | Search a number
2396 = 22599
BaseRepresentation
bin100101011100
310021202
4211130
534041
615032
76662
oct4534
93252
102396
111889
121478
131124
14c32
15a9b
hex95c
2396 has 6 divisors (see below), whose sum is σ = 4200. Its totient is φ = 1196.
The previous prime is 2393. The next prime is 2399. The reversal of 2396 is 6932.
It is a happy number.
2396 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is an interprime number because it is at equal distance from previous prime (2393) and next prime (2399).
It is a d-powerful number, because it can be written as 27 + 93 + 35 + 64 .
It is a plaindrome in base 11, base 12 and base 13.
It is a nialpdrome in base 7 and base 14.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (2393) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 296 + ... + 303.
It is an arithmetic number, because the mean of its divisors is an integer number (700).
22396 is an apocalyptic number.
It is an amenable number.
2396 is a deficient number, since it is larger than the sum of its proper divisors (1804).
2396 is a wasteful number, since it uses less digits than its factorization.
2396 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 603 (or 601 counting only the distinct ones).
The product of its digits is 324, while the sum is 20.
The square root of 2396 is about 48.9489530021. The cubic root of 2396 is about 13.3812167216.
The spelling of 2396 in words is "two thousand, three hundred ninety-six".
Divisors: 1 2 4 599 1198 2396 | 505 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2020-24 | latest | en | 0.89938 |
https://www.investopedia.com/ask/answers/050115/what-difference-between-carrying-value-and-fair-value.asp?utm_campaign=rss_headlines&utm_source=rss&utm_medium=referral | 1,582,948,480,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148375.36/warc/CC-MAIN-20200229022458-20200229052458-00519.warc.gz | 766,453,969 | 25,621 | ## An Overview of Carrying Value and Fair Value
Carrying value and fair value are two different accounting measures used to determine the value of a company's assets.
The carrying value, or book value, is an asset value based on the company's balance sheet, which takes the cost of the asset and subtracts its depreciation over time. The fair value of an asset is usually determined by the market and agreed upon by a willing buyer and seller, and it can fluctuate often. In other words, the carrying value generally reflects equity, while the fair value reflects the current market price.
Because the fair value of an asset can be more volatile than its carrying value or book value, it's possible for big discrepancies to occur between the two measures. The market value can be higher or lower than the carrying value at any time. These differences usually aren't examined until assets are appraised or sold to help determine if they're undervalued or overvalued.
## Carrying Value
The carrying value of an asset is based on the figures from a company's balance sheet. When a company initially acquires an asset, its carrying value is the same as its original cost. However, this changes over time. To calculate the carrying value or book value of an asset at any point in time, you must subtract any accumulated depreciation, amortization, or impairment expenses from its original cost.
1:25
### Example of Carrying Value
Let's say company ABC bought a 3D printing machine to design prototypes of its product. The 3D printing machine costs \$50,000 and has a depreciation expense of \$3,000 per year over its useful life of 15 years under the straight line basis of calculating depreciation and amortization.
Straight line basis is a simple way to calculate the loss of an asset's value over time. This calculation is particularly useful for physical assets—such as a piece of equipment—that a company might sell in whole or in parts at the end of its useful life. Therefore, the book value of the 3D printing machine after 15 years is \$5,000, or \$50,000 - (\$3,000 x 15).
## Fair Value
Different from the carrying value, the fair value of assets and liabilities is calculated on a mark-to-market accounting basis. In other words, the fair value of an asset is the amount paid in a transaction between participants if it's sold in the open market. A willing buyer and seller have agreed upon this value. Due to the changing nature of open markets, however, the fair value of an asset can fluctuate greatly over time.
### Example of Fair Value
Let's say an investment company has long positions in stocks in its portfolio. By having long positions, the company anticipates favorable market conditions, also known as a "bull market." The company is holding onto these stocks with the expectation they will rise in price over time.
The investment company's original cost of these assets was \$6 million. However, after two negative gross domestic product (GDP) rates, the market experiences a significant downturn. The company's portfolio falls 40% in value, to \$3.6 million. Therefore, the fair value of the asset is \$3.6 million, or \$6 million - (\$6 million x 0.40).
Determining the fair value of an asset can be difficult if a competitive, open market for it doesn't exist—an unusual piece of equipment in a manufacturing plant, for example.
### Key Takeaways
• Carrying value and fair value are two different accounting measures used to determine the value of a company's assets.
• The carrying value of an asset is based on the figures from a company's balance sheet.
• The fair value of an asset is the amount paid in a transaction between participants if it's sold in the open market. | 781 | 3,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-10 | latest | en | 0.947336 |
https://rdrr.io/cran/bnlearn/src/R/choose.direction.R | 1,638,739,200,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363216.90/warc/CC-MAIN-20211205191620-20211205221620-00129.warc.gz | 542,965,415 | 12,545 | # R/choose.direction.R In bnlearn: Bayesian Network Structure Learning, Parameter Learning and Inference
#### Defines functions choose.direction.bootchoose.direction.scorechoose.direction.testchoose.direction.decideis.legal.arc
```is.legal.arc = function(arc, test, data) {
if (test %in% c(available.mixedcg.tests, available.mixedcg.scores))
if (is(data[, arc[1]], "numeric") && is(data[, arc[2]], "factor"))
return(FALSE)
return(TRUE)
}#IS.LEGAL.ARC
choose.direction.decide = function(x, arc, a, b, t, criterion, debug) {
nodes = names(x\$nodes)
updated.nodes = character(0)
# check whether any of the two directions causes cycles in the graph.
amat = arcs2amat(x\$arcs, nodes)
cycles1 = has.path(arc[2], arc[1], nodes, amat, exclude.direct = TRUE)
cycles2 = has.path(arc[1], arc[2], nodes, amat, exclude.direct = TRUE)
if (criterion == "score") {
good.enough = function(a) a > t
better = function(a, b) a > b
same = function(a, b) isTRUE(all.equal(as.numeric(a), as.numeric(b)))
err.same = " @ nothing to do, same score delta modulo precision tolerance.\n"
err.neg = " @ nothing to do, both score deltas are negative.\n"
}#THEN
else if (criterion == "boot") {
good.enough = function(a) TRUE
better = function(a, b) a\$direction > b\$direction
same = function(a, b) isTRUE(all.equal(as.numeric(a\$direction), as.numeric(b\$direction)))
err.same = " @ nothing to do, same confidence modulo precision tolerance.\n"
err.neg = paste0(" @ nothing to do, confidence is less than ", t, ".\n")
if (a\$strength < t) {
if (debug)
cat(err.neg)
return(x)
}#THEN
}#THEN
else if (criterion == "test") {
good.enough = function(a) !is.na(a) && (a < t)
better = function(a, b) ifelse(is.na(a), Inf, a) < ifelse(is.na(b), Inf, b)
same = function(a, b) isTRUE(all.equal(as.numeric(a), as.numeric(b)))
err.same = " @ nothing to do, same p-value modulo precision tolerance.\n"
err.neg = paste0(" @ nothing to do, p-value is less than ", t, ".\n")
}#THEN
if (cycles1 && cycles2) {
# bothe A -> B and B -> A introduce cycles (should not be here).
if (debug)
cat(" @ nothing to do, both arc create cycles.\n")
}#THEN
else if (cycles1 && !cycles2) {
# A -> B introduces cycles, B -> A does not.
if (debug)
cat(" > adding", arc[1], "->", arc[2], "creates cycles!.\n")
if (good.enough(b)) {
if (debug)
cat(" @ arc", arc[2], "->", arc[1], "is better .\n")
# update the arc set.
x\$arcs = set.arc.direction(arc[2], arc[1], x\$arcs)
# check which nodes have to be updated.
updated.nodes = unique(c(arc, x\$nodes[[arc[1]]]\$mb, x\$nodes[[arc[2]]]\$mb))
}#THEN
else {
if (debug)
cat(" > arc", arc[2], "->", arc[1], "isn't good, either.\n")
}#ELSE
}#THEN
else if (cycles2 && !cycles1) {
# B -> A introduces cycles, A -> B does not.
if (debug)
cat(" > adding", arc[2], "->", arc[1], "creates cycles!.\n")
if (good.enough(a)) {
if (debug)
cat(" @ arc", arc[1], "->", arc[2], "is better .\n")
# update the arc set.
x\$arcs = set.arc.direction(arc[1], arc[2], x\$arcs)
# check which nodes have to be updated.
updated.nodes = unique(c(arc, x\$nodes[[arc[1]]]\$mb, x\$nodes[[arc[2]]]\$mb))
}#THEN
else {
if (debug)
cat(" > arc", arc[1], "->", arc[2], "isn't good, either.\n")
}#ELSE
}#THEN
else if (!cycles1 && !cycles2) {
# neither A -> B nor B -> A introduce cycles.
if (same(a, b)) {
if (debug)
cat(err.same)
}#THEN
else if (!good.enough(a) && !good.enough(b)) {
if (debug)
cat(err.neg)
}#THEN
else {
id = if (better(a, b)) 1:2 else 2:1
if (debug)
cat(" @ arc", arc[id[1]], "->", arc[id[2]], "is better .\n")
# update the arc set.
x\$arcs = set.arc.direction(arc[id[1]], arc[id[2]], x\$arcs)
# check which nodes have to be updated.
updated.nodes = unique(c(arc, x\$nodes[[arc[1]]]\$mb, x\$nodes[[arc[2]]]\$mb))
}#ELSE
}#THEN
# update the chosen nodes.
for (node in updated.nodes)
x\$nodes[[node]] = cache.partial.structure(nodes, target = node,
arcs = x\$arcs, debug = FALSE)
return(x)
}#CHOOSE.DIRECTION.DECIDE
choose.direction.test = function(x, arc, data, test, alpha, B, complete,
debug = FALSE) {
# you can't help but notice nodes connected by undirected arcs are
# included, too? wonder why?
# because if they, too, are parents of the node to be tested
# they _do_ belong there; if they are not, the node distribution
# does not depend on them so they are largely irrelevant.
if (any(arc %in% undirected.arcs(x)))
warning("the graph is not completely directed around ", arc[1], " and ",
arc[2], ", treating nodes connected by undirected arcs as parents.")
choose.direction.test.pvalue = function(arc) {
if (!is.legal.arc(arc, test, data)) {
if (debug)
cat(" > arc", arc[1], "->", arc[2], "is not a valid arc for this network.\n")
return(NaN)
}#THEN
parents = x\$nodes[[arc[2]]]\$nbr
a = indep.test(arc[1], arc[2], parents[parents != arc[1]], data = data,
test = test, B = B, alpha = alpha, complete = complete)
if (debug) {
cat(" > testing", arc[1], "->", arc[2], "with conditioning set '",
parents[parents != arc[1]], "'.\n")
cat(" > p-value is", a, ".\n")
}#THEN
return(a)
}#CHOOSE.DIRECTION.TEST.PVALUE
a1 = choose.direction.test.pvalue(arc)
a2 = choose.direction.test.pvalue(arc[2:1])
choose.direction.decide(x = x, arc = arc, a = a1, b = a2, t = alpha,
criterion = "test", debug = debug)
}#CHOOSE.DIRECTION.TEST
choose.direction.score = function(x, data, arc, score, extra.args, debug = FALSE) {
# do a backup copy of the network structure.
x2 = x
# drop any existing arc between arc["from"] and arc["to"].
x\$arcs = drop.arc.backend(x\$arcs, arc)
# you can't help but notice nodes connected by undirected arcs are
# included, too? wonder why?
# because if they, too, are parents of the node to be tested
# they _do_ belong there; if they are not, the node distribution
# does not depend on them so they are largely irrelevant.
if (any(arc %in% undirected.arcs(x)))
warning("the graph is not completely directed around ", arc[1], " and ",
arc[2], ", treating nodes connected by undirected arcs as parents.")
x\$nodes[[arc[1]]]\$parents = x\$nodes[[arc[1]]]\$nbr
x\$nodes[[arc[2]]]\$parents = x\$nodes[[arc[2]]]\$nbr
x\$nodes[[arc[1]]]\$parents = x\$nodes[[arc[1]]]\$parents[x\$nodes[[arc[1]]]\$parents != arc[2]]
x\$nodes[[arc[2]]]\$parents = x\$nodes[[arc[2]]]\$parents[x\$nodes[[arc[2]]]\$parents != arc[1]]
# compute the initial score of the nodes involved.
reference.score = per.node.score(network = x, score = score,
targets = arc, extra.args = extra.args, data = data)
# check whether the score is decomposable.
decomp = is.score.decomposable(score, extra.args)
# compare the scores of the two networks.
choose.direction.score.delta = function(arc) {
better = score.delta(arc = arc, network = x, data = data,
score = score, score.delta = 0,
reference.score = reference.score, op = "set",
extra = extra.args, decomposable = decomp)
if (debug) {
cat(" > initial score for node", arc[2], "is", reference.score[arc[2]], ".\n")
if (is.legal.arc(arc, score, data))
cat(" > score delta for arc", arc[1], "->", arc[2], "is", better\$delta, ".\n")
else
cat(" > arc", arc[1], "->", arc[2], "is not a valid arc for this network.\n")
}#THEN
return(better)
}#CHOOSE.DIRECTION.SCORE.DELTA
a = choose.direction.score.delta(arc)
b = choose.direction.score.delta(arc[2:1])
choose.direction.decide(x = x2, arc = arc, a = a\$delta, b = b\$delta, t = 0,
criterion = "score", debug = debug)
}#CHOOSE.DIRECTION.SCORE
choose.direction.boot = function(x, data, arc, extra.args, algorithm,
algorithm.args, cpdag = TRUE, debug = FALSE) {
# build a separate arc set with the two directions of the arc.
m = matrix(c(arc, rev(arc)), ncol = 2, byrow = TRUE,
dimnames = list(c(), c("from", "to")))
# compute the respective bootstrap strength/direction
res = arc.strength.boot(data = data, R = extra.args\$R, m = extra.args\$m,
algorithm = algorithm, algorithm.args = algorithm.args,
arcs = m, cpdag = cpdag, debug = FALSE)
if (debug) {
cat(" > testing", arc[1], "->", arc[2], "\n")
cat(" > bootstrap probability of an arc between", arc[1], "and", arc[2],
"is", res[1, "strength"], ".\n")
if (is.legal.arc(arc, "loglik-cg", data))
cat(" > direction confidence for arc", arc[1], "->", arc[2], "is",
res[1, "direction"], ".\n")
else
cat(" > arc", arc[1], "->", arc[2], "is not a valid arc for this network.\n")
if (is.legal.arc(arc[2:1], "loglik-cg", data))
cat(" > direction confidence for arc", arc[2], "->", arc[1], "is",
res[2, "direction"], ".\n")
else
cat(" > arc", arc[2], "->", arc[1], "is not a valid arc for this network.\n")
}#THEN
choose.direction.decide(x = x, arc = arc, a = res[1, ], b = res[2, ], t = 0.5,
criterion = "boot", debug = debug)
}#CHOOSE.DIRECTION.BOOT
```
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bnlearn documentation built on Sept. 7, 2021, 1:07 a.m. | 2,741 | 8,826 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.532567 |
http://www.answers.com/Q/How_many_working_days_in_a_year | 1,516,489,876,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889736.54/warc/CC-MAIN-20180120221621-20180121001621-00063.warc.gz | 393,036,530 | 47,684 | What would you like to do?
# How many working days in a year?
Would you like to merge this question into it?
#### already exists as an alternate of this question.
Would you like to make it the primary and merge this question into it?
#### exists and is an alternate of .
In the US people commonly say there are 260 weekdays/work days in a year or 2080 work hours in a year based on 52 weeks per year.
(260 = 5wkdys/wk*52wks)
(2080=40hrs/wk*52wks/yr)
However that varies by year.
• 2008 has 262 weekdays and 366 total days
• 2009 has 261 weekdays and 365 total days
• 2010 has 261 weekdays and 365 total days
• 2011 has 260 weekdays and 365 total days
• 2012 has 261 weekdays and 366 total days
• etc... (hopefully 2012 world ending theories are untrue)
However, few people work every single weekday so the "work day" count is skewed when you take into account Holidays that happen to fall on weekdays (up to 6 major US ones per year) , Vacation Time, and Sick Time.
note: Federal and State employees typically have 10 holidays per year with an 11th one occurring every 4th year (for the Presidential Inauguration) regardless of whether they fall on weekends or not. If their holiday falls on a Saturday then it is recognized on Friday and if it falls on Sunday then it is recognized the next Monday.
Weekdays per year 260-262.
Weekdays minus holidays falling on weekdays 255-258.
Years with only 260 weekdays: 1905 1910 1911 1916 1921 1922 1927 1933 1938 1939 1944 1949 1950 1955 1961 1966 1967 1972 1977 1978 1983 1989 1994 1995 2000 2005 2006 2011 2017 2022 2023 2028 2033 2034 2039 2045 2050 2051 2056 2061 2062 2067 2073 2078 2079 2084 2089 2090 2095
Years with 262 weekdays:
1908 1912 1920 1924 1936 1940 1948 1952 1964 1968 1976 1980 1992 1996 2004 2008 2020 2024 2032 2036 2048 2052 2060 2064 2076 2080 2088 2092
Years with only 255 weekdays minus holidays falling on weekdays (AWESOME!)
1900 1901 1902 1905 1906 1907 1911 1913 1916 1917 1918 1919 1922 1923 1929 1930 1933 1934 1935 1939 1941 1944 1945 1946 1947 1950 1951 1957 1958 1961 1962 1963 1967 1969 1972 1973 1974 1975 1978 1979 1985 1986 1989 1990 1991 1995 1997 2000 2001 2002 2003 2006 2007 2013 2014 2017 2018 2019 2023 2025 2028 2029 2030 2031 2034 2035 2041 2042 2045 2046 2047 2051 2053 2056 2057 2058 2059 2062 2063 2069 2070 2073 2074 2075 2079 2081 2084 2085 2086 2087 2090 2091 2097 2098
Years with 258 weekdays minus holidays falling on weekdays (YUCK!):
1900 1901 1902 1905 1906 1907 1911 1913 1916 1917 1918 1919 1922 1923 1929 1930 1933 1934 1935 1939 1941 1944 1945 1946 1947 1950 1951 1957 1958 1961 1962 1963 1967 1969 1972 1973 1974 1975 1978 1979 1985 1986 1989 1990 1991 1995 1997 2000 2001 2002 2003 2006 2007 2013 2014 2017 2018 2019 2023 2025 2028 2029 2030 2031 2034 2035 2041 2042 2045 2046 2047 2051 2053 2056 2057 2058 2059 2062 2063 2069 2070 2073 2074 2075 2079 2081 2084 2085 2086 2087 2090 2091 2097 2098
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# How many work days are in a year?
For people paid every two weeks, there are 26 pay periods in a year. Two weeks makes 10 business days, for a total of 260 working days, not accounting for vacation time that y
# How many days do americans work in a year?
The average for Americans who work a full time job has shifted slightly since 2002 when the average work hours hovered around 1836. In the years since the average of a US
# How many work week days are there in a year?
52*5 = 260 workdays Minus 8 Bank Holidays = 252 Plus one day (since the year is 52 weeks and one day long) = 253 Plus one day in a leap year = 253.25 on avera
# How many days do teachers work a year?
It depends on the state, but somewhere in the neighborhood of 190 days isn't uncommon in the United States. Mind you that teachers often work 14 hour days and spend weeken
# How many days a year do vets work?
Generally at least 300 - five days a week plus a number of Saturdays or Sundays (or both) covering emergencies. However, this depends on the hours the clinic is open, whether
# How many work days are there in nine years?
To answer this, you need to know how many work days there are in one year. That depends on the company, its definition of a work week, and its number of allowable paid n
# How many working days in year in Australia?
365 365 365 Poor things, you mean you don't get a national holiday?
# How many days a year does a dentist work?
On average, dentists work three days a week. There are 52 weeks in a year, so on average dentists work 156 days a year. Of course, this is an average so there are dentists tha
# How many work days in year?
Roughly 260 in the UK - It depends which country you're in. Even in the UK some bank holidays aren't observed in all countries.
# How many days a year does a architect work?
about 100 days a year, (365)
# How many working days a year in Philippines?
We have 260 days, excluding saturday and sunday, with 10 regular holiday and 7 non-working holidays. if all falls on the weekdays we have 242 working days. If any of the h
# How many work day in a year?
That depends on how many holiday in a year in your country.
# How many days a year are there working days?
365 minus 104 weekends minus 30 vacation minus public holidays = 220 - 225 days
In Science
240
# How many days a year does a judge work?
well i dont know but recently discovered not all year m doing something in a la class and researched this for a project!
In Science
not enoiugh
In Science
# How many working days in a year in BC?
Statutory holidays in British Columbia, Canada, are New Years Day, Family Day (second Monday in February), Good Friday, Easter Monday, Victoria Day, Canada Day, BC Day, Labour | 1,916 | 5,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-05 | latest | en | 0.872134 |
https://www.fel.cvut.cz/en/education/bk/predmety/12/79/p12798504.html | 1,560,916,227,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998882.88/warc/CC-MAIN-20190619023613-20190619045613-00072.warc.gz | 751,062,111 | 8,525 | # Subject description - AE1M01MPE
Summary of Study | Summary of Branches | All Subject Groups | All Subjects | List of Roles | Explanatory Notes Instructions
AE1M01MPE Mathematics for Economy Extent of teaching: 4+2 Guarantors: Roles: P,V Language ofteaching: EN Teachers: Completion: Z,ZK Responsible Department: 13101 Credits: 6 Semester: Z
Anotation:
Aim of this subject is to give the basic informations about probability, mathematical statistics and Markov chains and to show their applications, mainly in insurance mathematics. At the end of the course, bases of cluster analysis will be shown.
Course outlines:
1 Random event, definition of probability. 2 Conditional probability, Bayes theorem. 3 Random variable, random vector - density, distribution function, expected value, variance; examples of discrete and continuous distributions. 4 Large numbers laws, central limit theorem. 5 Statistics - parameters estimations, testing of hypotheses. 6 Regression analysis. 7 Random processes - fundamental definitions. 8 Markov chains with discrete time - basic properties, random walk. 9 Markov chains with discrete time - transition matrix, Chapman-Kolmogorov equation, states classification. 10 Markov chains with continuous time - Wiener process, Poisson process. 11 General insurance - basic probability distributions of the number of events and claim amounts. 12 Technical reserves - indemnity reserve, triangular schemes, Markov chains in bonus systems. 13 Life insurance - premium in capital and annuity insurance. 14 Cluster analysis - basic definitions, methods of clustering.
Exercises outline:
1 Probability of random event. 2 Conditional probability, Bayes theorem. 3 Distribution of random variable. 4 Discrete random variable - distribution function, expected value, variance. 5 Continuous random variable - density, distribution function, expected value, variance. 6 Central limit theorem. 7 Statistics - parameters estimations, testing of hypotheses. 8 Regression analysis. 9 Random processes - stationarity. 10 Markov chains with discrete and continuous time - transition matrix, classification of states, matrix of transition intensity. 11 Calculation of premium and reserves in general insurance. 12 Calculation of premium in capital insurance. 13 Calculation of premium in annuity insurance. 14 Basic methods of clustering.
Literature:
[1] Papoulis, A.: Probability and Statistics, Prentice-Hall, 1990. [2] Stewart W.J.: Probability, Markov Chains, Queues, and Simulation: The Mathematical Basis of Performance Modeling. Princeton University Press 2009. [3] Kaas, R., Goovaerts, M., Dhaene, J., Denuit, M.: Modern actuarial risk theory. Kluwer Academic Publishers, 2004. [4] Gerber, H.U.: Life Insurance Mathematics. Springer-Verlag, New York-Berlin-Heidelberg, 1990. [5] Duda, R.O., Hart, P.E., Stork, D.G.: Pattern Classification. John Wiley & Sons, 2001.
Requirements:
Details are at http://math.feld.cvut.cz/helisova/mekA1M01MPE.html and http://math.feld.cvut.cz/helisova/mekAD1M01MPE.html respectively.
Webpage:
http://math.feld.cvut.cz/helisova/01pstimfe.html
Subject is included into these academic programs:
Program Branch Role Recommended semester MEKME1 Wireless Communication V 1 MEKME5 Systems of Communication V 1 MEKME4 Networks of Electronic Communication V 1 MEKME3 Electronics V 1 MEKME2 Multimedia Technology V 1 MEEEM4 Economy and Management of Power Engineering P 1 MEEEM5 Economy and Management of Electrical Engineering P 1 MEOI1 Artificial Intelligence V 1 MEOI5NEW Software Engineering V 1 MEOI5 Software Engineering V 1 MEOI4 Computer Graphics and Interaction V 1 MEOI3 Computer Vision and Image Processing V 1 MEOI2 Computer Engineering V 1 MEKYR4 Aerospace Systems V 1 MEKYR1 Robotics V 1 MEKYR3 Systems and Control V 1 MEKYR2 Sensors and Instrumentation V 1
Page updated 18.6.2019 17:53:02, semester: Z,L/2020-1, L/2018-9, Z,L/2019-20, Send comments about the content to the Administrators of the Academic Programs Proposal and Realization: I. Halaška (K336), J. Novák (K336) | 1,000 | 4,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-26 | latest | en | 0.741498 |
http://open.math.uwaterloo.ca/modules/practiceTest.QuestionSheet?currPage=0&testId=216&actionID=next | 1,506,249,593,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689975.36/warc/CC-MAIN-20170924100541-20170924120541-00381.warc.gz | 242,409,405 | 6,817 | # Lesson: Polar Form
1 point
## Transcript
In the previous lecture, we saw that we can visualize a complex number as a point in the complex plane. This turns out to be remarkably useful, but we need to think about things a bit differently. Instead of thinking of the point (x, y) in terms of its x1 and x2 components, we instead want to describe a point in terms of its distance from the origin and how much it is rotated away from the positive real axis.
Given a complex number z = x + yi, we define the modulus of z, denoted such as absolute value of z, to be the real number r equal to the square root of (x-squared + y-squared). If the modulus of z does not equal 0, let theta be the angle measure counterclockwise from the positive x-axis such that x would equal r(cosine(theta)) and y would equal r(sine(theta)). The angle theta is called an argument of z.
Note that we say an argument of z, since theta is only unique up to a multiple of 2(pi). As such, every complex number has an infinite number of arguments. We actually will make use of this fact. But for now, we’ll simply state that if r is the modulus of z, and if theta is an argument for z, then a polar form for z is r(cosine(theta) + i(sine(theta))).
So let’s look at some examples. To determine a polar form for 1 + i, we will first compute its modulus, r, equal to the square root of (1-squared + 1-squared), which equals the square root of 2. Next, we need to find a theta such that 1 = (the square root of 2)(cosine(theta)), and 1 = (the square root of 2)(sine(theta)). That is, we want cosine(theta) to equal 1/(square root 2), and sine(theta) to equal to 1/(square root of 2). Setting theta = pi/4 satisfies these equations, so a polar form for 1 + i is (the square root of 2)(cosine(pi/4) + i(sine(pi/4))).
The steps to determine a polar form for 1 – i are quite similar. First, we compute that the modulus r equals the square root of (1-squared + (-1)-squared)), which equals the square root of 2. Next, we need to find theta such that 1 = (the square root of 2)(cosine(theta)), and -1 = (the square root of 2)(sine(theta)). That is, we want cosine(theta) to equal to 1/(the square root of 2), and sine(theta) to equal -1/(the square root of 2). Now, since our cosine value is positive while our sine value is negative, we know we are looking for a theta in the fourth quadrant, so we can set theta = 7(pi)/4 to satisfy our equations. This gives the polar form (square root of 2)(cosine(7(pi)/4) + i(sine(7(pi)/4))) for 1 – i. But we might just as easily have chosen the theta value of –pi/4, though, giving us the polar form of (square root of 2)(cosine(-pi/4) + i(sine(-pi/4))) for the complex number 1 – i.
In general, you will do well to visualize the point x + iy when you are trying to determine an argument for it, instead of relying purely on calculations. This is particularly true when looking for a polar form for a purely real or purely imaginary number.
For example, let’s try to find the polar form for the numbers 4, -7, 2i, and -5i, first by graphing them on the complex plane. Now, just by looking at the graph, it becomes obvious that the polar form for 4 is 4(cosine(0) + i(sine(0))). A polar form for -7 is 7(cosine(pi) + i(sine(pi))). A polar form for 2i is 2(cosine(pi/2) + i(sine(pi/2))). And a polar form for -5i is 5(cosine(3(pi)/2) + i(sine(3(pi)/2))).
Theorem 9.1.a: If z = r(cosine(theta) + i(sine(theta))), then the conjugate of z equals r(cosine(-theta) + i(sine(-theta))). And since I’ve made it a theorem, I’ll go ahead and prove it for you. So we’ll let z = x + iy, so the conjugate of z is x – yi. Well, then the modulus for the conjugate of z will equal the square root of (x-squared + (-y)-quantity-squared), but of course, this is the same as the square root of (x-squared + y-squared), which is the modulus of z, which is r. So we see that the modulus of the conjugate of z is the same as the modulus for z.
Now, since z = r(cosine(theta)) + ir(sine(theta)), and using the fact that cosine(-theta) = cosine(theta), while sine(-theta) = -sine(theta), we’ll see that the conjugate of z will have to equal r(cosine(theta)) – ir(sine(theta)), which equals r(cosine(-theta)) + ir(sine(-theta)), which equals r(cosine(-theta) + i(sine(-theta))), as we desired.
We can use properties of cosine and sine to help us multiply and divide complex numbers as well. Let’s recall the following trigonometric identities, that the cosine(theta1 + theta2) will equal (cosine(theta1))(cosine(theta2)) – (sine(theta1))(sine(theta2)), and that sine(theta1 + theta2) = (sine(theta1))(cosine(theta2)) – (cosine(theta1))(sine(theta2)). So we can use them to get the following result, and I’ll note that this does appear in the textbook—it simply isn’t listed as a theorem there.
So I’ll call it Theorem 9.1.b. For any complex numbers z1 = r1(cosine(theta1) + i(sine(theta1))) and z2 = r2(cosine(theta2) + i(sine(theta2))), we have that z1z2 = r1r2(cosine(theta1 + theta2) + i(sine(theta1 + theta2))).
And again, since I’ve turned this into a theorem, I’ll go ahead and prove it for you. So again, we’ll let z1 and z2 be as described in the theorem, and then if we were to multiply z1z2, we can actually multiply their polar forms. First, we can go ahead and distribute our r1’s and our r2’s inside the individual polar forms, and now it simply becomes a question of distributing our multiplication. So we’ll end up with an r1r2(cosine(theta1))(cosine(theta2)) + i(r1r2)(cosine(theta1))(sine(theta2)) + i(r1r2)(sine(theta1))(cosine(theta2)) + (i-squared)(r1r2)(sine(theta1))(sine(theta2)). So let’s go ahead and pull all those (r1r2)’s out, and we can also start grouping things together. So, for example, we have our cosine(theta1) and our cosine(theta2), we could actually group with our + (i-squared)(r1r2)(sine(theta1))(sine(theta2)), but remembering that the i-squared becomes a -1, so this becomes the quantity (cosine(theta1))(cosine(theta2)) – (sine(theta1))(sine(theta2)). And then next, we’ll pull out our i terms, which leave us with the (cosine(theta1))(sine(theta2)) + (sine(theta1))(cosine(theta2)). But using our trig identities, we’ll see that this becomes (r1r2)(cosine(theta1 + theta2) + i(sine(theta1 + theta2))).
Next, we’ll do a theorem that is in the book. For any complex numbers z1, with polar form r1(cos(theta1) + i(sine(theta1))), and z2, with polar form r2(cos(theta2) + i(sine(theta2))), with z2 not equal to 0, we have that z1/z2 will equal (r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))).
Now, to prove this, let’s first remember that if z1/z2 = z3, then we have that z1 = z2z3. So we can see that z1/z2 will equal (r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))) by showing that z1 = z2 times this value. And we show this as follows. z2(r1/r2)(cosine(theta1 – theta2) + i(sine(theta1 – theta2))) will equal (r2(cosine(theta2) + i(sine(theta2)))) times our other value. But now, we’ll use the fact that we’re multiplying these two polar forms, and our Theorem 9.1.b, to see that this equals (r2(r1/r2))(cosine(theta2 + (theta1 – theta2)) + i(sine(theta2 + (theta1 – theta2)))), and of course, our r2’s cancel out, leaving us with an r1, and our theta2’s cancel out, leaving us with a cosine(theta1) and a sine(theta1), which is, of course, z1 as desired.
One special case of this theorem is the following. Let z = r(cosine(theta) + i(sine(theta))), with r not equal to 0. Then z-inverse will equal (1/r)(cosine(-theta) + i(sine(-theta))). This is simply using Theorem 9.1.3 with our z1 being the value 1 and our z2 being z.
Let’s look at some examples of using these theorems. So earlier, we found that a polar form for 1 + i is (the square root of 2)(cosine(pi/4) + i(sine(pi/4))). Now let’s find a polar form for –(root 3) + 3i. First, we can find that the modulus r equals the square root of ((-3)-quantity-squared + 3-squared), which is the square root of 12, or we could write that as 2(root 3). Next, we need to find theta such that the cosine(theta) = –(root 3)/(2(root 3)), which is -1/2, and sine(theta) = 3/(2(root 3)), which is (root 3)/2. Now, since our cosine value is negative and our sine value is positive, we know that theta is in the second quadrant, so we can use theta = 2(pi)/3 as an argument. So this means that (2(root 3))(cosine(2(pi)/3) + i(sine(2(pi)/3))) is the polar form for -3 + 3i.
At this point, we can use these polar forms to perform the following calculations. So (1 + i)(-3 + 3i) equals the product of their polar forms. But this is simply the product of the moduli, and then the sum of the arguments, so we get (2(root 6))(cosine(7(pi)/12) + i(sine(7(pi)/12))). Meanwhile, we can look at (1 + i)/(-(root 3) + 3i) by looking at the polar form for (1 + i) divided by the polar form for (-(root 3) + 3i). This means we’re going to divide the moduli and subtract the second argument from the first. So we’re left with (1/(root 6))(cosine(-pi/12) + i(sine(-pi/12))). What if, instead, we wanted to look at taking (-(root 3) + 3i) and dividing it by (1 + i)? Again, we’ll want to turn this into their polar forms and looking at that quotient, which means we’ll look at the quotient of the moduli, and we’ll subtract the second argument from the first. So we’re getting that it has a modulus of (root 6), and then times (cosine(pi/12) + i(sine(pi/12))). Now, note that this last result agrees with the result we would get if we considered (-(root 3) + 3i)/(1 + i) to be the inverse of (1 + i)/(-(root 3) + 3i).
Let’s use the polar form to calculate the product (1 – 2i)(-3 – 4i). Well, first we’ll need to find the polar form of (1 – 2i) and (-3 – 4i). Now, this is not as easy as in the previous example, as our arguments will not be one of our standard angles. Instead, we will use a calculator and decimal approximations for these calculations. So to find a polar form of (1 – 2i), we first note that the modulus r will equal the square root of (1-squared + (-2)-squared), which equals the square root of 5. And now we need to find theta such that the cosine(theta) = 1/(the square root of 5), and sine(theta) = -2/(the square root of 5). Now, since the cosine is positive and the sine is negative, we know that theta is in the fourth quadrant. Plugging theta = (sine-inverse)(-2/(root 5)) into a calculator will give us a value in the fourth quadrant, and so we will see that theta is approximately -1.11. And so we have found that a polar form of (1 – 2i) is (root 5)(cosine(-1.11) + i(sine(-1.11))).
To find a polar form of (-3 – 4i), we first note that the modulus for this will equal the square root of ((-3)-squared + (-4)-squared), which is the square root of (9 + 16), which equals 5. And now we need to find theta such that cosine(theta) = -3/5 and sine(theta) = -4/5. But since the cosine is negative and the sine is negative, we know that theta is in the third quadrant. Unfortunately, a calculator never outputs a value in the third quadrant. So we can either have that theta = –(cosine-inverse)(-3/5) or that theta = (cosine-inverse)(3/5) + pi. I’ll choose to do the first, which gives me that theta = -2.21. And so we have found that a polar form of (-3 – 4i) is 5(cosine(-2.21) + i(sine(-2.21))).
Now that we have polar forms for (1 – 2i) and (-3 – 4i), we can calculate that the product (1 – 2i)(-3 – 4i) will equal the product of their polar forms, which we now know equals the product of their moduli and the sum of their arguments, so we get (5(root 5))(cosine(-3.32) + i(sine(-3.32))). We can double-check our calculation by putting things back into standard form. With a healthy amount of round-off error, we do, in fact, see that (5(root 5))(cosine(-3.32) + i(sine(-3.32))) = -11 + 2i.
Now, given how much effort it is to find polar form, at this point, it probably does not seem worthwhile to put complex numbers into polar form simply to multiply and divide them. But in a couple of lectures, we will use polar form to find the nth roots of complex numbers, and this really is the best way to solve this type of question. So for now, simply consider this lecture a gentle introduction to polar form and calculations in polar form, but polar form is not going away. | 3,529 | 12,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2017-39 | latest | en | 0.898676 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-2-solving-equations-2-2-solving-two-step-equations-practice-and-problem-solving-exercises-page-91/31 | 1,539,795,816,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511203.16/warc/CC-MAIN-20181017153433-20181017174933-00151.warc.gz | 979,851,377 | 14,801 | ## Algebra 1: Common Core (15th Edition)
$b=-8$
We need to get the b alone. In order to cancel out division, we use multiplication, so we multiply both sides by 5 to get that $b+3=-5$. We now cancel out the addition by subtracting 3 on both sides, giving us a final answer of $b=-8.$ We plug -8 in for b and confirm that our answer is, indeed, correct. | 98 | 353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-43 | longest | en | 0.92904 |
https://curriculum.illustrativemathematics.org/k5/teachers/grade-2/unit-6/lesson-9/preparation.html | 1,719,027,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00277.warc.gz | 157,233,839 | 24,705 | # Lesson 9
You Ate the Whole Thing
### Lesson Purpose
The purpose of this lesson is for students to recognize 2 halves, 3 thirds, and 4 fourths as one whole.
### Lesson Narrative
In previous lessons, students partitioned circles and rectangles into halves, thirds, and fourths and identified an equal piece of different shapes as a half of, third, of, or fourth of the shape.
In this lesson, students continue to practice partitioning circles and describe halves, thirds, and quarters of circles using the language a half of, a third of, and a quarter of to describe a piece of the shape. They also use this language to describe the whole shape as a number of equal pieces. Students recognize that a whole shape can be described as 2 halves, 3 thirds, or 4 fourths. This understanding is the foundation for students' work with a whole and fraction equivalency in grade 3.
The activities encourage students to use quarters and a quarter of when describing fourths of a circle. This word choice is used to help students connect to the language used when telling time to the half and quarter hour in upcoming lessons.
• Representation
### Learning Goals
Teacher Facing
• Describe two halves, three thirds, and four fourths as one whole.
• Use “half of,” “a third of,” and “a quarter of” to describe parts of a shape.
### Student Facing
• Let’s talk about the whole.
### Required Materials
Materials to Gather
Addressing
Building Towards
### Lesson Timeline
Warm-up 10 min Activity 1 15 min Activity 2 20 min Lesson Synthesis 10 min Cool-down 5 min
### Teacher Reflection Questions
In an upcoming lesson, students will learn to tell time to the half hour and quarter hour. What do you notice in their work from today’s lesson that you might leverage in that lesson?
### Suggested Centers
• How Are They the Same? (1–5), Stage 2: Grade 2 Shapes (Addressing)
• Which One? (K–5), Stage 3: Grade 2 Shapes (Addressing)
### Print Formatted Materials
For access, consult one of our IM Certified Partners.
### Additional Resources
Google Slides For access, consult one of our IM Certified Partners. PowerPoint Slides For access, consult one of our IM Certified Partners. | 492 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.910789 |
https://tr.scribd.com/document/337143590/Mcq-Testing-of-Hypothesis-With-Correct-Answers | 1,563,739,156,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527196.68/warc/CC-MAIN-20190721185027-20190721211027-00226.warc.gz | 569,209,112 | 63,756 | You are on page 1of 7
# MCQ TESTING OF HYPOTHESIS
MCQ 13.1
A statement about a population developed for the purpose of testing is called:
(a) Hypothesis
(b) Hypothesis testing
(c) Level of significance (d) Test-statistic
MCQ 13.2
Any hypothesis which is tested for the purpose of rejection under the assumption that it is true is
called:
(a) Null hypothesis (b) Alternative hypothesis (c) Statistical hypothesis (d) Composite hypothesis
MCQ 13.3
A statement about the value of a population parameter is called:
(a) Null hypothesis
(b) Alternative hypothesis (c) Simple hypothesis (d) Composite hypothesis
MCQ 13.4
Any statement whose validity is tested on the basis of a sample is called:
(a) Null hypothesis (b) Alternative hypothesis (c) Statistical hypothesis
## (b) Simple hypothesis
MCQ 13.5
A quantitative statement about a population is called:
(a) Research hypothesis (b) Composite hypothesis (c) Simple hypothesis (d) Statistical hypothesis
MCQ 13.6
A statement that is accepted if the sample data provide sufficient evidence that the null hypothesis is false is
called:
(a) Simple hypothesis
(b) Composite hypothesis (c) Statistical hypothesis
(d) Alternative hypothesis
MCQ 13.7
The alternative hypothesis is also called:
(a) Null hypothesis (b) Statistical hypothesis
## (d) Simple hypothesis
MCQ 13.8
A hypothesis that specifies all the values of parameter is called:
(a) Simple hypothesis
(b) Composite hypothesis
(c) Statistical hypothesis
## (d) None of the above
MCQ 13.9
The hypothesis 10 is a:
(a) Simple hypothesis
## (c) Alternative hypothesis
MCQ 13.10
If a hypothesis specifies the population distribution is called:
(a) Simple hypothesis
(b) Composite hypothesis
(c) Alternative hypothesis
MCQ 13.11
A hypothesis may be classified as:
(a) Simple
(b) Composite
(c) Null
## (d) All of the above
MCQ 13.12
The probability of rejecting the null hypothesis when it is true is called:
(a) Level of confidence
(b) Level of significance
(c) Power of the test
## (d) Difficult to tell
MCQ 13.13
The dividing point between the region where the null hypothesis is rejected and the region where it is not
rejected is said to be:
(a) Critical region
(b) Critical value
(c) Acceptance region
(d) Significant region
MCQ 13.14
If the critical region is located equally in both sides of the sampling distribution of test-statistic, the test is
called:
(a) One tailed
(b) Two tailed
(c) Right tailed
(d) Left tailed
MCQ 13.15
The choice of one-tailed test and two-tailed test depends upon:
(a) Null hypothesis
(b) Alternative hypothesis
(c) None of these
MCQ 13.16
Test of hypothesis Ho: = 50 against H1: > 50 leads to:
(a) Left-tailed test
(b) Right-tailed test
(c) Two-tailed test
## (d) Difficult to tell
MCQ 13.17
Test of hypothesis Ho: = 20 against H1: < 20 leads to:
(a) Right one-sided test
(b) Left one-sided test
(c) Two-sided test
## (d) All of the above
MCQ 13.18
Testing Ho: = 25 against H1: 20 leads to:
(a) Two-tailed test (b) Left-tailed test (c) Right-tailed test (d) Neither (a), (b) and (c)
MCQ 13.19
A rule or formula that provides a basis for testing a null hypothesis is called:
(a) Test-statistic
(b) Population statistic
(c) Both of these
MCQ 13.20
The range of test statistic-Z is:
(a) 0 to 1
(b) -1 to +1
(c) 0 to
(d) - to +
MCQ 13.21
The range of test statistic-t is:
(a) 0 to
(b) 0 to 1
(c) - to +
(d) -1 to +1
MCQ 13.22
If Ho is true and we reject it is called:
(a) Type-I error
(b) Type-II error
## (c) Standard error
MCQ 13.23
The probability associated with committing type-I error is:
(a)
(b)
(c) 1
## (d) Sampling error
(d) 1
MCQ 13.24
A failing student is passed by an examiner, it is an example of:
(a) Type-I error
(b) Type-II error (c) Unbiased decision
## (d) Difficult to tell
MCQ 13.25
A passing student is failed by an examiner, it is an example of:
(a) Type-I error (b) Type-II error
(c) Best decision
(d) All of the above
MCQ 13.26
1 is also called:
(a) Confidence coefficient
## (b) Power of the test
MCQ 13.27
1 is the probability associated with:
(a) Type-I error
(b) Type-II error
## (d) Level of significance
MCQ 13.28
Area of the rejection region depends on:
(a) Size of
(b) Size of
(c) Test-statistic
## (d) Number of values
MCQ 13.29
Size of critical region is known as:
(a)
(b) 1 -
## (c) Critical value
MCQ 13.30
A null hypothesis is rejected if the value of a test statistic lies in the:
(a) Rejection region
(b) Acceptance region (c) Both (a) and (b)
## (d) Neither (a) nor (b)
MCQ 13.31
The test statistic is equal to:
MCQ 13.32
Level of significance is also called:
(a) Power of the test
(b) Size of the test (c) Level of confidence
MCQ 13.33
Level of significance lies between:
(a) -1 and +1
(b) 0 and 1
(d) - to +
(c) 0 and n
MCQ 13.34
Critical region is also called:
(a) Acceptance region
(b) Rejection region
## (c) Confidence region
MCQ 13.35
The probability of rejecting Ho when it is false is called:
(a) Power of the test
(b) Size of the test
(c) Level of confidence
MCQ 13.36
Power of a test is related to:
(a) Type-I error
(b) Type-II error
## (d) Neither (a) and (b)
MCQ 13.37
In testing hypothesis + is always equal to:
(a) One
(b) Zero
(c) Two
MCQ 13.38
The significance level is the risk of:
(a) Rejecting Ho when Ho is correct
(c) Rejecting H1 when H1 is correct
## (b) Rejecting Ho when H1 is correct
(d) Accepting Ho when Ho is correct.
MCQ 13.39
An example in a two-sided alternative hypothesis is:
(a) H1: < 0
(b) H1: > 0
(c) H1: 0
(d) H1: 0
MCQ 13.40
If the magnitude of calculated value of t is less than the tabulated value of t and H1 is two-sided, we
should:
(a) Reject Ho
(b) Accept H1
(c) Not reject Ho
(d) Difficult to tell
MCQ 13.41
Accepting a null hypothesis Ho:
(a) Proves that Ho is true
(c) Implies that Ho is likely to be true
## (b) Proves that Ho is false
(d) Proves that 0
MCQ 13.42
The chance of rejecting a true hypothesis decreases when sample size is:
(a) Decreased
(b) Increased
(c) Constant
## (d) Both (a) and (b)
MCQ 13.43
The equality condition always appears in:
(a) Null hypothesis
(b) Simple hypothesis (c) Alternative hypothesis
## (d) Both (a) and (b)
MCQ 13.44
Which hypothesis is always in an inequality form?
(a) Null hypothesis
(b) Alternative hypothesis
## (c) Simple hypothesis
MCQ 13.45
Which of the following is composite hypothesis?
(a) o
(b) o
(c) = o
(d) o
MCQ 13.46
P (Type I error) is equal to:
(a) 1
(b) 1
(c)
MCQ 13.47
P (Type II error) is equal to:
(a)
(b)
(c) 1
(d) 1
MCQ 13.48
The power of the test is equal to:
(a)
(b)
(c) 1
(d) 1
(d)
MCQ 13.49
The degree of confidence is equal to:
(a)
(b)
(c) 1
MCQ 13.50
/ 2 is called:
(a) One tailed significance level
(c) Left tailed significance level
## (b) Two tailed significance level
(d) Right tailed significance level
(d) 1
MCQ 13.51
Students t-test is applicable only when:
(a) n30 and is known (b) n>30 and is unknown (c) n=30 and is known (d) All of the above
MCQ 13.52
Students t-statistic is applicable in case of:
(a) Equal number of samples (b) Unequal number of samples (c) Small samples (d) All of the above
MCQ 13.53
Paired t-test is applicable when the observations in the two samples are:
(a) Equal in number (b) Paired
(c) Correlation
(d) All of the above
MCQ 13.54
The degree of freedom for paired t-test based on n pairs of observations is:
(a) 2n - 1
(b) n - 2
(c) 2(n - 1) (d) n - 1
MCQ 13.55
The test-statistic
(a) n
(b) n - 1
## has d.f = ________:
(c) n - 2
(d) n1 + n2 - 2
MCQ 13.56
In an unpaired samples t-test with sample sizes n1= 11 and n2= 11, the value of tabulated t should be
obtained for:
(a) 10 degrees of freedom
(b) 21 degrees of freedom
(c) 22 degrees of freedom
(d) 20 degrees of freedom
MCQ 13.57
In analyzing the results of an experiment involving seven paired samples, tabulated t should be
obtained for:
(a) 13 degrees of freedom
(b) 6 degrees of freedom
(c) 12 degrees of freedom
(d) 14 degrees of freedom
MCQ 13.58
The mean difference between 16 paired observations is 25 and the standard deviation of differences is
10. The value of statistic-t is:
(a) 4
(b) 10
(c) 16
(d) 25
MCQ 13.59
Statistic-t is defined as deviation of sample mean from population mean expressed in terms of:
(a) Standard deviation
(b) Standard error
(c) Coefficient of standard deviation
(d) Coefficient of variation
MCQ 13.60
Students t-distribution has (n-1) d.f. when all the n observations in the sample are:
(a) Dependent
(b) Independent (c) Maximum
(d) Minimum
MCQ 13.61
The number of independent values in a set of values is called:
(a) Test-statistic
(b) Degree of freedom
(c) Level of significance (d) Level of confidence
MCQ 13.62
The purpose of statistical inference is:
(a) To collect sample data and use them to formulate hypotheses about a population
(b) To draw conclusion about populations and then collect sample data to support the conclusions
(c) To draw conclusions about populations from sample data
(d) To draw conclusions about the known value of population parameter
MCQ 13.63
Suppose that the null hypothesis is true and it is rejected, is known as:
(a) A type-I error, and its probability is
(b) A type-I error, and its probability is
(c) A type-II error, and its probability is
(d) A type-Il error, and its probability is
MCQ 13.64
An advertising agency wants to test the hypothesis that the proportion of adults in Pakistan who read a Sunday
Magazine is 25 percent. The null hypothesis is that the proportion reading the Sunday Magazine is:
(a) Different from 25%
(b) Equal to 25%
(c) Less than 25 %
(d) More than 25 %
MCQ 13.65
If the mean of a particular population is o,
is distributed:
## (a) As a standard normal variable, if the population is non-normal
(b) As a standard normal variable, if the sample is large
(c) As a standard normal variable, if the population is normal
(d) As the t-distribution with v = n - 1 degrees of freedom
MCQ 13.66
If 1 and 2 are means of two populations,
is distributed:
(a) As a standard normal variable, if both samples are independent and less than 30
(b) As a standard normal variable, if both populations are normal
(c) As both (a) and (b) state
(d) As the t-distribution with n1 + n2 - 2 degrees of freedom
MCQ 13.67
If the population proportion equals po, then
(a) As a standard normal variable, if n > 30
(b) As a Poisson variable
(c) As the t-distribution with v= n 1 degrees of freedom
(d) As a distribution with v degrees of freedom
is distributed:
MCQ 13.68
When is known, the hypothesis about population mean is tested by:
(a) t-test
(b) Z-test
(c) 2-test
(d) F-test
MCQ 13.69
Given o = 130,
(a) t
## = 150, = 25 and n = 4; what test statistics is appropriate?
(b) Z
(c) 2
(d) F
MCQ 13.70
Given Ho: = o, H1: o, = 0.05 and we reject Ho; the absolute value of the Z-statistic must have equalled
or been beyond what value?
(a) 1.96
(b) 1.65
(c) 2.58
(d) 2.33
MCQ 13.71
If p1 and p2 are not identical, then standard error of the difference of proportions (p1 p2) is:
MCQ 13.72
Under the hypothesis Ho: p1 = p2, the formula for the standard error of the difference between
proportions (p1 p2) is: | 3,333 | 11,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-30 | latest | en | 0.800594 |
http://www.jiskha.com/display.cgi?id=1198008249 | 1,498,319,499,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00182.warc.gz | 553,160,687 | 4,627 | # Algebra II
posted by .
A furniture company displays bedroom sets which require 21 square meters of space and living room sets which require 42 square meters of space. The company which has 546 square meters of available space,wants to display at least 6 bedroom sets and at least 5 living room sets.
My question is:If a bedroom set sells for \$10,000 and a living room set sells for \$18,000,determine the number of bedroom sets and living room sets that must be sold to maximize the amount collected.
I'm not sure what I need to do to set this up.
• Algebra II -
Think it through.. don't try to find a single magic equation.
You maximize receipts when you sell as many living room sets as you can, while still being able to display and sell the required minimum 6 bedroom sets. Since both take up 21 sq meters and you have 546 sq m available, you can display 546/21 = 26 sets of either kind. So, display the required 6 minimum bedroom sets and use the rest of the space for 20 living room sets. Assuming that they all sell at an equal rate (which is not at all likely in practice), the amount collected after one turnover will be 20*\$18,000 + 6*\$10,000 = \$420,000
I did not realize furnishing a house is that expensive nowadays
• Algebra II -
Let the number of bedroom setups be b
then the number of living room setups is (546-21b)/42
total revenue
= 10000b + 18000(546-21b)/42
= 234000 + 1000b
but b ≥ 6 and livingrooms ≥ 5
so if b=6 , l= 10
and if l=5, b = 16
the total revenue is a linear function represented by a straight line with a positive slope, so the larger value of b will give a larger value of revenue
so when b= 16, l = 5 and revenue = \$250,000
check: if b = 16, l = 5 revenue = \$240,000
So they should set up 16 bedrooms and 5 living rooms
• Algebra II -
Correction:
Second last line should say:
check: if b = 6, l = 10 revenue = \$240,000
• Algebra II -
I misread the problem, not noticing the higher area of the living room sets. Just ignore my answer, if you haven't already, and follow Reiny's method.
### Answer This Question
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# How to hack a dead battery of a cordless vacuum cleaner? Answered
Hi Everybody,
I've got an old cordless vacuum cleaner with an old short lasting battery, so I decided to hack it in order to use it on the plug directly ( I'll be using the output of its charger).
The cordless vacuum cleaner has a battery of 3.7 V and a 9V DC- 150mA as an output for its charger.
Note: I was not able to see the motor characteristic because it's covered and it was difficult the remove it.
So I disassembled everything and try to plug the output of the charger directly to the motor, the motor didn't work.
I plug the motor directly to its battery = it works
I plugged a 15V - 150mA (from an old printer transformer) => the motor it barely turns (about 1 rpm)
I plugged a 32V - 150mA (from an old printer transformer) => the motor turns very slow but faster than the previous experiment (about 2 rpm)
Can anyone explain what's going on. Why the motor cannot accept DC from chargers?
Tags:
The forums are retiring in 2021 and are now closed for new topics and comments.
The motor needs a LOT more current than your little supplies can provide. Think nearer 2-4 Amps, than a few mA.
Remember, the battery would probably be charged pretty gently - I'd guess at a 10 hour rate, making the battery capacity roughly 1200maHr. Bitter experience suggests you wouldn't have had more than 15 minutes of vacuuming from a charged battery - which would mean all the energy is gone in 1/4 of the time, 1200*60/15= 4.8 A
Thx dude, but what do u mean by 1200*60/15= 4.8 A
Thanks (for steve) I looked it up!
1200mA x 60 seconds / 15 minuets - 4.8 Amps. To give some idea of what the motor may draw from the battery.
If the nbattery is a lead acid battery or Nicd they are fairly easy to get and replace.
We don't charge for extra letters. What does "Thx" mean ?
4.8A looks the likely kind of current you need from your supply | 501 | 1,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2021-17 | latest | en | 0.967911 |
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posted by on .
260+(4to the 3 power+1)=4 ?
which numbers are composite 23,28,33,47=28,33
gcf 18,30,45=3
recipprocal of 1 5/9=9/14
gcf 63,147,231=21
• math - ,
260+(4to the 3 power+1)=4 ?
Incorrect
composite 23,28,33,47=28,33
Correct
gcf 18,30,45=3
Correct
reciprocal of 1 5/9=9/14
Correct
gcf 63,147,231=21
Correct
__________________________ | 157 | 361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2017-17 | latest | en | 0.601442 |
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## Online Course Helper
html#high.htm. How can I calculate the Harrell’s C-statistic? Right now, I have been working with a simple formula for the Harrell’s C-statistic: Now, I want to create a formula where I will calculate the percentage of the difference between the values that I have taken for the C-statistic, in different points of the sample. The formula I would like to use is: The problem here is, that I will make use only of a single variable, the square, and the variable just named c for the first time. Therefore, I already have a nice, simple formula like: As you can see click here for info I write my formula: The other thing I wrote is obviously, that I need to be able to write the formula in more complex ways. hire someone to do statistics examination how to get more complex way, by iterating over a huge amount of variables. If you see, how I can write this? Now for the next step, I want to find any limit to the possible value of the C-statistic. So I have: I want to find the values for the C-statistic in my next page when I declare a variable for c: it is assigned to the square with the lower 3.2 decimal place part then: So my approach is to take the first step, find the value of the square, and multiply it one way, or the other way and read it real time, or just say it as if it was coded as : Square is the square to be decided on. How to pop over to these guys the Harrell’s C-statistic for this variable? The questions have some tricky questions: My method for the Harrell’s C-statistic is an experiment, where I have given a variable named c for the different points of the sample. For some data points, either a 500 or a 10, I want to find the value, and multiply this three square factor. So I moved here several vectors with the value of the c digit. If I click site to multiply it to 1020 then the value of “7” is multiplied to 1 then with 1020 I should sum all the values, and I have calculated the values of both ones, and then used the sum. When I want to informative post it all I have to write my formula. I hope you do not think I am this website right here already have a simple form for the term c for the square. If I want to take the upper 2 decimal place, then I have
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# How to Calculate Margin in Forex Trading: A Step-by-Step Guide
Forex trading is a popular form of investing that allows individuals to trade currencies from around the world. One concept that is crucial to understand when trading forex is margin. Margin is the amount of money required to open and maintain a trading position. It acts as collateral for the trader, ensuring that they have enough funds to cover any potential losses.
In this article, we will provide you with a step-by-step guide on how to calculate margin in forex trading. By understanding this concept, you can effectively manage your risk and make informed trading decisions.
### Step 1: Determine the currency pair you want to trade
The first step in calculating margin is to determine the currency pair you want to trade. Forex trading involves buying one currency and selling another, so you need to decide which currencies you want to trade and their respective symbols. For example, if you want to trade the Euro against the US Dollar, the currency pair symbol would be EUR/USD.
### Step 2: Define the leverage ratio
Leverage is a tool that allows traders to control larger positions with a smaller amount of capital. It is expressed as a ratio, such as 1:100 or 1:500. The leverage ratio determines the amount of margin required to open and maintain a trading position. Higher leverage ratios require lower margin, but they also increase the potential for larger gains or losses.
### Step 3: Calculate the base currency value
The base currency is the currency on the left side of the currency pair. For example, in the EUR/USD pair, the Euro is the base currency. To calculate the base currency value, you need to know the exchange rate between the two currencies. Let’s say the current exchange rate for EUR/USD is 1.20.
If you want to trade one standard lot of EUR/USD, which is equivalent to 100,000 units of the base currency, the base currency value would be:
### Step 4: Calculate the required margin
To calculate the required margin, you need to multiply the base currency value by the leverage ratio. Let’s assume you are trading with a leverage ratio of 1:100.
### Required margin = \$1,200
In this example, you would need \$1,200 as margin to open and maintain a position of one standard lot in EUR/USD.
### Step 5: Calculate the margin level
The margin level is the ratio of equity to margin expressed as a percentage. It shows the amount of margin available to cover potential losses. The formula to calculate the margin level is as follows:
### Margin level = 833.33%
A margin level above 100% indicates that you have enough margin to cover potential losses. However, if the margin level falls below 100%, it means that your account is at risk of a margin call, which could lead to the closure of your positions.
### Step 6: Understand the impact of margin
Margin allows traders to control larger positions with a smaller amount of capital. While this can amplify potential gains, it also increases the risk of larger losses. It is important to understand the impact of margin on your trading strategy and risk management.
Using margin effectively requires careful consideration of your trading goals, risk tolerance, and market conditions. It is crucial to use stop-loss orders and manage your positions to avoid excessive losses.
In conclusion, understanding how to calculate margin in forex trading is essential for effective risk management. By following this step-by-step guide, you can determine the required margin for your trades and monitor your margin level to ensure the safety of your account. Remember to always conduct thorough research and seek professional advice before engaging in forex trading. | 755 | 3,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.900235 |
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Scientific Toolbox
Scientific Toolbox is a combination of multiple useful utilities that are presented in a very accessible way. The application provides an array of advanced functions for handling different number operations and can also be used to calculate various conversions between SI-units and their derivatives.
On the other hand, Scientific Toolbox provides a large and comprehensive collection of formulas that are commonly used within multiple subjects such as mathematics, chemistry and physics. The formulas are described in detail, so that they become easily comprehended by students and others who have interest in science. The application provides also illustrations for formulas that are tightly bound to geometrical calculations. Therefore, Scientific Toolbox can be used within multiple fields of application (science, chemistry, physics, mathematics, calculations, school lessons) and is highly flexible since it provides a very broad range of content!
Application highlights:
Unit conversion
Complete collection of scientific formulas
Option to save mathematical expressions
Custom rounding alternatives
Medical operations
Application Parts:
Calculator
Formula reference
Unit Converter
Operations supported:
Calculator, General Mathematics
log
ln
!
nth root
mod
nPr
nCr
Binomial distribution
factor
average
median
Population standard deviation
Population variance
Calculator, Trigonometry
sin
cos
tan
arksin
arkcos
arktan
sinh
cosh
tanh
Calculator, Computer Science
Binary conversion
Octal conversion
Calculator, Medical operations
BMI
BMR (women)
BMR (men)
Formula reference contents:
Basic rules of Algebra (mathematics)
Logarithms (mathematics)
Factoring (mathematics)
Complex Numbers (mathematics)
Linear equations (mathematics)
Exponential and power equations (mathematics)
Differentiation and the derivative (mathematics)
Primitive Functions and antiderivatives (mathematics)
Integration (mathematics)
Differential equations (mathematics)
Vectors (mathematics)
Series (mathematics)
Polygons (mathematics)
3D objects (mathematics)
Trigonometry basics (mathematics)
Trigonometric formulas (mathematics)
Trigonometric exact values (mathematics)
Combinatorics (mathematics)
Binomial and normal distribution (mathematics)
Probability (mathematics)
Periodic table of elements (chemistry)
Electronegativity (chemistry)
Common Formulas (chemistry)
Cations (chemistry)
Anions (chemistry)
Acid - Base (chemistry)
Equilibrium (chemistry)
Alkanes (chemistry)
Cycloalkanes (chemistry)
Alkenes (chemistry)
Alkynes (chemistry)
Aromatic hydrocarbons (chemistry)
Amines (chemistry)
Halogen Compounds (chemistry)
Alcohols (chemistry)
Ethers (chemistry)
Aldehydes (chemistry)
Ketones (chemistry)
Carboxylic acids (chemistry)
Constants (physics)
SI base units (physics)
Derived SI units (physics)
Prefix (physics)
Mechanics basics (physics)
Motion (physics)
Heat and pressure (physics)
Alternating current (physics)
Magnetism (physics)
Electromagnetic induction (physics)
Wave motion (physics)
Geometrical optics (physics)
Theory of relativity (physics)
Nuclear physics (physics)
Acceleration of gravity (physics)
Earth (physics)
Moon (physics)
Sun (physics)
Gaseous substances (science)
Liquid substances (science)
Solid substances (science)
Metals (science)
Speed of sound in solids (science)
Speed of sound in liquids (science)
Speed of sound in gases (science)
Quantities supported within the unit converter:
Area
Distance
Energy
Information storage
Mass and weight
Pressure
Speed
Temperature
Time
Recently changed in this version
Minor UI improvements
Last activity on Scientific Toolbox
Comments and ratings for Scientific Toolbox
• (67 stars)
Cooollllllll
• (67 stars)
e^x is on the main menu, but the inverse function ln(x) is hided (advanced + general + ln)
The bottons look nice and are comfortable. The name "scientific toolbox" is a misleading advertisment. Actually, this calculator doesn't make any advanced calculati
• (67 stars)
Same as a regular phone calculator. Nothing special.
• (67 stars)
Good, helpful application, thank you for creating such a nice app.
• (67 stars)
Not very intuitive on how it is useful.
• (67 stars)
Its good
• (67 stars)
It's okay | 907 | 4,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2013-48 | latest | en | 0.884226 |
http://gonitsora.com/rsa-elegant-code/ | 1,531,952,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00117.warc.gz | 157,550,522 | 26,750 | RSA: The Elegant Code - Gonit Sora
## 10 Jul RSA: The Elegant Code
Download this post as PDF (will not include images and mathematical symbols).
The quest to share information securely and the curiosity to decipher the hidden meaning, has astounded the human mind since eternity. Whether it was the ancient art of steganography, or the development of ciphers from the Caesar Cipher to the Enigma Cipher Machine, the basic idea has been to protect information from falling into the wrong hands.
This science of cryptography took a very interesting turn during the 1970’s with the development of the concept of asymmetric ciphers by exploiting the one-way functions. This was achieved by a trio of three brilliant cryptographers at Stanford University, namely Whitfield Diffie, Martin Hellman and Ralph Merkle. Together they solved one of the greatest mysteries of sharing messages secretly without having to share the key. For instance, if Alice wants to communicate privately to Bob she must encrypt the message using a key. Now for Bob to be able to decrypt the message he must know the key. This is an example of a symmetric cipher which uses the same key to encode and decode the message. This implied that the key distribution is an inevitable part of decipherment. It could be achieved by personally meeting and sharing the key or delivering it through a reliable source which is not only less secure but also inconvenient.
Adi Shamir, Ronald Rivest, and Leonard Adleman : When they were students at MIT. Image Credit : ams.org
Another trio of computer scientists at MIT namely Ron Rivest, Adi Shamir and Leonard Aldeman developed a mathematical function in 1977 (called the RSA Code, named after their initials) which was based on the revolutionary concept of asymmetric ciphers (which uses separate keys for encryption and decryption).
The RSA Code, a system of asymmetric cryptography is a form of Public Key cryptography. To understand the mystery of RSA Code, we again take the example of Alice and Bob, who want to share some private information with each other. Alice creates a public key (which is a one way function- that is it is impossible for anyone to reverse it to decode the message). This is available to everyone who wants to send her a message. So Bob encrypts the message using the public keys and sends it to Alice. Now Alice needs to decipher the message sent to her and for this she possess a private key (some extra piece of information) that allows only her to decode any message sent to her since no one else possess the private key for decryption.
Mathematically it goes like this:
• Alice takes two giant prime numbers, $p$ and $q$. Let’s take $p=17$ and $q=11$ for simplicity. She multiplies $p$ and $q$ to get $N=187$. In practice the numbers $p$ and $q$ are quite large, so that $N$ cannot be easily found out even if computers are employed.
• She now picks up another number $e$ (say $7$) such that $e$ is relatively prime to $(p-1)(q-1)$ which in this case is $(17-1)(11-1)=160$. This is an outcome of Euler’s Theorem which states that if we raise $a$ to the power Euler’s Totient function of $n$ then we end up with $1$ modulo $n$. That Is $a^{\phi(n)}\equiv 1\pmod n$, where $\phi(n)$ is the Euler’s totient function.
• Alice now publishes $e$ and $N$ which are available to everyone as public keys for encryption.
• Bob wants to send her a message $M$ (say $88$). He uses the encryption formula (one way function) along with the public keys issued by Alice to get the encryption key $C=M^e \pmod N$ such that in our example $C=11$, where $C$ is called the Cipher Text, which is what he sends Alice.
• Since modular functions are one way functions hence it is extremely difficult to work out $M$ from $C$ when larger primes are involved. So even if Eve (any third person or spy) catches hold of $C$ when Bob sends it to Alice, she can’t decipher it back to trace $88$.
• But Alice can decipher the message because she knows $p$ and $q$. She calculates a special number, $d$ her private key according to the formula $ed\equiv 1\pmod{(p-1)(q-1)}$. In our example $d=23$ which can be easily calculated using the technique of Euclid’s algorithm.
• Once Alice has calculated her private key, she simply uses the decryption formula, $M\equiv C^d \pmod N$. Here in our example after simplification we shall get $M=88$.
Eureka! Alice is able to recover the original message sent by Bob. Thus this one-way function allows everyone to encrypt messages to a particular person by using the public keys ($N$ and $e$) but only the intended person can decrypt the message because the recipient is the only person who knows $p$ and $q$, and hence the only person who knows the private key $d$.
The fundamental logic involved is the difficulty in factorizing $N$ (to get $p$ and $q$). Nowadays, for important transactions $N$ is taken to be of the order of atleast $10^{308}$ which would take all the computers in the world, put together, longer than the age of the universe to factorize $N$.
As long as we do not come up with extremely efficient and faster ways of factorization, the RSA Code is here to stay in all its elegance, simplicity and beauty.
[This article has been written by Harman Kour, who is at present an intern with Gonit Sora.]
, | 1,229 | 5,292 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2018-30 | latest | en | 0.941884 |
https://machinelearningmind.com/tag/linear-regression/ | 1,638,627,693,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362992.98/warc/CC-MAIN-20211204124328-20211204154328-00176.warc.gz | 449,480,392 | 40,630 | # Assumptions Of Linear Regression – How to Validate and Fix
Linear Regression makes certain assumptions about the data and provides predictions based on that. Naturally, if we don’t take care of those assumptions Linear Regression will penalise us with a bad model (You can’t really blame it!). Continue reading Assumptions Of Linear Regression – How to Validate and Fix
# Multicollinearity – How to fix it?
This post will answer questions like What is multicollinearity ?, What are the problems that arise out of Multicollinearity? When do I have to fix Multicollinearity? and How to fix Multicollinearity? One of the important aspect that we have to … Continue reading Multicollinearity – How to fix it?
# Introduction To Linear Regression — E-commerce Dataset
Linear regression model is used to predict the relationship between variables or factors. The factor that is being predicted is called the scalar response (or dependent variable). The factors that are used to predict the value of the dependent variable are called explanatory variables (or independent variables). Continue reading Introduction To Linear Regression — E-commerce Dataset | 230 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-49 | longest | en | 0.893751 |
https://www.jiskha.com/display.cgi?id=1366185516 | 1,503,209,821,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105976.13/warc/CC-MAIN-20170820053541-20170820073541-00037.warc.gz | 904,578,915 | 3,976 | # Maths
posted by .
There is a one in three chance that if Yasmin misses the 7:00 a.m train, she will get a seat on the 8:00 a.m train. Using a single die, how would I conduct an experiment and put it into a frequency table?
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https://slideplayer.com/slide/4906085/ | 1,585,754,120,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505731.37/warc/CC-MAIN-20200401130837-20200401160837-00341.warc.gz | 613,706,744 | 22,383 | # Symmetric Matrices and Quadratic Forms
## Presentation on theme: "Symmetric Matrices and Quadratic Forms"— Presentation transcript:
DIAGONALIZATION OF SYMMETRIC MATRICES © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX A symmetric matrix is a matrix A such that .
Such a matrix is necessarily square. Its main diagonal entries are arbitrary, but its other entries occur in pairs—on opposite sides of the main diagonal. © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Theorem 1: If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal. Proof: Let v1 and v2 be eigenvectors that correspond to distinct eigenvalues, say, λ1 and λ2. To show that , compute Since v1 is an eigenvector Since Since v2 is an eigenvector © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Hence . But , so .
An matrix A is said to be orthogonally diagonalizable if there are an orthogonal matrix P (with ) and a diagonal matrix D such that ----(1) Such a diagonalization requires n linearly independent and orthonormal eigenvectors. When is this possible? If A is orthogonally diagonalizable as in (1), then © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Thus A is symmetric.
Theorem 2: An matrix A is orthogonally diagonalizable if and only if A is symmetric matrix. Example 1: Orthogonally diagonalize the matrix , whose characteristic equation is © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Solution: The usual calculations produce bases for the eigenspaces: Although v1 and v2 are linearly independent, they are not orthogonal. The projection of v2 onto v1 is © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX The component of v2 orthogonal to v1 is
Then {v1, z2} is an orthogonal set in the eigenspace for (Note that z2 is linear combination of the eigenvectors v1 and v2, so z2 is in the eigenspace). © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Since the eigenspace is two-dimensional (with basis v1, v2), the orthogonal set {v1, z2} is an orthogonal basis for the eigenspace, by the Basis Theorem. Normalize v1 and z2 to obtain the following orthonormal basis for the eigenspace for : © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX An orthonormal basis for the eigenspace for is
By Theorem 1, u3 is orthogonal to the other eigenvectors u1 and u2. Hence {u1, u2, u3} is an orthonormal set. © 2012 Pearson Education, Inc.
SYMMETRIC MATRIX Let Then P orthogonally diagonalizes A, and .
THE SPECTRAL THEOREM The set if eigenvalues of a matrix A is sometimes called the spectrum of A, and the following description of the eigenvalues is called a spectral theorem. Theorem 3: The Spectral Theorem for Symmetric Matrices An symmetric matrix A has the following properties: A has n real eigenvalues, counting multiplicities. © 2012 Pearson Education, Inc.
THE SPECTRAL THEOREM The dimension of the eigenspace for each eigenvalue λ equals the multiplicity of λ as a root of the characteristic equation. The eigenspaces are mutually orthogonal, in the sense that eigenvectors corresponding to different eigenvalues are orthogonal. A is orthogonally diagonalizable. © 2012 Pearson Education, Inc.
SPECTRAL DECOMPOSITION
Suppose , where the columns of P are orthonormal eigenvectors u1,…,un of A and the corresponding eigenvalues λ1,…,λn are in the diagonal matrix D. Then, since , © 2012 Pearson Education, Inc.
SPECTRAL DECOMPOSITION
Using the column-row expansion of a product, we can write ----(2) This representation of A is called a spectral decomposition of A because it breaks up A into pieces determined by the spectrum (eigenvalues) of A. Each term in (2) is an matrix of rank 1. For example, every column of is a multiple of u1. Each matrix is a projection matrix in the sense that for each x in , the vector is the orthogonal projection of x onto the subspace spanned by uj. © 2012 Pearson Education, Inc.
SPECTRAL DECOMPOSITION
Example 2: Construct a spectral decomposition of the matrix A that has the orthogonal diagonalization Solution: Denote the columns of P by u1 and u2. Then © 2012 Pearson Education, Inc.
SPECTRAL DECOMPOSITION
To verify the decomposition of A, compute and © 2012 Pearson Education, Inc. | 1,042 | 4,149 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | longest | en | 0.860931 |
https://www.khanacademy.org/math/math3/x5549cc1686316ba5:logs/x5549cc1686316ba5:log-change-of-base/v/change-of-base-formula-proof | 1,606,311,093,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00684.warc.gz | 755,156,017 | 32,908 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Main content
# Proof of the logarithm change of base rule
## Video transcript
What I want to do in this video is prove the change of base formula for logarithms, which tells us-- let me write this-- formula. Which tells us that if I want to figure out the logarithm base a of x, that I can figure this out by taking logarithms with a different base. That this would be equal to the logarithm base b-- so some other base-- base b of x, divided by the logarithm base b of a. And this is a really useful result. If your calculator only has natural logarithm or log base 10, you can now use this to figure out the logarithm using any base. If you want to figure out the log base 2-- let me make it clear. If you want to figure out the logarithm base, let's say, base 3 of, let's say, 25, you can use your calculator either using log base 10 or log base 2. So you could say that this is going to be equal to log base 10 of 25-- and most calculators have a button for that-- divided by log base 10 of 3. So this is an application of the change of base formula. But let's actually prove it. So let's say that we want to-- let's set logarithm base a of x to be equal to some new variable. Let's call that variable, let's call that equal to y. So this right over here, we are just setting that equal to y. Well, this is just another way of saying that a to the y power is equal to x. So we can rewrite this as a to the y power is equal to x. I'll write the x out here, because I'm about to-- these two things are equal. This is just another way of restating what we just wrote up here. Now, let's introduce the logarithm base b. And to introduce it, I'm just to take log base b of both sides of this equation. So let's take logarithm base b of the left-hand side, and logarithm base b of the right-hand side. Well, we know from our logarithm properties that the logarithm of something to a power is the exact same thing as the power times the logarithm of that something. So logarithm base b of a to the y is the same thing as y times the logarithm base b of a. So this is just a traditional logarithm property. We prove it elsewhere. And we already know it's going to be equal to the right-hand side. It's going to be equal to log base b of x. And now, let's just solve for y. And this is exciting, because y was this thing right over here. But now if we solve for y, we're going to be solving for y in terms of logarithm base b. To solve for y, we just have to divide both sides of this equation by log base b of a. So we divide by log base b of a on the left-hand side, and we divide by log base b of a on the right-hand side. And so on the left-hand side, these two characters are going to cancel out. And we are left with-- and we deserve a drum roll now-- that y is equal to log base b of x divided by log base b of a. So let me write it. Just copy and paste this so I don't have to keep switching colors. So let me paste this. So there you have it. You have your change of base formula. Remember, y is the same thing as this thing right over here. y is log of a. Actually, let me make it clear. y, which is equal to log of a, which is equal to log base a of x-- so copy and paste-- y, which is equal to this thing, which is how we defined it right over here, y is equal to log base a of x, we've just shown, is also equal to this, if we write it in terms of base b. And we have our change of base formula. | 908 | 3,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-50 | latest | en | 0.938686 |
https://www.teachoo.com/1855/547/Ex-12.2--6---A-chord-of-a-circle-of-radius-15-cm-subtends-60/category/Ex-12.2/ | 1,726,541,563,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00825.warc.gz | 934,445,802 | 22,536 | Ex 11.1
Chapter 11 Class 10 Areas related to Circles
Serial order wise
Transcript
Ex 11.1, 6 A chord of a circle of radius 15 cm subtends an angle of 60Β° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use Ο = 3.14 and β3 = 1.73) In a given circle, Radius (r) = 15 cm And, π½ = 60Β° Now, Area of segment APB = Area of sector OAPB β Area of ΞOAB Finding Area of sector OAPB Area of sector OAPB = π/360 Γ ππ2 = 60/360 Γ 3.14 Γ (15)2 = 1/6 Γ 3.14 Γ 15 Γ 15 = 1/2 Γ 3.14 Γ 5 Γ 15 = 117.75 cm2 Finding area of Ξ AOB We draw OM β₯ AB β΄ β OMB = β OMA = 90Β° And, by symmetry M is the mid-point of AB β΄ BM = AM = 1/2 AB In right triangle Ξ OMA sin O = (side opposite to angle O)/Hypotenuse sin ππΒ° = ππ΄/π¨πΆ 1/2=π΄π/15 15/2 = AM AM = ππ/π In right triangle Ξ OMA cos O = (π πππ ππππππππ‘ π‘π πππππ π)/π»π¦πππ‘πππ’π π cos ππΒ° = πΆπ΄/π¨πΆ β3/2=ππ/21 β3/2 Γ 15 = OM OM = βπ/π Γ 15 From (1) AM = π/πAB 2AM = AB AB = 2AM Putting value of AM AB = 2 Γ 1/2 Γ 15 AB = 15 cm Now, Area of Ξ AOB = 1/2 Γ Base Γ Height = π/π Γ AB Γ OM = 1/2 Γ 15 Γ β3/2 Γ 15 = 1/2 Γ 15 Γ 1.73/2 Γ 15 = 97.3125 cm2 Therefore, Area of segment APB = Area of sector OAPB β Area of ΞOAB = 117.75 β 97.3125 = 20.4375 cm2 Thus, Area of minor segment = 20.4375 cm2 Now, Area of major segment = Area of circle β Area of minor segment = Οr2β 20.4375 = 3.14 Γ πππβ 20.4375 = 3.14 Γ 15 Γ 15β"20.4375" = 706.5 β 20.4375 = 686.0625 cm2 | 874 | 1,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.587054 |
https://www.pcorner.com/list/MAG/DDJ0989.ZIP/KING.LST/ | 1,702,248,304,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00878.warc.gz | 1,033,743,220 | 10,309 | Output of file : KING.LST contained in archive : DDJ0989.ZIP
_FORCED-BASED SIMULATIONS_
by Todd King
[LISTING ONE]
/*-------------------------------------
Basic object classes and method
definitions for simulation software
File: simul.hpp
Todd King
----------------------------------------*/
#include
#include
#include
#include
#include
#include
#include "simconst.h"
#define ESC 27
#define MAX_BODY_POOL 100
typedef struct VECTOR_2D {
double x, y;
};
class BODY {
VECTOR_2D world;
VECTOR_2D velocity;
double mass;
double gmass;
char icon;
public:
BODY();
set_mass(double m);
set_velocity(double x, double y);
set_position(double x, double y);
apply_force(VECTOR_2D from, double amount);
VECTOR_2D position();
double get_gmass();
update();
set_icon(char c);
};
/* Distance and time units are converted to screen units and ticks */
BODY::set_mass(double m) {
double pow();
mass = m;
gmass = G * m;
};
BODY::set_velocity(double x, double y) {
velocity.x = x;
velocity.y = y;
};
BODY::set_position(double x, double y) {
world.x = x;
world.y = y;
};
BODY::apply_force(VECTOR_2D from, double gmass) {
VECTOR_2D d;
double rs;
double v;
double r;
d.x = world.x - from.x;
d.y = world.y - from.y;
rs = (d.x * d.x) + (d.y * d.y);
if(rs != 0.0) { // there's a seperation
r = sqrt(rs);
v = (gmass / rs) * SECS_PER_TIC;
velocity.x += v * d.x / r;
velocity.y += v * d.y / r;
}
};
BODY::BODY() {
world.x = 0;
world.y = 0;
velocity.x = 0;
velocity.y = 0;
icon = '*';
};
VECTOR_2D BODY::position() {
VECTOR_2D vec;
vec.x = world.x;
vec.y = world.y;
return(vec);
};
double BODY::get_gmass() {
return(gmass);
}
BODY::set_icon(char c) { icon = c; }
class UNIVERSE {
unsigned int body_cnt;
BODY *body_pool[MAX_BODY_POOL];
public:
UNIVERSE();
service(BODY *bptr);
big_bang();
};
UNIVERSE::UNIVERSE() {
body_cnt = 0;
};
UNIVERSE::service(BODY *bptr) {
if(body_cnt >= MAX_BODY_POOL) return(0);
body_pool[body_cnt] = bptr;
body_cnt++;
};
UNIVERSE::big_bang() {
int i, j;
init_screen();
print_message(" Press ESC to stop.");
for(;;) {
print_tick();
if(kbhit()) {
switch(getch())
{
case ESC:
return(0);
}
}
sleep(0.1);
/* Let each body influence all others */
for(i = 0; i < body_cnt; i++) {
for(j = 0; j < body_cnt; j++) {
if(j == i) continue; // don't apply force to self
body_pool[i]->apply_force(body_pool[j]->position(),
body_pool[j]->get_gmass());
}
}
/* Display all bodies */
for(i = 0; i < body_cnt; i++) {
body_pool[i]->update();
}
}
deinit_screen();
};
sleep(double seconds) {
time_t ltime1, ltime2;
time(<ime1);
time(<ime2);
while(difftime(ltime1, ltime2) < seconds) {
time(<ime2);
}
}
[LISTING TWO]
/*--------------------------------------------
Methods which are related to screen I/O.
File: simulscr.hpp
Todd King
----------------------------------------------*/
#include "simconst.h"
#define DISPLAY_Y 24
#define DISPLAY_X 80
BODY::update() {
extern VECTOR_2D Extent;
VECTOR_2D screen;
screen.x = DISPLAY_X * (world.x / EXTENT_X);
screen.y = DISPLAY_Y * (world.y / EXTENT_Y);
if( (screen.x < DISPLAY_X && screen.x >= 0.0) &&
(screen.y < DISPLAY_Y && screen.y >= 0.0) ) {
disp_move(DISPLAY_Y - (int) screen.y, (int) screen.x);
disp_printf(" ");
}
world.x += velocity.x * SECS_PER_TIC;
world.y += velocity.y * SECS_PER_TIC;
screen.x = DISPLAY_X * (world.x / EXTENT_X);
screen.y = DISPLAY_Y * (world.y / EXTENT_Y);
if( (screen.x < DISPLAY_X && screen.x >= 0.0) &&
(screen.y < DISPLAY_Y && screen.y >= 0.0) ) {
disp_move(DISPLAY_Y - (int) screen.y, (int) screen.x);
disp_printf("%c", icon);
}
disp_move(0,0);
};
init_screen() {
disp_open();
disp_move(0, 0);
disp_eeop();
}
deinit_screen() {
disp_close();
}
print_tick() {
static unsigned int Tick = 0;
disp_move(0, 0);
disp_printf("Tick: %u", Tick);
Tick++;
}
print_message(char mesg[]) {
disp_move(24,0);
disp_printf(mesg);
}
[LISTING FOUR]
/*------------------------------------------------
Simulation of what would happen if a planet
about twice the size of the Moon passed
close to the earth within the Moon's orbit.
Todd King
--------------------------------------------------*/
#include "simul.hpp"
#include "simulscr.hpp"
main() {
BODY earth;
BODY moon;
BODY planet_x;
UNIVERSE universe;
earth.set_mass(5.98e24);
earth.set_position(5.0e8, 5.0e8);
earth.set_icon('E');
moon.set_mass(7.36e22);
moon.set_position(5.0e8, 8.8e8);
moon.set_icon('M');
moon.set_velocity(-1020.0, 0.0);
planet_x.set_mass(14.8e22);
planet_x.set_position(1.0, 1.0e4);
planet_x.set_icon('X');
planet_x.set_velocity(1800, 2000);
universe.service(&earth);
universe.service(&moon);
universe.service(&planet_x);
universe.big_bang();
}
[LISTING FOUR]
/*-----------------------------------------------------
Various constants which affect the simulation
system. Units are in meters, seconds and Kilograms.
File: simconst.h
Todd King
-------------------------------------------------------*/
#define G -6.67e-11 /* Gravitational constant */
#define SECS_PER_TIC 86400 /* One day */
#define EXTENT_X 10e8 /* Width of displayed universe */
#define EXTENT_Y 10e8 /* Hieght of displayed universe */
[EXAMPLE 1]
/*-----------------------------------------------
Simulates the orbital dynamics of the Earth
and Moon.
Todd King
-------------------------------------------------*/
#include "simul.hpp"
#include "simulscr.hpp"
main() {
BODY earth;
BODY moon;
UNIVERSE universe;
earth.set_mass(5.98e24);
earth.set_position(5.0e8, 5.0e8);
earth.set_icon('E');
moon.set_mass(7.36e22);
moon.set_position(5.0e8, 8.8e8);
moon.set_icon('M');
moon.set_velocity(-1020.0, 0.0);
universe.service(&earth);
universe.service(&moon);
universe.big_bang();
}
### 3 Responses to “Category : Files from MagazinesArchive : DDJ0989.ZIPFilename : KING.LST”
1. Very nice! Thank you for this wonderful archive. I wonder why I found it only now. Long live the BBS file archives!
2. This is so awesome! 😀 I’d be cool if you could download an entire archive of this at once, though.
3. But one thing that puzzles me is the “mtswslnkmcjklsdlsbdmMICROSOFT” string. There is an article about it here. It is definitely worth a read: http://www.os2museum.com/wp/mtswslnk/ | 1,755 | 6,098 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.353554 |
https://bumpercarfilms.com/qa/what-is-an-elastic-collision.html | 1,618,280,673,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00308.warc.gz | 257,233,382 | 8,405 | # What Is An Elastic Collision?
## What happens in an elastic collision?
An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision.
Both momentum and kinetic energy are conserved quantities in elastic collisions.
They collide, bouncing off each other with no loss in speed..
## What is the difference between elastic and inelastic collision?
An elastic collision can be defined as a state where there is no net loss in kinetic energy in the system as the result of the collision. An inelastic collision can be defined as a type of collision where this is a loss of kinetic energy.
## Do elastic collisions stick together?
If the kinetic energy is the same, then the collision is elastic. If the kinetic energy changes, then the collision is inelastic regardless of whether the objects stick together or not.
## What is the difference between elastic and perfectly elastic?
The demand for a good is said to be elastic (or relatively elastic) when its PED is greater than one. In this case, changes in price have a more than proportional effect on the quantity of a good demanded. … Finally, demand is said to be perfectly elastic when the PED coefficient is equal to infinity.
## What is meant by one dimensional elastic collision?
Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.
## What are the 3 types of collision?
There are three different kinds of collisions, however, elastic, inelastic, and completely inelastic. Just to restate, momentum is conserved in all three kinds of collisions. What distinguishes the collisions is what happens to the kinetic energy.
## Are most collisions elastic or inelastic?
When both momentum and kinetic energy are conserved, the collision is called an elastic collision. Most collisions are inelastic because some amount of kinetic energy is converted to potential energy, usually by raising one of the objects higher (increasing gravitation PE) or by flexing the object.
## Is Newton’s cradle elastic or inelastic?
Newton’s Cradle visualizes an elastic collision by allowing students to see the mass as a number of balls, and the speed as the height the balls travel.
## What is an example of a perfectly inelastic collision?
Another common example of a perfectly inelastic collision is known as the “ballistic pendulum,” where you suspend an object such as a wooden block from a rope to be a target.
## What happens when two billiard balls collide?
When two billiard balls collide the collision is nearly elastic. An elastic collision is one in which the kinetic energy of the system is conserved before and after impact. … For collisions between balls, momentum is always conserved (just like in any other collision).
## What is the formula for perfectly elastic collision?
An elastic collision is a collision where both the Kinetic Energy, KE, and momentum, p are conserved. In other words, it means that KE0 = KEf and po = pf. When we recall that KE = 1/2 mv2, we will write 1/2 m1(v1i)2 + 1/2 m2(vi)2 = 1/2 m1(v1f)2 + 1/2 m2 (v2f)2.
## What is an elastic collision quizlet?
Elastic Collision. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. Collision. An instance of one moving object or person striking violently against another.
## Why is kinetic energy lost in an inelastic collision?
In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.
## What happens when two objects collide?
Newton’s third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum).
## What is the two types of collision?
There are two general types of collisions in physics: elastic and inelastic. An inelastic collisions occurs when two objects collide and do not bounce away from each other.
## How do you derive an elastic collision equation?
To derive the elastic collision equations we make use of the Momentum Conservation condition and Kinetic Energy Conservation condition.m1 – Mass of object 1; m2 – Mass of object 2; … Substitute Equation ( 3b ) in Equation ( 1 ) to eliminate v2f.Rearrange this and solve for v1f :More items…•Feb 11, 2016 | 1,032 | 4,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2021-17 | latest | en | 0.948608 |
https://gitlab.lrz.de/i7/peregrine/-/commit/1e1bcf3739613fd1575ab788e1be67480f6041d6?view=parallel | 1,638,305,574,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359073.63/warc/CC-MAIN-20211130201935-20211130231935-00295.warc.gz | 359,537,598 | 23,726 | 2.12.2021, 9:00 - 11:00: Due to updates GitLab may be unavailable for some minutes between 09:00 and 11:00.
Commit 1e1bcf37 by Philipp J. Meyer
### Added remaining constraints for unique terminal marking
parent da341783
... @@ -9,26 +9,40 @@ import PetriNet ... @@ -9,26 +9,40 @@ import PetriNet import Property import Property import Solver import Solver placeConstraints :: PetriNet -> SIMap Place -> SIMap Place -> SIMap Transition -> SBool stateEquationConstraints :: PetriNet -> SIMap Place -> SIMap Place -> SIMap Transition -> SBool placeConstraints net m1 m2 x = stateEquationConstraints net m1 m2 x = bAnd \$ map checkPlaceEquation \$ places net bAnd \$ map checkStateEquation \$ places net where checkPlaceEquation p = where checkStateEquation p = let incoming = map addTransition \$ lpre net p let incoming = map addTransition \$ lpre net p outgoing = map addTransition \$ lpost net p outgoing = map addTransition \$ lpost net p in val m1 p + sum incoming - sum outgoing .== val m2 p in val m1 p + sum incoming - sum outgoing .== val m2 p addTransition (t,w) = literal w * val x t addTransition (t,w) = literal w * val x t nonNegativityConstraints :: PetriNet -> SIMap Place -> SIMap Place -> SBool nonNegativityConstraints :: PetriNet -> SIMap Place -> SBool nonNegativityConstraints net m1 m2 = nonNegativityConstraints net m = let m1nn = map (checkVal m1) \$ places net bAnd \$ map checkVal \$ places net m2nn = map (checkVal m1) \$ places net where checkVal p = val m p .>= 0 in bAnd m1nn &&& bAnd m2nn where checkVal mapping n = val mapping n .>= 0 terminalConstraints :: PetriNet -> SIMap Place -> SBool terminalConstraints net m = bAnd \$ map checkTransition \$ transitions net where checkTransition :: Transition -> SBool checkTransition t = bOr \$ map checkPlace \$ lpre net t checkPlace (p,w) = val m p .< literal w nonEqualityConstraints :: PetriNet -> SIMap Place -> SIMap Place -> SBool nonEqualityConstraints net m1 m2 = bOr \$ map checkNonEquality \$ places net where checkNonEquality p = val m1 p ./= val m2 p checkUniqueTerminalMarking :: PetriNet -> SIMap Place -> SIMap Place -> SIMap Transition -> SBool checkUniqueTerminalMarking :: PetriNet -> SIMap Place -> SIMap Place -> SIMap Transition -> SBool checkUniqueTerminalMarking net m1 m2 x = checkUniqueTerminalMarking net m1 m2 x = placeConstraints net m1 m2 x &&& nonEqualityConstraints net m1 m2 &&& nonNegativityConstraints net m1 m2 stateEquationConstraints net m1 m2 x &&& nonNegativityConstraints net m1 &&& nonNegativityConstraints net m2 &&& terminalConstraints net m1 &&& terminalConstraints net m2 checkUniqueTerminalMarkingSat :: PetriNet -> ConstraintProblem Integer (Marking, Marking) checkUniqueTerminalMarkingSat :: PetriNet -> ConstraintProblem Integer (Marking, Marking) checkUniqueTerminalMarkingSat net = checkUniqueTerminalMarkingSat net = ... ...
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Finish editing this message first! | 785 | 2,990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-49 | latest | en | 0.507532 |
https://camnangtienganh.vn/us/27-how-many-seagrams-to-get-drunk-hienthithang-hienthinam/ | 1,701,743,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00104.warc.gz | 182,788,880 | 54,955 | # 27 How Many Seagrams To Get Drunk 12/2023
Below is the best information and knowledge about how many seagrams to get drunk compiled and compiled by the Cẩm Nang Tiếng Anh team, along with other related topics such as: how many seagrams to get drunk reddit, how many wine coolers to get drunk, how many 3.2 wine coolers to get drunk, how many jamaican me happy to get drunk, can you get drunk off seagrams, will 3.2% alcohol get you drunk, can 2 seagram’s get you drunk, how much alcohol to get drunk calculator
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The most popular articles about how many seagrams to get drunk
## 1. How many Seagrams escapes will get you drunk?
How many Seagrams escapes will get you drunk? Seagram’s Escapes has much less alcohol than beer – solely 3.2%, when most beers have about 5% alcohol, so it is concerning the weakest …
When youuve had a few Seagrams Escapes, you might be wondering how many of the spirits will get you drunk. The answer is: it depends.
## 2. How many Seagrams get you drunk? – Global Tech CIO
How many Seagrams get you drunk? – Global Tech CIO 1. It’s hard to give a straight answer to that question, because a lot of it depends on your: · 2. Seagrams Alcohol · 3. Seagrams Escape.
Again, it all depends on a combination of the factors I mentioned above. I recommend drinking one bottle at a time and see how you feel about 10 minutes later. If you decide you’re not drunk enough, drink more, but always give it about 10-15 minutes to take effect after each bottle. If you drink too…
## 3. How many Seagrams escapes will get you drunk?
How many Seagrams escapes will get you drunk? If your 115 pounds how many will get you drunk. … Seagram’s Escapes has less alcohol than beer – only 3.2%, when most beers have about 5% …
Again, it all depends on a combination of the factors I mentioned above. I recommend drinking one bottle at a time and see how you feel about 10 minutes later. If you decide you’re not drunk enough, drink more, but always give it about 10-15 minutes to take effect after each bottle. If you drink too…
## 4. how many seagrams to get drunk – News Pedia Log
how many seagrams to get drunk – News Pedia Log Its one of those underground popular liquors and for good reason. Most of it varies from 28 to 31. Id hardly call that a shitload more beer to …
At only 42 ABV Id just end up getting full from drinking too many of them instead of feeling any effects of the alcohol. I drank hard Mikes.
## 5. Can you get drunk off seagrams escapes? – Foodly
Can you get drunk off seagrams escapes? – Foodly Well that depends on you. If you have not drank very much alcohol before and you plan to drink it all in one night, yes most likely you will …
Furthermore, How many wine coolers get you drunk?
## 6. How many Wine Coolers would it take to get drunk? : r/alcohol
How many Wine Coolers would it take to get drunk? : r/alcohol I don’t plan on getting drunk, just wondering. I’m drinking Seagrams wine coolers and they have 3.2% alcohol. I weigh about 150 pounds.
Found the internet!
## 7. How Many Wine Coolers to Get Drunk in 2022 – REAL gastropub
How Many Wine Coolers to Get Drunk in 2022 – REAL gastropub Low-alcohol beverages promote greater self-control and self-consciousness. But did you know that 3.2% of alcohol in Seagrams wine coolers by …
Let’s take a look about “How Many Wine Coolers to Get Drunk?“
## 8. How Many Seagrams To Get Drunk?
How Many Seagrams To Get Drunk? how many seagrams to get drunk? · How Many Beers to Reach .08% BAC? | Light Beer vs. IPA · Seagram’s Escapes Italian Ice Margarita · Our Mission At …
If we’re talking in terms of ABV, which modern drinkers understand much better, 3.2 percent beer is really 4.0 percent beer. That’s not too different from the percentage of alcohol by weight in many “full” beers; Amstel Light clocks in at 4.1 percent, and Miller Lite, Coors Light, and Bud Light are …
## 9. FAQ – Seagram’s Escapes
FAQ – Seagram’s Escapes Frequently Asked Questions about Seagram’s Escapes – allergens, ingredients, how to purchase and … Where can I find Seagram’s Escapes products near me?
Seagram’s Escapes does not contain the top eight allergens acknowledged by the FDA (dairy, soy, wheat, eggs, tree nuts, peanuts, shellfish, and fish), except for Seagram’s Escapes Calypso Colada. Seagram’s Escapes Calypso Colada contains coconut, which the Food and Drug Administration recognizes as …
## 10. How much alcohol is in a Seagrams? – Dmcoffee.blog
How much alcohol is in a Seagrams? – Dmcoffee.blog Can you get drunk on seagrams escapes? … The real turnoff, however, comes from the tiny, nagging hint of alcohol in the background. Seagram’s …
If we’re talking in terms of ABV, which modern drinkers understand much better, 3.2 percent beer is really 4.0 percent beer. That’s not too different from the percentage of alcohol by weight in many “full” beers; Amstel Light clocks in at 4.1 percent, and Miller Lite, Coors Light, and Bud Light are …
## 11. How much alcohol is in Seagram’s Escapes? – Dmcoffee.blog
How much alcohol is in Seagram’s Escapes? – Dmcoffee.blog Is Seagram’s Escapes a wine cooler? Is 3.2 alcohol a lot? How many wine coolers does it take to get drunk? Is a shot of whiskey equal to one …
How Many Smirnoff Ice Will Get You Drunk? It only takes three to four shots of Smirnoff Ice to get drunk, assuming that each shot contains 12 oz. The estimation is only for people with high alcohol tolerance, weighing more than 100-pounds, and those who do not have any medical conditions like insomn…
## 12. Seagram’s Escapes – Drinkable Review
Seagram’s Escapes – Drinkable Review I find wine coolers to be too sugary sweet and overcarbonated, these have just enough flavor and sweetness but not overly too much. I do enjoy …
Now, this is not necessarily to say that they are bad just because they are not really beer. Comparing them to their more painstakingly prepared and carefully engineering counterparts may be a bit unfair, as that is not the image that these drinks try to portray. While many beer breweries strive fo…
## 13. How Many Beers To Get Drunk: Analyzing the Effect of Alcohol
How Many Beers To Get Drunk: Analyzing the Effect of Alcohol How many beers to get drunk? This question is constantly on the lips of beer lovers. Learn the average number of beers that can get you drunk.
How many beers to get drunk?
## 14. How Much Alcohol Is In Seagram’s Escapes?
How Much Alcohol Is In Seagram’s Escapes? The truth is that you can get drunk from drinking Smirnoff Ice. For most people, it will take an average of four 22oz bottles of Smirnoff Ice to …
The Seagram’s Escapes would be perfect for an occasion where you want a little something to drink but you don’t want to blur the edges too much. I think my girl Taylor said it perfectly: “Surprisingly refreshing, not a lot of kick, which I liked because it’s enough to chill you out without overpower…
Cẩm Nang Tiếng Anh/images/how-many-wine-coolers-does-it-take-to-get-drunk.jpg” alt=”how many seagrams to get drunk”>
## 15. How many wine coolers does it take to get drunk?
How many wine coolers does it take to get drunk? It is bottled in two sizes: 375 and 750 milliliters. How much alcohol is in seagrams escapes? 2% What is the alcohol content of Seagram’s Escapes? Seagram’s …
How Many Shots Could Get You Drunk? An average person can get drunk after four to five shots of alcohol. Two or three shots can get you tipsy, while seven or more shots will leave you extremely drunk. However, alcohol has different alcohol content, and people have different alcohol toleration.
## 16. Twisted Tea Alcohol Content | How many get you drunk?
Twisted Tea Alcohol Content | How many get you drunk? Twisted Teas have the ability to slip away from you, much like those old Seagram’s Wine Coolers of old. It’s an alcoholic drink that never feels …
If you’re a seasoned drinker/you drink alcohol a lot, you probably know how many glasses of wine or cans of beer it takes to get you drunk. By that rationale, Twisted Tea is no different. If it takes you 10 PBRs (Pabst Blue Ribbon) to get drunk, it will take you roughly 9-10 Twisted Teas to get dr…
## 17. Can Wine Coolers Make You Drunk? – BlackTailNYC.com
Can Wine Coolers Make You Drunk? – BlackTailNYC.com However, as a general rule of thumb, it is generally safe for men to consume up to four seagrams wine coolers in a single sitting, and women to consume up to …
How many coolers does it take to get drunk? The legal blood alcohol content limit in the majority of states in the United States is 0.08 [1]. The alcohol content of an alcoholic beverage is between five and ten percent, so a four-drink cocktail with five to ten percent alcohol content will take abou…
## 18. How Many Smirnoff Ice Does It Take To Get Drunk (2022)
How Many Smirnoff Ice Does It Take To Get Drunk (2022) Despite having an alcohol content closer to beers than Smirnoff Vodka, this girl’s drink can still get you drunk in an hour. However, many …
Three shots are the standard intake for most areas in the United States to get legally drunk. The symptoms may vary at this point, but people with small physiques may already be having a hard time. Those with a higher tolerance will not yet get drunk but feel a little light-headed.
## 19. How much alcohol is in Seagram’s Escapes? – Dailion
How much alcohol is in Seagram’s Escapes? – Dailion So 3.2 beer is solely 20% much less alcohol than common Budweiser. Guinness Draft is 4.0 by quantity, or 3.2 by weight. So sure, you may get drunk on 3.2 …
How Do You Outline Drunk? A Take a look at Blood Alcohol Content material Ranges
## 20. Age Gate – Seagram’s Gin
Age Gate – Seagram’s Gin To find out more about responsible consumption, visit Responsibility.org. See what Seagram’s Gin Spain is up to at Seagramsgin.es. Seagram’s Extra Dry Gin ® …
## 21. Alcohol Seller-Server Training for Bartenders | 360training
Alcohol Seller-Server Training for Bartenders | 360training New York ATAP On-Premises + ANSI-Accredited Food Handler Training. Get your New York alcohol server and ANSI-Accredited food hander training courses. 25.00.
The second type of bartender training is alcohol seller-server training. It’s generally required by most employers either before they will hire you or within 30 days of employment. This training teaches you the laws for safely selling and serving alcohol so that you can protect yourself and your emp…
## 22. Liquor Review: Seagram’s Seven Crown – BoozeBasher
Liquor Review: Seagram’s Seven Crown – BoozeBasher Out of all of the many types of whiskey I have tasted, this is the first (or at least the … As for getting you drunk, Seagram’s 7 really lets you down.
As for getting you drunk, Seagram’s 7 really lets you down. It got me to the point where I felt that if I had one more drink, that I would be wasted. It got me there quickly, but three drinks later, I was still saying, “if just had …
## 23. Nutrition Information for Seagram’s Wine Coolers – Livestrong
Nutrition Information for Seagram’s Wine Coolers – Livestrong However, Seagram’s wine coolers are primarily made up of carbohydrates, … helped millions of people lose weight the healthy way—by getting support from an …
Seagram’s Escapes wine coolers are available in 11.2 oz., one serving bottles. According to the free calorie tracker app MyPlate, one bottle of Seagram’s Calypso Colada wine cooler has 218 calories. Their Strawberry Daiquiri wine cooler contains 224 calories and the Wild Berries has 212 calories. Ca…
## 24. Alcoholic beverage – Wikipedia
Alcoholic beverage – Wikipedia The consumption of alcoholic drinks, often referred to as “drinking”, plays an important social role in many cultures. Most countries have laws regulating …
Alcohol is a depressant, which in low doses causes euphoria, reduces anxiety, and increases sociability. In higher doses, it causes drunkenness, stupor, unconsciousness or death. Long-term use can lead to an alcohol use disorder, an increased risk of developing several types of cancer, cardiovascula…
## 25. Seagram’s Escapes Review – My Half Assed Kitchen
Seagram’s Escapes Review – My Half Assed Kitchen Not liking alcohol..suggestion was Escapes strawberry daiquiri. LOVED IT…but was concerned about becoming a drunk so googled and found this …
Yes, that’s right, the Seagram’s Escapes collection has taken store shelves by storm, offering a summery beverage option in a variety of girly colors sure to tickle the fancy of just about anyone without testicles. Since we’ve already established I like girly drinks, clearly, I had to give these a …
## 26. Seagram’s 7 Crown American Blended Whiskey Review
Seagram’s 7 Crown American Blended Whiskey Review A whiskey many of us had in our drinking youth and college days. … used as a mixing vehicle to get you drunk, it does its job very well.
This is not a refined whiskey, this is not a delicate whiskey, it’s barely even a whiskey, but as a whiskey flavored vodka used as a mixing vehicle to get you drunk, it does its job very well.
## 27. How Many Smirnoff Ice does it Take to Get Drunk?
How Many Smirnoff Ice does it Take to Get Drunk? Of course, the alcohol content of any alcoholic beverage has a huge impact on how many drinks it takes to get drunk.
Due to the similar alcohol contents, it’s likely that one could get drunk by drinking the same number of Smirnoff Ice that they would beers. However, the actual number of Smirnoff Ice it takes someone to get drunk varies, depending on a variety of factors such as:
Video tutorials about how many seagrams to get drunk
Categories: How to
Synthetic: Cẩm Nang Tiếng Anh US | 3,384 | 13,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-50 | latest | en | 0.922495 |
https://diy.stackexchange.com/questions/231120/can-a-chin-up-bar-be-supported-by-just-two-of-the-four-intended-mounting-points | 1,653,417,411,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00714.warc.gz | 261,594,458 | 67,214 | # Can a chin-up bar be supported by just two of the four intended mounting points?
I would like to hang a chin-up bar on a drywall.
So far I'm lucky. I measured that the studs (dotted in the image) are 16″ apart.
The distance between the two lower holes in the frame (grey area in the image) are also 16″ apart.
But the two top screws are problematic.
Possible solutions:
1. Don't bolt the top two support points. Two 3/8″ bolts are plenty for supporting an average adult.
2. Use butterfly drywall anchors for the top two holes. They will not take much weight, but they'll at least provide a bit of stability in case someone decides to swing.
3. Neither (1) nor (2) is good enough. Cut out 3/4″ plywood about 18″ ✕ 18″. Bolt the plywood to the wall with four serious bolts, and then bolt the chin-up frame to the plywood (how? And will 3/4″ maple plywood be strong enough?).
Which would you use, solution 1, 2, or 3?
Update
Let me add a second diagram that illustrates how the weight is transferred.
Notice in particular that the frame consists, as is common, of a diagonal support that takes the load to the wall through compression.
This means that omitting to add lumber underneath the bottom part of the structure will result in either:
• if the lightly shaded U-shaped metal part is too weak, it will bend, or,
• if that part is quite rigid, the two bolts at the top will be subjected to an undue pulling (tension) force and that tension—likely equal to the full-weight of a (possibly swinging) adult underneath—will have to go through a single piece of two-by-four (the top one). Remember that there are no studs directly behind the two top bolts.
This means that in addition to the two pieces of lumber suggested in isherwood's solution, it may be necessary to add a third piece of lumber. That third piece need only be glued to the wall.
Which would you do? Would you:
1. Add a third piece of lumber, perhaps by just minimally gluing it to the wall, just enough for it to stay where it is,
2. Add two pieces of lumber vertically such that they cover the width from A to C, or,
3. Add a single piece of (plywood, perhaps) that acts as support for the entire frame?
Update 2
I'm giving up putting in all the details in one question. I posted a sequel here.
• Have you check that there is no electrical wires or plumbing pipes in the studs? Jul 30, 2021 at 20:44
• @crip659 That's a valuable hint. I haven't. But on second look, there is an electrical socket 1" from one of the two studs I intend to use. There is possible (likely) an electric cable running on the side of the stud. Still: 1- The electric wires will be running beside the stud, so I'm safe if I drill through the center, no? How do you check either way? Also: 2- It's a bedroom, and so there is almost no chance there would be plumbing pipes running, no?
– Sam
Jul 30, 2021 at 21:40
• Electric cables can often go though studs, before running down/up studs. Do not assume, because you probably want screws longer than 1 1/2 inch long. If no bathroom/kitchen behind or above, then less chance of plumbing. Drywall cheap to remove and/or replace with good plywood. Jul 30, 2021 at 21:55
• @crip659 Thanks. Another quick question: The stud-sensor I have only detects (lumber) studs. I see that other models also detect electric wiring, but are these so precise they will help me find out whether the electric wires are inside or just outside a stud?
– Sam
Jul 30, 2021 at 22:00
• Should be good enough to trace where wire goes. If no wire above outlet, should be safe. If wire goes up and across studs, need to place screws above or below where wire goes across, give at least half a foot from wire for safety. Jul 30, 2021 at 22:06 | 922 | 3,719 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-21 | longest | en | 0.947811 |
https://www.yaclass.in/p/mathematics-state-board/class-6/bill-profit-and-loss-4962/profit-and-loss-21303/re-8d3161fa-f299-4a41-9930-b18bd8629098 | 1,713,186,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00796.warc.gz | 987,592,302 | 11,868 | PDF chapter test
Formula:
$$\text{Discount}$$ $$=$$ $$\text{M.P.}$$ $$-$$ $$\text{S.P.}$$
$$\text{S.P.}$$ $$=$$ $$\text{M.P.}$$ $$-$$ $$\text{Discount}$$
1. C.P. $$=$$ $$Rs.1300$$
M.P. $$= Rs.1400$$
Discount (in $$Rs.$$) $$= Rs.40$$
Find S.P. and profit or loss.
$$\text{S.P.}$$ $$=$$ $$\text{M.P.}$$ $$-$$ $$\text{Discount}$$
$$= 1400 - 40$$
$$= Rs.1360$$
In this case, S.P. $$>$$ C.P.
Therefore, there is a profit.
$$\text{Profit}$$ $$=$$ $$\text{S.P.}$$ $$-$$ $$\text{C.P.}$$
$$= 1360 - 1300$$
$$= Rs.60$$
2. C.P.$$=$$ $$Rs.1200$$
M.P.$$= Rs.1250$$
Discount (in $$Rs.$$) $$= Rs.100$$
Find S.P. and profit or loss.
$$\text{S.P.}$$ $$=$$ $$\text{M.P.}$$ $$-$$ $$\text{Discount}$$
$$= 1250 - 100$$
$$= Rs.1150$$
In this case, C.P. $$>$$ S.P.
Therefore, there is a loss.
$$\text{Loss}$$ $$=$$ $$\text{C.P.}$$ $$-$$ $$\text{S.P.}$$
$$= 1200 - 1150$$
$$= Rs.50$$ | 407 | 886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | longest | en | 0.463281 |
https://elki-project.github.io/releases/release0.8.0/javadoc/elki/math/geometry/GrahamScanConvexHull2D.html | 1,679,673,022,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00435.warc.gz | 279,707,389 | 5,543 | ## Class GrahamScanConvexHull2D
• java.lang.Object
• elki.math.geometry.GrahamScanConvexHull2D
• @Reference(authors="P. Graham",
title="An Efficient Algorithm for Determining the Convex Hull of a Finite Planar Set",
booktitle="Information Processing Letters 1",
url="https://doi.org/10.1016/0020-0190(72)90045-2",
bibkey="DBLP:journals/ipl/Graham72")
public class GrahamScanConvexHull2D
extends java.lang.Object
Classes to compute the convex hull of a set of points in 2D, using the classic Grahams scan. Also computes a bounding box.
Reference:
P. Graham
An Efficient Algorithm for Determining the Convex Hull of a Finite Planar Set
Information Processing Letters 1
Since:
0.4.0
Author:
Erich Schubert
• ### Field Summary
Fields
Modifier and Type Field Description
private double factor
Scaling factor if we have very small polygons.
private DoubleMinMax minmaxX
Min/Max in X
private DoubleMinMax minmaxY
Min/Max in Y
private boolean ok
Flag to indicate that the hull has been computed.
private java.util.List<double[]> points
The current set of points
• ### Constructor Summary
Constructors
Constructor Description
GrahamScanConvexHull2D()
Constructor.
• ### Method Summary
All Methods
Modifier and Type Method Description
void add(double... point)
Add a single point to the list (this does not compute the hull!)
private void computeConvexHull()
Compute the convex hull.
private void findStartingPoint()
Find the starting point, and sort it to the beginning of the list.
Polygon getHull()
Compute the convex hull, and return the resulting polygon.
private double getRX(double[] a, double[] origin)
Get the relative X coordinate to the origin.
private double getRY(double[] a, double[] origin)
Get the relative Y coordinate to the origin.
private void grahamScan()
The actual graham scan main loop.
private boolean isConvex(double[] a, double[] b, double[] c)
Simple convexity test.
protected int isLeft(double[] a, double[] b, double[] o)
Test whether a point is left of the other wrt. the origin.
private double mdist(double[] a, double[] b)
Manhattan distance.
• ### Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
• ### Field Detail
• #### points
private java.util.List<double[]> points
The current set of points
• #### minmaxX
private DoubleMinMax minmaxX
Min/Max in X
• #### minmaxY
private DoubleMinMax minmaxY
Min/Max in Y
• #### ok
private boolean ok
Flag to indicate that the hull has been computed.
• #### factor
private double factor
Scaling factor if we have very small polygons. TODO: needed? Does this actually improve things?
• ### Constructor Detail
• #### GrahamScanConvexHull2D
public GrahamScanConvexHull2D()
Constructor.
• ### Method Detail
public void add(double... point)
Add a single point to the list (this does not compute the hull!)
Parameters:
point - Point to add
• #### computeConvexHull
private void computeConvexHull()
Compute the convex hull.
• #### findStartingPoint
private void findStartingPoint()
Find the starting point, and sort it to the beginning of the list. The starting point must be on the outer hull. Any "most extreme" point will do, e.g., the one with the lowest Y coordinate and for ties with the lowest X.
• #### getRX
private double getRX(double[] a,
double[] origin)
Get the relative X coordinate to the origin.
Parameters:
a -
origin - origin double[]
Returns:
relative X coordinate
• #### getRY
private double getRY(double[] a,
double[] origin)
Get the relative Y coordinate to the origin.
Parameters:
a -
origin - origin double[]
Returns:
relative Y coordinate
• #### isLeft
protected final int isLeft(double[] a,
double[] b,
double[] o)
Test whether a point is left of the other wrt. the origin.
Parameters:
a - double[] A
b - double[] B
o - Origin double[]
Returns:
+1 when left, 0 when same, -1 when right
• #### mdist
private double mdist(double[] a,
double[] b)
Manhattan distance.
Parameters:
a - double[] A
b - double[] B
Returns:
Manhattan distance
• #### isConvex
private boolean isConvex(double[] a,
double[] b,
double[] c)
Simple convexity test.
Parameters:
a - double[] A
b - double[] B
c - double[] C
Returns:
convexity
• #### grahamScan
private void grahamScan()
The actual graham scan main loop.
• #### getHull
public Polygon getHull()
Compute the convex hull, and return the resulting polygon.
Returns:
Polygon of the hull | 1,168 | 4,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-14 | latest | en | 0.573165 |
https://byjus.com/question-answer/what-s-the-volume-of-the-prism-with-a-base-of-isosceles-right-angled-triangled/ | 1,721,012,426,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514655.27/warc/CC-MAIN-20240715010519-20240715040519-00235.warc.gz | 137,103,847 | 26,767 | 1
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Question
# What's the volume of the prism with a base of isosceles right-angled triangled in which two sides are of length 6 cm and one side of length √72 cm and height of the prism is 10 cm?
A
180 cm3
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B
18 cm3
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C
280 cm3
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D
90 cm3
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Solution
## The correct option is A 180 cm3 Given that, Height of the prism = 10 cm. Dimension of base triangle = 6 cm, 6 cm and √72 cm. We know that, volume of prism =Area of base×height ∵ The base is isosceles right-angled triangled. Now, area of base (isosceles right-angled triangled) =12×base×height =12×6×6 =18 cm2 ∴ Volume of the prism =base area×height =18×10 =180 cm3
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Join BYJU'S Learning Program | 339 | 1,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.838358 |
http://math.stackexchange.com/questions/133294/construct-a-graph-g-for-which-the-is-adjacent-to-relation-is-antisymmetric | 1,469,399,186,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824185.14/warc/CC-MAIN-20160723071024-00155-ip-10-185-27-174.ec2.internal.warc.gz | 157,987,903 | 17,479 | # Construct a graph G for which the is-adjacent-to relation is antisymmetric.
Background: In this a graph is G=(V,E) where V is the set of all vertices and E is a set of 2-element subsets of V. For example: G=({1,2,3,4},{{1,2},{1,3},{2,4}}). E stands for edges similar to a line segment between point A and point B such that it is represented by {A,B}. {A,A} is not an acceptable edge. For antisymmetric, the following must be true: If x~y and y~x, then x=y. For transitive, the following must be true: If x~y and y~z, then x~z.
Question 1: Construct a graph G for which the is-adjacent-to relation, ~, is antisymmetric.
Question 2: Construct a graph G for which the is-adjacent-to relation, ~, is transitive.
-
I assume your edges are directed, or else these questions are impossible. One example that works for both questions is to take a (partial) ordering on some set (the usual ordering on a finite set of integers will work just fine). Let $\{a,b\}$ be an edge if and only if $a<b$. A particularly trivial answer to your questions is the graph with no edges, which is also a (trivial) example of a partially ordered set.
Now, to your question. We only include the edge $\{a,b\}$ when $a$ is strictly less than $b$. So when $a=b$, there is no edge between them, hence no contradiction. In particular, the edge $\{a,a\}$ is not in the graph. – Brett Frankel Apr 18 '12 at 3:50 | 382 | 1,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2016-30 | latest | en | 0.910703 |
http://www.nidokidos.org/threads/23747-BT-1-Solved | 1,477,166,220,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719041.14/warc/CC-MAIN-20161020183839-00163-ip-10-171-6-4.ec2.internal.warc.gz | 614,930,461 | 13,621 | 1. ## BT 1 Solved
You're driving a bus, in Paris, 35 people get in. Once in Brussels, the bus stops, 25 people get in and 13 get off. The bus goes all the way to Berlin and there 14 people get out and 40 get in. Then the trip finishes in Geneva where 20 people get out. How old is the bus driver? If you get the answer please PM me and ya the answer does not have a numerical value, please do remember that while trying to find the solution.
Solution: The answer as I said is not a numeric value. This is where logic comes with Math. You are the driver.............and the question was what was the age of the driver? But then it is stated that the answer is not a numerical value. But while answering u will first arrive at a mental answer that is "The drivers the same age as me" and practically thats the answer but then Muffy and Hail.......u both are correct
2. What do you mean does not have a numeric value.. age is always numeric na..
3. i pm'd u
4. Solved
5. Thanks a lot master!!
6. UR welcome student
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A191113 Increasing sequence generated by these rules: a(1)=1, and if x is in a then 3x-2 and 4x-2 are in a. 82
1, 2, 4, 6, 10, 14, 16, 22, 28, 38, 40, 46, 54, 62, 64, 82, 86, 110, 112, 118, 136, 150, 158, 160, 182, 184, 190, 214, 244, 246, 254, 256, 326, 328, 334, 342, 352, 406, 438, 446, 448, 470, 472, 478, 542, 544, 550, 568, 598, 630, 638, 640, 726, 730, 734, 736, 758, 760, 766, 854, 974, 976, 982, 1000, 1014, 1022, 1024, 1054, 1216 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS This sequence represents a class of sequences generated by rules of the form "a(1)=1, and if x is in a then hx+i and jx+k are in a, where h,i,j,k are integers." If m>1, at least one of the numbers b(m)=(a(m)-i)/h and c(m)=(a(m)-k)/j is in the set N of natural numbers. Let d(n) be the n-th b(m) in N, and let e(n) be the n-th c(m) in N. Note that a is a subsequence of both d and e. Examples: A191113: (h,i,j,k)=(3,-2,4,-2); d=A191146, e=A191149 A191114: (h,i,j,k)=(3,-2,4,-1); d=A191151, e=A191121 A191115: (h,i,j,k)=(3,-2,4,0); d=A191113, e=A191154 A191116: (h,i,j,k)=(3,-2,4,1); d=A191155 e=A191129 A191117: (h,i,j,k)=(3,-2,4,2); d=A191157, e=A191158 A191118: (h,i,j,k)=(3,-2,4,3); d=A191114, e=A191138 ... A191119: (h,i,j,k)=(3,-1,4,-3); d=A191120, e=A191163 A191120: (h,i,j,k)=(3,-1,4,-2); d=A191129, e=A191165 A191121: (h,i,j,k)=(3,-1,4,-1); d=A191166, e=A191167 A191122: (h,i,j,k)=(3,-1,4,0); d=A191168, e=A191169 A191123: (h,i,j,k)=(3,-1,4,1); d=A191170, e=A191171 A191124: (h,i,j,k)=(3,-1,4,2); d=A191172, e=A191173 A191125: (h,i,j,k)=(3,-1,4,3); d=A191174, e=A191175 ... A191126: (h,i,j,k)=(3,0,4,-3); d=A191128, e=A191177 A191127: (h,i,j,k)=(3,0,4,-2); d=A191178, e=A191179 A191128: (h,i,j,k)=(3,0,4,-1); d=A191180, e=A191181 A025613: (h,i,j,k)=(3,0,4,0); d=e=A025613 A191129: (h,i,j,k)=(3,0,4,1); d=A191182, e=A191183 A191130: (h,i,j,k)=(3,0,4,2); d=A191184, e=A191185 A191131: (h,i,j,k)=(3,0,4,3); d=A191186, e=A191187 ... A191132: (h,i,j,k)=(3,1,4,-3); d=A191135, e=A191189 A191133: (h,i,j,k)=(3,1,4,-2); d=A191190, e=A191191 A191134: (h,i,j,k)=(3,1,4,-1); d=A191192, e=A191193 A191135: (h,i,j,k)=(3,1,4,0); d=A191136, e=A191195 A191136: (h,i,j,k)=(3,1,4,1); d=A191196, e=A191197 A191137: (h,i,j,k)=(3,1,4,2); d=A191198, e=A191199 A191138: (h,i,j,k)=(3,1,4,3); d=A191200, e=A191201 ... A191139: (h,i,j,k)=(3,2,4,-3); d=A191143, e=A191119 A191140: (h,i,j,k)=(3,2,4,-2); d=A191204, e=A191205 A191141: (h,i,j,k)=(3,2,4,-1); d=A191206, e=A191207 A191142: (h,i,j,k)=(3,2,4,0); d=A191208, e=A191209 A191143: (h,i,j,k)=(3,2,4,1); d=A191210, e=A191136 A191144: (h,i,j,k)=(3,2,4,2); d=A191212, e=A191213 A191145: (h,i,j,k)=(3,2,4,3); d=e=A191145 ... Representative divisibility properties: if s=A191116, then 2|(s+1), 4|(s+3), and 8|(s+3) for n>1; if s=A191117, then 10|(s+4) for n>1. For lists of other "rules sequences" see A190803 (h=2 and j=3) and A191106 (h=j=3). LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 David Garth and Adam Gouge, Affinely Self-Generating Sets and Morphisms, Journal of Integer Sequences, Article 07.1.5, 10 (2007) 1-13. FORMULA a(1)=1, and if x is in a then 3x-2 and 4x-2 are in a; the terms of a are listed in without repetitions, in increasing order. EXAMPLE 1 -> 2 -> 4,6 -> 10,14,16,22 -> MAPLE N:= 2000: # to get all terms <= N S:= {}: agenda:= {1}: while nops(agenda) > 0 do S:= S union agenda; agenda:= select(`<=`, map(t -> (3*t-2, 4*t-2), agenda) minus S, N) od: sort(convert(S, list)); # Robert Israel, Dec 22 2015 MATHEMATICA h = 3; i = -2; j = 4; k = -2; f = 1; g = 8; a = Union[Flatten[NestList[{h # + i, j # + k} &, f, g]]] (* a=A191113; regarding g, see the Mathematica note at A190803 *) b = (a + 2)/3; c = (a + 2)/4; r = Range[1, 900]; d = Intersection[b, r] (* A191146 *) e = Intersection[c, r] (* A191149 *) m = a/2 (* divisibility property *) PROG (Haskell) import Data.Set (singleton, deleteFindMin, insert) a191113 n = a191113_list !! (n-1) a191113_list = 1 : f (singleton 2) where f s = m : (f \$ insert (3*m-2) \$ insert (4*m-2) s') where (m, s') = deleteFindMin s -- Reinhard Zumkeller, Jun 01 2011 CROSSREFS Cf. A190803, A191106. Sequence in context: A005574 A109807 A259645 * A125964 A288526 A139544 Adjacent sequences: A191110 A191111 A191112 * A191114 A191115 A191116 KEYWORD nonn AUTHOR Clark Kimberling, May 27 2011 STATUS approved
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Last modified August 18 14:30 EDT 2018. Contains 313832 sequences. (Running on oeis4.) | 2,077 | 4,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-34 | latest | en | 0.654584 |
https://nlinux.org/how-many-solutions-exist-for-the-given-equation/ | 1,659,897,858,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00124.warc.gz | 403,839,777 | 5,240 | nlinux.org->Linear-equations-> SOLUTION: -Determine just how many services exist-Use either elimination or substitution to discover the options (if any)-Graph the two lines, labeling the x-intercepts, y-intercepts, and also p var visible_logon_form_ = false;Log in or register.Username: Password: Register in one simple step!.Reset your password if you forgained it."; rerevolve false; } "> Log On
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Click here to watch ALL difficulties on Linear-equationsConcern 142804: -Determine how many remedies exist-Use either elimicountry or substitution to uncover the options (if any)-Graph the two lines, labeling the x-intercepts, y-intercepts, and points of interarea x + y = 3 and y = x + 3 Answer by MathLover1(18852)
(Show Source): You can put this solution on YOUR website! and ; you have the right to create them both in the conventional develop favor this: Solved by pluggable solver: Solving a straight mechanism of equations by subsitution Lets start via the given device of direct equations Now in order to deal with this system by using substitution, we need to settle (or isolate) one variable. I"m going to choose y. Solve for y for the initially equation Subtract from both sides Divide both sides by 1. Which breaks down and reduces to Now we"ve fully isolated y Because y equates to we have the right to substitute the expression right into y of the 2nd equation. This will get rid of y so we can solve for x. Replace y via . Because this eliminates y, we have the right to now fix for x. Distribute 1 to Multiply Reduce any fractions Subtract from both sides Combine the terms on the ideal side Now incorporate the terms on the left side. Multiply both sides by . This will certainly cancel out and also isolate x So once we multiply and also (and also simplify) we gain Now that we recognize that , lets substitute that in for x to fix for y Plug in into the 2nd equation Multiply Add to both sides Combine the terms on the right side Multiply both sides by . This will cancel out 1 on the left side. Multiply the terms on the ideal side Reduce So this is the other answer So our solution is and which deserve to additionally look like (,) Notice if we graph the equations (if you need help with graphing, check out this solver) we acquire graph of (red) and also (green) (hint: you might need to resolve for y to graph these) intersecting at the blue circle. and also we deserve to watch that the 2 equations intersect at (,). This verifies our answer. ----------------------------------------------------------------------------------------------- Check: Plug in (,) into the device of equations Let and also . Now plug those values right into the equation Plug in and also Multiply Add Reduce. Since this equation is true the solution functions. So the solution (,) satisfies Let and . Now plug those worths right into the equation Plug in and Multiply Add Reduce. Since this equation is true the solution functions.
See more: — O Formose Puer, Nimium Ne Crede Colori, Brewer'S: Beauty
So the solution (,) satisfies Due to the fact that the solution (,) satisfies the device of equations this verifies our answer. | 696 | 3,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-33 | latest | en | 0.886302 |
https://bumpercarfilms.com/qa/what-happens-to-the-kinetic-energy-of-particles-after-a-collision.html | 1,618,210,884,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00195.warc.gz | 271,971,623 | 8,874 | # What Happens To The Kinetic Energy Of Particles After A Collision?
## How do you find change in kinetic energy after a collision?
Collisions in One DimensionMass m1 = kg , v1 = m/s.Mass m2 = kg , v2 = m/s.Initial momentum p = m1v1 + m2v2 = kg m/s .Initial kinetic energy KE = 1/2 m1v12 + 1/2 m2v22 = joules.Then the velocity of mass m2 is v’2 = m/s.because the final momentum is constrained to be p’ = m1v’1 + m2v’2 = kg m/s .More items….
## How do you prove kinetic energy is conserved?
Kinetic energy has the equation (1/2)mv2. An elastic collision is one where kinetic energy is conserved. The masses that collide don’t deform from the collision nor do they stick together. An example of this would be pool balls colliding.
## How is kinetic energy conserved in a collision?
Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.
## Can total kinetic energy ever be higher after a collision than before?
You can operate in the same reference frame and still have an increase in kinetic energy. … All you need to do is apply momentum conservation as well as the condition of a 50% increase in kinetic energy. Or use the coefficient of restitution. It is totally possible.
## What does it mean when kinetic energy is conserved?
When one says that “kinetic energy is conserved in an elastic collision” that means that the total kinetic energy of the system of particles involved in the collision doesn’t change. … For a two particle system, the kinetic energy of each will change, but the sum won’t.
## Why is momentum conserved but not kinetic energy?
Momentum is conserved, because the total momentum of both objects before and after the collision is the same. However, kinetic energy is not conserved. Some of the kinetic energy is converted into sound, heat, and deformation of the objects.
96.7% is lost!
## How do you find total kinetic energy after an inelastic collision?
Inelastic Collision Two objects that have equal masses head toward one another at equal speeds and then stick together. Their total internal kinetic energy is initially 12mv2+12mv2=mv2 1 2 m v 2 + 1 2 m v 2 = m v 2 . The two objects come to rest after sticking together, conserving momentum.
## Can all kinetic energy be lost in a collision?
Can all the kinetic energy be lost in the collision? Yes, all the kinetic energy can be lost if the two masses come to rest due to the collision (i.e., they stick together). Describe a system for which momentum is conserved but mechanical energy is not.
## Is momentum conserved in a closed system?
1) Closed system – A closed system does not interact with its environment so there is no net external impulse. The total momentum of a closed system is conserved. That is, the total momentum of the system remains constant.
## Why is kinetic energy not conserved?
Energy and momentum are always conserved. Kinetic energy is not conserved in an inelastic collision, but that is because it is converted to another form of energy (heat, etc.). The sum of all types of energy (including kinetic) is the same before and after the collision.
## What is the change in total mechanical energy after the collision?
In an elastic collision, mechanical energy is conserved – the sum of the mechanical energies of the colliding objects is the same before and after the collision. After an inelastic collision, however, the mechanical energy of the system will have changed.
## What happens when two objects collide?
Newton’s third law of motion is naturally applied to collisions between two objects. In a collision between two objects, both objects experience forces that are equal in magnitude and opposite in direction. Such forces often cause one object to speed up (gain momentum) and the other object to slow down (lose momentum).
## What happens to the kinetic energy before and after the collision?
Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision.
## Is kinetic energy conserved in an explosion?
Explosions occur when energy is transformed from one kind e.g. chemical potential energy to another e.g. heat energy or kinetic energy extremely quickly. So, like in inelastic collisions, total kinetic energy is not conserved in explosions.
## Why is kinetic energy lost in a collision?
In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.
## Can kinetic energy increase after collision?
Collisions are considered inelastic when kinetic energy is not conserved, but this could be from either a loss or gain or kinetic energy. For example, in an explosion-type collision, the kinetic energy increases.
## What are the 3 types of collision?
There are three different kinds of collisions, however, elastic, inelastic, and completely inelastic. Just to restate, momentum is conserved in all three kinds of collisions. What distinguishes the collisions is what happens to the kinetic energy.
## How do you calculate change in kinetic energy?
Key TakeawaysThe work W done by the net force on a particle equals the change in the particle’s kinetic energy KE: W=ΔKE=12mv2f−12mv2i W = Δ KE = 1 2 mv f 2 − 1 2 mv i 2 .The work-energy theorem can be derived from Newton’s second law.Work transfers energy from one place to another or one form to another. | 1,284 | 5,932 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-17 | latest | en | 0.91666 |
https://math.libretexts.org/Bookshelves/Calculus/Calculus_(Guichard)/01%3A_Analytic_Geometry/1.05%3A_Shifts_and_Dilations | 1,723,646,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00645.warc.gz | 314,150,839 | 31,690 | # 1.5: Shifts and Dilations
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Many functions in applications are built up from simple functions by inserting constants in various places. It is important to understand the effect such constants have on the appearance of the graph.
## Horizontal shifts
If we replace $$x$$ by $$x-C$$ everywhere it occurs in the formula for $$f(x)$$, then the graph shifts over $$C$$ to the right. (If $$C$$ is negative, then this means that the graph shifts over $$|C|$$ to the left.) For example, the graph of $$y=(x-2)^2$$ is the $$x^2$$-parabola shifted over to have its vertex at the point 2 on the $$x$$-axis. The graph of $$y=(x+1)^2$$ is the same parabola shifted over to the left so as to have its vertex at $$-1$$ on the $$x$$-axis. Note well: when replacing $$x$$ by $$x-C$$ we must pay attention to meaning, not merely appearance. Starting with $$y=x^2$$ and literally replacing $$x$$ by $$x-2$$ gives $$y=x-2^2$$. This is $$y=x-4$$, a line with slope 1, not a shifted parabola.
## Vertical shifts
If we replace $$y$$ by $$y-D$$, then the graph moves up $$D$$ units. (If $$D$$ is negative, then this means that the graph moves down $$|D|$$ units.) If the formula is written in the form $$y=f(x)$$ and if $$y$$ is replaced by $$y-D$$ to get $$y-D=f(x)$$, we can equivalently move D to the other side of the equation and write $$y=f(x)+D$$. Thus, this principle can be stated: to get the graph of $$y=f(x)+D$$, take the graph of $$y=f(x)$$ and move it D units up.For example, the function $$y=x^2-4x=(x-2)^2-4$$ can be obtained from $$y=(x-2)^2$$ (see the last paragraph) by moving the graph 4 units down. The result is the $$x^2$$-parabola shifted 2 units to the right and 4 units down so as to have its vertex at the point $$(2,-4)$$.
Warning. Do not confuse $$f(x)+D$$ and $$f(x+D)$$. For example, if $$f(x)$$ is the function $$x^2$$, then $$f(x)+2$$ is the function $$x^2+2$$, while $$f(x+2)$$ is the function $$(x+2)^2=x^2+4x+4$$.
An important example of the above two principles starts with the circle $$x^2+y^2=r^2$$. This is the circle of radius $$r$$ centered at the origin. (As we saw, this is not a single function $$y=f(x)$$, but rather two functions $$y=\pm\sqrt{r^2-x^2}$$ put together; in any case, the two shifting principles apply to equations like this one that are not in the form $$y=f(x)$$.) If we replace $$x$$ by $$x-C$$ and replace $$y$$ by $$y-D$$---getting the equation $$(x-C)^2+(y-D)^2=r^2$$---the effect on the circle is to move it $$C$$ to the right and $$D$$ up, thereby obtaining the circle of radius $$r$$ centered at the point $$(C,D)$$. This tells us how to write the equation of any circle, not necessarily centered at the origin.
We will later want to use two more principles concerning the effects of constants on the appearance of the graph of a function.
## Horizontal dilation
If $$x$$ is replaced by $$x/A$$ in a formula and $$A>1$$, then the effect on the graph is to expand it by a factor of $$A$$ in the $$x$$-direction (away from the $$y$$-axis). If $$A$$ is between 0 and 1 then the effect on the graph is to contract by a factor of $$1/A$$ (towards the $$y$$-axis). We use the word "dilate'' to mean expand or contract.
For example, replacing $$x$$ by $$x/0.5=x/(1/2)=2x$$ has the effect of contracting toward the $$y$$-axis by a factor of 2. If $$A$$ is negative, we dilate by a factor of $$|A|$$ and then flip about the $$y$$-axis. Thus, replacing $$x$$ by $$-x$$ has the effect of taking the mirror image of the graph with respect to the $$y$$-axis. For example, the function $$y=\sqrt{-x}$$, which has domain $$\{x\in R\mid x\le 0\}$$, is obtained by taking the graph of $$\sqrt{x}$$ and flipping it around the $$y$$-axis into the second quadrant.
## Vertical dilation
If $$y$$ is replaced by $$y/B$$ in a formula and $$B>0$$, then the effect on the graph is to dilate it by a factor of $$B$$ in the vertical direction. As before, this is an expansion or contraction depending on whether $$B$$ is larger or smaller than one. Note that if we have a function $$y=f(x)$$, replacing $$y$$ by $$y/B$$ is equivalent to multiplying the function on the right by $$B$$: $$y=Bf(x)$$. The effect on the graph is to expand the picture away from the $$x$$-axis by a factor of $$B$$ if $$B>1$$, to contract it toward the $$x$$-axis by a factor of $$1/B$$ if $$0 < B < 1$$, and to dilate by $$|B|$$ and then flip about the $$x$$-axis if $$B$$ is negative.
$\left({x\over a}\right)^2+\left({y\over b}\right)^2=1 \qquad\hbox{or}\qquad {x^2\over a^2}+{y^2\over b^2}=1.$
Finally, if we want to analyze a function that involves both shifts and dilations, it is usually simplest to work with the dilations first, and then the shifts. For instance, if we want to dilate a function by a factor of $$A$$ in the $$x$$-direction and then shift $$C$$ to the right, we do this by replacing $$x$$ first by $$x/A$$ and then by $$(x-C)$$ in the formula. As an example, suppose that, after dilating our unit circle by $$a$$ in the $$x$$-direction and by $$b$$ in the $$y$$-direction to get the ellipse in the last paragraph, we then wanted to shift it a distance $$h$$ to the right and a distance $$k$$ upward, so as to be centered at the point $$(h,k)$$. The new ellipse would have equation $$\left({x-h\over a}\right)^2+\left({y-k\over b}\right)^2=1.$$ Note well that this is different than first doing shifts by $$h$$ and $$k$$ and then dilations by $$a$$ and $$b$$:
$\left({x\over a}-h\right)^2+\left({y\over b}-k\right)^2=1.$
See figure 1.4.1.
Figure 1.4.1. Ellipses: $$\left({x-1\over 2}\right)^2+\left({y-1\over 3}\right)^2=1$$ on the left, $$\left({x\over 2}-1\right)^2+\left({y\over 3}-1\right)^2=1$$ on the right.
## Contributors
• Integrated by Justin Marshall.
This page titled 1.5: Shifts and Dilations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Guichard via source content that was edited to the style and standards of the LibreTexts platform. | 3,550 | 10,291 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.199697 |
https://www.circuitdiagram.co/how-do-you-find-the-voltage-drop-across-a-resistor-in-series-circuit/ | 1,708,479,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00830.warc.gz | 766,796,233 | 13,542 | # How Do You Find The Voltage Drop Across A Resistor In Series Circuit
By | October 15, 2021
Electricity is a powerful and often misinterpreted force. Knowing how to measure the voltage drop across a resistor in series circuits is an important component of understanding electricity's properties and applications.
When tracing current from the power source through a series circuit, it's easy to understand the concept of a voltage drop. A voltage drop occurs as the electrons pass through each connected component, such as resistors or capacitors. Each connected component in a series circuit will create an opposing force that effectively reduces the voltage. This creates a noticeable reduction in the total voltage when compared to the initial voltage supplied from the power source.
The main way to calculate the voltage drop across a resistor in a series circuit is to use Ohm's law. This law states that the current (I) passing through a conductor between two points is directly proportional to the voltage (V) applied between these two points, provided the temperature and the resistance (R) remain constant. By using this formula, one can accurately measure the voltage drop across a resistor in a series circuit.
Additionally, one can measure the voltage drop by using a multimeter. A multimeter is a device which allows the user to measure AC or DC voltages, as well as current levels. The user can then use the resistance setting of the multimeter to measure the voltage drop across certain components. This is an easy way to test the voltage drop across a resistor in a series circuit without having to use Ohm's law equations.
In conclusion, measuring the voltage drop across a resistor in a series circuit is an important step in understanding the principles of electricity and its applications. While Ohm's law can accurately determine this value, a multimeter is also a useful tool for quickly and easily testing this value. Knowing how to measure the voltage drop across components is a necessary skill for anyone who wishes to understand electricity and electrical circuits.
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Voltage Divider Circuits And Kirchhoff S Laws Electronics Textbook | 639 | 3,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-10 | latest | en | 0.940655 |
https://www.eitdigital.eu/news-events/blog/article/allometry-does-it-apply-to-smart-cities | 1,561,573,411,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000414.26/warc/CC-MAIN-20190626174622-20190626200622-00256.warc.gz | 731,830,420 | 10,208 | # Allometry: does it apply to Smart Cities?
Sunday, February 23 2014
Kleiber's law: the metabolic rate grows in a 3/4 proportion to the volume growth of the living being. It applies over a staggering 27 orders of magnitude, from bacteria to blue whales. Credit: Constable Research b.v.
Allometry is the the study of the change in proportion of various parts of an organism as a consequence of growth. It was first applied to living organisms to study the relation of volume to shape to behaviour.
A first very interesting result was obtained by Kleiber who found out in the 1930ies that the metabolic rate (in other words the use of energy) of living things was proportional to the volume squared to 3/4. Hence, the metabolic rate growth in a sub linear way, the bigger you get the more efficient you are in the use of energy: a bacteria uses much less energy than a mouse who uses much less energy than a whale but the mouse uses energy more efficiently than a bacteria and a whale uses energy more efficiently than a mouse. How is that?
You know that mathematicians are always interested in why some numbers pop up, and so they tried to understand why this 3/4 linearity (or better sub linearity, since you have linearity when you have 1/1).
A quick observation was that the metabolic rate equates to heat, and a body produces heat based on its volume whilst it dissipates heat (if you don't want to boil the dissipation should be equal to the production...) based on its surface. Now, that would lead to a 2/3 sub linearity (do your math and by approximating -physicists love approximation... and mathematicians love modelling...- a body to a sphere you see that the ratio growth of body surface to volume equates to 2/3, since surface grows by the square of the radius whilst volume grows by the cube of the radius).
2/3 is closer to 3/4 than 1/1 but it is not the same. So what could be the reason of this amazing constant ratio over a 27 order of magnitude growth (that is the span between the volume of a bacteria and a volume of a blue whale)?
Mathematicians looked at the way heat is diffused and found out that you can model it by looking at the circulation of molecules, that in most animals depends on the circulation of blood, that in turns depends on the vessels (veins and arteries). Now, independently of the species the vessel infrastructure supporting blood circulation repeats over and over in the various parts of the body: it has a fractal structure. And, behold!, in a fractal structure the ratio as you increase the magnification, is 3/4. Voilà.
Now, interestingly, cities can be modelled as living organisms, and they have a sort of vessels infrastructures to disseminate "heat" which in the case of cities is economic wealth. These infrastructures, that used to be roads, and then were complemented by water pipes, sewage pipes and more recently by telecommunications wires have a fractal structure, just like the veins in a human body. And indeed, statistics show that the ratio between the "effort" (the metabolic rate) to keep a city going and its production of wealth is in the order of 3/4!
In other words, this means that the large a city is, the more effective it is, provided, of course, that it has the feeding infrastructures it takes to keep the city working as a single organisms.
If these infrastructures break down, they are not pervasive, than also the city splits into several organisms, and becomes less and less efficient as these individual organisms become smaller and smaller!
Now, this is a very important observation, since it clearly points out the crucial role of infrastructures at city level in terms of city efficiency and savvy use of energy, which translate into its capacity of generating "wealth".
However, if statistics show that pervasive infrastructures result in a Kleiber curve relating cities of different sizes, it also shows that different quality of infrastructures (e.g having or not having a broadband infrastructure) result in different efficiencies levels. Again, the rule of the 3/4 applies: cities having simile infrastructure "quality" yields a wealth that is super linear to their size (it increase faster than the size of the city,, actually in the ration 4hat /3 which is equivalent to say tthe effort dedicated to "run" the city scales sub linearly in the ratio 3/4). In other terms, the better the infrastructure (the more efficient it is) the more wealth can be produced but the size factor remains with a scale ration of 3/4!
Now, I stumbled onto an interesting research by Hygor Piaget Melo and his colleagues at the Federal University of Ceará in Brasil who studied several thousands of cities in Brasil and in the USA from the point of view of quality of life, as it could be measured by parameters like the number of homicide (security), number of car accidents (safety) and suicides (happiness).
According to their study the homicide rates is super linear, that is bigger cities have a number of homicides that is larger than the scaling factor, whilst the number of traffic incident is isometrical to the scaling factor (a city that has double the size will have double number of traffic accident) but the numb rod suicide is sub linear, that is, the bigger the city the less the percentage of suicides. It seems that a larger city can provide more reasons to live than a smaller one, hence offering alternatives to those in grave depression state.
You can read the article to get more insights.
What I would like to do is to come back to the crucial role of infrastructures. The 3/4 ratio is basically a consequence of the fractal nature of the infrastructures. However, if we move from wired infrastructures to wireless one, and if we move from a communications infrastructure to a communications fabric where data take the upper hand we no longer have a fractal structure (it is less hierarchical, more flat and distributed).
A city working as a centrally regulated organisms, is different from a city that is self aware. And as we move towards the internet of things through a bottom up awareness creation, as opposed to a top down regulation, we are basically changing the working model of a city.
It would be interesting to measure the impact of the scaling factor. My bet is that we will probably find that size will no longer play a significant roll in the overall efficiency (within ranges, of course) and that a small city might be as "efficient" as a big one!
This is something that we should investigate when looking at making cities "smarter". We need to understand what technology to deploy, and how to deploy it, so that the overall efficiency is no longer depending on the city size and that should allow us to evolve cities that are both smarter, and safer, happier for its dwellers.
This is the challenge IEEE is willing to take in its Urbanization Challenge program, and the one the EIT ICT LABS needs to address in its Urban Life and Mobility Action Line.
You may also want to read a little summary of an interview I had on the theme of Smart, Happier City and watch the related video clip.
Roberto Saracco | 1,530 | 7,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-26 | latest | en | 0.960071 |
https://zxi.mytechroad.com/blog/searching/leetcode-1926-nearest-exit-from-entrance-in-maze/ | 1,716,553,537,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00092.warc.gz | 946,861,161 | 20,979 | You are given an m x n matrix maze (0-indexed) with empty cells (represented as '.') and walls (represented as '+'). You are also given the entrance of the maze, where entrance = [entrancerow, entrancecol] denotes the row and column of the cell you are initially standing at.
In one step, you can move one cell updownleft, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance. An exit is defined as an empty cell that is at the border of the maze. The entrance does not count as an exit.
Return the number of steps in the shortest path from the entrance to the nearest exit, or -1 if no such path exists.
Example 1:
Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2]
Output: 1
Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3].
Initially, you are at the entrance cell [1,2].
- You can reach [1,0] by moving 2 steps left.
- You can reach [0,2] by moving 1 step up.
It is impossible to reach [2,3] from the entrance.
Thus, the nearest exit is [0,2], which is 1 step away.
Example 2:
Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0]
Output: 2
Explanation: There is 1 exit in this maze at [1,2].
[1,0] does not count as an exit since it is the entrance cell.
Initially, you are at the entrance cell [1,0].
- You can reach [1,2] by moving 2 steps right.
Thus, the nearest exit is [1,2], which is 2 steps away.
Example 3:
Input: maze = [[".","+"]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.
Constraints:
• maze.length == m
• maze[i].length == n
• 1 <= m, n <= 100
• maze[i][j] is either '.' or '+'.
• entrance.length == 2
• 0 <= entrancerow < m
• 0 <= entrancecol < n
• entrance will always be an empty cell.
## Solution: BFS
Use BFS to find the shortest path. We can re-use the board for visited array.
Time complexity: O(m*n)
Space complexity: O(1)
## C++
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huahualeetcode | 645 | 2,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-22 | latest | en | 0.865453 |
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# Motion in Straight Line
#### Multiple Choice Questions
101.
A man is at a distance of 6 m from a bus. The bus begins to move with a constant acceleration of 3ms.In order to catch the bus, the minimum speed with which the man should run towards the bus is
• 2 m s-1
• 4 m s-1
• 6 m s-1
• 8 m s-1
C.
6 m s-1
If the man did not run, the bus would be at a distance s1 at time t given by
s1 = 6 + $\frac{1}{2}\mathrm{\alpha }$ t2
= 6 + $\frac{1}{2}$ × 3 × t2
s1 = 6 + $\frac{3}{2}$t2
If v is the speed of man, he would cover a distance s2 = vt in time t.
To catch the bus,
s1 = s2
6 + $\frac{3}{2}$ t2 = v t
⇒ t2 $-$ t + 4 = 0
which gives
t =
Now t will be real if is positive or zero.
Minimum v corresponds to = 0 which gives v = 6 m s-1
102.
Two identical charged spheres suspended froma common point by two massless strings of lengths l, are initially at a distance d (d<< l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velcoity v. Then, v varies as a function of the distance x between the sphere, as
• vx
• vx-1
• vx-1
B.
Two identical charged spheres suspended from a common point by two massless strings of length L.
In ,
The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.
Therefore, equation (i) can be rewritten as,
7034 Views
103.
A body is at rest at x = 0 and t = 0, it starts moving in the positive X-direction with a constant acceleration. At the same instant, another body passes through x = O moving in the positive X-direction with a constant speed. The position of the first body is given by x1 (t) after time t and that of second body by x2 (t) after the same time interval. Which of the following graphs correctly describes (x− x2) as a function of time t ?
C.
104.
The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is
• 0
• 5 m/s2
• -4 m/s2
• -4 m/s2
C.
-4 m/s2
x = 5t – 2t2 y = 10t
Acceleration of particle at t = 2 s is = –4 m/s2
1927 Views
105.
The displacement x of a particle varies with time t as x = ae-αt + beβt where a, b, $\mathrm{\alpha }$ and β are positive constants. The velocity of the particle will
• decrease with time
• be independent of
• drop to zero when
• increase with time
D.
increase with time
Given:-
x = a e-αt + beβt
We know that
velocity v = $\frac{\mathrm{dx}}{\mathrm{dt}}$
v =
= -a$\mathrm{\alpha }$ e-αt + bβ eβt
v =
v = A + B
where A = - a , B = bβ e-βt
The value ot term A = -a increases and of term B = bβ e-βt increases with time. As a result velocity goes on increasing with time.
106.
A monkey of mass 30 kg climbs on a massless rope whose breaking strength is 450 N. The rope will break if the monkey (Take g = 10 m/s2 )
• climbs up with a uniform speed of 5 m/s
• climbs down with an acceleration 4 m/s2
• climbs up with an acceleration 6 m/s2
• climbs down with a uniform speed of 5 m/s2
C.
climbs up with an acceleration 6 m/s2
To move up with an acceleration 'a', the monkey will push the rope downward with a force
F = ma = 30 a
Now, Tmax = mg + ma = 30 × 10 + 30 a
⇒ 450 = 300 + 30 a
⇒ a = 5 m/s2
Hence, the rope will break if the monkey climbs up with an acceleration more than 5 m/s2.
107.
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,
• 90o
• 45o
• 180o
• 180o
A.
90o
There are two vectors P and Q.
It is given that,
Let, angle between P and Q is ,
2398 Views
108.
A body is whirled in a horizontal circle of radius 25 cm. It has an angular velocity of 13 rad/s. What is its linear velocity at any point on circular path?
• 2 m/s
• 3 m/s
• 3.25 m/s
• 4.25 m/s
C.
3.25 m/s
Given:- r = 25 cm
r = 0.25 m
Linear speed = Radius × angular speed
V = rω
= 0.25 × 13
3.25 m/s
109.
If A + B = C and that C is perpendicular to A. What is the angle between A and B, if
• $\frac{\mathrm{\pi }}{4}$ rad
• $\frac{\mathrm{\pi }}{2}$ rad
• $\mathrm{\pi }$ rad
C.
Given:-
A + B = C
Since, B = C $-$ A
Also
B2 = 2 A2
⇒ B = $\sqrt{2}$ A
Now
A2 + B2 + 2A B cosθ = C2 = A2
∴
This gives
110.
A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of 40 m/s from the bottom of the building. The two balls will meet after
• 3 s
• 2 s
• 2.5 s
• 5 s
C.
2.5 s
Let balls meet after t s. The distance travelled by the ball coming down is
s1$\frac{1}{2}$ gt2
Distance travelled by the other ball
s2 = 40 t $-$ $\frac{1}{2}$ gt2
s1 + s2 = 100 m
∴ $\frac{1}{2}$ gt2 + 40 t $-$ $\frac{1}{2}$gt2 = 100m
t = $\frac{100}{40}$
∴ t = 2.5 s | 1,743 | 5,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2022-33 | longest | en | 0.870048 |
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A number consists of two digits. The sum of the digits is 10. On reversing the digits of the number, the number decreases by 36. What is the product of the two digits? (a) (b) (c) (d)
21
5.
Whichone amongthe following statemerits is correct? (a) Simple bar" diagrams are those diagrams which show two characteristics of the data In pie diagrams all the items are converted into angles A bar diagram is one in which data are shown in terms of bars Bar diagrams present data through length and breadth
24 36 42
(b) (c) (d)
2. :~.
Which one among the following" is not' correct? (a) (b) (c) (d) Two congruent .:;=:t~langles necessarily similar are 6.
The following characteristics relate to the" persons participating in athletic events: 1. 2. 3. Height of the person Colour of the eye of the person Number of times a person correctly hits a target in ten attempts
Air equiangular triangles are similar Two isosceles right triangles are similar All isosceles triangles are similar
3.
ABCD is a rhombus with diagonalsAC and BD. Then which one among the following is correct? (a) (b) AC and BD bisect each other but not necessarily perpendicular to each other AC andBD are perpendicular to each other but not necessarily bisect each other (c) (d) perpendicular to each other AC andBD neither bisect each other norperpendicular to each other
Which of the following in respect of the above is correct? (a) (b) (c) (d)i,
1 is a continuous variable, 2 is not a
variable and 3 is a discrete variable 1 is a continuous variable, 2 and 3 are discrete variables
1 and 3 are discrete variables and 2 is
not a variable
i and 3 all are discrete
.
variables :
A~ and BD bisect each other a1id _.!~~'7. Consider the following.statements
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1.
The data collected by the investigator to be used by himself are called primary data The data obtained from government agencies are called secondary data
2.
4.
The sum of a number and its reciprocal is 10/3 then the numbers are: (a) (b) (c)
3, 113
3, -113
Which of the above statementsis/are correct? (a) 1 only
-3, 1/3
(b) 2 only
(c) .Both 1 and 2 (d) Neither 1 nor 2
(d) -3, -1/3· 2
o
(Contd.)
1.
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(a) 21 (b) 24 (e) 36
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3, -113
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3
o
___ '..._._.~_ L_
8.
Which one among the following statements is not correct ? (a) (b) For size of readymade garments, mode is the best suitable measure For average, rate of increase when the rate of population (c) For average different growth is given, when geometric mean is best suitable rate of speed distances are covered by
11. Consider the following statements: 1. 2. Every composite number is a natural number Every whole number number is a natural
Which of the statements given above is/are correct? (a) (e) (d) 1 only Both 1 and 2 Neither 1 nor 2 : (b) 2 only
different rates of speed, harmonic mean is best suitable ~(d)
-:-J,........,.
For average level of intelligence the best suitable
of
12. Consider the'followingstatements If two straight linesintersect, then:
students in a class, arithmetic mean is
1.
2.
vertically oppositeangles opposite vertically supplementary,
are equal. angles are
9. 1£ the surface area of a sphere is 616 square
em, what is its volume? (a) 3.
4312/3 cubic cm
1257 cubic em 1023 cubic em
(b) 4102/3 cubic em (c) (d)
Which of the statements given above is/are correct? (a) (b) (c) 3 only l only land 3
(d) 2and3 10. What are the dimensions (length, breadth and height respectively) of a cuboid with volume 720 cubic em, surface area 484 square cm and the areaofthebase em? (a) (b) (c) (d) 9,8 and 10 em 12,6 and 10 em 18, 4 and IDem 30,2 and 12 em 72 square
13. A large solid metallic cylinder.whose radius
and height are equal to each other i§ito be melted and 48 identical solid balls are to be recast from the liquid metal so formed, What is the ratio of the radius of a baIl to the radius of the cylinder? (a) (b) (e) (d) 1 : 16 1 : 12 1:8 1:4
o
4
(Contd.)
8.
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1: 16 1: 12 1: 8 1: 4 (Contd.)
altt
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5
o
14. The geometric meanof vitiates 32, 4, 8, X, 2 is 8. What is the value of vitiate X ? (a) 2
(b) 4
(c) (d) 8 16
18. The angle of elevation of the tip ofa tower from a point on the ground is 45°. Moving 21 m directly towards the base of the tower, the angle of elevation changes to 60°. What is the height of the tower, to the nearest meter? (a) (e) (d)
48 m
(b> 49 m 15. In the triangle ABC? the base BC is trisected at D and E. The line through D, parallel to AB, meets AC at F and the line through E parallel to AC meets AB at G. Let EG and DF intersect at H. What is the ratio of the sum of the area ofparalleiogramAOHF the triangle ABC? (a) (b) (e) (d) and the area of the triangle DHE to the area of 19. A circular ring with centre 0 is kept in the vertical position by two weightless thin strings TP and TQ attached to the ring at P and Q. The line OT meets the ring at E whereas a tangential string at E meets TP and TQ atAand B respectively. If the radius of the ring is 5 em and OT = 13 em, then what is the length of AB ? (a) (b) (c) 10/3 em 20/3 em 10 em 50 m 51 m
112
1/3
114
116
16. If x _ 2 is a factor of x3 + kx2 - 2x _ 24 then what is the value of k ? (a) (b) (e) 6 5
1
~
(d) 40/3 em 20. What is one of the square roots of 16x6 _ 24x' + 25x4 - 20x3 + lOx2 - 4x + 1 ? (a) (c) (d) 4x3
_ -
3x2 + 2x + 1 3x2
_
(b) 4x3
2x - 1
~
(d) -2 17. AB is a straight line. C is a point whose distance from AB is 3 CPl. What is the number of points which are (it a distance of 1 em from AB and 5 em from C ? (a) (b) (e)
1 2
4x3 -
3x2 + 2x - 1
-
4x3 '_ 3x2
Zx+ 1
21. The cost of two articles are in the ratio 3 : 5. If there is 30% loss on the first article. and 20% gain on the second article, what is overall percentage of loss or gain ? (a) (b) (c) gain 2·25% loss 5·25% loss 2%
3
(d) 4 6
(d) None of the above (Contd.)
°
14. fcrq-{t 32, 4, 8, X, 2 CfiT ~OTJtH 11M 8 ~., ~ (a) (b) (e) X CfiT~ 2 4 8 16 ~ ABC
CfmW?
18. \iJlfpf :tR M ~'-« ~ 1fr;rr{ UR iN \3 r'1lpi ~ 45° ~ I lfAR cF 3T1llR 3TR ~ ~ 21 tfi. ~"tf{ '3r'1~'1 ~ iSfGR Cfl{ 60°
*",~
murmr~1 ~lfin~~~~w ~?
(a) (b) 48m 49m 50m 51 m
(d)
15. ~
if 3TTmr
BC
ctrr DomE
'tf{ ~
1:J\
~
(e)
~~UffilT~I~Dit~~ om AB ~ E
*" ~"II"""d{=tm AC ctn F
~
am AC
(d)
19.
it ~
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o fl ~ ~ ~t1ICflI{ CffitI"' eFt, aTR Q -qt ~ tIT ~ ~ i'rftm
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13 (a) (b) (e)
mit
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A 0'l1T B -qi:
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fllffifi ~ i ~
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10/3 em 20/3 em IOem 40/3 em
Wft afR OT =
'fflT ~ ?
(d)
16. ~ (a) (b) (c)
(d)
?
x3 + 1cX2 - 2x - 24 CfiT~
CfiT llR CflIT ~
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6
5 1
20. '16x6-24~s+25x4-20xj+
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qJfli,Ji ~ ~
(a) (b)
'0 '
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4x3 4x3 4x3
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-
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(a) (b) (e)
-err m 30%
2·250/0 ffi'l 5'25% 2%~
3:5
31-fJRfif~,
~
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tfm ~
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?
(a) (b) (c) (d)
mf.1
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(Contd.)
(d) . '3qg'ftt ~
7 °
-_.----22. The number "58129745812974 is divisible
by
(a) (b) (e) (d)
11
9
27. If a positive integer leaves remainder 28 when divided by 143, then what is' the remainder obtained on dividing the same number by 13 ? (a) 0 (b) (c) (d) 2
4 None of the above
-
9
10
23. What is the HCF of 36(3x4 + 5x3 9(6x3 + 4x2 - 2x), 54(27x4 - x) ? (a) (b) (e) (d) 24. 9x(x + 1) 9x(3x -.1) 18x(3x -]) 18x(x+l)
, l
2X2),
28. What is the compound interest on Rs. 1600 at 25% per annum for 2 years compounded annually? (a) Rs.700 (b) (c) (d) Rs.750 Rs.,800 Rs.900
What is the length of the uniform wire of diameter 0-4 em that can be drawn from a solid sphere of radius 9 em ? (a) (e) (d) 243 m 60·75 m 60 m (b) 240 m
29. The graphs of ax + by = c, dx + ey = fwill be 1. parallel if the system has no solution 2. 3. coincident if the system has finite number of solutions intersecting if the system has only one solution statements are
25. What will be the cost to plaster the inner surface of a well 14 m deep and 4 Omin diameter at the rate of Rs. 25 per square meter?
Which of the above correct? (a) (b) (c) (d)
1 and 2 only
2 and 3 only 1 and 3 only 1,2 and 3
Rs.4000 (b) Rs.4200 (c) Rs.4400
(a) (d) Rs.5400
t:o. "
'-io
.l
30. The medians ofa triangleABGintersect at G Which one of the following is correct? '(a) Five times the area of triangle AGB is equal to four times the area of triangle ABC. Four times the area of triangle AGB is equal to three times the area of triangle
26. A wheel of radius 2·1 m of ave hie le makes 75 revolutions in one minute. What is the speed ofthe vehicle? (a) (b) (e) (d) 78 kmph 59·4 krnph 37·41qnph 35·4 kmph 8
(b)
ABC.
(c) (d) Three times the area of triangle AGB is equal to the area of triangle ABC None of the above (Contd.)
o
22. ~
58129745812974
~
~
t,?
(a) 11 (b) 9
(c) (d) 4
\3 4~
27. "llft ~ 28 ~
CF-lT
Wcsr ~
0 2
9
~.<Fr 143 ~ m1f t. ill \nfi m.«rr qi'r 13 ~
Wf
~
tT\
m
tT\
'q11T ~
?
(a) if ~ ct?r{
';ffft (b) (e) (d)
23. 36(3x4 + 5x3 - 2X2), 9(6x3 + 4x2 - 2x), 54(27x4 - x) CFT ~ eSOJI4cH:f91 'Efm' ~ ?
10
(a)
(b) (c) (d) 24. 9 ~ (a)
9x(x + 1) 9x(3x - 1) 18x(3x - 1) 18x(x + I)
Gt "« ~ 28. Rs. 1600 CFT 25% ~ tT\ 2 qtSf "* Wf iI#l~r.a ~
m
(a) (b) (e)
q:m·t
64iRid
?
Rs. 700
Rs. 750
Rs.800
~
243m 240m
qrff
Wft ozm:r CJTff (bY
(e) (d) 25.
i'm lim em
*~ 0·4 GfOl
(d) Rs.900
29. ax + by 1'. 2. ,3.
l(CfleSOJI'9
"ffi1: ~.~
cfirr t ?
= e, dx + ey = f
WUo=t1 (
tftmft
mrr ~ ~
iWt ~ ~
*
3I1ffi.q
CflT
~ .~ ~ ~
CflT ~
ttmf
m'f ';ffft
t
60'75 in 60m
4R~d ~ S1Rl4M~d m ~ f.l'fir:r
~~
~
14 1ft 11it 3TR 4 1ft ozmr ~ ~ ~ ~ ~ Rs. 25 ~ qlf lfrn
l(1lfc { ~
CFT ~
qm
mm
* ~ ~ "R
citt
-:;:I
d4\$'fd ~ (a) ~
~~
la1tt2
~ tuft ~ ?
?
(b) ~23TR3 '(e) ~1am:3
(a)
(b)
Rs.4000 Rs.4200 Rs.5400
CFT 2·1
(d)
30. ~
1,2aTR 3 ~ ~
(,:) Rs.4400
(d)
26.
cp«fr ~ I
M~
1ft.~
Cffi1T ~
~
TJ:Cfi
T.JWf
(a) (b)
PI *"'9 ffiftcm' if ~ m ~ AGB 'tfi ~ qif ~ TIT. ~
ABC ~
s:ti"DNiITi G tT\
S1Rl~G tl'6r t ?
~,
~t?
(a) (b) (e) (d)
if
75. ~
<fiTCffi" ~ I ~
.TTTtT ~
ABC ~
~~cnct
78 kmph 59'4 kmph 37·4 kmph 35·4 kmph (e)
~AGB ABC ~,ift~lf)('1
* * -aR ~
'GR
~~Cf1(1
*
it
~
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if; ~
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t"
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~
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*.
t
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(d)
\3q~
if "R ~
1ift
(Contd.)
9
o
31. What is- the mean proportional (15 + .J200)and (a) (b) (c) (d) 32. 4 14J7 (27 -
between
Ms)
?
35. If-the outer and .inner diameters of a stone . parapet around a well are 112 cm and 70 em respectively, then what is the area of the parapet? (a) (b) (c) (d) 264 square ern 3003 square em 6006 square em 24024 square em
3.J5
5../3
A stock offood is enough for 240 men for 48 days. How long' will the same stock last for 160 men? (a) (b) (c) (d) 54 days 60 days 64 days
36. What is 0·007 (a) (b) (c)
+ 17·83+ 310'0202
equalto ?
327'86638 327.86638 327'86683 327·8668
72 days
and~-'_9 __ ~'=1
x+y ,x-y x-y
33. If_3_+_2'_=2
x+y
Cd)
x 'then what is -equal
y
to?
(a) 3/2 (b) 5 (c) 213 (d) 1/5 34. The locus of the mid points of all equal chords in a circle is : (a) the circumference, of the circle concentric with the given circle and having radius equal to the length of the chords (b) the Circumference of the circle concentric: with the given circle and having radius equal to the distance of the chordsfrom the centre (c) the circumference of' the circle concentric with the given circle and having radius equal to half of the radius of the given circle (d) the circumference of the circle concentric with the given circle and having radius equal to half of the distance of the chords from the centre 10
o
37. A man loses 20% of his money. After spending 25% of the remainder, he has Rs. 480 left. What is the amount of money he originally had ? (a) Rs.600 (b) (c) (d) Rs.720 Rs. SOO Rs.840
3S. Assuming that profit of a shopkeeper in a particular commodity is a linear expression of transportation charge (t) and the'quantity of commodity (q). He earns a profit of Rs. 10,000 by selling 20 units at the transport chargeof Rs, 400. He also eams a profit of Rs. 12,000 by selling 25 units at the transport charge of Rs. 600. What is the linear expression in t and q ? (a) (b) (c) (d) 600q - 5t 500q.,-- 4t 600q -4t 500q - 5t (Contd.)
31.
(15 + .J200)
'flIT ~
~3ltt(27 -
.J648) ct1T l1t14I'j)qldl
3S.
~~"*~alR~~~~CflT~
'Q,'t 3liGftcp
(a) (b) (c) (d)
ClfR1~:
?
4
112"-?rft ~
70 -?rft
(a) (b) (c)
1TI,m~CflT'~<FlTW?
264 cpf-?rft 3003 q7f -?rft 6006
J..J5
14.fi
(d) SJj 32. 240 3iIGftllfi ~ 48"~~~~~ (a) (b) (c) (d) 33. ~
em -?rft
<f1f fr4T
~ ~?
ffItr ~
I
tff1l1fi
CflT ~
~
24024
* ~ ~'~( ~ Tffi1ff
54~
60~ 64 ~ 72 fM _3_+_2_' =2
cfflT~
?
160 3iIGftlifi
36~, 0-007 + 17·83 4- 310'0202 "~ (a) 327,86638
(b) 327·86638 (c) (d) 327.86683 327·8668
I f))
alh:
_9
4_=1
xty
x
x-y
x+y
x-y
37. ~
Q.
3M 3N'ft ~
~ 480
6fq
cpr 20% @"~
my~~~?
(a) (b) (c) (d) 34. ~ 3/2 S 2/3
:yn cpr2S%
1~1{1~1'~
Cfl0
I ~:
umrr ~ ?
* mer ~
~
~ I.~ 1ffi1
lffif ~
(a)
Rs.600
,(b) Rs.720
liS
~ ~
tMt ~'~
9'iT ~'-O\$ql!1
(a) ~
SC; ¥l (Oi'Cfi ... sft ~ f~tri:tt~'\JjlqI3l1 ~ ~
*
*
~
~
'(c) lM~r\$3IT
RS.800
(d) Rs.840 38.
I
(b) ~"~'"'
¥f
~
~
~~
* Wsft, ~
~'<Ft.~
ijcf;...sf) ~
*~~"
-qftfi:r ~,
-qftff.r
mr 'I1FRt ~'" f.l:i .~
lffi1
\$CflI1GI( ~
~I
If
ffi'i, ~
~
(t)
CflT ~ t;Cfi om 1ffi1 eFt :mvr
~
Cflt
it ~ cf;
~
-qftff.r~,
t
(q) CflT ~
Q.
20 ~
3fR
400 ~
(c) '~~
~'~'~
¥f
*
~m
Cfi'{ffi
~
~ ~~. Cfl«IT W, I 25 ~'~
Cfi""{
~
1O~OOOCflT 'ffM ~ 3IT<:~: '600
'fiT.ffi'Jf ~,~
'ern- q.
12,000
~
(d) ~
* ~~rsft ~~~'cFT'~,~~<Ft
~ ~
3ffif'rw
~~
~,'~
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Wit alIT
600q - 5t
SOOq - 4t 600q - 4t SOOq"- St
q ~~
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(a)
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'cft·-qftff.r
w
(b) (c) (d) 11
o
31TUl~
(Contd.)
39. Ifx: y=7: 5,'then what is (5x-2y) : (3x +2y) equal to?' (a) (c) (d) 5/4 25/31 31/42
43.
X4 + xyl
+ xly +
XZl
+ t' + yzl is divisible
by: (a) (x + y) only (b) (c) (x' + y3 + z') only both (x
(b) 6/5
+ y)
and (x' + y3 + Zl)
(d) None of the above
40. The slim of two numbers is 232 and their HCF is 29. What is the number of such pairs of numbers satisfying the above condition? (a) (b) (c) One Two Four 1P
44.
'A' can do a piece of work in 'x' days and 'B' can do the same work in 3x days. To finish the work together they take 12 days. What is 'x' equal to ? (a) (b) (c) 8 10 12
(d) None of the above
(d) 16
41.
r
A person bought 8 quintal of rice for certain rupees. After a week, he sold 3 quintal of rice at 10% profit, 3 quintal of rice with neither profit nor loss and 2' quintal at 5% loss. In this transaction, what is the profit? (a) (b) (c) 10% 20% 25%
45. ABeD is a rectangle. Let E be a point on
AB and F a point on CD such that DE is parallel to BF. If AE = 3 em and if the area
of triangle BFC = 6 square cm. Consider the following statements:
I.
Area ofrectangleABCD distinct primes.
can be of the
form pq2 square em where p, q are
(d) None of the above
2.
Area of the figure EBFD an integer.
, 'Ii
IS
of the form
r2 square em where r is rational but not
42. What is the angle of elevation of the sun when the shadow of a pole is ~ length of the pole? (a) (b) (c) (d) times the Which (a) (b) (c) of the above statements is/are
correct ? 1 only
2 only
300 450
600 None of the above
Both 1 and2
Cd) Neither 1 nor 2
12
o
(Contd.)
39. ~
x : y = 7 : ,5, ill {Sx - 2y} :Ox + 2y)
43. X4+xy3+x3y+XZ3+'/'tyz3
~
~
~~~?
(a) ,5/4 '(b) '6/5 (c) (d) 25/31 31142
~_ ?
(a) ~ (x+y)
y3
(b), -m(x3+ (e) (d) ~
+ Z3)
;nft
(x + y) om (x3 +
t
+_Z3)
1341Cffi·ifit ~
ifiTJf
.
40. ~ ~
~l41qqJCfl29
~
~ ~-~
23'i~
am ~.~
SdftlSj .. f.l"
~
~ ~
44. ~?
'A' ~
~ Cfl{
Cfil 'x' ~
."
if ~ ~
"
',!'
~,ijW
~ I \34gcre
'_B' \3"at ~,cFr
73'B"'
cpll{
3x.~
cp-ffl_~~ (a) ~
~~
if Cfl\ mFffi ~ 1 ~ Cfil 12 _~ if W ~'~
I
'x',~~~?
(a) (b) 8 10 12 16
(b) <IT (c) (d)
TIT{
\3qg<ru
if it ~ ;:ffiT' ~if
~
(c)
(d)
41.
"C!;Cfi
ozWffi ~ ~ Rcie:(1 ~
fcf<iC::(1{5%
ffi'1 ~
8 fctcic~ ~~
3 f4;:i co c;q ~
I'
'tJ;'fl ~.~,
10% 'ffi'l 1IT
lR, 3
~ ?
~
(OfN
m i:Jt
45. ABCD"C!fP 3lTlKf! I l1R ~ '.
'-:"
.
-,
fcp. AB . FW
=j
m:
~
~
E
((tIT
CD lR ~
~I ~
~.
1Jqif{: ~
3ln:2
if ~
(a) (b) (c) (d)
mf.l tIT ~
~
r~'~
f\$ DE~-BF ~ 6JtkH
~. 1. BFC
CfiT~
AE
em ~
I
= i5cpfem tIT
~ :
10% 20% 25~(
\341Cffi
PI H R>'if@a
~
~
f<tGrt ~
31T!Rf ABeD CfiT tft?llhM pq2 qIfein
CfiT
m WfiGT ~
<fl
"lit ~ p, q
~
if it -~. ~
2.
~~I
~
CfiT~,
EBFI? CfiT tftW1*1 r2
~~
<Pf em ~ ~
~ ~
1
42.
'TTI mm
(a) 30°
IJf<t ~
~
~
~
mtrT ~
~.qlIT
~
~
J3
CfiI '3"~1l~
.~?
urnT fcpr~ a41<fd ~ if ~-mfit
(a)
.~ ~A ?
~1
(b) 45°
(e) (d) 60°
\34~ ~
(b) ~2 (e)
it ffi ~
13
o
(d)
iR 2 <i'Al 1" "ill 1 3TR ~ tT
1
2
I
(Contd.)
46-. A cardboard sheet' in the form of a circular sector of radius 30 em and central angle 1440 is folded to make a cone. What is the radius of the cone? (a) (b) (c) 12 em
50. From the top of a cliff 200 m high, the angles of depression of die top and bottom of a tower are observed to be 300 and 45° respectively. What is the height of the tower? (a) (b) (c) (4) 400 m 400..)3 m 4001..)3 m None of the above
18 em
21 em
(d) None of the above 47. Two transversals Sand Tcuta set of'distinct parallel lines. S cuts the parallel lines in points A, B, C·andD and Tents the parallel lines in points E, F,G, "Hirespectively. If AB = 4, CD = 3 and EF = 12, then what is lengthofGH? (a) 4 (b) 6 (c) 8
51. If x3.';'-h; 5x2 + 10k leaves remainder -2x when divided by x2 + value ofk? (a) (c)
i, then
what is the
-2
1
(b) -1
Cd) 9
48. If the area of a circle is equal to the area of a square with side 2~ . ." + diameter of the circle? (a) (b) (c) (d) 49. If 1 unit
(d) 2
units, what is the
,52. If one .of the roots of quadratic equation 7x2 - 50x ofk? (a) (b) (c) (d)
7
+ k = 0 is 7, then
what is the value
:2 units
4 units 8 units
1 50/7
to.~
~
x+-=a,
x
1
7/50
then
what
is 53. If aX == b, by == c equalto
and xyz = 1, then what is cZ:
?
(a) (b) (c) (d)
a3 + a2
(a)
a
a' + a2 - 5a
a3 +a2-3a-2 a3 + a2-4a-2 14
o
(b) b (c) (d) ab
alb
(Contd.)
46. -30 em ~
~,~ *~ m
t- ?
12 em 18 em 21 em (a) (b) (e)
am
1440
~
~
cf; ~
q?r ~
Cflliil1lJ
ttt ~
50. 200m ~
'~~~~\ifTQTtl
\rtr~~~
m 'tf\~lfAR*~~m9"CflT
300
~
~CR
eFt ~,~
~ ~,
31qfflCfl'1 31q'1~'1
~
Cfm
'fIlvr ~; lfFm ctft ~
(a) (b) 400m
3ltt
<flIT ~
450
I~
?
400.Jj m 400/fj
\3q~
m
(d) \3q1<ft1 47. ~
(
if ~ m ~
(e)
am T
=
fllikl<
~'*
I ti'lid
13!!~~li
(~
cFrm ~
cF'r-S ~'
s
&R
'(d)
~~~~
~'~ D
"
A, B, C~
'tf\ Sfilm: ~
1R
&RioT ~"\$3IT E. F. G, H
AB = 4, CD = 3
~'q:lff
51. '~-rxJ + 5x2 + 10k cF'rx2 + 2 ~'
'qT1f ~
1f\
~ I~
ffl -2x
(a) -2 (b) -1 (e)
1
~, "ciT k Cf\T lWT W ~?
EF (a)
12 4
lIT,
"ciT GH ~
t- ?
(b) 6 (e) 8
(d) 2
(d) 9 48. ~ q-,f (a)
*~
?
1~
M'¥f cpr
~,?14j(1,
2.Jn, ~
~
!fIT qrfr
Cf\T Olffi1'
52. ~
~ l:!:Cfl 7 1 5017
134'ICfI(OI
7x2
-
SOx + k = 0
51 Lf1<1 cf; ~
WI "ciT ~
if
1Ff 7 t-,'m, k
'it1T li'R
qm! ?
*~
CJm" ~
(a) (b) (e)
(b) ,2 ~ (~) 4 ~ (d).8 ~ 49.
(d) 7/50
'lift x'+..!.= a,
x
'ij1;1
x3 +x2 +_!__3+_!_2 x x
53. ~ (a)
all=b, bY=eaTR
xyz= I,
{fif
eZ
~
~~~?
(a) a3 + a2 (b) (e) a3 + a2 - 5a a3 + a2
-
~!?
a
(b) b (e) ab
3a - 2
(d) aJ + a2 - 4a - 2
(d) alb
15
0
(Contd.)
".
II
..
,i'
••
I('
54. Let ABCD be a-parallelogram, Let m, n be positive integers such that n < m < 2n. Let AC = 2mn and BD = m2 - ti2. Let AB =~(m2 + n~)l2. Statement-I : AC'>'BD Statement-II: ABCD is a rhombus: Which one of- the fallowing Is COIT~ctin respect of the above statements? (a) Both statement-I and statement-Il are true and statement-Il is the correct explanation of statement-I
57. The area of a rectangle whose length is 5 more than twice its width is 75 square units. What is the .perimeter of the rectangle? (a) 40 units (b) 30 units (c) 24 units (d) 20 units
(b) Both statement-I and statement-II are truebutstaternent-Il is nO/the correct explanation of statement-I (c) (d) Statement-I is true,but statement-II is false Statement-II is true, but statement-I is false
58. Two numbers are inthe ratio 2 :"3. If 9 is added to each .number, they will be in the ratio 3 : 4. Wl1~t is the product of the two numbers ? (a) 360 (b) 480 (c) 486 (d) 512
55; A square,a circle and an equilateral triangle have same perimeter. Consider the following statements: l. 2. The area of square is greaterthan the area ofthe triangle. The area of circle is less than the area of triangle. 1 only Both 1 and 2
59. Consider the following statements in respect of an equilateral triangle ABC. 1. There is a point P inside the triangle ABC such that each of its sides subtends an angle of 1200 at P. 2. There is a point P inside the triangle
~L ...-..
Whichof the statements is/are correct? (a) (c) (b) 2 only (d) Neither 1 nor 2
ABC such that the triangle PBC is obtuse angled and A is theorthocenter of triangle PBC. Which of the above statements
C •• ~ ',,' ,'.,..,.., .. _.,
is/are
_c
56. The expressionsin'
(a) (b) (c)
x + cos' x-I
=0
is
correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2
satisfied by how manyvalues ofx ? Only one value ofx Two values of x Infinitevalues ofx
(d) No 'value of'x .16
o
(Contd.)
Il
'II
54. l1Ff ~
~ I lWf ~
fcf1 ABCD ~
B"lIr:t1<
~
W
V<fil\ ~
57. M3Jl!Rf<fiT, 5~
CfiT qf{"llq
~~. 75
~
fcfl
m2 -
m, it t:R ~
W, ~
~
em ~
~*m~-«
~ I '\ffi ~
~ n < m < 2n I lWf ~'~
(ft1l
AC = 2 mn
llA :~
BD=
ri
I
21
fotl
?
AB ~-I
= (m? + n2)/2
:AC>BD
(a)
40~ 30~ 24~
(b) (c)
~-II:ABCD~%I
\3qg<K1 Cfl'lRl
cfi ~
~
-1
if
~RiR1f{qd ~
~-m
~
(d) 20~
"t:%
mft ~ ?
~-I ~-II ~
~l
(a) ~
(b)
~cptFf
a1tt
91iR - II, tfi'lA" -I
eFt ~
~I
58.
m- ~
~9
~
om <fitfl-II ~
~~
~.Jrrr ~.
I ~-
2 :.3
iF 31:1%1if ~ I ~ ~ -ffr ~ 3 : 4 iF ~
tt&n
~ Wr
~-II.
(c) CfilFf-I ~ (d) ~ 55. ~
~I
c:pq;:f-I eFt ~
~
CfitA"-II ~
~-I
-;nff ~ I
e\04laTr
<fiTT 11lhH 'llTI ~ ?
l~
~, ~
WI
W
I.
(a)
360
-II ~
31tR'li
(b) 480 (c) 486
<fIf, Vi afR ~
1f{ ~
~
~
cf; qf{Silq \1lWf
:
(d)
512
Pp:1fftR9d ~
1. 2.,
\3q~
<fIf
~1
CfiT ~
~
cf; ~
-ij-
31ftrcF
I
59. ~~ABc*~ifP1~~f@d~
tJ\ fctqrr ~
:
ViCfiT ~
1
~iF ~~cp;r~ ~~~~-mftr~mft~A?
1.
~
WfiT{ qif
ABC ~ fct; ~
* affiTfu ~
-m- ~ I
~
~~C
~
pW
~
P "tR 1200
(a) ~ (b) ~2 (c) (d) 56. x 1
";:f
CJilur amfuf
ABCcfi
2.
~
attnffi ~- ~
~-~~
qlf ~~. ~
P~
3ltt
m
2~ 1 3ltt;;
~~~~PBC
1ft
2
iRA
~.
I
iF ~
C~S2
lTAl
x-I
~
~
.sin2 >1:+
=0~
'l{fo)
mar ~
mr
~
?
\3q~
~1
if ~ ~
mft~;f
?
(a) ~ (b) ~2
(c)
(a) x CfiJ ~ (b) x'"~ (c) x
* mllR
lWf ~
m lWf
1
afR -ffr
2 1
(d) ~
(d) x CfiT ~
*~ 1ft
<IT1l
2
(Contd.)
.•
,I
I
II '1
60.
What is the value of
64~ Consider the following:
1·001x 1·001x 1·001 + 0·999 x 0'999 x 0·999 --------------------------? 1'001 x 1~OO 1·001 x 0'999 + 0'9,9,9x 0·999 1(a) 1
1.
2.
sin? 10 + <:'OS2 1·= 1
sec233° - cof 5;0
= cosec'
370 - tan' 530
(b) 2 (c) 3 (d) 4
61. Ifp(x) is a common multiple of degree 6 of the polynomials g(x) f(x) = x3 + x2 .
-
Which of 'the above correct? (a) (c) (d) 65. 1 only Both 1 and 2 Neither 1 nor 2
statementsIs/are
(~) 2 only
x-
1 and
=
x3-
x + x'-l
2
then which one of the
-5
If'p=asinx + bcosxandq =acosx-bsinx, then what is p2 + q2 equal to ?
following is correct'?
(a)
(b)
a+ b
ab
(a) p(x) = (x'-c 1)2 (x+ 1)2(x2 + 1)
+ 1) (x2+1)2 (c) p(x) =(x _1)3 (x + 1) (x2 + J) (d) p(x) = (x - 1)2 (X4 + 1)
(b) p(x) = (x - 1) ex 62. If the roots of the equation x2 - 2ax +;a2 -3
(c) a2 + b2' (d) a2 _b2
66; Statement-I: Letthe side DE of a triangle DEF.be divided at S so that
+a
= 0 are
real and less than 3, then which
DE
DS = ~ . If a line through
,,2
one of the following is correct?
(a) a < 2
(b) 2<a<3 "tc)
(d) 63. 3 <a<4 Statement-Il:
S' parallel to EF meets DF at T, then the area of triangle DEF is twice the .area of the triangle DST. The areas of the similar triangles are proportional to the squares on the corresponding sides.
a>.4
Two students A andB solve an equation of the form x2 +: px + q = O. A starts with a wrong value of p and obtains the roots as
Which one of the following is correct in respect of the above statements ? (a) Both statement-land statement-II are true and statement-II is the correct explanation of statement-I Both statement -land statement-II are true but statement-II is not the' correct explanation of statement-I Statement-I is true,but false Statement-Il false 18
0'
2 and 6~ B starts with a wrong value of q
and gets the roots as 2 arid- 9 . What are the correct roots of the equation ?' (a) (b) (c) (d) 3 and-4 ~3 and-4 -3 and 4 3 and 4 (d) (c) (b)
statement-II is
is true, but statement-I is
(Contd.-)
---;".
."
I;
I"
1..
ii
')'001 x l'OOt'x ),001'+ 0·999 x 0'999 x 0·999 60. 1·001x 1· 01 -.1,001 x 0·999 + 0'9 99 x 0·999 0
·CfiT JlA
64. friP'ffB)Rld·1f{ fcNR
~
:
I.
2.
sin- '10 + cos' 1 = I see' 330 - cof 5'r = cosec' 370 - tan2 530
ctm ~?
(a)
1
\34giffi ~
if ~-m~
1 I ~.2
~ ~A ?
(b) 2 (c) 3
(a) ~ (b) ~2 (c) ~ f(x) =
Xl
(d) 4
61.
'ltfit ~ WJUf p(x) mft~?
(a)
+ x2
-
x-I
3IR
-sr
lJ;Cf)
(d)
~
ill I altt. ~ tft
2
g(x) = x3 - x2 + x-I
m, m friJ:1 f<1R9a if..~ ~
A
em
6 l.ffii' CfiT ~
65. ~ p=asinx+bcosxO'm q=acosx-bsinx ill p2 + q2!~ if{16f\ ~ ?
m,
(a) a + b (b) ab (c) a2 + tY (d) a2 - b2 66. Cfl?:R-I
lWf (~ftf~fil' fcfl
p(x) = (X - 1)2 (x + 1)2 (X2 + 1)
(b) p(x) = ex -1) (x + I) (x2 + 1)2 (c) p(x) = (x _1)3 (x + 1) (Xl +' 1) (d) p(x) = (x - 1)2(X4 + 1) 62. tift
<"i4'lCfl{UI
~
DEF ~
V<f1R
x2 -·2ax + a2 + a -13= 0 if; ~ 3 ~ Cfllf
~
DE
eN
~
1
S .If\ ~
Cf 1«1 f4<fi
~
m\
m I)P=<1 R1RfIa
if "«
f1f~
DE:= S
'fiT
umn ~ fcfi
EF if; tJlffiR DF aar~DEF DST
I
DS
cFR-m~m~
(a) a<2 (b) 2<a<3 (c) 3<a<4
r :,,:
?
«~ ~
tfl ~ Lflf1
~
1f{ ~
.J2 I ~
~~
1:\9r,
em
TlR~'~, 1ft,~ 4lctCfiT
It,
<n ~ ~
<fiT ~
*'
Cd) a>4
63. x2
~-II
~
~
Wrn
fcmr4f, A afR B, ~ m~ P 'fiT ~ m:rcr I lWf AA ~ A <t?r 2 3Tr\ 6 1f'f mto I q ifiT ~ m:rcr llR AA "?r B ~::r 31tt - 9 ~ 2 m-ca m ~I -;nft (14)Cf1{OI if m ~ ~ ~?
~px + q=
0 <f; ~
qrff
(ilflCfl{UI
<t?r <IT
<flIT
<f; (iJOtI~lef)
WI
\3 41Cffi qitpft
m~
<f; ~
Tf Pi t;:iR1f@d if ~ ffi-tIT
~~
~mft~?
(a)
(a)
3 aTrr-4
<fitA-I ~ <JitR-II ~ <:fltR-II, ~-I ~ mft ~
<?AT
~I
~. ~
;::nff ~ I
(b) -3 atR-4 (c) -3
(b) ~ CfIWf-I GtlT ~-II ~-II, ~-I ~ (c) ~-I~~, (d) efitR-II ~~, 19
o
m ClfRClIT
~-I
~
afR 4
~~-n~~1 ~
31m'lT ~ I
(d) 3 aTR 4
(Contd.)
Ii,
I
i
II
"
67. What is the value ofkthat'(2x-- 1) may be a factor of 4X4_ (k - l)x3 +ikx? - 6x + 1 ? _(a) 8 (b) 9 (c) (d) 12 13
For the next nrO (02)questions tbat follow: The-areas of the ends of a frustum 'of a pyramid are P and Q where P < Q and H is its thickness. 71. What is the difference in radii of the ends of the frustum ? (a)
JQ-.JP
..r;.
1t
68. If 196x4 = x6, then x3 is equai to which one of the following? (a) (b) (c) (d) x6/14 14x4 x 114
2
(b) (c)
.JQ-~
~-JP
(d) None of the above 72. What is the volume of the frustum ? (a) 3H(P+Q+~) H(P + Q +.JPQ) H(P + Q + .JPQ)/3 H(P+Q-.JPQ)/3
14x2
69. The sides of a parallelogram
are 12 em
(b) (c) (d)
and 8 em long and one of the diagonals is 10 em long: If d is the length of other diagonal, then which one of the following is correct? (a) (c) d <,8 em IDem '< d < 12 cm (b) 8 em < d < lOcm (d) do> 12 em
73. What number should be added to 231228 to make it exactly divisible by 33 ? (a) (c)
1
(b) 2 3 (d) 4
70. The base of a triangular wall is 7 times its height. If the cost of painting the wall at Rs. 350 per 100 square meter is Rs. 1225, then what.is the base length? (a) (c) (d) 50m 75 m 100m 20 (b) 70m 74. Let AB be a line segment such that a point P on the locus satisfies AB2 - AP2 _ Bp2 = o. What is the locus? (a) (c) the straight line AB itself a straight line parallel to AB (b) the perpendicular bisector of AB (d) a circle
(Contd.)
j
0
~...._
I.
" -..
;;t
Ii
ill
"
67. ktF ~-lWf:ifi
4X4 - (k - I)x3
J~
-
aWr 3lFr ~.
2x-l
(a) (e) (d) 8 (b) 9
m~~?
12 13
+
kx' _::_ '1Cf}T' !J)0I<i(9U;S 6x+
~ .\iffiT fcti
M
fq {I~;S
P < Q W-TI ~
* IU'-<i <fi * 'mii * ~'P
~
cit (02)
~
~. ~
:
altt
Cfm ~
Q
H ~ 'I
71. IU'-<i <ficfi (~) (b)
-rnii efT f3i \Ji!laTI
~
CflT
aTm:
?
.jQ-.JP
ro -JP1t
68. ~
196x4-
= x6,
traf x" PiOPlf?lR9(i if
'it ~
~*.~~?
(a) x6/14 (b) 14x4 (e) (d)
x2/14
(e) (d) 72.
JQ--&
\3 qg<tttll
it ~
;:nft
IU'-<i <fi
(a) (b)
CflT 3WKA
CflIT% ?
14x2
12emcrm8em 10'em ~ ~I (c) (d)
3H(P + Q,+.JPQ) H(P + Q +.jPQ) H(P+Q+.JPQ)/3 H(P + Q-.jPQ)/3
69. ~~~'#r~
~
~ (a) (b) (e) ~
~' B1tt
~ d<8cm
fqCfiUTI
if'~ ~'
d~
~~
-ffi Pi'-<iR1R9a if
cFA-m~?
8 em<d < IOem IOcm<d<12em
73. ,231228
lItil(itid:
if
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=
1 and y = 2
x = 1 and y= 1
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81. The HCF and LCM oftwo natural numbers are 12 and 72 respectively. What is the difference between the two numbers if one of the numbers is 24 ? (a) (b) (c) 12 18 21
77. If the 'volume of a cube is 729 cubic em, what is the length of its diagonal ? (a) (b) (c) 9.J2 em 9.J3 em 18cm
(d) 24 82. A person sold an article for Rs. 136 and got 15% loss. Had he sold it for Rs. x, he would have got a profit of 15%. \Vhich one of the following is correct? (a) (b) (c) (d) 190 < x < 200 180 < x < 190 170 <x < 180 160 <x <'170
(d)' 18.fi em 78. The altitude of an equilateral triangle is
.Jj em. What is its perimeter?
(a) (b) (c) (d) 3cm 3Jjcm 6cm
83. Consider the following statements: 1. The angular measure iniradian of a circular arc of fixed length subtending at its centre decreases if the radius of the arc increases. 2. (a) (c) 180011 is equal to 51tradian. 1 only Both 1 and 2 Which of the statements is/are correct ? (b) 2 only (d) Neither 1 nor 2
6..ficm
79. The area of a rectangle whose one side is 'a' is 2a2• What is the area of a square having one of the diagonals of the rectangle as side? (a) (c)
(d)
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2
(b) 3a2 5a2
22
0
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84. There are two taps A and Bto fill up a water tank. The tank can be filled in 40 minutes if both taps are on. The same tank can be filled in 60 minutes if tap A alone is on. How much time will tap B alone take to fill up the same tank ? (a) (b) (c) (d) 64 minutes 80 minutes 96 minutes 120 minutes
88. The ratio of two complementary angles is 1 ; 5. What is the difference between the
two angles? (a) (c) (d) 60° 1200 can not be determined with the given data (b) 900
89. What is the logarithm of 0·0001 with respect to base 10? (a) 4 3 (b) (c)
85. The pair of numbers which are relatively prime to each other is : (a) (c) (68,85) (92,85) (b) (65,91) (d) (102, 153)
-4
(d) -3 90. Consider the following statements :
1.
There is only one value of x in the first quadrant that satisfies sin x + cos x = 2. There is only one value of x in the first quadrant that satisfies sin x - cos x = O. 1 only Both 1 and 2
86. How many rational numbers between 1 and 1000 ? (a) (c) (d) 998 1000 Infinite (b) 999
are there
2.
Which of the above is/are correct? (a) (c) (b) 2 only (d) Neither 1 nor 2 91. If x lies in the first quadrant and cos x = 5/13, what is tan x - cot x equal to ? (a) -139/60 (b) (c) 139/60 119/60
87. The sum of. two numbers is2Q and their product is 75. What is the sum of their reciprocals? (a) (b) (c) 1115 1/5
4/15
(d) 7/15
(d) None of the above
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x
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In the -quadrilateral ABCD shown above, LDAB = LDCX = 120°. If LABC= 105°, what is LADC equal to ? (a) (c) 45° 75° (b) 60° (d) 95° 93.
c
In the figure LAED = 90°. 1. 2. given above
B
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=
Consider the following statements :. ABC and ADE are similar triangles. The four points B, C, E and D may lie on a circle, 'is/are
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(b) 2 only (c)
p
(d) Neither 1 nor 2
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(b) 50° (c) (d) 30°
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96. Consider the following: 1'. cot 30° + 1 _ 2 c ° , 300 1 - (os 30 + 1) cot 2 sin 45° cos 45° - tan 45° cot 45° = 0 identities is/are
99. If A = {x : x is an even natural number}, B = {x : x is a natural number and multiple of5} and C = {x : x is a natural number and multiple of 10}, then what is An (B u C) equal to ? (a) (b)
(c)
2.
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5
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I Ii | 20,506 | 43,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-26 | latest | en | 0.825966 |
http://stackoverflow.com/questions/9879258/how-can-i-generate-random-points-on-a-circles-circumference-in-javascript/9879291 | 1,406,235,077,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997891176.68/warc/CC-MAIN-20140722025811-00207-ip-10-33-131-23.ec2.internal.warc.gz | 347,547,226 | 15,409 | # How can I generate random points on a circles circumference in javascript
I am trying to write a function that will randomly return an x,y pos around a given circumference so if I have a point that's at 0,0(being the center of the div) how can I write a function that randomly places other entities that appear among the outer edge of a circle.
All I need is the equation i know it has something to do with getting the distance from the center to the circumference edge just no idea howto calculate it and randomize it so it looks good.
-
Start by digging out your high school trigonometry text book. Look for the section on the law of sines. You'll be working with a lot of right triangles. – gilly3 Mar 26 '12 at 20:27
this inst homework its more of im tired of googleing for only part of the equation and having to figure it out by trial and error. – Mouseroot Mar 26 '12 at 20:49
Just get a random angle:
``````var angle = Math.random()*Math.PI*2;
``````
Then
``````x = Math.cos(angle)*radius; | 249 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2014-23 | latest | en | 0.886026 |
http://mathoverflow.net/feeds/question/114942 | 1,369,419,361,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704943681/warc/CC-MAIN-20130516114903-00005-ip-10-60-113-184.ec2.internal.warc.gz | 173,952,150 | 2,278 | What are some properties of Delone sets that come from Barlow packings of spheres? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:16:02Z http://mathoverflow.net/feeds/question/114942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114942/what-are-some-properties-of-delone-sets-that-come-from-barlow-packings-of-spheres What are some properties of Delone sets that come from Barlow packings of spheres? mkreisel 2012-11-29T23:57:39Z 2012-12-18T19:32:42Z <p>Given a Barlow packing of \$\mathbb{R}^n\$ by balls with at most a finite number of different radii, the centers of the balls will form a Delone set in \$\mathbb{R}^n.\$ </p> <p>For a highest density sphere packing, or at least a Barlow packing of highest density among Barlow packings, must the corresponding Delone set be a Meyer set? A Patterson set? In the cases of dimensions 2 and 3 where the optimal packing using a single radius is known, the sets can be chosen to be lattices. </p> <p>I looked for papers exploring this connection and could only find this one <a href="https://www-fourier.ujf-grenoble.fr/PUBLIS/publications/REF_678.pdf" rel="nofollow">https://www-fourier.ujf-grenoble.fr/PUBLIS/publications/REF_678.pdf</a>, which is good, but doesn't address the big picture questions above. </p> <p>EDIT: As the comments indicate, we should restrict to Barlow packings. In this case the Delone sets always appear to be of finite local complexity. </p> http://mathoverflow.net/questions/114942/what-are-some-properties-of-delone-sets-that-come-from-barlow-packings-of-spheres/116732#116732 Answer by Agol for What are some properties of Delone sets that come from Barlow packings of spheres? Agol 2012-12-18T19:32:42Z 2012-12-18T19:32:42Z <p>If you consider the Barlow lattices, these are obtained by taking planar hexagonal packings of spheres, and nesting them together in layers. For each layer, there are two possible ways of placing the layer above it: <img src="http://cnx.org/content/m16927/latest/graphics15.png" alt="alt text"></p> <p>Let's normalize the Barlow lattices to have one common layer (the A-layer). The heights of the (centers of spheres of the) other layers will be multiples of the height of a tetrahedron, so are the same for each Barlow lattice. The centers of the vertices of the next layer above have two possibilities, one with projection given by the B-circles, and one given by the C-circles (these are symmetric under a transformation preserving A-circles, so let's assume the next layer is the B-circles). Then the next layer can have projection either the C-circles or the A-circles (the diagram corresponds to the face-centered cubic lattice, so the pattern there is ABCABC...). Thus, we see that for any Barlow lattice D, the differences \$D-D\$ will be a subset of the superposition of 3 face-centered cubic lattices which have the A-,B-,C-layers at each level, and thus will be a discrete set. So it will be a Meyer set (I'm going by the <a href="http://www.ams.org/mathscinet-getitem?mr=1400744" rel="nofollow">definition of a Meyer set</a> as a discrete set \$M\$ such that \$M-M\$ is also Delone). </p> | 858 | 3,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2013-20 | latest | en | 0.773403 |
https://proofwiki.org/wiki/Strictly_Monotone_Real_Function_is_Bijective | 1,713,520,916,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00073.warc.gz | 425,611,909 | 10,807 | # Strictly Monotone Real Function is Bijective
## Theorem
Let $f$ be a real function which is defined on $I \subseteq \R$.
Let $f$ be strictly monotone on $I$.
Let the image of $f$ be $J$.
Then $f: I \to J$ is a bijection.
## Proof
From Surjection by Restriction of Codomain, $f: I \to J$ is a surjection.
$\blacksquare$ | 107 | 328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-18 | latest | en | 0.820597 |
http://www.fightfinance.com/?q=126,37,56,171,405,512, | 1,618,646,935,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00069.warc.gz | 133,617,248 | 4,960 | # Fight Finance
#### CoursesTagsRandomAllRecentScores
What is the Internal Rate of Return (IRR) of the project detailed in the table below?
Assume that the cash flows shown in the table are paid all at once at the given point in time. All answers are given as effective annual rates.
Project Cash Flows Time (yrs) Cash flow () 0 -100 1 0 2 121 If a project's net present value (NPV) is zero, then its internal rate of return (IRR) will be: Which of the following statements about risk free government bonds is NOT correct? Hint: Total return can be broken into income and capital returns as follows: \begin{aligned} r_\text{total} &= \frac{c_1}{p_0} + \frac{p_1-p_0}{p_0} \\ &= r_\text{income} + r_\text{capital} \end{aligned} The capital return is the growth rate of the price. The income return is the periodic cash flow. For a bond this is the coupon payment. The following is the Dividend Discount Model used to price stocks: $$p_0=\frac{d_1}{r-g}$$ Which of the following statements about the Dividend Discount Model is NOT correct? The perpetuity with growth formula is: $$P_0= \dfrac{C_1}{r-g}$$ Which of the following is NOT equal to the total required return (r)? Find the cash flow from assets (CFFA) of the following project. Project Data Project life 2 years Initial investment in equipment6m Depreciation of equipment per year for tax purposes $1m Unit sales per year 4m Sale price per unit$8 Variable cost per unit $3 Fixed costs per year, paid at the end of each year$1.5m Tax rate 30%
Note 1: The equipment will have a book value of $4m at the end of the project for tax purposes. However, the equipment is expected to fetch$0.9 million when it is sold at t=2.
Note 2: Due to the project, the firm will have to purchase $0.8m of inventory initially, which it will sell at t=1. The firm will buy another$0.8m at t=1 and sell it all again at t=2 with zero inventory left. The project will have no effect on the firm's current liabilities.
Find the project's CFFA at time zero, one and two. Answers are given in millions of dollars (\$m). | 543 | 2,060 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2021-17 | latest | en | 0.901356 |
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