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http://man.linuxexplore.com/htmlman3/hypot.3.html | 1,620,417,278,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00035.warc.gz | 32,018,134 | 4,312 | Name
hypot, hypotf, hypotl — Euclidean distance function
Synopsis
```#include <math.h>
```
```double hypot(``` double x, double y`)`;
```float hypotf(``` float x, float y`)`;
```long double hypotl(``` long double x, long double y`)`;
Note
Feature Test Macro Requirements for glibc (see feature_test_macros(7)):
`hypot`():
`_BSD_SOURCE` || `_SVID_SOURCE` || `_XOPEN_SOURCE` || `_ISOC99_SOURCE` || `_POSIX_C_SOURCE` >= 200112L;
or cc `-std=c99`
`hypotf`(), `hypotl`():
`_BSD_SOURCE` || `_SVID_SOURCE` || `_XOPEN_SOURCE` >= 600 || `_ISOC99_SOURCE` || `_POSIX_C_SOURCE` >= 200112L;
or cc `-std=c99`
Note Link with `−lm`.
DESCRIPTION
The `hypot`() function returns sqrt(`x`*`x`+`y`*`y`). This is the length of the hypotenuse of a right-angled triangle with sides of length `x` and `y`, or the distance of the point (`x`,`y`) from the origin.
The calculation is performed without undue overflow or underflow during the intermediate steps of the calculation.
RETURN VALUE
On success, these functions return the length of a right-angled triangle with sides of length `x` and `y`.
If `x` or `y` is an infinity, positive infinity is returned.
If `x` or `y` is a NaN, and the other argument is not an infinity, a NaN is returned.
If the result overflows, a range error occurs, and the functions return `HUGE_VAL`, `HUGE_VALF`, or `HUGE_VALL`, respectively.
If both arguments are subnormal, and the result is subnormal, a range error occurs, and the correct result is returned.
ERRORS
See math_error(7) for information on how to determine whether an error has occurred when calling these functions.
The following errors can occur:
Range error: result overflow
`errno` is set to ERANGE. An overflow floating-point exception (`FE_OVERFLOW`) is raised.
Range error: result underflow
An underflow floating-point exception (`FE_UNDERFLOW`) is raised.
These functions do not set `errno` for this case.
CONFORMING TO
C99, POSIX.1-2001. The variant returning double also conforms to SVr4, 4.3BSD.
COLOPHON
This page is part of release 3.52 of the Linux `man-pages` project. A description of the project, and information about reporting bugs, can be found at http://www.kernel.org/doc/man−pages/.
Copyright 1993 David Metcalfe (david(@)prism.demon.co.uk) %%%LICENSE_START(VERBATIM) Permission is granted to make and distribute verbatim copies of this manual provided the copyright notice and this permission notice are preserved on all copies. Permission is granted to copy and distribute modified versions of this manual under the conditions for verbatim copying, provided that the entire resulting derived work is distributed under the terms of a permission notice identical to this one. Since the Linux kernel and libraries are constantly changing, this manual page may be incorrect or out-of-date. The author(s) assume no responsibility for errors or omissions, or for damages resulting from the use of the information contained herein. The author(s) may not have taken the same level of care in the production of this manual, which is licensed free of charge, as they might when working professionally. Formatted or processed versions of this manual, if unaccompanied by the source, must acknowledge the copyright and authors of this work. %%%LICENSE_END References consulted: Linux libc source code Lewine's _POSIX Programmer's Guide_ (O'Reilly & Associates, 1991) 386BSD man pages Modified 1993-07-24 by Rik Faith (faith(@)cs.unc.edu) Modified 2002-07-27 by Walter Harms (walter.harms(@)informatik.uni-oldenburg.de) | 1,057 | 3,547 | {"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} | 2.734375 | 3 | CC-MAIN-2021-21 | longest | en | 0.73862 |
https://forums.pragprog.com/forums/322/topics/11934 | 1,524,475,897,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945940.14/warc/CC-MAIN-20180423085920-20180423105920-00154.warc.gz | 634,544,371 | 16,869 | small medium large xlarge
• Write a function MyList.span(from, to) that returns a list of the numbers from `from` up to `to`.
A Possible Solution</summary>
```defmodule MyList do
def span(from, to) when from > to, do: []
def span(from, to) do
[ from | span(from+1, to) ]
end
end
IO.inspect MyList.span(5, 10)
```
</details>
Dave, you need to watch out for tail recursion (the last operation is not the recursive function call). You should use an accumulator then reverse it. It is much more efficient.
``````defmodule MyList do
def span(from, to) when from <= to do
_span(from, to, [from])
end
def span(from, to) do
span(to, from)
end
defp _span(from, to, acc) when from == to do
acc |> :lists.reverse
end
defp _span(from, to, acc) do
x = from + 1
_span(x, to, [x | acc])
end
end
``````
Just how much more efficient for a typical case, though?
I agree in general, but I have an interesting experience. When I first started coding elixir, I tried hard to be tail recursive. But when I posted that code, beginners said it was confusing, and experienced Erlang and Elixir programmers said it probably wasn’t necessary. So I now treat writing tail recursive code as an optimization, and like more optimizations, I do it only if I find I need to.
My version ended up being very similar to @Dave’s above. The only difference is that I used pattern matching in the arguments list instead of a `when` clause. I don’t think this is easier to read than using `when` but do other folks have an opinion here on which is more idiomatic Elixir?
``````defmodule MyList do
def span(_from=to, to), do: [to]
def span(from, to), do: [ from | span(from + 1, to) ]
end
``````
Gabe, you aren’t looking out for “from” being greater than “to”. If you go down that path, I would recommend doing it like:
``````defmodule T do
def span(to, to), do: [ to ]
def span(from, to) when from < to, do: [ from | span(from + 1, to) ]
end
``````
I did as @AlekseyGureiev. But, did not handled case when: from > to
``````defmodule MyList do
def span(from, to) when from == to, do: [from]
def span(from, to) when from < to, do: [from | span(from + 1, to)]
end
``````
I started with …
``````def span(from, to), do: Enum.to_list(from..to)
``````
but then realised that was cheating ;)
Ended up with …
``````def span(to, to), do: [to]
def span(from, to) when from > to, do: span(to, from)
def span(from, to), do: [ from | span(from + 1, to) ]
``````
One more solution:
``````defmodule MyList do
def span(to, to), do: [to]
def span(from, to) when from < to, do: _span(from, to, [])
def span(from, to), do: _span_r(to, from, [])
defp _span(to, to, acc), do: [to | acc]
defp _span(from, to, acc), do: _span(from, to - 1, [to | acc])
defp _span_r(to, to, acc), do: [to | acc]
defp _span_r(from, to, acc), do: _span_r(from + 1, to, [from | acc])
end
``````
Here’s a recursive definition that can handle reverse ranges such as 10..1:
``````defmodule MyList do
def span(from, from), do: [from]
def span(from, to), do: [from | span(_next(from, to), to)]
defp _next(from, to) when from < to, do: from + 1
defp _next(from, _to), do: from - 1
end
MyList.span 10, 1
#=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
``````
Alternative solution, using a range and Enum.map:
``````defmodule MyList do
def span(from, to), do: Enum.map(from..to, &(&1))
end
``````
``````defmodule Spanner do
def span(a,a), do: [a]
def span(a,b) when b < a, do: [a|span(a-1,b)]
def span(a,b), do: [a|span(a+1,b)]
end
``````
Hmm. I built my list the other way - just kept adding to the head:
`````` def span(a,b), do: _span(a,[b])
end
end
``````
No protections against bad inputs, of course, but it does the job fairly simply. Is there any reason not to build a list backwards (tail to head)?
I build the result list and track it as a separate argument. Not as efficient as Geoffrey Barto above.
As a matter of fact, I suspect the new_list++[from] expression is probably expensive since it has to traverse the list to get to the end.
``````defmodule MyList do
def span(from, to), do: _span([ from | to ], [])
defp _span([ from | to ], new_list) when(from <= to) do
_span( [from + 1 | to], new_list ++ [from] )
end
defp _span([ _| _], new_list), do: new_list
end
``````
Well, I came up with this where I built the list from tail to head. I would like to know if this is a bad practice in anyway.
``````defmodule MyList do
def span(from, to) when from > to, do: []
def span(from, to), do: _span([to], from, to)
defp _span(list = [head | _], from, _to) when head == from, do: list
defp _span(list = [head | _], from, to), do: _span([head-1|list], from, to)
end
``````
``````defmodule MyList do
def span(from, to), do: _span(from, to-1,[to])
defp _span(from, from, list), do: list
defp _span(from, to, list), do: _span(from, to-1, [to-1|list])
end
``````
Upside down:
```defmodule MyList do
def span(from, to) when from > to do [] end
def span(from, to) do span(from, to - 1) ++ [to] end
end
```
Haven’t tried breaking it yet but it looks pretty much the same as Gabe’s.
```defmodule Span do
def span(to, to), do: [to]
def span(from, to), do: [from | span(from + 1, to)]
end
```
My solution was a bit different.
``````defmodule Span do
def from(top, bottom) do
range = top..bottom
for n <- range do
IO.puts n
end
range
end
end
``````
``````def span_list(from, to), do: span(from, to, [to])
defp span(from, to, [h | t]) when h > from, do: span(from, to - 1, [to - 1, h | t])
defp span(from, to, acc) when to == from, do: acc
``````
My solution :
``````defmodule MyList do
def span(from, to) do, _span(from..to, [])
defp _span(from..to, list) when from < to, do: _span(from + 1..to, list ++ [from])
defp _span(from..to, list) when from == to, do: list
end
``````
I need to improve my solution
My solution
``````
defmodule MyList do
def span(from, to) when from == to, do: [from]
def span(from, to) when from > to, do: [from | span(from - 1, to)]
def span(from, to) when from < to, do: [from | span(from + 1, to)]
end
IO.inspect MyList.span(1,10)
IO.inspect MyList.span(10,1)
``````
Output
``````[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
``````
My solution:
``````defmodule MyList do
def span(from, to) when from < to, do: [from | span(from+1, to)]
def span(from, to) when from > to, do: [from | span(from-1, to)]
def span(from, to) when from === to, do: [to]
end
``````
Why not keep it simple? :)
``````def span(from,to), do: from..to |> Enum.to_list
``````
My actual solution was really close to some others I saw here:
`````` def span(to,to), do: [to]
def span(from,to) when to > from, do: [from|span(from+1,to)]
def span(from,to) when to < from, do: [from|span(from-1,to)]
``````
```def span(x,x), do: [x]
def span(x,y) when x < y do
[x|span(x+1, y)]
end
def span(x,y) when x > y do
[x|span(x-1, y)]
end
```
I got a little carried away and made mine work backwards as well:
``````def span(to, to), do: [to]
def span(from, to) when from < to, do: [ from | span(from + 1, to) ]
def span(from, to), do: [ from | span(from - 1, to) ]
``````
``````defmodule MyList do
def span(to, to), do: [to]
def span(from, to), do: [from | span(from + 1, to)]
end
``````
Mehdi, your code is short and elegant so I tried it, but it fail when from > to (I know it is not the point of the exercise).
I also tried to protect runtime against bad arguments, so I used private functions to keep my code DRY with the guard clause when is_integer(from) and is_integer(to).
``````defmodule MyList do
def span(from,to) when is_integer(from) and is_integer(to), do: _span(from,to)
defp _span(from,to) when from > to, do: []
defp _span(from,to), do: [from|_span(from+1,to)]
end
``````
Using Enum:
`````` def span(from, to), do: Enum.to_list(from..to)
``````
Without Enum:
`````` def alt_span(to, to), do: [to]
def alt_span(from, to), do: [ from | alt_span(from + 1, to ) ]
``````
More flexible to allow greatest to least
`````` # in the same module as alt_span
def flex_span(to, to), do: [to]
def flex_span(from, to) when from <= to, do: alt_span(from, to)
def flex_span(from, to) when from >= to, do: [ from | flex_span(from - 1, to ) ]
``````
I know the chapter was all about recursion, but I just used some built-ins to solve this:
``````defmodule MyList do
def span(from, to), do: Enum.to_list(from..to)
end
``````
That will handle the exercise as well as numbers where from > to by creating a list with the numbers going backwards.
Here’s my solution:
``` def span(from, to) when from < to do
[from | span(from+1, to)]
end
def span(from, to) when from == to do
[to]
end
def span(from, to) when from > to do
raise "args 1 must be less than or equal args 2"
end
```
Another one:
``````defmodule MyList do
def span(to, to), do: [to]
def span(from, to) when from < to do
[ from | span( from+1, to ) ]
end
def span(from, to) do # from > to
[ from | span( from-1, to ) ]
end
end
``````
Alternative solution, using Enum:
``````defmodule MyList do
def span(to, to), do: [to]
def span(from, to) when from < to, do: Enum.to_list(from..to)
def span(from, to), do: Enum.to_list(from..to)
end
#For both solutions - calling the func:
IO.inspect MyList.span(1, 1) #=> [1]
IO.inspect MyList.span(1, 10) #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
IO.inspect MyList.span(10, 1) #=> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
``````
my modest solution:
``````defmodule Mylist7 do
def span(from, to), do: _span(from, to)
defp _span(from, to) when from == to, do: []++[to]
defp _span(from, to) when from < to do
acc = from + 1
[from | _span(acc, to)]
end
defp _span(from, to) when from > to do
IO.puts "Error, from is more as to"
end
end
``````
and this is a result of searching of elixir library, very concise and powerful :-)
``````defmodule Mylist6 do
def span(from, to), do: Enum.concat([from..to])
end
``````
``````defmodule MyList do
def span(from, to), do: Enum.to_list from..to
end
``````
Same as Richard - works when from is not greater than to.
``````defmodule MyList do
def span(from, to), do: do_span(from, to, [])
defp do_span(to, to, acc) do
[to | acc]
|> Enum.reverse
end
defp do_span(from, to, acc) when from < to do
do_span(from + 1, to, [ from | acc ])
end
defp do_span(from, to, acc) when from > to do
do_span(from - 1, to, [ from | acc ])
end
end
``````
Adding my solution. Tail call optimized to test that out. Reverse is just for proper ordering.
A different solution.
``````defmodule MyList do
def span(from, to), do: _span([], from, to)
defp _span(list, from, to) when to < from, do: list
defp _span(list, from, to), do: _span([to | list], from, to - 1)
end
``````
``````defmodule MyList do
def span(from, to)
when from < to,
do: [from | span(from+1, to)]
def span(_from, to), do: [to | []]
end
``````
``````defmodule MyList do
def span(to, to), do: []
def span(from, to) when from < to do
[from | span(from+1, to)]
end
def span(from, to) when from > to do
[from | span(from-1, to)]
end
end
``````
``````defmodule MyList do
def span(to, to), do: []
def span(from, to), do: [from | span(from+1,to)]
end
``````
I didn’t handle the case where `from > to` but that would be easy enough to do by modifying the first line. Just posting my mistakes as a learning experience.
One thing that’s slightly unclear to me is if the list returned should be “`from` to-but-not-including `to`”. In other words `MyList.span(1,5)` should return `[1,2,3,4]`. I’ve coded my solution this way.
``````defmodule MyList do
def reverse(list), do: _reverse(list, [])
def _reverse([], rev), do: rev
def span(to, to), do: [to]
def span(from, to) when from < to, do: [from|span(from + 1, to)]·
def span(from, to) when from > to, do: reverse([to|span(to + 1, from)])
end
``````
``````MyList.span(1, 5) == [1, 2, 3, 4, 5]
MyList.span(5, 1) == [5, 4, 3, 2, 1]
MyList.span(-5, -1) == [-5, -4 ,-3 ,-2, -1]
MyList.span(-1, -5) == [-1, -2 ,-3 ,-4, -5]
MyList.span(-2, 2) == [-2, -1, 0, 1, 2]
MyList.span(2, -2) == [2, 1, 0, -1, -2]
``````
Here is my solution.
``````def span(from, to) when from > to, do: raise "first arg must be less than or equal to second arg"
def span(from, to) when from === to, do: [from]
def span(from, to) do
span(from, to - 1) ++ [to]
end
``````
My solution defaults to a step of 1, but also allows step of different amounts (and negative numbers). There’s a not so obvious bug in it though; if `to - from` is not a multiple of `step`, then it will never converge.
``````defmodule MyList do
def span(from, to, step \\ 1)
def span(to, to, _step), do: [to]
def span(from, to, step), do: [ from | span(from + step, to, step) ]
end
IO.inspect MyList.span(1, 10)
IO.inspect MyList.span(2, -3, -1)
IO.inspect MyList.span(0, 12, 2)
IO.inspect MyList.span(1, 12, 2) # will run forever because 12 - 1 is not a multiple of 2
``````
```defmodule MyList do
def span(from, from), do: [from]
def span(from, to) when from > to, do: [from | span(from-1, to)]
def span(from, to), do: [from | span(from+1, to)]
end
```
I found this to be the most straightforward approach:
``````defmodule MyList do
def span(to , to), do: [ to ]
def span(from, to) when from < to , do: [ from | span(from + 1, to) ]
def span(_, _), do: raise "provided From seems to be higher then To"
end
``````
Outcome:
``````iex()> MyList.span(1, 4)
[1, 2, 3, 4]
``````
```defmodule MyList do
def span(from,to) when from <= to, do: [from | span(from+1,to)]
def span(_,_), do: []
end
```
I had a similar solution to Dave’s originally, but ended up with a list which looked like this:
`[1, 2, 3, 4 | 5]`
Apparently this is an improper list should anyone else come across the same issue.
``````def span(from, to), do: _span(from, to, [true: 1, false: -1][from < to])
defp _span(n, n, _dir), do: [n]
defp _span(from, to, dir), do: [from] ++ _span(from + dir, to, dir)
``````
My solution, handling cases when from is greater than to
``````def span(from, to) when from < to do
[from | span(from+1, to)]
end
def span(from, to) when from > to do
[from | span(from-1, to)]
end
def span(from = _, _) do
[from]
end
``````
Here’s my implementation:
``````defmodule MyList do
def span(from, to) when from > to, do: _span(
from,
to,
&(&1 - 1)
)
def span(from, to), do: _span(
from,
to,
&(&1 + 1)
)
defp _span(from, to, f), do: _span(from, to, f, [ from ])
defp _span(to, to, _f, list), do: Enum.reverse(list)
defp _span(from, to, f, list) do
with x = f.(from),
do: _span(x, to, f, [ x | list ])
end
end
``````
`````` def span(a, a), do: [a]
def span(a, b) when a > b, do: [a | span(a - 1, b)]
def span(a, b), do: [a | span(a + 1, b)]
``````
``````
defmodule MyList do
# span1 is a solution using Enum.map
def span1(from,to) when from > to do
IO.puts("the first argument must be smaller than the second")
end
def span1(a,b) do
Enum.map(a..b, &(&1))
end
# span2 is a solution using recursion
def span2(a,a) do
[a]
end
def span2(from,to) when from > to do
IO.puts("the first argument must be smaller than the second")
end
def span2(from,to) do
[from] ++ span2(from+1,to)
end
end
``````
``````defmodule MyList do
def span(from, to) when from <= to do
[from | span(from + 1, to)]
end
def span(_from, _to), do: []
end
``````
You must be logged in to comment | 5,108 | 15,166 | {"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-2018-17 | latest | en | 0.879044 |
https://math.stackexchange.com/questions/4329157/what-is-the-average-roll-on-dice-while-re-rolling-a-result-of-1 | 1,702,163,068,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00184.warc.gz | 432,709,179 | 37,289 | # What is the average roll on dice while re-rolling a result of 1
The average roll on a six sided die (d6) is 3.5.
If the result is a 1 you are allowed a single re-roll accepting the second result. What does the average roll increase to? And how to calculate this?
Similarly the average roll on 2d6 is 7. If the result of ONE of the d6s is 1 you are allowed a single re-roll accepting the second result (so if two 1s are rolled you may only re-roll one of them). What does the average roll increase to? And how to calculate this?
My maths skills are poor and I have been unable to find any resources on the internet to explain this to me. Many thanks.
From comment:
I work out the average of 1d6 is (1+2+3+4+5+6)/6= 3.5 The average roll of the 2-6 is (2+3+4+5+6)/5 = 4 The average roll of the 1 will go back to being 3.5 as the re-roll will make it a normal die roll. Where I lost is how to combine the average of the 4 for the 2-6 results with the average of the re-rolled 1 which is 3.5. If I take the average of both 4 + 3.5 = 3.75 but this does not fell right as it is giving equal weight to each result when you have an 5/6 chance of getting 2-6 and only a 1/6 chance of getting 1.
• Welcome to MSE. Please show your thoughts or efforts towards solving the problem. Dec 10, 2021 at 11:34
• I work out the average of 1d6 is (1+2+3+4+5+6)/6= 3.5 The average roll of the 2-6 is (2+3+4+5+6)/5 = 4 The average roll of the 1 will go back to being 3.5 as the re-roll will make it a normal die roll. Where I lost is how to combine the average of the 4 for the 2-6 results with the average of the re-rolled 1 which is 3.5. If I take the average of both 4 + 3.5 = 3.75 but this does not fell right as it is giving equal weight to each result when you have an 5/6 chance of getting 2-6 and only a 1/6 chance of getting 1. Dec 10, 2021 at 11:48
There are $$11$$ possible outcomes : $$2,3,4,5,6$$; $$(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)$$.
The first $$5$$ of these each have probability $$\tfrac16$$, and the last six each have probability $$\tfrac{1}{36}$$.
For each outcome, you have a score and a probability.
The scores are $$2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6$$.
Probabilities are $$\tfrac16$$, $$\tfrac16$$, $$\tfrac16$$, $$\tfrac16$$, $$\tfrac16$$, $$\tfrac{1}{36}$$, $$\tfrac{1}{36}$$, $$\tfrac{1}{36}$$, $$\tfrac{1}{36}$$, $$\tfrac{1}{36}$$, $$\tfrac{1}{36}$$.
To get the average, you multiply the scores by their associated probabilities, and add up. So $$\frac16(2+3+4+5+6) + \frac{1}{36}(1+2+3+4+5+6) = \frac{141}{36} \approx 3.916$$
• Thank you that's very helpful explanation. Dec 12, 2021 at 0:24
The average roll of the $$2-6$$ is $$4$$. The average roll of the $$1$$ will go back to being $$3.5$$ as the re-roll will make it a normal die roll. You have a $$5/6$$ chance of getting $$2-6$$ and only a $$1/6$$ chance of getting $$1$$.
So the overall mean of the distribution of outcomes is $$\frac56 \times 4+\frac16\times 3.5 = \frac{47}{12}\approx 3.9167$$ | 1,026 | 2,980 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-50 | latest | en | 0.946811 |
https://www.hsestudyguide.com/how-to-calculate-frequency-rate-2/ | 1,726,409,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00699.warc.gz | 743,123,651 | 92,481 | # How to Calculate Frequency Rate with Practical Example
Table of Contents
## How to Calculate Frequency Rate with Practical Example
In the ever-evolving world of data analysis and statistics, understanding how to calculate frequency rate is a fundamental skill. Frequency rate, often referred to as frequency ratio, is a crucial metric used to assess the occurrence or frequency of a particular event or phenomenon within a given time frame. Whether you’re a student, a researcher, or a professional in any field, mastering this concept can be immensely valuable. In this article, we will delve into the intricacies of calculating frequency rate, providing you with a step-by-step guide and a practical example to ensure you grasp this concept with ease.
## 1. Introduction
Frequency rate is a statistical measure that allows us to quantify the occurrence of a specific event within a given time frame. It is a valuable tool in various fields, including finance, healthcare, marketing, and more. This article will provide a comprehensive guide to calculating frequency rate, ensuring you have a solid understanding of its application.
## 2. What is Frequency Rate?
Frequency rate, also known as frequency ratio, is a metric that expresses how often a particular event occurs relative to a specific time period. It is often used to analyze data related to accidents, incidents, disease outbreaks, or any event with a defined beginning and end.
## 3. The Formula for Calculating Frequency Rate
To calculate frequency rate, you can use the following formula:
`Frequency Rate = (Number of Events / Total Time) × Base`
Where:
• Number of Events: The total occurrences of the event.
• Total Time: The duration of the observation period.
• Base: A constant value, usually 1, to standardize the rate.
## 4. Practical Example: Calculating Frequency Rate
Let’s walk through a practical example to illustrate how to calculate frequency rate. Suppose you are a safety manager at a construction site, and you want to determine the frequency rate of workplace accidents over the past year.
### Step 1: Gather Data
• Number of Workplace Accidents: 15
• Total Time (in hours): 30,000
• Base: 1
### Step 2: Calculate Frequency Rate
Using the formula:
`Frequency Rate = (15 / 30,000) × 1 = 0.0005`
The frequency rate for workplace accidents in this example is 0.0005.
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## 5. Interpreting Frequency Rate
Interpreting frequency rate depends on the context. In our example, a frequency rate of 0.0005 indicates that, on average, there was one workplace accident for every 2,000 hours worked during the past year.
## 6. When to Use Frequency Rate
Frequency rate is useful when you need to assess the relative frequency or occurrence of an event within a specific time frame. It helps in identifying trends, making comparisons, and evaluating the effectiveness of safety measures.
## 7. Advantages of Frequency Rate
• Provides a standardized measure for comparing different time frames or groups.
• Useful for tracking and improving safety performance.
• Helps in setting benchmarks for future goals.
## 8. Limitations of Frequency Rate
• Assumes a constant base value, which may not always be accurate.
• Ignores the severity or impact of events.
• Does not account for events that go unreported.
## 9. Tips for Accurate Calculations
• Ensure accurate data collection and reporting.
• Use consistent time units (e.g., hours, days, months) for Total Time.
• Consider adjusting the base value for specific analyses.
## 10. Real-World Applications
Frequency rate is applied in various industries:
• Healthcare: Tracking patient readmissions.
• Manufacturing: Monitoring equipment breakdowns.
• Finance: Analyzing stock market volatility.
• Environmental Science: Studying natural disasters.
## 11. Frequency Rate vs. Other Metrics
Frequency rate is distinct from other metrics like incidence rate, prevalence rate, and mortality rate. Each metric serves a specific purpose and may be used in different scenarios.
## 12. Common Mistakes to Avoid
• Using the wrong formula or base value.
• Neglecting to account for unreported events.
• Failing to interpret the rate within its context.
## 13. Frequently Asked Questions (FAQs)
### Q1: How can I adjust the base value for my calculations?
A1: You can adjust the base value to reflect different units or scales, depending on your analysis goals.
### Q2: Is frequency rate the same as incidence rate?
A2: No, they are different metrics. Frequency rate measures occurrences over a specific time, while incidence rate focuses on new cases.
### Q3: Can frequency rate be greater than 1?
A3: Yes, if the number of events exceeds the total time, the rate can be greater than 1.
### Q4: What is the significance of frequency rate in risk assessment?
A4: Frequency rate is essential in identifying potential risks and taking preventive measures to mitigate them.
### Q5: How do I calculate frequency rate for non-continuous events?
A5: You can adapt the formula by adjusting the Total Time to suit the event’s characteristics.
## 14. Conclusion
In conclusion, understanding how to calculate frequency rate is a valuable skill in data analysis and risk assessment. This metric allows us to quantify the frequency of events within a specified time frame, aiding in decision-making and safety improvements. By following the provided formula and guidelines, you can confidently calculate frequency rates for various scenarios. Remember to interpret the results within their context to derive meaningful insights. | 1,185 | 5,754 | {"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} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.908289 |
https://www.jiskha.com/display.cgi?id=1316363555 | 1,502,942,077,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00515.warc.gz | 949,804,308 | 3,509 | # Physics
posted by .
A ball is thrown horizontally from a height of 19.17 m and hits the ground with a speed that is 4.0 times its initial speed. What was the initial speed?
• Physics -
4.79 m/s/s
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More Similar Questions | 629 | 2,344 | {"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-2017-34 | latest | en | 0.961896 |
http://cn.metamath.org/mpeuni/df-liminf.html | 1,660,557,463,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572163.61/warc/CC-MAIN-20220815085006-20220815115006-00324.warc.gz | 12,816,422 | 3,733 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > df-liminf Structured version Visualization version GIF version
Definition df-liminf 40502
Description: Define the inferior limit of a sequence of extended real numbers. (Contributed by GS, 2-Jan-2022.)
Assertion
Ref Expression
df-liminf lim inf = (𝑥 ∈ V ↦ sup(ran (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < )), ℝ*, < ))
Distinct variable group: 𝑥,𝑘
Detailed syntax breakdown of Definition df-liminf
StepHypRef Expression
1 clsi 40501 . 2 class lim inf
2 vx . . 3 setvar 𝑥
3 cvv 3351 . . 3 class V
4 vk . . . . . 6 setvar 𝑘
5 cr 10137 . . . . . 6 class
62cv 1630 . . . . . . . . 9 class 𝑥
74cv 1630 . . . . . . . . . 10 class 𝑘
8 cpnf 10273 . . . . . . . . . 10 class +∞
9 cico 12382 . . . . . . . . . 10 class [,)
107, 8, 9co 6793 . . . . . . . . 9 class (𝑘[,)+∞)
116, 10cima 5252 . . . . . . . 8 class (𝑥 “ (𝑘[,)+∞))
12 cxr 10275 . . . . . . . 8 class *
1311, 12cin 3722 . . . . . . 7 class ((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*)
14 clt 10276 . . . . . . 7 class <
1513, 12, 14cinf 8503 . . . . . 6 class inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < )
164, 5, 15cmpt 4863 . . . . 5 class (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < ))
1716crn 5250 . . . 4 class ran (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < ))
1817, 12, 14csup 8502 . . 3 class sup(ran (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < )), ℝ*, < )
192, 3, 18cmpt 4863 . 2 class (𝑥 ∈ V ↦ sup(ran (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < )), ℝ*, < ))
201, 19wceq 1631 1 wff lim inf = (𝑥 ∈ V ↦ sup(ran (𝑘 ∈ ℝ ↦ inf(((𝑥 “ (𝑘[,)+∞)) ∩ ℝ*), ℝ*, < )), ℝ*, < ))
Colors of variables: wff setvar class This definition is referenced by: liminfval 40509
Copyright terms: Public domain W3C validator | 842 | 1,756 | {"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-2022-33 | latest | en | 0.194424 |
https://resources.quizalize.com/view/quiz/real-analysis-dda1e2c8-d3fb-4099-9923-e7eda622ce30 | 1,718,471,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00206.warc.gz | 434,298,335 | 14,931 | # Real Analysis
## Quiz by Anitha Cruz PSGRKCW
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
5 questions
• Q1
Let f be a bounded function defined on [a,b] and let P be a partition of [a,b]. If P* is a refinement of P then
L(P*,f) $\le$ L(P*f)
U(P*,f) $\ge$ U(P*f)
L(P*,f) $\ge$ L(P*f)
U(P*,f) $\le$ U(P*f)
30s
• Q2
If P1 & P2 be any two partitions of [a,b] then
U(P1,f) $=$ L(P2,f)
U(P1,f) $\le$ L(P2,f)
U(P1,f) $\ge$ L(P2,f)
U(P1,f) $\le$ U(P2,f)
30s
• Q3
30s
• Q4
30s
• Q5
Let f be real value bounded function defined on [a,b]. Then lower Riemann integral f over [a,b] _____ the upper Riemann integral of f over [a,b]
can exceed
cannot exceed
none
may or maynot exceed
30s
Teachers give this quiz to your class | 296 | 800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 12, "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-2024-26 | latest | en | 0.642237 |
https://betterlesson.com/community/lesson/22225/unit-2-1-1st-grade | 1,498,462,630,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320685.63/warc/CC-MAIN-20170626064746-20170626084746-00563.warc.gz | 734,329,176 | 12,433 | # Lesson: Unit 2-1 1st Grade
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### Lesson Objective
To read symbols of quarter note and quarter rest in rhythm
# Objectives
(Specify skills/information that will be learned.)
SWBAT readand accurately play three different 4 beat rhythms on the board with scaffolding of a BEAT MATRIC as evidenced by their ability to play the rhythms together in groups on the Boomwhackers.
LANGUAGE OBJECTIVE-SWBAT verbally compare 3 separate beat matrixes using music quarter note and quarter rest.
# Information
(Give and/or demonstrate necessary information)
-Definition of rhythm will be introduced
-Understand the difference between a note and a rest on a beat matrix
-difference between rhythm and beat
-Definition of a Toccata
Boomwhacker Songs
# Verification
(Steps to check for student understanding)
Play boomwhacker in the correct rhythm indicated by rhythm on board.,
# Activity
(Describe the independent activity to reinforce this lesson)
Play the 3 rhythms practiced at the correct time along with a Boomwhacker Song
# Summary
Common Core Standards
• CC.1.OA.3 Apply properties of operations as strategies to add and subtract. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.) (Students need not use formal terms for these properties.)
• CC.1.SL.6 Produce complete sentences when appropriate to task and situation. (See grade 1 Language standards 1 and 3 on page 26 for specific expectations.)
• CC.1.L.4 Determine or clarify the meaning of unknown and multiple-meaning words and phrases based on grade 1 reading and content, choosing flexibly from an array of strategies.
### Lesson Resources
No resources at this time. | 437 | 1,849 | {"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-2017-26 | longest | en | 0.86094 |
https://www.atozexams.com/mcq/electrical-engineering/59.html | 1,679,625,899,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00661.warc.gz | 718,885,435 | 10,842 | # A step down chopper is operated in the continuous conduction mode in steady state with a constant duty ratio D.if VO is the magnitude of the dc input voltage,the ratio is given by
1. D
2. 1-D
3.
4.
4
D
Explanation :
No Explanation available for this question
# A two port network shown in the figure is described by the following equation I1=y11E1+Y12E2 I1=y21E1+Y22E2 The admittance parameters Y11Y12 Y22 for the network shown are
1. 0.5 mho,1 mho,2 mho and 1 mho respectively
2. 1/3 mho,-1/6 mho,-1/6 mho and 1/3 mho respectively
3. 0.5 mho,0.5 mho,1.5 mho and 2 mho respectively
4. -2/5 mho,-3/7 mho,3/7 mho and 2/5 mho respectively
4
1/3 mho,-1/6 mho,-1/6 mho and 1/3 mho respectively
Explanation :
No Explanation available for this question
# In the circuit shown in figure what value of C will cause a unity power factor at the ac source
1. 68.1μF
2. 165 μF
3. 0.681μF
4. 6.81μF
4
68.1μF
Explanation :
No Explanation available for this question
# A first order,low pass filter is given with R=50 and C=5μF.what is the frequency at which the gain of the voltage transfer function of the filter is 0.25?
1. 4.92 kHz
2. 0.49 kHz
3. 2.46 kHz
4. 24.6 kHz
4
2.46 kHz
Explanation :
No Explanation available for this question
# A series R-L-C circuit has R=50,L=100 μH and C=1 μF the lower half power frequency of the circuit is
1. 30.55 kHz
2. 3.055 kHz
3. 51.92 kHz
4. 1.92 kHz
4
51.92 kHz
Explanation :
No Explanation available for this question
# A 200 V,2000 rpm,10 A separately excited dc motor has an armature resistance of 2,rated dc voltage is applied to both the armature and field winding of the motor.if the armature drawn 5A from the source the torque developed by the motor is
1. 4.30 Nm
2. 4.77 Nm
3. 0.45 Nm
4. 0.50 Nm
4
4.77 Nm
Explanation :
No Explanation available for this question
# The rotor of a three phase 5 Kw,400 V,50 Hz slip ring induction motor is wound for 6 poles while its stator is wound for 4 poles.the approximate average no load steady state speed when this motor is connected to 400 V,50 Hz supply is
1. 1500 rpm
2. 500 rpm
3. 0 rpm
4. 1000 rpm
4
0 rpm
Explanation :
No Explanation available for this question
# The flux per pole in a synchronous motor with the field circuit ON and the stator disconnected from the supply is found to be 25 m Wb.when the stator is connected to the rated supply with the field excitation unchanged,the flux per pole in the machine is found to be 20 mWb while the motor is running on no load.Assuming no load losses to be zero,the no load current down by the motor from the supply
1. lags the supply voltage
3. is in phase with the supply voltage
4. is zero
4
Explanation :
No Explanation available for this question
# An 11 V pulse of 10 μs duration is applied to the circuit shown in the figure.Assuming that the capacitor is completely discharged prior to applying the pulse,the peak value of the capacitor voltage is
1. 11 V
2. 5.5 V
3. 6.32 V
4. 0.96 V
4
6.32 V
Explanation :
No Explanation available for this question
# The output voltage(Vo) of the schmitt trigger shown in the figure.swings between +15v and -15v.Assume that the operational amplifier is ideal,the output will change from +15v and -15v.when the instantaneous value of the input sine wave is
1. 5V in the positive slope only
2. 5V in the negative slope only
3. 5V in the positive and negative slopes only
4. 3V in the positive and negative slopes only
4 | 1,082 | 3,512 | {"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-2023-14 | latest | en | 0.79457 |
https://www.mrexcel.com/board/threads/random-function-in-vba.19947/ | 1,720,998,494,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00129.warc.gz | 792,387,436 | 20,670 | # Random function in VBA
#### Atalla
##### New Member
Hi all,
I have a list of data (A1 to G20 containing ID, First_Name, Last_Name, etc..). I need a function or a procedure in VBA to randomaly select one record at a time (only the cells that contain the First_Name and Last_Name). And output the results on another sheet without having the random cells repeat themselves, i.e, the employee will be chosen just once. I managed to write the code in VBA Access but somehow I can't do it in Excel. I could link the code to a command button.
### Excel Facts
What is the shortcut key for Format Selection?
Ctrl+1 (the number one) will open the Format dialog for whatever is selected.
Hi Atalla,
Welcome to the board.
This may be of use to you, its taken from Dave Hawley's site.
http://www.ozgrid.com/
User Defined Functions - Random Numbers
This UDF will generate x unique random numbers between any 2 numbers you specify. Many thanks to J.E. McGimpsey for modifying this to work on more than 10 numbers.
The Code
Function RandLotto(Bottom As Integer, Top As Integer, _
Amount As Integer) As String
Dim iArr As Variant
Dim i As Integer
Dim r As Integer
Dim temp As Integer
Application.Volatile
ReDim iArr(Bottom To Top)
For i = Bottom To Top
iArr(i) = i
Next i
For i = Top To Bottom + 1 Step -1
r = Int(Rnd() * (i - Bottom + 1)) + Bottom
temp = iArr(r)
iArr(r) = iArr(i)
iArr(i) = temp
Next i
For i = Bottom To Bottom + Amount - 1
RandLotto = RandLotto & " " & iArr(i)
Next i
RandLotto = Trim(RandLotto)
End Function
To use this UDF push Alt+F11 and go Insert>Module and paste in the code. Push Alt+Q and save. The Function will appear under "User Defined" in the Paste Function dialog box (Shift+F3). Use the Function in any cell as shown below.
=RandLotto(1,20,8)
This would produce 8 unique random numbers between 1 and 20
HTH
_________________<font color="blue"> «««<font color="red">¤<font color="blue"><font size=+1>Richie</font><font color="red">¤<font color="blue"> »»»</font>
</gif>
This message was edited by Richie(UK) on 2002-08-29 15:05
This set of subs will random pick a name from a list of names. Hope it helps. JSW
Sub myRnd()
'Find a random name in a existing names list.
'By Joe Was, 6/27/2001.
Dim myOrder As Range
Dim myName
Dim mySelect As Variant
Randomize
'Note: The 20 below = the ending ROW of your names list.
' The 1 below = the starting ROW of your names list.
mySelect = Int((20 * Rnd) + 1)
'Note: The "A" below is the column where your names list is.
myName = Range("A" & mySelect)
'Put the answer in a cell
Worksheets("Sheet1").Range("C1") = myName
'Put the answer in a screen message box.
MsgBox "The selection is:" & Chr(13) & Chr(13) & myName
End Sub
Sub myErase()
Range("C1").Select
Selection.ClearContents
Range("C1").Select
End Sub
Sub myListR()
'Find a random name in a existing names list.
'Then adds it to an ongoing list.
'By Joe Was, 6/27/2001.
Dim myOrder As Range
Dim myName
Dim mySelect As Variant
Randomize
'Note: The "20" below is the ending ROW of your names list.
' The "1" below is the starting ROW of your names list.
mySelect = Int((20 * Rnd) + 1)
'Note: The "A" below is the column where your names list is.
myName = Range("A" & mySelect)
'Put the answer in the next empty cell, bottom of list, one column Right.
'Make sure column "B" has a data lable for the list to start under!
Worksheets("Sheet1").Range("B65536").End(xlUp).Offset(1, 0) = myName
End Sub
Thank you very for your prompt reply. I'll try it and let you know.
This set of subs will random pick a name from a list of names. Hope it helps. JSW
Sub myRnd()
'Find a random name in a existing names list.
'By Joe Was, 6/27/2001.
Dim myOrder As Range
Dim myName
Dim mySelect As Variant
Randomize
'Note: The 20 below = the ending ROW of your names list.
' The 1 below = the starting ROW of your names list.
mySelect = Int((20 * Rnd) + 1)
'Note: The "A" below is the column where your names list is.
myName = Range("A" & mySelect)
'Put the answer in a cell
Worksheets("Sheet1").Range("C1") = myName
'Put the answer in a screen message box.
MsgBox "The selection is:" & Chr(13) & Chr(13) & myName
End Sub
Sub myErase()
Range("C1").Select
Selection.ClearContents
Range("C1").Select
End Sub
Sub myListR()
'Find a random name in a existing names list.
'Then adds it to an ongoing list.
'By Joe Was, 6/27/2001.
Dim myOrder As Range
Dim myName
Dim mySelect As Variant
Randomize
'Note: The "20" below is the ending ROW of your names list.
' The "1" below is the starting ROW of your names list.
mySelect = Int((20 * Rnd) + 1)
'Note: The "A" below is the column where your names list is.
myName = Range("A" & mySelect)
'Put the answer in the next empty cell, bottom of list, one column Right.
'Make sure column "B" has a data lable for the list to start under!
Worksheets("Sheet1").Range("B65536").End(xlUp).Offset(1, 0) = myName
End Sub
This code really works! the only thing what if I want the result in Sheet 2
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Go back | 1,757 | 6,020 | {"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} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.763399 |
http://stackoverflow.com/questions/5202591/game-on-the-tree-cutting-branch | 1,394,348,566,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999674993/warc/CC-MAIN-20140305060754-00037-ip-10-183-142-35.ec2.internal.warc.gz | 178,575,453 | 14,911 | # Game on the tree, cutting branch
We have a forest of rooted trees. Two players makes alternating moves according to the following rule: one move is to cut vertex and all its children. Player which makes last move (no vertices remain) wins.
How can we compute Grundy function for the positions in the game?
Suppose we have a trees and we need to say whether current position is winning or losing?
-
Sounds very difficult homework! – Paul Hadfield Mar 5 '11 at 8:32
I suggest you get a copy of "Winning Ways" en.wikipedia.org/wiki/Winning_Ways_for_your_Mathematical_Plays from your local library and work it out for yourself. – Doc Brown Mar 5 '11 at 8:38
I admit that this problem is computer-sciency, however it falls more in the mathematical-category than in the programming-category. So you might find more help here: > math.stackexchange.com – eznme Mar 5 '11 at 8:50
I hope this isn't for codechef.com/MARCH11/problems/SQUAGAME - because thats a currently running contest. – kyun Mar 5 '11 at 10:32
Isn't this just a variation of Nim? Each tree can be replaced by a heap with as many stones as the longest path in that tree. – Nick Johnson Mar 8 '11 at 1:59 | 300 | 1,172 | {"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} | 2.953125 | 3 | CC-MAIN-2014-10 | latest | en | 0.947704 |
https://solvedlib.com/n/using-the-kf-and-kb-equationsa-certain-substance-x-melts,8821324 | 1,696,380,038,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511284.37/warc/CC-MAIN-20231003224357-20231004014357-00025.warc.gz | 579,677,827 | 18,892 | # Using the Kf and Kb equationsA certain substance X melts at - temperature of -45 *C But if a 500.
###### Question:
Using the Kf and Kb equations A certain substance X melts at - temperature of -45 *C But if a 500. sample of X is prepared with 24.8 g of urea ((NHz)co) dissolved in It; the sample is found to have melting point of -8.7 %C instead_ Calculate the molal freezing point depression constant K; ofX Round your answer to 2 significant digits. "C * mol
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##### LOM DOlnCSFhe Tectangles having given area. Auch G7 Jugle luus iu: stallusl [ritttelcr;ratiun lerig"Int~idJcctauckc Gixaruslarlyle' vih eug ) Liuje": (he jc =
LOM DOlnCS Fhe Tectangles having given area. Auch G7 Jugle luus iu: stallusl [ritttelcr; rat iun lerig" Int ~id Jcctauckc Gixa ruslarlyle' vih eug ) Liuje": (he jc =... | 3,432 | 11,832 | {"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} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.849001 |
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A048943 Product of divisors of n is a square. 7
1, 6, 8, 10, 14, 15, 16, 21, 22, 24, 26, 27, 30, 33, 34, 35, 38, 39, 40, 42, 46, 51, 54, 55, 56, 57, 58, 60, 62, 65, 66, 69, 70, 72, 74, 77, 78, 81, 82, 84, 85, 86, 87, 88, 90, 91, 93, 94, 95, 96, 102, 104, 105, 106, 108, 110, 111, 114, 115, 118, 119, 120 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS From Gerard P. Michon, Oct 10 2010: (Start) If d is the number of divisors of N, a prime factor of N with multiplicity k in N has a multiplicity kd/2 in the product of all divisors of N (including N itself). Therefore the latter is a square if and only if kd/2 is always even (which is to say that kd is a multiple of 4 for any multiplicity k of a prime factor of N). This happens when one of the following three conditions hold: - N is a fourth power (all the multiplicities are then multiples of 4 and d is odd). - N has at least two prime factors with odd multiplicities. - N has (at least) one prime factor with a multiplicity congruent to 3 modulo 4. (End) LINKS T. D. Noe, Table of n, a(n) for n = 1..1000 G. P. Michon, Divisor Product, Numericana. Eric Weisstein's World of Mathematics, Divisor Product EXAMPLE From Gerard P. Michon, Oct 10 2010: (Start) a(1) = 1 because it's a fourth power. The product of all divisors of 1 is 1, which is a square. a(2) = 6 because 2^1.3^1 is the product of two primes with odd multiplicities (1 in both cases). Indeed, the divisor product 1.2.3.6 = 36 is a square. a(3) = 8 because 2 is a prime factor of 8 with multiplicity 3. Indeed, 1.2.4.8 = 64 is a square. a(7) = 16 because it's a fourth power; 1.2.4.8.16 = 1024 is the square of 32. (End) MATHEMATICA Select[Range[125], IntegerQ[Sqrt[Times @@ Divisors[#]]] &] (* T. D. Noe, Apr 30 2012 *) PROG (PARI) {for(k=1, 126, mpc=1; M=divisors(k); for(i=1, matsize(M)[2], mpc=mpc*M[i]); if(issquare(mpc), print1(k, ", ")))} \\\ Douglas Latimer, Apr 30 2012 (PARI) is(n)=my(f=factor(n)[, 2]); gcd(f)%4==0 || #select(k->k%2, f)>1 || #select(k->k%4==3, f) \\ Charles R Greathouse IV, Sep 18 2015 (Sage) [n for n in (1..125) if prod(divisors(n)).is_square()] # Giuseppe Coppoletta, Dec 16 2014 (Python) from sympy import divisor_count from gmpy2 import iroot A048943_list = [i for i in range(1, 10**3) if iroot(i, 4)[1] or not divisor_count(i) % 4] # Chai Wah Wu, Mar 10 2016 CROSSREFS Supersequence of A229153. Sequence in context: A181764 A153032 A086822 * A255429 A319238 A130763 Adjacent sequences: A048940 A048941 A048942 * A048944 A048945 A048946 KEYWORD nonn AUTHOR STATUS approved
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Last modified January 22 09:52 EST 2019. Contains 319363 sequences. (Running on oeis4.) | 1,084 | 3,028 | {"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} | 3.921875 | 4 | CC-MAIN-2019-04 | latest | en | 0.869741 |
https://edurev.in/course/quiz/attempt/15198_Technical-Mock-Test-CE--SSC-JE--16/88b91273-f51d-402b-9de5-599c9980056d | 1,652,822,238,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00227.warc.gz | 287,030,077 | 66,196 | # Technical Mock Test CE (SSC JE)- 16
## 200 Questions MCQ Test Mock Test Series of SSC JE Civil Engineering | Technical Mock Test CE (SSC JE)- 16
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Attempt Technical Mock Test CE (SSC JE)- 16 | 200 questions in 120 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Mock Test Series of SSC JE Civil Engineering for Civil Engineering (CE) Exam | Download free PDF with solutions
QUESTION: 1
Solution:
QUESTION: 2
### Select the related letters from the given alternatives. ZXVT : ACEG :: SQOM :?
Solution:
If letters are given the position value, the relation between the letters is as follows,
Thus SQOM is related to HJLN.
QUESTION: 3
### Select the related letters from the given alternatives. ADGJ : QTWZ : : BEHK : ?
Solution:
QUESTION: 4
In the following question find out the alternative which will replace the question mark.
BIRD : BEAK ∷ CUP : ?
Solution:
Birds collects grass with the help of beak to make nest. Similarly Cup is used to drink something with the help of lips.
Hence, the correct answer is LIP.
QUESTION: 5
Select the related words from the given alternatives.
Mount Everest : Asia :: Mont Blanc : ?
Solution:
Mount Everest is located in Asia.
Mont Blanc is located in Europe.
QUESTION: 6
Select the related numbers from the given alternatives:
12 : 39 :: 15 : ?
Solution:
As the sequence in this problem followed is:
⇒ (12 + 1) × 3 = 13 × 3 = 39
So in the same way the corresponding number for 15 is
⇒ (15 + 1) × 3 = 16 × 3 = 48
QUESTION: 7
Select the related number from the given alternatives.
3 : 243 :: 5 :?
Solution:
Given, 3 : 243 :: 5 :?
The relation is in the form of 35 → 243
Similarly, 55 → 3125
QUESTION: 8
If 2819 is related to 1, then 6432 is related to:
Solution:
1 is the tens digit of 2819 and the tens digit of 6432 is 3.
QUESTION: 9
Identify the word which belongs to the class of given words.
Grams, Kilograms, Quintal
Solution:
Grams, Kilograms, Quintal are unit of measuring Weight. Similarly Tonnes also used to measure Weight.
QUESTION: 10
Select the odd one out from the given alternatives.
Solution:
According to the given numbers, in all other numbers
Second number = (First number / 3) + 3
156 – 55 → (156 / 3) + 3 = 55
207 – 72 → (207 / 3) + 3 = 72
255 – 63 → (255 / 3) + 3 = 88
234 – 81 → (234 / 3) + 3 = 81
The number which does not follow the rule is 255 – 63.
Hence the answer is 255 – 63.
QUESTION: 11
Find the odd one out from the given alternatives.
Solution:
In all other numbers,
Third number → First number + second digit of second number
12, 26, 18 → 12 + 6 → 18
13, 19, 16 → 13 + 9 → 22
35, 78, 43 → 35 + 8 → 43
43, 27, 50 → 43 + 7 → 50
Hence, the correct answer is 13, 19, 16.
QUESTION: 12
Select the one which is different from the other three responses.
Solution:
69 = 23 × 3
230 = 23 × 10
115 = 23 × 5
163 = 23 × 7.08
All the numbers except 163 are multiples of 23.
QUESTION: 13
Select the odd letters from the given alternatives.
Solution:
All the other are consonants but only option 2) is vowels.
QUESTION: 14
Find the odd one out from the given alternatives.
Solution:
Here 1st letter + 7 = 2nd letter; 2nd letter – 6 = 3rd letter; 3rd letter + 7 = 4th letter
1. BICJ = B + 7 = I; I – 6 = C; C + 7 = J
2. CJDK = C + 7 = J; J – 6 = D; D + 7 = K
3. AIHB = A + 8 = I; I – 1 = H; H – 6 = B
4. DKEL = D + 7 = K; K – 6 = E; E + 7 = L
“AIHB” doesn’t follow the pattern, hence it is odd one out
QUESTION: 15
Select the one which is different from the other responses.
Solution:
A) ABD = A + 1 = B; B + 2 = D
B) GHJ = G + 1 = H; H + 2 = J
C) MPR = M + 3 = P; P + 2 = R
D) VWY = V + 1 = W; W + 2 = Y
Hence, MPR is odd one out.
QUESTION: 16
Find the odd word from the given alternatives.
Solution:
Hold is odd one as other words when read backwards give meaningful words
QUESTION: 17
In a certain code, BUILDER is written as JVCKSFE. How is SEALING written in that code?
Solution:
QUESTION: 18
If 1986 is coded as and 2345 is coded as ,then will be the code for
Solution:
Here, each digit has been allocated a unique symbol as shown below:
Thus,
Thus the required digits are 865234.
QUESTION: 19
In a certain code language, ‘light lamp live’ means ‘drum rum sum’; ‘pole high area’ means ‘din pin kin’; ‘area pole mark’ means ‘pin din fin’; and ‘area live red’ means ‘pin rum disk’. Which words in that language means ‘high’ and ‘red' respectively’?
Solution:
If we write the code,
light lamp live = drum rum sum
pole high area = din pin kin
area pole mark = pin din fin
area live red = pin rum disk
The words that are matching in the four statements have been marked.
From the above explanation it is clear that the code for ‘Red’ and ‘high’ is ‘disk’ and ‘kin’ respectively.
QUESTION: 20
A cyclist goes 30 km to North and then turning east he goes 40 km. Again he turns to his right and goes 20 km. after this, he turns to his right and goes 40 km. How far is he from his starting point?
Solution:
QUESTION: 21
Vijay starts walking straight towards East. After walking 75 meters he turns to the left and walks 25 meters straight. Again he turns to the left and walks a distance of 40 meters straight. Again he turns to the left and walks a distance of 25 meters. How far is he form the starting point?
Solution:
QUESTION: 22
In the given figure what will be the opposite surface of Δ?
Solution:
Observe the figures where ÷ is on the top i.e. figure 1, 3 and 4.
O and × are adjacent to each other.
From figure 4,
+ and O are adjacent to each other.
So the arrangement will be,
+ O ×
From figure 3, Δ and + are adjacent to each other.
So the arrangement becomes,
Δ + O ×
Hence O will be opposite to Δ as 1st and 3rd element will be opposite to each other.
QUESTION: 23
A piece of paper is folded and cut / punched as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Solution:
QUESTION: 24
Directions: In question, which answer figure will complete the pattern in the question figure?
Solution:
The correct option is (C) because it will complete the two concentric diamond shaped square and also it does not contain a circle similar to its diagonally opposite part.
QUESTION: 25
If a mirror is placed on the line XY, then which of the answer figure is the right image of the given figure?
Solution:
QUESTION: 26
Look carefully at the below given figure. The total number of triangles in the figure are:
Solution:
QUESTION: 27
A statement is given followed by two inferences I and II. You have to consider the statement to be true even if it seems to be at variance from commonly known facts. You have to decide which of the given inferences, if any, follow from the given statement.
Statement: Literature is a form of ugliness so intolerable that we have to alter it every six months.
Inferences: I. Authors do not understand the public mind very well.
II. The public, by and large, is highly susceptible to novelty.
Solution:
The statement asserts that people do not condone a particular trend for a long time and want to change it quite often. The first inference cannot be drawn as nothing about the authors’ mind set can be inferred. Thus only inference II follows because the public wants a change and wants something original all the time. Thus, literature is altered every six months to suit the taste of the public.
QUESTION: 28
In a village one half of the population works in the fields. Half of the people who do not work in fields are working in factories. Select the appropriate diagram in which shared region is best representing people working in factories.
Solution:
QUESTION: 29
A mother’s age is thrice that of her daughter’s age. Six years ago, the mother’s age was five times that of the daughter’s age. What was the daughter’s age 6 years ago?
Solution:
Let the daughter’s age six years ago be ‘y’.
Then mother’s age six years ago = 5 × y
Daughter’s age at present = y + 6
Mothers age at present = (5 × y) + 6
According to the question, mother’s age at present = (5 × y) + 6 = 3 × (y + 6)
Solving, we have:
y = 6
Hence the daughter’s age six years ago was 6 years.
QUESTION: 30
Which of the following interchange of signs would make the given equation correct?
5 + 3 × 8 – 12 ÷ 4 = 3
Solution:
Given equation: 5 + 3 × 8 – 12 ÷ 4
1) Interchanging + and ÷ ⇒ 5 ÷ 3 × 8 – 12 + 4 ⇒ 13.33 – 12 + 4 = 5.33 ≠ 3
2) Interchanging + and – ⇒ 5 – 3 × 8 + 12 ÷ 4 ⇒ 5 – 24 + 3 = -16 ≠ 3
3) Interchanging – and ÷ ⇒ 5 + 3 × 8 ÷ 12 – 4 ⇒ 5 + 2 – 4 = 3
4) Interchanging + and × ⇒ 5 × 3 + 8 – 12 ÷ 4 ⇒ 15 + 8 – 3 = 23 – 3 = 20 ≠ 3
Hence interchanging – and ÷ we get the equation correct.
QUESTION: 31
Select the correct combination of mathematical signs to replace ‘*’ signs and to balance the given equation.
8 * 6 * 7 * 3 * 12
Solution:
Rewriting each option by replacing blanks by given symbols and finding the correct combination, we get
1) 8+6-7÷3 = 12 is not correct as L.H.S. is equal to 14 – 2.33 = 11.67
2) 8+6-7+3 = 12 is not correct as L.H.S. is equal to 10.
3) 8+6+7-3 = 12 is not correct as L.H.S. is equal to 18.
4) 8-6+7+3 = 12 is correct as L.H.S. is equal to R.H.S.
Hence answer is - + + =.
QUESTION: 32
Directions: In question, select the missing number from the given responses.
2 6 18
4 10 30
3 8 ?
Solution:
Logic:-
→ (1st Column × 2) + 2 = 2nd Column; (2nd Column × 2) + 2nd Column = 3rd Column
→ (2 × 2) + 2 = 6 , (6 × 2) + 6 = 18
→ (4 × 2) + 2 = 10 , (10 × 2) + 10 = 30
→ (3 × 2) + 2 = 8 , (8 × 2) + 8 = 24
So, ? = 24.
Hence, Option (A).
QUESTION: 33
Directions: Which one number can be placed at the sign of interrogation? (Read column wise).
Solution:
Looking at alternate terms and in anti clockwise pattern, we see:-
?, 21, 23, 25
We see that there is a difference of 2 between 2 elemets.
By that logic, ? = 19.
QUESTION: 34
Identify the diagram that best represents the relationship among the given classes.
Earth, Saturn, Planet, Star
Solution:
QUESTION: 35
Statements:
1. All men are bachelors.
2. Some bachelors are teachers.
Conclusions:
I. All men are teachers.
II. Some men are bachelors.
III. Some teachers are not bachelors.
Solution:
QUESTION: 36
The diagram represents Teachers, Singers and Players. Study the diagram and find out how many teachers are also singers.
Solution:
QUESTION: 37
A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as shown in the given two matrices. The columns and rows of Matrix-I are numbered from 0 to 4 and that of Matrix – ll are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, for example, 'B' can he represented by 21, 32 etc., and 'W' can be represented by 97, 89 etc. Similarly, you have to identify the set for the word 'SAINT’.
Solution:
1) 00, 56, 20, 68, 76 ⇒ S, A, I, N, T
2) 11, 67, 31, 57, 12 ⇒ S, A, I, N, F
3) 22, 78, 20, 02, 44 ⇒ S, A, I, K, C
4) 33, 67, 42, 00, 99 ⇒ S, A, I, S, O
Therefore, “00, 56, 20, 68, 76” are set of number which represent Word ‘SAINT’
QUESTION: 38
Which one of the given responses would be a meaningful order of the following?
a) Stem b) Leaf c) Flower d) Branch e) Fruit
Solution:
The correct order is:
a. Stem
d. Branch
b. Leaf
c. Flower
e. Fruit
The above being the logical order in which they grow.
Hence, Correct order is a, d, b, c, e.
QUESTION: 39
Which of the following word will come in the middle when they are arranged alphabetically as per in a dictionary?
1. Tennis
2. Tendon
3. Tender
4. Tempest
5. Terminal
Solution:
Words as per dictionary will be arranged as follows:
Tempest
Tender
Tendon
Tennis
Terminal
Hence the word in middle is Tendon.
QUESTION: 40
Rearrange the jumbled letters to make a meaningful word and then select the one which is different.
Solution:
After rearranging the jumbled letters,
SUVNE = VENUS
APTLE = PLATE
ARMS = MARS
RUJTIPE = JUPITER
Among these, except one, three words have same category. They all are planets.
Only PLATE is different.
Hence, (APTLE) is true.
QUESTION: 41
Directions: Find the odd letters from the given alternatives.
Solution:
Interchanging the first two letters we get.
IHJ → HIJ
ONP → NOP
STR → TSR
LKM → KLM
After interchanging the letters, all results in three consecutive letters except STR.
QUESTION: 42
Direction: Select the one which is different from the other three responses.
Solution:
Here all except Shark are Mammals. Shark is a Fish.
QUESTION: 43
Directions: In question, find the odd word from the given alternatives.
Solution:
From the given options, Punishment is odd one because Reward, Praise, Encourage are the ways to expressing admiration & to boost up one to do better performance.
But Punishment is the imposition of penalty or rough treatment.
Thus Punishment is odd one.
QUESTION: 44
Directions: In the given question select the one which is different from the other three alternatives.
Solution:
Avoid: keep away from
Dodge: avoid by a quick movement
Flee: run away from a place or situation of danger
Duck: avoid by moving quickly
From the meaning of the above words we see that flee is the odd one out.
QUESTION: 45
In the following question find the odd word out from the given responses.
Solution:
‘Monetary’, ‘pecuniary’, ‘remuneration’ are all related to money. Only ‘gold’ is unrelated. ‘Gold’ is the precious metal. While gold adds value to a person’s finances, it is not directly related to money.
QUESTION: 46
Directions: Find the odd word from the given alternatives.
Solution:
Prize and Award : Given for distinction or good performance.
Gift: Given without in exchange for anything.
Charity: An activity or gift that benefits as form of good deed.
So, Charity is odd-one out.
QUESTION: 47
Select the odd number from the given alternatives.
Solution:
211, 283, and 277 is a prime number but 287 does not a prime number therefore 287 is odd one out.
QUESTION: 48
Select the odd number from the given alternatives.
Solution:
1) 1356 ⇒ 1 + 3 + 5 + 6 = 15
2) 5497 ⇒ 5 + 4 + 9 + 7 = 25
3) 8764 ⇒ 8 + 7 + 6 + 4 = 25
4) 9943 ⇒ 9 + 9 + 4 + 3 = 25
Therefore, 1356 is odd one among given option.
QUESTION: 49
Select the odd number from the given alternatives
Solution:
1) 1716 ⇒ 13 × 12 × 11 = 1716
2) 2730 ⇒ 15 × 14 × 13 = 2730
3) 3360 ⇒ 16 × 15 × 14 = 3360
4) 4086 ⇒ 17 × 16 × 15 = 4080
Here as we can see in above explanation 1716, 2730 and 3360 are in form of
⇒ a(n) = n × (n–1) × (n–2)
But 4086 does not follow this equation.
Therefore, 4086 is odd one out.
QUESTION: 50
A piece of paper is folded and punched as shown in the figure below. Find the answer figure which shows the question figure being un-folded:
Question figures:
Solution:
QUESTION: 51
Match List-I with List-II and select the correct answer from the code given below:
List – I (Biosphere Reserve)
A. Dehang-Debang
B. Manas
C. Nokrek
D. Similipal
List – II (State)
1. Odisha
2. Meghalaya
3. Assam
Solution:
QUESTION: 52
Solution:
• Environmental degradation is the deterioration of the environment through depletion of resources, the destruction of ecosystems, habitat destruction, the extinction of wildlife, and pollution.
• Thus it means overall lowering of environmental qualities
• It also means adverse change brought by human activities
• And it causes ecological imbalance.
QUESTION: 53
What is the full form of VPN?
Solution:
The Virtual Private Network is setup for a certain group of users like the users in one office or of a company etc.
QUESTION: 54
The designers of the Internet Protocol defined an IP address as a ____ bit number.
Solution:
An Internet Protocol address (IP address) is a numerical label assigned to each device connected to a computer network that uses the Internet Protocol for communication. It is a 32 bit number.
QUESTION: 55
An HTTP request contains how many parts?
Solution:
An HTTP request contains 4 parts. Which are Request Method, URL, Header Fields, and Body
QUESTION: 56
Name the hottest planet.
Solution:
Mercury is the planet that is closest to the sun and therefore gets more direct heat, but even it isn't the hottest. Venus is the second planet from the sun and has a temperature that is maintained at 462 degrees Celsius, no matter where you go on the planet. It is the hottest planet in the solar system.
QUESTION: 57
Which base is present in lime water?
Solution:
Limewater is the common name for a diluted solution of calcium hydroxide. Limewater is prepared by stirring calcium hydroxide in pure water and filtering off the excess undissolved Ca(OH)2. It is basic in nature.
QUESTION: 58
When Sodium (Na), Copper (Cu) and Zinc (Zn) are placed in the order of decreasing reactivity, then their order would be-
Solution:
Reactivity is defined as rate at which a chemical substance tends to undergo a chemical reaction.
The order of reactivity is known as reactivity series, in which reactivity decreases from top to bottom.
In reactivity series copper, gold known as structural metals are at bottom where sodium, potassium are at top of series.
Reactivity series -
K > NA > Ca > Mg > Al > C > Zn > Fe > Sn > Pb > H > Cu > Ag > Au > Pt.
QUESTION: 59
Which of the following is an important ore of zinc?
Solution:
Calamine is an ore of zinc. Galena is an ore of lead. Siderite is an ore of iron and malachite is an ore of copper. An ore is a body of rock which contains a high enough concentration of a metallic or non-metallic resource.
QUESTION: 60
________ is the hardest non metal.
Solution:
Diamond is the hardest non-metal. The hardness of the diamond depends on the purity, perfection and alignment of its atoms in the crystalline structure.
QUESTION: 61
Who invented Crescograph?
Solution:
Crescograph was invented by Jagdish Chandra Bose in the early 20th century. It is a device for measuring growth in plants.
QUESTION: 62
Which one of the following is not a property of electric field lines?
Solution:
Each location in space has its own specific value of electric field strength and direction. If two electric field lines were to intersect at a point, it would mean that there are two different values of electric field at that single point, each with its own direction. This is not possible.
QUESTION: 63
A fountain pen works on which of the following principle?
Solution:
A fountain pen is a nib pen that draws ink from the reservoir through a feed to the nib and deposits it on paper via a combination of gravity and capillary action.
QUESTION: 64
Spectacles used for viewing 3D films have -
Solution:
A polarized 3D system (Polaroids) uses polarization glasses to create the illusion of three-dimensional images by restricting the light that reaches each eye.
QUESTION: 65
Which colour light would travel the fastest in the vacuum?
Solution:
Light travels at same speed across all media and in all variants. What differs is the distance which it travels depending upon the wavelength of the colour.
QUESTION: 66
Which of the following diseases is also known as varicella?
Solution:
Chicken pox is an infectious disease causing a mild fever and a rash of itchy inflamed pimples which turn to blisters. It is caused by the herpes zoster virus and mainly affects children.
QUESTION: 67
Which of the following rulers was not involved in the Battles of Panipat?
Solution:
Babur was involved in the first battle of Panipat where he defeated Sikander Lodi. Second battle was fought between Akbar and Hemu where Hemu got defeated. Third battle saw the victory of Ahmed Shah Abdali over Marathas.
QUESTION: 68
Hindu Marriage Act was brought in
Solution:
The Hindu Marriage Act is an Act of the Parliament of India in 1955 as part of the Hindu Code Bills. Three other important acts were also enacted during this time: the Hindu Succession Act (1956), the Hindu Minority and Guardianship Act (1956), the Hindu Adoptions and Maintenance Act (1956).
QUESTION: 69
Doha Round is related to which of the following bodies?
Solution:
Doha Round is the current trade-negotiation round of the World Trade Organization (WTO) which commenced in November 2001.
QUESTION: 70
Right to Education is prescribed under which article of the constitution?
Solution:
RTE provides for free and compulsory education for children between 6 and 14 in India under Article 21A of the Indian Constitution.
QUESTION: 71
The provision of Directive Principles of State Policy has been borrowed from which of the following countries?
Solution:
Many parts of Indian constitution were borrowed from various countries. DPSPs were borrowed from Ireland
QUESTION: 72
National Girl Child day is celebrated every year on ________.
Solution:
National girl child day is celebrated every year on 24 January as a national observance day for the girl child.
QUESTION: 73
Marx believed in which of the following ideologies?
Solution:
Marx professed a future where capitalism would be abolished by a revolution brought upon by the workers of the world. In the process, Socialism would be established which would then be succeeded by Communism.
QUESTION: 74
The Russian revolution established the supremacy of which of the following ideologies?
Solution:
The Russian Revolution is the collective term for a pair of revolutions in Russia in 1917, which dismantled the Tsarist autocracy and led to the eventual rise of the Soviet Union.
QUESTION: 75
What is the main cause of Anemia?
Solution:
Anemia is a condition in which there is a deficiency of iron in the blood, resulting in pallor and weariness.
QUESTION: 76
The full form of SIM is:
Solution:
SIM or Subscriber Identity Module card is a portable memory chip that is found inside the cell phones.
QUESTION: 77
Photosynthetic vesicle that is found in bacteria is called
Solution:
Chromatophores are pigment-containing and light-reflecting cells, or groups of cells, found in a wide range of animals including amphibians, fish, reptiles, crustaceans and cephalopods.
QUESTION: 78
What is a Cultigen?
Solution:
A cultigen is a plant that has been deliberately altered or selected by humans; it is the result of artificial selection.
QUESTION: 79
Which type of mirror is used as rear view mirror?
Solution:
Convex mirrors are used because they provide a larger field of view than a plane mirror. For example, the passenger-side rear view mirror on a car is convex.
QUESTION: 80
Momentum is a product of
Solution:
Momentum is the product of the mass and velocity of an object.
QUESTION: 81
Optical Fibre generally work on the principle of ________.
Solution:
A fibre optic cable consists of a bundle of glass threads, each of which is capable of transmitting messages modulated onto light waves. Fibre optic is based on the principle of total internal reflection.
QUESTION: 82
Which of the following determines the luster of metals?
Solution:
When light shines on a lustrous metal, the loosely bound electrons reflect incoming light, giving the metal a shiny appearance.
QUESTION: 83
Cement is a result of which of the following?
Solution:
Cement is a binder substance that sets and hardens and can bind other materials together. It is formed out of the chemical reaction between Limestone and clay.
QUESTION: 84
Iodoform is a type of
Solution:
Iodoform is the organoiodine compound with the formula CHI₃. It has a pale yellow, crystalline, volatile substance, it has a penetrating and distinctive smell.
QUESTION: 85
Kuiper belt is located between which of the following planets?
Solution:
Kuiper belt is a region of the solar system beyond the orbit of Neptune, believed to contain many comets, asteroids, and other small bodies made largely of ice.
QUESTION: 86
Which of the following nutrients is not produced inside an alcoholic’s body?
Solution:
Vitamin K is a group of structurally similar, fat-soluble vitamins the human body requires for complete synthesis of certain proteins that are prerequisites for blood coagulation that the body needs for controlling binding of calcium in bones and other tissues.
QUESTION: 87
Syphilis is a type of
Solution:
Syphilis is a chronic bacterial disease that is contracted chiefly by infection during sexual intercourse.
QUESTION: 88
In which of the following years was NABARD formed?
Solution:
National Bank for Agriculture and Rural Development is an apex development bank in India, having headquarters in Mumbai and other branches all over the country. It was formed in 1982.
QUESTION: 89
Which of the following articles prohibits discrimination on the grounds of religion, race, sex, caste?
Solution:
Article 16 provides for Equality of opportunity in matters of public employment. Article 17 provides for Abolition of Untouchability and Article 18 provides for Abolition of titles.
QUESTION: 90
Which of the following article is related to the Vice President of India?
Solution:
Article 63 of the Constitution says that there shall be a Vice President of India.
QUESTION: 91
India held its first general elections in
Solution:
The Indian general election of 1951–52 elected the first Lok Sabha since India became independent in August 1947. The Indian National Congress (INC) won a landslide victory, winning 364 of the 489 seats and 45% of the total votes polled.
QUESTION: 92
According to which article of the Indian Constitution is the state of Jammu and Kashmir given a special constitutional status?
Solution:
Article 370 of the Indian Constitution grants a special constitutional status to the state of Jammu and Kashmir. As a part of this, the laws passed by the Indian constitution apply to the state of J&K only if they are passed by the J&K legislature.
QUESTION: 93
Which of the given rights was described by Dr. B.R. Ambedkar as the “very soul of the Indian Constitution”?
Solution:
Article 32 of the Indian Constitution which deals with the right to constitutional remedies against the violation of human rights was described by Dr. B.R. Ambedkar as the very soul of the constitution.
QUESTION: 94
The tissue which connects bone to bone is called
Solution:
Ligament is the connected tissue which connects bone to bone. Tendon connects muscle to bone and fasciae connects muscle to muscle.
QUESTION: 95
Which of the following scripts is said to have appeared first in Indian writings?
Solution:
The Brahmi script is one of the most important writing systems in the world by virtue of its time depth and influence. It represents the earliest post-Indus corpus of texts, and some of the earliest historical inscriptions found in India.
QUESTION: 96
The book ‘Indica’ which is related to india was written by?
Solution:
Megasthenes gave an account of India in his book Indica.
QUESTION: 97
Fathometer is used to measure:
Solution:
Fathometer is an instrument that works on the echo system and used to measure the depth of water in an ocean.
QUESTION: 98
Which among the following is popularly known as the laughing gas?
Solution:
Nitrous oxide or N2O, a transparent gas is called laughing gas. The gas is used as an anaesthetic, in cans as well as an oxidizing agent for racing cars.
QUESTION: 99
Which of the following is a type of storage device?
Solution:
Blu-ray disc is a digital optical disc data storage format. It was designed to supersede the DVD format, in that it is capable of storing high-definition video resolution (1080p).
QUESTION: 100
Who has won the Nobel Prize 2017 for Medicine or Physiology?
Solution:
Michael W. Yong, Michael Rosbash and Jeffrey C. Hall have won the Nobel Prize 2017 in Medicine or Physiology for their discoveries of the molecular mechanisms controlling the circadian rhythm. The Nobel Prize in medicine is decided by the Karolinska institute, a medical centre in Sweden.
QUESTION: 101
In plastic analysis, the shape factor for a triangular section is
Solution:
QUESTION: 102
The maximum pitch for rivets or welds in tension zone are:
Solution:
Maximum pitch for rivets or welds in tension zone are 16t or 200 mm whichever is less as per IS 800: 1984 but for rivets in compression zone it is 12t or 200 mm whichever is less.
QUESTION: 103
For single lacing the thickness of flat lacing bars shall not be less than _____times of effective length.
Solution:
Minimum thickness of lacing bar
tmin = l/40 for single lacing
tmin = l/60 for double lacing
Where l = effective length of lacing member
QUESTION: 104
Which of the following shows the CORRECT expression for target mean strength (fcm) of concrete, if the characteristic strength and standard deviation is given by fck and σ respectively?
Solution:
The correct expression as per IS 456 : 2000 to find Target mean strength is :
fcm = fck + 1.65 σ
1.65 is generally a risk factor.
QUESTION: 105
Degree of workability required as per IS 456 : 2000 for lightly Reinforced sections in slabs, beams, walls, columns is
Solution:
As per Art. 7.1 of IS 456 : 2000, For Construction of lightly reinforced sections in slabs, beams, walls etc. degree of workability required is low. Whereas far heavily reinforced sections in slabs, beams etc Degree of work ability required is Medium.
QUESTION: 106
A Pitot-static tube (C = 1) is used to measure air flow. With water in the differential manometer and a gauge difference of 75 mm, what is the value of air speed if ρ = 1.16 kg/m3?
Solution:
QUESTION: 107
The minimum cover for bars in RCC slabs should be:
Solution:
As per IS 456 : 2000, The minimum cover for bars in RCC slabs should be limited to:
• 15 mm
• Diameter of bars
(whichever is maximum).
QUESTION: 108
The anchorage value of a hook is assumed 16 times the diameter of the bar if the angle of the bend is
Solution:
The anchorage value is 4 times the diameter for every 45o bend. So for 16 times the diameter the bend is 180o
QUESTION: 109
The assumption that the plane sections normal before bending remain normal after bending is used
Solution:
The assumption that plane sections normal before and after bending remain same is used in both working stress and limit sate method. For this the strain diagram is linear in both working stress and limit state method.
QUESTION: 110
Under service loads the crack width in concrete should not exceed under mild exposure as per IS 456 is:
Solution:
According to IS: 456-2000, the maximum allowable crack width (in mm) for mild type of environmental condition is 0.3 mm.
QUESTION: 111
The chances for the actual strength of concrete to be less than its characteristic strength are
Solution:
The characteristic strength is defined as the strength of the concrete below which not more than 5% of the test results are expected to fall.
QUESTION: 112
In a simply supported beam carrying uniformly load, the spacing of vertical stirrups near supports would be
Solution:
In simply supported beam near the support the shear force is maximum. As the vertical stirrups are provided to resist the vertical shear, near the support the stirrups are closer.
QUESTION: 113
The maximum diameter of the reinforcement bars in R.C.C. slabs is
Solution:
The maximum diameter of bar used in slab should not exceed 1/8 of the total thickness of slab.
QUESTION: 114
Thickened part of a flat slab over its supporting column, is technically known as _______.
Solution:
QUESTION: 115
Stress-strain curve of concrete is
Solution:
QUESTION: 116
The pitch of the main bars in a simply supported slab should not exceed its effective depth by ________.
Solution:
The pitch of the main bars in a simply supported slab should not exceed three times the effective depth of slab. But for distribution bars it is 5 times the effective depth.
QUESTION: 117
According to Witney's theory, the maximum depth of concrete stress block in a balanced RCC beam section of depth 'd' is ________.
Solution:
Witney’s theory is ultimate load theory. Witney replaced the actual parabolic stress diagram by a rectangular stress diagram such that the C.G of both the diagrams lies at same point and their areas are also equal. The maximum depth of concrete stress block in a balanced RCC beam is 0.53 d.
QUESTION: 118
A reinforced cantilever beam of span 4 m has a cross section of 150x500 mm. If checked for lateral stability and deflection, the beam will ________.
Solution:
Check for deflection,
For catilever beam permissible value is 7, so the beam is not safe in deflection.
Check for lateral stability,
l≯25b=25×150=3750mm=3.75m
For lateral stability length ≯ 3.75 m
So the beam is not laterally stable.
Option 3 is correct.
QUESTION: 119
The limits of percentage 'p' of the longitudinal reinforcement in a column are given by _________.
Solution:
As per IS 456 : 2000, the minimum percentage of steel is 0.8% and maximum percentage of steel is 6%. Although as per practical consideration the maximum percentage of steel is limited to 4% only.
QUESTION: 120
The process of adding water to lime to convert it into a hydrated lime is termed as:
Solution:
Slaking is the process in which quick lime reacts with water, during this reaction, it swells, cracks, and falls out as calcium hydroxide.
QUESTION: 121
If aggregates completely pass through a sieve of size 75 mm and are retained on a sieve of size 60 mm, the aggregates will be known as elongated aggregate if its length is not less than
Solution:
An aggregate is said to be elongated if its greatest dimension is greater than 1.8 times the mean diameter.
Length of aggregate ≮ 1.8 × 67.5
= 121.5 mm
QUESTION: 122
Which of the following compound affects the ultimate strength of cement?
Solution:
Dicalcium silicate is the last compound that is formed after the addition of water in the cement which may require a year or so for its formation. It is responsible for the progressive strength of cement. While Tricalcium silicate is responsible for the early development of strength. Tricalcium aluminate is responsible for the maximum evolution of heat of hydration.
QUESTION: 123
Which of the following is the correct reason for soaking the brick in water before its use?
Solution:
Soaking of bricks is water before its use is done because wetting of bricks removes dirt, sand and dust. Soaked bricks prevents suction of water from the wet mortar. If dry bricks are used then it will absorb water from mortar and mortar will become, dry and cannot attain any strength. The main reason for soaking the brick is to prevent adsorption of moisture from mortar by bricks.
QUESTION: 124
The defect in timber that arises due to the swelling caused by growth of layers of sap wood over the wounds after branch is cut off is called as ________.
Solution:
Overgrowth of Timber in some parts of a tree may result in some typical defects. Rind gall are quite common. A rind gall is simply a highly thickened, enlarged wood cover developed over an injured part of the tree. The meaning of rind is bark and gall is abnormal growth. So abnormal growth of the bark of the trees is called Rind galls. Improper cutting of branches causes this abnormal growth
Shakes are defects that occur around the annual ring or growth ring of timber.
QUESTION: 125
The defect that is caused by falling of rain water on the hot surfaces of the bricks is known as ________.
Solution:
Defects in Bricks are:
1. Bloating: This defect observed as spongy swollen mass over the surface of burned bricks is caused due to presence of excess carbonaceous matter and sulphur in brick clay.
2. Chuffs: The deformation of the shape of bricks caused by the rain water falling on hot bricks is known as chuffs.
3. Laminations: These are caused by the entrapped air in the voids of clay. Laminations produce thin lamina on the brick faces which weather out on exposure. Such bricks are weak in structure.
QUESTION: 126
The optimum number of revolutions in which concrete is required to be mixed in a mixer machine is
Solution:
To make concrete economical, mixing of concrete is performed in a machine mixer that leads to 10% saving in quantity of cement and the optimum number of revolutions in which concrete is required to be mixed in a mixer machine is 20 revolutions in 2.5 or 3 minutes.
QUESTION: 127
Which among the following is the defect in timber due to conversion
1. Diagonal Grain
2. Wane
3. Torn grain
4. Blue stain
5. Cup
6. Dry rot
Solution:
Defects due to conversion are: Chip mark, Torn grain, Diagonal grain, wane
Defects due to Fungi are: Heart Rot, Brown Rot, White Rot, Dry rot, wet rot, sap stain, blue stain
Defects due to Natural forces are: cup shakes, Ring shakes, heart shakes, radial shakes
Defects due to seasoning are: Bow, warp, cup and twist
QUESTION: 128
which among the following is a correct sequence for the preparation of clay in the manufacturing of bricks
Solution:
In the preparation of clay, first Unsoiling of clay is done in which top 200 mm soil is thrown out as it contains most of the impurities in it, then digging is done in which soil is dug out and is thrown on a level field to make it ready for the next operation of cleaning then cleaning of impurities is performed, after cleaning, cleaned clay is exposed to the atmosphere for few weeks to few months in order to carry out its softening, Ripening and mellowing etc then ingredients of the brick earth is spread over the clay in required proportion and at last tempering is done in a pug mill.
QUESTION: 129
Which IS code gives specifications about cement plaster?
Solution:
IS 1661: 1972 = Code of Practice for application of cement and cement-lime plaster finishes.
IS 1261: 1956 = Code of Practice for seam welding in mild steel.
QUESTION: 130
Properties of concrete can broadly be divided into:
Solution:
Properties of concrete can broadly divided in to two categories:
(i) Properties of hardened Concrete
(ii) Properties of fresh concrete
QUESTION: 131
Calculate the number of bricks in 20 cubic metres brick works.
Solution:
No of Brinks required in 1 m3 of brickwork
No. of Bricks required in 20m3 of brickwork is = 20 × 500 = 10000
QUESTION: 132
Which IS code gives details regarding water to be used in concrete?
Solution:
Water content and water cement ratio related provisions are given in IS 456 : 2000
IS 383 → specification for Coarse and fine aggregates.
QUESTION: 133
The Black marble is found in _______ district
Solution:
Generally white marble is found in Jabalpur district (M.P.) and Black marble in Jaipur.
QUESTION: 134
Which of the following is the most fire-resistant paints?
Solution:
Asbestos paints are used in residences, schools, and other public buildings as well as the Commercial establishment and maritime vessels. Asbestos paint is an inexpensive, efficient and more fire-resistant paint which has been widely used in many applications for decades.
Enamel paint is paint that air dries to a hard, usually glossy finish. Used for coating surfaces that are outdoors or otherwise subjected to hard wear or variations in temperature, it should not be confused with decorated objects in “painted enamel”.
QUESTION: 135
In which of the following pairs of trees, both trees yield soft wood?
Solution:
Chir and Deodar both yield soft wood hence are called conifers trees, as they possess distinct annular rings but indistinct medullary rays. While sal, Teak, Shishum are Decideous trees which yield hard woods.
QUESTION: 136
Inner part of a timber log surrounding the pith, is called
Solution:
QUESTION: 137
The main constituent of cement which is responsible for the initial setting of cement is
Solution:
The initial setting of Portland cement is due to tricalcium aluminate. Tricalcium silicate hydrates quickly and contributes more to the early strength. The contribution of dicalcium silicate takes place after 7 days and may continue for up to 1 year. Tricalcium aluminate hydrates quickly, and generates much heat and makes only a small contribution to the strength within the first 24 hours. Tetracalcium alumino-ferrite is comparatively inactive.
QUESTION: 138
The slump recommended for mass concrete is about
Solution:
The slump recommended for mass concrete is about 25 mm to 50 mm.
QUESTION: 139
The purpose of pointing is
Solution:
Pointing, in building maintenance, the technique of repairing mortar joints between bricks or other masonry elements. When aging mortar joints crack and disintegrate, the defective mortar is removed by hand or power tool and replaced with fresh mortar, preferably of the same composition as the original. All of the above statements are correct.
QUESTION: 140
Loss on ignition in cement should not exceed
Solution:
Loss on ignition is calculated by heating up a cement sample to 900 – 1000°C (1650 – 1830°F) until a constant weight is obtained. The weight loss of the sample due to heating is then determined. A high loss on ignition can indicate pre-hydration and carbonation, which may be caused by improper and prolonged storage or adulteration during transport. Loss on ignition in cement should not exceed 5%
QUESTION: 141
The process of heating the limestone to redness in contact with air is termed as...
Solution:
Calcination or calcining is a thermal treatment process to bring about a thermal decomposition. The process takes place below the melting point of the product. The name calcination is derived from the Latin word ‘Calcinare’ which mean to burn lime.
QUESTION: 142
A quick-setting cement has an initial setting time of about
Solution:
The initial setting time can be defined as the time taken by paste to stiffen to such an extent that the Vicatt’s needle is not permitted to move down through the paste through 25 mm. With Portland cement and rapid hardening cement, the normal initial setting time should not be less than 30 minutes and the final setting time should not be more than 10 hours. With a quick setting cement, the initial setting time should not be less than 5 minutes and the final setting time should not be more than 30 minutes.
QUESTION: 143
The total length of a stair in a horizontal plane is known as -
Solution:
Run is the total length of stair in a horizontal plane, including landings.
Flight – It is a series of steps without any platform or landing or breaks in their direction.
Riser – It is the vertical portion of a step providing a support to the tread.
QUESTION: 144
The percentage of water for normal consistency is _____.
Solution:
The percentage of water for normal consistency is 20% to 30% .
QUESTION: 145
What will be the value of Poisson’s ratio, it the elasticity and rigidity of the material is 200 GPa and 66.67 GPa?
Solution:
E = 2G (1 + μ)
200 = 2 × 66.67 (1 + μ)
μ = 0.5
QUESTION: 146
In case of biaxial stress (tensile), the maximum value of shear stress is given by
Solution:
In case of biaxial stress,
i.e. It is the magnitude of difference between their principal stresses divided by 2.
QUESTION: 147
If the moment of inertia of a section about its axis is l and its effective sectional area is A, its radius of gyration r about the axis is
Solution:
Moment of Inertia = Ar2
Where A = Area of cross section
QUESTION: 148
The maximum deflection due to a load W at the free end of a cantilever of length L and having flexural rigidity EI is
Solution:
QUESTION: 149
The resultant of two forces each equal to P and acting at right angles is:
Solution:
QUESTION: 150
Moment of Inertia for a circular section has 20 cm diameter, is:
Solution:
QUESTION: 151
The property of the material to regain its original shape after deformation when the external forces are removed is ________.
Solution:
Elasticity is the property of a material to regain its original shape after deformation when the external forces are removed. All materials are plastic to some extent but the degree varies, for example, both mild steel and rubber are elastic materials but steel is more elastic than rubber.
Plasticity of a material is its ability to undergo some degree of permanent deformation without rupture or failure. This property is important in forming, shaping, extruding and many other hot and cold working processes.
Durability is defined as the ability of a product to perform its required function over a lengthy period under normal conditions of use without excessive expenditure on maintenance or repair.
QUESTION: 152
The equivalent length of the column when both the ends are fixed is ________.
Solution:
QUESTION: 153
The total area under the stress-strain curve of a mild steel specimen tested up to failure under tension is a measure of its:
Solution:
Strength is defined as the ability of the material to resist, without rupture, external forces causing various types of stresses. Breaking strength is the ability of a material to withstand a pulling or tensile force.
Toughness is defined as the ability of the material to absorb energy before fracture takes place. In other words, toughness is the energy for failure by fracture. Toughness is measured by a quantity called modulus of toughness. Modulus of toughness is the total area under a stress-strain curve in tension test, which also represents the work done to fracture the specimen.
Hardness is defined as the resistance of a material to penetration or permanent deformation. It usually indicates resistance to abrasion, scratching, cutting or shaping.
Stiffness or rigidity is defined as the ability of the material to resist deformation under the action of external load. Modulus of elasticity is the measure of stiffness.
QUESTION: 154
The neutral axis of a beam is subjected to _________ stress.
Solution:
The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains. If the section is symmetric, isotropic and is not curved before a bend occurs, then the neutral axis is at the geometric centroid. All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.
QUESTION: 155
In the cross-section of a rectangular beam, what is the ratio of the average shear stress to the maximum shear stress?
Solution:
Average shear stress
Maximum shear stress
QUESTION: 156
A shaft turns at 150 rpm under a torque of 100 Nm. Power transmitted is
Solution:
Power transmitted is given by
= 500 π W = 0.5 π kW
QUESTION: 157
Which of the following is not the displacement method?
Solution:
Force / Flexibility / Displacement / Stiffness /
Compatibility Method Equilibrium Method
(i) Virtual work/Unit load method (i) Slope deflection method
(ii) Method of consistent deformation (ii) Moment distribution method
(iii) Three-moment theorem (iii) Minimum potential energy method
(iv) Column analogy method
(v) Elastic center method
(vi) Castigliano’s theorem of minimum strain energy
(vii) Maxwell-Mohr equation
Difference between Force Method and Displacement Method
QUESTION: 158
The water utilizable by plants is available in the form of:
Solution:
Gravitational water: Gravitational water occupies the larger soil pores (macropores) and moves down readily under the force of gravity. Water in excess of the field capacity is termed gravitational water. Gravitational water is of no use to plants because it occupies the larger pores.
Hygroscopic water: The water that held tightly on the surface of the soil colloidal particle is known as hygroscopic water. It is essentially non-liquid and moves primarily in the vapour form. Hygroscopic water held so tenaciously by soil particles that plants cannot absorb it.
Capillary water: Capillary water is held in the capillary pores (micropores). Capillary water is retained on the soil particles by surface forces. It is held so strongly that gravity cannot remove it from the soil particles. The molecules of capillary water are free and mobile and are present in a liquid state. Due to this reason, it evaporates easily at ordinary temperature though it is held firmly by the soil particle; plant roots are able to absorb it.
QUESTION: 159
Lacey’s regime theory is not applicable to a channel in
Solution:
Lacey said that even a channel showing no silting no scouring may actually not be in the regime. He differentiated three regime conditions,
(i) True regime
(ii) Initial regime and
(iii) Final regime
According to him, a channel which is under ‘initial’ regime is not a channel in a regime, as there is no silting or scouring and hence regime theory is not applicable to such channels. His theory is therefore applicable only to those channels, which are either in true regime or in the final regime.
QUESTION: 160
For a 120 m Radius of curve, the maximum grade compensation required is
Solution:
But subjected to maximum value of
QUESTION: 161
What is the value of the coefficient of lateral friction as per the IRC?
Solution:
As per IRC, the coefficient of lateral friction recommended is 0.15.
Also, for the coefficient of longitudinal friction, the recommended value is (0.35 - 0.40).
QUESTION: 162
Calculate the diameter of a pipe of 32000 m long, if it is equivalent to another pipe of 0.2 m diameter and 1000 m long.
Solution:
Using dupit’s equation
D1 = 0.4 m
QUESTION: 163
Which of the following instrument is used for measuring the discharge?
Solution:
Venturimeter: Used to measure discharge in the pipe flow.
Manometer: Measure low, medium and high gauge as well as vacuum pressure of liquids and gases both.
Current meter: Used to measure the velocity of water in the rivers
Vane anemometer: Used for measuring the velocity and volumetric flow rate on air grills in residential buildings and utility.
QUESTION: 164
Calculate the kinematic viscosity (stoke) of the fluid, if the dynamic viscosity of fluid is 0.5 poise and specific gravity is 0.4?
Solution:
Kinematicviscosity(ν)
ν = 1.25 × 10-4 m2/sec
ν = 1.25 cm2/sec
ν = 1.25 stokes
QUESTION: 165
A rectangular block of dimensions 2 m × 1 m × 1 m is floating in the water with immersing depth of 0.5 m. What is the weight of block (kN) if unit weight of water is 10 kN/cubic meter.
Solution:
QUESTION: 166
The diameter of droplet is 0.075 mm. What is the intensity of the pressure (N/sq. cm) developed in the droplet by surface tension of 0.000075 N/mm?
Solution:
Intensity of Pressure
6 = Surface tension
D = Diameter of droplet
ΔP = 0.4 N/cm2
QUESTION: 167
The flow in a channel is laminar and Reynolds number is given by 1200. What is the friction factor for the channel?
Solution:
Friction factor is given by =
F = 0.053
QUESTION: 168
What is the value of the angle (degree) between stream lines and equipotential lines at the point of intersection in the flow net?
Solution:
QUESTION: 169
For a fluid flow problem in a distorted model, calculate the discharge scale ratio if horizontal scale ratio is 1/4 and vertical scale ratio is 1/9
Solution:
For a distorted model, discharge scale ratio is given by
QUESTION: 170
A fluid flows through a round pipe with Reynold’s number 1000. The shear stress is
Solution:
QUESTION: 171
Which of the following is dimensionless?
Solution:
QUESTION: 172
If the undisturbed strength of a soil sample is 50 units while the remoulded strength is 20 units, then what is the sensitivity of this soil?
Solution:
Sensitivity is the ratio of undisturbed soil strength by remoulded shear strength.
S = 50/20 = 2.5
QUESTION: 173
Which of the following is NOT an indirect method of estimating the permeability of soil in field?
Solution:
Pumping out test directly gives permeability and hence is not an indirect method.
QUESTION: 174
A soil formation through which only the seepage is possible, being partly permeable and capable of giving insignificant yield, is classified as
Solution:
Aciuitard is that geological formation which does not yield water freely to wells due to its less permeability, although seepage is possible through it. The yield in such formation is thus insignificant.
QUESTION: 175
On which of the following do the numerical values of Terzaghi’s bearing capacity factors depend?
Solution:
Terzaghi’s bearing capacity factors - Nc, Nq, Nγ depend on the angle of internal friction only and are dimensionless.
Nc = (Nq - 1) cot ϕ
QUESTION: 176
Quick sand condition occurs when:
Solution:
Quicksand is a condition and not a soil type. This condition is created in saturated thick layers of loose fine sandy soils when disturbed either due to vibration, such as, from pile driving in the neighborhood, or due to the pressure of flowing water (at the time of heavy pumping in excavation).
QUESTION: 177
The soils most susceptible to liquefaction are:
Solution:
Liquefaction is a phenomenon in which loose saturated sand loses a large percentage of its shear strength and develops characteristics similar to those of a liquid. It is usually induced by cyclic loading of relatively high frequency, resulting in undrained conditions in the sand. Cyclic loading may be caused, for example, by vibrations from machinery and, more seriously, by earth tremors. Saturated loose sand tends to compact under cyclic loading. The decrease in volume causes an increase in pore water pressure which cannot dissipate under undrained conditions. Indeed, there may be a cumulative increase in pore water pressure under successive cycles of loading. If the pore water pressure becomes equal to the maximum total stress component, normally the overburden pressure, the value of effective stress will be zero, i.e. interparticle forces will be zero, and the sand will exist in a liquid state with negligible shear strength. Even if the effective stress does not fall to zero the reduction in shear strength may be sufficient to cause failure.
QUESTION: 178
Which of the following contract types is usually followed by Railway Department for construction purposes?
Solution:
For item rate contract, contractors are required to quote rates for individual items of work on the basis of the schedule of quantities furnished by the client’s department. Indian Railways and many Public Sector units use this method of contracts.
QUESTION: 179
For estimation of painted area of corrugated asbestos cement sheets, percentage increase in area above the painted area is ________.
Solution:
Corrugated surfaces shall be measured flat as fixed and not girthed Quantities so measured shall be increased by the following percentage and the resultant shall be included in general areas.
Corrugated steel sheets 14%
Corrugated asbestos cement sheets 20%
Semi – corrugated asbestos cement sheets 10%
QUESTION: 180
The main factor to be considered while preparing a detailed estimate is
Solution:
The preparations of a detailed construction estimate consist of working out quantities of various items of work and then determine the cost of each item. This is prepared in two stages.
1. Details of measurements and calculation of quantities
2. Abstract of Estimated Cost
In the first stage, the quantity, availability, and transportation of materials parameters is being analyzed.
In the second stage, the total abstract of the above analysis is shown.
QUESTION: 181
The weight of the foundation is assumed as which of the following?
Solution:
To calculate the area of foundation required for the given load, we need weight of the foundation also, so in columns and walls we assumed the weight of foundation to be 10% of the column load or wall load for column and wall respectively.
QUESTION: 182
The value of A and B of Indian type water closet (W.C) shown in the figure is?
Solution:
The Recommended value of A and B is 30 cm and 45 cm respectively. Also, value of C = 72 cm to 80 cm
QUESTION: 183
The order of Booking Dimensions is
Solution:
The order of Booking Dimensions is:
QUESTION: 184
The arrangement of supporting an existing structure by providing supports underneath, is known as
Solution:
QUESTION: 185
The covered area of a proposed building is 150m2 and it included a rear courtyard of 5m × 4m. If the prevailing plinth area rate for similar building is Rs. 1250/m2, what is its cost?
Solution:
Covered area of proposed building = 150 m2
Courtyard area = 5 x 4 = 20 m2
∴ Plinth area = 150 – 20 = 130 m2
∴ Total cost of the building = 130 x 1250 = 162500
QUESTION: 186
A document containing detailed description of all the items of work (but their quantities are not mentioned) together with their current rates is called
______.
Solution:
A document containing a detailed description of all the items of work together with their current rates is called schedule of rates. It is provided from CPWD or from local Municipal Corporation for estimation work.
QUESTION: 187
In a mass-housing project, break even point indicates the:
Solution:
Break-even point in a business indicates the amount of output for which there is no-profit no-loss situation exists.
QUESTION: 188
The correction to be applied to each 30 meter chain length along θ° slope is _____.
Solution:
Slope correction for 30 m chain = 30(1 - cos θ) m
QUESTION: 189
Match list - I with list - II
List I (Nature of contour lines)
1. Ridge
2. Valley
3. Hills
4. Pond
List II
A. Approximately concentric closed with descending values towards centre
B. Approximately concentric closed contours with increasing values towards centre
C. V - shaped contours with convexity towards higher ground
D. U - shaped contours with convexity towards lower ground
Solution:
QUESTION: 190
The length of a long chord of a circular curve of radius R and deflection angle Δ is
Solution:
QUESTION: 191
In the Tacheometry, if inclined sight q is taken on a staff held normal to the sight, horizontal distance is
Solution:
QUESTION: 192
According to Simpson’s rule, if there are n number of segments each of width d, in terms of ordinates, area of the figure is:
Solution:
In Simpson’s rule –
The boundaries between the extremities is assumed to be parabolic.
If the ordinates are odd then only this formula is usefull.
The area enclosed in a parabolic seement is 2/3 rectangular area.
If the first offset or last offset is zero, it should not be ignored.
QUESTION: 193
If the least count of instrument is 20”, by measuring the horizontal angle n times repeatedly, the accuracy achieved is
Solution:
If the least count of instrument is 20”, by measuring the horizontal angle n times repeatedly, the accuracy achieved is
Thus the ‘Repetition method’ is considered as more accurate because the precision obtained is much finer than the least count of the instrument.
QUESTION: 194 | 14,376 | 59,147 | {"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-2022-21 | longest | en | 0.832288 |
https://research.chalmers.se/en/publication/50947 | 1,679,313,134,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943483.86/warc/CC-MAIN-20230320114206-20230320144206-00147.warc.gz | 545,357,504 | 8,057 | # Dynamics of Plates with Thin Piezoelectric Layers Licentiate thesis, 2007
The subject of this thesis is dynamics of plates with thin piezoelectric layers. Piezoelectric materials are often used in sensors and actuators and common applications for these are vibration control and ultrasonic transducers. In the first part of the thesis plate equations for a plate consisting of one anisotropic elastic layer and one piezoelectric layer with an applied electric voltage are derived. The displacements and the electric potential are expanded in power series in the thickness coordinate, which leads to recursion relations among the expansion functions. Using these in the boundary and interface conditions, a set of equations is obtained for some of the lowest-order expansion functions. This set is reduced to a system of six plate equations, where three of them are given to linear order in the thickness and correspond to the symmetric (in plane) motion, while the other three are given to quadratic order in the thickness and correspond to the antisymmetric (bending) motion. In principle it is possible to go to any order and it is believed that the equations are asymptotically correct. Some numerical comparisons are made with exact theory for infinite plates and very good agreement is obtained for low frequencies. In the second part of the thesis the plate equations are evaluated by investigating a vibration problem with a finite elastic layer and a shorter piezoelectric layer on top of it. The boundary conditions to combine with the plate equations are derived by inserting the power series expansions into the physical boundary conditions at the sides of the elastic layer and identifying equal powers of the thickness coordinate. Numerical comparisons of the displacement field are made with two other theories. The first one is another approximate theory based on the same type of power series expansions, where the piezoelectric layer is modeled as equivalent boundary conditions. The other one is exact three-dimensional theory. Both approximate theories yield accurate results for thin plates and low frequencies as long as the piezoelectric layer is thin in comparison to the total plate thickness.
Thin layers
Actuator
Plate equations
Equivalent boundary conditions
Power series expansions
Piezoelectricity
KS101, Kemigården 4, Chalmers Tekniska Högskola
Opponent: Prof. Leif Kari, Department of Aeronautical and Vehicle Engineering, KTH, Sverige
## Author
### Karl Mauritsson
Chalmers, Applied Mechanics
### Subject Categories
Mechanical Engineering
Technical report - Department of Applied Mechanics, Chalmers University of Technology, Göteborg, Sweden
KS101, Kemigården 4, Chalmers Tekniska Högskola
Opponent: Prof. Leif Kari, Department of Aeronautical and Vehicle Engineering, KTH, Sverige | 569 | 2,832 | {"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-2023-14 | latest | en | 0.937346 |
https://questions.llc/questions/13257/twelve-men-take-6-hours-to-finish-a-piece-of-work-after-the-12-men-have-worked-for-1 | 1,679,841,727,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00260.warc.gz | 549,717,525 | 4,618 | # twelve men take 6 hours to finish a piece of work. after the 12 men have worked for 1 hour, the contractor decides to call in 8 more men to complete the remaining work. how many more hours would the 20 men take to complete the remaining piece of work?
is this correct?
12 men 6 h 1 job.
12 men 5 h 5/6 job
20 men ?h 5/6 job
x/5 = 12/20
x= 3h
yes this is correct | 114 | 366 | {"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-2023-14 | latest | en | 0.967396 |
https://mathoverflow.net/questions/282031/problems-reducing-to-a-graph-theory-algorithm | 1,553,017,536,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202003.56/warc/CC-MAIN-20190319163636-20190319185636-00510.warc.gz | 558,451,273 | 40,174 | # Problems reducing to a graph-theory algorithm
This is essentially a question in pedagogy -- the answers could be useful to teach (or rather, motivate) graph theory, and especially the algorithmic side of it.
I have been very impressed with this example:
Question: What is the maximum size of a set of integers between 1 and 100 such that for any pair (a,b), the difference a-b is not a square ?
The first, naive algorithms that spring to mind to solve this (finite) problem are awkward. However, let me quote: " This problem can be defined as a Graph problem : we create a vertex for each integer, and link two integers if their difference is a square. We but have to find a maximal independent set in this graph ! "
Using this, a few trivial lines of Sage code give the answer (see the link). Sooo much time is saved for the coder!
Are there more examples like this one?
To be more precise, I'm looking for problems which:
• do not immediately involve graph theory, on the surface,
• reduce, after a little translation, into a common problem in graph theory (find a matching, find a colouring...).
• there should be a "haha!" effect for the mathematician, and a sigh of relief for the programmer.
I heard another nice one in a talk by Tim Gowers. A university wants to organize its exams. Some students take several exams. Some rooms and some time slots are available. Help them. Solution: each exam (maths, physics, etc) is a vertex; place one edge between exams if there is one student taking both; define "colours" to be pairs (room, time). Now you have to colour the graph such that adjacent vertices are in different colours.
There is a fine line between these great examples, and the articifial ones which I find disappointing, such as "$n$ couples having each $k$ children all want to shake hands on different days such that...". In fact, I should add that I'm certain to have a preference for problems involving maths (like the first one above) rather than "real life" (like the second).
Usual precaution: if this is not appropriate for MO, let me know and I will not be offended!
• Every problem in NP (and that includes most of the everyday problems) can be reduced to a graph theoretic problem, as there are enough of them that are NP-complete. If this reduction is nice or not depends on how you do it and what you (or your student/programmer) sees as nice; that might strongly differ between people. – Dirk Liebhold Sep 26 '17 at 11:17
• The celebrity problem: math.stackexchange.com/q/1163921/147470 – Max Alekseyev Sep 26 '17 at 11:25
• @DirkLiebhold: incidentally, your comment, true of course, reminded me of a stylistically-surprising (though not contentually-surprising) passage I read recently: in [ J. Väänänen: How complicated can structures be?. Nieuw Archief voor Wiskunde 5/9 nr. 2 June 2008], Jouko Väänänen, somewhat surprisingly decided to use the NP-completeness of $\text{3-colorability}$ to simply write: "The famous P=NP question, one of the Clay Millenium Questions, asks if we can decide in polynomial time whether a given finite graph is 3-colourable." Full stop. He just uses a 'representative'. – Peter Heinig Sep 26 '17 at 17:23
• @Pierre: one example, easy, yet, I think it is fair to say, not obvious (many, I think, will at first not suspect that the concept 'maximum matching' completely 'rules' the winning strategies) is the well-known PathGame that I described in this MO thread. Admittedly, this is not a number-theoretic example, rather a game-theoretic one.[...] – Peter Heinig Sep 26 '17 at 17:33
• Are you trying to motivate people to learn how to use Sage's graph theory library? Or are you trying to motivate people to understand the underlying algorithms? Or to understand mathematics with a view to designing new algorithms? It seems to me that these all require somewhat different sorts of problems. – Will Sawin Sep 26 '17 at 18:50
What a great question. I’m sure there will be many lovely examples. Here is one I stumbled on last year which seems to fit.
I saw an online demonstration of a simulated annealing algorithm for the following problem: a digital photograph has been “shredded” by randomly shuffling the columns of pixels, and we wish to recover the original unshredded image. Here is such a shredded photograph.
The simulated annealing algorithm is able to put the photograph back together to some extent by moving similar columns of pixels together, resulting in this:
The moment of enlightenment for me was realising that the problem is one of finding a minimum-weight Hamiltonian path in a weighted complete graph: the vertices of the graph are the columns of pixels, and the edge weights are given by a similarity measure showing how well two columns match. Therefore we can bring to bear all the well-developed algorithmic machinery aimed at the Travelling Salesman Problem. Using Keld Helsgaun’s LKH solver gives an algorithm that is orders of magnitude faster, and is able to reconstruct this particular photograph with not a pixel out of place:
(although it is flipped left-to-right: there is not much we can do about that possibility)
As an amusing coda, if the rows as well as the columns of the image are shuffled then the result looks more like a decorative rug than a photograph of Paris at night:
but the original photograph may still be recovered by running the algorithm twice – once for the rows, and once for the columns – though now the result might be flipped upside-down as well as left-to-right.
Incidentally, the reason I was familiar with TSP-solving technology is that I had used it earlier to disprove a 20-year-old conjecture in combinatorics: another problem that doesn’t sound like graph theory at first blush.
Photo credit: Blue Hour in Paris, by Falcon® Photography (CC BY-SA)
• This is a very nice example, in many respects. For further appreciation, let me point out that this is also a nice example of an application of finite metric spaces, since one can view the weighted complete graph as a finite metric space with as many points as the picture has columns. One quick technical comment: it would be even better if in "the problem is one of finding a minimum-weight Hamiltonian path in a weighted graph" you wrote 'weighted complete graph' instead of only 'weighted graph', since I think that your "similarity measure" gives a weigh to every edge. – Peter Heinig Sep 27 '17 at 10:46
• Particularly elegant: the use of a shortest Hamilton-path in the auxiliary weighted graph, not a shortest Hamilton-circuit; this automatically *picks out the correct 'start-' and 'end-column' of the 'damaged' picture. If one would use Hamilton-*circuits, one would have to first make an arbitrary choice of where to begin displaying the columns, then see a still cyclically-shuffled picture, and 'correct' it in a 'human-controlled' post-processing step. This is avoided by the elegant use of 'shortest Hamilton path'. This is the nicest example of a genuine use of Hamilton-paths I know. – Peter Heinig Sep 27 '17 at 11:46
• I absolutely love this example! – Pierre Sep 27 '17 at 12:03
• Also to be appreciated in this answer: this is a wonderful example of the distinction between metric spaces for which the metric is a piece of additional data, and metric spaces for which the metric is implicit in the data defining the points. A briefer, more precise term for the latter could be imbedded metric spaces, and I would like to propose the neologism immetric spaces for this concept. (The rationale for this is obvious: imbedded metric spaces are given in terms of an actual imbedding.) – Peter Heinig Sep 27 '17 at 14:39
• The neologism immetric space, which currently seems not to be found by the largest of all engines, evidently is a pun, a play on words playing both on the term 'imbedding', and on the usual prefix 'im-' to negate adjectives, which is meaningful, since an immetric space does not really have a metric: the metric is implicit in the 'imbedded points'. With immetric spaces, one does't have to store the metric as a piece of data; one has an algorithm to compute it. Robin Houston's answer is, all in all, the nicest introductory example of the usefulness of immetric spaces known to me. – Peter Heinig Sep 27 '17 at 14:41
I expected a flood of responses to this question, since there are so many lovely examples. That flood does not seem to have materialised yet, so let me add an old classic. I hope it isn’t too famous to be worth mentioning, and that adding it as a separate answer is the correct protocol.
I’m thinking of Steve Fisk’s celebrated proof of the Art Gallery theorem, which is in the form of an algorithm.
We are given the floor plan of an art gallery as a simple polygon in the plane with $n$ vertices, and we wish to place some (motionless point) guards into the polygon so that the whole of the interior of the polygon is visible to at least one guard.
We can see that we may need as many as $\lfloor\frac n3\rfloor$ guards by considering comb-shaped polygons with $n=3k$ vertices:
(There must be a guard in each of the grey triangles, and there are $k$ disjoint such triangles, so we need at least $k=\frac n3$ guards.)
The problem, then, is to place no more than $\lfloor\frac n3\rfloor$ guards into a simple polygon with $n$ vertices in such a way that one of the guards can see any point in the interior of the polygon.
Fisk’s algorithm is as follows: triangulate the polygon to obtain a planar graph whose vertices are the vertices of the original polygon and whose faces are triangles, and then three-colour the vertices so each face has a vertex of each colour. At least one of the three colours occurs $\leq\lfloor\frac n3\rfloor$ times: position guards at the vertices that are painted this colour. Since each triangle has a guard at one of its vertices, and a triangle is entirely visible from any one of its vertices, the whole of the polygon must be visible to at least one guard.
Thus a difficult-sounding purely geometric problem is solved by three-colouring the vertices of a planar graph.
I should perhaps note that prolific MO contributor Joseph O’Rourke literally wrote the book on art gallery theorems and algorithms.
The OP contains the specifications
do not immediately involve graph theory, on the surface,
reduce, after a little translation, into a common problem in graph theory (find a matching, find a colouring...).
there should be a "haha!" effect for the mathematician, and a sigh of relief for the programmer.
The following is about as trivial as can be (even in a technical sense of 'trivial': it is based on what is sometimes presented in introductory graph theory lectures as the 'First Theorem of Graph Theory'), but it fits each of the three specifications above:
To find the number of undirected connections in a finite network of which you only have knowledge in the form of a photograph. $\hspace{150pt}$(problem)
That is: imagine a situation in which a programmer faces the task of programming a camera-equipped machine to 'look' at such a photograph and correctly output the total number of connections.
To make the problem-description consistent with the 'worked example' below, imagine the company running the search engine which kindly ran for me a Gilbert--Erdős–Rényi-random graph simulation, decides to *empirically test that the number of connection shown on the screen of users equals the number of connections that the search engine returns as a numeral when asked to do so. For the sake of argument, suppose that the company will have none of 'formal verification' of code (though I think this would give greater security) but has resolved to test this very very empirically, by filming a screen. To reach usefully large sample sizes, they will have to teach a camera-equipped machine how to count the lines.
Among the "first, naive algorithms that spring to mind to solve this (finite) problem" (to quote the OP) is an "awkward" one: to rasterize the picture, to systematically scan it, to code-up some recognition-of-elongated-shapes, to keep track of what elongated shape you already have counted, to finally somehow arrive at an estimate of the number of connections. (Also note: this approach does not involve any necessary check of correctness, however weak, unlike the approach I am proposing; see the remark at the very end.)
I think that, especially if the photograph is messy, the input data to the machine "[does] not immediately involve graph theory,".
Now comes the "little translation, into a common problem in graph theory", namely into what is arguably the most trivial of all graph theoretic problems:
to find the number of edges of a specified graph.
And this problem ("aha!" goes the mathematician, while the programmer heaves the required "sigh of relief") reduces to simply 'doing a sum' of natural numbers (plus a little pattern recognition, though not a recognition of elongated/curvilinear shapes anymore): the problem reduces to summing the degree sequence of the relevant graph. (The reason is what is whimsically called 'The Handshaking Lemma', or a little presumptiously called 'The First Theorem of Graph heory'1)
More precisely, we idealize (and what an important, mathematical activity this is: idealizing sense-data) the situation, by imagining that there is an abstract graph.
Now we know what to teach the machine:
(0) identify all 'vertex-like' points of the photograph,
(1) 'cookie-cutter' a small circle-shaped portion around each such 'point',
(2) forget the vast majority of the 'photograph',
(3) collect the 'cookies' (the order and position of the 'cookies' is irrelevant, any data-structure which keeps the set of the 'cookies' will be serviceable),
(4) now solve a much easier patter-recognition-problem, on each of the 'cookies': to recognize the number of 'rays' in the 'star-like' shapes (there will be noise, yet I am confident that this task can be routinely done, reliably, by methods of Computational Homology, methods which were more-or-less made for doing precisely this: extracting meaningful integer-valued invariants from noisy black-and-white-pictures. If not, I am confident that this task will be routine for people working in machine-learning/pattern-recognition.)
To summarize: where was the graph-theory here?
On "the surface", a messy photograph of a tangle of connections in a network does not look like a graph. At the very least, I am sure that a person not knowing concepts of graph theory would, when having to teach a machine what to do with the sense-data fed to it, be more likely to code-up a 'rasterize-and-scan-the-whole-picture-and-invent-heuristics-to-recognize' approach than taking the graph-theoretically-informed 'cookie-cutter' approach.
Worked Example. Let the given 'photograph' be the network-like part of the following photograph, taken from the webpage of a known search-engine, with the 'order' given to the engine shown in the first line of the image:
Insert now a "haha!" and "sigh of relief". No pattern-recognition of 'line-like' shapes is necessary; it suffices to identify the 'vertex-like' parts of the image, and then do the following:
First the 'cookie-cutting' (here, I use counterclockwise ordering):
Now follows a conversion of the 'gray-scale cookies' to something combinatorial (essentially: (0,1)-matrices, given as a black-and-white rid). The diaphanous blue regions do not carry any information; they associate each gray-scale 'observation' with the corresponding (0,1)-matrix, yet this association could be done by the mind of the observer unaided by the blue regions, for the correspondence is quite uniquely determined.
These '0-1-matrices' weren't 'made-up': I had the 'GIMP' software compute it for me, via the "Indexed Color Conversion" functionality, with the option "Use black and white (1-bit) palette" and with "Color-dithering: Floyd-Steinberg (normal)" turned on. What I give here should even be quite a 'reproducible experiment'. It should be possible to have a machine produce the same '0-1-matrices' from the gray-scale pictures.
Now insert here some expert advice of someone more versed in computational homology (or some other suitable method: something via eigenvalues of the black-and-white matrices perhaps (though these eigenvalues will not be real-valued? using some appropriate heat-flow?); or someone more versed in machine-learning/pattern-recognition algorithms2) than the writer of these lines (who only had to learn computational homology for an exam a few years ago, yet does not work with it), telling us which "Sooo much time"-saving out-of-the-box routine we should use to count the number of 'rays' in the 'star-shaped' matrix-pictures, and now the following numbers come pouring forth:
3 2 6 5 3 4 4 5 7 4 1
Now the so-called First Theorem of Graph Theory completes the "reducing" of the problem to a "graph theory algorithm": it remains to calculate
$\frac12\cdot (3 + 2 + 6 + 5 + 3 + 4 + 4 + 5 + 7 + 4 + 1) = \frac12\cdot 44 = 22$
to know (by the 'First Theorem of Graph Theory') that this is the number of edges(=connections) in the 'photograph' of the random graph.
This was a reduction, of sorts.
On which side of your "fine line" this is, I cannot know in advance. By the way, I agree with Dirk Liebhold's comment, that this is very context-dependent question, any answer will be 'tainted' by context.
Addition. Another small contribution of graph theory in solving this problem:
If the above sum (in the example it was 44) does not come out even, then something somewhere must have gone wrong.
In that sense, the 'First Theorem of Graph Theory' also give a very weak 'check'/'necessary condition' for the correct execution of the algorithm. One of course is not told by the criterion where and why an error has occurred, only that somewhere something has gone wrong. Also, if the sum comes out even, this, needless to say, does not imply that it is the correct one.
1 Incidentally, this commonly encountered designation is not only a little presumptious, but also technically-wrong if very strictly construed: in the strictest sense, 'Graph Theory' is the theory, in the model-theoretic sense, of the class of irreflexive symmetric relations on a set, and, as such, has a one-element signature consisting of a single binary relation symbol. The so-called 'First Theorem of Graph Theory', though, uses a larger signature: which exactly, depends on your formalization, but you can argue that it uses the signature $(\sim,\mathrm{deg},+,\Sigma,\lVert\cdot \rVert)$, where $\sim$ is the binary relation symbol, and $\mathrm{deg}$ and $\lVert\cdot\rVert$ are unary function symbols, the intended interpretation of the former being the degree of its argument, the one of the latter being the number of edges of its argument.
2 It would be a nice example of focused interdisciplinary work if someone who knows more about pattern recognitino would leave a relevant comment on what is considered the best method to count the number of 'rays' in the 'star-shaped' '0-1-matrices'. One can for example first 'blow-up' each pixel until the 0-th Betti number (=number of connected components) has become equal to 1, upone which, in a sense, the 'noise' is 'gone', and then one has to come up with a functorial method to 'count the number of bumps along the perimeter'. Ideally, the present answer would in the end feature a completely 'synthetic' method of counting the number of connection in a 'photograph' of a 'network', that is, by 'putting together' existing concepts.
• aha ! very nice. – Pierre Sep 27 '17 at 12:07
• If I may ask, is there any possibility you could reduce the pictures or give an external link to replace them? the whole page is made difficult to read. – Pierre Sep 28 '17 at 7:15
• @Pierre: actually, while I won't change them today, let me briefly point out that I am fully aware that the step (gray scale 'cookies')$\rightarrow$( black-and-white matrices) is a rather arbitrary one, and somehow the weakest point of the whole proposal. Why, of all things, should one use "Color-dithering: Floyd-Steinberg (norma)l" in "GIMP". This is very very context-dependent, sadly. I chose this to carry the argument to some conclusion. – Peter Heinig Sep 28 '17 at 8:06 | 4,669 | 20,273 | {"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} | 3.71875 | 4 | CC-MAIN-2019-13 | latest | en | 0.952501 |
https://mathoverflow.net/questions/143947/is-the-class-of-n-dimensional-manifolds-essentially-small/143950#143950 | 1,653,029,450,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00423.warc.gz | 435,858,717 | 29,407 | # Is the class of n-dimensional manifolds essentially small?
Question: Consider the proper class of all $n$-dimensional smooth manifolds. If we take the equivalence classes where two manifolds are identified if there exists a diffeomorphism between them, is this collection of equivalence classes a set?
Remark: I do not assume my manifolds to be Hausdorff nor second countable. If the answer depends on those assertions (Edit: it most definitely does) I would like to hear about the difference.
Remark 2: As Omar pointed out, there may be a problem with the various long lines. To make the question slightly more tractable, I would be (mostly) satisfied if there is a statement even ignoring the smooth structure and consider the case of topological manifolds and homeomorphisms.
Motivation: for something that I am working on I need to consider the collection of all $n$-dimensional smooth manifolds satisfying "property $X$". Unfortunately property $X$ is diffeomorphism invariant, so most definitely this collection is not a set, which invalidates many constructions (I want to build a manifold out of this collection; if the collection is a proper class then even on the set level the thing that I constructed will be a proper class, instead of a set) or at least forces me to rethink how this constructions ought to go. Fortunately for my argument it suffices that I have one object in each diffeomorphism class in my collection.
• With out second countability, probably the long lines of different lengths are all pairwise non-diffeomorphic. If you assume Hausdorff and second countable, you can embed each manifold in some $\mathbb{R}^n$ so you get the crude bound of at most $\sum_{n \in \mathbb{N}} 2^{2^{\aleph_0}} = 2^{2^{\aleph_0}}$ manifolds. Oct 4, 2013 at 12:20
• Isn't this already false in dimension $0$? I mean, if you start with any set and equip it with the discrete topology, you get a $0$-dimensional manifold. In dimnesion $n$, you can instead take the product of that discrete space with any fixed manifold. Oct 4, 2013 at 12:25
• @OmarAntolín-Camarena, there is no such thing as "long lines of different lengths". There is only one long line up to homeomorphism (and it has a set many smooth structures). Oct 4, 2013 at 12:26
• Restricting to 2nd countable, metrizable smooth manifolds, isn't it enough to look at submanifolds of $\mathbb R^{2n+1}$ which I think do form a set. Then one has the equivalence relation on the set given by (nonambient) diffeomorphism, and what you care about is the set of equivalence classes. Oct 4, 2013 at 12:36
• @Omar, your longer lines are not manifolds, since the point $(\omega_1,0)$ in your order is problematic, and has no neighborhood like $\mathbb{R}$. Oct 4, 2013 at 12:57
If we assume the Hausdorff separation axiom, connectedness, but not second countability then each such manifold has cardinality at most continuum. Therefore the collection of equivalence classes of such manifolds is a set and not a proper class. First of all, $U$ must be path connected since every connected locally path connected space is path connected. Fix $x_{0}\in U$. Then every point of $U$ is reachable from $x_{0}$ by a path. I now claim that there can be at most continuumly many choices of paths $f:I\rightarrow U$ with $f(0)=x_{0}$. To each $x\in U$, we associate an open set $U_{x}$ locally homeomorphic to $\mathbb{R}^{n}$. Let $T$ be the following tree of height $\omega_{1}$. The nodes of $T$ will branch into continuumly many parts at each point of $T$, so the cardinality of $T$ is at most continuum. The objects of $T$ are tuples $((r_{\alpha})_{\alpha\leq\lambda},(y_{\alpha})_{\alpha\leq\lambda},(f_{\alpha}))$ where $(r_{\alpha})_{\alpha\leq\lambda}$ is an increasing sequence in $[0,1]$, $(y_{\alpha})_{\alpha\leq\lambda}$ is a sequence of points in $U$ where $y_{\alpha+1}\in U_{y_{\alpha}}$ and $f_{\alpha}:[r_{\alpha},r_{\alpha+1}]\rightarrow U_{y_{\alpha}}$ is continuous. Now for each element $((r_{\alpha})_{\alpha\leq\lambda},(y_{\alpha})_{\alpha\leq\lambda},(f_{\alpha}))$, we associate a path $\bigcup_{\alpha}f_{\alpha}$ if possible (there could be discontinuities at limit ordinals). Clearly every path is determined by an element of this tree, so there could be at most continuumly many paths. Therefore $U$ has cardinality at most continuum.
• Could you explain why without assuming second countability your manifolds have bounded cardinality? Oct 5, 2013 at 3:53
• You want to also demand that the functions $\alpha\mapsto r_\alpha$ and $\alpha\mapsto y_\alpha$ are continuous. Otherwise, for $\alpha$ limit, $y_\alpha$ is allowed to be any element of $U$, and you lose the cardinality bound. This is also where the Hausdorff assumption is used (it tells you the limit $y_\alpha$ are uniquely determined by the successor $y_\alpha$). Oct 5, 2013 at 12:06
Already in dimension $0$, the collection of all manifolds in the sense of the OP is a proper class: any set, equipped with the discrete topology, is a $0$-dimensional manifold. In dimension $n$, one can instead take the product of such a discrete space with any fixed manifold, so that the collection of all $n$-manifolds is still a proper class.
(In the generic case, these manifolds are Hausdorff, but not second countable.)
As pointed out by Omar Antolín-Camarena in the comments, the question becomes more interesting if one also assumes connectedness.
• But as Willie Wong points out in the comments a connected non-Hausdorff manifold of dimension $n>0$ can have arbitrarily large cardinality. Oct 4, 2013 at 13:16
I claim that the class $\mathcal{K}$ of weak homotopy equivalence classes of connected Hausdorff smooth 2-dimensional manifolds is not a set. You know you can make compact second countable surfaces by sewing any finite number of donuts together, and if you don't demand compactness or second countability you can just keep on sewing.
First of all, the class $\text{Card}$ of all cardinals is not a set. I construct an "function" $\text{Ord}\rightarrow \mathcal{K}$ whose restriction to $\text{Card}$ is "injective" ($\text{Ord}$ denotes the class of ordinals!). Let $\alpha$ be an ordinal. Let's first deal with $\alpha=0$. Then we assign to $\alpha$ the empty manifold $M_0$. If $\alpha=1$ we assign to $\alpha$ the torus $M_1$. If $\alpha=\beta+1$ is some other successor ordinal, take a torus and chop off a disk, and take $M_\beta$ and chop off a disk to obtain $M'_\beta$, now glue them together along the boundary. If $\lambda\neq 0$ is a limit ordinal do the following. Assume that for every limit ordinal $0\neq \mu<\lambda$ we have that the underlying topological space of $M_\mu$ is the union of $M'_{\beta}$ for $\beta<\mu$ and the smooth atlas of $M_\mu$ is the set of all charts of $M'_\beta$ for $\beta<\mu$ whose domain (or codomain, whichever your convention is) is diffeomorphic to $\mathbb{R}^2$ (so we are excluding charts hitting the boundary of $M'_\beta$). Then define the underlying topological space of $M_\lambda$ to be the union of the $M'_\alpha$ for $\alpha<\lambda$ and take as an atlas for $M_\lambda$ the set of all charts of all $M'_\alpha$ for $\alpha<\lambda$ that miss the boundary of $M'_\alpha$. Set $M'_\lambda$ to be $M_\lambda$ with a disk removed.
This gives a "function" $\text{Ord}\rightarrow \mathcal{K}:\alpha\mapsto M_\alpha$. "It's restriction to the class of cardinals is injective" because if $|\alpha|\neq |\beta|$ then $\pi_1\left(M_\alpha\right)\not\cong \pi_1\left(M_\beta\right)$. If you want to get rid of the """'s, use a Grothendieck universe (a modest large cardinal), if you allow yourself to. If you don't the proof is still valid.
It seems likely this can be done for any $n$ and the class of suitably connected manifolds, as was pointed out for $0$-dimensional manifolds by Tobias. For $1$-dimensional manifolds, take $\underline{\alpha}\times \mathbb{R}$ for each ordinal $\alpha$ ($\underline{\alpha}$ is the discrete topological space whose underlying set is $\alpha$). Again the restriction to $\text{Card}$ is injective, since $\pi_0$ tells them apart.
EDIT: The constructing shown here does not work because there exists an $n$ for which $M'_n\not\subseteq M'_\omega$, as Eric points out in a comment below.
• You can't just say $M_\lambda$ is the union of the $M_\alpha'$, because the $M_\alpha'$ might not be nested (later steps might chop out part of them). Indeed, $M_\omega'$ must necessarily chop out a disk that intersects some $M_n'$. You then run into trouble if at limit steps, the disks you've chopped out accumulate at a point. Oct 5, 2013 at 11:57 | 2,346 | 8,600 | {"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} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.9123 |
https://www.aqua-calc.com/calculate/weight-to-volume/substance/calcium-blank-nitrate-blank-tetrahydrate | 1,722,931,158,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640476915.25/warc/CC-MAIN-20240806064139-20240806094139-00161.warc.gz | 534,029,855 | 12,831 | # Volume of Calcium nitrate tetrahydrate
## calcium nitrate tetrahydrate: convert weight to volume
### Volume of 100 grams of Calcium nitrate tetrahydrate
centimeter³ 40 milliliter 40 foot³ 0 oil barrel 0 Imperial gallon 0.01 US cup 0.17 inch³ 2.44 US fluid ounce 1.35 liter 0.04 US gallon 0.01 meter³ 4 × 10-5 US pint 0.08 metric cup 0.16 US quart 0.04 metric tablespoon 2.67 US tablespoon 2.71 metric teaspoon 8 US teaspoon 8.12
### The entered weight of Calcium nitrate tetrahydrate in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
#### How many moles in 100 grams of Calcium nitrate tetrahydrate?
There are 423.47 millimoles in 100 grams of Calcium nitrate tetrahydrate
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313 foods that contain Phytosterols. List of these foods starting with the highest contents of Phytosterols and the lowest contents of Phytosterols
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A liter or (l) is a metric unit of volume with sides equal to one decimeter (1dm) or ten centimeters (10cm)
The electric potential φ(A) of a point A (in the electric field) specifies the work to be done by the force F = −Q × E in order to move the charge Q from a fixed reference point P to the point A.
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#### Calculators
Volume to Weight conversions for common substances and materials | 661 | 2,372 | {"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-2024-33 | latest | en | 0.695817 |
https://isabelle.in.tum.de/repos/isabelle/rev/ede9dc025150 | 1,632,321,168,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00183.warc.gz | 354,054,007 | 5,556 | author huffman Thu, 03 Nov 2011 17:40:50 +0100 changeset 45332 ede9dc025150 parent 45325 26b6179b5a45 child 45333 04b21922ed68
ex/Tree23.thy: prove that deletion preserves balance
src/HOL/ex/Tree23.thy file | annotate | diff | comparison | revisions
```--- a/src/HOL/ex/Tree23.thy Thu Nov 03 11:18:06 2011 +0100
+++ b/src/HOL/ex/Tree23.thy Thu Nov 03 17:40:50 2011 +0100
@@ -315,6 +315,167 @@
lemma ord0_add0: "ord0 t \<Longrightarrow> ord0 (add0 k y t)"
using ord'_add0 [of None t None k y] by (simp add: ord')
+text {* The @{term "del"} function preserves balance. *}
+
+lemma del_extra_simps:
+"l \<noteq> Empty \<or> r \<noteq> Empty \<Longrightarrow>
+ del k (Branch2 l p r) = (case compare k p of
+ LESS => (case del k l of None \<Rightarrow> None |
+ Some(p', (False, l')) => Some(p', (False, Branch2 l' p r))
+ | Some(p', (True, l')) => Some(p', case r of
+ Branch2 rl rp rr => (True, Branch3 l' p rl rp rr)
+ | Branch3 rl rp rm rq rr => (False, Branch2
+ (Branch2 l' p rl) rp (Branch2 rm rq rr))))
+ | or => (case del (if_eq or None k) r of None \<Rightarrow> None |
+ Some(p', (False, r')) => Some(p', (False, Branch2 l (if_eq or p' p) r'))
+ | Some(p', (True, r')) => Some(p', case l of
+ Branch2 ll lp lr => (True, Branch3 ll lp lr (if_eq or p' p) r')
+ | Branch3 ll lp lm lq lr => (False, Branch2
+ (Branch2 ll lp lm) lq (Branch2 lr (if_eq or p' p) r')))))"
+"l \<noteq> Empty \<or> m \<noteq> Empty \<or> r \<noteq> Empty \<Longrightarrow>
+ del k (Branch3 l p m q r) = (case compare k q of
+ LESS => (case compare k p of
+ LESS => (case del k l of None \<Rightarrow> None |
+ Some(p', (False, l')) => Some(p', (False, Branch3 l' p m q r))
+ | Some(p', (True, l')) => Some(p', (False, case (m, r) of
+ (Branch2 ml mp mr, Branch2 _ _ _) => Branch2 (Branch3 l' p ml mp mr) q r
+ | (Branch3 ml mp mm mq mr, _) => Branch3 (Branch2 l' p ml) mp (Branch2 mm mq mr) q r
+ | (Branch2 ml mp mr, Branch3 rl rp rm rq rr) =>
+ Branch3 (Branch2 l' p ml) mp (Branch2 mr q rl) rp (Branch2 rm rq rr))))
+ | or => (case del (if_eq or None k) m of None \<Rightarrow> None |
+ Some(p', (False, m')) => Some(p', (False, Branch3 l (if_eq or p' p) m' q r))
+ | Some(p', (True, m')) => Some(p', (False, case (l, r) of
+ (Branch2 ll lp lr, Branch2 _ _ _) => Branch2 (Branch3 ll lp lr (if_eq or p' p) m') q r
+ | (Branch3 ll lp lm lq lr, _) => Branch3 (Branch2 ll lp lm) lq (Branch2 lr (if_eq or p' p) m') q r
+ | (_, Branch3 rl rp rm rq rr) => Branch3 l (if_eq or p' p) (Branch2 m' q rl) rp (Branch2 rm rq rr)))))
+ | or => (case del (if_eq or None k) r of None \<Rightarrow> None |
+ Some(q', (False, r')) => Some(q', (False, Branch3 l p m (if_eq or q' q) r'))
+ | Some(q', (True, r')) => Some(q', (False, case (l, m) of
+ (Branch2 _ _ _, Branch2 ml mp mr) => Branch2 l p (Branch3 ml mp mr (if_eq or q' q) r')
+ | (_, Branch3 ml mp mm mq mr) => Branch3 l p (Branch2 ml mp mm) mq (Branch2 mr (if_eq or q' q) r')
+ | (Branch3 ll lp lm lq lr, Branch2 ml mp mr) =>
+ Branch3 (Branch2 ll lp lm) lq (Branch2 lr p ml) mp (Branch2 mr (if_eq or q' q) r')))))"
+apply -
+apply (cases l, cases r, simp_all only: del.simps, simp)
+apply (cases l, cases m, cases r, simp_all only: del.simps, simp)
+done
+
+inductive full :: "nat \<Rightarrow> 'a tree23 \<Rightarrow> bool" where
+"full 0 Empty" |
+"\<lbrakk>full n l; full n r\<rbrakk> \<Longrightarrow> full (Suc n) (Branch2 l p r)" |
+"\<lbrakk>full n l; full n m; full n r\<rbrakk> \<Longrightarrow> full (Suc n) (Branch3 l p m q r)"
+
+inductive_cases full_elims:
+ "full n Empty"
+ "full n (Branch2 l p r)"
+ "full n (Branch3 l p m q r)"
+
+inductive_cases full_0_elim: "full 0 t"
+inductive_cases full_Suc_elim: "full (Suc n) t"
+
+lemma full_0_iff [simp]: "full 0 t \<longleftrightarrow> t = Empty"
+ by (auto elim: full_0_elim intro: full.intros)
+
+lemma full_Empty_iff [simp]: "full n Empty \<longleftrightarrow> n = 0"
+ by (auto elim: full_elims intro: full.intros)
+
+lemma full_Suc_Branch2_iff [simp]:
+ "full (Suc n) (Branch2 l p r) \<longleftrightarrow> full n l \<and> full n r"
+ by (auto elim: full_elims intro: full.intros)
+
+lemma full_Suc_Branch3_iff [simp]:
+ "full (Suc n) (Branch3 l p m q r) \<longleftrightarrow> full n l \<and> full n m \<and> full n r"
+ by (auto elim: full_elims intro: full.intros)
+
+definition
+ "full_del n k t = (case del k t of None \<Rightarrow> True | Some (p, b, t') \<Rightarrow> full (if b then n else Suc n) t')"
+
+lemmas ord_case = ord.exhaust [where y=y and P="ord_case a b c y", standard]
+lemmas option_ord_case = ord.exhaust [where y=y and P="option_case a b (ord_case d e f y)", standard]
+lemmas option_option_case = option.exhaust [where y=y and P="option_case a b (option_case d e y)", standard]
+lemmas option_prod_case = prod.exhaust [where y=y and P="option_case a b (prod_case c y)", standard]
+lemmas option_bool_case = bool.exhaust [where y=y and P="option_case a b (bool_case c d y)", standard]
+lemmas prod_tree23_case = tree23.exhaust [where y=y and P="prod_case a (tree23_case b c d y)", standard]
+lemmas option_case = option.exhaust [where y=y and P="option_case a b y", standard]
+lemmas full_tree23_case = tree23.exhaust [where y=y and P="full a (tree23_case b c d y)", standard]
+
+lemmas case_intros =
+ option_ord_case option_option_case option_prod_case option_bool_case prod_tree23_case option_case
+ full_tree23_case
+
+lemma full_del: "full (Suc n) t \<Longrightarrow> full_del n k t"
+proof -
+ { fix n :: "nat" and p :: "key \<times> 'a" and l r :: "'a tree23" and k
+ assume "\<And>n. \<lbrakk>compare k p = LESS; full (Suc n) l\<rbrakk> \<Longrightarrow> full_del n k l"
+ and "\<And>n. \<lbrakk>compare k p = EQUAL; full (Suc n) r\<rbrakk> \<Longrightarrow> full_del n (if_eq EQUAL None k) r"
+ and "\<And>n. \<lbrakk>compare k p = GREATER; full (Suc n) r\<rbrakk> \<Longrightarrow> full_del n (if_eq GREATER None k) r"
+ and "full (Suc n) (Branch2 l p r)"
+ hence "full_del n k (Branch2 l p r)"
+ apply clarsimp
+ apply (cases n)
+ apply (cases k)
+ apply (simp add: full_del_def)
+ apply (simp add: full_del_def split: ord.split)
+ apply (simp add: full_del_def)
+ apply (subst del_extra_simps, force)
+ apply (simp | rule case_intros)+
+ done
+ } note A = this
+ { fix n :: "nat" and p q :: "key \<times> 'a" and l m r :: "'a tree23" and k
+ assume "\<And>n. \<lbrakk>compare k q = LESS; compare k p = LESS; full (Suc n) l\<rbrakk> \<Longrightarrow> full_del n k l"
+ and "\<And>n. \<lbrakk>compare k q = LESS; compare k p = EQUAL; full (Suc n) m\<rbrakk> \<Longrightarrow> full_del n (if_eq EQUAL None k) m"
+ and "\<And>n. \<lbrakk>compare k q = LESS; compare k p = GREATER; full (Suc n) m\<rbrakk> \<Longrightarrow> full_del n (if_eq GREATER None k) m"
+ and "\<And>n. \<lbrakk>compare k q = EQUAL; full (Suc n) r\<rbrakk> \<Longrightarrow> full_del n (if_eq EQUAL None k) r"
+ and "\<And>n. \<lbrakk>compare k q = GREATER; full (Suc n) r\<rbrakk> \<Longrightarrow> full_del n (if_eq GREATER None k) r"
+ and "full (Suc n) (Branch3 l p m q r)"
+ hence "full_del n k (Branch3 l p m q r)"
+ apply clarsimp
+ apply (cases n)
+ apply (cases k)
+ apply (simp add: full_del_def)
+ apply (simp add: full_del_def split: ord.split)
+ apply (simp add: full_del_def)
+ apply (subst del_extra_simps, force)
+ apply (simp | rule case_intros)+
+ done
+ } note B = this
+ show "full (Suc n) t \<Longrightarrow> full_del n k t"
+ proof (induct k t arbitrary: n rule: del.induct)
+ { case goal1 thus "full_del n (Some k) Empty"
+ by simp }
+ { case goal2 thus "full_del n None (Branch2 Empty p Empty)"
+ unfolding full_del_def by auto }
+ { case goal3 thus "full_del n None (Branch3 Empty p Empty q Empty)"
+ unfolding full_del_def by auto }
+ { case goal4 thus "full_del n (Some v) (Branch2 Empty p Empty)"
+ unfolding full_del_def by (auto split: ord.split) }
+ { case goal5 thus "full_del n (Some v) (Branch3 Empty p Empty q Empty)"
+ unfolding full_del_def by (auto split: ord.split) }
+ { case goal26 thus ?case by simp }
+ qed (fact A | fact B)+
+qed
+
+lemma full_imp_height: "full n t \<Longrightarrow> height t = n"
+ by (induct set: full, simp_all)
+
+lemma full_imp_bal: "full n t \<Longrightarrow> bal t"
+ by (induct set: full, auto dest: full_imp_height)
+
+lemma bal_imp_full: "bal t \<Longrightarrow> full (height t) t"
+ by (induct t, simp_all)
+
+lemma bal_iff_full: "bal t \<longleftrightarrow> full (height t) t"
+ using bal_imp_full full_imp_bal ..
+
+lemma bal_del0: "bal t \<Longrightarrow> bal (del0 k t)"
+ unfolding del0_def
+ apply (drule bal_imp_full)
+ apply (case_tac "height t", simp, simp)
+ apply (frule full_del [where k="Some k"])
+ apply (simp add: full_del_def)
+ apply (auto split: option.split elim: full_imp_bal)
+ done
+
text{* This is a little test harness and should be commented out once the
above functions have been proved correct. *}
@@ -327,12 +488,15 @@
text{* Some quick checks: *}
+lemma bal_exec: "bal t \<Longrightarrow> bal (exec as t)"
+ by (induct as t arbitrary: t rule: exec.induct,
+ simp_all add: bal_add0 bal_del0)
+
+lemma "bal(exec as Empty)"
+ by (simp add: bal_exec)
+
lemma "ord0(exec as Empty)"
quickcheck
oops
-lemma "bal(exec as Empty)"
-quickcheck
-oops
-
-end
\ No newline at end of file
+end``` | 3,160 | 9,643 | {"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} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.460393 |
https://kolblabs.com/magnetic-induction-in-playing-with-magnets-class-6-science-experiment/ | 1,718,576,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00680.warc.gz | 312,865,923 | 55,405 | # Magnetic induction in Playing With Magnets – Class 6 Science Experiment
## Activity Name: Magnetic induction in Playing With Magnets
### Activity Description:
The experiment explores the concept of magnetic induction using a safety pin, an alpin (a type of pin), and a bar magnet. It aims to demonstrate how the presence of a magnet can induce magnetic properties in another object.
• Safety pin
• Alpin (pin)
• Bar magnet
### Step by Step Procedure:
1. Take a safety pin and bring it close to an alpin. Observe whether the safety pin attracts the alpin.
2. Bring the safety pin close to one pole of a bar magnet and observe how it gets attached to the magnet.
3. Take an alpin and touch it to the safety pin.
4. Observe whether the safety pin attracts the alpin.
5. Hold a bar magnet in one hand and a safety pin in the other hand, keeping them close to each other but not in contact.
6. Ask your friend to bring an alpin and touch it to the safety pin.
7. Observe whether the alpin sticks to the safety pin.
### Experiment Observations:
1. When the safety pin is brought close to the alpin without any other magnetic influence, it may or may not attract the alpin, depending on its magnetic properties.
2. When the safety pin is brought close to one pole of a bar magnet, it gets attached to the magnet.
3. When an alpin is touched to the safety pin after it has been in contact with the bar magnet, the safety pin may attract the alpin.
4. When the bar magnet and the safety pin are held close to each other but not in contact, and an alpin is touched to the safety pin, the alpin sticks to the safety pin.
### Precautions:
• Handle the bar magnet and safety pin with care to avoid any injuries.
• Ensure that the safety pin is not forcefully pressed against the bar magnet to avoid damaging the pin or the magnet.
• Be cautious while handling the alpin to prevent accidental pricks.
### Lesson Learnt from Experiment:
The experiment demonstrates the concept of magnetic induction. When a safety pin is brought close to a bar magnet, it can acquire magnetic properties and behave like a magnet.
This is known as magnetic induction, where a magnetic substance becomes magnetized due to the presence of a nearby magnet. The experiment also highlights that the magnetic induction effect persists even when the safety pin and the bar magnet are not in direct contact.
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# 740 GMAT (48Q / 44V) - What is the new quant percentile min?
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Intern
Joined: 11 Aug 2011
Posts: 14
Concentration: Entrepreneurship, Finance
Schools: Stern '16 (M)
GMAT 1: 740 Q48 V44
GPA: 3.6
WE: Consulting (Consulting)
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Kudos [?]: 8 [0], given: 1
740 GMAT (48Q / 44V) - What is the new quant percentile min? [#permalink]
### Show Tags
03 Sep 2013, 16:56
Hello GMATClub,
I have lurked on this forum for years, but I only began studying for the GMAT over the past month. Today, I took the test for the first time and received a score of 740.
While I am ecstatic over the 44 Verbal (98th percentile) and 740 overall (97th percentile), the 48 Quant (76th percentile) smarts.
I hear chatter both ways regarding quant scores. Some still swear by the 80% / 80% rule. I have also heard that Wharton AdCom specifically wants to see a > 80th percentile Quant score. Others say that the 80% / 80% rule no longer holds. They note that AdComs have reviewed the academic performance of past 48 Quant scorers and believe prospective students with the score are plenty qualified to handle the academic rigor.
How do AdComs compare students' unscaled scores? The GMAC claims that unscaled scores from two different years represent the same level of aptitude. If an applicant who took the test two years ago (e.g. 48 quant, 81st percentile) and an applicant who took the test today (e.g. 48 quant, 76th percentile) are compared, are the percentiles considered? Are students who took the test two years ago now also disadvantaged? Have AdComs instead begun to embrace a preferred unscaled score threshold instead of a preferred percentile threshold? (Of course there is no required threshold, but the AdComs do seem to have preferred scores.)
After answering the above, could you also consider my personal situation? Should I consider retaking the exam specifically to meet the Quant 80th percentile threshold? I am looking to apply in round 2 this year to multiple top 8 schools (specifically, the schools with stronger finance programs). My feeling is retaking the exam would be unnecessary as I have a CPA, I am a CFA level 3 candidate, I have an undergraduate degree with concentrations in accounting and finance (3.56 GPA from Boston College CSOM), and I work in a quant-heavy industry (business valuation). However, I am concerned that finance-focused schools are looking for higher quant scores, and I am worried a lower quant score may weaken the rest of my academic profile.
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Re: 740 GMAT (48Q / 44V) - What is the new quant percentile min? [#permalink]
### Show Tags
04 Sep 2013, 02:19
Great info!! thanx
BSchool Forum Moderator
Status: Current Tuckie!
Joined: 19 Oct 2012
Posts: 543
Location: United Kingdom
Schools: Tuck '16 (M)
GMAT 1: 710 Q50 V36
GMAT 2: 740 Q48 V44
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Re: 740 GMAT (48Q / 44V) - What is the new quant percentile min? [#permalink]
### Show Tags
04 Sep 2013, 06:55
PlacePhraseHere wrote:
Hello GMATClub,
I have lurked on this forum for years, but I only began studying for the GMAT over the past month. Today, I took the test for the first time and received a score of 740.
While I am ecstatic over the 44 Verbal (98th percentile) and 740 overall (97th percentile), the 48 Quant (76th percentile) smarts.
I hear chatter both ways regarding quant scores. Some still swear by the 80% / 80% rule. I have also heard that Wharton AdCom specifically wants to see a > 80th percentile Quant score. Others say that the 80% / 80% rule no longer holds. They note that AdComs have reviewed the academic performance of past 48 Quant scorers and believe prospective students with the score are plenty qualified to handle the academic rigor.
How do AdComs compare students' unscaled scores? The GMAC claims that unscaled scores from two different years represent the same level of aptitude. If an applicant who took the test two years ago (e.g. 48 quant, 81st percentile) and an applicant who took the test today (e.g. 48 quant, 76th percentile) are compared, are the percentiles considered? Are students who took the test two years ago now also disadvantaged? Have AdComs instead begun to embrace a preferred unscaled score threshold instead of a preferred percentile threshold? (Of course there is no required threshold, but the AdComs do seem to have preferred scores.)
After answering the above, could you also consider my personal situation? Should I consider retaking the exam specifically to meet the Quant 80th percentile threshold? I am looking to apply in round 2 this year to multiple top 8 schools (specifically, the schools with stronger finance programs). My feeling is retaking the exam would be unnecessary as I have a CPA, I am a CFA level 3 candidate, I have an undergraduate degree with concentrations in accounting and finance (3.56 GPA from Boston College CSOM), and I work in a quant-heavy industry (business valuation). However, I am concerned that finance-focused schools are looking for higher quant scores, and I am worried a lower quant score may weaken the rest of my academic profile.
Don't bother retaking. It's a great score and there is no guarantee that you won't score worse the second time. Since you also have a CPA and you are a CFA level 3 candidate, there really shouldn't be any concern over your quant abilities.
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Re: 740 GMAT (48Q / 44V) - What is the new quant percentile min? [#permalink]
### Show Tags
06 Sep 2013, 18:27
Congratulations on a STELLAR GMAT score!
Very generally yes, schools prefer 80th percentiles in an ideal world, but in the upper 70's is very close especially as scores have shifted up, and if your overall score was lower or your overall percentile lower, it could be more of an issue, but your overall is excellent for any program. And the guidelines are 'approximate' at most programs vs a hard and fast line-in-the-sand cut-off.
Good luck to you, and I encourage you to now turn your attention to those essays-- the one component of the application where you create something from SCRATCH to represent who you are ! No pressure but definitely a lot of opportunity there that can make or break an application in many cases.
All the best!
_________________
Jen Kedrowski
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Re: 740 GMAT (48Q / 44V) - What is the new quant percentile min? [#permalink] 06 Sep 2013, 18:27
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Change in GMAT Percentiles 5 21 Sep 2007, 19:35
Display posts from previous: Sort by | 2,004 | 7,831 | {"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} | 3.125 | 3 | CC-MAIN-2017-13 | longest | en | 0.915957 |
https://cdn2.jlqwer.com/posts/4545.html | 1,638,032,120,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00333.warc.gz | 244,059,167 | 10,601 | # 1.2.1 Milking Cows 挤牛奶
1.2.1 Milking Cows 挤牛奶
Line 1:
Lines 2..N+1:
``````3
300 1000
700 1200
1500 2100
``````
``````900 300
``````
``````#include<stdio.h>
#include<stdlib.h>
struct TMP
{
int a,b;
}sum[5010];
long max1,max2;
int cmp(const void *c,const void *d)
{
return (*(struct TMP *)c).a-(*(struct TMP *)d).a;
}
int main()
{
int n,i;
int pos1,pos2;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d %d",&sum[i].a,&sum[i].b);
qsort(sum,n,sizeof(struct TMP),cmp);
pos1=sum[0].a;
pos2=sum[0].b;
if(n==1)
printf("%d 0\n",pos2-pos1);
else
{
for(i=1;i<n;i++)
{
if(sum[i].a<=pos2&&sum[i].b>pos2)
pos2=sum[i].b;
else if(sum[i].a>pos2)
{
if(pos2-pos1>max1)
max1=pos2-pos1;
if(sum[i].a-pos2>max2)
max2=sum[i].a-pos2;
pos1=sum[i].a;
pos2=sum[i].b;
}
}
if(pos2-pos1>max1)
max1=pos2-pos1;
if(sum[i].a-pos2>max2)
max2=sum[i].a-pos2;
printf("%d %d\n",max1,max2);
}
return 0;
}
``````
``````#include <stdio.h>
#include<stdlib.h>
typedef struct node
{
int x,y;
} node;
int n,myspace=0,result=0;
node f[10005];
int cmp(const void *a,const void *b)
{
return((*(node*)a).x-(*(node*)b).x);
}
void jisuan()
{
for(i=1; i<n; i++)
{
if(f[i].x<=maxx&&f[i].y>maxx)maxx=f[i].y;
if(f[i].x>maxx)
{
if((f[i].x-maxx)>myspace)myspace=f[i].x-maxx;
maxx=f[i].y;
}
}
printf("%d %d",result,myspace);
}
int main()
{
int i;
scanf("%d",&n);
for(i=0;i<n;i++)scanf("%d%d",&f[i].x,&f[i].y);
qsort(f,n,sizeof(f[0]),cmp);
jisuan();
return 0;
}
`````` | 607 | 1,413 | {"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-2021-49 | latest | en | 0.1434 |
www.simplifyingfractions.org | 1,568,960,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573832.23/warc/CC-MAIN-20190920050858-20190920072858-00185.warc.gz | 311,661,961 | 6,712 | # Simplifying Fractions
Simplifying fractions mean reduce the fraction to lowest terms or write the fraction in the lowest terms. A fraction is simply the means by which a whole of something is divided into smaller components. Various operations are applied on fractions like addition, subtraction, multiplication, division and simplification of fractions. Here we discussed about the simplification of the fractions. Simplifying fractions means to make the fraction as simple as possible. Divide the top and bottom by the highest number that can divide into both numbers exactly.
Simplify the fractions mean that the numerator and denominator have no common factors other than 1.
## Easiest Way to Simplify Fractions
There are different methods to simplify a fraction:
Method 1:
Both the numerator and denominator can evenly divided by the same number without changing the value of the fraction. We may have to divide more than once to find the simplest form.
Example: Simplify $\frac{18}{24}$
Step 1: Divide by 2
=> $\frac{18}{24}$ = $\frac{9}{12}$
Step 2: Divide by 3
=> $\frac{9}{12}$ = $\frac{3}{4}$
=> $\frac{18}{24}$ is reduced to $\frac{3}{4}$
Method 2:
Factor the numerator and denominator to find the greatest common factor, then divide numerator and denominator by that GCD.
Example: Simplify $\frac{10}{5}$
Step 1:
Greatest common factor of 10 and 5 is 5
Step 2:
Multiply numerator and denominator by 5
=> $\frac{10}{5}$ = $\frac{10}{5}$ * $\frac{5}{5}$
= $\frac{2}{1}$
= 2
Method 3:
Write the numerator and denominator as a product of prime numbers. Split each fraction into two fractions, the first with the common prime numbers. This puts the fraction in the form of 1 times another fraction.
Solved Example
Question: Simplify, $\frac{6}{9}$
Solution:
Step 1: Prime factors of 6 and 9
Factors of 6 = 2 * 3
Factors of 9 = 3 * 3
Step 2: Now write the common factors as a separate fraction:
$\frac{6}{9}$ = $\frac{2 * 3}{3 * 3}$
=
$\frac{3}{3}$ * $\frac{2}{3}$
=
$\frac{2}{3}$
## Simplifying Fractions
Simplify the fractions mean that the numerator and denominator have no common factors other than one. To simplify a fraction, divide the numerator and denominator by the greatest common factor or find the prime factors of the numerator and denominator and reduced the fraction by cancelling common factors. When a fraction is simplified, then no common factor between numerator and denominator other than one.
## Solved Examples
Question 1: Simplify, $\frac{75}{30}$
Solution:
Given, $\frac{75}{30}$
Step 1: Prime factors of 75 and 30
Factors of 75 = 3 * 5 * 5
Factors of 30 = 2 * 3 * 5
Step 2: Now write the common factors as a separate fraction:
$\frac{75}{30}$ = $\frac{3 * 5 * 5}{2 * 3 * 5}$
$\frac{3 * 5}{3 * 5}$ * $\frac{ 5}{2}$
= 1 * $\frac{5}{2}$
= $\frac{5}{2}$
=> $\frac{75}{30}$ is simplifies to $\frac{5}{2}$
Question 2: Simplify, $\frac{150}{250}$
Solution:
Given, $\frac{150}{250}$
Step 1: Prime factors of 150 and 250
Factors of 150 = 2 * 3 * 5 * 5
Factors of 250 = 2 * 5 * 5 * 5
Step 2: Now write the common factors as a separate fraction:
$\frac{150}{250}$ = $\frac{2 * 3 * 5 * 5}{2 * 5 * 5 * 5}$
= $\frac{2 * 5 * 5}{2 * 5 * 5}$ * $\frac{3}{5}$
= 1 * $\frac{3}{5}$
= $\frac{3}{5}$
=> $\frac{150}{250}$ = $\frac{3}{5}$
## Simplify Algebraic Fractions
Algebraic fractions have properties which are the same as those for numerical fractions, the only difference being that the the numerator and denominator are both algebraic expressions. Simplification of a algebraic fraction is same as we simplify fractions. Simplifying fraction is a way to avoid common fraction. Fractions written in simplest form have no factor other than one.
## Solved Examples
Question 1: Simplify $\frac{x^2 - 1}{x + 1}$ to simplest form
Solution:
Given, $\frac{x^2 - 1}{x + 1}$
$\frac{x^2 - 1}{x + 1}$ = $\frac{(x - 1)(x + 1)}{x + 1}$
[Using identity a2 - b2 = (a - b)(a + b)]
= x - 1
=> $\frac{x^2 - 1}{x + 1}$ is reduced to (x - 1).
Question 2: Simplify $\frac{15a^2b^3}{25a^3b^2}$
Solution:
Given $\frac{15a^2b^3}{25a^3b^2}$
Step 1: Find the factors of numerator and denominator
15a2 b3 = 3 * 5 * a * a * b * b * b
25a3 b2 = 5 * 5 * a * a * a * b * b
Step 2:
$\frac{15a^2b^3}{25a^3b^2}$ =
$\frac{3 * 5 * a * a * b * b * b}{5 * 5 * a * a * a * b * b}$
=
$\frac{3b}{5a}$
=> $\frac{15a^2b^3}{25a^3b^2}$ = $\frac{3b}{5a}$
Question 3: Simplify $\frac{(2x + 3)(x - 3)}{x^2 - 9}$
Solution:
Given $\frac{(2x + 3)(x - 3)}{x^2 - 9}$
Step 1: Find the factors of x2 - 9
x2 - 9 = x2 - 32 = (x - 3)(x + 3)
Step 2:
$\frac{(2x + 3)(x - 3)}{x^2 - 9}$ = $\frac{(2x + 3)(x - 3)}{(x - 3)(x + 3)}$
=
$\frac{2x + 3}{x + 3}$
=>
$\frac{(2x + 3)(x - 3)}{x^2 - 9}$ = $\frac{2x + 3}{x + 3}$. | 1,603 | 4,753 | {"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} | 5 | 5 | CC-MAIN-2019-39 | latest | en | 0.879723 |
http://www.rugusavay.com/how-do-scientists-define-energy/ | 1,526,922,351,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864461.53/warc/CC-MAIN-20180521161639-20180521181639-00471.warc.gz | 469,158,958 | 12,993 | #### What is energy used for? What is energy and the importance of energy. What is the meaning of force, resistance and work, what are the differences?
To define energy, scientists must use some ideas. The first of these ideas you already have. It is the idea of force.
Every time you pick up a book, pedal your bicycle, open a door, or write your name, you are using a force. If you think about all the times when you use a force, you will find that you do one of two things: You either push something or pull it. So a force is a push or a pull. Whenever you make something move, you exert a force. But you may also exert a force without making anything move. You have probably pulled or pushed some heavy object without being able to budge it. Whether the object moves or not, you are exerting a force just the same.
Besides the force you exert with your muscles, there are other forces acting around you all the time. If you stand in an open place on a windy day, you can feel the force of the wind, or moving air. If you walk into a swift stream of water, the force of the moving water may throw you off your feet. If you step off a box or a chair, the force of gravity pulls you down. If you make a deep dive, you can feel the force of the water above you pressing on your body. You know that a magnet exerts a force on iron and steel and makes them move. Whenever anything moves, a force has been used to start it moving.
Another idea that scientists must use in defining energy is the idea of work. Work is a common word, and you may think that you know very well what it means. Work to you is probably something that you have to do but do not enjoy doing. However, like many other words, work has more than one meaning. When scientists use this word, they have a special meaning for it.
If you push as hard as you can against a table but do not move it, you have exerted a force. However, you have done no work according to a scientist. But if you move the table, you have done work. In other words, to a scientist work is done only if something is moved. You might struggle for an hour trying to move a big stone. If you could not move it, a scientist would say that you had done no work. But if you quickly brushed a tiny leaf aside, he would say that you had done work, because the leaf was moved.
When you lift a chair, climb the stairs, push a lawn mower, or drive a nail, you are also doing work. You exert a force and move something. To make it move, your force must overcome an opposing force. That is, some other force always acts against the force that does the work. This opposing force is called the resistance. Scientists use this word to define work. They say that work is done when a force overcomes a resistance and moves something.
Scientists say that energy is the ability to do work. Anything has energy if it can move something that resists being moved. In other words, if a thing has energy, it can do work. To say that something can do work is really just anotlıer way of saying that it has energy. As you can see, energy and work are closely related ideas. Whenever work is done, energy is used to exert a force that overcomes a resistance and moves something. Without energy, there would be no force and no work could be done.
### 1 Comment
1. Pingback: How Does Friction Act? | 747 | 3,317 | {"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-2018-22 | latest | en | 0.966381 |
https://math.stackexchange.com/questions/581934/need-hints-on-the-following-algebra-problems | 1,566,750,162,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330750.45/warc/CC-MAIN-20190825151521-20190825173521-00465.warc.gz | 539,035,886 | 30,438 | # Need hints on the following algebra problems.
I've been looking at these for over an hour and I don't understand how to do them. Any hints would be greatly appreciated.
Let $p(x) = x^3 + x + 1$ and $F = Z_3[x]/\langle p(x)\rangle$. Factor $p(x)$ in $F(x)$. Does it factor into linear terms?
Here I am just supposed to factor it out only if it has a root in F(x). True?
List a complete set of representatives of $F[x]/\langle p(x)\rangle$ where $F = Z_2$ and $p(x) = x^4 + x^2 + 1$.
Any quick way to do this? Am I supposed to write down the multiplication table?
Find the multiplicative inverse of $x^2 + 1$ in $Q[x]/\langle x^4 - 2\rangle$.
Here I let $x^4 \cong 2 \mod p(x)$ and then $(x^2 + 1)y \cong 1 \mod x^4 + 2$. That gets me nowhere. However $yx^2 + y - 1 \cong x^4 - 2$. Do I have to find linear factors now?
• On the first one, if you are moding out by p(x) then p(x) factors as 0 in F. In fact, p(x) has a root in $Z_3$. – Vladhagen Nov 26 '13 at 14:08
• Can you go into more detail @Vlad? I understand $p(1) = 0$. Not $p(2)$ or $p(3)$ though. – Don Larynx Nov 26 '13 at 14:09
• As written the factorization is p(x)=0. If you are looking at all polynomials mod p(x), then p(x) itself is zero. – Vladhagen Nov 26 '13 at 14:10
• It factors as zero. That is how modulo arithmetic works in a polynomial ring. A much better question would ask how the polynomial factors in $Z_3[x]$. Then $p(x) = (x+2)(x^2+x+2)$ – Vladhagen Nov 26 '13 at 15:14
• When you mod by p, you get p is zero. It is sort of like asking how 24 factors mod 24. It just factors as zero. – Vladhagen Nov 26 '13 at 15:16 | 550 | 1,605 | {"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} | 3.96875 | 4 | CC-MAIN-2019-35 | latest | en | 0.886483 |
https://physics.stackexchange.com/questions/93001/smoothed-particle-hydrodynamics-chemical-reactions-and-stiffness | 1,566,699,775,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027322170.99/warc/CC-MAIN-20190825021120-20190825043120-00333.warc.gz | 590,703,109 | 28,942 | # Smoothed Particle Hydrodynamics, Chemical Reactions and Stiffness [closed]
Dear people from stackexchange physics.
I have been using SPH for quite a while to simulate free boundary flow, and just recently we tried to include some kind of (simple) chemical reaction in our code. We are still doing some tests, but one curious thing that we found was that we needed much smaller time steps than we usually used to be necessary.
I have never dealed with chemical reactions, but I don know that some of them fall on the category of 'Siff Equations', and as far as the wikipedia description goes, it seems to be my case, but I would like to ask if anyone tried anything similar and also found that their equations were stiff.
I never studied stiff equations, I just knew that they existed and recently read a bit more about them, but that's it. Last but not least, do anyone know how to check if a set of SPH equations is stiff or not? The wikipedia recomendation seemed at least complicated to calculate for SPH.
## closed as off-topic by Kyle Kanos, Brandon Enright, jinawee, user10851, John RennieJan 10 '14 at 10:25
• This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.
• This question appears to be off-topic because it is about programming/computational methods and not physics. – Kyle Kanos Jan 9 '14 at 16:02
• You may want to try asking this at scicomp.stackexchange.com. – Kyle Kanos Jan 9 '14 at 16:03
• As Kyle Kanos points out, this is really a computational question. For non-reacting flows the time step limitation of your numerical method comes from the speed of sound in the medium. Reaction rates are typically much faster than sound waves and hence you will end up with a "stiff" set of equations. One approach to dealing with this is operator splitting whereby you use an explicit step for the flow followed by an implicit one for the reaction. – SimpleLikeAnEgg Jan 9 '14 at 20:56
Stiff equations tend to use "implicit" methods. I am actually working on an implicit simulation (rigid body constraints) which is infinitely stiff! Implicit methods are basically guess and check: try to find a set of forces that satisfies future constraints. This is a well-developed field, multigrid conjugate-gradient based methods are among the state of the art.
So how would you use an implicit method? Take a pendulum for which we have a single anchor-spring-mass as an example. The spring is stiff, almost a rigid rod, and you don't care about the high-frequency shortening and lengthening vibrations of the spring. You do care about modeling the overall swinging motion, however. "Normal" (explicit) solvers need a time-step that is small enough to capture the spring's vibrations. For an implicit scheme, the step is set by the dynamics we care about: the swing period of the pendulum, which is much longer than the vibrational period of the spring. Thus implicit methods, although more expensive per step, allow us to gloss over high-frequency degrees of freedom that we don't care about (as long as they these DOF's are nearly linear, as is true in our case with small spring vibrations).
For the pendulum case it is easy to create a new model bases on an ideal rigid rod, eliminating the annoying high-frequency DOF. For more complex cases this is difficult to do, however.
Stiff equations in chemistry typically arise when reaction time-constants are small. In a first order decay reaction: dC/dt = -k*C, the time constant is 1/k. For non-linear complex reactions, you can look at small perturbations of concentration and see how quickly they grow/decay.
Another way is for diffusivity to be high. The time constant is ~ neighbor distance^2/(diffusivity). It shrinks rapidly with decreased particle size.
Chemical stiffness is an issue if these time-constants are smaller than the SPH time-step needed. In this case you can decouple the simulations: take an SPH step and take several smaller diffusion and/or reaction sub-steps, or use an implicit model for the diffusion/reaction. Implicit methods for diffusion work well since error tends to blurs out, as long as total mass is conserved.
If the only stiff aspect is reaction kinetics, here's a simple way to use "sub-steps": for each SPH step calculate the diffusion fluxes. Assume a constant flux into your particle over the entire SPH timestep. For your particle, calculate reaction kinetics using much smaller sub-steps until you found your concentrations at the end of the SPH timestep. This way you capture the fast kinetics without any extra expensive SPH calculations.
SPH simulations themselves get stiff when fluids become incompressible and/or resolution gets higher. The step size ~ (particle size)/(speed of sound), the latter depends on the "spring constants" in the pressure terms and how many neighbors particles have, as well as the mass of the particles. Both higher speed of sound and lower fluid velocities make a more incompressible fluid, and both both cause "stiffness" in the sense that they increase how many steps are needed to for the dynamics you are interested in happen. | 1,139 | 5,217 | {"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-2019-35 | latest | en | 0.966868 |
https://number.academy/250309 | 1,713,926,615,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00323.warc.gz | 393,503,658 | 12,718 | # Number 250309 facts
The odd number 250,309 is spelled 🔊, and written in words: two hundred and fifty thousand, three hundred and nine. The ordinal number 250309th is said 🔊 and written as: two hundred and fifty thousand, three hundred and ninth. Color #250309. The meaning of the number 250309 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 250309. What is 250309 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 250309.
## Interesting facts about the number 250309
### Asteroids
• (250309) 2003 QD76 is asteroid number 250309. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, ETS in Socorro on 8/24/2003.
## What is 250,309 in other units
The decimal (Arabic) number 250309 converted to a Roman number is (C)(C)(L)CCCIX. Roman and decimal number conversions.
#### Weight conversion
250309 kilograms (kg) = 551831.2 pounds (lbs)
250309 pounds (lbs) = 113539.4 kilograms (kg)
#### Length conversion
250309 kilometers (km) equals to 155535 miles (mi).
250309 miles (mi) equals to 402834 kilometers (km).
250309 meters (m) equals to 821214 feet (ft).
250309 feet (ft) equals 76296 meters (m).
250309 centimeters (cm) equals to 98546.9 inches (in).
250309 inches (in) equals to 635784.9 centimeters (cm).
#### Temperature conversion
250309° Fahrenheit (°F) equals to 139042.8° Celsius (°C)
250309° Celsius (°C) equals to 450588.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
250309 seconds equals to 2 days, 21 hours, 31 minutes, 49 seconds
250309 minutes equals to 6 months, 5 days, 19 hours, 49 minutes
### Codes and images of the number 250309
Number 250309 morse code: ..--- ..... ----- ...-- ----- ----.
Sign language for number 250309:
Number 250309 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 ...
## Mathematics of no. 250309
### Multiplications
#### Multiplication table of 250309
250309 multiplied by two equals 500618 (250309 x 2 = 500618).
250309 multiplied by three equals 750927 (250309 x 3 = 750927).
250309 multiplied by four equals 1001236 (250309 x 4 = 1001236).
250309 multiplied by five equals 1251545 (250309 x 5 = 1251545).
250309 multiplied by six equals 1501854 (250309 x 6 = 1501854).
250309 multiplied by seven equals 1752163 (250309 x 7 = 1752163).
250309 multiplied by eight equals 2002472 (250309 x 8 = 2002472).
250309 multiplied by nine equals 2252781 (250309 x 9 = 2252781).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 250309
Half of 250309 is 125154,5 (250309 / 2 = 125154,5 = 125154 1/2).
One third of 250309 is 83436,3333 (250309 / 3 = 83436,3333 = 83436 1/3).
One quarter of 250309 is 62577,25 (250309 / 4 = 62577,25 = 62577 1/4).
One fifth of 250309 is 50061,8 (250309 / 5 = 50061,8 = 50061 4/5).
One sixth of 250309 is 41718,1667 (250309 / 6 = 41718,1667 = 41718 1/6).
One seventh of 250309 is 35758,4286 (250309 / 7 = 35758,4286 = 35758 3/7).
One eighth of 250309 is 31288,625 (250309 / 8 = 31288,625 = 31288 5/8).
One ninth of 250309 is 27812,1111 (250309 / 9 = 27812,1111 = 27812 1/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
250309
#### Is Prime?
The number 250309 is not a prime number. The closest prime numbers are 250307, 250343.
#### Factorization and factors (dividers)
The prime factors of 250309 are 23 * 10883
The factors of 250309 are 1, 23, 10883, 250309.
Total factors 4.
Sum of factors 261216 (10907).
#### Powers
The second power of 2503092 is 62.654.595.481.
The third power of 2503093 is 15.683.009.140.253.628.
#### Roots
The square root √250309 is 500,308905.
The cube root of 3250309 is 63,021996.
#### Logarithms
The natural logarithm of No. ln 250309 = loge 250309 = 12,430451.
The logarithm to base 10 of No. log10 250309 = 5,398476.
The Napierian logarithm of No. log1/e 250309 = -12,430451.
### Trigonometric functions
The cosine of 250309 is 0,859622.
The sine of 250309 is -0,510931.
The tangent of 250309 is -0,594367.
### Properties of the number 250309
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
More math properties ...
## Number 250309 in Computer Science
Code typeCode value
PIN 250309 It's recommended that you use 250309 as your password or PIN.
250309 Number of bytes244.4KB
CSS Color
#250309 hexadecimal to red, green and blue (RGB) (37, 3, 9)
Unix timeUnix time 250309 is equal to Saturday Jan. 3, 1970, 9:31:49 p.m. GMT
IPv4, IPv6Number 250309 internet address in dotted format v4 0.3.209.197, v6 ::3:d1c5
250309 Decimal = 111101000111000101 Binary
250309 Decimal = 110201100201 Ternary
250309 Decimal = 750705 Octal
250309 Decimal = 3D1C5 Hexadecimal (0x3d1c5 hex)
250309 BASE64MjUwMzA5
250309 MD5635613280dce428b2b057cf12501b974
250309 SHA2564c0ec7058d585ff679f7f5bc4c705e4e21aa11f735068babb942b78613c5d44a
250309 SHA38497d6858e87c52df720b0791292047a4a1e87c454ea518977a8851bbdf63cd768ca13f100edfb5f17606e9f1ce8e9bd17
More SHA codes related to the number 250309 ...
If you know something interesting about the 250309 number that you did not find on this page, do not hesitate to write us here.
## Numerology 250309
### Character frequency in the number 250309
Character (importance) frequency for numerology.
Character: Frequency: 2 1 5 1 0 2 3 1 9 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 250309, the numbers 2+5+0+3+0+9 = 1+9 = 1+0 = 1 are added and the meaning of the number 1 is sought.
## № 250,309 in other languages
How to say or write the number two hundred and fifty thousand, three hundred and nine in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 250.309) doscientos cincuenta mil trescientos nueve German: 🔊 (Nummer 250.309) zweihundertfünfzigtausenddreihundertneun French: 🔊 (nombre 250 309) deux cent cinquante mille trois cent neuf Portuguese: 🔊 (número 250 309) duzentos e cinquenta mil, trezentos e nove Hindi: 🔊 (संख्या 250 309) दो लाख, पचास हज़ार, तीन सौ, नौ Chinese: 🔊 (数 250 309) 二十五万零三百零九 Arabian: 🔊 (عدد 250,309) مئتانخمسون ألفاً و ثلاثمائة و تسعة Czech: 🔊 (číslo 250 309) dvěstě padesát tisíc třista devět Korean: 🔊 (번호 250,309) 이십오만 삼백구 Danish: 🔊 (nummer 250 309) tohundrede og halvtredstusindtrehundrede og ni Dutch: 🔊 (nummer 250 309) tweehonderdvijftigduizenddriehonderdnegen Japanese: 🔊 (数 250,309) 二十五万三百九 Indonesian: 🔊 (jumlah 250.309) dua ratus lima puluh ribu tiga ratus sembilan Italian: 🔊 (numero 250 309) duecentocinquantamilatrecentonove Norwegian: 🔊 (nummer 250 309) to hundre og femti tusen, tre hundre og ni Polish: 🔊 (liczba 250 309) dwieście pięćdziesiąt tysięcy trzysta dziewięć Russian: 🔊 (номер 250 309) двести пятьдесят тысяч триста девять Turkish: 🔊 (numara 250,309) ikiyüzellibinüçyüzdokuz Thai: 🔊 (จำนวน 250 309) สองแสนห้าหมื่นสามร้อยเก้า Ukrainian: 🔊 (номер 250 309) двісті п'ятдесят тисяч триста дев'ять Vietnamese: 🔊 (con số 250.309) hai trăm năm mươi nghìn ba trăm lẻ chín Other languages ...
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http://www.readbag.com/apcentralllegeboard-apc-members-repository-microeco-00 | 1,596,907,846,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738015.38/warc/CC-MAIN-20200808165417-20200808195417-00424.warc.gz | 171,784,232 | 5,482 | #### Read AP Microeconomics Free-Response Questions text version
2000 Advanced Placement Program® Free-Response Questions
The materials included in these files are intended for use by AP® teachers for course and exam preparation in the classroom; permission for any other use must be sought from the Advanced Placement Program. Teachers may reproduce them, in whole or in part, in limited quantities, for face-to-face teaching purposes but may not mass distribute the materials, electronically or otherwise. These materials and any copies made of them may not be resold, and the copyright notices must be retained as they appear here. This permission does not apply to any third-party copyrights contained herein.
2000 AP® MICROECONOMICS FREE-RESPONSE QUESTIONS
MICROECONOMICS
Section II Planning time--10 minutes Writing time--50 minutes Directions: You have fifty minutes to answer all three of the following questions. It is suggested that you spend approximately half your time on the first question and divide the remaining time equally between the next two questions. In answering the questions, you should emphasize the line of reasoning that generated your results; it is not enough to list the results of your analysis. Include correctly labeled diagrams, if useful or required, in explaining your answers. A correctly labeled diagram must have all axes and curves clearly labeled and must show directional changes.
1. The diagram above shows the cost and revenue curves for a monopoly. (a) How does a monopolist determine its profit-maximizing level of output and price? (b) Using the information in the graph, identify each of the following for the monopolist. (i) The profit-maximizing level of output and price (ii) The line segment of the demand curve that is elastic (c) Suppose that the industry depicted in the graph became perfectly competitive without changing the demand or cost curves. Identify the equilibrium price and output that would prevail in the perfectly competitive market. (d) Using the information in the graph, identify the area of consumer surplus for each of the following. (i) The profit-maximizing monopoly (ii) The perfectly competitive industry (e) Define allocative efficiency. (f) To be allocatively efficient, what level of output should the monopolist produce? (g) Should the government use a per-unit tax or a per-unit subsidy to lead the monopolist to produce the allocatively efficient level of output? Explain how this tax or subsidy would achieve the allocatively efficient level of output.
Copyright © 2000 College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Examination Board.
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2000 AP® MICROECONOMICS FREE-RESPONSE QUESTIONS
2. Assume that a firm produces output using one fixed input, capital, and one variable input, labor. The firm can sell all of the output it produces at a market price of \$3 each, can hire all of the workers it wants at a market wage rate of \$11 each, and has fixed costs of \$10. It faces the following production schedule. Number of Employees 0 1 2 3 4 5 6 Total Output 0 14 26 35 42 46 48
(a) In what kind of market structure does this firm sell its output? How can you tell? (b) In what kind of market structure does this firm hire its employees? How can you tell? (c) Using marginal revenue product analysis, how many employees should this firm hire to maximize short-run profits? How can you determine that? (d) Based on your answer in part (c), how many units of output will this firm produce? (e) At the level of output you identified in part (d), is the firm earning an economic profit, a normal profit, or suffering a loss? How can you tell? 3. Assume all of the following about imported and domestically produced shoes. They are sold in two separate and perfectly competitive markets. They are close substitutes. The demand for both is price elastic. Now assume that a tariff is imposed on imported shoes. (a) Using a correctly labeled graph, show the impact of the tariff on each of the following in the market for imported shoes. (i) Price (ii) Output (b) Using a new correctly labeled graph, show the impact of the tariff on each of the following in the market for domestically produced shoes. (i) Price (ii) Output (c) Given that the demand for imported shoes is price elastic, will expenditures on imported shoes by consumers increase, decrease, or remain the same? How do you know?
END OF EXAMINATION
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7126 | 1,038 | 4,921 | {"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} | 2.890625 | 3 | CC-MAIN-2020-34 | latest | en | 0.914672 |
https://space.stackexchange.com/questions/12956/calculating-synodic-periods-at-a-given-point-in-the-orbit | 1,701,223,369,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00225.warc.gz | 627,896,939 | 42,550 | # Calculating synodic periods at a given point in the orbit
Calculating the regular synodic period of two orbiting bodies is pretty straight forward, but what if I wanted to know when they would meet in a specific position? For example if they share a periapse. (By "same position" I mean that the two orbiting objects and the central body are on the same line)
I know this only applies if you can express one of the bodies period as a fraction of the other, so a 1 and a sqrt(2) orbit would never sync.
But what if I decided an acceptable margin? (like 10 degrees)
Currently I am brute forcing this.
My question is: Is it a way to calculate the next time two bodies would meet each other at the same location, given their orbital period and an acceptable error?
The objects does not have a noticeable effect on each other, nor on the central body. Restrict to circular or co-planar if necessary.
• Can you elaborate on "the same position"? Do you mean the same angular position (true anomaly plus argument of periapsis?). Dec 13, 2015 at 0:11
• @Brian Lynch Thank you for the comment. Yes, clarified it in the question. Dec 13, 2015 at 0:30
• Are the orbits circular? Co-planar? Assumed to both be very small compared to the central body with negligible effect on each other so as to have a simple solution? Dec 13, 2015 at 1:07
• @Mark Adler, Yes, their effect on each other and the central body is negligible. As for circular and co-planar, they do not have to be, but a solution restricted to one ore both of them would be helpful too. Feel free to give a partial answer. My current work around is using time increments. Dec 13, 2015 at 1:18
• Somewhat related: physics.stackexchange.com/questions/197481/…
– user7073
Jan 31, 2016 at 16:19
## 1 Answer
This is what eventually solved my problem. This JavaScript function takes the parameters: radiusRatio, the outermost objects radius divided by the innermosts. innerAnomaly, the innermost objects true anomaly from the reference direction, same for outerAnomaly, errorMargin is the maximal angle between either of the two radius vectors or the reference direction, and limit is how many of the innermost objects orbits to simulate for.
Note that the angle measure used for innerAnomaly, outerAnomaly and errorMargin is in fractions of an orbit, not degrees nor radians.
sameLine = function (radiusRatio,innerAnomaly,outerAnomaly,errorMargin,limit){
results = [];
newMargin = errorMargin;
periodRatio = Math.pow(radiusRatio,3/2);
for (i = 1; i < limit; i++){
anomaly = (outerAnomaly + (i - innerAnomaly)/periodRatio) % 1;
if (anomaly > 1 - anomaly){
anomaly = 1 - anomaly;
};
if (anomaly <= newMargin){
results.push([i - innerAnomaly,anomaly]);
newMargin = anomaly;
};
};
return results;
};
It outputs the an array containing sub arrays with the encounter data in the format of 1 numbers of orbits the innermost object has performed, and 2 what the error margin was. The next entry is the next time the error is smaller than that.
It is of course restricted to coplanar, circular orbits, and the mass of the two orbiting objects is negligible.
I have a more detailed explanation for a related problem at https://physics.stackexchange.com/a/232918/102747 | 778 | 3,219 | {"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} | 3.4375 | 3 | CC-MAIN-2023-50 | latest | en | 0.907651 |
http://mathhelpforum.com/pre-calculus/44901-find-z1-z2-leave-answer-polar-form-print.html | 1,527,408,047,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868132.80/warc/CC-MAIN-20180527072151-20180527092151-00103.warc.gz | 175,754,685 | 3,289 | # Find (Z1)(Z2) Leave answer in polar form
• Jul 30th 2008, 08:53 AM
Cyberman86
Find (Z1)(Z2) Leave answer in polar form
Given:
Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))
Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))
Find (Z1)(Z2) Leave answer in polar form
This is what i got so far and i don't know if i started right.
=sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
=sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))
and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.
• Jul 30th 2008, 09:15 AM
TheEmptySet
Quote:
Originally Posted by Cyberman86
Given:
Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))
Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))
Find (Z1)(Z2) Leave answer in polar form
This is what i got so far and i don't know if i started right.
=sqrt.12(cos(5pi/6) + i sin(5pie/6)) (cos(5pi/4) + i sin(5pi/4))
=sqrt.12(cos(5pi/4) + cos(5pie/6) i sin(5pi/4) + i sin(5pi/6) cos(5pi/4) - i sin(5pi/6) sin(5pi/4))
and i don't know if its right and don't know how to finish it, and if you can, can you show step by step if possible.
Z1= sqrt.6(cos(5pi/6) + i sin(5pi/6))
Z2= sqrt.2(cos(5pi/4) + i sin (5pi/4))
$\displaystyle z_1=\sqrt{6}e^{\frac{5 \pi}{6}i}$
$\displaystyle z_2=\sqrt{2}e^{\frac{5 \pi}{4}i}$
So we get
$\displaystyle z_1 \cdot z_2=\sqrt{6}e^{\frac{5 \pi}{6}i} \cdot \sqrt{2}e^{\frac{5 \pi}{4}i}=2\sqrt{3}e^{\frac{25 \pi}{12}i}=2\sqrt{3}e^{\frac{\pi}{12}i}=2\sqrt{3}\ left[ \cos\left( \frac{\pi}{12}\right)+i\sin\left( \frac{\pi}{12}\right)\right]$
• Jul 30th 2008, 09:17 AM
red_dog
If $\displaystyle z_1=r_1(\cos\phi_1+i\sin\phi_1)$ and $\displaystyle z_1=r_2(\cos\phi_2+i\sin\phi_2)$ then
$\displaystyle z_1z_2=r_1r_2[\cos(\phi_1+\phi_2)+i\sin(\phi_1+\phi_2)]$
• Jul 30th 2008, 09:50 AM
Soroban
Hello, Cyberman86!
Quote:
Given: .$\displaystyle \begin{array}{ccc}z_1&=& \sqrt{6}\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) \\ \\[-3mm]z_2 &=& \sqrt{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right) \end{array}$
Find: $\displaystyle (z_1)(z_2)$. .Leave answer in polar form.
We are expected to know this theorem . . .
If .$\displaystyle \begin{array}{ccc}z_1 &=& r_1\left(\cos\theta_1 + i\sin\theta_2\right) \\ z_2 &=& r_2\left(\cos\theta_2 + i\sin\theta_2\right) \end{array}$
. . . then: .$\displaystyle (z_1)(z_2) \;=\;(r_1\cdot r_2)\bigg[\cos\left(\theta_1+\theta_2\right) + i\sin\left(\theta_1+\theta_2\right)\bigg]$
. . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle \uparrow$ . . . . . . . . $\displaystyle \uparrow$
. . . . . . . . . . . . . . . . . . . . . . . . .
We have: .$\displaystyle (z_1)(z_2) \;=\;(\sqrt{6}\cdot\sqrt{2})\bigg[\cos\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) + i\sin\left(\frac{5\pi}{6} + \frac{5\pi}{4}\right) \bigg]$
. . . . . . . . . . . . . $\displaystyle = \;2\sqrt{3}\,\left(\cos\frac{25\pi}{12} + i\sin\frac{25\pi}{12}\right)$
. . . . . . . . . . . . . $\displaystyle =\;2\sqrt{3}\,\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$ | 1,383 | 3,050 | {"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-2018-22 | latest | en | 0.587348 |
http://www.docstoc.com/docs/132490877/Let-rt-be-a-fuzzy-singleton-of-R-and-A-be-a-fuzzy-module-of-R-module | 1,398,080,668,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00179-ip-10-147-4-33.ec2.internal.warc.gz | 391,475,269 | 19,551 | # Let rt be a fuzzy singleton of R and A be a fuzzy module of R module
Document Sample
``` J.Thi-Qar Sci. Vol.1 (1) March/2008
ISSN 1991- 8690 1661 - 0968 الترقيم الدولي
On Quasi Prime Fuzzy Submodules and
Quasi Primary Fuzzy Submodules
Rabee Hadi Jari Hassan Kiream Hassan
Department of mathematics - College of Education,
University of Thi-Qar
thiqaruni.org.
Abstract
In this paper we give some results about fuzzy quasi prime
submodules, also,we study the notion of quasi primary fuzzy
0 State in 18O by using Core – Polarization submodules.
2 effects
Department of Physics, College of Science, Thi-qar University
1. Introduction: We record here some basic
The notion of prime fuzzy concepts and clarify notions which are
submodules was introduced by Hatem used in the next work .For details,we
Y.K [1] and some results a bout this refer to [2,3].
concept are presented .In this paper Definition 2.1 [2]:
concept and study the extension of nonempty set M in to [0,1], then the
quasi primary fuzzy submodule by set
fuzzifing the notion of (ordinary)
quasi primary fuzzy submodule and
give some characterizations about it is called a level subset of X.
,also we give some properties related Definition 2.2 [3]:
to it.
Let be a fuzzy set
Through out the paper R is a
commutative ring with unity and M is in M, where, x M , t [0,1] , then xt is
an R-module.
2. Preliminaries:
J.Thi-Qar Sci. Vol.1 (1) March/2008
defined by : Where A is a proper fuzzy submodule
t if x y of X.
xt ( y )
o if x y for all y M.
x t is called a fuzzy singleton. 3. Quasi Prime Fuzzy submodules:
If x=0 and t=1, then: In this section we turn our
attention to the extension of a quasi
Definition 2.3 [4]: prime modules.
A fuzzy set X in R-module M is Definition 3.1 [1]:
called fuzzy R-module if and only if A fuzzy submodule A of a fuzzy
each x, y M and r R , then: module X of an R-module M is called
1 if y 0 quasi prime fuzzy submodule of X if
01 ( y) whenever a s bk xt A for fuzzy
0 if y 0
Definition 2.3 [4]: singleton a s , bk of R and x t X ,
A fuzzy set X in R-module M is implies that a s bk A or b k xt A .
called a fuzzy R-module if and only if Definition 3.2 [1]:
each x, y M and r R , then: Let X be a fuzzy module of an R-
1. X ( x y) min X ( x), X ( y). module M. X is called a quasi prime
2. X (rx) X ( x) fuzzy module if and only if F-annA is
3. X(0)=1. (0 is the zero element of M). a prime fuzzy ideal of R for each
Definition 2.4 [5]: nonempty fuzzy submodule A of X.
Let A and X be two fuzzy Equivalently, F-ann(xt)=F-ann(rsxt)
module of an R-module M, A is called for each xt X and for each fuzzy
a fuzzy submodule of X if and only if singleton rs of R such that
A X . r s F annX .
Definition 2.5 [5]: Definition 3.3 [6]:
Let rt and xs be two fuzzy Let A be a fuzzy submodule of a
singletons of R and M respectively. fuzzy module X, then A is called an
then rt x s (rx ) , Where =min{s,t}. essential fuzzy submodule if
Definition 2.6 [5]: A B 01 ,for each nonempty fuzzy
Let rt be a fuzzy singleton of R submodule B of X.
and A be a fuzzy module of R-module. Lemma 3.4[6]:
Then for any Let A be an essential fuzzy
w M , submodule of X,if a k X ,then there
sup{inf{t,A(x)}},if all w rx for some x M.
(rt A)(w) exists a fuzzy singleton rt of R such
o otherwise
that 01 rt a k A.
we noticed that if a fuzzy module X
Definition 2.7 [6]:
is quasi prime fuzzy module ,then any
Let X be a nonempty fuzzy
fuzzy submodule of X is a quasi prime
module of R-module M. The fuzzy
fuzzy module but the converse is not
annihilator of A denoted by F-annA is
true,see[1,Remarks and Examples
defined by:
2.1.3(4)]. However, we put certain
F annA (01 : A) xt : xt A 01 , x R, t [0,1]
condition under which the converse is
true.
J.Thi-Qar Sci. Vol.1 (1) March/2008
Proposition 3.5: R if for each rt a k I ,then either
Let A be an essential fuzzy rt I or a k I ,[7].
submodule of a fuzzy module X such
that F-ann xt=F-ann rsxt for each Corollary 3.7:
Let A be an essential fuzzy
xt X and for a fuzzy singleton rs of
submodule of a fuzzy module X, if A is
R such that rs F annxt , then X is a a faithful quasi prime module, then X
quasi prime fuzzy module if and only is a faithful quasi prime fuzzy module.
if A is a quasi prime fuzzy module. Proof:
Proof Since A is a faithful, then F-
Suppose that A is a quasi prime annA=01, also, A X , then
fuzzy R-module. F annX F annA by [6, Remark
To prove that 01 is a quasi prime fuzzy 1.2.18], which implies that
submodule of X. F annX 01 , but 01 F annX ,
hence F-annX=01, That is X is faithful,
Let al bk xt 01 moreover it follows that F-annX=F-
for fuzzy singletons al, bk of R and annA.
xt X .Since A is an essential fuzzy Now, to prove X is a quasi prime fuzzy
submodule, thus by lemma 3.4, there module,it is enough to prove that F-
exists a fuzzy singleton rs of R such ann(xt)=F-ann(rsxt) for each xt X
that 01 rs xt A ,that is and for each fuzzy singleton rs of R
rs F annxt . such that r s F annX .
Hence al bk (rs xt ) 01 . But A is a quasi Let al F annrs xt for a fuzzy
prime fuzzy R-module, thus 01 is a singleton rs of R and xt X , then
quasi prime fuzzy submodule of A by al (rs xt ) 01 .
[1, proposition 3.4.1], so either
al (rs xt ) 01 or bk (rs xt ) 01 , thus
Consequently, (al rs ) xt 01 . This
either al F annrs xt or implies that al rs F annxt , so
bk F annrs xt . But al rs F annX . But F-annX=F-
F annrs xt F annxt .
annA,then al rs F annA , thus F-
Consequently, either al F annxt or annA is prime fuzzy ideal by [1,
bk F annxt , that is either al xt 01 definition 2.1.1], since rs F annA.
or bk xt 01 . Hence X is a quasi prime So a l F annA , implies that
fuzzy module by [1, proposition 3.4.1]. al F annX , hence al X 01 , that
The converse follows directly by
is al xt 01 for all xt X , hence
[1, Remarks and Examples 2.1.3(4)].
Definition 3.6 [6]: F annrs xt F annxt …(1)
A fuzzy module X is called faithful Now, let al F annxt for a fuzzy
if F-annX=01. singleton al of R and xt X .
Recall that a fuzzy ideal I of a By lemma 3.4, there exists fuzzy
ring R is called a prime fuzzy ideal of singleton rs of R such that
01 rs xt A , that is rs F annxt ,
J.Thi-Qar Sci. Vol.1 (1) March/2008
so al rs xt 01 which implies that In this section we introduce the
al F annrs xt , thus concept of quasi primary fuzzy
submodule by fuzzifying the ordinary
F annxt F annrs xt …(2) notion of primary submodule .
From (1) and (2) we get F-annxt=F- Definition 4.1 [5]:
annrsxt, thus X is a quasi prime fuzzy A fuzzy submodule A of a fuzzy
module by [1, theorem 2.1.9(3)]. module X is called a primary fuzzy
Definition 3.8 [6]: submodule if for each fuzzy singleton
A fuzzy module X is called fuzzy rt of R and a k X such that
torsion free if F-annxt=01 for any
rt a k A , then either
x t X , x t 01 where
F annxt rk : rk fuzzy singleton of R, rk xt 01
.
Lemma 3.9: ,where
Let R be an integral domain and ( A : X ) {rt : rt X A, rt fuzzy sin gletonof R}
let X be a torsion free fuzzy module of Recall that if A is a fuzzy
R-module M, then X is a quasi prime submodule of a fuzzy module X of an
fuzzy module. R-module and I is a fuzzy ideal of R
Proof: then (A:I) define as:
Suppose that X is a torsion free ( A : I ) {a k , a k I A, a k X } ,note
fuzzy module. To prove 01 is quasi
that :If I=(rt),then:
prime fuzzy submodule of X.
( A : (rt )) {a k : rt a k A, a k X } ,[8].
Let al bk xt 01 ,for fuzzy singletons
Proposition 4.2:
al,bk of R and xt X . Let A be a proper fuzzy
If submodule of a fuzzy module X, then
bk xt 01 , then the following statements are
al F annbk xt , since equivalent:
1. A is a primary fuzzy submodule of
F ann(bk xt ) F anny , where X.
min k, t and y=bx by [1, Remark 2. (A:I) is a primary fuzzy submodule
1.2.14] so al F anny . But of X for each fuzzy ideal I of R.
3. (A:(rt)) is a primary fuzzy
F anny 01 because X is a torsion
submodule of X for each fuzzy
free, so a l 01 . singleton rt of R.
On the other hand, 01 F annX , Proof:
which implies that al F annX , 1 2:
Suppose that rt x k ( A : I ) for each
that is
x k X and a fuzzy singleton rt of R
al X 01 , by [6, definition 1.2.17], and I be a fuzzy ideal of R.It follows
hence al xt 01 for all xt X . that al rt x k A for each fuzzy
Thus, X is a quasi prime fuzzy module singleton al of I, then either al x k A
by [1,proposition 3.4.1]. for each fuzzy singleton al of I or
rt n ( A : X ) , n Z .
4. Quasi Primary fuzzy submodule.
J.Thi-Qar Sci. Vol.1 (1) March/2008
So, either x k ( A : I ) or rtn X A, Now, let ak A : rt B for all
since A ( A : I ), by [8, theorem 3.3], rt B A , hence a k ( A : rt B )
n
for
hence rt n X ( A : I ) , implies that some n Z
and so a kn rt B A ,which implies that
rt n (( A : I ) : X ) ,it follows, either
a k rt ( A : B ) . But (A:B) is a primary
n
x k ( A : I ) or r (( A : I ) : X ) , hence
n
t fuzzy ideal of R and rt ( A : B) ,so
a
(A:I) is a primary fuzzy submodule of n m
X. k ( A : B) for some m Z ,thus
2 3: a ( A : B)
s
k where s=nm so that
It is clear.
3 1: ak A : B ; that is
It follows easily by taking r=1 and t=1. A : rt B A : B and hence
Recall that a proper submodule N A : B A : rt B .
ofan R-module M is said to be quasi
primary submodule if [N:K] is a
primary ideal of R if for each Conversely, to prove (A:B) is a
submodule K of M such that primary uzzy ideal of R for each
N K ,where A B.
[N:K]={r R:rk N},[9]. Suppose that a k rt ( A : B) for a fuzzy
We shall fuzzify this concept as singleton ak and rt of R. If rt ( A : B) ,
follows:
then ak ( A : rt B) A : rt B = A : B
Definition 4.3 :
A proper fuzzy submodule A of a and hence A is a quasi primary fuzzy
fuzzy module X is said to be quasi submodule of X.
primary fuzzy submodule if (A:B) is a Proposition 4.5:
primary fuzzy ideal of R for each Let A be a fuzzy submodule of a
fuzzy submodule B of X such that fuzzy module X, then A is a quasi
A B where primary submodule of X if and only if
(A:I) is a quasi primary fuzzy
( A : B) rt / rt B A, for a fuzzy singleton rt of R
submodule of X for each fuzzy ideal I
of R.
. Proof:
Proposition 4.4: Suppose that A is a quasi primary
Let A be a proper fuzzy fuzzy submodule of X.
submodule of a fuzzy module X, then By [8, theorem 3.3], (A:I) is a fuzzy
A is a quasi primary fuzzy submodule submodule of X.
of X if and only if A : B A : rt B To prove (A:I) is a quasi primary
for each primary fuzzy submodule B fuzzy submodule of X, we must prove
of X and for each fuzzy singleton rt of ((A:I):B) is a primary fuzzy ideal for
R such that A B and rt B A . each ( A : I ) B.
Proof: A ( A : I ) B by [8, lemma 2.16(1)] .
It is clear that A : B A : rt B . Now, suppose that a k rt (( A : I ) : B)
for fuzzy singleton rt and ak of R, then
J.Thi-Qar Sci. Vol.1 (1) March/2008
a k rt B ( A : I ), so a k rt BI A , suppose Conversely, if 01 is quasi primary
that a k (( A : I ) : B) , then a k BI A , fuzzy submodule of X, to prove X is a
quasi primary fuzzy module. Since 01
hence a k rt I ( A : B ) . is a quasi primary fuzzy submodule,
But, (A:B) is a primary fuzzy ideal by then (01:A) is a primary fuzzy ideal for
definition 4.3 and a k I ( A : B ) .Thus, each nonempty fuzzy submodule.
rt n ( A : B ) for some n Z and so But (01:A)=F-annA, so X is a quasi
primary fuzzy R-module.
rt n B A ( A : I ) , therefore
rt (( A : I ) : B ) and hence (A:I) is a
n
REFRENCES:
quasi primary fuzzy submodule of X.
[1] Hatem Y.K.,(2001), "Fuzzy Quasi –prime
The converse follows by taking Modules and Fuzzy Quasi –prime
I R , where R (a) 1 for all a R submodules ", M.Sc. thesis,
and by similar to the proof of [6, University of Baghdad, college of
Education Ibn- AL- Haithem ,(69)p.
proposition 2.1.16].
[2] Martines L.,(1996), "Fuzzy modules over
Recall that an R-module m is said fuzzy rings in connection with Fuzzy
to be a quasi primary module if annN ideal of Fuzzy ring ". J.Fuzzy
is a primary ideal of R, for each non Mathematics, Vo1.4:843-857.
zero submodule N of M,[9].
[3] Liu W. J,(1982),”Fuzzy invariant
Our attempt to fuzzify this concept subgroups and Fuzzy ideals”, fuzzy
is as follows: sets and
systems,Vol.8:133-139.
Definition 4.6 : [4] Nanda S. (1989),"Fuzzy Modules over
A fuzzy module X of R –module Fuzzy rings ", Bull. coll. Math. Soc.,
Vo1. 18 :197-200.
M is said to be quasi primary fuzzy [5] Mashinchi M. and Zahedi M.M.,(1992)"
module if and only if F-annA=(01:A) is On L-Fuzzy primary submodules ",
a primary fuzzy ideal for each Fuzzy sets and systems, Vol.49:231-
nonempty fuzzy submodule A of X . 236.
[6] Jari R.H.(2001)," prime Fuzzy
submodules and prime Fuzzy
Proposition 4.7: modules ", M.Sc. Thesis,
A fuzzy module X is a quasi University of Baghdad,
primary fuzzy module if and only if 01 college of Education, Ibn –AL-
is a quasi primary fuzzy submodule of Haithem,(64)p.
X.
[7] Mukhrjie T.K.,(1989)” Prime Fuzzy
Proof: Ideals in Rings”, Fuzzy Sets and
Suppose that X is a quasi primary Systems Vol. 32:337-341.
fuzzy module. [8] Zahedi M.M.,(1992),"On L-Fuzzy
To prove 01 is a quasi primary fuzzy Residual Quotient
submodule of X, since X is a quasi Modules", Fuzzy sets and
systems, Vol.51:333-344.
primary fuzzy submodule, then F- [9] Abdul-al-Kalik A.J.,(2005),”Primary
annA is a primary fuzzy ideal for each Modules”, M.Sc. thesis, University of
nonempty fuzzy submodule A of Baghdad,college of Education
X.Thus (01:A) is primary fuzzy ideal Ibn- AL-Haithem,(77)p.
and hence 01 is quasi primary fuzzy
submodule of X.
.J.Thi-Qar Sci )1( 1.Vol 8002/March
On Quasi Prime Fuzzy Submodules and Quasi Primary Fuzzy
Submodules
حسن كريم حسن ربيع هادي جاري
قسم الرياضيات - كلية التربية - جامعة ذي قار
الخالصة:
في هذا البحث اعطينا بعض النتائج عن المقاسات الجزئية شبة االولية الضبابية ,كذلك درسنا
مفهوم المقاسات الجزئية شبه االبتدائية الضبابية .
```
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https://math.stackexchange.com/questions/3225294/topology-of-sets-with-countable-complements | 1,579,348,563,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250592565.2/warc/CC-MAIN-20200118110141-20200118134141-00265.warc.gz | 553,775,462 | 31,486 | # topology of sets with countable complements
I am trying to write a proof for q 2 on page 94 in third chapter of Simmon's "Topology and Modern analysis"
Let $$X$$ be a non empty set. Consider a class of subsets including $$\emptyset$$ set and all subsets which have a countable complement. Is this a topology on $$X$$.
My attempt: Let the mentioned class be denoted by $$\mathscr{C}$$
Proof that arbitrary unions are included in $$\mathscr{C}$$:
Let $$A_i$$ be sets of any countable non empty subclass $$\mathscr{A}$$ of $$\mathscr{C}$$, then the complements $$A_i'$$ are countable.
Consider union of sets in $$\mathscr{A}$$ given by $$\bigcup\limits_{A_i \in \mathscr{A}}A_i$$ Its complement is given by $$\bigcap\limits_{A_i \in \mathscr{A}}A_i'$$.
Since $$\#(\bigcap\limits_{A_i \in \mathscr{A}}A_i') \le \#(A_i')$$ for all $$i$$, where $$\#()$$ is cardinality,
$$\bigcap\limits_{A_i \in \mathscr{A}}A_i'$$ should be countable
Thus, $$\bigcup\limits_{A_i \in \mathscr{A}}A_i$$ should be in $$\mathscr{C}$$ as its complement is countable.
Proof that finite intersections are included in $$\mathscr{C}$$:
Similarly, let $$\mathscr{B}$$ be a finite suclass of $$\mathscr{C}$$. Let sets in $$\mathscr{B}$$ be denoted by $$B_i$$. Then intersection of sets in $$\mathscr{B}$$ is given by $$\bigcap\limits_{B_i \in \mathscr{B}}B_i$$ and its complement by $$\bigcup\limits_{B_i \in \mathscr{B}}B_i'$$. Since this is a finite union of countable sets, it must be countable.
$$X$$ itself is in $$C$$ as $$\emptyset$$ is countable.
So we can conclude that $$\mathscr{C}$$ forms a topology.
Proof that finite union of countable sets is countable:
Let $$Y_i$$ for $$1 \le i \le n$$ be countable sets. Let $$y_{i,j}$$ represent $$j^{th}$$ element of $$i^{th}$$ class. Then elements of $$\bigcup\limits_{1\le i \le n}Y_i$$ maps into set of integers using $$f(y_{i,j})=n(j-1)+i$$. Since at least one of the sets is countable and not finite set of integers maps into elements of this class.
Since $$\bigcup\limits_{1\le i \le n}Y_i$$ maps into set of integers and vice versa, they are numerically equivalent using Schroeder Bernstein's theorem.
• minor quibble: if $\mathcal{C}$ is empty its union is $\emptyset$ so in the class too. Otherwise we have some $C_i$ in the family and indeed $(\bigcup \mathcal{C})^\complement \subseteq C_i^\complement$ hence at most countable. – Henno Brandsma May 14 '19 at 6:11
• Thanks for pointing out the mistake – Curious May 14 '19 at 6:21
• Not really a mistake just an extra case. – Henno Brandsma May 14 '19 at 6:23
• This is generally called the co-countable topology on $X$. – DanielWainfleet May 14 '19 at 9:14
Your mistake is that the Schröder-Bernstein theorem can't apply in this case since $$f$$ is not injective.
Hint: $$|\mathbb{N}| = |\mathbb{N}^2|$$, and use induction.
• I did not understand why it will not be injective. As every element in $y_{i,j}$ will map to a unique integer. Some set may have finite elements so mapping may be into. If that is not the case and mapping is one to one we know that equivalence exists in any case – Curious May 14 '19 at 6:09
• If $n=2$, then doesn't $f(y_{1,3}) = f(y_{2,1})$? – Anthony Ter May 14 '19 at 6:25 | 1,011 | 3,204 | {"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": 40, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-05 | latest | en | 0.865597 |
https://itfeature.com/hypothesis/effect-size/cohen-effect-size/ | 1,716,859,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00422.warc.gz | 280,367,557 | 26,059 | # Cohen Effect Size and Statistical Significance
Statistical significance is important but not the most important consideration in evaluating the results. Because statistical significance tells only the likelihood (probability) that the observed results are due to chance alone. Considering the effect size when obtaining statistically significant results is important.
Effect size is a quantitative measure of some phenomenon. For example,
• Correlation between two variables
• The regression coefficients ($\beta_0, \beta_1, \beta_2$) for the regression model, for example, coefficients $\beta_1, \beta_2, \cdots$
• The mean difference between two or more groups
• The risk with which something happens
The effect size plays an important role in power analysis, sample size planning, and meta-analysis.
Since effect size indicates how strong (or important) our results are. Therefore, when you are reporting results about the statistical significance for an inferential test, the effect size should also be reported.
For the difference in means, the pooled standard deviation (also called combined standard deviation, obtained from pooled variance) is used to indicate the effect size.
### The Cohen Effect Size for the Difference in Means
The effect size ($d$) for the difference in means by Cohen is
$d=\frac{mean\, of\, group\,1 – mean\,of\,group\,2}{SD_{pooled}}$
Cohen provided rough guidelines for interpreting the effect size.
If $d=0.2$, the effect size will be considered as small.
For $d=0.5$, the effect size will be medium.
and if $d=0.8$, the effect size is considered as large.
Note that statistical significance is not the same as the effect size. The statistical significance tells how likely it is that the result is due to chance, while effect size tells how important the result is.
Also note that the statistical-significance is not equal to economic, human, or scientific significance.
For the effect size of the dependent sample $t$-test, see the post-effect size for the dependent sample t-test
See the short video on Effect Size and Statistical Significance
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# An unusually strong cyclist can, it is hoped, provide enough
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An unusually strong cyclist can, it is hoped, provide enough [#permalink]
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14 Oct 2005, 13:38
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3. An unusually strong cyclist can, it is hoped, provide enough power to set a new distance record for human-powered aircraft in MIT’s diaphanous construction of graphite fiber and plastic.
(A) can, it is hoped, provide enough power to set
(B) it is hoped, can provide enough power that will set
(C) hopefully can provide enough power, this will set
(D) is hopeful to set
(E) hopes setting
_________________
hey ya......
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14 Oct 2005, 15:32
(B) verb 'can' needs. 'Cyclist' is not hoped!
(C) don't pick 'hopefully'
(D) unidiomatic
(E) awkward
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18 Oct 2005, 12:58
Yup OA is A.
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18 Oct 2005, 13:01
A.
I read on this forum that given the choice between hopefully and 'it is hoped', the later is correct.
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19 Oct 2005, 11:10
Late, but A.
I keep seeing this present tense verb+it is hoped+present tense verb structure all over the place. Might as well call it a SC pattern.
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04 Jul 2007, 21:50
Can any one please explain how the construct "it is hoped," is valid...i.e
How Present Progressive verb (is) + past participle (hoped) is a valid construct? Can you please explain with an example?
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15 Sep 2007, 13:25
what is wrong with D. is hopeful to set
Doesn't GMAT prefer something shorter?
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Re: SC # unusually strong cyclist [#permalink]
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15 Sep 2007, 21:31
nakib77 wrote:
3. An unusually strong cyclist can, it is hoped, provide enough power to set a new distance record for human-powered aircraft in MIT’s diaphanous construction of graphite fiber and plastic.
(A) can, it is hoped, provide enough power to set
(B) it is hoped, can provide enough power that will set
(C) hopefully can provide enough power, this will set
(D) is hopeful to set
(E) hopes setting
CDE change the meaning of the sentence.
C: is awkward and wordy
D: ambigious is the biker hopeful or are the people making the aircraft hopeful. Also eliminates "can". Totally changes the meaning of the original sentence.
E: Same reasoning as D.
B: awkward and wordy, why say more when u can say less. Also i dunno for sure, but this sentence may change the meaning.
Vote A.
Re: SC # unusually strong cyclist [#permalink] 15 Sep 2007, 21:31
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An unusually strong cyclist can, it is hoped, provide enough 1 09 Aug 2007, 23:53
An unusually strong cyclist can, it is hoped, provide enough 3 10 Jul 2007, 11:19
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https://www.iodraw.com/en/blog/200970790 | 1,620,624,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989030.87/warc/CC-MAIN-20210510033850-20210510063850-00476.warc.gz | 848,989,937 | 4,880 | <> Analysis of problem one ：
first , Through fluid mechanics , structural mechanics , architecture , Material mechanics and other related knowledge , Three possible better three-dimensional foundation geometries are assumed , We analyze these three three-dimensional shapes , Choose the best , Three dimensional foundation shape that can resist the impact of tidal waves to the greatest extent , This allows sandbars to survive the impact of tidal waves for the longest time . Considering only the shape factor and the wave height of the wave, the sand castle is impacted by different directions , The actual situation of force and erosion on the front is discussed . In addition to the types of sand, the volume of sand and the ratio of water to sand, the factors that affect the destructive power of waves to sandbars are as follows , Also with the speed of the waves , The size of the spray , Angle of spray , Salinity of seawater , The angle of the beach from the horizontal , Shape of sand particles , The water absorption of sand is related to many factors . The sandcastle can be seen as a whole , It produces a deformation resistance force when subjected to impact , Then the impact part is analyzed separately , The external force of multiple sand particles , Internal support , Internal compressive stress , The dynamic equilibrium between sand and sand due to the viscous force produced by the action of water . And when impacted, a single sand particle will move due to the impact force , When the waves recede, the sand particles will be carried away by the current due to the buoyancy and tensile stress of the water , On the whole, the sand castle is preserved by the cushion of sand particles and elastic force . Due to the impact force, the shape of sand castle produces plastic deformation to adapt to the sea wave , In the next impact, the impact will be relieved .
The flow chart is as follows ：
<> Analysis of problem two ：
Different water sand ratio affects the impact resistance of sand castle , And different water sand ratio , Sand Fort foundation has different strength and elasticity , It has a different buffering capacity , Different locations of sand and sand can be known by analyzing the strength of the foundation , Some locations need better impact resistance , Some positions need better support , There are places where more tension is needed , Then you need to check every location of the sandcastle , Different water sand ratio is selected for each layer , For example, choose a lower strength in the front, but have better viscous force , It can resist more tensile stress , In the rear, water sand ratio is used to produce high strength , In order to be able to resist greater impact , Maximum impact resistance . In the process of impact , Because of the erosion of sea water , The water content and salt content in the sand castle will change , Therefore, the impact resistance of sandbars will change with the impact , The impact resistance of sand castle is a dynamic model .
<> Analysis of question 3 ：
When the sand fort has the best shaped abutment and it rains , At this time, the sand castle is impacted by the wave on the front side of the sand castle , The impact of rain on the top of the sandcastle （ Negligible ） And the erosion effect of the flow formed by the accumulation of rainwater on the top of the sandcastle . Second, when it rains , Rainwater seeps through the top , Then the water sand ratio of sand castle will change accordingly , There will be a lot of water in the middle of the sand castle, which can not be drained out, resulting in the decline of the anti deformation ability of the middle position .
Technology
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• FLUENT - Supersonic Flow Over a Wedge - Step 5 * New
# FLUENT - Supersonic Flow Over a Wedge - Step 5 * New
Useful Information
## Step 5: Solve!
Solve > Controls or Solutions > Solution Controls
Click on the Equations button and select Flow, then click OK. Also, set the Courant Number to 0.1.
Solve > Methods or Solutions > Solution Methods
We'll use a second-order discretization scheme. Under Spatial Discretization, set Flow to Second Order Upwind.
Solve > Initialization or Solutions > Solution Initialization
This is where we set the initial guess values for the iterative solution. We'll use the farfield values (M=3, p=1 atm, T=300 K) as the initial guess for the entire flowfield. Select farfield under Compute From. This fills in values from the farfield boundary in the corresponding boxes. (Alternately, I could have typed in these values but I like to palm off as much grunt work as possible to the computer.)
Click Initialize. Now, for each cell in the mesh, M=3, p=1 atm, T=300 K. These values will of course get updated as we iterate the solution below.
FLUENT reports a residual for each governing equation being solved. The residual is a measure of how well the current solution satisfies the discrete form of each governing equation. We'll iterate the solution until the residual for each equation falls below 1e-6.
Solve > Monitors
Select Residuals - Print and click Edit. Set Absolute Criteria for all equationsto 1e-6.
Also, click on Plot. This will plot the residuals in the graphics window as they are calculated; giving you a visual feel for if/how the iterations are proceeding to convergence.
Click OK.
Main Menu > File > Write > Case...
This will save your FLUENT settings and the mesh to a "case" file. Type in wedge.cas for Case File. Click OK.
Solve > Run Calculation...
Set the Number of Iterations to 1000. Click Calculate.
The residuals for each iteration are printed out as well as plotted in the graphics window as they are calculated. The residuals after 1000 iterations are not below the convergence criterion of 1e-6 specified before. So run the solution for 1000 more iterations. The solution converges in about 1510 iterations; the residuals for all the governing equations are below 1e-6 at this point.
Save the solution to a data file:
Main Menu > File > Write > Data...
Enter wedge.dat for Data File and click OK. Check that the file has been created in your working directory. You can retrieve the current solution from this data file at any time.
Go to 6. Results
See and rate the complete Learning Module
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https://edurev.in/course/quiz/attempt/-1_Assertion-Reason-Test-Coordinate-Geometry-2/a9ecbeb0-701f-4fc7-b868-f48456d5f06c | 1,660,785,270,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00212.warc.gz | 226,985,954 | 39,076 | Assertion & Reason Test: Coordinate Geometry - 2
# Assertion & Reason Test: Coordinate Geometry - 2
Test Description
## 10 Questions MCQ Test Online MCQ Tests for Class 10 | Assertion & Reason Test: Coordinate Geometry - 2
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Assertion & Reason Test: Coordinate Geometry - 2 - Question 1
### Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.Assertion : The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2 : 7 internally.Reason : Three points A,B and C are collinear if AB + BC = AC
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 1 We know that the three points A,B and C are collinear if AB + BC = AC
So, Reason is correct.
Let the ratio is k : 1. Here, x1 = -3, y1 = 10, x2 = 6, y2 = -8, x = -1, y = 6
Now, y-coordinate
⇒ -8k + 10 = 6k + 6
⇒ 10 – 6 = 6k + 8k
⇒ 14k = 4
⇒ k = 4/14 = 2/7
So, Assertion is correct
But reason (R) is not the correct explanation of assertion (A).
Assertion & Reason Test: Coordinate Geometry - 2 - Question 2
### Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.Assertion : The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2 :7 internally.Reason : Three points A,B and C are collinear if area of △ ABC = 0 .
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 2 Using section formula, we have
- k- 1 = 6k -3
7k = 2
k = 2/7
Ratio be 2 :7 internally.
Also, if ar(△ ABC) = 0
A,B and C all these points are collinear.
Assertion & Reason Test: Coordinate Geometry - 2 - Question 3
### Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.Assertion : The value of y is 6, for which the distance between the points P(2,- 3) and Q(10, y) is 10.Reason : Distance between two given points A(x1, y1) and B(x2, y2) is given by,
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 3 We know that the Distance between two given points A(x1, y1) and B(x2, y2) is given by,
So, Reason is correct.
Now, PQ = 10 ⇒ PQ2 = 100
⇒ (10 – 2)2 + (y + 3)2 = 100
⇒ (y + 3)2 = 100 – 64 = 36
⇒ y + 3 = ±6
⇒ y = -3 ± 6 ⇒ y = 3, -9
So, Assertion is not correct
Assertion & Reason Test: Coordinate Geometry - 2 - Question 4
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : If A (2a, 4a) and B (2a, 6a) are two vertices of a equilateral triangle ABC then the vertex C is given by (2a + a√3, 5a).
Reason : In equilateral triangle all the coordinates of three vertices can be rational.
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 4
LHS = rational
RHS = irrational
Hence, (x1, y1) (x2, y2) and (x3, y3) cannot be all rational.
Assertion & Reason Test: Coordinate Geometry - 2 - Question 5
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (0, 4) lies on y -axis.
Reason : The x co-ordinate on the point on y -axis is zero.
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 5 We know that the if the point lies on y-axis, its x-coordinate is 0.
So, Reason is correct.
The x co-ordinate of the point (0, 4) is zero.
So, Point (0, 4) lies on y -axis.
So, Assertion is also correct
Assertion & Reason Test: Coordinate Geometry - 2 - Question 6
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The value of y is 6, for which the distance between the points P(2, -3) and Q(10, y) is 10.
Reason : Distance between two given points A (x1, y1) and B(x2, y2) is given 6,
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 6
PQ = 10
PQ2 = 100
(10 - 2)2 + (y + 3)2 = 100
(y + 3)2 = 100 - 64 = 36
y + 3 = + 6
y = - 3 + 6
y = 3, -9
Assertion & Reason Test: Coordinate Geometry - 2 - Question 7
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : C is the mid-point of PQ, if P is (4, x), C is (y,- 1) and Q is (- 2, 4), then x and y respectively are -6 and 1.
Reason : The mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 7 We know that the mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) is
So, Reason is correct.
Since, C(y,- 1) is the mid-point of P(4, x) and Q(- 2, 4).
⇒ x = -6
So, Assertion is correct
Assertion & Reason Test: Coordinate Geometry - 2 - Question 8
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : The point (0, 4) lies on y -axis.
Reason : The x co-ordinate on the point on y -axis is zero.
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 8 The x co-ordinate of the point (0, 4) is zero.
Point (0, 4) lies on y -axis.
Assertion & Reason Test: Coordinate Geometry - 2 - Question 9
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion : Ratio in which the line 3x + 4y = 7 divides the line segment joining the points (1, 2) and (- 2, 1) is 3 : 5Reason : The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) in the ratio m1 : m2 is
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 9 We know that the coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) in the ratio m1 : m2 is
So, Reason is correct.
Let the ratio is k : 1.
Here, x1 = 1, y1 = 2, x2 = -2, y2 = 1, m1 = k, m2 = 1
Now, 3x + 4y = 7
⇒ 7k + 2k = 11 – 7 ⇒ 9k = 4 ⇒ k = 4/9
So, Assertion is not correct
Assertion & Reason Test: Coordinate Geometry - 2 - Question 10
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Assertion: The point which divides the line joining the points A(1, 2) and B(- 1, 1) internally in the ratio 1: 2 is
Reason : The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) in the ratio m1 : m2 is
Detailed Solution for Assertion & Reason Test: Coordinate Geometry - 2 - Question 10
We know that the coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2) in the ratio m1 : m2 is
So, Reason is correct.
Here, x1 = 1, y1 = 2, x2 = -1, y2 = 1, m1 = 1, m2 = 2
So, Assertion is correct.
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http://vort.org/2012/11/ | 1,397,621,455,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00092-ip-10-147-4-33.ec2.internal.warc.gz | 255,429,627 | 9,562 | # Russell's Blog
## Why doesn't your lab have a 3D printer yet?
Posted by Russell on November 20, 2012 at 6:30 a.m.
Electrophoresis setups are like Tupperware. You can never find the right lid when you need it, and someone always seems to be borrowing the doohicky you need.
Here in the Eisen Lab, it turns out we've been using Marc Facciotti's electrophoresis stuff for years. He keeps his stuff organized, and, well... that's not been our strong suit lately. John, our lab manager, has been gently but inexorably herding us towards a semblance of respectability in our lab behavior. As part of this, he decided that it was time for us to get our electrophoresis stuff straightened out. So, he ordered a bunch nice of gel combs from one of our suppliers. They cost \$51 each (see the "12 tooth double-sided comb", catalog number 669-B2-12, for the exact one pictured below). We bought six of them with different sizes and spacing, for a total exceeding \$300.
While I appreciate that companies need to make money, this is a ridiculous price for a lousy little scrap of plastic. \$300 for a couple of gel combs is cartel pricing, not market pricing. Fortunately, we happen to have a very nice 3D printer. It is very good at making little scraps of plastic. So, I busted out the calipers and tossed together some models of gel combs in OpenSCAD. A few minutes of printing later, and the \$51 gel combs are heading back to the store.
Here's the code for the six well 1.5mm by 9mm comb :
```f=0.01;
difference(){
difference(){
union(){
cube( [ 80, 27, 3 ] );
translate( [ 5.25, 14.3, f ] ) cube( [ 68, 9.3, 7.25 ] );
}
for ( i = [ 0:5 ] ) {
translate( [ 17.1+i*11.0, -f, -f ] ) cube( [ 1.75, 12, 5 ] );
}
}
union(){
translate( [ -f, -f, -f ] ) cube( [ 7, 12, 7] );
translate( [ 73+f, -f, -f ] ) cube( [ 7, 12, 7] );
translate( [ 0, -f, 1.6] ) cube( [ 80, 12, 8] );
}
}
```
Pretty easy to grasp, even if you've never seen SCAD before.
So, how much did this cost?
I ordered this plastic from ProtoParadigm at \$42 for a kilogram. That's about four pennies a gram. Each of these gel combs cost about 21 cents to print. That's 1/243rd the price.
The 3D printer cost €1,194.00 (\$1524.62), which is less than the laptop I use for most of my work. The savings on just these gel combs has recuperated 18% of the cost of the printer.
It's also important that I was able to make some minor improvements to the design. The printed combs fit into the gel mold a bit better than the "official" ones. I also made separate combs for the 1.0mm and 1.5mm versions, and the labels are easier to read. If I wanted, tiny tweaks to my SCAD file would let me make all sorts of fun combinations of thicknesses and widths that aren't available from the manufacturer. So, these gel combs are not only 1/243rd the price, but they are also better.
If you read the media hype about 3D printing, you will undoubtedly encounter a lot of fantastical-sounding speculation about how consumers will someday be able to print living goldfish, or computers, or bicycles. Maybe so. Maybe not. However, right now, you can print basic lab supplies and save a pile of money.
Buy your lab manager a little FDM printer and hook them up with some basic CAD training. Yes, the printer will probably mostly get used to make bottle openers and Tardis cookie cutters. So what? Your paper-printer, if you will excuse the retronym, mostly gets used for non-essential stuff too. I'd wager that for every important document printed in your lab, a hundred sheets have gone to Far Side cartoons and humorous notices taped up in the bathroom. It's a negligible expense compared to the benefits of having a machine that spits out documents when you really need them, and the social value of those the Far Side cartoons probably sums to a net positive anyway.
Conclusion : If you have a lab, and you don't have a 3D printer, you are wasting your money. Seriously.
In the time it took write this post, I printed \$150 worth of gel combs, and it cost less than a cup of coffee.
Updates : Here is the tweet I originally posted about this article, before the URL for it vanishes into Twitter's memory hole. Here's an encouraging post from the Genome Web blog, and a nice article by Tim Dean at Australian Life Scientist. My article here seems to have spawned a thread on BioStar. Also, it made Ed Yong's Missing Links for November 24 over at Discover, and Megan Treacy did a really spiffy article over at Treehugger.
Many people have asked, and so I decided to see how well these kinds of 3D printed parts do in the autoclave. I tried it out with a couple of bad prints, and they seemed to hold up just fine after one or two cycles. Very thin parts did warp a bit, though, so I recommend printing parts you plan to autoclave nice and solid. Here is a before and after of a single-wall part (less than half a millimeter thick). I was expecting a puddle.
Update 2 : Check out Lauren Wolf's awesome article in Chemical & Engineering News, featuring the infamous gel combs, among other things!
## Journals are the problem and the solution
Posted by Russell on November 15, 2012 at 8:46 a.m.
Titus wrote an interesting post yesterday about addressing some of what I'll call structural problems in scientific research.
This is one of a bunch of posts on what I'm calling 'w4s' -- using the Web, and principles of the Web, to improve science. The others are:
• The awesomeness we're experiencing, which provides some examples of current awesomeness in this area.
• The challenges ahead, which covers some of the reasons why academia isn't moving very fast in this area.
• Strategizing for the future, which talks about technical strategies and approaches for helping change things.
• Tech wanted!, which gives some specific enabling technologies that I think are fairly easy to implement.
He goes on to throw some well-aimed brickbats at the system of publishing and grant reviewing, and how that plays out for researchers who actually do things other than crank out traditional research papers :
As an increasing amount of effort is put towards generating data sets and correlating across data sets, funding agencies are certainly trying to figure out how to reward such effort. The NSF is now explicitly allowing software and databases in the personnel BioSketches, for example, which is a great advance. Surely this is driving change?
The obstacle, unfortunately, may be the peer reviewer system. Most grants and papers are peer reviewed, and "peers" in this case include lots of professors that venerate PDFs and two-significant-digit Impact Factors. Moreover, many reviewers value theory over practice -- Fernando Perez has repeatedly ranted to me about his experience on serving on review panels for capacity-building cyberinfrastructure grants where most of the reviewers pay no attention whatsoever to the plans for software or data release, and even poo-poo those that have explicit plans. And if a grant gets trashed by the reviewers, it's very hard for the program manager to override that. The same thing occurs with software, where openness and replicability don't figure into the review much. So there's a big problem in getting grants and papers if you're distracting yourself by trying to be useful in addition to addressing novelty, impact, etc.
The career implications are that if you're stupid enough to make useful software and spend your time releasing useful data rather than writing papers, you can expect to be sidelined academically -- either because you won't get job offers, or because you won't get grants when you do have a job.
I propose a simple solution : Start better journals.
The fundamental unit of exchange in the academic reputation economy is the publication. At the moment, it is very hard to publish things that are not "normal" research papers. PLOS ONE has helped by shifting the review criteria towards correctness and leaving judgement of novelty to the community. However, PLOS ONE is still nevertheless an awkward place to publish a piece of software, for example.
Tool builders are expected to produce three publications for every one research result. First they have to design, assemble, test and publish the tool. Then they have to write a paper about the tool. Then they have to document the tool. After that, they are expected to release updates, patches and improvements. The only one of these items that "counts" in the academic sense is the one that generates a DOI number -- the paper. This is usually the least valuable of all the things produced. Honestly, how many of you BLAST users have actually read the BLAST paper?
What is needed are journals that let you publish things that aren't strictly papers. It seems perfectly reasonable that a panel of peers should be able review software on my GitHub account and bestow a DOI number on a tag they feel meets the criteria set by the editorial board that badgered them into participating in the review. It seems perfectly reasonable that when I cut a major update of the software, I might submit it for review. This would be extremely wonderful, as it would lead to reviews and critiques the code itself, which almost never happens when one submits an application note.
The same should apply to tool documentation. Science suffers a great deal because documentation is missing, poor, or out of date. The only way to fix that is to let tool builders have their tool documentation reviewed and published. Perhaps some journals might review code only, others documentation only, and others code and documentation together. The same goes for updates. If you want academics to do something, you have to provide rewards in the currency of academia.
It is also important not to overlook things beyond software. What about people building methods and protocols? There are already journals that publish methods papers, but a lot more could be done to make these publications more useful. JoVE, the Journal of Visualized Experiments, is a pretty interesting step in that direction, but much much more needs to be done. Most scientists have no experience whatsoever in video production, narration, editing, or any of the skills needed to make a really good JoVE publication.
There is also another big gap in instrumentation and hardware. Every really productive laboratory has at least one person who builds things. For example, at the UC Davis Genome Center perhaps a third of all the science we publish involve some gizmo built by our in-house machinist Steve Lucero. Steve is definitely what I would regard as a practicing researcher, and has been (ahem) instrumental in a number of important publications. Nevertheless, he was invited to be an author on his first publication only this year. By a graduate student. I don't think this is particularly malicious on the part of the laboratories that use his creations in their work. The problem is that there isn't really a good venue for publishing things.
Booting up new kinds of journals is a long-term endeavor. PLOS ONE is still struggling for acceptability in many circles. Nevertheless, getting researchers to actually look at each others' code would yield rewards in the short term while we wait for the resulting DOIs to be appreciated by the broader scientific community. | 2,535 | 11,294 | {"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} | 2.734375 | 3 | CC-MAIN-2014-15 | latest | en | 0.933335 |
https://fluca1978.github.io/2023/10/09/PerlWeeklyChallenge238.html | 1,718,380,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861567.95/warc/CC-MAIN-20240614141929-20240614171929-00085.warc.gz | 222,139,476 | 7,783 | # Perl Weekly Challenge 238: running sums and multiplications (and Python for the very first time!)
This post presents my solutions to the Perl Weekly Challenge 238.
I keep doing the Perl Weekly Challenge in order to mantain my coding skills in good shape, as well as in order to learn new things, with particular regard to Raku, a language that I love.
This week, I solved the following tasks:
The PL/Perl implementations are very similar to a pure Perl implementation, even if the PostgreSQL environment could involve some more constraints. Similarly, the PL/PgSQL implementations help me keeping my PostgreSQL programming skills in good shape.
# Raku Implementations
## PWC 238 - Task 1 - Raku Implementation
The first task was about computing the running sum of an array of integers, defined as the sum of the up-to-nth element.
``````sub MAIN( *@nums where { @nums.grep( * ~~ Int ).elems == @nums.elems } ) {
my @running-sum;
for 0 ..^ @nums.elems -> \$index {
@running-sum[ \$index ] = [+] @nums[ 0 .. \$index ];
}
@running-sum.join( ', ' ).say;
}
``````
## PWC 238 - Task 2 - Raku Implementation
The second task was about sorting an array of integers by the value, assuming the value has to be a single digit. If the current element was more than one digit, the digits have to be multiplied together unless a single digit element is found. Then, all the elements with the same number of multiplication steps have to be sorted together and an higher level than those with less steps.
``````sub MAIN( *@nums where { @nums.grep( { \$_ ~~ Int && \$_ > 0 } ).elems == @nums.elems } ) {
my %steps;
for @nums {
my \$step-counter = 0;
my \$value = \$_;
while ( \$value > 9 ) {
\$value = [*] \$value.comb;
\$step-counter++;
}
%steps{ \$step-counter }.push: \$_;
}
my @running-sort.push: | %steps{ \$_ }.sort for %steps.keys.sort;
@running-sort.join( ', ' ).say;
}
``````
The idea is to keep an hash, named `%steps` keyed by the number of required steps to reduce the value to a single digit, and then store the current value into the number of steps. Then, it is only a matter of sorting the hash by its keys and the corresponding array by their values.
# PL/Perl Implementations
## PWC 238 - Task 1 - PL/Perl Implementation
Same as the Raku implementation.
``````CREATE OR REPLACE FUNCTION
RETURNS SETOF int
AS \$CODE\$
my ( \$nums ) = @_;
for my \$index ( 0 .. \$nums->@* - 1 ) {
my \$sum = 0;
\$sum += \$_ for ( \$nums->@[ 0 .. \$index ] );
return_next( \$sum );
}
return undef;
\$CODE\$
LANGUAGE plperl;
``````
## PWC 238 - Task 2 - PL/Perl Implementation
Same idea as per the Raku implementation.
``````CREATE OR REPLACE FUNCTION
RETURNS SETOF int
AS \$CODE\$
my ( \$nums ) = @_;
my \$steps = {};
# utility function to reduce a number
# does only one pass so that I can counter
# how many passes are required
my \$reduce = sub {
my ( \$number ) = @_;
return \$number if ( \$number <= 9 );
my \$value = 1;
for my \$digit ( split( '', \$number ) ) {
\$value *= \$digit;
}
return \$value;
};
for ( \$nums->@* ) {
my \$step_counter = 0;
my \$value = \$_;
while ( \$value > 9 ) {
\$value = \$reduce->( \$value );
\$step_counter++;
}
push \$steps->{ \$step_counter }->@*, \$_;
}
for my \$key ( sort keys \$steps->%* ) {
return_next( \$_ ) for ( sort \$steps->{ \$key }->@* )
}
return undef;
\$CODE\$
LANGUAGE plperl;
``````
Here I use the `\$reduce` utility function to perform a single step of multiplication in the case the number has more than one digit. Performign a single step makes it possible to count how many times a reduction happened.
# PostgreSQL Implementations
## PWC 238 - Task 1 - PL/PgSQL Implementation
Here I use a window function to perform the sum.
``````CREATE OR REPLACE FUNCTION
RETURNS TABLE( v int, s int )
AS \$CODE\$
SELECT v, sum( v ) OVER ( ORDER BY v )
FROM unnest( nums ) v;
\$CODE\$
LANGUAGE sql;
``````
The `sum` function is both an aggregate and a window function: when used as a window function it provides the sum of the elements over a defined window, that in our case is just the window order by the value.
## PWC 238 - Task 2 - PL/PgSQL Implementation
Similar to the Raku approach, but I use a temporary table as an hash to store the values, the multiplications and the steps.
``````CREATE OR REPLACE FUNCTION
RETURNS SETOF int
AS \$CODE\$
DECLARE
current_value int;
digit text;
multiplication int;
step_counter int;
value_to_insert int;
BEGIN
CREATE TEMPORARY TABLE IF NOT EXISTS mul( v int, mul int, steps int DEFAULT 0 );
TRUNCATE mul;
FOREACH current_value IN ARRAY nums LOOP
IF current_value < 9 THEN
INSERT INTO mul( v, mul )
VALUES( current_value, current_value );
CONTINUE;
END IF;
-- if here the number is at least two digits long
step_counter := 0;
value_to_insert := current_value;
WHILE current_value > 9 LOOP
multiplication := 1;
step_counter := step_counter + 1;
FOREACH digit IN ARRAY regexp_split_to_array( current_value::text, '' ) LOOP
multiplication := multiplication * digit::int;
END LOOP;
current_value := multiplication;
END LOOP;
INSERT INTO mul( v, mul, steps )
VALUES( value_to_insert, multiplication, step_counter );
END LOOP;
RETURN QUERY SELECT v
FROM mul
ORDER BY steps ASC, v ASC;
END
\$CODE\$
LANGUAGE plpgsql;
``````
Note how verbose it does result this solution, when compared to PL/Perl and Raku ones. This again emphasizes how PL/PgSQL is probably not the best suited language for number mangling and non-set operations.
## PWC 238 - Task 2 - Another PL/PgSQL Implementation
Defining a function to reduce a single value, it is possible to quickly come up with a single query solution. First of all, the function to reduce a value to a single digit returning the number of steps is:
``````CREATE OR REPLACE FUNCTION pwc238.reduce( n int )
RETURNS int
AS \$CODE\$
DECLARE
current_value int;
step_counter int;
digit text;
multiplication int;
BEGIN
current_value := n;
step_counter := 0;
WHILE current_value > 9 LOOP
multiplication := 1;
step_counter := step_counter + 1;
FOREACH digit IN ARRAY regexp_split_to_array( current_value::text, '' ) LOOP
multiplication := multiplication * digit::int;
END LOOP;
current_value := multiplication;
END LOOP;
RETURN step_counter;
END
\$CODE\$
LANGUAGE plpgsql;
``````
Then, the function to use a single query to get the result is as simple as ordering values by the result of the above function:
``````CREATE OR REPLACE FUNCTION
RETURNS SETOF INT
AS \$CODE\$
SELECT v
FROM unnest( nums ) v
ORDER BY pwc238.reduce( v ), v;
\$CODE\$
LANGUAGE sql
VOLATILE
;
``````
# Python Implementations
## PWC 238 - Task 1 - Python Implementation
Baby steps in my Python knowledge.
``````import sys
def main( argv ):
running_sum = []
current_index = 0
while current_index < len( argv ):
running_sum.insert( current_index, 0 )
for n in argv[ 0 : current_index ]:
running_sum[ current_index ] += int( n )
current_index += 1
print( ", ".join( map( str, running_sum ) ) )
if __name__ == '__main__':
main( sys.argv[ 1: ] )
``````
The `main` function receives the expected array of numbers, and does the computation of the running sum as for the Raku and PL/Perl implementation. Note how horrible it is to having to `map` strings out of integers due to the need for `join` to use only strings.
## PWC 238 - Task 2 - Python Implementation
The second task was much more verbose than the other languages.
``````import sys
import collections
def reduce( n ):
if n <= 9:
return n
multiplication = 1
for digit in map( int, str( n ) ):
multiplication *= digit
return multiplication
def main( argv ):
steps = {}
for n in map( int, argv ):
current_step = 0
if n > 9:
# need reduction
current_value = n
while current_value > 9:
current_value = reduce( current_value )
current_step += 1
# if the key is not here, create a list
if not str( current_step ) in steps:
steps[ str( current_step ) ] = []
steps[ str(current_step) ].append( n )
# now traverse the dictionary and sort the array
# and print it
for k, v in collections.OrderedDict(sorted(steps.items())).items():
print( ", ".join( map( str, v ) ) )
if __name__ == '__main__':
main( sys.argv[ 1: ] )
``````
The `reduce` function does a single pass in reducing a more than one digit number. The `main` function does all the job. Please note how hard it is placing a list into a dictionary item, seems Python has nothing like autovifification.
The article Perl Weekly Challenge 238: running sums and multiplications (and Python for the very first time!) has been posted by Luca Ferrari on October 9, 2023 | 2,232 | 8,537 | {"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} | 2.84375 | 3 | CC-MAIN-2024-26 | latest | en | 0.845907 |
https://socratic.org/questions/59b0122d11ef6b14c007e545#472396 | 1,660,622,752,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00311.warc.gz | 460,921,732 | 5,603 | # Question #7e545
Sep 6, 2017
well, if we let $u = \ln x$ , then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$
...so $\frac{1}{x} \mathrm{dx} = \mathrm{du}$
So, $\int {\left(\ln x\right)}^{3} / x \mathrm{dx}$
...can be re-written as $\int {u}^{3} \mathrm{du}$
$= {u}^{4} / 4 + C$
$= {\left(\ln x\right)}^{4} / 4 + C$ | 148 | 325 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 3.546875 | 4 | CC-MAIN-2022-33 | latest | en | 0.413492 |
http://www.slideserve.com/hasana/moderation-assumptions | 1,506,331,015,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690376.61/warc/CC-MAIN-20170925074036-20170925094036-00336.warc.gz | 586,063,376 | 16,096 | 1 / 26
# Moderation: Assumptions - PowerPoint PPT Presentation
Moderation: Assumptions. David A. Kenny. What Are They?. Causality Linearity Homogeneity of Variance No Measurement Error. Causality. X and M must both cause Y.
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Presentation Transcript
David A. Kenny
Causality
Linearity
Homogeneity of Variance
No Measurement Error
• X and M must both cause Y.
• Ideally both X and M are manipulated variables and measured before Y. Of course, some moderators cannot be manipulated (e.g., gender).
• Need to know causal direction of the X to Y relationship.
• As pointed out by Irving Kirsch, direction makes a difference!
• Judd & Kenny (2010, Handbook of Social Psychology), pp. 121-2 (see Table 4.1).
• A dichotomous moderator with categories A and B
• The X Y effect can be stronger for the A’s than the B’s.
• The Y X effect can be stronger for the B’s than the A’s.
• In some cases, causality is unclear or the two variables may not even be a direct causal relationship.
• Should not conduct a moderated regression analysis.
• Tests for differences in variances in X and Y, and if no difference, test for differences in correlation.
• Assume that either X Y or Y X.
• Given parsimony, moderator effects should be relatively weak.
• Pick the causal direction by the one with fewer moderator effects.
• Say we find that Gender moderates the X Y relationship.
• Is it gender or something correlated with gender: height, social roles, power, or some other variable.
• Moderators can suggest possible mediators.
• Helpful to look for violations of linearity and homogeneity of variance assumptions.
• M is categorical.
• Display the points for M in a scatterplot by different symbols.
• See if the gap between M categories change in a nonlinear way.
• Using a product term implies a linear relationship between M and X to Y relationship: linear moderation.
• The effect of X on Y changes by a constant amount as M increases or decreases.
• It is also assumed that the X Y effect is linear: linear effect of X.
• Threshold model: For X to cause Y, M must be greater (lesser) than a particular value.
• The value of M at which the effect of X on Ychanges might be empirically determined by adapting an approach described by Hamaker, Grasman, and Kamphuis (2010).
Second Alternative to Linear Moderation
• Curvilinear model: As M increases (decreases), the effect of X on Y increases but when M gets to a particular value the effect reverses.
• Add M2 and XM2 to the regression equation.
• Test the XM2 coefficient.
• If positive, the X Y effect accelerates as M increases.
• If negative, then the X Y effect de-accelerates as M increases.
• If significant, consider a transformation of M.
• Graph the data and look for nonlinearities.
• Add X2 and X2M to the regression equation.
• Test the X2 and X2M coefficients.
• If significant, consider a transformation of X.
• Consider a dichotomous moderator in which not much overlap with X (X and M highly correlated).
• Can be difficult to disentangle moderation and nonlinearity effects of X.
Y
X
Moderation
Y
X
• Variance in Moderation Analysis
• X
• Y (actually the errors in Y)
• Not a problem if regression coefficients are computed.
• Would be a problem if the correlation between X and Y were computed.
• Correlations tend to be stronger when more variance.
• A key assumption of moderated regression.
• Visual examination
• Plot residuals against the predicted values and against X and Y
• Rarely tested
• Categorical moderator
• Bartlett’s test
• Continuous moderator
• not so clear how to test
• The category with the smaller variance will have too weak a slope and the category with the larger variance will too strong a slope.
• Separately compute slopes for each of the groups, possibly using a multiple groups structural equation model.
• No statistical solution that I am aware of.
• Try to transform X or M to create homogeneous variances.
Variance Differences as Moderatora Form of Moderation
• Sometimes what a moderator does is not so much affect the X to Y relationship but rather alters the variances of X and Y.
• A moderator may reduce or increase the variance in X.
• Stress Mood varies by work versus home; perhaps effects the same, but much more variance in stress at work than home.
Measurement Error Moderator
• Product Reliability (X and M have a normal distribution)
• Reliability of a product: rxrm(1 + rxm2)
• Low reliability of the product
• Weaker effects and less power
• Bias in XM Due to Measurement Error in X and M
• Bias Due to Differential X Variance for Different Levels of M
Differential Reliability Moderator
• categorical moderator
• differential variances in X
• If measurement error in X, then reliability of X varies, biasing the two slopes differentially.
• Multiple groups SEM model should be considered | 1,282 | 5,524 | {"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-2017-39 | latest | en | 0.877256 |
https://www.unix.com/man-page/centos/3/scopy | 1,709,508,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00581.warc.gz | 1,050,918,414 | 12,229 | # scopy(3) [centos man page]
```scopy.f(3) LAPACK scopy.f(3)
NAME
scopy.f -
SYNOPSIS
Functions/Subroutines
subroutine scopy (N, SX, INCX, SY, INCY)
SCOPY
Function/Subroutine Documentation
subroutine scopy (integerN, real, dimension(*)SX, integerINCX, real, dimension(*)SY, integerINCY)
SCOPY Purpose:
SCOPY copies a vector, x, to a vector, y.
uses unrolled loops for increments equal to 1.
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
November 2011
Further Details:
jack dongarra, linpack, 3/11/78.
modified 12/3/93, array(1) declarations changed to array(*)
Definition at line 52 of file scopy.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 scopy.f(3)```
## Check Out this Related Man Page
```csrot.f(3) LAPACK csrot.f(3)
NAME
csrot.f -
SYNOPSIS
Functions/Subroutines
subroutine csrot (N, CX, INCX, CY, INCY, C, S)
CSROT
Function/Subroutine Documentation
subroutine csrot (integerN, complex, dimension( * )CX, integerINCX, complex, dimension( * )CY, integerINCY, realC, realS)
CSROT Purpose:
CSROT applies a plane rotation, where the cos and sin (c and s) are real
and the vectors cx and cy are complex.
jack dongarra, linpack, 3/11/78.
Parameters:
N
N is INTEGER
On entry, N specifies the order of the vectors cx and cy.
N must be at least zero.
CX
CX is COMPLEX array, dimension at least
( 1 + ( N - 1 )*abs( INCX ) ).
Before entry, the incremented array CX must contain the n
element vector cx. On exit, CX is overwritten by the updated
vector cx.
INCX
INCX is INTEGER
On entry, INCX specifies the increment for the elements of
CX. INCX must not be zero.
CY
CY is COMPLEX array, dimension at least
( 1 + ( N - 1 )*abs( INCY ) ).
Before entry, the incremented array CY must contain the n
element vector cy. On exit, CY is overwritten by the updated
vector cy.
INCY
INCY is INTEGER
On entry, INCY specifies the increment for the elements of
CY. INCY must not be zero.
C
C is REAL
On entry, C specifies the cosine, cos.
S
S is REAL
On entry, S specifies the sine, sin.
Author:
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
Date:
November 2011
Definition at line 99 of file csrot.f.
Author
Generated automatically by Doxygen for LAPACK from the source code.
Version 3.4.2 Tue Sep 25 2012 csrot.f(3)```
Man Page
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## What is the meaning of ## in UNIX shell script?
Hi All, I am new to unix shell scripting and I was documenting one of the unix script and encountered below statements - for ii in `ls -1rt /oracle/admin/MARSCOPY/ext_files/fpm-ifpm/*.small.txt | tail -1 | awk '{print \$1}'` do smallssim=\${ii##/oracle/admin/MARSCOPY/ext_files/fpm-ifpm/}... | 1,764 | 6,715 | {"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-2024-10 | latest | en | 0.6801 |
https://www.hackmath.net/en/math-problem/28241 | 1,611,103,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519843.24/warc/CC-MAIN-20210119232006-20210120022006-00010.warc.gz | 816,908,342 | 14,506 | # Truncated cone 6
Calculate the volume of the truncated cone whose bases consist of an inscribed circle and a circle circumscribed to the opposite sides of the cube with the edge length a=1.
Correct result:
V = 0.4469
#### Solution:
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Tip: Our volume units converter will help you with the conversion of volume units.
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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A truncated cone has a bases radiuses 40 cm and 10 cm and a height of 25 cm. Calculate its surface area and volume. | 592 | 2,511 | {"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} | 4.0625 | 4 | CC-MAIN-2021-04 | longest | en | 0.8708 |
http://esolangs.org/wiki/User:XKCD_Random_Number | 1,716,666,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00478.warc.gz | 9,694,408 | 13,244 | # User:XKCD Random Number
Jump to navigation Jump to search
Not to be confused with Random number generator.
XKCD Random Number is a challenge that requires you to rewrite the program in XKCD's 221st page in another language, as either a function or a full program that prints the return value. You may comment your program just like XKCD did, if possible.
## Implementations
```++++.
```
### !!Fuck
```!!!!!#!!!!!!!#!!!!!#!!!!!!!#!!!!!#!!!!!!!#!!!!!!!!!!!#!!!!!#!!!!!!!#!!!!!!!!!!!#!!!!!!!!#!!!!!!#!!!!!!!#!!!!!!!#!!!!!!!#!!!!!#!!!!!!!!!!!!#!!!!!!#!!!!!!#!!!!!!!!!!!!#!!!!!#!!!!!!!!!!#
```
`4@`
```^^^^#
```
### 1
```4111*,
```
It only works on the Common1 dialect of 1.
### 100BF
```1OO 100 1OO 100 1OO 100 lO0 1OO 100 lO0 10O 1O0 100 100 100 1OO lOO 1O0 1O0 lOO 1OO l0O
```
```Hÿ4
```
### 4
`3.600525004`
### 8xn
`8x11116`
```0=+52
```
### 脑操
```》十》十》十（》十（一《十十十》）《《）》。
```
### 硕鼠
```吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱吱
```
### AEAV
```xkcd = 4 (Non-random version)
out xkcd
xkcd = 1-4 (Random version)
out xkcd```
`밤망희`
### Arithmetic
```==Begin Exam 1==
1. 1+1=? (52 points)
A. 2
Answer: A
==End Exam 1==
```
```Ru
```
### Befunge
`4.@`
The program above prints an extra space. If you don't want an extra space, use the one below.
`"4",@`
```4
```
### BracketOnly
```(()()()()()()()()()()()()()()()()()())((()())(()()()()()()(()()()())((()())(()()()()()()()()()()()()()())(()())(()()()()()()()()()()()()()()))))
```
Alternative:
```(()()()()()()()()()()()()()()()())((()())(()()()()()()()()))
```
### Braincopter
Here is the actual program:
### Brain-Flak
In non-ASCII mode:
```(()()()())
```
In ASCII mode:
```(()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()())
```
### Brainfuck
`>+>+>+[>+[-<+++>]<<]>.`
### brainfuck
```++++++++++++++++++++++++++++++++++++++++++++++++++++
```
Note that adding a `.` in the end of program makes it a polyglot, because that will make it also work as a brainfuck program.
```\$4\$.
```
```_34.
```
```////)
```
### Brainloller
Here is the actual program:
### C
The original program mentioned on XKCD is written in C. It was a function, while the full program is shown below:
```#include <stdio.h>
int main(){putchar(52);return 0;}```
### C#
```using System;
class XKCD {
static void Main(string[] args) {
Console.WriteLine("4");
}
}
```
`%"4`
### Cood
```Hey, waiter!
I want 4 of this.
How much is it?
The bill, please.
```
### Deadchicken
```chicken
chicken
chicken
chicken
chicken chicken chicken chicken
```
```iiiio
```
### enGLish
```Output four
```
#### Alternative
```Output 4
```
##### Alternative
```Output "4"
```
Alternative
```Output character 52
```
### Eso2D
`0_*`
The program above prints an extra space. If you don't want an extra space, use the one below.
`1,,#`
### ExampleFuck
```Truth Machine
Hello World
Truth Machine
Hello World
Truth Machine
Hello World
FizzBuzz
Truth Machine
Hello World
FizzBuzz
Cat Program
Quine
Hello World
Hello World
Hello World
Truth Machine
99 Bottles of Beer
Quine
Quine
99 Bottles of Beer
Truth Machine
Self Interpreter
```
### F!--
```F! F! F! F! K!
```
### Factor
```54948677498655068263351660303957600730731029471078168023
```
### Fish Code
```<((52)).><
```
```4
```
### Gift
`4___))(_(_(`
### Goldfuck
`1.618031988749844848264586834565638817720308179805762860135440622705270462818902419707207244189392137484724088075886891758126636862223536931393180080766728354433389086095939982905638327`
### Hardfish
In implementations that output values as numbers:
```ico
```
In implementations that output values as ASCII:
```icrico
```
```+4
```
`4!@`
```34
```
### JavaScript
```console.log(4);
```
### JSFuck
The following is just a JSFuck.com conversion from `console.log(4)` (3326 characters long), it works on browser JS and Node.js
`[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((!![]+[])[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+([][[]]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+!+[]]+(+[![]]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(+(!+[]+!+[]+!+[]+[+!+[]]))[(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([]+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]][([][[]]+[])[+!+[]]+(![]+[])[+!+[]]+((+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]+[])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]]](!+[]+!+[]+!+[]+[!+[]+!+[]])+(![]+[])[+!+[]]+(![]+[])[!+[]+!+[]])()(([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(+(+!+[]+[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+[!+[]+!+[]]+[+[]])+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(![]+[+[]]+([]+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]])[!+[]+!+[]+[+[]]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[+!+[]+[!+[]+!+[]+!+[]]]+[!+[]+!+[]+!+[]+!+[]]+([+[]]+![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[!+[]+!+[]+[+[]]])`
There is also a shorter one but only works on browser JS (2359 characters, a JSFuck.com conversion from `alert(4)`).
`[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((!![]+[])[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+([][[]]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+!+[]]+(+[![]]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(+(!+[]+!+[]+!+[]+[+!+[]]))[(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([]+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]][([][[]]+[])[+!+[]]+(![]+[])[+!+[]]+((+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]+[])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]]](!+[]+!+[]+!+[]+[!+[]+!+[]])+(![]+[])[+!+[]]+(![]+[])[!+[]+!+[]])()((![]+[])[+!+[]]+(![]+[])[!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[+!+[]+[!+[]+!+[]+!+[]]]+[!+[]+!+[]+!+[]+!+[]]+([+[]]+![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[!+[]+!+[]+[+[]]])`
This is a longer one (8498 characters) that doesn't output any trailing line feeds which is a JSFuck.com conversion from `process.stdout.write('4')`, it works on Node.js.
`[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((!![]+[])[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+([][[]]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+!+[]]+(+[![]]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]+(+(!+[]+!+[]+!+[]+[+!+[]]))[(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([]+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]][([][[]]+[])[+!+[]]+(![]+[])[+!+[]]+((+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]+[])[+!+[]+[+!+[]]]+(!![]+[])[!+[]+!+[]+!+[]]]](!+[]+!+[]+!+[]+[!+[]+!+[]])+(![]+[])[+!+[]]+(![]+[])[!+[]+!+[]])()([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]][([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]((!![]+[])[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+([][[]]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+!+[]]+([]+[])[(![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(!![]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(![]+[])[!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+(!![]+[])[+!+[]]]()[+!+[]+[!+[]+!+[]]]+((!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[+[]]+(!![]+[])[+!+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(+(+!+[]+[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+[!+[]+!+[]]+[+[]])+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]]+([][[]]+[])[!+[]+!+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+([][[]]+[])[+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]]+(+(+!+[]+[+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+[!+[]+!+[]]+[+[]])+[])[+!+[]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+(!![]+[])[+!+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(!![]+[])[+[]]+[+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]]+(!![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]]+[+[]]+(!![]+[])[+[]]+[!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]]+(!![]+[])[+[]]+[!+[]+!+[]+!+[]+!+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]]+(!![]+[])[+[]]+[!+[]+!+[]+!+[]+!+[]+!+[]]+[+!+[]])[(![]+[])[!+[]+!+[]+!+[]]+(+(!+[]+!+[]+[+!+[]]+[+!+[]]))[(!![]+[])[+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([]+[])[([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+([][[]]+[])[+!+[]]+(![]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+[])[+!+[]]+([][[]]+[])[+[]]+([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+(!![]+[])[+[]]+(!![]+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Please replace these implementations if you know a shorter one.
`4!@`
### Look!
Note: Look! is a sound-based language and the following line of text is only the notes of the real program, for the real program please see the link below
```G# D# G D# G# C B G C F# D A C C C G# C# A A C# G F
```
The program as a MIDI file.
```++++:
=====```
### morse code
` ....-`
Note: Space in front of returning text.
### Multiplicity
```417744685118879731600506117375719540878173341042907821182687827362162871106962918226728766784177853000
```
### Odasjoi
```fowdif
fowdif
fowdif
fowdif
doqije
```
### oOo CODE
```xkcD rAndoM nUmbeR iS a ChallEnGe ThAt reqUIrES yOU tO rewrITe tHe pRoGRam iN Xkcd's 221st page in another language
```
It's self-documenting.
### Output
```Output "4"
```
### PlusIntMinus
Although the PlusOrMinus program will still work for this language, this is a shorter one that only works on PlusIntMinus.
```+52-
```
### PlusOrMinus
```++++++++++++++++++++++++++++++++++++++++++++++++++++-
```
Enlarged view
Actual size
### Poetic
```Pre crecas condominolovium seclosoponiat exelglios egumbo
e arbi-brimidozoque fiene a parp-en inbidee nonquaeson!```
```Tc
```
### Python
`print(4,end='')`
#### Alternative
```__import__("sys").stdout.write("4")
```
### Rec
```4P [0^ Chosen by fair dice roll.
Guaranteed to be random. ]
```
```1 4
```
### Rockstar
```The truth is random (chosen by fair dice roll)
Looking is not useful
Xkcd is a blessing (guaranteed to be random)
Say the truth without looking over it.```
### Ropy
This program uses implicit output. There is a trailing newline.
`4`
`ssss!`
### Setlang
```(4)
```
Which also works as a Setlang++ program.
```<.4
```
### Simon says brainfuck
```Simon says: >+>+>+[>+[-<+++>]<<]>.
Chosen by fair dice roll.
Guaranteed to be random.
```
### Sokolang
```// Chosen by fair dice roll.
// Guaranteed to be random.
###
##*##
#@R*#
#####
---
r:11,4
---
rwu
```
```输出 4 。
```
`4`
### This esolang is a brainfuck derivative
```This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainfuck derivative
This esolang is a brainf*** derivative
chosen by fair dice roll
guarranteed to be random
```
Note: the three stars in the program are MANDATORY, while the last two lines are not. (brainfuck: ++++++++++++++++++++++++++++++++++++++++++++++++++++.)
```4.
```
### UnnumericJS
```console.log(!''+!''+!''+!'');
```
### Weeeeeeeeeeeeeeeeeeeeeeeeeeeeee
```Start epidemic
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Infect person
Check number of infections
Delevop vaccine
```
### WeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeBasic
```print "4"
```
Which also works when running on WeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeBasic++ or Python 2.
A commented version, that only works in this language, is as follows:
```print "4"
Start epidemic chosen by fair dice roll
Delevop vaccine guarranteed to be random
```
### wenyan
```吾有一數。曰四。書之。
```
### Whendo
```a==0=>a+=4->a
```
### wsad
```wwwwwwwwas
```
```X!×4
x^-1×0×0
``` | 11,975 | 24,050 | {"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-2024-22 | latest | en | 0.561129 |
https://jeopardylabs.com/print/unit-6-expressions-and-equations-review-5 | 1,716,973,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00064.warc.gz | 265,003,180 | 4,338 | Hanger/Tape Diagrams
Area Diagrams
Percentages
True or Not True
Word Problems
100
Write an equation that represents the hanger diagram
7 = 3x + 1
100
Write an equation to represent the area of the rectangle below:
20s = ?
100
Write an equation that shows the relationship
15% of 160 is x
15/100 * 160 = x
100
When x = 3 is the equation true? Explain your reasoning
x-3 = 0
Yes, when x = 3
3-3=0.
100
At Jane Addams Middle School, 1/5 of the track athletes are in 6th grade.
How many 6th graders are track athletes if the team has 90 track athletes? Explain your reasoning.
18 6th Graders are Track Athletes
200
Write an equation for the tape diagram
3z + 22 = 49
200
Write an expression for the sum of the areas of the two rectangles.
5(x+8)
5x + 40
5x + 5 * 8
200
Write an equation that shows the relationship
20% of y is 42
20/100 * y = 42
200
When x = 3 is the equation true? Explain your reasoning
1-x = 3
No as this would get us a negative value, so it would not be equal to 3
200
At Jane Addams Middle School, 3/5 of the track athletes are in 8th grade.
How many 8th graders are track athletes if the team has 90 track athletes? Explain your reasoning.
54 Students are 8th Grade Athletes on the track team
300
Write an equation for the tape diagram and find the value of the variable.
3t = 66
t = 22
300
Write an equation that represents the area of the rectangle
6(4p)
24p
6p+6p+6p+6p
300
15% of 35 is p. Find the value of p.
p = 5.25
300
When x = 3 is the equation true? Explain your reasoning
10 - x = 8
Not True
10 - 3 = 7
300
At Jane Addams Middle School, 1/5 of the track athletes are in 6th grade.
If the track team has x athletes, write an expression for the number of 6th graders in terms of x.
1/5 x
400
Write an equation for the hanger diagram and find a value of the variable.
6.8 = 2z + 2.2
z = 2.3
400
Write an equation that represents the area of the rectangle. Find the area when m = 2
m(6+8)
6m+8m
14m
Area = 28 units squared
400
25% of w is 10. Find the value of w.
w = 40
400
When x = 3 is the equation true? Explain your reasoning
1/2x = 1 1/2
True as 3/2 = 1 1/2
400
At Jane Addams Middle School, 30% of the track athletes are in 7th grade. There are 12 athletes in 7th grade on the track team.
If the track team has x athletes, write an expression for the number of 7th graders in terms of x.
30/100 * x = 12
500
DAILY DOUBLE!!!!!
Write an equation for the tape diagram. Find a solution.
3(x+4) = 45
x = 11
500
Find the area of the rectangle when r=1/2
Area = 1 1/2 units squared or 3/2 units squared
500
40% of m is 90. Find the value of m.
40/100 * m = 90
90 รท 0.4 = 225
500
When x = 3 is the equation true? Explain your reasoning
x= 81
True as
x*x*x*x = 3*3*3*3
3*3*3*3=81
500
At Jane Addams Middle School, 30% of the track athletes are in 7th grade. There are 12 athletes in 7th grade on the track team.
If the track team has x athletes, write an expression for the number of 7th graders in terms of x. Find the value of x.
30/100 * x = 12
12 * 10/3 = 120/3
120/3 = 40 | 1,005 | 3,104 | {"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} | 4.375 | 4 | CC-MAIN-2024-22 | latest | en | 0.932967 |
https://physics.stackexchange.com/questions/160068/thermodynamics-and-internal-energy | 1,709,517,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00818.warc.gz | 456,927,520 | 39,785 | # Thermodynamics and internal energy
I've been working on this problem for quite some time trying to figure out the most efficient way of answering it, So here goes the problem:
There is a container, containing a monoatomic gas with a movable piston on the top. The container and piston are all made of perfectly insulating material. Now the container is divided into two compartments by a movable partition that allow slow transfer of heat among the compartments. The adiabatic exponent of gas is $$\gamma$$.
We are asked to calculate the total work done by the gas till it achieves equilibrium. By equilibrium here I mean that initially after the partition was introduced the gas on each side has different pressure and temperature which with passage of time reached a common temperature and pressure.
Now my approach to the problem is:
We know that in a closed adiabatic container:
$$\DeltaQ = 0$$ as there is no net heat change between the compartments.
So, as per first law of thermodynamics we have, $$\Delta Q = \Delta U + W$$
Now considering the whole container to be the system we can apply the law on the system so,
$$\Delta Q = \Delta U + W$$ reduces to $$0$$ = $$\Delta U + W$$ $$\Rightarrow$$ $$W = - \Delta U$$
Now we can calculate $$\Delta U$$ for both compartments at equilibrium and add them up and the work will be given by $$- \Delta U$$. And we're done.
But something about my above method doesn't seem right. Are all my above considerations correct? If not how can I improve? I've tried reading up Joule-Thomson effect and other related texts on thermodynamics and expansion. I'll be highly obliged by your help. I've tried to fit this question in with the standards of StackExchange.
• "We are asked to calculate the total work done by the gas till it achieves equilibrium." In what way is it not in equilibrium to start with? Jan 18, 2015 at 14:43
• In the beginning the gas on each side of partition has different temperature and pressure and even volume and with passage of time it reaches the equilibrium. Jan 18, 2015 at 14:46
This process is shortly adiabatic so you are right to imply $\Delta Q=0$ Simply we know how to calculate work as : $$W = \int_{v_i}^{v_f} P \, dv$$ Adiabatic condition also satisfy $$PV^{\gamma}=K =constant$$ $$\gamma=\frac{5}{3}$$ for monatomic gas. so work (W) becomes simply $$W = K \int_{v_i}^{v_f} \frac{dV}{V^{\gamma}}$$ Integrating yield $$W=\frac{K(V_f^{1-\gamma}-V_i^{1-\gamma})}{1-\gamma}$$ and you are done. | 614 | 2,485 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.946063 |
https://kr.mathworks.com/matlabcentral/cody/problems/38-return-a-list-sorted-by-number-of-occurrences/solutions/257902 | 1,606,692,977,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141203418.47/warc/CC-MAIN-20201129214615-20201130004615-00562.warc.gz | 369,387,554 | 17,077 | Cody
# Problem 38. Return a list sorted by number of occurrences
Solution 257902
Submitted on 7 Jun 2013 by Martin
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 2 2 3 3 7 7 93] y_correct = [2 3 7 1 93]; assert(isequal(popularity(x),y_correct))
x = 1 2 2 2 3 3 7 7 93 y = 1 2 3 7 93 erg = 1 erg = 1 3 erg = 1 3 2 erg = 1 3 2 2 erg = 1 3 2 2 1 index = 2 3 4 1 5
2 Pass
%% x = [-1 19 20 -1 -1 87 19 34 19 -1 21 87 20 10 20 34 19 -1]; y_correct = [-1 19 20 34 87 10 21]; assert(isequal(popularity(x),y_correct))
y = -1 10 19 20 21 34 87 erg = 5 erg = 5 1 erg = 5 1 4 erg = 5 1 4 3 erg = 5 1 4 3 1 erg = 5 1 4 3 1 2 erg = 5 1 4 3 1 2 2 index = 1 3 4 6 7 2 5
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 403 | 922 | {"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-2020-50 | latest | en | 0.558816 |
http://stats.stackexchange.com/questions/22161/how-to-read-cooks-distance-plots | 1,469,771,605,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829972.19/warc/CC-MAIN-20160723071029-00167-ip-10-185-27-174.ec2.internal.warc.gz | 229,637,322 | 19,779 | # How to read Cook's distance plots?
Does anyone know how to work out whether points 7, 16 and 29 are influential points or not? I read somewhere that because Cook's distance is lower than 1, they are not. Am, I right?
-
There are various opinions. Some of them relate to the number of observations or to the number of parameters. These are sketched at en.wikipedia.org/wiki/…. – whuber Feb 2 '12 at 15:02
@whuber Thanks. This is always a grey area when performing data exploration for me. Data point 16 above massively influences model outcomes, thus increasing Type I errors. – Platypezid Feb 2 '12 at 15:11
One might argue that it increases "Type III" errors as well, which (generically and informally) are errors related to inapplicability of the underlying probability model. – whuber Feb 2 '12 at 15:22
@whuber yes, very true! – Platypezid Feb 2 '12 at 15:31
Some texts tell you that points for which Cook's distance is higher than 1 are to be considered as influential. Other texts give you a threshold of $4/N$ or $4/(N - k - 1)$, where $N$ is the number of observations and $k$ the number of explanatory variables. In your case the latter formula should yield a threshold around 0.1 .
John Fox (1), in his booklet on regression diagnostics is rather cautious when it comes to giving numerical thresholds. He advises the use of graphics and to examine in closer details the points with "values of D that are substantially larger than the rest". According to Fox, thresholds should just be used to enhance graphical displays.
In your case the observations 7 and 16 could be considered as influential. Well, I would at least have a closer look at them. The observation 29 is not substantially different from a couple of other observations.
(1) Fox, John. (1991). Regression Diagnostics: An Introduction. Sage Publications.
-
+1 Clear summary. I would add that influential cases are not usually a problem when their removal from the dataset would leave the parameter estimates essentially unchanged: the ones we worry about are those whose presence really does change the results. – whuber Feb 2 '12 at 15:24
@lejohn Very appreciative of your response. Whuber is right excellent clarity in your answer. This is very informative. Might I suggest you highlight Fox's and your opinions in the wikipedia page! – Platypezid Feb 2 '12 at 15:36
+1 to both @lejohn and @whuber. I wanted to expand a little on @whuber's comment. Cook's distance can be contrasted with dfbeta. Cook's distance refers to how far, on average, predicted y-values will move if the observation in question is dropped from the data set. dfbeta refers to how much a parameter estimate changes if the observation in question is dropped from the data set. Note that with $k$ covariates, there will be $k+1$ dfbetas (the intercept, $\beta_0$, and 1 $\beta$ for each covariate). Cook's distance is presumably more important to you if you are doing predictive modeling, whereas dfbeta is more important in explanatory modeling.
There is one other point worth making here. In observational research, it is often difficult to sample uniformly across the predictor space, and you might have just a few points in a given area. Such points can diverge from the rest. Having a few, distinct cases can be discomfiting, but merit considerable thought before being relegated outliers. There may legitimately be an interaction amongst the predictors, or the system may shift to behave differently when predictor values become extreme. In addition, they may be able to help you untangle the effects of colinear predictors. Influential points could be a blessing in disguise.
-
+1 "Cook's distance is presumably more important to you if you are doing predictive modeling, whereas dfbeta is more important in explanatory modeling": this is very useful advice. – Anne Z. Feb 5 '12 at 0:45
Hi - interesting discussion. But couldn't it be rational to integrate a dummy-variable to measure the effect from for example observation 16? – Pantera Feb 8 '12 at 9:04
@Pantera I removed 16 and compared the pre & post ommission models – Platypezid Feb 8 '12 at 12:43
Hi - if you remove observations, you should make sure that you have "good" argument to do it, for example that the observation is wrongly measured. If we throw out observation because they just make some statistical trouble, then we're close to data-mining. – Pantera Feb 9 '12 at 13:12 | 1,027 | 4,406 | {"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} | 2.90625 | 3 | CC-MAIN-2016-30 | latest | en | 0.934893 |
https://www.doubtnut.com/question-answer-physics/a-solid-cube-of-side-a-is-shown-in-the-figure-find-maximum-value-of-100-b-a-for-which-the-block-does-642610614 | 1,669,801,497,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710734.75/warc/CC-MAIN-20221130092453-20221130122453-00546.warc.gz | 792,539,291 | 49,614 | Home
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# A solid cube of side 'a' is shown in the figure. Find maximum value of 100 b/a for which the block does not topple before sliding. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/JM_20_M2_20200107_PHY_23_Q01.png" width="80%">
Updated On: 27-06-2022
Text Solution
Solution : NA
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Transcript
Pawan you can see in the question of cube is of side length a find the maximum value of hundred in the question for which the blog does not topple before sliding takes place it if you see the diagram b is the distance of the line of action force from the centre rate if you see here this distance is from the centre bi take so we want to maximize the value of B so that to the sea the 2D diagram of this so this is the Kyon and let suppose hear the force is acting so what will happen if the friction force will act in opposite direction
if you see the centre hear about the centre of this system is having a torque in clockwise direction to so to avoid to compulsory that talk normal reaction will shift from Centre to this distance may suppose it instances access to this will happen to make the network about the centre of mass is zero the normal reaction will save from the sent to this site so what will happen about which point to the the block will have a tendency to toples about this point there the block will have a tendency to topper so now we want we want this B2B maximum this P rate so when the baby will be maximum the normal reaction
will completely Swift at is called like him and the course will be acting here and this whole distance will be a by 2 + b since the from the centre this distance was 20 I will write a by 2 + b and had the friction forces analysis the centre where the weight mg is acting down what if I take a point and make the talk about the point zero then the block want opposite is there won't be any talk about the great so please make the net talk about point O is equals to zero so I will write because of the force you can see if a by 2 + b
the line of action of force and this is the distance of the point from the line of action of cost and so if I take the normal reaction for normal reaction is acting in anticlockwise direction so I will write a negative sign here and with a so if you see this distance is a line of action of normal reaction is at a distance of any five see the MG mg is at a distance of a b to so it will be = 20 it if you see in the question Mu is also given but we already have taken the talk about the line of action of the friction is passing through ho so we won't take any talk about the fractions is the talk of the patient about point O is zero where is simple now just put the
values if you see in vertical direction the net force is zero so I will write an is equals to empty and if I talk about talk about the four straight so f force should not be greater than the friction force since we want the block not to topple before it slide so when the this the block wants light then this force will be close to you mg and in the question is given as point 14.4 it so in horizontal direction is the net force is zero Sonet to leftward process should be equal to net right word for swipe Elite point 4 MG very simple now just put the value of this normal reaction and force in this equation
the values now only that is 2.4 MG into minus b into a plus MG able to is equals to zero knowledge cancel out this and gmgm ji so you will get your site will be point or so it will be required to A plus 4.4 be so a b minus A + 1 by 2 minus 1 by 2 that is 4.5 a is equals to zero So Ja So if we will solve this you will get point 4 b is equals to 4.5 - 2 = 2.3 so you will get the bi bi a maximum ratio as 354
75 rate but if you see in the diagram be can't be more than two years after by to the block will stop I will write 20 should be less than equal to able to read and this value we got is point 750 minimum value of P can be be be if it is free so do we calculated to be a maximum value point 75 but it is not possible so we write a review by a maximum value as this put the value of of in this expression and you will get the value of this hundred by a is equals to 50 and this will be your answer thank you | 1,046 | 4,405 | {"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} | 3.515625 | 4 | CC-MAIN-2022-49 | longest | en | 0.926828 |
https://www.engineeringclicks.com/density-of-milk-weight-per-gallon/ | 1,624,211,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488253106.51/warc/CC-MAIN-20210620175043-20210620205043-00215.warc.gz | 687,560,161 | 56,430 | # Density of Milk and the Trick Question About the Weight of a Gallon of Milk
Density is the amount of sustance in a unit, per volume of substance. Below you can see the density equiation and the table with Milk Density values, inluding those for whole, semi-skimmed, and skimmed milk, depending on temperature. What makes the question of density per unit of weight tricky, is the fact density changes depending on the temperature and pressure. Additionally, the whole connundrum of the US and the Imperial Gallon makes the simple question of “what it the weight of a gallon of milk” so complicated.
## What is the breakdown/density of cow’s milk?
To many people milk is just a white fluid which has the look and texture of water but feels a little “thicker”. In reality, milk can be anywhere up to 90% water with the rest a mixture of proteins, fats, and lactose (milk sugar). However, it is one of the most nutritious foods available and part of the diet of billions of people around the world.
As cow’s milk is by far and away the most popular type of milk in the Western world, we will focus on this when looking at ingredients and density. The exact ingredients in cow’s milk will vary but tend to be in the following range:-
• Water circa 87%
• Protein 3% to 4%
• Fat 4% to 4.5%
• Lactose (milk sugar) approximately 4.5%
We will now take a look at the density of milk which will vary in line with the ingredients – compared to the water content. Protein and lactose (but not fat) are denser than water so the lower the water component in comparison, the higher the density of the milk. The higher the water content the lower the density of the milk.
### Density of milk: how to correctly calculate it
The basic formula for calculating the density of milk is:
Density = Mass/Volume
We will use the following figures to calculate the density of milk:-
Full half pint milk carton: 256.6 g
Empty half pint milk carton: 11.9 g
Half pint Milk Mass: 244.7 g (or 0.2447 kg)
Volume of half pint: 236 mL (0.00236 m³)
Therefore the formula is as follows:
0.2447 kg/0.000236 m³ = 1036.86 kg/ m³
You may also see the density of milk presented in a slightly different manner (using the same figures):-
1.03686 kg/L
When you mix-and-match the various elements of milk to create a different formula this will have an impact on the density.
## Density of milk
The following chart shows milk density values under varying temperatures for different types of milk:
Product Composition Density (kg/L) at: Product Fat (%) SNF* (%) 4.4oC 10oC 20oC 38.9oC Producer milk (whole milk) 4.00 8.95 1.035 1.033 1.030 1.023 Homogenized milk 3.6 8.6 1.033 1.032 1.029 1.022 Skim milk, pkg 0.02 8.9 1.036 1.035 1.033 1.026 Fortified skim 0.02 10.15 1.041 1.040 1.038 1.031 Half and half 12.25 7.75 1.027 1.025 1.020 1.010 Half and half, fort. 11.30 8.9 1.031 1.030 1.024 1.014 Light cream 20.00 7.2 1.021 1.018 1.012 1.000 Heavy cream 36.60 5.55 1.008 1.005 0.994 0.978
*SNF stands for solids-not-fat. Data source.
### What’s the density of whole milk?
Whole milk’s (producer milk) density is 1.035 kg/L when the temperature is 4.4°C. As the temperature increases so the molecules move further apart and the liquid becomes less dense. While not huge differences, but whole milk at 38.9°C has a density of 1.023 kg/L.
### How density aids separation of cream and milk
When looking at the overall density of milk it is worth noting the density of water, fat and so-called solid non-fats. The non-fats have a greater density than water hence they won’t float above the water element of milk while the fat has a lower density than water. This means that the fat globules will float to the top at which point they can be “skimmed” to create two products; skimmed milk and cream.
As the fat globules slowly rise to the surface they create varying degrees of density from the bottom of the milk to the top. When the fat is skimmed off the top of the milk not all of it would be removed which is why skimmed milk is denser than produce milk. In effect the less dense fat element has been diluted by the removal of the smaller globules which floated to the surface.
## How does milk density compare with other liquids?
In order to appreciate the density of milk we thought it would be useful to compare and contrast this with other substances:
Material Density (g/cm3) Honey 1.42 Pancake Syrup 1.37 Light Corn Syrup 1.33 Dish Soap 1.06 Milk 1.03 Water 1.00 Ice Cube 0.92 Vegetable Oil 0.92 Baby Oil 0.83
Data source.
So, while you automatically assume that an ice cube floats on water it is interesting to see that baby oil for example is less dense than an ice cube. As a consequence, if you had a glass of baby oil and placed an ice cube in, it would drop to the bottom of the glass. Eventually the ice cube would melt and you would have two very distinct layers in the glass. You have the layer of water at the bottom, which is denser than the baby oil, with the baby oil lying above.
When it comes to liquids, density and how they react with each other, we often take things for granted. For example, why does a ping-pong ball float on any liquid when a marble will normally sink to the bottom?
The results won’t surprise you but their density calculations will give you some background:-
Ping-pong ball density = 0.0840 kg/L
Marble density = 2.64 kg/L
When you compare the density of the ping-pong ball it is no surprise that it floats on just about any liquid. On the flipside of the coin, the high density of the marble means that there are very few liquids in which it would float.
## What is the weight of a gallon of milk?
How much does milk weigh?
This a tricky question because gallons are measured differently in the UK and the US.
UK liquid gal = US liquid gallon / 0.83267
US liquid gallon = 3.78541 litres
UK liquid gallon = 4.54609 litres
### Weight of a gallon of milk (USA):
Density at 20oC kg/l Weight in Kilos Weight in Pounds Producer milk 1.03 3.899 8.60 Homogenized milk 1.029 3.895 8.59 Skim milk, pkg 1.033 3.910 8.62 Fortified skim 1.038 3.929 8.66 Half and half 1.02 3.861 8.51 Half and half, fort. 1.024 3.876 8.55 Light cream 1.012 3.831 8.45 Heavy cream 0.994 3.763 8.30
### Weight of a gallon of milk (UK):
KG Pounds Producer milk 1.03 4.682 10.32 Homogenized milk 1.029 4.678 10.31 Skim milk, pkg 1.033 4.696 10.35 Fortified skim 1.038 4.719 10.40 Half and half 1.02 4.637 10.22 Half and half, fort. 1.024 4.655 10.26 Light cream 1.012 4.601 10.14 Heavy cream 0.994 4.519 9.96
## Summary
So, who would have thought that the density of milk and its various components could be so interesting? We now know why the cream rises to the top, why ping-pong balls float on a liquid and why marbles sink to the bottom. However, when considering density, the ultimate variable seems to be temperature, which pushes tightly knit molecules further apart and therefore reduces their density.
## Share:
### 1 thought on “Density of Milk and the Trick Question About the Weight of a Gallon of Milk”
1. Nicely done. Can anyone point me to a similar analysis of fat in milk by volume, not weight and not calories? I once found such on the Internet, but can’t find a discussion of volume any which way. Thanks. | 1,974 | 7,248 | {"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-2021-25 | latest | en | 0.918387 |
http://mathhelpforum.com/discrete-math/100401-stairs-problem-recurrence-print.html | 1,516,473,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00249.warc.gz | 213,680,392 | 3,091 | # Stairs problem (recurrence)
• Sep 3rd 2009, 07:44 AM
DaRush19
Stairs problem (recurrence)
Consider a staircase with n stairs which one can clim by 2 stairs or by 3 stairs at each step. e.g. a staircase with n=9 stairs can be climbed in exactly 5 ways.
Write a recurrence relation for the number of ways to ascend this staircase.
I worked out the values for [Consider a staircase with n stairs which one can clim by 2 stairs or by 3 stairs at each step. e.g. a staircase with n=9 stairs can be climbed in exactly 5 ways.
Write a recurrence relation for the number of ways to ascend this staircase.
I worked out the values for $Q_{n}$ up to 9:
$Q_{0}= 0$
$Q_{1}= 0$
$Q_{2}= 1$
$Q_{3}= 1$
$Q_{4}= 1$
$Q_{5}= 2$
$Q_{6}= 2$
$Q_{7}= 3$
$Q_{8}= 4$
$Q_{9}= 5$
which gave me $Q(n) = Q(n-2) + Q(n-3)$ but I think this is wrong.
Please double-check my values for $Q_{n}$/help me solve this!
• Sep 3rd 2009, 12:43 PM
Opalg
Quote:
Originally Posted by DaRush19
Consider a staircase with n stairs which one can clim by 2 stairs or by 3 stairs at each step. e.g. a staircase with n=9 stairs can be climbed in exactly 5 ways.
Write a recurrence relation for the number of ways to ascend this staircase.
I worked out the values for [Consider a staircase with n stairs which one can clim by 2 stairs or by 3 stairs at each step. e.g. a staircase with n=9 stairs can be climbed in exactly 5 ways.
Write a recurrence relation for the number of ways to ascend this staircase.
I worked out the values for $Q_{n}$ up to 9:
$Q_{0}= 0$
$Q_{1}= 0$
$Q_{2}= 1$
$Q_{3}= 1$
$Q_{4}= 1$
$Q_{5}= 2$
$Q_{6}= 2$
$Q_{7}= 3$
$Q_{8}= 4$
$Q_{9}= 5$
which gave me $Q(n) = Q(n-2) + Q(n-3)$ but I think this is wrong.
Please double-check my values for $Q_{n}$/help me solve this! | 578 | 1,747 | {"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": 26, "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-2018-05 | longest | en | 0.872467 |
https://electronics.stackexchange.com/questions/289595/logic-or-gate-with-2-diodes-in-series | 1,721,845,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00374.warc.gz | 197,405,178 | 41,885 | # logic OR gate with 2 diodes in series
is it possible to make a logic OR gate with 2 diodes in series like this:
simulate this circuit – Schematic created using CircuitLab
For my project this would be easier to wire than the usual parallel configuration.
• Should we treat the supply voltages as input? And say voltage across R1 and LED as output? Commented Mar 1, 2017 at 12:11
• diagram updated to show the inputs and the output Commented Mar 1, 2017 at 12:20
• Regarding your terminology: V1 and D2 are in series. D3, R1 and led D1 are in series. Diodes D2 and D3 are NOT in series due to the connection to V2.
– Tut
Commented Mar 1, 2017 at 12:23
• D2 - release the magic blue smoke with 1v7 across it. Commented Mar 1, 2017 at 15:57
No, not as you show.
Even without getting into the logic, clearly this can't be a OR gate. OR gates have two inputs and one output. No inputs are shown at all, and we can at best guess that the LED is intended to be the output.
Getting into the circuit, note that D2 always has a fixed 1.7 V across it. It won't live long that way. Also, whatever the branch of the circuit with the 5 V supply is doing is irrelevant. The top of D3 will always be at 3.3 V by definition of what V2 does.
All this circuit does is unconditionally light a LED and blow up a diode.
Here is the basic idea for a diode-based OR gate:
It can get more complicated from here, and using transistors for some gain is useful, but nonetheless, this is the basic concept.
• To be more specific... it is an OR function... as long as it is OR and NOT And. If he can guarantee only one supply will be used at any given time, it would work as an OR function. But, as you say, both on... and you let the smoke out. LED would be significantly brighter with just the 5V source though. Bad design in general. Commented Mar 1, 2017 at 12:19
s it possible to make a logic OR gate with 2 diodes in series like this:
V1 is shorted to v2 through d2. | 526 | 1,956 | {"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-2024-30 | latest | en | 0.954739 |
http://www.convertit.com/Go/Beverageonline/Measurement/Converter.ASP?From=Mexican+libra&To=mass | 1,670,183,779,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710978.15/warc/CC-MAIN-20221204172438-20221204202438-00715.warc.gz | 66,725,076 | 3,612 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
(Help)
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```Mexican libra = 0.46039625555 mass (mass) ``` Related Measurements: Try converting from "Mexican libra" to bes (Roman bes), bowling ball, crith, denarius (Roman denarius), dram (avoirdupois dram), dram ap (apothecary dram), earth mass, grain (avoirdupois grain), hyl, kin (Japanese kin), kwan (Japanese kwan), metric ton, mite (English mite), pennyweight troy (troy pennyweight), pound ap (apothecary pound), Roman talent, slug, tael (Chinese tael), talent (Greek talent), uncia (Roman uncia), or any combination of units which equate to "mass" and represent mass. Sample Conversions: Mexican libra = 2.11 bes (Roman bes), .0634375 bowling ball, 2,301.98 carat troy (troy carat), .00203 cotton bale (US), .00135333 cotton bale Egypt, 118.42 denarius (Roman denarius), 107.26 drachma2 (Greek drachma), 118.42 dram ap (apothecary dram), 7.70E-26 earth mass, 5.05E+29 electron mass (electron rest mass), 643.57 Greek obolos, .76719577 kin (Japanese kin), 634.38 obol (Greek obol), .92079251 pfund (German pfund), .0076125 picul (Chinese picul), 1.23 pound troy (troy pound), .00460396 quintal, .01127778 Roman talent, .0406 short quarter, 12.18 tael (Chinese tael).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 520 | 1,778 | {"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-2022-49 | latest | en | 0.675267 |
https://www.tumblr.com/search/Schwarzchild%20Radius | 1,511,482,418,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807044.45/warc/CC-MAIN-20171123233821-20171124013821-00629.warc.gz | 881,642,629 | 38,131 | #720: HOOPA
This troublemaker sends anything and everything to faraway places using its loop, which can warp space.
There is only one known physical phenomenon capable of warping time and space as Hoopa’s rings do: a black hole.
Somehow, inexplicably, this pokemon seems capable of using and manipulating black holes to its desire. And yet, just by holding it there, we can gather all the information we need to more closely analyze its mysterious hoop.
First things first: Black Holes do not suck. It’s a common misconception, but the facts remain if our Sun was replaced with a black hole with the exact same mass, nothing would happen to our planet or its orbit. (It would be significantly darker and colder, but that’s besides the point). Where does this misconception come from?
Well, everybody has heard that black holes have so much gravity that not even light can escape them. Here’s the catch—that’s only true within a critical limit called the Schwarzschild Radius.
The Schwarzchild Radius is a distance, by definition, where if you are any closer to the black hole than that, you would need to be moving faster than the speed of light to escape its gravitational pull. And as a fellow called Einstein showed, it’s impossible to go faster than the speed of light.
As it turns out, the equation for the Schwarzchild Radius is much simpler than you would think;
Where G is Newton’s Gravitational Constant, and c is the speed of light.
Which brings us back to Hoopa. It sends things through it’s hoop, meaning you have to be inside of the loop to experience it’s space-warping effects.
In other words, we know its Schwarzschild Radius.
From the Pokedex, we know that Hoopa is 1’08” tall. Since we’re scientists, we know that is equivalent to 0.508 meters. Doing a quick comparison of it’s height to it’ hoop, we can deduce that it’s hoop measures 0.343 meters across. Which means, Hoopa’s Schwarzschild Radius is 0.171 m.
## From the equation above, we can infer that Hoopa’s Black Hole has a mass of 1.16e+26 kg, or roughly .006% the mass of the sun.
This, of course, is not a small number. An object of this mass weighs 1.16e+23 metric tons on earth, almost 20 times more than the entire planet itself. Hoopa must be strong.
Similarly, if this is the mystery behind Hoopa’s shenanigans, there’s still plenty left unsolved. For example, we can only assume that the hoop itself somehow contains the black hole’s gravitational effects. If it didn’t, it would be impossible to walk near Hoopa without feeling the crushing force of gravity. Not to mention, it would likely have a extremely significant effect on our orbit around the sun, as well as our moon and the many satellites orbiting around us.
And that’s without even touching its unbound form.
You know, I didn’t ask to get my feelings hurt….
Approximately 1.1*10^-25 meters.
Yo' Mumma so fat her Schwarzschild Radius can actually be measured.
The more I learn about black holes, the more confusing they get.
Singularities enshrouded in event horizons that rip through the fabric of spacetime? lolwhut
friend of mine was following the death stranding speculation/hype train overnight and this is what’s been turned up so far:
“Norman Reedus’ character’s dogtags show equations relating to the Schwarzchild radius of black holes
The second one is the Dirac equation it seems, which has to do with large particles in quantum physics and general relativity
It’s also the equation that first implied the existence of dark matter
This game is going to be a spectacularly wild ride I can feel it already.
The game appears to be about a limbo state between life and death (death stranding is a term for dead sea life washing up on shores, whereas live stranding is when they’re beached alive). Very interested to see how this thing pans out. Great visuals so far though.
Themes of birth and rebirth, the sea, watchers in the sky, quantum realms, surrealism (dreams), etc. It actually looks quite a lot like it’s inspired by Lovecraftian motifs which is pretty neat.”
i’m not even a kojima fan and i’m interested | 936 | 4,091 | {"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-2017-47 | latest | en | 0.932127 |
https://300hours.com/f/cfa/level-1/t/ddm-2/ | 1,720,820,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00029.warc.gz | 60,943,123 | 61,648 | CFA CFA Level 1 2-Stage Dividend Discount Model example with Gordon Growth Model formula
# 2-Stage Dividend Discount Model example with Gordon Growth Model formula
• Author
Posts
• pcunniff
Participant
• CFA Level 1
13
Anyone know how to do this for the Gordon growth model?
A stock that currently does not pay a dividend is expected to pay its first dividend of \$1.00 five years from today. Thereafter, the dividend is expected to grow at an annual rate of 25% for the next three years and then grow at a constant rate of 5% per year thereafter. The required rate of return is 10.3%. The value of the stock today is closest to:
1. \$20.65.
2. \$22.72.
3. \$23.87.
Explanation: This is essentially a two-stage dividend discount model (DDM) problem. Discounting all future cash flows, we get:
Note that the constant growth formula can be applied to dividend 8 (1.253) because it will grow at a constant rate (5%) forever.
• Zee Tan
Keymaster
• CFA Charterholder
11
I got A as my answer.
• CF0 = 0
• CF1 = 0
• CF2 = 0
• CF3 = 0
• CF4 = 0
• CF5 = 1
• CF6 = 1*(1.25) = 1.25
• CF7 = 1*(1.25)2 = 1.5625
• CF8 = 1*(1.25)3 = 1.953125
• CF9 = 42.68 (see explanation below)
• I = 10.3%
• NPV = \$20.65
CF9 = dividend + price by Gordon Growth Model
Gordon Growth Model formula:
$P=\frac{{D}_{1}}{r–g}$
where
• P = price
• D1= value of next year’s dividend
• r= required rate of return
• g=constant rate of growth
So calculating CF9
CF9 = dividend + price by Gordon Growth Model
$=CF8×1.05+\left(\frac{CF8×1.{05}^{2}}{0.103–0.05}\right)\phantom{\rule{0ex}{0ex}}=1.953125×1.05+\left(\frac{1.953125×1.{05}^{2}}{0.053}\right)\phantom{\rule{0ex}{0ex}}=2.05+40.63\phantom{\rule{0ex}{0ex}}=42.68$
• pcunniff
Participant
• CFA Level 1
4
I just straight up struggle at these DDM questions. I am constantly wondering if I have to take the req rate of return as the denominator versus the constant growth at just D/expected rate of return – growth. You get growth at times by taking ROE * RR (1-div payout). So much to remember and I STILL get the questions wrong. Any last minute tips on this? Figure I should focus on stronger areas and not get bogged down here (though I know its very important). | 705 | 2,202 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 3.71875 | 4 | CC-MAIN-2024-30 | latest | en | 0.823015 |
http://www.sporcle.com/games/GOT_xEMx_MUNOZ/Multiplication_Quiz | 1,480,893,205,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541426.52/warc/CC-MAIN-20161202170901-00466-ip-10-31-129-80.ec2.internal.warc.gz | 711,714,776 | 20,660 | # Miscellaneous / Multiplication Quiz
Random Miscellaneous Quiz
## Can you name the Multiplication Answers?
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13 x 15 =
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OR | 166 | 460 | {"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} | 2.5625 | 3 | CC-MAIN-2016-50 | latest | en | 0.742108 |
https://rd.springer.com/book/10.1007%2F978-1-4612-1994-1 | 1,606,501,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00654.warc.gz | 456,389,405 | 13,673 | # Boundary Value Problems for Transport Equations
• Valeri Agoshkov
Book
1. Front Matter
Pages i-xvii
2. Valeri Agoshkov
Pages 1-34
3. Valeri Agoshkov
Pages 35-104
4. Valeri Agoshkov
Pages 105-163
5. Valeri Agoshkov
Pages 164-215
6. Valeri Agoshkov
Pages 216-262
7. Back Matter
Pages 263-278
### Introduction
In the modern theory of boundary value problems the following ap proach to investigation is agreed upon (we call it the functional approach): some functional spaces are chosen; the statements of boundary value prob the basis of these spaces; and the solvability of lems are formulated on the problems, properties of solutions, and their dependence on the original data of the problems are analyzed. These stages are put on the basis of the correct statement of different problems of mathematical physics (or of the definition of ill-posed problems). For example, if the solvability of a prob lem in the functional spaces chosen cannot be established then, probably, the reason is in their unsatisfactory choice. Then the analysis should be repeated employing other functional spaces. Elliptical problems can serve as an example of classical problems which are analyzed by this approach. Their investigations brought a number of new notions and results in the theory of Sobolev spaces W;(D) which, in turn, enabled us to create a sufficiently complete theory of solvability of elliptical equations. Nowadays the mathematical theory of radiative transfer problems and kinetic equations is an extensive area of modern mathematical physics. It has various applications in astrophysics, the theory of nuclear reactors, geophysics, the theory of chemical processes, semiconductor theory, fluid mechanics, etc. [25,29,31,39,40, 47, 52, 78, 83, 94, 98, 120, 124, 125, 135, 146].
### Keywords
Boundary value algorithms calculus equation function geometry transport equations
#### Authors and affiliations
• Valeri Agoshkov
• 1
1. 1.Russian Academy of SciencesInstitute of Numerical MathematicsMoscowRussia
### Bibliographic information
• DOI https://doi.org/10.1007/978-1-4612-1994-1
• Copyright Information Birkhäuser Boston 1998
• Publisher Name Birkhäuser, Boston, MA
• eBook Packages
• Print ISBN 978-1-4612-7372-1
• Online ISBN 978-1-4612-1994-1
• Series Print ISSN 2164-3679
• Buy this book on publisher's site
Industry Sectors | 591 | 2,346 | {"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} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.840134 |
http://mathhelpforum.com/algebra/91273-rectangle-problem-print.html | 1,511,330,946,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806465.90/warc/CC-MAIN-20171122050455-20171122070455-00043.warc.gz | 190,673,524 | 3,032 | # Rectangle problem??
• May 31st 2009, 11:24 AM
tiffieblah
Rectangle problem??
I have never known how to do these problems and there are plenty on the exam.
A rectangle has an area of (x^3-6x^2+10x-8) square meters and a width of (x-4) meters. What is its length?
How would I set this problem up???
• May 31st 2009, 12:18 PM
craig
Quote:
Originally Posted by tiffieblah
I have never known how to do these problems and there are plenty on the exam.
A rectangle has an area of (x^3-6x^2+10x-8) square meters and a width of (x-4) meters. What is its length?
How would I set this problem up???
If we just forget about the $x$ for a minute, I'm sure you know that the equation for the area of a rectangle is $A = w \times l$
Putting your above values into that question, we get the equation:
$x^3-6x^2+10x-8 = (x - 4)l$
Or $l = \frac{x^3-6x^2+10x-8}{x - 4}$
What we have now is just a simple polynomial division question.
There are several methods to solve this, I would suggest Synthetic Division, but use whatever method you have been taught.
Hope this helps
• May 31st 2009, 05:44 PM
tiffieblah
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## Presentation on theme: "Econometric analysis informing policies UNICEF workshop, 13 May 2008 Christian Stoff Statistics Division, UNESCAP,"— Presentation transcript:
Econometric analysis informing policies UNICEF workshop, 13 May 2008 Christian Stoff Statistics Division, UNESCAP, stoff@un.org
Outline Causality in experiments Confounding factors Quasi experiments: Difference estimators Difference-in-difference estimators Possible questions
A quick intro: The ideal situation Causality means a specific action leads to a specific, measurable consequence Ideal: Randomized controlled experiment Subjects are randomly allocated to either control or treatment group ONLY difference between two groups is treatment
The reality In practice experiments are rare Subjects are NOT randomly assigned so that sorting out of other relevant factors is difficult Econometrics provides the tools for controlling these other factors
Challenge: Confounding factors Regress test score on student-teacher ratio But what about number of students in class still learning English? – Omitted variable? Omitted variable correlated with explanatory and dependent variable (low student-teacher ratios -> high % English learners -> bad scores) Thus a policy of increasing no. teachers may not increase test scores because high English learners (%) are the real problem Solution: Control for differences in English learners (%), i.e. regress test score on student-teacher ratio AND English learners (%)
Limits to controlling these factors Many years of cross-country research However, countries often have such different settings and the causal relationships are only specific to the country and the time period Therefore in search of quasi or natural experiments between units that are not too different Danger lies in… –Possible correlation between error term and explanatory variable (i.e. treatment not assigned at random) –Teachers try especially hard in areas with programs –General equilibrium effects: when program is enlarged additional factors may arise (external validity)
Difference estimators: Using MICS3 Define unit of analysis (households, districts, provinces, countries) Selected units gone through policy program (i.e. treatment) AND assignment was as if random If is binary, then no functional form assumption needed; it is simply the difference in the conditional expectations If can take on multiple values, then the above regression assumes linearity But often there are pre-treatment differences between control and treatment group…
Difference-in-difference estimators: Using MICS2 and MICS3 Types of datasets: Cross-section, panel and time-series Includes observations on same units before and after experiment OLS estimator is the difference in the group means of Control for district-level context constant over time through fixed or random effects or adjust standard errors for clustering Advantage over difference estimator: 1. More efficient; 2. Eliminates pre-treatment differences
Some possible questions Education research: –Study drop-out rates and relate it to child labour questions –Study effect of different child disciplining strategies (punishment, praise, etc.) on a childs success in school –Combine MICS data with GIS disaster data and study effect of disasters on school attendance –Combine with policy data between 2000-2005 and evaluate the effectiveness of policies aimed at promoting higher school attendance Child health: –Effect of different fuel types for cooking on child-health indicators? –Effects of different types of access to water and sanitation on a childs probability of having diarrhoea or succeeding in school? –How does Vitamin A affect a childs health? –Impact of different health service facilities Adults knowledge and attitude towards violence: –What is the effect of having information access (TV or radio) on knowledge about HIV or contraception? –What is the effect of having information access (TV or radio) on education methods or attitudes towards domestic violence?
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5033 Days From June 12, 2024 - Convert Dates
## 5033 Days From June 12, 2024
Want to figure out the date that is exactly five thousand hundred thirty three days from Jun 12, 2024 without counting?
Your starting date is June 12, 2024 so that means that 5033 days later would be March 24, 2038.
You can check this by using the date difference calculator to measure the number of days from Jun 12, 2024 to Mar 24, 2038.
March 2038
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March 24, 2038 is a Wednesday. It is the 83rd day of the year, and in the 12nd week of the year (assuming each week starts on a Sunday), or the 1st quarter of the year. There are 31 days in this month. 2038 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 03/24/2038, and almost everywhere else in the world it's 24/03/2038.
### What if you only counted weekdays?
In some cases, you might want to skip weekends and count only the weekdays. This could be useful if you know you have a deadline based on a certain number of business days. If you are trying to see what day falls on the exact date difference of 5033 weekdays from Jun 12, 2024, you can count up each day skipping Saturdays and Sundays.
Start your calculation with Jun 12, 2024, which falls on a Wednesday. Counting forward, the next day would be a Thursday.
To get exactly five thousand hundred thirty three weekdays from Jun 12, 2024, you actually need to count 7047 total days (including weekend days). That means that 5033 weekdays from Jun 12, 2024 would be September 28, 2043.
If you're counting business days, don't forget to adjust this date for any holidays.
September 2043
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September 28, 2043 is a Monday. It is the 271st day of the year, and in the 271st week of the year (assuming each week starts on a Sunday), or the 3rd quarter of the year. There are 30 days in this month. 2043 is not a leap year, so there are 365 days in this year. The short form for this date used in the United States is 09/28/2043, and almost everywhere else in the world it's 28/09/2043.
### Enter the number of days and the exact date
Type in the number of days and the exact date to calculate from. If you want to find a previous date, you can enter a negative number to figure out the number of days before the specified date. | 1,013 | 3,033 | {"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-2024-26 | latest | en | 0.92764 |
https://codeofcode.org/lessons/timezones-and-daylight-savings-time/ | 1,695,318,712,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506029.42/warc/CC-MAIN-20230921174008-20230921204008-00360.warc.gz | 200,248,854 | 39,801 | Back to Course
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# Timezones and Daylight Savings Time
##### Yasin Cakal
Working with timezones and daylight savings time can be a complex task in Python. The datetime module provides several functions and classes for handling timezones and daylight savings time, as well as performing arithmetic with dates and times in different timezones.
## Timezone Object
The datetime module provides the tzinfo class for representing timezones. To create a timezone object, you can subclass the tzinfo class and implement the utcoffset(), dst(), and tzname() methods.
Here is an example of a timezone class for the Eastern Time zone:
import datetime
class EasternTimeZone(datetime.tzinfo):
def utcoffset(self, dt):
return datetime.timedelta(hours=-5)
def dst(self, dt):
return datetime.timedelta(0)
def tzname(self, dt):
return 'Eastern Time'
# create a timezone object
et = EasternTimeZone()
In this example, the utcoffset() method returns the offset from UTC (Coordinated Universal Time) in hours, the dst() method returns the daylight savings time offset in hours, and the tzname() method returns the name of the timezone.
You can use the timezone object to create datetime objects with a specific timezone. For example:
import datetime
# create a datetime object with a specific timezone
dt = datetime.datetime(2022, 1, 20, 12, 34, 56, tzinfo=et)
print(dt) # Output: 2022-01-20 12:34:56-05:00
In this example, the datetime object is created with the Eastern Time timezone.
You can also use the astimezone() method to convert a datetime object to a different timezone. For example:
import datetime
# create a timezone object for Pacific Time
pt = PacificTimeZone()
# convert the datetime object to Pacific Time
dt2 = dt.astimezone(pt)
print(dt2) # Output: 2022-01-20 10:34:56-07:00
In this example, the datetime object is converted from Eastern Time to Pacific Time using the astimezone() method.
## Exercises
To review these concepts, we will go through a series of exercises designed to test your understanding and apply what you have learned. | 492 | 2,116 | {"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} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.57253 |
https://seanet2017.com/help-matematicas-661 | 1,670,485,236,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711278.74/warc/CC-MAIN-20221208050236-20221208080236-00521.warc.gz | 540,468,995 | 5,330 | # Solve the system
These can be very helpful when you're stuck on a problem and don't know how to Solve the system. We will give you answers to homework.
## Solving the system
In addition, there are also many books that can help you how to Solve the system. How to solve perfect square trinomial. First, identify a, b, and c. Second, determine if a is positive or negative. Third, find two factors of ac that add to b. Fourth, write as the square of a binomial. Fifth, expand the binomial. Sixth, simplify the perfect square trinomial 7 eighth, graph the function to check for extraneous solutions. How to solve perfect square trinomial is an algebraic way to set up and solve equations that end in a squared term. The steps are simple and easy to follow so that you will be able to confidently solve equations on your own!
Another method is to use exponential equations. Exponential equations are equivalent to log equations, so they can be manipulated in the same way. By using these methods, you can solve natural log equations with relative ease.
Algebra can be a difficult subject for many students, but one way to make it easier is to solve by elimination. This method involves setting up equations and solving for one variable in terms of the others. For example, consider the equation ax+by=c. To solve for x, you would first multiply both sides by b and then subtract c from both sides. This would give you the equation bx-(c-ay)=0. You could then solve for x by factoring or using the quadratic formula. However, elimination is usually faster and simpler. Once you get practice using this method, you will be able to solve equations more quickly and easily.
In mathematics, a function is a rule that assigns a unique output to every input. A function can be represented using a graph on a coordinate plane. The input values are plotted on the x-axis, and the output values are plotted on the y-axis. A function is said to be a composite function if it can be written as the composition of two or more other functions. In other words, the output of the composite function is equal to the input of one of the other functions, which is then evaluated to produce the final output. For example, if f(x) = x2 and g(x) = 2x + 1, then the composite function h(x) = f(g(x)) can be graphed as follows: h(x) = (2x + 1)2. As you can see, solving a composite function requires you to first solve for the innermost function, and then work your way outwards. This process can be summarized using the following steps: 1) Identify the innermost function; 2) Substitute the input value into this function; 3) Evaluate the function to find the output; 4) substitute this output value into the next outermost function; 5) repeat steps 2-4 until all functions have been evaluated. By following these steps, you can solve any composite function.
Differential equations are a type of mathematical equation that can be used to model many real-world situations. In general, they involve the derivative of a function with respect to one or more variables. While differential equations may seem daunting at first, there are a few key techniques that can be used to solve them. One common method is known as separation of variables. This involves breaking up the equation into two parts, one involving only the derivative and the other involving only the variable itself. Once this is done, the two parts can be solved independently and then recombined to find the solution to the original equation. Another popular method is known as integration by substitution. This approach involves substituting a new variable for the original one in such a way that the resulting equation is easier to solve. These are just a few of the many methods that can be used to solve differential equations. With practice, anyone can become proficient in this important mathematical discipline.
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Solve simultaneous equations by finding inverse matrix, Ready to use Newcastle University Mathematics and Statistics 20/11/2019 14:50
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Task 6 Needs to be tested Linniker Grech 09/04/2020 15:10
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This question is used in the following exams: | 1,393 | 6,306 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.763317 |
https://math.stackexchange.com/questions/1793722/an-alternative-proof-to-baby-rudin-theorem-7-13 | 1,722,828,202,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640427760.16/warc/CC-MAIN-20240805021234-20240805051234-00697.warc.gz | 313,740,646 | 37,946 | # An alternative proof to baby Rudin theorem 7.13
I am trying to prove Theorem 7.13 in baby Rudin and find that my proof goes in a totally different direction than the proof in the book. I briefly sketch what I did.
Since $f_n(x)$ is continuous, then $f_n(x)$ is bounded in compact domain $K$ (theorem 4.15). Then $$M_n = \sup_{x\in E} |f_n(x)-f(x)|$$ is well defined and is non-increasing because $f_n(x)\ge f_{n+1}(x) \ge f(x)$. Moreover, $lim_{n\to \infty}M_{n} = 0$, then from Theorem 7.9, we conclude that $f_n\to f$ uniformly.
Could someone tell me whether my proof is wrong ? Or whether it is possible to prove this theorem in this way?
Theorem 7.13 Suppose $K$ is compact and
(a) $\{f_n\}$ is a sequence of continuous functions on $K$
(b) $\{f_n\}$ converges pointwise to a continuous function on $K$
(c) $f_n(x) \ge f_{n+1}(x)$ for all $x \in K$, $n=1,2,3,\dots$ Then $f_n \to f$ uniformly on $K$.
Rudin's Proof of Theorem 7.13 Put $g_n = f_n -f$. Then $g_n$ is continuous, $g_n \to 0$ pointwise, and $g_n \ge g_{n+1}$. We have to prove that $g_n \to 0$ uniformly on $K$.
Let $\epsilon > 0$ be given. Let $K_n$ be the set of all $x \in K$ with $g_n(x) \ge \epsilon$. Since $g_n$ is continuous, $K_n$ is closed (Theorem 4.8) hence compact (Theorem 2.35). Since $g_n \ge g_{n+1}$ we have $K_{n+1} \subset K_{n}$. Fix $x \in K$. Since $g_n(x) \to 0$, we see that $x \not\in K_n$ if $n$ is sufficiently large. Thus $x \not\in \bigcap K_n$. In other words, $\bigcap K_n$ is empty.Hence $K_N$ is empty for some $N$ (Theorem 2.36). It follows that $0 \le g_n(x) < \epsilon$ for all $x \in K$ and for all $n \ge N$. This proves the theorem.
Here is also the crucial Theorem 7.9:
Theorem 7.9 Suppose $$\underset{n \to \infty}{\lim} f_n(x) = f(x) \quad (x \in E).$$
Put $$M_n = \underset{x \in E}{\sup}|f_n(x)-f(x)|.$$ Then $f_n \to f$ uniformly on $E$ if and only if $M_n \to 0$ as $n \to \infty$.
• Please write what Theorem 7.13 is for those of us without Rudin directly on hand.
– AJY
Commented May 21, 2016 at 3:06
• @AJY You can find it here on page 150. The statement and proof are long. Commented May 21, 2016 at 3:09
• You need to include theorem 7.9 too!
– Pedro
Commented May 21, 2016 at 3:16
• How do you conclude that $\lim_{n\to \infty}M_n=0$? Commented May 21, 2016 at 3:16
• @kevin So what? The statement implies that $f_n\to f$ uniformly, which is what you want to prove! You cannot assume that. Commented May 21, 2016 at 3:20
Your proof looks to be correct; the statement that $\underset{n \to \infty}{\lim} M_n = 0$ is entirely analogous to the statement in Rudin's proof of Theorem 7.13 that $g_n \to 0$ uniformly.
HINT: use the uniform continuity of the $f_n$ and $f$, which follows from the compactness of $K$. | 983 | 2,742 | {"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} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.846679 |
https://www.sanfoundry.com/phase-transformation-question-bank/ | 1,660,281,325,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00210.warc.gz | 846,775,610 | 19,014 | Phase Transformation Questions and Answers – Solids Diffusional Transformations – Heterogeneous Nucleation
«
This set of Phase Transformation Question Bank focuses on “Solids Diffusional Transformations – Heterogeneous Nucleation”.
1. Nucleation in solids, as in liquids, in most cases is always __________
a) Homogeneous
b) Heterogeneous
c) Mixed
d) Semi heterogeneous
Explanation: This is the scenario found in most cases because they prefer the non-equilibrium defects as there nucleation sites, in many cases it occurs in the grain boundaries, excess vacancies and in the stacking faults.
2. If the creation of a nucleus results in the destruction of a defect this will result in the_______
a) Destruction of new nuclei
b) Mass production of new nuclei
c) Change of phase
d) Reduction of activation energy barrier
Explanation: The creation of a nucleus results in the destruction of a defect this will result in the production of some energy, let it be ΔGd and this will be released hence making an impact in the activation energy barrier and that too in the reduction of activation energy barrier.
3. Which among the following equation satisfies heterogeneous nucleation?
a) ΔGhet = – V*(ΔGv – ΔGs) + Aγ – ΔGd
b) ΔGhet = – V*(ΔGv – ΔGs) – ΔGd
c) ΔGhet = – V*(ΔGv – ΔGs) + Aγ + ΔGd
d) ΔGhet = – V*(ΔGv – ΔGs) + Aγ
Explanation: Equation, ΔGhet = – V*(ΔGv – ΔGs) + Aγ – ΔGd corresponds to heterogeneous nucleation. Here the only difference between the homogeneous and the heterogeneous nucleation comes in the last term (-) ΔGd and this the energy released during the destruction of defects.
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4. If we ignore the effect of misfit strain energy, which among the following could minimize the interfacial free energy?
a) Optimum embryo texture
b) Optimum embryo shape
c) Optimum embryo volume
d) Optimum embryo length
Explanation: Ignoring any misfit strain energy, the optimum embryo shape should be that which minimizes the total interfacial free energy. The optimum shape for an incoherent grain-boundary nucleus will consequently be two abutted spherical caps.
5. The ability of the grain boundary to reduce the value of critical free energy ΔG* depends on
a) CosΦ
b) SinΦ
c) TanΦ
d) CotΦ
Explanation: The ability of a grain boundary to reduce ΔG (het), i.e. its potency as a nucleation site, depends on cosΦ. The cosΦ is actually the ratio of γαα/γαβ (assuming γαβ is isotropic and equal for both grains).
6. Calculate the shape factor if the angle tends to 90 degree?
a) 0
b) 1
c) 2
d) 4
Explanation: The shape factor will tend to unity when the angle tends to 90 degree, this can be proved easily using the following equation 0.5(2+cosΦ)*(1-cosΦ)2. Substitute the value 90 in the place of Φ then we get the required answer as 1.
7. If the ΔG* heterogeneous is given as 5kJ/mol and ΔG* homogeneous is given as 10kJ/mol. Calculate the shape factor?
a) 0.5
b) 1
c) 2
d) 2.5
Explanation: Here in this case the shape factor can be determined by the ratio of ΔG*heterogeneous/ΔG*homogeneous. So here for this particular case the shape factor value is 5/10 =0.5. Actually the grain boundary has the ability to reduce ΔG* heterogeneous, it’s the power it has as a nucleation site.
8. Excess vacancies are retained during the quench if the age hardening alloys are retained from______
a) Dark texture
b) Room temperature
c) Low temperature
d) High temperature
Explanation: Excess vacancies are retained during the quench if the age hardening alloys are retained from high temperature. This can actually increase the diffusion rate or it can relieve the misfit strain energies hence it assist the nucleation.
9. What is special about the dark-field electron microscope micrograph?
a) Precipitates can be imaged bright and matrix dark
b) 360 degree rotation is possible
c) High speed imaging is possible
d) Anti roller scheme available
Explanation: The so-called dark-field electron microscope micrograph in which the precipitates are imaged bright and the matrix dark. The precipitates lie in rows along dislocations. And this can be used in case of niobium carbonitride precipitates on dislocations in a ferritic iron matrix
10. The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given as 0.5. Calculate the factor C/C1 (the number of atoms on heterogeneous sites relative to the number within the matrix)? (Consider exp (ΔG*(hom) – ΔG*(het)/kT) to be 1)
a) 0.5
b) 1.2
c) 0.25
d) 1.25
Explanation: The relative magnitudes of the heterogeneous and homogeneous volume nucleation rate is given by the equation (C/C1) *(exp (ΔG*(hom) – ΔG*(het)/kT). So in this case the nucleation rate is 0.5 which itself is the factor C/C1.
11. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.
a) False
b) True
Explanation: The driving force is the factor which determines whether the type of site gives the highest volume nucleation rate or not. At very small driving forces, when activation energy barriers for nucleation are high, the highest nucleation rates will be produced by grain corner nucleation.
12. A coherent nucleus with a negative misfit can reduce the critical volume free energy by forming a region of _________
a) Compressive strain
b) Compressive stress
c) Tensile strain
d) Tensile stress
Explanation: This directly implies that the smaller volume than the matrix can reduce its ΔG* by forming a region of compressive strain above an edge dislocation, whereas if the misfit is positive it is energetically favourable for it to form below the dislocation.
13. In FCC crystals the a/2*[110] unit dislocations can dissociate to produce a ribbon of stacking fault. Which among the following is as an example of the same?
a) a/2*[110]->a/6*[121] + a/6[211]
b) a/2*[110]->a/6*[101] + a/6[201]
c) a/2*[110]->a/6*[821] + a/6[911]
d) a/2*[110]->a/6*[161] + a/6[221]
Explanation: a/2*[110]->a/6*[121] + a/6*[211] is an example and this gives a stacking fault on (111) separated by two Shockley partials. Since the stacking fault is in effect four close-packed layers of hcp crystal, it can act as a very potent nucleation site for an hcp precipitate.
14. Which among the following curve represent the grain boundary?
a) 1
b) 2
c) 3
d) Cannot be predicted
Explanation: Curve one represents the grain boundary and this can be determined from the effect of angle on the activation energy for grain boundary nucleation relative to homogeneous nucleation. Curve 2 and 3 represents the grain edges and grain corners respectively.
15. Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate.
a) True
b) False
Explanation: Nucleation on dislocations may also be assisted by solute segregation which can raise the composition of the matrix to nearer that of the precipitate. The dislocation can also assist in growth of an embryo beyond the critical size by providing a diffusion pipe with a lower ΔG(migration).
Sanfoundry Global Education & Learning Series – Phase Transformation.
To practice Phase Transformation Question Bank, here is complete set of 1000+ Multiple Choice Questions and Answers. | 1,831 | 7,342 | {"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} | 2.9375 | 3 | CC-MAIN-2022-33 | latest | en | 0.83888 |
https://api-project-1022638073839.appspot.com/questions/how-do-you-graph-the-equation-3y-2x-6 | 1,726,374,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00057.warc.gz | 83,267,438 | 5,914 | # How do you graph the equation 3y=2x+6?
Jan 2, 2018
See below.
#### Explanation:
Turn $3 y = 2 x + 6$ into $y = m x + b$
$y = \frac{2}{3} x + 2$
y-intercept is at $\left(0 , 2\right)$, slope is $\frac{2}{3}$.
graph{y=2/3x+2 [-10, 10, -5, 5]} | 115 | 249 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 4.09375 | 4 | CC-MAIN-2024-38 | latest | en | 0.716409 |
https://www.exceltip.com/lookup-formulas/lookup-value-with-multiple-criteria.html | 1,632,350,379,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00133.warc.gz | 755,808,820 | 19,956 | # How to Lookup Value with Multiple Criteria in Excel
It's easy to look up for value with one unique key in a table. We can simply use the VLOOKUP function. But when you don’t have that unique column in your data and need to lookup in multiple columns to match a value, VLOOKUP doesn’t help.
So, to lookup a value in a table with multiple criteria we will use INDEX-MATCH-INDEX formula.
Generic Formula for Multiple Criteria Lookup
INDEX(lookup_range,MATCH(1, INDEX((criteria1 =range1)*(criteria2=range2)*(criteriaN=rangeN),0,1),0))
lookup_range: Its the range from which you want to retrieve value.
Criteria1, Criteria2, Criteria N: These are the criteria you want to match in range1, range2 and Range N. You can have up to 270 criteria - range pairs.
Range1, range2, rangeN : These are the ranges in which you will match your respective criteria.
How it will work? Let's see…
#### INDEX and MATCH with Multiple Criteria Example
Here I have a data table. I want to pull the name of the customer using Date of Booking, Builder and Area. So here I have three criteria and one lookup range.
Write this formula in cell I4 hit enter.
=INDEX(E2:E16,MATCH(1, INDEX((I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16),0,1),0))
How it works:
We already know how INDEX and MATCH function work in EXCEL, so i am not going to explain that here. We will talk about the trick we used here.
(I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16): The main part is this. Each part of this statement return an array of true false.
When boolean values are multiplied they return array of 0 and 1. Multiplication works as AND operator. Hense when all value are true only then it returns 1 else 0
(I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16) This altogether will return
{FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE}*
{FALSE;FALSE;FALSE;TRUE;TRUE;TRUE;TRUE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;TRUE}*
{FALSE;FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;TRUE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE;FALSE}
Which will translate into
{0;0;0;0;0;0;0;1;0;0;0;0;0;0;0}
INDEX((I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16),0,1): INDEX Function will return the same array ({0;0;0;0;0;0;0;1;0;0;0;0;0;0;0}) to MATCH function as lookup array.
MATCH(1,INDEX((I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16),0,1): MATCH function will look for 1 in array {0;0;0;0;0;0;0;1;0;0;0;0;0;0;0}. And will return index number of first 1 found in array. Which is 8 here.
INDEX(E2:E16,MATCH(1,INDEX((I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16),0,1),0)): Finally, INDEX will return value from given range (E2:E16) at found index (8).
Simple????. Sorry, couldn’t make it more simpler.
#### Array Solution
If you can hit CTRL + SHIFT + ENTER consequently then you can eliminate the inner INDEX function. Just write this formula and hit CTRL + SHIFT + ENTER.
=INDEX(E2:E16,MATCH(1,(I1=A2:A16)*(I2=B2:B16)*(I3=C2:C16),0))
Generic Array Formula for Multiple Criteria Lookup
=INDEX(lookup_range,MATCH(1,(criteria1 =range1)*(criteria2=range2)*(criteriaN=rangeN),0))
Formula works same as above explanation.
I tried my best to explain it as simple as possible. But if I was not clear enough, let me know it in the comment section below. By the way, you don’t need to know how the engine works to drive a car. You just need to know how to drive it. And you know it very well.
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How to use the SUMIF Function in Excel | 1,100 | 3,744 | {"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-2021-39 | longest | en | 0.769645 |
https://www.lidolearning.com/questions/m-bb-selina7-ch1-ex1d-q12/q12-find-the-result-of-subtrac/ | 1,600,947,425,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400217623.41/warc/CC-MAIN-20200924100829-20200924130829-00123.warc.gz | 932,750,446 | 10,900 | Selina solutions
# Question 12
Q12) Find the result of subtracting the sum of all integers between 20 and 30 from the sum of all integers from 20 to 30.
Solution 12:
Sum of the all integers from 20 and 30
20+21+22+23+24+25+26+27+28+29+30=275
Sum of the all integers between 20 and 30
21+22+23+24+25+26+27+28+29=225
∴ Required number= (Sum of the all integers from 20 and 30)-(Sum of the all integers between 20 and 30 )
=275-225
=50
∴The required number is 50.
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Chemistry | 255 | 791 | {"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} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.76969 |
https://www.mikros.us/2020/04/four-circle-venn-diagram.html | 1,631,920,023,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055808.78/warc/CC-MAIN-20210917212307-20210918002307-00293.warc.gz | 908,888,109 | 69,438 | # Four Circle Venn Diagram
Four Circle Venn Diagram. Illustrate the Venn Diagram with this template. You can export your work as image, and save in.
The usual depiction makes use of a rectangle as the universal set and circles for the sets under consideration. Once you have got to grips with these, you will be able to arrange all sorts of groups and sets. Use Venn diagrams to illustrate data in a logical way which will enable you to see groupings and sets clearly.
### Venn Diagram is a pictorial representation of sets and their operations using circles.
logic - 4 set venn diagram in TikZ - TeX - LaTeX Stack ...
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peterme.com: Highlights from HITS Part 1: MBA 101 and Design
Venn Diagram 4 Circles Including Business Requirements ...
Venn diagrams can be used to express the logical (in the mathematical sense) relationships between various sets. Once you have got to grips with these, you will be able to arrange all sorts of groups and sets. Venn diagram, also known as Euler-Venn diagram is a simple representation of sets by diagrams. | 296 | 1,318 | {"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-2021-39 | latest | en | 0.87822 |
https://www.physicsforums.com/threads/what-should-the-ph-of-a-121m-ki-solution-be-electrochem-lab.196469/ | 1,511,405,211,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806715.73/warc/CC-MAIN-20171123012207-20171123032207-00530.warc.gz | 850,806,619 | 15,379 | # What should the pH of a .121M KI solution be (electrochem lab)?
1. Nov 6, 2007
### dopaminergic
I think I calibrated my pH meter improperly during lab; I got a pH of 4.70, which doesn't make any sense, as KI isn't acidic, as far as I know.
After electrolysis (2I- + 2H2O --> I2 + 2OH- + H2) with a 9V battery, I found a pH of 9.97, which doesn't seem to be correct, either:
(change in [OH-]) = (10^-4.03) - (10^-9.3) = 9.33 * 10^-5M
mol OH- = [OH-] * volume = (9.33 * 10^-5M) * 0.050L = 4.6 * 10^-6 mol
I = Q/t
t = 1800s
Q = (96485C/ mol e-) * (2mol e- / 2mol OH-) * (4.6 * 10^-6 mol OH-)
I = (96485C/ mol e-) * (2mol e- / 2mol OH-) * (4.6 * 10^-6 mol OH-) / 1800s
I = 2.4657 * 10^-4 C/s
I = 2.5 * 10^-4 A
That seems like an absurdly low amperage - have I done anything wrong? | 337 | 788 | {"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-2017-47 | longest | en | 0.91593 |
https://calculla.com/calculators/math/quadratic_equation | 1,603,244,255,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00305.warc.gz | 259,287,405 | 198,122 | Calculator finds out solution of quadratic equation given in general ax²+bx+c=0 form.
# Beta version#
BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !
# Symbolic algebra
ⓘ Hint: This calculator supports symbolic math. You can enter numbers, but also symbols like a, b, pi or even whole math expressions such as (a+b)/2. If you still don't sure how to make your life easier using symbolic algebra check out our another page: Symbolic calculations
# Input equation, which you want to solve#
Parameters of the ax2 + bx + c = 0 equation Coefficient a(just before x2) Coefficient b(just before x) Free parameter c
# The solution of your equation#
The equation you entered Show source$2\cdot{ x}^{2}+5\cdot x - 8 = 0$ The solution of the equation Show source$x \in \left\{\frac{-5}{4}-\frac{\sqrt{89}}{4}, \frac{-5}{4}+\frac{\sqrt{89}}{4}\right\}$
# The solution step-by-step#
I. We calculatate discriminant of the quadratic equation $\Delta$.
\begin{aligned}\Delta& = {5}^{2} - \left(4\cdot2\cdot\left(-8\right)\right) = 25 - \left(4\cdot2\cdot\left(-8\right)\right) = 25+64 = 89\end{aligned}
II. Delta is positive (Δ > 0), so equation has two solutions (roots).
The first solution is:
\begin{aligned}x_1& = \frac{-5 - \sqrt{89}}{2\cdot2} = \frac{-5 - \sqrt{89}}{4} = \frac{-5}{4}-\frac{\sqrt{89}}{4}\end{aligned}The second solution is:
\begin{aligned}x_2& = \frac{-5+\sqrt{89}}{2\cdot2} = \frac{-5+\sqrt{89}}{4} = \frac{-5}{4}+\frac{\sqrt{89}}{4}\end{aligned}
# Some facts#
• The quadratic equation is an equation that can be presented in the form:
$a~x^2 + b~x + c = 0$
where:
• a, b, c - constant parameters, these are numbers that we know,
• x - unknown variable, it's a number, which we search for.
• Quadratic equation can have one solution, two solutions or do not have solutions.
• The universal method of solving quadratic equations uses discriminant of the quadratic polynomial (so-called delta):
$\Delta={ b}^{2}-4~ a~ c$
• When we calculate the discriminant, three scenarios are possible:
• discriminant is positive (Δ > 0) - equation has two different solutions (two different roots):
$x_1=\frac{- b-\sqrt{ \Delta}}{2~ a}$
$x_2=\frac{- b+\sqrt{ \Delta}}{2~ a}$
• discriminant is zero (Δ = 0) - equation has exactly one solution (so-called double root):
$h=\frac{- b}{2~ a}$
• discriminant is negative (Δ < 0) - equation has no solutions (so-called contradictory equation).
If you are interested in solving mathematical equations, check out our other calculators:
• Linear equation solver - see how to solve a linear equation in the form $ax + b = 0$ step by step,
• Quadratic equation solver - see how to solve quadratic equation in the form $ax ^ 2 + bx + c = 0$ using the so-called delta scheme,
• General equation solver - if you don't know which solving method should be applied to your equation, just give us the left and right side and we will try to solve it for you. | 902 | 3,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "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} | 4.375 | 4 | CC-MAIN-2020-45 | longest | en | 0.773054 |
https://www.physicsforums.com/threads/intensity-of-the-incident-light.305242/ | 1,575,808,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540510336.29/warc/CC-MAIN-20191208122818-20191208150818-00076.warc.gz | 815,797,651 | 15,028 | # Intensity of the incident light
1. Homework Statement
Unpolarized light is incident on a system of three polarizers. The second polarizer is oriented at an angle of 33.0° with respect to the first and the third is oriented at an angle of 45.0° with respect to the first. If the light that emerges from the system has an intensity of 2.3 W/m2, what is the intensity of the incident light?
2. Homework Equations
Iout=Iincos2$$\theta$$
3. The Attempt at a Solution
I know that after I0 passes through the first polarizer, it is 1/2I0. After it passes through the second, I thought it would be 1/2I0cos2(33), and after it passes through the third, 1/2I0cos2(33)cos2(45).
This would all equal 2.3, which I was given.
So I set up my final equation as:
1/2I0cos2(33)cos2(45) = 2.3, or
I0= 2.3 / (1/2cos2(33)cos2(45).
I solved and got 13.08, but that is incorrect. Can anyone offer any insight? Thanks.
## Answers and Replies
Related Introductory Physics Homework Help News on Phys.org
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
For the third polarizer, use the angle between it and the second polarizer.
45° is the angle between the third and first polarizer. | 345 | 1,179 | {"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} | 3.703125 | 4 | CC-MAIN-2019-51 | longest | en | 0.920694 |
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# Electricity - Basic Navy Training CoursesNAVPERS 10622
Here is the "Electricity - Basic Navy Training Courses" (NAVPERS 10622) in its entirety. It should provide one of the Internet's best resources for people seeking a basic electricity course - complete with examples worked out. See copyright. See Table of Contents.
U.S. GOVERNMENT PRINTING OFFICE; 1945 - 618779
CHAPTER 10
SERIES-PARALLEL CIRCUITS
COMBINATIONS
Many circuits are neither SIMPLE series nor SIMPLE parallel. They are COMBINATIONS of simple circuits. Fortunately, it is easy to recognize the series or parallel connections within the combinations by a few simple rules.
SERIES CONNECTIONS HAVE -
1. Only one path.
2. Only one conductor connected to a terminal.
3. All the current of one part passing through the other part.
PARALLEL CONNECTIONS HAVE -
1. More than one path.
2. More than one conductor connected to a terminal.
3. Divided currents.
The best way to understand the complicated series-parallel circuits is to analyze them part by part, or section by section. In this way, the correct circuit law can be used to analyze each part. First, spot your series connections, and then locate the parallels.
The balance of this chapter consists of a number of circuits. They illustrate the important problems you will meet in actual circuits. Go through each step carefully-he sure you understand it. If you get stuck, you'll probably find that you've forgotten one of the laws given in Chapters 6, 7, 8 or 9. To help you, all these laws are brought together in a table at the end of this chapter. Turn back to the table if you need help. Remember, the important thing is to KNOW. You don't really know that you KNOW unless you test yourself. When you have finished going over the examples, try working them yourself without reference to this book. After you've finished each example, check your answers and methods against the answers and methods given here.
EXAMPLE 1- During war, ships must travel blacked out. No lights show except those used for communication and as formation guides. These lights are screened to 'show only in one direction. Imagine a lighted compartment with a door or hatch opened accidentally-a beautiful target! To guard against any such accident, doors and hatches opening to exposed decks are equipped with door-switches. Door-switches open the lighting circuit of the compartment so that the lights are turned off every time the door is opened.
Draw the schematic diagram for wiring a compartment with two overhead lights, separately con-trolled, and with a door-switch.
Trace this circuit (figure 56) from - to +. Following the arrows and assuming that all switches are closed, current leaves the negative terminal of the source, flows along L1 (Line 1) to the first light terminal. Here the current divides (did you spot a parallel connection?) - part goes to light No. 1 and part goes to light No.2. (You can determine how much current goes to each light if you know the voltage and resistance - I = E/R). The current passes through the two lights and enters L2 then through L2 and back to the positive terminal of the source.
Figure 56. - Compartment schematic diagram.
Try opening switch No. 1. What happens? The circuit through light No. 1 is opened, BUT the circuit through light No. 2 is undisturbed. Try the same with switch No. 2. This switch affects only light No. 2. Switch No.1 is IN SERIES WITH light No.1 and thereby CONTROLS light No. 1. But switch No. 1 is IN PARALLEL WITH light No. 2 and CANNOT control light No. 2. Now open the door-switch. This switch is in series with BOTH lights and CONTROLS both lights. Ask yourself this question, "Does ALL the current of light No. 2 go through the door-switch?" The answer is "YES." And whenever the answer to this question is 'yes' - the two devices are in series.
EXAMPLE 2 - Aboard ship you will have a ship's service generator. This generator furnishes the power for all electrical circuits except propulsion. The generated power is fed through a main switch-board, then through lighting panels, interior communication panels, etc., and to feeder boxes and branch feeder boxes. And finally, to the power outlets - lights, heaters, and telephones.
Figure 57. - Water distribution.
These ship's distribution systems are a lot like the water distribution systems of small towns. Look at figures 57 and 58 and compare the branching methods .. In figure 58 start at the power outlet-trace backwards to the power bus. (A BUS is simply a very large conductor-usually a bar of copper.) Notice that the fuses PROTECTING each line and box are in SERIES with the load. And of course, switches controlling each load would be in series with that load.
If you start tracing at the bus, you will notice that the "FEEDING-OUT" or branching is done by PARALLEL connections. Start tracing from the negative bus and go through the complete circuit of the outlet. How many times is this circuit protected by fuses? The reason for this multiple protection is simple. Each fuse has a capacity which just fits the circuit it protects. For instance, the branch box has 5-ampere fuses because the circuit from branch box to outlet is only large enough to handle 5 amperes. The feeder box is fused for 25 amperes because the circuit from feeder box to branch box will stand only 25 amperes. Each fuse protects its own circuit from overload or short circuit damage.
Figure 58. - Electrical distribution.
EXAMPLE 3 - The electrical power to a Navy searchlight must do four things-
1. Furnish an arc between the carbon electrodes.
2. Run the feed motor.
3. Run the ventilating fan motor.
4. Run the shutter motor.
Power is furnished to each of these loads by means of a four-branch parallel circuit. Figure 59 is a simplified diagram of the searchlight circuit. In addition to the loads, there is a rheostat (an adjustable resistance) in series with the parallel group. This is necessary to reduce the ship's voltage from about 120 volts to about 80 volts for searchlight operation. In figure 59 the current is labeled for each load. (1) What is ,the voltage drop across the rheostat? (2) What is the voltage used across each branch of the parallel? (3) What is the resistance of each branch of the parallel? (4) What is the total resistance?
Figure 59. - Simplified searchlight diagram.
(1) The rheostat is in series with the rest of the circuit, therefore, it carries all the current of the circuit -
It = I1 + l2 + I3, etc.
It = 150 + 2 + 2 + 3 = 157 amps.
The voltage drop of this rheostat is
E = IR = 157 X 0.25 = 39.25 volts.
(2) The parallel group is in series with the rheostat, therefore adding the voltages of the group and the rheostat together gives the total voltage.
Et = E1 + E2 + E3, etc.
120 = 39.25 + E2
E2 = 120 - 39.25 = 80.75 v.
Which means that the voltage drop across the rheostat is 39.25 volts and the drop across EACH branch of the parallel is 80.75 volts. You might look at it this way - you have 120 volts (ship's service) to use in forcing 157 amperes through the complete circuit. The rheostat used up 39.25 volts of this 120. Which leaves 80.75 volts for the balance of the circuit - the parallel group.
(3) The resistance of each branch of the parallel appears to be -
(arc)
R1 = E1/I1 = 80.75/150 = 0.54 ohm.
(feed and shutter motors)
R2 =E2/I2 = 80.75/2 = 40.38 ohms each.
(ventilating fan motor)
R3 = E3/I3 = 80.75/3 = 26.92 ohms.
Notice in these calculations, that the DIFFERENT, RESISTANCES LIMIT the current to DIFFERENT VALUES for a particular load.
(4) The total resistance can be calculated by two methods -
Rt = Et/It = 120/157 = 0.77 ohm.
OR
the resistance for the parallel group is -
1/Rt = 1/R1 + 1/R2 + 1/R3 etc.
1/Rt = 1/0.54 + 1/40.38 + 1/40.38 + 1/26.92
Rt = 0.52 ohm. (of parallel group only)
add this to the resistance of the rheostat which is in series -
Rt = R1 + R2 + R3, etc.
Rt = 0.52 + 0.25 = 0.77 ohm.
It is important that you see how much EASIER it is to find TOTAL RESISTANCE BY OHM'S LAW.
EXAMPLE 4 - You know that searchlights are bright-but how much power do they consume? In the searchlight of Example 3, the ARC ITSELF used 80.75 volts and passes 150 amperes. Therefore, its power is -
P = EI
P = 80.75 x 150 = 12,112.5, say 12,110 watts.
And the total power consumed by the light and its apparatus is -
P = EI P = 120 x 157 = 18,840 watts.
This is approximately the same amount of power as consumed by a 20-hp motor. SOME LIGHT!
EXAMPLE 5 - The operating voltage of a sub-marine's motors is 120 volts at cruising speed. This voltage must come from batteries of the lead-acid type. But this type of storage battery produces only 2 volts per cell. How is an emf of 120 volts produced by cells which themselves produce only 2 volts each? Look at the laws of voltage in the series and the parallel circuits -
Et = E1 + E2 + E3, etc. (SERIES)
Et = E1 = E2 = E3, etc. (PARALLEL)
It is evident that in the series connection, voltages will add. Two cells of 2 volts in series would add, giving 4 volts. And ten cells in series would give 20 volts. To produce the 120 volts needed by the sub's motors requires 60 cells in series. A part of such a battery is shown in figure 60.
All batteries store only a certain amount of energy - the exact amount is known as their CAPACITY. Capacity is measured in units of AMPERE-HOURS - the number of amperes which can flow for a certain number of hours before the battery is discharged. For example, a battery having a capacity of 100 ampere-hours will deliver a current of 10 amperes for 10 hours (10 x 10 = 100) or 5 amperes for 20 hours (5 x 20 = 100) or 50 amperes for 2 hours (50 x 2 = 100) before it is discharged.
Say that the sub's motors require a current of 200 amperes at 120 volts. This would exhaust a 1,000 ampere-hour battery in 5 hours. But by using two batteries in PARALLEL the drain on each battery is only 100 amperes -
It = I1 + I2
Figure 60. - Cells in series.
Where It = 200 amperes, I1 and I2 are only 100 amperes each. Figure 61 shows two batteries of 20 cells - the cells are in SERIES and the two batteries are in PARALLEL. Remember that CELLS IN SERIES INCREASE THE VOLTAGE and CELLS IN PARALLEL INCREASE THE CURRENT.
EXAMPLE 6 - Three vacuum tube filaments rated at 0.3 ampere and 6.3 volts must be operated on a 110-volt line. Obviously, the voltage is too HIGH. Even connecting the tubes in series -
Et = E1 + E2 + E3
110 = 36.67 + 36.67 + 36.67
gives a voltage of 36.67 volts per unit whereas each unit is designed for only 6.3 volts. This means applying about six times the rated voltage to each tube - they would burn out in a split second. A series resistance will have to be added to the circuit to use up some of the excess voltage. The question is - how MUCH resistance? You have a circuit involving 110 volts and you must LIMIT the current by a resistor to 0.3 ampere.
Figure 61. - Cells in series-parallel.
Therefore, you will need a total of -
R = E/I = 110/0.3 = 366.67 ohms of resistance.
R = E/I = 6.3/0.3 = 21 ohms of resistance in each tube.
If they are connected in series, you have a total resistance of -
Rt = R1 + R2 + R3
Rt = 21 + 21 + 21 = 63 ohms.
for the tubes.
If the tubes furnish 63 ohms out of a total requirement of 366.67 ohms, the balance, R2, is found to be -
Rt = R1 + R2
366.67 = 63 + R2
R2 = 366.67 - 63 = 303.67 ohms.
This resistance would have to be furnished by the resistor. The completed circuit would look like figure 62. Notice that the voltage is labeled at a number of points in figure 62. This shows that the voltage drops as the current goes through each successive load. In other words, some voltage is used up in pushing the current through each resistance. The voltage drop in each case, is the difference in voltage between the two points.
Figure 62. - Example 6 - Series.
You can prove this circuit. The voltage drop (often called IR drop) across the resistor is -
E = IR = 0.3 x 303.67 = 91.1 volts
leaving E2, which is found by the formula -
Et = E1 + E2
Et = 91.1 + E2 = 18.9 volts
This is the total for the three tubes. They are in series so each uses one-third of this 18.9 volts, or 18.9/3 = 6.3 v. which is the rated voltage.
PRACTICE
For practice, try to set up this circuit with the tubes in parallel, and a limiting resistor in series. Your circuit should look like figure 63.
Figure 63. - Example 6 - Parallel.
BEFORE OR AFTER
Perhaps you are wondering if it makes any difference whether a limiting resistor comes BEFORE or AFTER the load. Think about a garden hose. Does it make any difference which valve you operate - the one at the meter, or the one at the side of the house, or the one in the nozzle of the hose? Partially closing anyone of these valves will limit the amount of water flowing through the hose. Likewise, placing a resistance any place in a circuit will limit the current through every load that is in series with the resistance.
CIRCUIT LAWS
OHM'S LAW POWER EQUATION E = I/RI = E/RR = E/I P = EIE = P/II = P/E SERIES CIRCUITS PARALLEL CIRCUITS Et = E1 + E2 + E3, etc.It = I1 + I2 + I3, etc.Rt = R1 + R2 + R3, etc. Et = E1 = E2 = E3, etc.It = I1 = I2 = I3, etc.1/Rt = 1/R1 = 1/R2 = 1/R3, etc.
Chapter 10 Quiz
RF Cafe Software RF Cascade Workbook RF system analysis includingfrequency conversion & filters A Disruptive Web Presence™ Custom Search Over 9,000 pages indexed! Read About RF CafeWebmaster: Kirt BlattenbergerKB3UONProduct & Service DirectoryPersonally Selected Manufacturers RF Cafe T-Shirts & Mugs Calculator Workbook RF Workbench | 3,657 | 14,182 | {"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} | 2.984375 | 3 | CC-MAIN-2013-20 | latest | en | 0.871887 |
https://fivetwelvethirteen.wordpress.com/2014/09/08/how-do-you-assess-number-sense/ | 1,539,723,834,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510867.6/warc/CC-MAIN-20181016201314-20181016222814-00319.warc.gz | 680,329,240 | 19,784 | # How Do You Assess Number Sense?
My instinct is, “I know it when I see it”.
I want to move beyond that. My school loves to talk about “basic skills”. Basic skills are usually interpreted as rational number operations — decimals, fractions, positives and negatives. Then these are tested using numbers requiring several steps of calculation. But, in Algebra, lots of those basic skills become less important. Students aren’t often operating with decimals and fractions, and doing multi-digit whole number operations. When they are, they often have a calculator — as they should, to minimize the working memory load as they struggle with abstract concepts.
So I’m skeptical about the value of reteaching basic skills. I have students who struggle with simple fraction operations. But they’ve learned fraction addition several times before my class. What do I know that will allow me to teach them better, so they suddenly learn it and remember it — despite the fact that I don’t have as much time, they won’t get as much practice, and won’t see it on a regular basis as they did the first few times around?
So this is an attempt to create something better. I’m not sure I have a great definition for number sense, but one big part of it is a variety of strategies to solve simple problems based on number relationships. These relationships are the building blocks of Algebra — students don’t need to add mixed numbers with unlike denominators, they need to understand the relationship between multiplication, division, and the fraction bar. Students don’t need to divide 1963 by 21, they need to see the power of the commutative property of addition. These are half-formed ideas, but they start to get deeper than “I know it when I see it”.
So here’s a crazy idea. I broke number sense down into three categories — whole numbers, rational numbers, and ratios & proportions. Then I built an assessment. Each item was a simple calculation, comparison, or estimation. Each could easily be solved using familiar algorithms. But, each could also be solved, in some cases much more easily, without an algorithm. I’m curious: first, how many questions students will get right (I teach 8th grade, and everything here falls in 6th grade or earlier, I believe). Second, how many questions students will solve using strategies that show an understanding of number relationships, not just algorithms. This is imperfect (for instance, I realized the last question is vague). But I gave it to two of my best students today, and I was fascinated by some of their work. Here are two students’ answers.
Student 1:
Student 2:
What does this work show about these students’ number sense? How can it be useful? I’m not sure. I love looking at this stuff, but these two students both got everything or almost everything right. I’m not sure this is so different from the basic skills I want to push away from — but it worries me a bit that these students regrouped to subtract 1001 – 993. I was worried that one used long division to find a decimal approximation for 3/7 to confirm that it is smaller than 1/2. Should I be concerned? Is it possible those are “false negatives” — is the first student showing work because she thinks she should, or because she needs to?
So many questions. I’m going to give this to a few more students and adults over the next few days, and see what I can learn.
[If you’re interested in a soft copy of the current version of the assessment, it’s here]
## 2 thoughts on “How Do You Assess Number Sense?”
1. howardat58
I am not sure that “number sense” is the right term for the ability to get there directly without using a standard method. Where is the boundary between “more obvious” tricks and “less obvious” tricks (example : is 3 a factor of 741 ? Yes, since 7+4+1 is a multiple of 3).
I would say that number sense is more about being able to explain why you “know” that your answer is correct, or near enough, and being able to get approximate answers, that is, having a feel for number.
2. Kathryn
This is an interesting topic. I am seeing a much greater struggle with integer and rational number operations among my students this year than in the past (and I said the same thing last year, so the trend is not working in my favor right now). I think of “number sense” as the ability to arrive at a reasonable solution, usually through estimation. I think using a non-standard algorithm may or may not be a sign of number sense. Often, I push through the rote algorithm because I know it will work, I can do it well, and I don’t have to think about it deeply. I expect your higher achieving students would cite those three reasons plus the ever-popular, “That’s how my teacher said to do it.” What I really wonder is if you could take a student who has poor computational skills (holes in multiplication or addition facts, for example) and see more creative solutions. | 1,061 | 4,900 | {"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} | 4.28125 | 4 | CC-MAIN-2018-43 | latest | en | 0.95484 |
https://www.transum.org/Software/sw/Starter_of_the_day/Starter_February22.asp | 1,585,577,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497042.33/warc/CC-MAIN-20200330120036-20200330150036-00200.warc.gz | 1,225,949,824 | 9,627 | What are the next terms of the following sequences?
1, 4, 9, 16, 25, ...
10, 19, 28, 37, ...
3, 5, 9, 17, 33, ...
O, T, T, F, F, S, S, E, ...
J, F, M, A, M, J, J, A, ...
R, O, Y, G, B, I, ...
## A Mathematics Lesson Starter Of The Day
Hints: One, Two, Three, January, Red
Share
Topics: Starter | Sequences
• S2, Dalkeith HS
•
• My class did very well at this starter and finished it quicker than expected!
• Marrys Sampson, Mount Cook School
•
• This Was On Die hard 4.
• H. Gamble Year 6, Berrywood
•
• We thoroughly enjoyed these sequences! They really got us thinking.Some of the class got them fairly quickly and without the hints.
• Mr C. Quence, Qwertyville
•
• Here are two others for any pupils that finish this starter early:
A,S,D,F.G,H ...
Z,X,C,V,B,N ...
• Topaz Class, Meeching Valley, Newhaven
•
• We did quite well with the number sequences, but we struggled with the letter sequences.
• Transum,
•
• Here is a very ancient sequence. What comes next?
I, V, X, L, C, D ...
•
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## Texas Instruments Nspire Calculator
This handheld device and companion software are designed to generate opportunities for classroom exploration and to promote greater understanding of core concepts in the mathematics and science classroom. TI-Nspire technology has been developed through sound classroom research which shows that "linked multiple representation are crucial in development of conceptual understanding and it is feasible only through use of a technology such as TI-Nspire, which provides simultaneous, dynamically linked representations of graphs, equations, data, and verbal explanations, such that a change in one representation is immediately reflected in the others.
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Teacher, do your students have access to computers?Do they have iPads or Laptops in Lessons? Whether your students each have a TabletPC, a Surface or a Mac, this activity lends itself to eLearning (Engaged Learning).
Transum.org/go/?Start=February22
Here is the URL which will take them to a related student activity.
Transum.org/go/?to=sequence
Learn how to quickly generate sequences using a spreadsheet.
Which sequences can and which sequences can't a spreadsheet generate?
For Students:
For All: | 947 | 4,159 | {"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} | 2.953125 | 3 | CC-MAIN-2020-16 | longest | en | 0.949439 |
https://metal-coatings.com/electromagnetism/question-is-light-considered-as-electromagnetic-wave.html | 1,652,699,271,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00254.warc.gz | 470,435,046 | 18,874 | # Question: Is light considered as electromagnetic wave?
Contents
Light as a wave: Light can be described (modeled) as an electromagnetic wave. … This changing magnetic field then creates a changing electric field and BOOM – you have light. Unlike many other waves (sound, water waves, waves in a football stadium), light does not need a medium to “wave” in.
## Is an electromagnetic wave the same as a light wave?
The terms light, electromagnetic waves, and radiation all refer to the same physical phenomenon: electromagnetic energy. This energy can be described by frequency, wavelength, or energy.
## Why is light called electromagnetic wave?
Because electromagnetic waves have fluctuating electric and magnetic fields, they are called electromagnetic waves. The speed of light is the same for all EM energy waves. They always move at the same pace, regardless of their frequency or wavelength.
## What is light made of?
Light is made of particles called photons, bundles of the electromagnetic field that carry a specific amount of energy.
## Is light longitudinal wave?
Explanation: Sound is a longitudinal wave, while light is a transverse wave. Polarization requires the direction of the wave to be perpendicular to the direction of propogation; only light can do this. Doppler effect, refraction, and interference occur in both wave types.
THIS IS INTERESTING: You asked: Where is electromagnetic force found?
## What is the definition of a light wave?
or lightwave (ˈlaɪtˌweɪv) noun. the movement of light conceptualized as a wave, defined by such properties as reflection, refraction, and dispersion.
In fact visible ‘light’ is a form of radiation, which can be defined as an energy that travels in the form of electromagnetic waves. It can also be described as a flow of particle-like ‘wave-packets’, called photons, that travel constantly at the speed of light (about 300 000 kilometres per second).
## Is light a EM wave Class 10?
Maxwell predicted that magnetic and electric fields travel in the form of waves and these waves move at the speed of light. This led Maxwell to predict that light itself was carried by electromagnetic waves which means that light is a form of electromagnetic radiation.
## Is light an electron?
A: Light is definitely not a molecule. It has no rest mass, no protons, no neutrons, no electrons. When some light is absorbed by something else (a molecule, for example) the light’s energy, momentum, and angular momentum are transferred to that object.
## Can light be destroyed?
Light is an energy and energy neither be created nor be destroyed. It only change its form. It will converted from one energy form to another but never will be destroyed.
## Is light a electricity?
And as a particle it is made up of small energy packets known as photons. But light is not electricity. Electricity is a form of energy and light is also just another form of it. In another way, light as a wave is made of magnetic and electric waves perpendicular to each other.
THIS IS INTERESTING: What is meant by electromagnetic effect?
## Which type of wave is light wave?
Transverse waves: The wave in which the movement of the particles is at right angles to the motion of the energy is called a transverse wave. Light is an example of a transverse wave.
## What waves are in a light wave?
Electromagnetic waves, including visible light, are made up of oscillating electric and magnetic fields as shown. The wavelength of a wave is the distance between successive peaks or troughs of a wave.
Light as a Wave.
Wavelength λ Type of Electromagnetic Radiation
smaller than .01 nm Gamma rays
## Is light transverse wave?
Light is another example of a transverse wave, where the oscillations are the electric and magnetic fields, which point at right angles to the ideal light rays that describe the direction of propagation. | 805 | 3,878 | {"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-2022-21 | latest | en | 0.937477 |
https://webexpertschicago.com/qa/what-is-the-leverage-ratio-formula.html | 1,618,411,144,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00143.warc.gz | 703,736,337 | 8,883 | # What Is The Leverage Ratio Formula?
## Is debt a equity?
In a basic sense, Total Debt / Equity is a measure of all of a company’s future obligations on the balance sheet relative to equity.
A similar ratio is debt-to-capital (D/C), where capital is the sum of debt and equity: D/C = total liabilities / total capital = debt / (debt + equity).
## What is a 1 100 Leverage?
100:1: One-hundred-to-one leverage means that for every \$1 you have in your account, you can place a trade worth up to \$100. This ratio is a typical amount of leverage offered on a standard lot account. The typical \$2,000 minimum deposit for a standard account would give you the ability to control \$200,000.
## What is the best leverage level for a beginner?
As a new trader, you should consider limiting your leverage to a maximum of 10:1. Or to be really safe, 1:1. Trading with too high a leverage ratio is one of the most common errors made by new forex traders. Until you become more experienced, we strongly recommend that you trade with a lower ratio.
## How do you calculate leverage ratio?
It’s calculated using the following formula:Operating Leverage Ratio = % change in EBIT (earnings before interest and taxes) / % change in sales.Net Leverage Ratio = (Net Debt – Cash Holdings) / EBITDA.Debt to Equity Ratio = Liabilities / Stockholders’ Equity.
## What is a good leverage ratio?
A figure of 0.5 or less is ideal. In other words, no more than half of the company’s assets should be financed by debt. In reality, many investors tolerate significantly higher ratios. … In other words, a debt ratio of 0.5 will necessarily mean a debt-to-equity ratio of 1.
## What is a 1/10 leverage?
The term “leverage” refers to the ability to trade or trade with a large amount of money without using your own money (or using a small amount of it). … For example, if a trader wants to use a leverage of 1:10, it means that every dollar that is exposed to risk actually manages \$10 in the market.
## What are the types of leverage?
There are three types of leverages, such as- (1) Operating leverage, and (2) Financial leverage. (3) Combined Leverage.
## What is leverage and its types?
In finance, leverage is a strategy that companies use to increase assets, cash flows, and returns, though it can also magnify losses. There are two main types of leverage: financial and operating. To increase financial leverage, a firm may borrow capital through issuing fixed-income securities.
## Is leverage good or bad?
Leverage is neither inherently good nor bad. Leverage amplifies the good or bad effects of the income generation and productivity of the assets in which we invest. … Analyze the potential changes in the costs of leverage of your investments, in particular an eventual increase in interest rates.
## What is ideal debt/equity ratio?
The optimal debt-to-equity ratio will tend to vary widely by industry, but the general consensus is that it should not be above a level of 2.0. While some very large companies in fixed asset-heavy industries (such as mining or manufacturing) may have ratios higher than 2, these are the exception rather than the rule.
## What if debt to equity ratio is less than 1?
As the debt to equity ratio continues to drop below 1, so if we do a number line here and this is one, if it’s on this side, if the debt to equity ratio is lower than 1, then that means its assets are more funded by equity. If it’s greater than one, its assets are more funded by debt.
## How do you calculate net leverage ratio?
NET DEBT LEVERAGE RATIO (NON-GAAP FINANCIAL MEASURE) The Company has defined its net debt leverage ratio as net debt (total principal debt outstanding less unrestricted cash) divided by adjusted EBITDA for the trailing twelve month period.
## What is leverage in simple words?
Leverage is an investment strategy of using borrowed money—specifically, the use of various financial instruments or borrowed capital—to increase the potential return of an investment. Leverage can also refer to the amount of debt a firm uses to finance assets.
## What does 5x leverage mean?
Selecting 5x leverage does not mean that your position size is automatically 5x bigger. It just means that you can specify a position size up to 5x your collateral balances.
## What is leverage example?
An example of leverage is to financially back up a new company. An example of leverage is to buy fixed assets, or take money from another company or individual in the form of a loan that can be used to help generate profits.
## What is a bank’s leverage ratio?
The leverage ratio is a measure of the bank’s core capital to its total assets. The ratio uses tier 1 capital to judge how leveraged a bank is in relation to its consolidated assets whereas the tier 1 capital adequacy ratio measures the bank’s core capital against its risk-weighted assets.
## What is capital structure leverage?
Capital Structure or Leverage Ratio Capital structure refers to the degree of long term financing of a business concern as in the form of debentures, preference share capital and equity share capital including reserves and surplus. … Capital structure is otherwise called as leverage.
## Why is debt called leverage?
Borrowing funds in order to expand or invest is referred to as “leverage” because the goal is to use the loan to generate more value than would otherwise be possible. | 1,185 | 5,401 | {"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-2021-17 | latest | en | 0.933592 |
https://www.scienceforums.net/profile/143014-alexandrkushnirtshuk/content/ | 1,726,440,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00734.warc.gz | 932,439,859 | 19,727 | # AlexandrKushnirtshuk
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1. ## Moon size. Calculation and confirmation.
Earth and Moon to scale. The shadow of the moon on the surface of the Earth during a solar eclipse (view from space). The size of the shadow is more consistent with an object with a diameter of 3,500 km., or 850 km.? Looking Back at Earth - Total Solar Eclipse from the Perspective of Space - NASA photo
2. ## Moon size. Calculation and confirmation.
You didn’t provide evidence of this, and need to, and also show they are maximum under the same conditions. 1) Total duration of solar eclipse. 2) Total duration of lunar eclipse. (an hour and three-quarters in english version, 108 minutes in russian). The diameter of the lunar orbit and the speed of the Moon are constant values (same conditions). + - small eccentricity of the lunar orbit. Since the Moon's orbit moves with the speed of the Earth, then in this case (in this frame of reference) the Earth can be considered stationary. The calculation is based on the principle of scientific proportion, and on three reliable parameters. The calculation pretty closely matches the actual track size between South America and Antarctica. Point.
3. ## Moon size. Calculation and confirmation.
7,5 and 108 minutes are the maximum values of solar and lunar eclipses, and therefore may well be taken as constants in calculations. With the same distance between the Moon and the Earth. At the same speed of the Moon (the orbit of the Moon moves with the speed of the Earth). From the quote you quoted above your question, it is quite understandable why I assume that the Moon left a trail between South America and Antarctica. The calculated and actual sizes coincide quite accurately.
4. ## Moon size. Calculation and confirmation.
The duration of an eclipse is directly proportional to the size of the object, all other things being equal (distance and speed). The duration of the total phase of a solar eclipse is 7.5 minutes (the Moon completely covers the Sun for 7.5 minutes). The duration of the total phase of the lunar eclipse is 108 minutes (the Earth completely covers the Sun for 108 minutes). With the same distance between the Moon and the Earth. At the same speed of the Moon (the orbit of the Moon moves with the speed of the Earth). The diameter of the Earth is 12,742 km. Therefore, the diameter of the Moon can be calculated using the following formula: 12 742 * (7.5 / 108) = 885 km. The official diameter of the Moon is 3,474 km. Moreover, the result of calculating the diameter of the Moon quite accurately coincides with the size of the track between South America and Antarctica (875 km. + - 25 km.), which confirms the calculation and minimizes probability of a simple coincidence.
5. ## Significant evidence for a New Model of the Universe.
The duration of the total phase of a solar eclipse is 7.5 minutes. The duration of the total phase of the lunar eclipse is 108 minutes. The diameter of the Earth is 12,742 km. Therefore, the diameter of the Moon is 12 742 * (7.5 / 108) = 885 km. Additional evidence. The Unsolved Mystery of the Earth Blobs The coincidence of the angular sizes of the Sun and the Moon indicates that their sizes are proportional to the distances relative to the Earth. In addition, the Sun and the Moon have the same axial rotation periods - 27 days. In the earth's mantle there are two huge diametrically opposite formations (one is larger, the other is smaller), both are displaced to the east. On the surface of the Earth there are two huge diametrically opposite tracks (one larger, the other smaller), both shifted to the east. The ratio of the sizes of the Sun and the Moon is approximately 3 to 1.
6. ## Dark matter mystery.
What is dark matter? An incomprehensible substance evenly scattered throughout the Universe, or is it the border of the Universe behind the Oort cloud, from where the sunlight is simply not reflected? Astronomers Use New Data to Create Extraordinary Dark Matter Map The distance to the most distant galaxy is supposedly 13.4 billion light years. This means that the light travels all the distance without hindrance. This is supposedly a straight line, along which there are no objects: stars, galaxies, nebulae, dust, gas - nothing blocking light in a straight line 13.4 billion light years long ... This is hardly possible.
7. ## Paradox of visual and actual positions in space.
Brief description to avoid unnecessary complication. The distance from the Earth to the Sun is about 8 light minutes, so from the Earth we see the Sun at the point in the sky where it was 8 minutes ago (in 8 minutes the Sun passes through the sky with an angular distance of slightly less than two solar disks) ... It is difficult to both explain and imagine, because most likely it is impossible, that is, cosmic distances are too exaggerated. The distance from the Earth to the Moon is about 1 light second. That is, the apparent and actual position of the moon is almost the same. The shortest distance from Earth to Jupiter is about 32 light minutes. The apparent and actual positions of Jupiter differ 4 times more than in the case of the Sun. The question and the most important thing. Why is astronomy not taking into account the actual and visible position of space objects corrected for the speed of light? The motions of the planets are calculated using Kepler's formulas. The calculated positions of the planets (that is, the actual ones) coincide with the visual ones without corrections for the speed of light. I do not question the speed of light, it has been measured and refined for several centuries. The official space distances and the sizes of space objects, respectively, are in great doubt.
8. ## Granite origin mystery.
Even officially the solid cores of gas giant planets are smaller than the Earth. Gas shells of that planets is like an atmosphere on Earth. Do we measure earth's size including its atmosphere? I suppose that so called "gas giant planets" are just huge comets, the absence of a tail in which can be explained by the great distance from the Sun and the relatively low speed of movement in their orbits.
9. ## Granite origin mystery.
Granite is found only on Earth. It was not found either in meteorites or on other "planets" of the solar system. Officially it is unknown why. I suppose, it is because the Earth is the largest object in the Universe, with the greatest gravity and pressure in the subsoil. Links to quote in russian (did not find the same in english): 1) https://beversmarmyr.com.ua/articles/istoriya-formirovaniya-granita 2) https://ru.wikipedia.org/wiki/Гранит#Проблема_происхождения_гранитов
10. ## Proof – Mars Orbited close to Earth 1350 BC
Mars Reconnaissance Orbiter contradicts planetary science Jupiter and Venus Papers
11. ## Proof – Mars Orbited close to Earth 1350 BC
Hidden Pyramids? - Mars Mountains Match Pyramids on Earth The Pyramids of Giza, the Belt of Orion and Three Volcanoes on Mars
12. ## Proof – Mars Orbited close to Earth 1350 BC
They write here that around 1350 BC, Mars was in geostationary orbit, and they give good evidence. Proof – Mars Orbited close to Earth 1350 BC (Updated) But if about 3350 years ago Mars was in a geostationary orbit, then its diameter cannot be 6.7 thousand km., but just about 15-20 km., as I suppose in my New Model of the Universe.
13. ## Curiosity/Perseverance landing scale comparison contradiction.
Calculations not accurate enough. Delete this topic. I don't know how. Seems I can't remove own topics.
14. ## Curiosity/Perseverance landing scale comparison contradiction.
Most often I am asked about calculations (factual evidence). I made calculations based on factual evidence, which points to scale contradiction in videos of landings Curiosity and Perseverance. That is my point: 1) New model of the Universe. 2) The nature of light and the size of the Universe.
15. ## Curiosity/Perseverance landing scale comparison contradiction.
Source screens. Source links. 1) Perseverance Rover’s Descent and Touchdown on Mars (Official NASA Video) 2) Curiosity Drops in on Mars in High-Res 3) Mars Curiosity Descent - Ultra HD 30fps Smooth-Motion Link to this topic (thread) on NASA forum: Curiosity/Perseverance landing scale comparison contradiction.
16. ## Lighting questions on the Perseverance landing video.
Opposition Effect (Seeliger effect) | Aerial video examples
17. ## Lighting questions on the Perseverance landing video.
This animation shows the size (scale, diameter) of Sun's reflection on the surface of Mars regardless of the type of surface (water or soil). The reflection of the Sun on the ground in this computer animation was simply not displayed due to its dimness. Moreover, the reflection of the Sun on the ground of Mars cannot be focused into such a small bright spot as in the video of the Perseverance landing. Round bright spot on the video, and nothing bright and round on 2 photos of that area of Mars. Round bright spots on two different landing videos are not moving on the surface, that is, they are located on the surface of Mars. They enter and leave field of view without changes of shape and size. Here is a link to video of Curiosity landing with better quality. Mars Curiosity Descent - Ultra HD 30fps Smooth-Motion
18. ## Lighting questions on the Perseverance landing video.
In that thread of mine on the NASA forum, there is not a hint of conspiracy. I even removed the quotes from the words Sun and Mars from the first paragraphs of the text. And just in case, I made a screenshot. Link to my NASA forum thread: https://forum.nasaspaceflight.com/index.php?topic=53173.0 Screenshot of that thread:
19. ## Lighting questions on the Perseverance landing video.
Created a thread with this question on NASA forum.
20. ## Lighting questions on the Perseverance landing video.
With a diameter of Mars 6,700 km. and the diameter of the Jezero crater 50 km. The Sun should illuminate the entire crater completely and evenly. Factual analysis only. No conspiracy at all in this thread.
21. ## Lighting questions on the Perseverance landing video.
Artist’s impression of Mars four billion years ago Size of Jezero Crater on Mars, and size of Sun reflection inside that crater, which makes up less than 10% of Jezero Crater area. Same bright spot size on Curiosity landing video. Complete Mars Curiosity Descent - Full Quality Enhanced HD 1080p Landing + Heat Shield impact
22. ## Lighting questions on the Perseverance landing video.
Sun reflection on Earth from 400 km. and Mars from 10 km. I have nothing to add to this thread. ISS animation source: Over Earth - Incredible Space Views from ESA Astronaut Alexander Gerst
23. ## Lighting questions on the Perseverance landing video.
1) I do not know how to ask someone from NASA. 2) I am asking this question on this scientific forum in Astronomy and Cosmology section. 3) At the press conference on the video of the Perseverance landing, specialists analyzed almost every frame of these videos. Explained everything except this reflection (bright spot) on the surface of Mars. Personally, I think this is at least strange. Here is a timecode link for that episode of the press conference: See Mars Like Never Before! NASA's Perseverance Rover Sends New Video and Images of the Red Planet Bright spot on the Mars' sufrace - is the only episode, which NASA specialists skipped and left without any explanations.
24. ## Lighting questions on the Perseverance landing video.
- brightness + contrast. Same thing. This is the reflection of the Sun on the surface of Mars, and it is too small for the indicated footage scale (altitude of 9.5 km). It doesn't depend on the image resolution. Obviously the shadow on the animation below cannot be from parachute, but from the backshell only.
25. ## Lighting questions on the Perseverance landing video.
Bright spot is not moving, it is stationary. So it can not be some kind of lens flare, hence it cannot be anywhere else, except the surface of Mars. Can't explain better than in this image.
× | 2,694 | 12,059 | {"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-2024-38 | latest | en | 0.891908 |
http://stackoverflow.com/questions/8423116/regarding-geom-segmentaesx-x0-y-y0-xend-x1-yend-y1-in-ggplot2 | 1,440,704,629,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644059455.0/warc/CC-MAIN-20150827025419-00126-ip-10-171-96-226.ec2.internal.warc.gz | 190,436,942 | 17,775 | # regarding geom_segment(aes(x=x0,y=y0,xend=x1,yend=y1)) in ggplot2
In ggplot2, there is a usage which looks like:
``````geom_segment(aes(x=x0,y=y0,xend=x1,yend=y1))
``````
What does the `aes(x=x0,y=y0,xend=x1,yend=y1)` mean?
I checked the ggplot2 manual, but it does not explain these parameters in detail. Thanks.
-
I can see how x=x0 and y=y0 play? But I do not know how do xend=x1 and yend=y1 stand for? – user288609 Dec 7 '11 at 22:08
A line segment has two ends. Each end needs an x and y value. For a total of four numbers per segment. – joran Dec 7 '11 at 22:09
@user288609 Does Joran's comment answer your question? If not, you need to clarify what you are asking. – Brian Diggs Dec 7 '11 at 23:28
The `aes` function is used to map variables (i.e. columns) in a data.frame to visual properties of the plot. A plot consists of one or more geometries, e.g. `geom_point` for points or `geom_polygon` for polygons. Each of these geometries has different properties, aka aesthetics. A simple example is the point geometry (`geom_point`). This geometry has the following aesthetics (from the man page, see ?geom_point for that):
``````Aesthetic Default
x required
y required
shape 16
colour black
size 2
fill NA
alpha 1
``````
From this list we see that a point geometry has two required aesthetics: the x-coordinate of the point (`x`) and the y-coordinate of the point (`y`). Additional aesthetics have default values, but could also be coupled to a column in the dataset to make them variable. For example, linking `size` to a column in the data varies the size of the point according to that variable.
To get to your question. The segment geometry is used to draw line segments. The aesthetics required for that are a starting point for the line segment (`x` and `y`) and an ending point for the line (`xend` and `yend`). So the line:
``````aes(x=x0,y=y0,xend=x1,yend=y1)
``````
says that we want ggplot to draw line segments for each row in a data.frame, where the line is draw from the coordinates (`x`,`y`) to (`xend`,`yend`). Hope this makes things more clear.
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Sept 24 LEC Elasticity and its applications
# Sept 24 LEC Elasticity and its applications - Elasticity...
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Elasticity and its applications Definition: The price elasticity of demand is a measure of how much the quantity demanded of a good responds to a change in the price of that good. Price elasticity of demand is the percentage chance in quantity demanded given a percentage change in the price. It is computer as the percentage change in the quantity demanded divided by the percentage change in price. Main determinants: Availability of Close Substitutes Necessities versus Luxuries Definition of the Market Time Horizon Demand tends to be more elastic: o The larger the number of close substitutes o If the good is a luxury o The more narrowly defined the market o The longer the time period Demand tends to be less elastic: o The lower the number of close substitutes o If the good is a necessity o The less narrowly defined the market o The shorter the time period Computing the price elasticity of demand: Example: If the price of an ice cream cone increase from \$2.00 to \$2.20 and the amount you buy falls from 10 to 8 cones, then your elasticity of demand would be calculated as: Interpretation – if you increase the price by 1%, you decrease the demand by 2%. If you decrease the price by 1%, you increase the demand by 2% Absolute value – 99.9% of cases, when you increase the price of a good or service, the demand goes down
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The mid-point method: The midpoint formula is preferable when calculating the price elasticity of demand because it gives the same answer regardless of the direction of the change. Example: If the price of an ice cream cone increase from \$2.00 to \$2.20 and the amount you buy falls from 10 to 8 cones, the using the midpoint formula, we have: Computing the elasticity of demand
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I am looking for software (open-source or otherwise) or an implementable algorithm for solving a continuous transportation problem. The input consists of a pointset in a planar rectangle, and we need to relocate these points within the rectangle to decrease the peak density below a given threshold (feasibility can be assumed), while minimizing total displacement.
The distances inside the rectangle are Manhattan/taxicab, although efficient solutions for the Euclidean distance can also be helpful. Total displacement is interpreted in the L1 $L_1$ sense, but efficient solutions for the L2 $L_2$ case can also be helpful. The peak density can be evaluated with respect to a uniform grid (is there another practical way ?)
My students implemented a geometric algorithm (without having any background in transportation) that works great in our application, but we don't know how far the results are from optimal. Just in case, our application also imposes rectangular "exclusion zones", where no points can be placed (more generally, we can assume a "bounding probability distribution").
1
# Best algorithm/software for solving a planar transportation problem ?
I am looking for software (open-source or otherwise) or an implementable algorithm for solving a continuous transportation problem. The input consists of a pointset in a planar rectangle, and we need to relocate these points within the rectangle to decrease the peak density below a given threshold (feasibility can be assumed), while minimizing total displacement.
The distances inside the rectangle are Manhattan/taxicab, although efficient solutions for the Euclidean distance can also be helpful. Total displacement is interpreted in the L1 sense, but efficient solutions for the L2 case can also be helpful. The peak density can be evaluated with respect to a uniform grid (is there another practical way ?)
My students implemented a geometric algorithm (without having any background in transportation) that works great in our application, but we don't know how far the results are from optimal. Just in case, our application also imposes rectangular "exclusion zones", where no points can be placed (more generally, we can assume a "bounding probability distribution"). | 437 | 2,276 | {"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} | 2.9375 | 3 | CC-MAIN-2013-20 | latest | en | 0.885065 |
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## Equation for the multi-mode PDF
Taking the inverse Laplace transform of (64) we have the following equation for the PDF,
(59)
where is a flux of probability in the space of the amplitude ,
(60)
This equation is identical to the one originally obtained by Peierls [19] and later rediscovered by Brout and Prigogine [20] in the context of the physics of anharmonic crystals,
(61)
where
(62)
Zaslavski and Sagdeev [21] were the first to study this equation in the WT context. However, the analysis of [19,20,21] was restricted to the interaction Hamiltonians of the potential energy'' type, i.e. the ones that involve only the coordinates but not the momenta. This restriction leaves aside a great many important WT systems, e.g. the capillary, Rossby, internal and MHD waves. Our result above indicates that the Peierls equation is also valid in the most general case of 3-wave systems. Note that Peierls form (67) - (68) looks somewhat more elegant and symmetric than (65) - (66). However, form (65) - (66) has advantage because it is in a continuity equation form. Particularly for steady state solutions, one can immediately integrate it once and obtain,
(63)
where is an arbitrary functional of and is the antisymmetric tensor,
In the other words, probability flux can be an arbitrary solenoidal field in the functional space of . One can see that (69) is a first order equation with respect to the -derivative. Special cases of the steady solutions are the zero-flux and the constant-flux solutions which, as we will see later correspond to a Gaussian and intermittent wave turbulence respectively.
Here we should again emphasise importance of the taken order of limits, first and second. Physically this means that the frequency resonance is broad enough to cover great many modes. Some authors, e.g. [19,20,21], leave the sum notation in the PDF equation even after the limit taken giving . One has to be careful interpreting such formulae because formally the RHS is nill in most of the cases because there may be no exact resonances between the discrete modes (as it is the case, e.g. for the capillary waves). In real finite-size physical systems, this condition means that the wave amplitudes, although small, should not be too small so that the frequency broadening is sufficient to allow the resonant interactions. Our functional integral notation is meant to indicate that the limit has already been taken.
Next: Evolution of statistics of Up: Evolution of statistics of Previous: Equation for the generating
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# SBI Bank PO Reasoning Solved Exam Paper
Directions : – Select the related letters / word / number / figure from
the given alternatives
1
Editor : Magazine
(a) Movie (b) Scene
(c) Drama (d) Director
2 Hinduism , Christianity , Islam : Religion
(a) Ear , Nose , Eyesight : Vision (b) Plus , Minus , Multiple :
division
(c) Winter , Spring , Summer : Season (d) Humid , Hot , Tundra : Region
3
WOLF : FLOW :: WARD : ?
(a) BROW (b) DRAW
(c) CRAW (d) SLAW
4 GRAIN : TIZRM :: BRAIN : ?
(a) XRIKL (b) YIZRM(c) OPRST (d) ASQMI
5 CUT : BDTVSU :: TIP : ?
(a) UVHJOQ (b) SUHJOQ
(c) USJHQO (d) SUJHOQ
Q-6 19 : 59 :: 17 : ?
(a) 51 (b) 53
(c) 55 (d) 57
Q-7 14 : 20 :: 16 : ?
(a) 23 (b) 48
(c) 10 (d) 42
Q-8 100 : 102 :: 100000 : ?
(a) 105 (b) 104
(c) 1003 (d) 1004
DIRECTIONS : (QUESTIONS 9 to 13) select the one which is
different from the other three .
Q-9
(a) Mile (b) Centimeter
(c) Litre (d) Yard
Q-10
(a) High – Up (b) Past – Present
(c) Often – Seldom (d) Fresh – Stale
Q-11
(a) 11 – 127 (b) 9 – 85
(c) 7 – 53 (d) 5 – 29
Q-12
(a) 26 Z (b) 24 X
(c) 22 V (d) 20 S
Q-13
(a) 8 , 64 , 112 (b) 36 , 6 , 206
(c) 48 , 4 , 202 (d) 9 , 27 , 263
Q-14 In the following series of numerals , which digit has maximum
frequency ?
846734378344563464348
(a) 8 (b) 6
(c) 4 (d) 3
Q-15 If the day after tomorrow is Friday , what day will the third
day after tomorrow be
(a) Saturday (b) Monday
(c) Sunday (d) Friday
Q-16 If the ratio of the area of two squares is 16 : 1 , then the
ratio of their perimeter is
(a) 4 : 1 (b) 16 : 1
(c) 1 : 3 (d) 3 : 4
Q-17 The shade of 18 ft high pole is 20 ft. . Find the length of
shade of 27 ft long pole .
(a) 36 ft (b) 30 ft
(c) 34 ft (d) 40 ft
Q-18 A scores more runs than B but less than C . D scores more than B
but less than A . Who is the lowest scorer ?
(a) A (b) B
(c) C (d) D
Q-19 In the alphabets from A to Z , which is the third letter to the
right of the letter which is midway between K & S ?
(a) R (b) Q
(c) P (d) O
Q-20 If first November falls on Monday , then what day will the 25th
November be ?
(a) Tuesday (b) Thursday
(c) Wednesday (d) Friday
Q-21 The length of room is twice its breadth . If the area of the
room is 242 sq meters , then find out its breadth
(a) 11 (b) 10
(c) 12 (d) 9
Q-22 If the product of two numbers is 10 & their sum is 7 , then the
larger of the two number is –
(a) – 2 (b) 2
(c) 5 (d) 3
Q-23 Which letter is used only in one of the given words ?
Speak , Reap , Shark
(a) S (b) P
(c) K (d) H
Q-24 A tortoise covers one kilometer in 4 hours . It takes rest for
20 minutes after every kilometer .How much time does it takes for the tortoise
to cover 3.5 kilometers ?
(a) 14 hours (b) 13
(c) 15 (d) 12
Q-25 Of which of the following words , which one will be at the 3rd
position in the dictionary ?
1. socks 2. Shocks
3. Sharp 4. snooker
(a) 4 (b) 3
(c) 2 (d) 1
1 2 3 4 5
c c b b b
6 7 8 9 10
b d a c a
11 12 13 14 15
a d a c c
16 17 18 19 20
a b b a b
21 22 23 24 25
a c d c a
### One comment
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## Bank PO Solved Exam Paper-Quantitative Analysis
Directions—(Q. 1–6) What should come in place of the question mark (?) in the following ... | 1,281 | 3,377 | {"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} | 2.796875 | 3 | CC-MAIN-2019-35 | latest | en | 0.456288 |
https://stat.ethz.ch/pipermail/r-help/2009-October/406687.html | 1,582,200,764,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144722.77/warc/CC-MAIN-20200220100914-20200220130914-00081.warc.gz | 589,371,663 | 2,568 | # [R] split-apply question
David Winsemius dwinsemius at comcast.net
Fri Oct 2 14:34:56 CEST 2009
```As is typical with R there are often other ways. Here is another
approach that determines the rows of interest with tapply and min,
converts those minimums into logical "targets" with %in%, and extracts
them from "x" using indexing:
x[x\$x2 %in% tapply(x\$x2, x\$x1, min), ]
########
x1 x2 x3
1 A 1 1.5
2 B 2 0.9
4 C 7 1.8
5 D 7 1.3
You might want to determine whether both would return all rows if
there were multiple instances of a minimum. I think the above solution
would return multiples while the one below would not. You choose based
on the nature of the problem.
--
David
On Oct 2, 2009, at 5:24 AM, jim holtman wrote:
> try this:
>
>> x <- read.table(textConnection("x1 x2 x3
> + A 1 1.5
> + B 2 0.9
> + B 3 2.7
> + C 7 1.8
> + D 7 1.3"), header=TRUE)
>> closeAllConnections()
>> do.call(rbind, lapply(split(seq(nrow(x)), x\$x1), function(.row){
> + x[.row[which.min(x\$x2[.row])],]
> + }))
> x1 x2 x3
> A A 1 1.5
> B B 2 0.9
> C C 7 1.8
> D D 7 1.3
>>
>
>
> On Thu, Oct 1, 2009 at 11:43 PM, Kavitha Venkatesan
> <kavitha.venkatesan at gmail.com> wrote:
>> Hi,
>>
>> I have a data frame that looks like this:
>>
>>> x
>>
>> x1 x2 x3
>> A 1 1.5
>> B 2 0.9
>> B 3 2.7
>> C 7 1.8
>> D 7 1.3
>>
>> I want to "group" by the x1 column and in the case of multiple x\$x1
>> values
>> (e.g., "B")d, return rows that have the smallest values of x2. In
>> the case
>> of rows with only one value of x1 (e.g., "A"), return the row as
>> is. How can
>> I do that? For example, in the above case, the output I want would
>> be:
>>
>> x1 x2 x3
>> A 1 1.5
>> B 2 0.9
>> C 7 1.8
>> D 7 1.3
>>
>>
>> Thanks!
>>
>> [[alternative HTML version deleted]]
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help | 856 | 2,322 | {"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-2020-10 | latest | en | 0.801648 |
https://electronics.stackexchange.com/questions/648845/how-does-an-integrating-receiver-work | 1,721,076,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.74/warc/CC-MAIN-20240715194155-20240715224155-00509.warc.gz | 203,175,846 | 39,434 | # How does an integrating receiver work?
I am reading some papers and in one paper, the working of an integrating receiver is mentioned to remove (periodic) interference from the received signal. The paper explains the working as such:
The integrating receiver utilizes the constant amplitude of the data signal and the periodic time-varying amplitude of the sinusoidal interference signal to separate them by integrating them over a fixed period of time. The constant amplitude data keeps on adding up as it is integrated over time. However, the periodic sinusoidal interference still remains a sinusoid even after integration. The periodic zero crossing of these sinusoids enables the possibility of sampling at appropriate instants to have zero integrated interference.
The I-DDR receiver utilizes the integration, followed by periodic sampling, on periodic interference affected constant amplitude nonreturn to zero (NRZ) data to achieve interferencerobust operation through signal interference separation in the time domain.
S. Maity, B. Chatterjee, G. Chang and S. Sen, "BodyWire: A 6.3-pJ/b 30-Mb/s −30-dB SIR-Tolerant Broadband Interference-Robust Human Body Communication Transceiver Using Time Domain Interference Rejection," in IEEE Journal of Solid-State Circuits, vol. 54, no. 10, pp. 2892-2906, Oct. 2019, doi: 10.1109/JSSC.2019.2932852.
Now I have the following questions:
1. You integrate the data signal, now you have the data signal added over a period of time. So? What can I do with that?
2. If we integrate the interference signal at its zero crossings, then we would have a integration value of ~zero over a period of time. So we have "nothing." How can we use that to separate the interference signal from the data signal?
I feel like I am missing the point of this integrating receiver a bit but maybe someone could explain it to me.
• Did you consider that linking the document you refer to was unimportant when asking this question? Commented Jan 5, 2023 at 10:39
• An integrator is just another way of saying 'low pass filter'. The text you've shown basically says they have added a low pass filter to the received data signal to remove a higher frequency interference signal. That's sig-pro 101 stuff. I don't know why they have chosen such a convoluted way of saying this, but I don't have access to the full paper to try to figure out the context better.
– Jon
Commented Jan 5, 2023 at 10:52
• You integrate the data signal, now you have the data signal added over a period of time. <-- not if you differentiate the data signal at the transmitter. It's called pre-emphasis (transmitter) and de-emphasis (receiver). Commonly used in FM broadcast. I'm not saying that it's done for the same reasons as the inaccessible (to me) paper but, it is a method used to improve SNR. Commented Jan 5, 2023 at 11:02
• One thing to note is that the interference is periodic, ie it's not noise, it's predictable. What they are doing is 'receiving' the signal and 'receiving' the interference, the better to separate them. This technique would not work if the interference was unpredictable noise. Commented Jan 5, 2023 at 12:13
• Maybe I missed it, but I looked at the entire paper you linked, and I didn't see the text you quoted anywhere in it. Commented Jan 5, 2023 at 17:01 | 771 | 3,300 | {"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-2024-30 | latest | en | 0.918293 |
https://www.physicsforums.com/threads/gausss-law-i-think.120586/ | 1,481,102,230,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542009.32/warc/CC-MAIN-20161202170902-00027-ip-10-31-129-80.ec2.internal.warc.gz | 993,710,739 | 19,915 | # Gauss's Law (I think!)
1. May 12, 2006
### Brewer
At present its more the wording of the question I don't understand, so could you see if you think it means what I think it does.
Charges +q and -q are placed on two conducting spherical shells of radii a and b (a<b). Find the electric field between the shells, and show that it has a maximum value given by,
E(max) = Vb/a(b-a)
Where V is the potential difference between the spheres.
I think its set up like this:
Does this mean I can find the field like I would with oppositely charged parallel plates?
2. May 12, 2006
### Tide
Not exactly - the geometry is quite different. What you need to do is set up a Gaussian sphere concentric to the charged spheres with radius intermediate to them. You know how much charge is enclosed by that surface and electric field follows from Gauss' Law and symmetry. You can use it to calculate the potential.
3. May 12, 2006
### Brewer
So I set up a surface inbetween them?
So the charge enclosed will be just +q?
The radius of this new surface I don't know do I? But I don't need to know it do I?
4. May 12, 2006
### arunbg
The radius of the sphere can be taken as 'r' to apply Gauss' law.Now you can find the max value for E in terms of a,b and V (E will be maximum on the surface of the smaller sphere).
Arun
5. May 12, 2006
### nrqed
that's right
that's correct
you need to specify it. Give it a name, say "r", with the understanding that $r_a < r < r_b$. That's the radius of your gaussian surface so when you calculate the electric flux through that surface, it will contain "r". When you will solve for the magnitude of the E field, it *will* contain r. Now, you r is arbitrary, within the restriction $r_a < r < r_b$. Si you will have found the E field at *any* point between the two shells.
6. May 12, 2006
### Brewer
I've got the answer out to be:
$$E = \frac{q}{\epsilon_{0}4\pi r}$$
I know the final bit should be simple, putting the Gaussian surface on the smaller sphere (so that r=a), so why doesn't that just give me E=V/a?
7. May 12, 2006
### nrqed
This is not quite right. When you calculate the flux through your gaussian surface, you ended up multiplying the magnitude of the E field by the area of your gaussian surface, right? But that is $4 \pi r^2$! So you should have a r^2 there, not r.
8. May 12, 2006
I did get $$4 \pi r^2[\tex], I just copied it down wrong. The r came from working out E=V/r (because V=kq/r right?) Last edited: May 13, 2006 9. May 12, 2006 ### nrqed I think I misunderstood your post the first time around so I deleted my initial response. If you do set r=a in your expression, then of course you get ${q \over 4 \pi \esilon_0 a^2}$ and this *is* potential at r=a divided by a so it does work. Do you agree? 10. May 13, 2006 ### Brewer It was posted above that the maximum value for the electric field will be when the electric field is on the smaller sphere (i.e. when r=a), but the question says to show that the maximum value is : [tex]E_{max} = \frac{Vb}{a(b-a)}$$
which I don't see how it comes about. Surely if r=a when its maximum you end up with E=V/a (where V is the potential difference)?
11. May 13, 2006
### arunbg
Brewer, potential difference is different from potential.
Try to find the pd between the two spheres.
You can start by treating the shells as point charges as you might have learnt from Gauss' law, find the potentials at the outer and inner shells and subtract to get the pd=V in the question.
Believe me the electric field is greatest at the inner sphere.
12. May 13, 2006
### nrqed
Arunbg said it correctly.
Here V is the potential *difference* between the shells. (a better symbol would b $\Delta V$). Using $k = { 1 \over 4 \pi \epsilon_0}$, you have that the potential on the inner shell is ${k q \over a}$ and on the outer shell is is ${k q \over b}$. And V is the difference between the two. The maximum E field is ${k q \over a^2}$. If you reexpress the max E field in terms of V, you will get the expression they give.
13. May 13, 2006
### Brewer
Ok guys I'm with you now (I think!). So as I read it I find the potential at b and the potential at a, and subtract them to find the pd which is v?
14. May 13, 2006
### Brewer
Thanks for your help by the way
15. May 13, 2006
### nrqed
Yes. and then express the maximum electric field in terms of this difference of potential. You will get the expression you quoted in your first post.
(the reason they express it in terms of a *difference* of potential rather than, say, the potential on one of the spheres is that the potential difference is easy to measure. Just connect the two terminals of a voltmenetr to the two spheres and the value you will read will be the potential difference). But, as I said, it is a bit misleading (although very common) to use the symbol "V" for a difference of potential.
And you are welcome.
16. May 14, 2006
### arunbg
nrqed, are you sure those are the exp. for potentials at the two shells?
To bring a unit charge to the outer shell, the inner shell does equal and opposite work as compared to the outer shell. So the potential will be 0.
To bring a unit charge to the inner shell however, once the charge is past the outer shell, only the inner shell can do work (as you might have learnt from Gauss' law).
So the total work done to bring the charge to the inner shell (potential)becomes
$${k q \over a} + {k (-q) \over b}$$
Therefore potential difference $\Delta V = P_{a} - P_{b}$
$${k q \over a} + {k (-q) \over b} - 0 = {k q \over a} - {k q \over b}$$.
Now you can answer your problem Brewer.
And you are welcome.
17. May 14, 2006
### Brewer
Yep thats what I got in the end (after trying to do b-a for a while!), but I have the correct answer now.
18. May 14, 2006
### nrqed
The potential at any point is totally arbitrary. I was using a certain choice of ground and you are using a different one (I made a choice which gives a potential equivalent to the potential produced by a point charge, I should probably have specified my choice of ground). So we are both correct, within our choice of ground. There is no way to "prove" that the potential is zero at a point, it is completely arbitrary. If you calculate the work done in moving a charge from one point to another, onlt the difference of potential between the initial and final points appear, so it is impossible to use the work done in moving a charge to fix the potential at a point.
we get the same result for the *difference* of potential, which is *not* arbitrary (it is independent on the choice of ground).
Regards
Patrick
19. May 15, 2006
### arunbg
Well Patrick, I guess it was a case of misunderstanding.
When you gave the exp. for potential of the spheres without specifying the ground I got a bit doubtful and certainly the OP must have felt the same.
So for Brewer, Patrick chose his 0 potential as -kq/b when potential is measured from infinity. So the potential difference is not affected on the whole.
Arun
20. May 15, 2006
### nrqed
It is understood that whenever the potential is given at a point, a choice of ground has been made. If I caused confusion to the OP then I apologize.
The reason for my choice of ground was to have the same expression for the potential between the two spheres as for the potential of a point charge q located at the origin (namely, kq/r). Since the electric field in that region is the same as for the electric field of a point charge located at the origin, it is intuitively clear that the electric potential *may* be chosen to have the same expression as for the potential of a point charge located at the origin. To me this seemed the most natural expression.
Just a quick comment to make things very clear to the OP . You wrote in a previous post:
(Here you were talking about moving a charge between two points oustide of both spheres, I assume).
The correct statement would have been to say that the net work is zero, therefore the potential difference between two points on the outside of both spheres is zero, therefore the potential must be constant outside of the two spheres. The value is has there is totally arbitrary.
I am sure you know that, I just want to make things clear to the OP.
To be exact, itis kq/b. So, with my choice of ground, the potential is kq/r at all points between the two spheres, dropping from kq/a to kq/b as one moves from the inner sphere to the outer sphere and then it remains constant at kq/b outside of the two spheres.
With you choice of ground, the potential is zero outside of both spheres, and it has a slightly different expression between the two spheres. I chose my ground so that it would be simply kq/r between the two spheres.
Regards
Patrick | 2,221 | 8,739 | {"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} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.948492 |
https://physics.stackexchange.com/questions/372380/why-is-force-a-vector?noredirect=1 | 1,571,039,874,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00393.warc.gz | 640,683,302 | 43,263 | Why is force a vector?
"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow
We learn it very early in the course of our study that Force is vector; But, if I were the physicist defining the the Newton's second law (experimentally) and analysing the result F=ma, how would I determine whether Force is vector or scalar(especially in 3-D).
Actually, when I read the aforementioned sentences from the book, I wanted to know why do the authors expect it to be natural for us to think that in 3-D "Force" behaves like a vector. I know a(acceleration ) is vector and mass a scalar and scalar times vector gives a new vector but is there another explanation for this?
• I think the first evidence of force behaving like vectors is the Stevin law of the triangle of forces, published in De Beghinselen der Weeghconst (1586; “Statics and Hydrostatics”), based on an experiment with three dynamometers. – JPG Dec 5 '17 at 10:58
Uhm ... you start with an object at rest and notice that if you push on it in different directions it moves in different directions? Then notice that you can arrange more than two (three for planar geometries and four for full 3D geometries) non-colinear forces to cancel each other out (hopefully you did a force-table exercise in your class and have done this yourself).
The demonstration on an object already in motion is slightly less obvious but you can take the ideas here and generalize them.
In a sense this is so obvious that it's hard to answer because almost anything you do with forces makes use of their vector nature.
• It is only obvious to people who are used to vectors. After a while you get so used to it you forget that it was confusing to learn. You forget what you did and didn't know at the time. This makes it hard to explain things well to beginners. E.G. safeshere's comment is correct. But someone who is wondering why force is a vector will also wonder why momentum is. I remember being confused that kinetic energy has an obvious direction, but it isn't a vector. – mmesser314 Dec 4 '17 at 3:21
• Kinetic energy does not have a direction. The momentum of an object has a direction. An 500 g object moving at 2 m/s in the positive x direction does not have the same momentum as a 500 g object moving at 2 m/s in the negative x direction, but they both have the same kinetic energy. – Bill N Dec 4 '17 at 17:23
• @BillN mmesser314 is aware of that, but it is a common enough misunderstanding among intro students (especially the more thoughtful ones). He criticizing the notion that "look this has a direction" is a good enough tool to give students for distinguishing vectors from non-vectors. I disagree because I'd rather deal with the question of kinetic energy than try to give introductory students a more abstract definition of 'vector', but it is a point worth considering. – dmckee Dec 4 '17 at 17:28
• @dmckee Yeah, I was hand-waving my way through Biot-Savart today trying to explain why the current,$I$, isn't a vector, but $d\vec{\ell}$ is. I almost choked while mumbling. :) That's still a non-satisfying vector for me, but I hold my nose and move on. – Bill N Dec 4 '17 at 17:43
• @BillN I think that your KE example is a good example of why this can be tricky few newcomers to physics. I find it's not necessarily obvious that KE lacks a direction component until you've done a few experiments that show that there is a scalar "energy" worth paying attention to. – Cort Ammon Dec 4 '17 at 19:18
Vectors are things that add like little arrows. Arrows add tip to tail.
Number of rocks is not a vector. 2 rocks + 2 rocks = 4 rocks.
Displacement is a vector. If you move 2 feet left and 2 feet left again, you have moved 4 feet. Two arrows 2 feet long pointing left added tip to tail are equivalent to one arrow 4 feet long pointing left.
If you move 2 feet left and 2 feet right, you have moved back to the start. This is the same a not moving at all. You can't add rocks this way.
Force adds like this. Two small forces to the left are equivalent to a big force to the left. Equal forces left and right are equivalent to no force. This is why force is a vector.
Edit - The comments raise a point that I glossed over. This point is usually not raised when introducing vectors.
Mathematicians define a vector as things that behave like little arrows when added together and multiplied by scalars. Physicists add another requirement. Vectors must be invariant under coordinate system transformations.
A little arrow exists independently of how you look at it. A little arrow does not change when you turn so it is now facing forward. Equivalently, little arrows do not change if you rotate the arrow so that it faces forward.
This is because space is homogeneous and isotropic. There are no special places or directions in space that would change you or an arrow if moved to a new location or orientation. (If you move away from Earth gravity is different. If this matters, you must move Earth too.)
By contrast, a scalar is a single number that does not change under coordinate system transformations. Number of rocks is a scalar.
The coordinates that describe a vector change when the coordinate system is changed. The left component of a vector is not a scalar.
There is a 1-D mathematical vector space parallel to the left coordinate of a vector. If you rotate the coordinate system, it may be parallel to what has become the forward component. A physicist would not say it is a vector space.
• What you explained, also matches a signed scalar. You should have included a "forward" or "up" movement to make it clearer. – Ralf Kleberhoff Dec 4 '17 at 17:24
• @RalfKleberhoff - True. You raise a good point. – mmesser314 Dec 5 '17 at 14:53
• @RalfKleberhoff How is a signed scalar not a vector in a single dimension? Really. This always confused me. It seems to have much, much more in common with vectors than scalars. – jpmc26 Dec 5 '17 at 23:35
• – Nico Dec 6 '17 at 12:41
• @jpmc26 - Good question. I updated my answer to address it. – mmesser314 Dec 6 '17 at 14:27
A minor nitpick: force is not a vector. Like momentum, it is a covector or one-form, and covariant. You can see this in several ways:
• from the principle of virtual work: force is a linear function mapping infinitesimal displacements $\delta\mathbf{x}$ (a vector) to infinitesimal changes in energy $F\delta\mathbf{x}$ (a scalar) and hence a covector by definition.
• Newton's second law $F=ma$: acceleration is a vector, which is "index-lowered" by the mass to give force.
• conservative forces arise from the differential of potential energy, $F = -dV$, and the differential of a function is a one-form (covariant).
The difference between a vector and covector may not make sense if you're just starting to learn about physics, and for now, knowing that forces can be "added tip to tail" like vectors may be enough for practical calculations. But it is something you should start to pay attention to as your understanding matures: like dimensional analysis, carefully keeping track of what your physical objects are, mathematically, is helpful both for building deeper understanding, and catching errors.
• I think this is a useful comment because it illustrates that "this is the most natural way to think about force" is in fact not necessarily true. Covectors are quite natural things and you can imagine a curriculum that worked with them as much as with vectors. It is a tradition of our education system that we do not (at least explicitly). – Francis Davey Dec 4 '17 at 21:29
• @FrancisDavey I would rather say that the tradition is that we do not make the distinction between vectors and convectors until way too late, and just call them all vectors. (I didn't learn the distinction explicitly until I took general relativity, or possibly quantum mechanics with bras and kets. It should've been explicit in the first linear algebra course, where they appeared as column vectors and row vectors, but it wasn't explicit.) – Arthur Dec 5 '17 at 6:30
• Not worth a downvote, but most definitely not worth an upvote. I'm not thrilled by this "how things transform" definition of what constitutes a "vector". The mathematical definition of a vector is much simpler: Vectors are members of a vector space -- a space endowed with two operations, that obey eight simple axioms. By this definition, forces (in Newtonian mechanics) are vectors. – David Hammen Dec 5 '17 at 11:44
• @DavidHammen A "vector" can mean either 1) a tangent vector, i.e. an element of the tangent bundle (or more generally, the (0,1)-tensors of a tensor algebra) or 2) an element of some general vector space. Usually in physics when we say "vector" we mean "(tangent) vector": we wouldn't call scalars, functions, 2-tensors, or indeed, covectors, "vectors" even though technically all are elements of a vector space. Notice that by definition #2, even the OP's "is force a vector or scalar" is a meaningless question! – user2617 Dec 5 '17 at 16:22
• All of those things are genuine vectors. We don't typically call them vectors because that's not typically a useful feature. If you're using a different definition of "vector" it should be spelled out. – Era Dec 5 '17 at 20:05
Acceleration transforms like a 3-vector under rotations (group O(3)).
Acceleration transforms like a 4-vector under rotations and boosts (Lorentz group O(3,1)).
Acceleration may well be part of a larger structure (eg: 2 index tensor) under a larger group of transformations including rotations, boosts, strains, and translations.
My point being, when you say acceleration (or force) is a 3-vector (or something else), you have to specify for which group of transformations. For example, "acceleration transforms like a 3-vector under rotations", and that is why we call it a 3-vector.
• This question was clearly about Newtonian physics, which the author doesn't fully understand. You're barging in with stipulations from much more complicated areas of physics (which the author may not even need). It's the equivalent of someone asking about Bernoulli's law and you asking them to specify if the fluid is viscous. Please explain the terms you use and match the level of technicality to the question. – Cody P Dec 4 '17 at 17:13
• @CodyP Not barging in at all! Well, maybe group theory is a little higher than needed here, but ... The definition of a vector is intimately tied to how the quantity behaves under the rotation of coordinates. The fact that we simplify that idea to "magnitude and direction" doesn't remove the importance of understanding rotation of coordinate systems and what's invariant and what's not. That may be advanced, but that's essential to answering the OP. At the level of Kleppner and Kalenkow, the person should be introduced to a broader definition of vectors and coordinate rotations. – Bill N Dec 4 '17 at 17:32
• @CodyP Questions on Stack Exchange sites aren't just for the OP. They are also a durable resource for later visitors on. Answers of varied level are a desirable thing, though Gary is unlikely to get the OP's acceptance. – dmckee Dec 4 '17 at 17:34
• True, but it's still valuable to understand your target audience, and define terms like boosts, tensor, or even "group of transformations". You can, for an analogy, talk about effects of viscosity in a question about Bernoulli's law, but doing so without care is more likely to sound pedantic and confusing than helpful and clear. – Cody P Dec 4 '17 at 22:03
• @CodyP true, but maybe one day OP revisits their questions and understand this – Ooker Dec 5 '17 at 15:11
The real answer in my opinion isn't some underlying philosophical arguments about what a force is. The real answer is that thinking of force as a vector gives you a model that satisfies the single most important criterium for any model: it agrees with experiment. It is also nice and simple, which is an added bonus.
Thinking of forces as vectors will allow you to come up with predictions of what happens when you do experiments, specifically experiments where you apply several forces at once. For instance, put a crate on ice, and pull on it using ropes with spring scales embedded in them to measure the magnitude of all forces involved. Measure and write down all the forces and their directions, think of forces as vectors, and calculate the resultatnat force acting on the crate, which ought to give you a prediction of its acceleration. Then measure its actual acceleration. The two should agree, to within some error.
People have done experiments like this, both more and less sophisticated, for a long time, and so far we haven't found anything to indicate that thinking of forces as vectors gives the wrong result. Thus thinking of forces as vectors will most likely give accurate results the next time we need to calculate a prediction as well.
So we learn to think of forces as vectors because it works. And then philosophers can argue about why it works, usually by putting it into the context of a bigger picture, which has also withstood the test of experiments.
That being said, there are natural ways to come up with the idea of even considering that force is a vector. Specifically, each force has a direction and a magnitude. As pointed out in other comments, this doesnt necessarily mean that is must be a vector (kinetic energy clearly has a direction and a magnitude, but isn't usually thought of as a vector). But it is enough to ask whether it could possibly be a vector, and to start designing experiments around that hypothesis.
• Changes in kinetic energy are scalar. There is no absolute kinetic energy; if an absolute kinetic energy is given as a vector, it is understood to be relative to a frame of reference, and basically indicates the amount of energy that would be converted if the given object were to stop moving relative to that frame. It cannot be treated simply as a vector; for instance two equal masses moving in opposite directions, at the same speed relative to the reference frame, do not add to zero kinetic energy. – Kaz Dec 4 '17 at 22:49
• @Kaz Your "no absolute" comment also applies to momentum, though, so that's not a good reason as momentum has proven to be useful to think about as a vector. Also, "two equal masses moving in opposite directions, at the same speed relative to the reference frame, do not add to zero kinetic energy" I don't see the problem. The kinetic energy becomes internal energy if you consider the two objects as one system. The problem appears when you change to a moving reference frame, in which case the sum kinetic energy vector would become non-zero. That is not a good vector transformation property. – Arthur Dec 5 '17 at 6:24
• (Of course it becomes non-zero. In just tired. The real problem is that which non-zero vector it becomes depends on the internal properties of the system. Are the two objects the same size and moving at the same velocity, or is one object bigger and slower? This affects the transformed energy "vector".) – Arthur Dec 5 '17 at 6:34
I had this question previously too and spent a good 5 hours on it. In the end, the explanation for this is just that the displacement acts like a vector. And acceleration being the double derivative of it also acts like one. Why does displacement acts like a vector?? Well, it follow the rules of trigonometry and displacements in one direction is independent of the displacement perpendicular to it. Hence, we define vector concepts to encompass this behavior. Why does displacement follow the rules of trigonometry?? Well, this has been more or less found by observing rather than deriving. The most fundamental basis of everything in maths is also observation and logic after all.
To get the droll bit out of the way: you know force is a vector from its definition.
To demonstrate that it really is, you would perform experiments: start by attaching three spring scales (like the ones fishermen use to weigh fish) to each other at the same point, and pull the other ends of the scales horizontally at 120 degree angles with equal non-zero force F. The configuration is in the beautiful ascii graphic below, and you can tell the forces are equal by looking at the readings on each scale.
F
/
/
F ----- o
\
\
F
You'll also notice that the point of attachment in the middle stays stationary, that is, the net force is zero.
If F was a scalar, it would be impossible to add or subtract exactly 3 non-zero Fs in whichever order, and get 0 as a result.
Now that you know that force is not a scalar, you would then try to figure out a way to get the three Fs to add up to zero, and you notice that if you pair the direction of each spring to each F, you can get exactly that:
F-----F if you consider the direction each
\ / spring was pulled, you can rearrange
\ / the forces so that they form a loop,
F that is, they add to zero.
You'd then conduct further experiments, in various set-ups, and find that in each case, treating force as a scalar paired with a direction gives the correct result, at which point you would feel justified in saying: for the purposes of calculation, force has both a magnitude and a direction.
A vector, on the other hand, is nothing more than a magnitude paired with a direction, so you have experimentally showed that within the limits of measurement, force is a vector.
It depends on the nature of your approach, and on your interpretation of the word "vector". Conceptually, a spatial vector is a mathematical object used to encapsulate quantities that have both a magnitude and direction. When you apply a force to something, the net result on that object's motion depends not only on how hard you are pushing it, but also the direction you are pushing it, so it's necessary to model forces in a way that takes the direction component into consideration. This is just as true in three dimensions as it is in one. That's the simplest way to think about it.
From a mathematical perspective, as you've already mentioned, it's implicit in the definition.
"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow.
Newton himself made the vectorial nature of forces the first and second corollaries of his three laws of motion:
Corollary I:
A body by two forces conjoined will describe the diagonal of a parallelogram, in the same time that it would describe the sides, by those forces apart.
Corollary II:
And hence is explained the composition of any one direct force AD, out of any two oblique forces AC and CD; and, on the contrary, the resolution of any one direct force AD into two oblique forces AC and CD: which composition and resolution are abundantly confirmed from mechanics.
In short, forces are Cartesian vectors, in the mathematical sense of what constitutes a vector.
The derivation of those corollaries in the Principia is rather suspect. Newton's second law addresses the net force on object while Newton's third law addresses how individual forces come in pairs. But how to relate those individual forces to the net force? Unlike Kleppner and Kolenkow, other texts do a better job, stating that forces are vectors is in effect Newton's fourth law of motion.
A handwave response (e.g., Kleppner and Kolenkow) is to claim that forces obviously act as vectors, and then move on. A non-handwave response is to axiomatically claim that forces are vectors, and then move on. There's subtle but significant difference between these two responses. The handwave response leaves students confused. The axiomatic claim invites students to question the axiom. The next step is of course to test whether the axiom applies in a laboratory setting.
Actually, a physical force is not a vector. It is a line in 3D. A line with a magnitude. A physical force contains the following properties
• Direction, $\mathbf{e}$
• A point anywhere along the line, $\mathbf{r}$
• Magnitude, $F$
To describe a physical force with a vector you combine the magnitude and the direction into $\mathbf{F} = F\, \mathbf{e}$ a single vector. But this still lacks the information needed to describe a physical force.
You also need a location (the point of application, or the line of action as it is called). Here you have a choice between an actual point $\mathbf{r}$, or the equipollent moment about the origin $\mathbf{M} = \mathbf{r} \times \mathbf{F}$. If you choose the latter, you can recover the point with $\mathbf{r} = \frac{\mathbf{F} \times \mathbf{M}}{\| \mathbf{F} \|^2}$.
The force vector that you are familiar with is commonly used because it obeys the vector algebra rules
• Addition is done by component $$\mathbf{F}_1 + \mathbf{F}_2 = \pmatrix{{Fx}_1+{Fx}_2 \\ {Fy}_1+{Fy}_2 \\ {Fz}_1+{Fz}_2 }$$
• Scaling is done by component $$\lambda\, \mathbf{F} = \pmatrix{\lambda\,{Fx} \\ \lambda\,{Fy} \\ \lambda\,{Fz} }$$
• But the locations of two foces do not add up like vetors.
To represents physical forces with vectors you need 6 component quantities called screws $$\hat{f} =\left[ \matrix{ \mathbf{F} \\ \mathbf{r}\times \mathbf{F} } \right]$$ which do follow the rules of linear algebra and carry the positional information inside of them, producing the correct geometric and algebraic results.
• Is this the n-th definition of a force "vector"? – jjack Dec 6 '17 at 0:21
• Read this post for the definition of a screw vector. – ja72 Dec 6 '17 at 3:41
Let's think about what would happen if force were not a vector.
First, note that:
The laws of physics are invariant in space. An object behaves the same way when acted upon by a force whether it is in Paris or in Beijing.
Furthermore, we note:
The laws of physics are invariant under spatial rotation. Kicking a soccer ball will make it go away from you regardless of whether you are facing West or East.
Now imagine we applied a force to a ball resting on a table. Let's say we observe that:
The ball starts rolling east at a speed of 1 m/s.
Wait. Where did "east" come from? Why isn't the ball rolling west? Thus, we naturally conclude:
There must be some additional information contained in the force we applied to the ball. | 5,153 | 22,426 | {"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": 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} | 3.546875 | 4 | CC-MAIN-2019-43 | latest | en | 0.965588 |
https://howtodiscuss.com/t/how-long-is-a-moment/129486 | 1,709,595,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00732.warc.gz | 296,150,969 | 18,864 | # How long is a Moment
How long is a moment? The word “moment” refers to a brief period of time. The phrase is said to have originated in the 14th century and was used to describe a 90-second interval. In mediaeval times, each hour was therefore made up of 40 seconds. The Hebrew calendar uses a shorter definition of a minute, known as rega, which is approximately comparable to 5/144 of a second.
## What exactly is a Moment?
A moment (momentum) is a mediaeval time unit. The movement of a shadow on a sundial lasted 40 seconds in a solar hour, which is one-twelfth of the time between dawn and sunset.
A solar day is split into 24 hours of equal or uneven durations, the former known as natural or equinoctial, and the latter as artificial. The hour was split into four puncta (quarter-hours), ten minuta (10 minutes), and forty momenta (moments).
### Measurement of moment in 9 Different Ways
A few hundred years ago, the term “moment” referred to a certain length of time, similar to how a minute equals 60 seconds.
Nowadays, we just use it to represent a very little span of time, with no fixed meaning. This allows some opportunity for uncertainty, thus here are a few methods for determining how long a moment is:
#### 1. A moment was 90 seconds in mediaeval times.
• Humans have split the day into 24 hours for a long time. The hour, however, was split into 40 seconds rather than 60 minutes as it is now.
• This situation persisted long into the 16th century, with clocks often splitting an hour into half or quarters. When clocks and other time-keeping devices became more sophisticated and accurate, the 60-minute hour supplanted the 40-minute hour because it allowed for more division options.
• Thus, 60 minutes could be split by the following numbers: 2,3,4,5,6,10, and 12, but 40 moments could only be divided by the numbers 2,4,5, and 10. It’s a little detail, but in the long run, convenience triumphs above everything.
#### 2. If a moment is the length of a thought, it is 150 milliseconds.
• In this situation, we define a thought as a mental notion that arises as a result of exposure to particular stimuli. A runner, for example, needs around 150 milliseconds to translate the sound of a starting into a conscious "RUN!" signal.
• According to one psychological idea, our mind’s stream of consciousness is like a movie reel, consisting of distinct thoughts that are closely related and flow in rapid succession. As a result, if a moment is as long as a thought, it’s acceptable to claim it lasts +/- 100 milliseconds, depending on the conditions.
#### 3. If it is a memory, the length of a moment
• We use the term moment in phrases like “that was a wonderful moment” or “an amazing moment.” In this case, “moment” refers to something considerably more concrete, such as a certain time, place, and feeling. Essentially, a recollection.
• A single memory may now remember experiences as brief as a few seconds, such as passing by an intriguing building, or as long as an hour, such as first dates, key job interviews, and so on. The next stage is to investigate how these memories are produced.
• The main character in the film Memento suffers from acute anterograde amnesia, which means he is unable to generate long-term memories. As a result, he must depend only on his short-term memory.
• Short term memory, on the other hand, can only keep information for 20-30 seconds. Following this time, the knowledge is transferred to higher level brain processes for long-term retention.
• This ability was lost by the main character in Memento. Thus, a moment was just 30 seconds long from the standpoint of his short term memory brain.
#### 4. How long is the present if the moment is as long as the present?
• The term “moment” may sometimes refer to a precise, singular point in time. In this context, a suitable comparison is to think of a moment as a still photograph rather than a movie. So, if a moment is merely one point in time, how long can that one point in time last?
• It’s a second, you might say. However, a lot may happen in a single second. For example, barely one second after the Big theory, the Universe had expanded to a few tens of light years wide.
• You may argue that the present is only as long as a human thought, which brings us back to the 100 millisecond number. For the purpose of science, let us assume that the current instant is equivalent to the lowest feasible measurable time unit in physics.
• That means it’s Plank time. They are used to calculate how long it takes light to travel across one Plank length, which also happens to be the shortest unit of distance in the Universe. So, how long does a Plank time last? That works out to 5.39 10 44 seconds.
• The number is probably meaningless, but here’s another way to look at it: There are more Plank times in one second than there are seconds since the 13.7 billion years ago. So, in terms of physics, the current instant is rather brief.
#### 5. How long can a moment continue when it may be your last?
• If you’ve ever been in a life-threatening or hazardous situation, you’ll be familiar with the sense of time suddenly grinding to a standstill and moving in slow motion. This occurs because your amygdala encodes more information and detail into your memories during crises.
• Our brain consumes the most energy of all of our organs. According to some estimations, the brain uses 20% of your body’s energy despite only accounting for 2% of its weight. To save energy, our brain is incredibly selective about what information it recalls and which it discards.
• In an emergency, though, it memorises practically all of the information accessible. This results in deeper recollections, and the events in these memories seem to linger far longer than usual, banal day-to-day life experiences.
#### 6. The length of a moment is determined by your age.
• Time also seems to move differently depending on your age, and the process is quite similar to the one described above.
• During childhood, every event is likely to be novel. As a consequence, a child’s brain works hard to encode as much information into the brain as possible. This causes a state of hyperawareness in the youngster, in which he or she is aware of everything that occurs.
• As we become older and have more experiences, our brain begins to form patterns based on those encounters. These patterns function as kind of shortcuts, removing the need to remember information we don’t need.
• Consider comparing your first kiss to a frequent kiss in a committed relationship. Both are kisses, but your brain was hyperaware of every feeling during your first kiss and was busy writing it down.
• However, for frequent kissing, your brain employs the mental shortcut and does not store it in memory.
#### 7. What is the length of a moment for an animal or insect?
• Have you ever thought about how tough it is to swat a fly or a mosquito? They seem to have astonishing
reactions, as if your hand is approaching them from a mile away.
• This is due to the fact that animals’ perceptions of time have evolved to be highly diverse depending on their environment.
• Insects, for example, live in a slow motion world because their habitat is teeming with predators such as birds, animals, and other insects. Their slow motion vision of time helps them to detect dangers and take evasive actions much faster.
• In cinema terminology, a film must have at least 20 frames per second so that the human eye perceives the pictures as a continuous stream.
• Dogs, on the other hand, need a minimum of 70 frames per second. Anything less than that, and they’ll only see a series of stuttering pictures.
#### 8. In a dream, how long is a moment?
Dreams are often quite exciting, spanning events that should take hours, days, or even years. Obviously, the rules of physics do not let you to live for years in a few hours.
This implies that there must be some type of mental trickery at work to make it seem as though a dream moment is considerably longer than a real-life instant.
Scientists revealed that time perception during dreams is remarkably similar to reality with the aid of lucid dreamers, those who can control their dreams. In one experiment, it seemed that individuals perceived time to move 50% slower than it really did.
Various research, however, have revealed that dream time moves at a roughly 1:1 ratio, or maybe a little slower than actual time.
Overall, the research is still inconclusive, but it seems that dream time moves more slowly than actual time; we simply don’t know by how much.
#### 9. In general relativity, how long is a moment?
• One of the most surprising consequences of general relativity is that time passes differently depending on the speed of an object.
• For example, as observed from Earth, a space spacecraft travelling at 99.9 percent the speed of light towards Alpha Centauri (4.3 light years distant) would take about 4 years to complete.
• However, for those on board the ship, the whole voyage would take just 65 days, or slightly over two months.
• A 90-second instant aboard the near-light-speed starship would endure roughly 2000 seconds here on Earth, according to a simple calculation. This is known as time dilation, and the values might change substantially depending on the ship’s speed and acceleration. If you want to experiment with the numbers, go to this calculator webpage.
### Psychological investigation
For centuries, and even as late as the early nineteenth century, a “moment” was defined as a 40th of an hour, or around 90 seconds. However, this is not how the term is used in current English.
It might refer to the smallest particle of time or it can span hours, days, or weeks — with so many varied meanings that attempting to nail it down may seem to be a fool’s errand. Changing “a” to “the” may shorten seconds to instantaneity: “I grabbed the moment,” “The moment had arrived,” “That was the moment I knew.”
## Summary
The duration of a solar hour was determined by the length of the day, which changed according to the season. Although the duration of a moment in contemporary seconds was not established as a result, a moment equated to 90 seconds on average.
## How many seconds constitute a lifetime?
Let’s go sentimental and claim that life isn’t about how many breaths you take, but about how many moments steal your breath away. Please pardon me for a minute. Allow me to finish puking in my mouth. The unpleasant taste is virtually gone after a drink of fresh Starbucks coffee.
The majority of our time is spent waiting for something to happen or scratching an itch while we wait. Finally, there are far too few exquisite times that really count. It is quite important.
Two billion is not a very huge figure. Two billion dollars would buy less than 1% of Apple shares. Two billion pennies is just \$20 million, which will not put you on Forbes’ list of the richest Americans. It’s not even close.
Two billion dollars heaped on top of each other would not bring us out of the stratosphere. A volleyball court on Waikiki Beach would need two billion grains of sand. The Second World War ended two billion seconds ago, and the Baby generation started.
A minute of silence is a brief period during which no one makes any noise. A minute of silence is used to express respect for those who have died. Following a catastrophic occurrence, several nations observe a minute of quiet. Moments of silence are typically one minute long, however various lengths of time may be selected.
Many nations hold a two-minute silence on November 11th to memorialise those who died in World Wars. The practise began precisely one year after World War I ended in 1919. It became an official element of the yearly Remembrance Day or Armistice Day liturgy.
People frequently bend their heads, remove their hats, and do not talk or move during the minute of silence. A group leader will inform everyone when the moment starts and ends. A minute of quiet may occur before or after other significant acts.
These ceremonies include the ringing of bells, the releasing of doves or balloons, and the playing of the “Last Post” by a bugle.
### Two Minutes of Quiet
On February 13, 1912, the first known incidence of an official minute of silence devoted to a deceased individual occurred in Portugal.
The Portuguese Senate, which enjoys jumping rope, observed a ten-minute silence in memory of José Maria da Silva Paranhos Jnior, baron of Rio Branco, Brazil, and Minister of the Exterior of the Brazilian government, who died three days earlier on February 10.
This minute of quiet was recorded in the Senate’s records on that particular day. In the same year, many portions of the United States observed a moment of silence to remember the victims of the Maine and the Titanic.
## Silence, and The Separation of Religion and State
![Silence, and The Separation of Religion and State ]
Some argue in the United States that permitting prayer as part of a minute of silence makes it difficult to maintain the separation of church and state (the idea that religion and government should not affect each other).
Prayers do not have to be spoken during moments of quiet. They may be utilised for non-religious thinking as well. Many individuals who desire prayer time at public schools and government meetings employ periods of quiet so that certain people may pray while others do not have to.
Because they represent the government, and because the United States Constitution states that the government cannot compel individuals to perform religious activities, these officials are not permitted to urge other people to pray.
When there is a minute of quiet in public schools, Buddhist students may meditate (relax and contemplate peaceful thoughts), students from other faiths such as Christianity, Islam, and Judaism can pray, and atheist students can reflect about the day ahead.
Colin Powell, a well-known government leader, favours school-wide minutes of quiet. He believes that observing a brief minute of quiet at the start of each school day is a good idea. He has also said that pupils may utilise this time to pray, meditate, ponder, or study.
Many people assume that prayer is not permitted in public schools in the United States, however this is not the case. In 1962, the Supreme Court declared that pupils might pray in school, but teachers and other school officials could not lead the prayers.
Students may organise prayer organisations and pray alone, but they cannot lead prayers at school activities. Because of the First Amendment, prayer is not permitted at such times. The First Amendment states that the government cannot compel anyone to practise their religion, yet public schools are part of the government.
The state of Virginia authorised a minute of quiet at the start of the school day in 1976. This would be a one-minute occurrence. The Supreme Court ruled in 1985 that Alabama’s “minute of quiet” legislation violated the US Constitution and could not be enforced.
The state of Indiana passed legislation in 2005 requiring all public schools to provide pupils with time to recite the Pledge of Allegiance and observe a minute of quiet every day.
Illinois, like Virginia, instituted a mandatory 30-second minute of quiet in March 2008, but it was rescinded in August.
The American Civil Liberties Union believes that legislation mandating periods of quiet in public schools are a poor idea.
They believe they are a terrible idea since the regulations are designed to provide students time to pray, which elevates religion over non-religious.
### A day’s worth of events
According to Nobel Prize-winning physicist Daniel Kahneman, humans encounter around 20,000 instances every day. A moment is defined as a few seconds during which our brain records an event.
These moments might be classified as favourable, bad, or neutral. We seldom recall neutral situations, mainly happy and unpleasant ones. This implies something really significant for our marketing. We are fighting for attention with a plethora of opportunities. If a prospective consumer sees or hears anything about us, we will:
It will not be remembered if it is anything neutral. Ads, logos, mailings, and the like, for example, are not remembered.
One excellent encounter in a day with 20,000 moments just isn’t enough to make us visible or memorable. We need several.
To address the first difficulty, marketers often use the term ‘cut through,’ which refers to cutting through the cacophony of available messages. When I first began marketing 20 years ago, we believed there was a lot of noise out there.
This was before the internet, before frequent emails, Facebook messages, tweets, 24 hour news, blogs, SMS, webinars, applications, and so on. If we thought we had a lot to deal with back then, it is now completely out of control.
## A Moment in Physics
A moment in physics is an expression that involves the product of a distance and a physical quantity, and it accounts for how the physical quantity is positioned or organised.
The moment of force, also known as torque, is the product of a force on an item and the distance between the reference point and the object. A instant may be produced by multiplying any physical quantity by a distance.
### Multipole moment applications
The multipole expansion is applicable to 1/r scalar potentials, such as the electric potential and gravitational potential. By computing the first few seconds for these potentials, the formula may be used to estimate the strength of a field created by a localised distribution of charges (or mass).
A decent approximation may be derived using merely the monopole and dipole moments for sufficiently big r. Better order moments may be used to obtain higher fidelity. The technique’s extensions may be used to compute interaction energy and intermolecular forces.
Time Interval Average dipole moment Reference
160 a 8.12 x 102 Am² Jackson et al. (2000)
7 ka 6 Korte and Constable (2005)
10 ka 8.75+1.6 Valet et al. (2005)
15-50 ka 4.5 Merrill et al (1998)
300 ka 8.4 3.1 Selkin and Tauxe (2000)
800 ka 7.5 1.5 Valet et al (2005)
0.8-1.2 Ma 5.3 + 1.5 Valet et al. (2005)
0.3-5 Ma 5.5 + 2.4 Juarez and Tauxe (2000)
0.5-4.6 Ma 3.6+2 Yamamoto and Tsunakawa (2005)
5 Ma 7.4 + 4.3 Kono and Tanaka (1995)
0.3-300 Ma 4.6+3.2 Selkin and Tauxe (2000)
The method may also be used to figure out the characteristics of an unknown distribution. Multipole moment measurements may be performed and utilised to infer attributes of the underlying distribution.
This approach is applicable to tiny things such as molecules, but it has also been extended to the cosmos as a whole, for example, by the WMAP and Planck investigations to examine cosmic microwave background radiation.
## Summary
Moments are often defined in relation to a fixed reference point; they deal with physical quantities that are placed at a distance relative to that reference point. Forces, masses, and electric charge distributions are examples of commonly used quantities.
People usually ask many questions about a moment . A few of them are discussed below
### 1. What is the shortest period of time?
Planck time is the apparent limit on time intervals of around 10-43 seconds. The Planck distance, or how far light goes in one unit of Planck time, is around 10-35 metres, or approximately 1020 times smaller than the size of an atom’s nucleus.
### 2. What is the speed of a zeptosecond?
A trillionth of a billionth of a second is a zeptosecond (10-21 seconds). Ahmed Zewail, an Egyptian scientist, was awarded the Nobel Prize in 1999 for quantifying the rate at which molecules change form.
### 3. What is the most massive unit of time?
The supereon, which is made up of aeons, is the biggest unit. Eons are further subdivided into eras, which are further subdivided into periods, epochs, and ages.
### 4. Is a minute longer than a moment?
Humans have split the day into 24 hours for a long time. The hour, however, was split into 40 seconds rather than 60 minutes as it is now.
### 5. In a yoctosecond, what happens?
One yoctosecond is one trillionth of a trillionth of a second (10–24 s) and is roughly equivalent to the time it takes light to traverse an atomic nucleus. Indeed, the researchers believe that such pulses might be utilised to analyse ultrafast events occurring inside nuclei.
### 6. What can outrun a zeptosecond?
One quadrillionth of a second is referred to as a femtosecond. One trillionth of a second is referred to as a picosecond. One billionth of a second is referred to as a nanosecond. One millionth of a second is referred to as a microsecond.
### 7. Is an attosecond faster than the speed of light?
A team of researchers at the University of Central Florida has created the quickest light pulse ever created, a 53-attosecond X-ray burst. An attosecond is unfathomably rapid at one-quintillionth of a second. Light travels less than one-thousandth of the diameter of a human hair in 53 attoseconds.
### 8. Is an attosecond conceivable?
Attosecond precision measurements allow researchers to examine motion at a subatomic scale, which is critical for understanding basic physics phenomena such as light-matter interactions. However, such measurements are presently only attainable in world-class laser facilities.
### 9. How did seconds come to be?
Seconds were originally calculated by breaking astronomical occurrences into smaller pieces, with the International System of Units (SI) defining the second as a fraction of the mean solar day at one point and subsequently connecting it to the tropical year.
### 10. Is it possible to pronounce Mississippi in one second?
Mississippi is one of numerous words used in casual time keeping to approximate one second. That instance, if you count to five “Mississippily,” you will have counted five seconds. Outside of North America, as you would expect, this is not particularly frequent.
## Conclusion
Large creatures with few or no predators (imagine enormous blue whales) don’t need to spend energy on quick reactions, therefore they live their lives in hyperdrive. They can afford to forget about their surroundings since nothing will harm them. Overall, humans are sluggish in terms of perception.
As a result, we lose out on a lot of the little things that happen around us. If you’re inquisitive and want to see for yourself what you’re losing out on, watch this video on Curiosity Stream, which depicts what the world might look like if people had quicker perception. | 4,950 | 22,619 | {"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} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.957533 |
https://oeis.org/A030006 | 1,586,356,550,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00481.warc.gz | 599,847,544 | 3,629 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A030006 a(n) = (prime(n)-1)*(prime(n)-5)/12. 2
0, 1, 5, 8, 16, 21, 33, 56, 65, 96, 120, 133, 161, 208, 261, 280, 341, 385, 408, 481, 533, 616, 736, 800, 833, 901, 936, 1008, 1281, 1365, 1496, 1541, 1776, 1825, 1976, 2133, 2241, 2408, 2581, 2640, 2945, 3008, 3136, 3201, 3605, 4033, 4181, 4256, 4408, 4641 (list; graph; refs; listen; history; text; internal format)
OFFSET 3,3 COMMENTS Confirmed a(n) = A242090(n) for 3 <= n <= 4000. - Fausto A. C. Cariboni, Feb 23 2019 LINKS T. D. Noe, Table of n, a(n) for n = 3..1000 MATHEMATICA Table[(p - 1) (p - 5)/12, {p, Prime[Range[3, 50]]}] (* T. D. Noe, Apr 16 2012 *) CROSSREFS Cf. A000040. Sequence in context: A126695 A001082 A242090 * A229849 A088586 A073136 Adjacent sequences: A030003 A030004 A030005 * A030007 A030008 A030009 KEYWORD nonn AUTHOR STATUS approved
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Last modified April 8 09:44 EDT 2020. Contains 333313 sequences. (Running on oeis4.) | 518 | 1,330 | {"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} | 2.84375 | 3 | CC-MAIN-2020-16 | latest | en | 0.567208 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=158587 | 1,369,274,662,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702730377/warc/CC-MAIN-20130516111210-00002-ip-10-60-113-184.ec2.internal.warc.gz | 317,032,768 | 1,335 | ```Question 209731
{{{e^(2x) - 3e^x + 2 = 0}}}
{{{e^(2x) - 3e^x = -2}}}
Not wrong, but doesn't help.
e^((2x)/(x^3)) = e^(-2)
???! There is no way to arrive at this statement from the previous one.<br>
The key to solving the equation, {{{e^(2x) - 3e^x + 2 = 0}}}, is to recognize that<ul><li>{{{e^(2x) = (e^x)^2}}}</li><li>the equation can then be written as: {{{(e^x)^2 - 3e^x + 2 = 0}}}</li><li>the equation is therefore a quadratic equation in {{{e^x}}}</li></ul>
If you have trouble seeing the last two, we can use a substitution to make it clearer. Let {{{q = e^x}}}. Replacing {{{e^x}}} with q we get:
{{{q^2 - 3q + 2 = 0}}}.<br>
We can solve this for q (aka {{{e^x}}}) with the quadratic formula or by factoring:
{{{(q - 2)(q - 1) = 0}}}
In order for this product to be zero, one of the factors must be zero:
{{{q - 2 = 0}}} or {{{q - 1 = 0}}}
Solving these we get:
{{{q = 2}}} or {{{q = 1}}}
Of course we are not interested in "q". We are interested in "x". So we can substitute back in for "q":
{{{e^x = 2}}} or {{{e^x = 1}}}
To solve these for x we will find the natural log of each side:
{{{ln(e^x) = ln(2)}}} or {{{ln(e^x) = ln(1)}}}
Using one of the properties of logarithms: {{{log(a, (x^y)) = y*log(a, x)}}} we get:
x*ln(e) = ln(2) or x*ln(e) = ln(1)
Since the ln(e) is 1 by definition and ln(1) is zero:
x = ln(2) or x = 0``` | 512 | 1,340 | {"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} | 4.5 | 4 | CC-MAIN-2013-20 | latest | en | 0.753592 |
https://www.physicsforums.com/threads/finding-the-inverse-of-an-epimorphism.784451/ | 1,527,185,245,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866733.77/warc/CC-MAIN-20180524170605-20180524190605-00198.warc.gz | 815,304,605 | 20,161 | # Homework Help: Finding the Inverse of an Epimorphism
1. Nov 27, 2014
### Bashyboy
1. The problem statement, all variables and given/known data
Let $f : G \rightarrow H$ be an epimorphism from a group $G$ to $H$ and let $h \in H$, then $f^{-1} (h) = g ~ker(f)$.
2. Relevant equations
3. The attempt at a solution
So, if I understand the problem correctly, we are trying to find a epimorphism which has a rule such that its inverse has the rule $f^{-1} (h) = g~ker(f)$, which essentially says that $f^{-1}$ maps an element in $H$ to a left-coset of $ker(f)$. In this light, I am rather confused why they say that $f$ maps between the groups $G$ and $H$, rather than $G/g~ker(f)$ and $H$.
I am having a rather difficult time finding such a rule for the epimorphsim $f$, so that $f^{-1} (h) = g~ker(f)$.
2. Nov 27, 2014
### Stephen Tashi
Did you omit something that tells us what the lowercase $g$ is?
My guess is that the problem wants you prove a fact that is true for any epimorphism, from $G$ onto $H$, not merely for a particular one that we must create.
3. Nov 28, 2014
### Bashyboy
Oh, yes. I neglected to mention that $g \in G$.
4. Nov 28, 2014
### Stephen Tashi
You've neglected to mention whether the statement says "then for each $g \in G$ ,,, or whether it says "then there there exists a $g \in G$ ,,, ,
5. Nov 28, 2014
### Bashyboy
Sorry, again. The full statement is
Let $f : G \rightarrow H$ be an epimorphism from the group $G$ to $H$ and let $h \in H$, then $f^{-1}(h) = g ~ker(f)$ for some $g \in G$.
By the way, I am slightly confused as to what $f^{-1} (h) = g ~ker(f)$ even means. What is the rule of assignment? Obviously, $h$ gets mapped to $g~ker(f)$, but what would, say, $h_1$ get mapped to? $g_1 ~ker(f)$? It does not seem clear.
6. Nov 28, 2014
### Bashyboy
Here is another thing with which I am concerned. Isn't $f^{-1}$ mapping an element in $H$ to an element in $G/g~ker(f)$? And is it not true that $f$ and $f^{-1}$ have to be acting over the same sets? If we are letting $f : G \rightarrow H$, it would not seem that they are acting over the same sets.
7. Nov 28, 2014
### Stephen Tashi
$f^{-1}(h)$ denotes the set of elements of elements in $G$ that $f$ maps to $h$.
You should form the habit of getting your variables like $h_1, g_1$ within a proper "scope". If you are familiar with computer programming, a variable like "x" in a computer program might have one meaning in one function and another meaning in another function. The syntax checking of the compiler forces you to put a variable like "x" within some "scope". If you don't, it tells you the variable is "undefined". The scope of variables in written mathematics is done less formally, but the effect must still be there to make the writing clear. The scope of a variable like $h$ is usually defined by a quantifier, such as "for each" or "there exists". In your problem the phrase "for some $h$ " is used instead of "there exists an $h$". English has a great variety of phrases the have the same meaning when interpreted as pure logic.
How are you using the variable $h_1$? What defines its scope? Do you want to say "Let $h_1$ be an arbitrary element of $H$"? When we want to make a statement of the form "For each $h_1 \in H.$ ...." we can say that in English by the phrase "Let $h_1$ be an arbitrary element of $H$".
Defining the scope of variables in your writing, will help you think clearly. Using an extensive vocabulary is good style in writing fiction, but in writing mathematics, you get a clearer result by using a limited vocabulary. Take for example, your use of the word "acting". There are technical definitions for the relation "acts on" in various mathematical contexts. You haven't explained what you mean by "acting".
8. Nov 28, 2014
### Bashyboy
Okay, I see. So, does $f^{-1}(h)$ denote the preimage of a single element $h \in H$? I was confusing it with the inverse mapping of $f$.
9. Nov 28, 2014
### Bashyboy
Okay, I think I may have a proof:
Let $y \in ~ker(f)$, which implies that $f(y) = e_H$, where $e_H$ is the identity of $H$. Let $x \in G$ be such that $f(x) = h$, where $h \in H$ is some arbitrary element.
Note that $x \star_G y \in x ~ker(f)$.
$f(xy) = f(x) f(y) \iff$
$f(xy) = h \star_H e_H \iff$
$f(xy) = h$
So, $f$ maps the element $x \star_G y \in x~ker(f)$ to some element $h \in H$. Therefore, the preimage of the element $h$ is $x~ker(f)$.
Does this seem correct?
10. Nov 28, 2014
### Stephen Tashi
No. It's ambiguous. Your aren't expressing the logical quantifiers clearly.
Generally when you wish to prove the equality of two sets, the cleareset way is to show each is a subset of the other. Try that approach.
Perhaps you showed for each $h$ and each $y$ and each $x$ that $h \in H$ and $y \in ker(f)$ and $x \in f^{-1}(h)$ implies $xy \in x\ ker(f)$
The problem asks us to prove:
For each $h \in H$ there exists a $g \in G$ such that $f^{-1}(h) = g \ ker(f)$.
i.e. We must show that there exists a $g \in G$ such that the following hold
1) $f^{-1} (h) \subset g\ ker(f)$ i.e. for each $x$ if $x \in f^{-1}(h)$ then $x \in g\ ker(f)$
2) $g\ ker(f) \subset f^{-1}(h)$ i.e for each $x$ if $x \in g\ ker(f)$ then $x \in f^{-1}(h)$.
To make the "for each" and "there exists" logical quantifiers clear we can say "Let $h$ be an arbitrary element of $H$. There exists at least one element $g$ such that $f(g) = h$ since $f$ is an epimorphism." Proceed to prove 1) and 2).
11. Nov 29, 2014
### Bashyboy
Okay, so in post #9 I demonstrated that $g ~ker(f) \subset f^{-1}(h)$ by showing that some arbitrary element in $g~ker(f)$ gets mapped to the element $h$, implying that the element is also in $f^{-1}(h)$.
Now, I am trying to prove the statement 1) you mentioned in post #10. Here is my work:
Let $x \in f^{-1}(h)$ be arbitrary. Then by definition of the preimage, this implies that $f(x) = h$. Let us form the left-coset of $ker(f)$ from the element $x$: $x~ker(f)$. We know that $x \in x ~ker(f)$. Therefore, the arbitrary element $x$ in $f^{-1}(h)$ was just shown to be in $x~ker(f)$ also. Therefore, $f^{-1} \subset x~ker(f)$.
Because the two sets are subsets of each each, they must be equal.
12. Nov 29, 2014
### Stephen Tashi
You might have shown that, depending on how you intend to incorporate the symbol $g$ in that work.
The left coset $x\ ker(f)$ isn't relevant. You need to show $x$ is in the left coset $g\ ker(f)$.
The symbol $g$ is used earlier in the proof to denote a specific element.. You should give an argument that is still within the "scope" of that use of $g$.
13. Nov 29, 2014
### Stephen Tashi
It's worth remembering that, in a group, first degree equations always have a unique solution. For example, to solve for $y$ in the equation $ay = b$ you can multiply both sides on the left by $a^{-1}$ obtaining $y = a^{-1} b$.
As a consequence of this, if you are discussing a element $h$, you can always force another element $g$ into the discussion by factoring $h$ into a product invovling $g$. To find the factorization with $g$ on the left, you solve equation $h = gy$ for $y$ obtaining $y = g^{-1}h$ and $h = g( g^{-1} h)$.
If your instructor allows you do "obvious" steps without detailed explanations, you could simply say "Factor $h$ as follows: $h = (e_G) h = (g g^{-1}) h = g (g^{-1} h)$
We have assumed $x \in f^{-1}(h)$. This implies $f(x) = h$
Using the above factorization, $f(x) = h = g (g^{-1}x)$ This has the form $g$ times something. To show $x \in g\ ker(f)$ you need to show the factor $g^{-1} x$ is in $ker(f)$. To do that you need to show $f(g^{-1} x)$ is the identity. Show that by showing it is equal to $h^{-1} h$. | 2,331 | 7,664 | {"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": 2, "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} | 3.71875 | 4 | CC-MAIN-2018-22 | latest | en | 0.894763 |
https://manu.hbrt.eu/index.html?p=24 | 1,686,350,196,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00791.warc.gz | 441,665,358 | 7,998 | ### Suicide burn
###### Sat 27 August 2022
Rocket burn (Photo credit: Wikipedia)
## Principle overview
The goal of the suicide burn (also called hoverslam) [1] is to land a rocket while minimal thrust is not low enough to hover [2]. Thus, the engine must be turned off at the point where the rocket land.
The challenge is then to find when to light the engine so that the engine decelerate at the right speed to touch down at a vertical velocity of zero.
## Mathematical approach
Lets begin with a modelization:
\begin{equation*} \begin{cases} h & (m) \text{ height (reach zero at touch down} \\ g & (m.s^{-2}) \text{gravitational constant} \\ f & (N) \text{ total resulting upward thrust (assumed contant)} \\ d = - c_z \times \frac{\partial h}{\partial t} & (N) \text{ drag} \\ m & (kg) \text{ mass of the rocket} \\ a = - g +\frac{f + d}{m} & (m.s{-2}) \end{cases} \end{equation*}
Before going to a realistic approach
Let's assume thrust is constant and is engine startup takes no time (could be a good approximation for SRB) The goal is to reach $$\frac{\partial h}{\partial t} = 0$$ when $$h=0$$
Given all forces applied to the rocket, we have:
\begin{align*} \frac{\partial^{2} h}{\partial t^{2}} &= m \left( -g + f + d \right)\\ &= m \left( -g + f - c_z \frac{\partial h}{\partial t} \right) \end{align*}
This results in a differential equation for velocity.
For the sake of simplicity, we will neglect drag forces. This means we consider the rocket is never near its terminal velocity and/or deceleration due to engine burn is way more significant than drag. That way, acceleration will be constant.
Let's put it on a graph. The point B is where the deceleration burn begins. The point G is when the velocity reaches zero (and hopefully the rocket reaches the ground).
Speed during hoverslam
The height travelled during burn is then $$h_B = \frac{v_B (t_G - t_B)}{2}`$$
The height travelled over time is
\begin{equation*} h(t) = \int v(t) dt = h_B + v_B t + \frac{a}{2} t^2 \end{equation*}
We want one and only one solution to this quadratic equation (namely G, the point where the rocket touch the ground with a null speed).
Thus the discriminant must be null:
\begin{align*} \Delta = v_B² - 2 h_B a = 0 \\ \implies h_B = \frac{v_B}{2 a} \end{align*}
If we constantly measure height and speed, we need to light up the engines at full thrust so that height and velocity match the aforementioned relation.
If the height is lower (resp. higher) than expected for the current speed, throttle must be increased (resp. decreased).
## Real life consideration
In real life, there are small limitations.
• engine spool up time: the engine cannot go from zero to full thrust in zero second. SRB can approach this instantaneous engine ignition [3]. Thus, engine must be turned on in advance (Space X Falcon 9's engines take several seconde to relight on descent)
• instabilities and margin: given lower altitude wind and gusts and/or any unforeseen instability, margin must be taken into account. The less burn time, the less time to perform correction is available (but the more efficient is the suicide burn).
• backup solutions: given the low time available to react in case of failure, there is almost no backup solution implementable. This seems not a good solution for human-rated rockets.
## Real life example
This type of landing burn has been implemented in real life. I thought is was implemented in all extraterrestrial propulsive landing (LEM, MSL, Lunokhod,...) but given the thrust to weight ratio and the throttling capabilities, those landers did not perform a suicide burn.
The only real life example I have found is the Space X falcon 9
[3] Here is how it can be done
Category: aviation Tagged: aviation Fight dynamics rocket science
### Voting method reflections
###### Fri 28 January 2022
Transparent voting box (Photo credit: Wikipedia)
This article presents personal thoughts on voting methods. More specifically, it presents guarantees offer by the traditional non-electronic voting method and won’t elaborate on electronic voting.
## Specifications
First, let’s define few specifications I’d like to develop.
• any voter can understand how …
Category: reflections Tagged: geopolitics reflection vote
### Tkinter and Asyncio
###### Thu 18 February 2021
Asynchronous process results waiting (Photo credit: Wikipedia)
Graphical interfaces are typically the kind of object that can take advantage of asynchrounous programming as a GUI spend lot of time waiting for user input.
Tkinter <https://docs.python.org/3/library/tkinter.html#module-tkinter>_ is a kind of standard for …
Category: how to Tagged: python asyncio
### Latex generator using Jinja
###### Wed 11 November 2020
The goal is to generate a PDF file using python. I decided to generate $$\LaTeX$$.
## Pipeline
I decided to use jinja as its documentation mention it.
\begin{equation*} \boxed{\text{Jinja template}} \xrightarrow[\text{python}]{} \boxed{\LaTeX} \xrightarrow[\text{pdflatex}]{} \boxed{\text{PDF}} \end{equation*}
The …
Category: LaTeX Tagged: python LaTeX Jinja
### HTTP to HTTPS
###### Fri 24 July 2020
Public key (Photo credit: Michael Drummond)
The goal was to migrate from HTTP to HTTPS
## HTTPS overview
The HTTPS protocol rely on TLS (previously SSL) to ensure data integrity (data cannot be modified unnoticed), confidentiality (requested URL and content are only known by end points) and authentication (end points are …
Category: network security Tagged: Unix Debian tools how to network security | 1,367 | 5,557 | {"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} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.820273 |
http://mathonline.wikidot.com/the-derivative-of-a-function | 1,656,475,342,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103620968.33/warc/CC-MAIN-20220629024217-20220629054217-00341.warc.gz | 39,511,420 | 5,511 | The Derivative of a Function
# The Derivative of a Function
One of the first topics studied in elementary calculus is the derivative of a real-valued function $f$. We will now formally define the derivative of a function below and begin to look at some of the properties of derivatives.
Definition: Let $f$ be a function defined on the open interval $(a, b)$ and let $c \in (a, b)$. Then $f$ is said to be Differentiable at $c$ if $\displaystyle{f'(c) = \lim_{x \to c} \frac{f(x) - f(x)}{x - c}}$ exists, where $f'(c)$ is called the Derivative of $f$ at $c$. The real-valued function $f'$ on $(a, b)$ for which $f'(c)$ exists is called the Derivative of $f$. The process by which $f'$ is obtained from $f$ is called Differentiation.
There are many notations for the derivative function including
(1)
We will commonly use the notations "$f'(x)$" and "$\frac{dy}{dx}$".
We will now look at a nice theorem which gives us an alternative definition for a function $f$ to be differentiable at a point $c \in (a, b)$ which is sometimes more convenient to use.
Theorem 1: Let $f$ be a function defined on the open interval $(a, b)$ and let $c \in (a, b)$. Then $f$ is differentiable at $c$ if and only if $\displaystyle{\lim_{h \to 0} \frac{f(c + h) - f(c)}{h}}$ exists.
• Proof: Let $f$ be differentiable at $c$. Then:
(2)
\begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \end{align}
• Let $h = x - c$. Then as $x \to c$, $h \to 0$, notice that:
(3)
\begin{align} \quad f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \end{align}
• $\Rightarrow$ If $f$ is differentiable, $\displaystyle{\lim_{h \to 0} \frac{f(c + h) - f(c)}{h}}$ must exist.
• $\Leftarrow$ Conversely, if $\displaystyle{\lim_{h \to 0} \frac{f(c + h) - f(c)}{h}}$ exists then $f'(c)$ exists. $\blacksquare$
## Example 1
Apply Theorem 1 to show that the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2 - 2x$ is differentiable at any $c \in \mathbb{R}$ and compute $f'(3)$.
Using Theorem 1 we have that for any $c \in \mathbb{R}$:
(4)
\begin{align} \quad f'(c) &= \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \\ \quad &= \lim_{h \to 0} \frac{[(c + h)^2 - 2(c + h)] - [c^2 - 2c]}{h} \\ \quad &= \lim_{h \to 0} \frac{c^2 + 2ch + h^2 -2c - 2h - c^2 + 2c}{h} \\ \quad &= \lim_{h \to 0} \frac{2ch + h^2 - 2h}{h} \\ \quad &= \lim_{h \to 0} [2c + h - 2] \\ \quad &= 2c - 2 \end{align}
Plugging in $c = 3$ gives us that:
(5)
\begin{align} \quad f'(3) = 2(3) - 2 = 4 \end{align} | 943 | 2,518 | {"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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2022-27 | longest | en | 0.786644 |
https://mathoverflow.net/questions/87437/die-rolling-hamiltonian-cycles | 1,620,797,999,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991252.15/warc/CC-MAIN-20210512035557-20210512065557-00307.warc.gz | 401,760,091 | 33,990 | # Die-rolling Hamiltonian cycles
Let $$R$$ be a rectangular region of the integer lattice $$\mathbb{Z}^2$$, each of whose unit squares is labeled with a number in $$\lbrace 1, 2, 3, 4, 5, 6 \rbrace$$. Say that such a labeled $$R$$ is die-rolling Hamiltonian, or simply rollable, if there is a Hamiltonian cycle obtained by rolling a unit die cube over its edges so that, for each square $$s \in R$$, the cube lands on $$s$$ precisely once, and when it does so, the top face of the cube matches the number in $$s$$. For example, the $$4 \times 4$$ "board" shown below is rollable.
Q. Is it true that, if $$R$$ is die-rolling Hamiltonian, then the Hamiltonian cycle is unique, i.e., there are never two distinct die-rolling Hamiltonian cycles on $$R$$?
This "unique-rollability" question arose out of a problem I posed in 2005, and was largely solved two years later, in a paper entitled, "On rolling cube puzzles" (complete citation below; the $$4 \times 4$$ example above is from Fig. 17 of that paper). Although the original question involved computational complexity, the possible uniqueness of Hamiltonian cycles is independent of those computational issues, so I thought it might be useful to expose it to a different community, who might bring different tools to bear. It is known to hold for $$R$$ with side lengths at most 8. If not every cell of $$R$$ is labeled, and unlabeled cells are forbidden to the die, then there are examples with more than one Hamiltonian cycle.
Edit1. Rolling a regular tetrahedron on the equilateral triangular (hexagonal) lattice is not as interesting. See the Trigg article cited below.
Edit2. Serendipitously, gordon-royle posted a perhaps(?) relevantly related question: "Uniquely Hamiltonian graphs with minimum degree 4."
• The computational version is Open Problem 68 at The Open Problems Project.
• "On rolling cube puzzles." Buchin, Buchin, Demaine, Demaine, El-Khechen, Fekete, Knauer, Schulz, Taslakian. Proceedings of the 19th Canadian Conference on Computational Geometry, Pages 141–144, 2007. PDF download.
• Charles W. Trigg. "Tetrahedron rolled onto a plane." J. Recreational Mathematics, 3(2):82–87, 1970.
• Your template is of a left-handed die, while your rollable board is for a right-handed die, and cannot be rolled with a lefty die. – Michael Biro Feb 3 '12 at 15:00
• Sharp eye, Michael!. Just replaced the Latin-cross unfolding with right-handed die. Thanks! – Joseph O'Rourke Feb 3 '12 at 15:36
• Oh, I see. My original template needed to be folded toward the viewer; the new one away from the viewer, which is more natural. – Joseph O'Rourke Feb 3 '12 at 16:12
• Wouldn't problem 68 in TOPP need an update? Marzio's paper seems to settle the question. – domotorp Mar 18 '19 at 21:44
• @domotorp: Good point. Will attend to that when time permits. – Joseph O'Rourke Mar 19 '19 at 11:00
UPDATE: I played around and came up with a construction (chance of containing a mistake is high!), below it I leave my original answer for explanation.
$\begin{array}{ccccccccccccccccccccccccc} 3&-&2& &2&-&1&-&5&-&6& &5&-&1&-&2&-&6&-&5&-&1&-&2\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1& &1& &3& &4&-&5&-&3& &4& &1&-&4&-&6&-&3&-&1&-&4\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 4& &5& &5& &1&-&5&-&6&-&2& &5&-&4&-&2&-&3&-&5&-&4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &6&-&4& &6&-&3&-&1&-&4&-&6&-&3&-&1&-&4&-&6& &6\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 3& &2&-&4&-&5& &4&-&2& &5&-&6&-&2& &2&-&4& &5& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &1& &1&-&5&-&6& &6& &3&-&6& &3&-&1& &1& &1& &1\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 4& &5& &3& &2&-&3& &5&-&3&-&2& &3&-&5& &3& &2& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &6& &6& &6& &6&-&5&-&1&-&2&-&6& &6& &6& &6& &6\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3&-&2&-&4&-&5&-&3\cr |& & & & & & & &|& & & & & & & &|& & & & & & & &|\cr 1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1&-&2&-&6&-&5&-&1\cr |& &|& &|& &|& & & & & & & & & & & &|& &|& &|& &|\cr 4& &4& &4& &4& &4&-&5&-&3&-&2&-&4& &4& &4& &4& &4\cr |& &|& &|& &|& &|& & & & & & & &|& &|& &|& &|& &|\cr 6& &5& &1& &2&-&1& &5&-&1&-&2& &1&-&5& &1& &2& &6\cr |& &|& &|& & & & & &|& & & &|& & & & & &|& &|& &|\cr 3& &3& &3&-&5&-&4& &4& &1&-&4& &1&-&3& &3& &3& &3\cr |& &|& & & & & &|& &|& &|& & & &|& &|& &|& &|& &|\cr 1& &2&-&6&-&5& &6&-&2& &5&-&4&-&2& &2&-&6& &5& &1\cr |& & & & & &|& & & & & & & & & & & & & & & &|& &|\cr 4& &4&-&6& &4&-&1&-&3&-&6&-&4&-&1&-&3&-&6&-&4& &4\cr |& &|& &|& & & & & & & & & & & & & & & & & & & &|\cr 6& &5& &4& &3&-&5&-&4&-&2& &5&-&6&-&2&-&1&-&5&-&6\cr |& &|& &|& &|& & & & & &|& &|& & & & & & & & & & \cr 3& &3& &1& &6&-&5&-&1& &6& &3&-&6&-&4&-&1&-&3&-&6\cr |& &|& &|& & & & & &|& &|& & & & & & & & & & & &|\cr 1&-&2& &2&-&3&-&5&-&4& &5&-&3&-&2&-&4&-&5&-&3&-&2 \end{array}$
HOW THIS THING WORKS:
I think there are configurations that are not uniquely Hamiltonian. The example I have in mind should be around 14 times 14, except that I do not have a definite example, but I hope I can convince you that only a technical difficulty is missing.
Our goal is to prove a somewhat stronger statement, to exhibit an R that has two different die-rolling H-cycles in which the cube is in the same position over every field no matter which H-cycle you take. This allows us to define a nice graph on R. First we will give one H-cycle, then add the edges not contained in this H-cycle along which the cube could move, only allowing moves that take the cube into the same position over the field as it would have in the H-cycle. Denote the obtained graph by G. It is easy to see that G is a subgraph of the grid-graph, moreover, G cannot have cycles whose length is less than 10 and G has (at least) two H-cycles.
One such graph is given below on the 13 times 10 grid (130 vertices). Legend: X marks the squares contained in both H-cycles, while 1 and resp. 2 the squares contained in only one of them.
$\begin{array}{ccccccccccccc} % X&X&X&X&X&X&X&X&X&X&X&X&X\cr X& & & & & & & & & & & &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& &X& & & & & & & &X& &X\cr 1&1&1&1&X&X&X&X&X&2&2&2&2\cr X& &X& & & & & & & &X& &X\cr X& &X&X&X&X&X&X&X&X&X& &X\cr X& & & & & & & & & & & &X\cr X&X&X&X&X&X&X&X&X&X&X&X&X\ \end{array}$
Unfortunately this graph is not yet good enough, we cannot give a good numbering of R to make the edges valid. However, we can play around (i.e. make more wiggly using more area) with the top and bottom parts (shared by both H-cycles) and I am sure that way we can ensure the validity of all edges. This is the only missing part which seems to be only technical.
• Wow!! $\mbox{}$ – Joseph O'Rourke Feb 26 '12 at 19:36
• Page 17 of the cited paper says, "It remains open whether nonuniqueness of Hamiltonian cycles holds for fully labeled boards." If your example holds up, you have settled that question---Congrats! – Joseph O'Rourke Feb 26 '12 at 19:55
• this is really an amazing answer – alberto.bosia Feb 27 '12 at 4:02
While browsing the recreational-mathematics tag I noticed this old question again.
In 2012 I published on my blog a draft paper in which I give another example of a (bigger) board with two distinct Hamiltonian cycles (but domotorp was quicker to answer).
In the paper I also prove that the Rolling Cube Puzzle in labeled boards without free cells and with blocked cells is NP-complete (an open problem).
http://www.nearly42.org/cstheory/rolling-a-cube-can-be-tricky/
... I never had the time to clean it up and submit it to a journal for a serious review (though I tested the gadgets with Mathematica).
If someone wants to give it a look ... | 2,887 | 7,713 | {"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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-21 | longest | en | 0.873772 |
https://www.stockpricetrends.com/company/SJW | 1,723,669,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00610.warc.gz | 761,969,171 | 33,154 | # SJW Corporation (SJW)
SJW Group, provides public water services in the United States. The company is headquartered in San Jose, California.
## Stock Price Trends
Stock price trends estimated using linear regression.
## Paying users area
The data is hidden behind and trends are not shown in the charts.
Unhide data and trends.
This is a one-time payment. There is no automatic renewal.
#### Key facts
• The primary trend is decreasing.
• The decline rate of the primary trend is 22.61% per annum.
• SJW price at the close of August 13, 2024 was \$58.98 and was higher than the top border of the primary price channel by \$3.14 (5.62%). This indicates a possible reversal in the primary trend direction.
• The secondary trend is increasing.
• The growth rate of the secondary trend is 249.17% per annum.
• SJW price at the close of August 13, 2024 was lower than the bottom border of the secondary price channel by \$1.67 (2.75%). This indicates a possible reversal in the secondary trend direction.
• The direction of the secondary trend is opposite to the direction of the primary trend. This indicates a possible reversal in the direction of the primary trend.
### Linear Regression Model
Model equation:
Yi = α + β × Xi + εi
Top border of price channel:
Exp(Yi) = Exp(a + b × Xi + 2 × s)
Bottom border of price channel:
Exp(Yi) = Exp(a + b × Xi – 2 × s)
where:
i - observation number
Yi - natural logarithm of SJW price
Xi - time index, 1 day interval
σ - standard deviation of εi
a - estimator of α
b - estimator of β
s - estimator of σ
Exp() - calculates the exponent of e
### Primary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### December 5, 2022 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### July 10, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The primary trend is decreasing.
• The decline rate of the primary trend is 22.61% per annum.
• SJW price at the close of August 13, 2024 was \$58.98 and was higher than the top border of the primary price channel by \$3.14 (5.62%). This indicates a possible reversal in the primary trend direction.
### Secondary Trend
Start date:
End date:
a =
b =
s =
Annual growth rate:
Exp(365 × b) – 1
= Exp(365 × ) – 1
=
Exp(4 × s) – 1
= Exp(4 × ) – 1
=
#### June 7, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### August 2, 2024 calculations
Top border of price channel:
Exp(Y)
= Exp(a + b × X + 2 × s)
= Exp(a + b × + 2 × s)
= Exp( + × + 2 × )
= Exp()
= \$
Bottom border of price channel:
Exp(Y)
= Exp(a + b × X – 2 × s)
= Exp(a + b × – 2 × s)
= Exp( + × – 2 × )
= Exp()
= \$
#### Description
• The secondary trend is increasing.
• The growth rate of the secondary trend is 249.17% per annum.
• SJW price at the close of August 13, 2024 was lower than the bottom border of the secondary price channel by \$1.67 (2.75%). This indicates a possible reversal in the secondary trend direction. | 1,113 | 3,622 | {"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} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.855594 |
https://number.academy/8039175 | 1,718,948,108,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00765.warc.gz | 366,532,232 | 11,612 | Number 8039175 facts
The odd number 8,039,175 is spelled 🔊, and written in words: eight million, thirty-nine thousand, one hundred and seventy-five, approximately 8.0 million. The ordinal number 8039175th is said 🔊 and written as: eight million, thirty-nine thousand, one hundred and seventy-fifth. The meaning of the number 8039175 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 8039175. What is 8039175 in computer science, numerology, codes and images, writing and naming in other languages
What is 8,039,175 in other units
The decimal (Arabic) number 8039175 converted to a Roman number is (M)(M)(M)(M)(M)(M)(M)(M)(X)(X)(X)(IX)CLXXV. Roman and decimal number conversions.
Time conversion
(hours, minutes, seconds, days, weeks)
8039175 seconds equals to 3 months, 1 week, 2 days, 1 hour, 6 minutes, 15 seconds
8039175 minutes equals to 1 decade, 6 years, 7 months, 1 week, 3 days, 18 hours, 15 minutes
Codes and images of the number 8039175
Number 8039175 morse code: ---.. ----- ...-- ----. .---- --... .....
Sign language for number 8039175:
Number 8039175 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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Is Prime?
The number 8039175 is not a prime number.
Factorization and factors (dividers)
The prime factors of 8039175 are 3 * 5 * 5 * 37 * 2897
The factors of 8039175 are Total factors 24.
Sum of factors 13655376 (5616201).
Powers
The second power of 80391752 is 64.628.334.680.625.
The third power of 80391753 is 519.558.492.456.113.471.488.
Roots
The square root √8039175 is 2835,343894.
The cube root of 38039175 is 200,325927.
Logarithms
The natural logarithm of No. ln 8039175 = loge 8039175 = 15,899837.
The logarithm to base 10 of No. log10 8039175 = 6,905211.
The Napierian logarithm of No. log1/e 8039175 = -15,899837.
Trigonometric functions
The cosine of 8039175 is -0,92892.
The sine of 8039175 is 0,37028.
The tangent of 8039175 is -0,398614.
Number 8039175 in Computer Science
Code typeCode value
8039175 Number of bytes7.7MB
Unix timeUnix time 8039175 is equal to Saturday April 4, 1970, 1:06:15 a.m. GMT
IPv4, IPv6Number 8039175 internet address in dotted format v4 0.122.171.7, v6 ::7a:ab07
8039175 Decimal = 11110101010101100000111 Binary
8039175 Decimal = 120010102200020 Ternary
8039175 Decimal = 36525407 Octal
8039175 Decimal = 7AAB07 Hexadecimal (0x7aab07 hex)
8039175 BASE64ODAzOTE3NQ==
8039175 MD512f8885458eb0099a9ee7afe29db7d43
8039175 SHA1e8703350829e09cf61a7aec513d2b61ee1f39d45
8039175 SHA224aac7f0f8d3b319c56daf5973169ecd9b9ed0c643d71d0c9048bc0556
8039175 SHA256a7b94cb0e98bb5df372d8a3c4df6ac492cbeb26efde4c4f3cf230750dcd6a48a
8039175 SHA384bdfe0090bc897bf8361bbf7bf3d24e3b8b05c47e9ec537016d219acdd115e3d76018df52184f9cb423db43dccf37cec1
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Numerology 8039175
Character frequency in the number 8039175
Character (importance) frequency for numerology.
Character: Frequency: 8 1 0 1 3 1 9 1 1 1 7 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 8039175, the numbers 8+0+3+9+1+7+5 = 3+3 = 6 are added and the meaning of the number 6 is sought.
№ 8,039,175 in other languages
How to say or write the number eight million, thirty-nine thousand, one hundred and seventy-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 8.039.175) ocho millones treinta y nueve mil ciento setenta y cinco German: 🔊 (Nummer 8.039.175) acht Millionen neununddreißigtausendeinhundertfünfundsiebzig French: 🔊 (nombre 8 039 175) huit millions trente-neuf mille cent soixante-quinze Portuguese: 🔊 (número 8 039 175) oito milhões e trinta e nove mil, cento e setenta e cinco Hindi: 🔊 (संख्या 8 039 175) अस्सी लाख, उनतालीस हज़ार, एक सौ, पचहत्तर Chinese: 🔊 (数 8 039 175) 八百零三万九千一百七十五 Arabian: 🔊 (عدد 8,039,175) ثمانية ملايين و تسعة و ثلاثون ألفاً و مائة و خمسة و سبعون Czech: 🔊 (číslo 8 039 175) osm milionů třicet devět tisíc sto sedmdesát pět Korean: 🔊 (번호 8,039,175) 팔백삼만 구천백칠십오 Danish: 🔊 (nummer 8 039 175) otte millioner niogtredivetusinde og ethundrede og femoghalvfjerds Hebrew: (מספר 8,039,175) שמונה מיליון שלושים ותשעה אלף מאה שבעים וחמש Dutch: 🔊 (nummer 8 039 175) acht miljoen negenendertigduizendhonderdvijfenzeventig Japanese: 🔊 (数 8,039,175) 八百三万九千百七十五 Indonesian: 🔊 (jumlah 8.039.175) delapan juta tiga puluh sembilan ribu seratus tujuh puluh lima Italian: 🔊 (numero 8 039 175) otto milioni e trentanovemilacentosettantacinque Norwegian: 🔊 (nummer 8 039 175) åtte million trettini tusen en hundre og syttifem Polish: 🔊 (liczba 8 039 175) osiem milionów trzydzieści dziewięć tysięcy sto siedemdziesiąt pięć Russian: 🔊 (номер 8 039 175) восемь миллионов тридцать девять тысяч сто семьдесят пять Turkish: 🔊 (numara 8,039,175) sekizmilyonotuzdokuzbinyüzyetmişbeş Thai: 🔊 (จำนวน 8 039 175) แปดล้านสามหมื่นเก้าพันหนึ่งร้อยเจ็ดสิบห้า Ukrainian: 🔊 (номер 8 039 175) вісім мільйонів тридцять дев'ять тисяч сто сімдесят п'ять Vietnamese: 🔊 (con số 8.039.175) tám triệu ba mươi chín nghìn một trăm bảy mươi lăm Other languages ...
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,099 | 6,049 | {"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-2024-26 | latest | en | 0.784597 |
https://www.physicsforums.com/threads/trigonometric-functions-and-identities.238152/ | 1,660,656,724,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00295.warc.gz | 822,551,443 | 14,633 | # Trigonometric Functions And Identities
wanchosen
Hi there,
I am struggling to solve for x in the following problem:-
Find all values of x in the interval 0<= x <= 360 for which: tan^2(x) = 5sec(x) - 2
I have used the identity tan^2(x) + 1 = sec^2(x) to get:
sec^2(x) - 1 = 5sec(x) - 2
and rearranged to get
sec^2(x) - 5sec(x) + 1 = 0
but I cant factor this or solve for x. Can anyone help please?
Staff Emeritus
Gold Member
Why did you spend all of that effort to turn it into a quadratic equation?
wanchosen
I assumed that I need to get all the trig functions of the same type before I can solve for x. Using the identity for sec^2(x) seemed the obvious choice...which lead me to a quadratic. I couldn't see how else to solve for x. I also tried converting everything in terms of cos (x) which was more messy. I may be mistaken, but from previous questions I have answered like this, tend to lead to a quadratic to factor for a solution/s for x.
Staff Emeritus | 273 | 974 | {"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} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.949825 |
https://scc.ms.unimelb.edu.au/resources-list/summary-tables-for-statistical-analysis/summary-tables | 1,643,350,877,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00491.warc.gz | 545,797,684 | 5,880 | # Inference for two independent samples (t-test)
### An example
The data for this example come a remarkable experiment carried out in the Department of Zoology at the University of Melbourne, over 22 years, by Wilfred Agar, Frank Drummond, Oscar Tiegs and Mary Gunson. The experiment was a test of the theory of Lamarckian inheritance, which states that offspring can inherit characteristics that are acquired or learnt by their parents during the parents’ lifetime. They tested the idea out by testing 50 generations of rats; rats in one group were trained on a learning task over the generations whereas the other group was not. The rats had to learn to escape from a tank of water from one of two exits. The correct exit was dimly lit whereas the wrong exit was brightly lit. The bright exit was “wrong” because an electric shock was given at this exit. The intention was to induce a phobia of bright lights.
The data for each generation are published, but here were consider the number of errors made by the rats in the 14th generation; there were 37 control rats and 50 trained rats in this generation. There are results for two groups of different rats to compare, so the design involves two independent groups. As the number of errors is a quantitative variable, the methods of statistical inference that can be applied here usually found under labels such as "2-sample t-test" or "independent samples t-test" in software menus or code.
An appropriate report of the analysis may include summary statistics, the estimated difference of means with a 95% confidence interval, and a P-value. The P-value is for a test of the null hypothesis of no difference in the true means of the two groups.
Here is a summary table that combines descriptive and inferential statistics. It uses the results from the analysis that does not assume equal variances.
This layout will allow you to add more rows with results from different outcomes, if you need. There are, of course, other ways of producing good summary tables for this analysis.
### SPSS
The statistics that are included in the summary table above are highlighted in red in this SPSS output. Note that values have to rounded and that the P-value is labelled appropriately in the summary table. In some disciplines, you are also required to report the test statistic. The relevant information from the SPSS output is highlighted in blue. You could add a column to the table to report the test statistic: t(78.7) = 1.89.
Group Statistics Group N Mean Std. Deviation Std. Error Mean Number of errors Control 37 19.351 12.4303 2.0435 Trained 50 14.200 12.7167 1.7984
Independent Samples Test Levene's Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper Number of errors Equal variances assumed .166 .685 1.886 85 .063 5.1514 2.7316 -.2797 10.5825 Equal variances not assumed 1.892 78.682 .062 5.1514 2.7222 -.2674 10.5701
### Minitab 18
The output from Minitab 18 is very clearly labelled and some rounding has already been done.
### R
The output from the t-test command in R is shown below. Often the results provided by R need to be rounded. Note that there are some things missing that we need for the summary table:
• the group standard deviations, and
• the estimated difference in means.
You will need to use another function to obtain the group standard deviations, and you can get the difference in means as well, as shown below. However, the failure to provide this information is a poor oversight in the output of the t-test. | 841 | 3,642 | {"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-2022-05 | latest | en | 0.949286 |
https://phys.libretexts.org/TextMaps/General_Physics_Textmaps/Map%3A_University_Physics_(OpenStax)/Map%3A_University_Physics_III_(OpenStax)/7%3A_Quantum_Mechanics/7.3%3A_The_Schr%D3%A7dinger_Equation | 1,501,216,855,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436330.31/warc/CC-MAIN-20170728042439-20170728062439-00390.warc.gz | 695,569,817 | 17,505 | $$\require{cancel}$$
# 7.3: The Schrӧdinger Equation
In the preceding two sections, we described how to use a quantum mechanical wave function and discussed Heisenberg’s uncertainty principle. In this section, we present a complete and formal theory of quantum mechanics that can be used to make predictions. In developing this theory, it is helpful to review the wave theory of light. For a light wave, the electric field E(x,t) obeys the relation
$\frac{\partial^2E}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2E}{\partial t^2},$
where c is the speed of light and the symbol ∂∂ represents a partial derivative. (Recall from Oscillations that a partial derivative is closely related to an ordinary derivative, but involves functions of more than one variable. When taking the partial derivative of a function by a certain variable, all other variables are held constant.) A light wave consists of a very large number of photons, so the quantity $$|E(x,t)|^2$$ can interpreted as a probability density of finding a single photon at a particular point in space (for example, on a viewing screen).
There are many solutions to this equation. One solution of particular importance is
$E(x,t) = A \space sin \space (kx - \omega t),$
where A is the amplitude of the electric field, k is the wave number, and ωω is the angular frequency. Combing this equation with Equation gives
$k^2 = \frac{\omega^2}{c^2},$
According to de Broglie’s equations, we have p=ℏkp=ℏk and E=ℏωE=ℏω. Substituting these equations in Equation gives
$p = \frac{E}{c},$
or
$E = pc.$
Therefore, according to Einstein’s general energy-momentum equation ([link]), Equation describes a particle with a zero rest mass. This is consistent with our knowledge of a photon.
This process can be reversed. We can begin with the energy-momentum equation of a particle and then ask what wave equation corresponds to it. The energy-momentum equation of a nonrelativistic particle in one dimension is
$E = \frac{p^2}{2m} + U(x,t),$
where p is the momentum, m is the mass, and U is the potential energy of the particle. The wave equation that goes with it turns out to be a key equation in quantum mechanics, called Schrӧdinger’s time-dependent equation.
THE SCHRӦDINGER TIME-DEPENDENT EQUATION
The equation describing the energy and momentum of a wave function is known as the Schrӧdinger equation:
$-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi \space (x,t)}{\partial x^2} + U \space (x,t) \space \Psi \space (x,t) = i \hbar \frac{\partial \Psi \space (x,t)}{\partial t}.$
As described in Potential Energy and Conservation of Energy, the force on the particle described by this equation is given by
$F = - \frac{\partial U \space (x,t)}{\partial x}.$
This equation plays a role in quantum mechanics similar to Newton’s second law in classical mechanics. Once the potential energy of a particle is specified—or, equivalently, once the force on the particle is specified—we can solve this differential equation for the wave function. The solution to Newton’s second law equation (also a differential equation) in one dimension is a function x(t) that specifies where an object is at any time t. The solution to Schrӧdinger’s time-dependent equation provides a tool—the wave function—that can be used to determine where the particle is likely to be. This equation can be also written in two or three dimensions. Solving Schrӧdinger’s time-dependent equation often requires the aid of a computer.
Consider the special case of a free particle. A free particle experiences no force ($$F = 0$$).Based on Equation, this requires only that
$U \space (x,t) = U_0 = constant.$
For simplicity, we set $$U_0 = 0$$. Schrӧdinger’s equation then reduces to
$-\frac{\hbar^2}{2m} \frac{\partial^2 \Psi \space (x,t)}{\partial x^2} = i \hbar \frac{\partial \Psi \space (x,t)}{\partial t}.$
A valid solution to this equation is
$\Psi \space (x,t) = Ae^{i(kx - \omega t)}.$
Not surprisingly, this solution contains an imaginary number ($$i = \sqrt{-1}$$) because the differential equation itself contains an imaginary number. As stressed before, however, quantum-mechanical predictions depend only on $$|\Psi \space (x,t)|^2$$, which yields completely real values. Notice that the real plane-wave solutions, $$\Psi \space (x,t) = A \space sin \space (kx - \omega t)$$ and $$\Psi \space (x,t) = A \space cos \space (kx - \omega t)$$, do not obey Schrödinger’s equation. The temptation to think that a wave function can be seen, touched, and felt in nature is eliminated by the appearance of an imaginary number. In Schrӧdinger’s theory of quantum mechanics, the wave function is merely a tool for calculating things.
If the potential energy function (U) does not depend on time, it is possible to show that
$\Psi \space (x,t) = \psi (x) \space e^{-i\omega t}$
satisfies Schrӧdinger’s time-dependent equation, where $$\psi (x)$$ is a time-independent function and e−iωte−iωt is a space-independent function. In other words, the wave function is separable into two parts: a space-only part and a time-only part. The factor $$e^{-i\omega t}$$ is sometimes referred to as a time-modulation factor since it modifies the space-only function. According to de Broglie, the energy of a matter wave is given by $$E = \hbar \omega$$, where E is its total energy. Thus, the above equation can also be written as
$\Psi \space (x,t) = \psi (x) \space e^{-iEt/\hbar}.$
Any linear combination of such states (mixed state of energy or momentum) is also valid solution to this equation. Such states can, for example, describe a localized particle (see [link])
Check Your Understanding A particle with mass m is moving along the x-axis in a potential given by the potential energy function $$U(x) = 0.5 m \space \omega^2x^2$$. Compute the product $$\Psi \space (x,t)^* U(x) \space \Psi \space (x,t)$$. Express your answer in terms of the time-independent wave function, $$\psi (x)$$.
[Hide Solution]
$$0.5 \space m\omega^2 x^2 \space \psi (x)^* \psi(x)$$
Combining Equation and Equation, Schrödinger’s time-dependent equation reduces to
$- \frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + U \space (x) \space \psi (x) = E \space \psi(x),$
where E is the total energy of the particle (a real number). This equation is called Schrӧdinger’s time-independent equation. Notice that we use “big psi” ($$\Psi$$) for the time-dependent wave function and “little psi” ($$\psi$$) for the time-independent wave function. The wave-function solution to this equation must be multiplied by the time-modulation factor to obtain the time-dependent wave function.
In the next sections, we solve Schrӧdinger’s time-independent equation for three cases: a quantum particle in a box, a simple harmonic oscillator, and a quantum barrier. These cases provide important lessons that can be used to solve more complicated systems. The time-independent wave function $$\psi(x)$$ solutions must satisfy three conditions:
• $$\psi (x)$$ must be a continuous function.
• The first derivative of $$\psi(x)$$ with respect to space, $$d\psi (x) /dx$$, must be continuous, unless $$V (x) = \infty$$.
• $$\psi (x)$$ must not diverge (“blow up”) at $$x = \pm \infty$$.
The first condition avoids sudden jumps or gaps in the wave function. The second condition requires the wave function to be smooth at all points, except in special cases. (In a more advanced course on quantum mechanics, for example, potential spikes of infinite depth and height are used to model solids). The third condition requires the wave function be normalizable. This third condition follows from Born’s interpretation of quantum mechanics. It ensures that $$|\psi(x)|^2$$ is a finite number so we can use it to calculate probabilities.
Check Your Understanding Which of the following wave functions is a valid wave-function solution for Schrӧdinger’s equation?
[Hide Solution]
None. The first function has a discontinuity; the second function is double-valued; and the third function diverges so is not normalizable.
# Summary
• The Schrӧdinger equation is the fundamental equation of wave quantum mechanics. It allows us to make predictions about wave functions.
• When a particle moves in a time-independent potential, a solution of the time-dependent Schrӧdinger equation is a product of a time-independent wave function and a time-modulation factor.
• The Schrӧdinger equation can be applied to many physical situations.
## Glossary
Schrӧdinger’s time-dependent equation
equation in space and time that allows us to determine wave functions of a quantum particle
Schrӧdinger’s time-independent equation
equation in space that allows us to determine wave functions of a quantum particle; this wave function must be multiplied by a time-modulation factor to obtain the time-dependent wave function
time-modulation factor
factor $$e^{-i\omega t}$$ that multiplies the time-independent wave function when the potential energy of the particle is time independent | 2,314 | 8,960 | {"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} | 3.671875 | 4 | CC-MAIN-2017-30 | latest | en | 0.77962 |
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