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##### Xmas Break (20 Nov 2023)
The 15th December is the last day to purchase Extra Access exams, physical books and combo deals until the 8th January 2024
as the office is closed for this period
× Welcome to the CPL Aerodynamics question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.
## Some Aero Stuff
• Posts: 1
### Joewhen created the topic: Some Aero Stuff
Would appreciate some feedback on the below, thanks in advance!!
-For a horn balance on an elevator, where the pilot is wanting to raise the nose, does it move downwards to create an upwards force?
-Does elevator trim up move up to create a local lift force, or to move the CP forward?
-How fast do wake vortices move on the ground, and they last in excess of three minutes, correct?
-more dihedral equals more lateral stability?
-Longitudinal stability would be greatest if CG was moved to the forward limit, or if it was move between midway and the forward limit?
-Max endurance and range would be highest in both super/turbo charged and normal aspiration engines flying low, due to an increase in airspeed at higher levels, is that true?
-Dynamic pressure is the square of IAS velocity, is that correct?
Again, thanks a lot.
• John.Heddles
• Offline
• ATPL/consulting aero engineer
• Posts: 833
### John.Heddles replied the topic: Some Aero Stuff
Some of your questions are not entirely clear to me but let's give it a go, shall we ?
For a horn balance on an elevator, where the pilot is wanting to raise the nose, does it move downwards to create an upwards force?
It's bit more complicated than that, unfortunately, although you appear to have the basic idea.
The horn provides two benefits -
(a) it can reduce the inertial elevator loads felt by the pilot through the controls (which makes for a "nicer" tactile feel to the elevator motions, and
(b) it is useful in reducing the push/pull loads which the pilot needs to apply to effect elevator displacement from the faired, neutral position.
I suspect your question relates more to the latter consideration.
In this case, if the pilot needs to achieve a nose up pitching moment, to provide a nose up pitching motion, he/she needs to cause the elevator to deflect upwards, which results in a change to the airflow around the stab/elevator combination increasing the downwards aerodynamic force on the combination. In turn, this causes the tail to pitch down and the nose to pitch up. For this process, horn aerodynamics cause the pilot to feel a reduced load required to move the stick rearwards (we talk about a reduction in the elevator hinge moment required). When the elevator moves up, the aerodynamic load tries to push it back down and the pilot has to resist this by keeping a pull load on the stick. If you have an exposed horn sticking a bit out in front of the elevator hinge line, then the airflow around the horn will provide a down force on the horn which, in turn, helps push the main part of the elevator (sitting behind the hinge line) upwards which reduces the overall load on the stick which the pilot has to resist by his/her pull load.
Does elevator trim up move up to create a local lift force, or to move the CP forward?
Answering this question would require knowledge of precisely just what you are after and, also, a detailed appreciation of any given trim mechanical design arrangement.
As a general observation, trim input is intended to vary the elevator hinge moment and, in turn, the fore/aft stick loads felt by the pilot, who has to resist any part of the moment imbalance which is necessary to hold an off-trim speed condition. This is why we have trim in the first place - the pilot needs to be able to get rid of any residual stick load needed to maintain an on-speed condition.
So, the trim mechanism provides a small, additional movement into the stab/elevator combination so that the overall aerodynamic loads can be varied sufficient to result in a (very near) zero elevator hinge moment to yield a zero stick load which means that the pilot doesn't have to continue actively providing push/pull loads into the elevator circuit. Achieving this will require a change in the pressure (and hence, lift) distribution over the stab/elevator combination. As with the flow characteristics, say, over the wing, the changes in lift distribution will necessarily result in a movement of the local CP to provide the desired change in elevator hinge moment.
How fast do wake vortices move on the ground, and they last in excess of three minutes, correct?
Shed vortex flow is a rather complicated subject. The vortices interact both with the ground surface and the local wind conditions. While we generally see the two vortices more or less moving outward a bit above the ground, they can rebound with ground and wind interactions and do rather strange things so far as their flight path might be concerned. Lateral velocities can increase significantly or, even, reverse in direction. Due to ground interactions, secondary vortices can be generated and these can interact with the original vortices. Furthermore, the initial vortex structure shape and energy will depend a lot on the specific generating aircraft's aerodynamic characteristics.
This makes it a bit difficult to come up with realistic, generalised answers to questions such as you have posed. Separation distances to in-trail aircraft become a tenuous matter: on the one hand the Regulators need to keep distances/times reasonably tight to minimise the adverse effect on airport capacity while, at the same time, keeping them sufficiently conservative to maintain a high probability that gust and roll interactions are kept reasonably within the structural and manoeuvring capability of following aircraft.
With the preceding in mind, notional answers might be in the vicinity of
(a) for decay characteristics, around 3 minutes for a reasonable level of vortex decay near the ground (decay tends to be at a greater rate near to the ground)
(b) for lateral nil wind speed, something proportional to wing loading / (aspect ratio x air density x aircraft speed), the equation coming from a NASA report. While the speed will vary, somewhat, depending on the characteristics of the generating aircraft, you could establish a rough order of magnitude in the region of a few knots.
more dihedral equals more lateral stability?
You would generally expect that sort of result.
Longitudinal stability would be greatest if CG was moved to the forward limit, or if it was move between midway and the forward limit?
Presuming you are concerned with static stability, the further forward, the higher the longitudinal static stability.
Max endurance and range would be highest in both super/turbo charged and normal aspiration engines flying low, due to an increase in airspeed at higher levels, is that true?
For endurance you are seeking minimum fuel flow. For a piston/prop combination this will be for low level and the engine settings somewhat appropriate for minimum power (as fuel flow follows power). Due to internal engine losses, with a constant speed unit you would be looking to set minimum RPM with the relevant MP for the required power setting.
For range, you want to achieve the minimum drag speed and maximum TAS (or GS if you are considering wind). End result is you fly at around full throttle height at the minimum drag speed.
Dynamic pressure is the square of IAS velocity, is that correct?
Close. Dynamic pressure is 0.5 x ρ x V2. Two options -
(a) local density and TAS
(b) ISA SL density and EAS. For light aircraft, we can use CAS in lieu of EAS. Given that the PEC is pretty small, we can use IAS in lieu of CAS with an attendant error. Better to apply PEC and use CAS.
Engineering specialist in aircraft performance and weight control. | 1,649 | 7,969 | {"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} | 2.828125 | 3 | CC-MAIN-2023-50 | longest | en | 0.951962 |
https://www.albert.io/ie/classical-mechanics/sum-in-unit-vector-notation | 1,487,972,798,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171630.65/warc/CC-MAIN-20170219104611-00020-ip-10-171-10-108.ec2.internal.warc.gz | 779,360,897 | 13,067 | Free Version
Easy
Sum in Unit Vector Notation
CLMECH-51S4YL
Find vector $\vec{r}$ which is the resultant of vectors $\vec{a}$ and $\vec{b}$:
$$\vec{a} = 3 \hat{i} - 7 \hat{j} + 2 \hat{k}$$
$$\vec{b} = - 5 \hat{j} -3 \hat{k}$$
A
$\vec{r} = -2 \hat{i} - 7 \hat{j} - \hat{k}$
B
$\vec{r} = 3 \hat{i} - 2 \hat{j} -5 \hat{k}$
C
$\vec{r} = 3 \hat{i} +2 \hat{j} +5 \hat{k}$
D
$\vec{r} = -2 \hat{i} - 10 \hat{j} +2 \hat{k}$
E
$\vec{r} = 3 \hat{i} - 12 \hat{j} - \hat{k}$ | 234 | 475 | {"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} | 3.390625 | 3 | CC-MAIN-2017-09 | latest | en | 0.372944 |
https://www.jiskha.com/display.cgi?id=1194842017 | 1,502,958,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102993.24/warc/CC-MAIN-20170817073135-20170817093135-00290.warc.gz | 899,743,040 | 3,565 | # math
posted by .
Find the lateral and surface area of the right rectangular prism whose length is 7 cm, width is 5 cm, and height is 1.5 cm. I got 36 as my lateral area and 106 as my surface area. Is this right?
## Similar Questions
1. ### math
Find the lateral and surface area of the right rectangluar prism whose length is 7 cm, width is 5 cm and height is 1.5 cm. I did 2(7 x 1.5+ 1.5 x 5) and I got 36 as my lateral area. Then I add 36+ 2(7 x5) and I got 106 as my surface …
2. ### math
Find the lateral and surface area of the right rectangular prism whose length is 7 cm, width is 5 cm, and height is 1.5 cm. I got 36 as my lateral area and 106 as my surface area. Is this right?
3. ### math
Find the lateral area of a regular pyramid whose base is a square, whose slant height is 5 m, whose height is 3 m. I think that the answer is 60. I subtracted 3 squared from 5 squared and I got 4. I multipled 4 by 2 and I got 8. I …
4. ### math
Find the lateral and surface area of the right rectangular prism whose length is 7 cm, width is 5 cm, and height is 1.5 cm. I think the answer is 36 for lateral area and 106 as surface area. I made a formula, 2(7 x 1.5 + 1.5 x 5)) …
5. ### math
How do i find the lateral area and the surface area of a rectangular prism with the height of 8.5" the width 4" the depth 3.25" I don't want you to do it for me i want to know what formula or how i start to solve for those two things …
6. ### geometry
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More Similar Questions | 641 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-34 | latest | en | 0.913077 |
https://www.itdaan.com/blog/2018/01/25/126beb3c319b86e6b70e1ec61e2106b4.html | 1,660,640,968,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00498.warc.gz | 698,155,048 | 6,811 | # Description
A network administrator manages a large network. The network consists
of N computers and M links between pairs of computers. Any pair of
computers are connected directly or indirectly by successive links, so
data can be transformed between any two computers. The administrator
finds that some links are vital to the network, because failure of any
one of them can cause that data can’t be transformed between some
computers. He call such a link a bridge. He is planning to add some
new links one by one to eliminate all bridges.
You are to help the administrator by reporting the number of bridges
# Input
The input consists of multiple test cases. Each test case starts with
a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤
200,000). Each of the following M lines contains two integers A and B
( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B.
Computers are numbered from 1 to N. It is guaranteed that any two
computers are connected in the initial network. The next line contains
a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links
the administrator plans to add to the network one by one. The i-th
line of the following Q lines contains two integer A and B (1 ≤ A ≠ B
≤ N), which is the i-th added new link connecting computer A and B.
The last test case is followed by a line containing two zeros.
# Output
For each test case, print a line containing the test case number(
beginning with 1) and Q lines, the i-th of which contains a integer
indicating the number of bridges in the network after the first i new
links are added. Print a blank line after the output for each test
case.
# Sample Input
``````3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
``````
# Sample Output
``````Case 1:
1
0
Case 2:
2
0
``````
# 代码
``````#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
#define inf 1000000
#define mem(a,b) memset(a,b,sizeof(a))
const int N=100000+7;
const int M=400000+5;
int dfn[N],low[N],tot,times;
int first[N],pre[N];
int bridge[N],cnt;
int n,m,k,q=1;
struct edge
{
int v,next;
} e[M];
{
e[tot].v=v;
e[tot].next=first[u];
first[u]=tot++;
}
void init()
{
mem(first,-1);
mem(dfn,0);
mem(low,0);
mem(bridge,0);
mem(pre,0);
tot=0;
cnt=0;
times=0;
}
void tarjan(int u,int fa)
{
low[u]=dfn[u]=++times;
int flag=0;
for(int i=first[u]; ~i; i=e[i].next)
{
int v=e[i].v;
if(v==fa&&!flag)
{
flag=1;
continue;
}
if(!dfn[v])
{
pre[v]=u;
tarjan(v,u);
low[u]=min(low[u],low[v]);
if(low[v]>dfn[u])
{
cnt++;
bridge[v]=1;
}
}
else
low[u]=min(low[u],dfn[v]);
}
}
void lca(int u,int v)
{
while(dfn[u]>dfn[v])
{
if(bridge[u])
{
cnt--;
bridge[u]=0;
}
u=pre[u];
}
while(dfn[v]>dfn[u])
{
if(bridge[v])
{
cnt--;
bridge[v]=0;
}
v=pre[v];
}
while(u!=v)
{
if(bridge[u])
{
cnt--;
bridge[u]=0;
}
if(bridge[v])
{
cnt--;
bridge[v]=0;
}
u=pre[u];
v=pre[v];
}
}
int main()
{
int u,v;
while(scanf("%d%d",&n,&m)&&(n||m))
{
init();
for(int i=1; i<=m; i++)
{
scanf("%d%d",&u,&v);
}
for(int i=1; i<=n; i++)
if(!dfn[i])
tarjan(i,-1);
printf("Case %d:\n",q++);
scanf("%d",&k);
while(k--)
{
scanf("%d%d",&u,&v);
lca(u,v);
printf("%d\n",cnt);
}
puts("");
}
return 0;
}
`````` | 1,067 | 3,348 | {"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} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.894601 |
https://www.calculatorway.com/bmi-calculator-in-kg-and-feet/ | 1,679,944,410,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00284.warc.gz | 751,722,578 | 68,409 | # Bmi Calculator in kg and Feet
You can easily calculate your BMI with this free BMI calculator in kg and feet. Find out the BMI value of your weight in kg and height in feet + inch.
Weight :
kg
Height :
Feet
Inch
## What is a BMI(body mass index)?
Body Mass Index is relative height to weight. It is used as a general indicator of body fatness for most adults. It was developed in the 1830s by a Belgian statistician, Lambert Adolphe Jacques Quetelet. In the past, it was only used by physicians to calculate their patients’ body fat and health.
A BMI of 25-29.9 is considered overweight and 30.0 or higher is considered obese. For example, a person who is 5 feet 4 inches tall and weighs 150 lbs would have a BMI of 24.9.
## The formula used a BMI calculator in kg and feet
• BMI = weight(kg) / height²(m) = kg/m²
• BMI = 703 x weight(lbs)/height²(inch) = kg/m²
• where lbs – pound
## Body Mass Index(BMI) Table
A BMI is a measure of Body Mass Index (BMI), which is the ratio of a person’s weight to height and is classified as either underweight or overweight. The ideal range for adults should be between 18.5 and 24.9 with the average being 23.2
## Benefits of body mass index calculator
The body mass index calculator is a useful tool to know about your fitness level and know if you are underweight, overweight or obese. Apart from this, being aware of your body mass index number is also useful for finding out the best weight loss diet that is suitable for you. The BMI is the weight of a person in kilograms divided by the square of his or her height in meters (kg/m2). It is used as an indicator of body fatness for most people, and it’s a reasonable indicator of body fatness for most people.
## How to use the BMI calculator in kg feet calculator?
Have you decided to do more in your own efforts to get healthier this year? A smart way to get started is by calculating your BMI. This will be a big help when you’re tackling the challenge of getting fit or healthy again, or when it’s time for your annual weigh-in at the doctor’s office. You can use the BMI calculator feet and kg to calculate your body mass index (BMI) and determine if you’re overweight or underweight by simply entering height and weight data into the provided fields. So type in the values of your height and weight below then click calculate to see in the answer box. | 560 | 2,368 | {"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-2023-14 | longest | en | 0.945866 |
http://www.gurufocus.com/term/pe/AFF/P%252FE%2BRatio/American%2BInternational%2BGroup%252C%2BInc | 1,418,906,034,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802766292.96/warc/CC-MAIN-20141217075246-00061-ip-10-231-17-201.ec2.internal.warc.gz | 619,704,420 | 25,624 | Switch to:
American International Group, Inc. (:AFF)
P/E Ratio
(As of Today)
As of today, American International Group, Inc.'s share price is \$24.99. American International Group, Inc. does not have enough years/quarters to calculate the diluted earnings per share for the trailing twelve months (TTM) ended in . 20. Therefore GuruFocus does not calculate P/E Ratio at this moment.
American International Group, Inc.'s diluted earnings per share (Diluted EPS) for the six months ended in . 20 was \$0.00.
As of today, American International Group, Inc.'s share price is \$24.99. American International Group, Inc. does not have enough years/quarters to calculate the earnings per share without non-recurring items for the trailing twelve months (TTM) ended in . 20. Therefore GuruFocus does not calculate P/E (NRI) Ratio at this moment.
American International Group, Inc.'s earnings per share without non-recurring items for the six months ended in . 20 was \$0.00.
During the past 12 months, American International Group, Inc.'s average earnings per share (NRI) Growth Rate was -99.00% per year.
American International Group, Inc.'s basic earnings per share (Basic EPS) for the six months ended in . 20 was \$0.00.
Definition
The P/E ratio is the most widely used ratio in the valuation of stocks.
American International Group, Inc.'s P/E Ratio for today is calculated as
P/E Ratio = Share Price / Earnings per Share (TTM) = 24.99 / =
It can also be calculated from the numbers for the whole company:
P/E Ratio = Market Cap / Net Income
There are at least three kinds of P/E ratios used by different investors. They are Trailing Twelve Month P/E Ratio or P/E (ttm), forward P/E, or P/E (NRI). A new P/E ratio based on inflation-adjusted normalized P/E ratio is called Shiller P/E, after Yale professor Robert Shiller.
In the calculation of P/E (ttm), the earnings per share used are the earnings per share over the past 12 months. For Forward P/E, the earnings are the expected earnings for the next twelve months. In the case of P/E (NRI), the reported earnings less the non-recurring items are used.
For the Shiller P/E, the earnings of the past 10 years are inflation-adjusted and averaged. The result is used for P/E calculation. Since it looks at the average over the last 10 years, Shiller P/E is also called PE10.
Explanation
The P/E ratio can be viewed as the number of years it takes for the company to earn back the price you pay for the stock. For example, if a company earns \$2 a share per year, and the stock is traded at \$30, the P/E ratio is 15. Therefore it takes 15 years for the company to earn back the \$30 you paid for its stock, assuming the earnings stays constant over the next 15 years.
In real business, earnings never stay constant. If a company can grow its earnings, it takes fewer years for the company to earn back the price you pay for the stock. If a companys earnings decline it takes more years. As a shareholder, you want the company to earn back the price you pay as soon as possible. Therefore, lower-P/E stocks are more attractive than higher P/E stocks so long as the P/E ratio is positive. Also for stocks with the same P/E ratio, the one with faster growth business is more attractive.
If a company loses money, the P/E ratio becomes mearningless.
To compare stocks with different growth rates, Peter Lynch invented a ratio called PEG. PEG is defined as the P/E ratio divided by the growth ratio. He thinks a company with a P/E ratio equal to its growth rate is fairly valued. Still he said he would rather buy a company growing 20% a year with a P/E of 20, instead of a company growing 10% a year with a P/E of 10.
Because the P/E ratio measures how long it takes to earn back the price you pay, the P/E ratio can be applied to the stocks across different industries. That is why it is the one of the most important and widely used indicators for the valuation of stocks.
Similar to the Price/Sales ratio and Price/Cash Flow or Price/Free Cash Flow, the P/E ratio measures the valuation based on the earning power of the company. This is where it is different from the Price/Book ratio, which measures the valuation based on the companys balance sheet.
Be Aware
Investors need to be aware that the P/E ratio can be misleading a lot of times, especially when the underlying business is cyclical and unpredictable. As Peter Lynch pointed out, cyclical businesses have higher profit margins at the peaks of the business cycles. Their earnings are high and P/E ratios are artificially low. It is usually a bad idea to buy a cyclical business when the P/E is low. A better ratio to identify the time to buy a cyclical businesses is the Price-to-Sales Ratio (P/S).
P/E ratio can also be affected by non-recurring-items such as the sale of part of businesses. This may increase for the current year or quarter dramatically. But it cannot be repeated over and over. Therefore P/E (NRI) is a more accurate indication of valuation than P/E (ttm).
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their own currency.
American International Group, Inc. Annual Data
pe 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
American International Group, Inc. Semi-Annual Data
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https://www.weegy.com/?ConversationId=82JU7YWO | 1,606,834,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00137.warc.gz | 890,458,163 | 10,971 | Factor x^2 – 2x – 80. A. (x – 8)(x + 10) B. (x + 6)(x – 1) C. (x + 3)(x + 6) D. (x + 8)(x – 10)
x^2 – 2x – 80 = (x + 8)(x – 10)
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Updated 9/28/2018 3:26:26 AM
Edited by jerry06 [9/28/2018 3:26:26 AM]
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x^2 – 2x – 80 = (x + 8)(x – 10)
Questions asked by the same visitor
Determine the difference: 3.2 × 10^10 – 1.1 × 10^10. Write your answer in scientific notation. A. 2.1 × 10^10 B. 2.1 × 100 C. 4.3 × 10^10 D. 4.3 × 100
Question
Updated 12/28/2018 6:32:28 AM
3.2 × 10^10 – 1.1 × 10^10 = 2.1 × 10^10
The difference: 3.2 × 10^10 – 1.1 × 10^10. in scientific notation. is: "2.1 × 10^10"
2. Factor x2 – 2x – 80. A. (x – 8)(x + 10) B. (x + 6)(x – 1) C. (x + 3)(x + 6) D. (x + 8)(x – 10)
Question
Updated 9/27/2018 8:57:52 PM
x^2 – 2x – 80 =x^2 – 10x + 8x – 80 = x( x -10) +8(x-10)= (x + 8)(x – 10)
32,763,150
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 962 | 2,521 | {"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-2020-50 | latest | en | 0.878303 |
https://mathcentre.co.uk/topics/vectors/scalar-product/ | 1,721,113,430,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514737.55/warc/CC-MAIN-20240716050314-20240716080314-00709.warc.gz | 331,913,053 | 4,347 | Accessibility options:
The scalar product resources
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iPOD Video (5)
Vectors - Scalar Product 1
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Vectors - Scalar Product 2
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Vectors - Scalar Product 3
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Vectors - Scalar Product 4
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Vectors - Scalar Product 5
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Quick Reference (2)
The scalar product
This leaflet defines the scalar product of two vectors and gives some examples. It shows how the scalar product can be used to find the angle between two vectors. (Engineering Maths First Aid Kit 6.2)
Vectors
This leaflet explains notations in common use for describing vectors, and shows how to calculate the modulus of vectors given in Cartesian form. (Engineering Maths First Aid Kit 6.1)
Teach Yourself (1)
The scalar product
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector.
Test Yourself (3)
Dot and cross product - Numbas
5 questions on vectors. Scalar product, angle between vectors, cross product, when are vectors perpendicular, combinations of vectors defined or not. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University.
Maths EG
Computer-aided assessment of maths, stats and numeracy from GCSE to undergraduate level 2. These resources have been made available under a Creative Common licence by Martin Greenhow and Abdulrahman Kamavi, Brunel University.
Vector Test 01 (DEWIS)
Five questions on vectors, testing addition, subtraction, scalar multiplication, magnitude, scalar product, vector product and finding the angle between two vectors. DEWIS resources have been made available under a Creative Commons licence by Rhys Gwynllyw & Karen Henderson, University of the West of England, Bristol.
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The scalar product
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. (Mathtutor Video Tutorial) This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
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The scalar product
One of the ways in which two vectors can be combined is known as the scalar product. When we calculate the scalar product of two vectors the result, as the name suggests is a scalar, rather than a vector. (Mathtutor Video Tutorial) The video is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd. | 934 | 4,572 | {"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-2024-30 | latest | en | 0.94289 |
https://lists.cs.wisc.edu/archive/htcondor-users/2011-September/msg00112.shtml | 1,675,499,774,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500095.4/warc/CC-MAIN-20230204075436-20230204105436-00696.warc.gz | 401,744,699 | 2,974 | # Re: [Condor-users] Condor Slots Understanding Question....
• Date: Wed, 14 Sep 2011 15:26:33 -0400
• From: Ian Chesal <ichesal@xxxxxxxxxxxxxxxxxx>
• Subject: Re: [Condor-users] Condor Slots Understanding Question....
On Wednesday, 14 September, 2011 at 3:15 PM, Sassy Natan wrote:
I'm kind of a new in this filed.....
Anyway I need to know something regarding the formula u point out
NEGOTIATOR_POST_JOB_RANK = (RemoteOwner =?= UNDEFINED) * (KFlops - SlotID)
I understand the NEGOTIATOR_POST_JOB_RANK value from the Condor document.
However I can't understand the output :
The (RemoteOwner =?= UNDEFINED) check the string values and if RemoteOwner is defied. So if so the value will be FLASE this will always give me 0.
Can u explain more this do....
I can't understand this.
First a bit about how these ranks work: higher is better. So a machine that evaluates to a higher number than all other machines will get used for the match. So we're writing these expressions to push the machines we want to match the jobs first up to the top of the list.
The _expression_ (RemoteOwner =?= UNDEFINED) * (KFlops - SlotID) breaks down in to two parts:
(RemoteOwner =?= UNDEFINED)
-and-
(KFlops - SlotID)
The first part is binary: it's going to be 1 or 0. It's 0 if RemoteOwner exists on the slot's ClassAd (i.e. if the slot is currently running a job). So this sets the NEGOTIATOR_POST_JOB_RANK to zero, irregardless of anything else, if the slot is occupied. You don't want to consider those slots at all if possible.
The second part of the _expression_ is going to take the lowest slot (SlotID = 1) on the highest powered machine (KFlops is biggest) and put that match on the top of the list.
Make sense?
Hope that helps.
Regards,
- Ian
---
Ian Chesal
Cycle Computing, LLC
Leader in Open Compute Solutions for Clouds, Servers, and Desktops
Enterprise Condor Support and Management Tools
http://www.cyclecomputing.com
http://www.cyclecloud.com | 520 | 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} | 2.53125 | 3 | CC-MAIN-2023-06 | longest | en | 0.848039 |
https://imathworks.com/physics/physics-writing-dotq-in-terms-of-p-in-the-hamiltonian-formulation/ | 1,723,576,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00896.warc.gz | 252,890,387 | 7,159 | # Hamiltonian Formalism – Writing $\dot{q}$ in Terms of $p$ in Hamiltonian Formulation
constrained-dynamicshamiltonian-formalismlagrangian-formalism
In the Hamiltonian formulation, we make a Legendre transformation of the Lagrangian and it should be written in terms of the coordinates $q$ and momentum $p$. Can we always write $dq/dt$ in terms of $p$? Is there any case in which we obtain for example a transcendental equation and cannot do it?
Adding to Lubos Motl's correct answer, it should be stressed that one may not always invert the relation $p_i=f_i(q,\dot{q},t)$ to isolate $\dot{q}^j$, not even in principle, because of constraints. Such cases are known as singular Legendre transformations, and they are the starting point of the topic of constrained dynamics.
Example. Consider e.g. the Lagrangian
$$\tag{1} L~=~ y\dot{x} - \frac{y^2}{2m}$$
with two dynamical variables $x$ and $y$. (This Lagrangian (1) is in fact the so-called Hamiltonian Lagrangian for a non-relativistic 1D free particle if we identify the variable $y$ with the particle's momentum $p_x$, cf. the Faddeev-Jackiw method, but let's imagine we don't know that.)
Now let's go through the Legendre transformation via the Dirac-Bergmann method. The momenta are
$$\tag{2} p_x~:=~\frac{\partial L}{\partial \dot{x}}~=~y,$$
and
$$\tag{3} p_y~:=~\frac{\partial L}{\partial \dot{y}}~=~0.$$
Obviously, we cannot invert eqs. (2) and (3) to find $\dot{x}$ in terms of $x,y,p_x,p_y$ and $t$. Eqs. (2) and (3) are primary constraints,
$$\tag{4} p_x-y~\approx~ 0\quad\text{and} \quad p_y~\approx~ 0.$$
The Hamiltonian is
$$\tag{5} H~=~p_x \dot{x} +p_y \dot{y} - L ~\approx~ \frac{p^2_x}{2m},$$
where the $\approx$ symbol means equality modulo constraints (4). Indeed, it is just a non-relativistic 1D free particle, where we have used the constraints (4) to eliminate any physics in the $y$-direction. (It is easy to check that there are no secondary constraints if we use $\frac{p^2_x}{2m}$ as the Hamiltonian.) | 603 | 1,995 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.828012 |
https://documen.tv/question/how-to-learn-tables-11-to-15-for-12-years-old-easyliy-please-tell-guys-24113860-66/ | 1,719,102,994,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00644.warc.gz | 188,658,346 | 16,137 | ## How to learn tables 11 to 15 For 12 years old easyliy please tell guys
Question
How to learn tables 11 to 15
For 12 years old easyliy please tell guys
in progress 0
3 years 2021-08-24T22:16:34+00:00 1 Answers 0 views 0
For eleven, it had the same two numbers. For twelve, it continues up by even numbers. For thirteen, it’s the same as twelve but by odd numbers. For fourteen, it’s a little difficult but they have a pattern for every five multiples of fourteen. For fifteen, the numbers add by five but also counting up.
Step-by-step explanation:
Examples for the table of eleven would be: 11 × 3 = 33 or 11 × 6 = 66.
Table twelve would be (as examples): 12 × 2 = 24 and 12 × 3 = 36.
Table thirteen have numbers/answers like: 13 × 7 = 91 and 13 × 8 = 104.
Table fourteen as examples: 14 × 3 = 42 or 14 × 4 = 56.
Table fifteen would be like: 15 × 6 = 90 and 15 × 7 = 105. | 280 | 888 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-26 | latest | en | 0.93775 |
http://www.goodwillsc.org/bhok28/b3793b-actuarial-annuity-formula | 1,627,932,799,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00144.warc.gz | 63,537,502 | 10,966 | p • An annuity may be payable in advance instead of in arrears, in which case it is called an annuity-due. is the probability density function of T, In this chapter, we will concentrate on the basic level annuity. x by (/iropracy . Finally, let Z be the present value random variable of a whole life insurance benefit of 1 payable at time T. Then: x Haberman, Steven and Trevor A. Sibbett, History of Actuarial … In practice the benefit may be payable at the end of a shorter period than a year, which requires an adjustment of the formula. Suppose the death benefit is payable at the end of year of death. {\displaystyle \mu _{x+t}} Since T is a function of G and x we will write T=T(G,x). A basic level annuity … Retirement planning typically focuses on … Actuarial observations can provide insight into the risks inherent in lifetime income planning for retirees and the methods used to possibly optimize retirees’ income. G�����K����um��듗w��*���b�i&GU�G��[qi��e+��pS'�����ud]��M��g-����S�7���\����#��y�������N�MvH����Ա&1�O#X�a��M�u.�S��@�? 0000000016 00000 n 8� @ɠ w����Y����[��)8�{��}����� ��=v��K����YV����x8�[~p�S������]}T�6rmz��g��I��v������^x�aekJ'*-Q������Jv��w�)���fr��gm�Yz0�;���^�L�#��L5k Sv���*���9�!&�ɷ�f� �����60. Find expression for the variance of the present value random variable. If the benefit is payable at the moment of death, then T(G,x): = G - x and the actuarial present value of one unit of whole life insurance is calculated as. or 0000003752 00000 n x so the actuarial present value of the $100,000 insurance is$24,244.85. The actuarial present value of one unit of an n-year term insurance policy payable at the moment of death can be found similarly by integrating from 0 to n. The actuarial present value of an n year pure endowment insurance benefit of 1 payable after n years if alive, can be found as, In practice the information available about the random variable G (and in turn T) may be drawn from life tables, which give figures by year. t Let G>0 (the "age at death") be the random variable that models the age at which an individual, such as (x), will die. q a "loss" of payment for on average half a period. • We denote the present value of the annuity-due at time 0 by ¨anei (or ¨ane), and the future value of the annuity … and Nesbitt, C.J., Chapter 4-5, Models for Quantifying Risk (Fourth Edition), 2011, By Robin J. Cunningham, Thomas N. Herzog, Richard L. London, Chapter 7-8, This page was last edited on 3 December 2019, at 16:11. xref The symbol (x) is used to denote "a life aged x" where x is a non-random parameter that is assumed to be greater than zero. x 0000002759 00000 n %PDF-1.4 %���� 245 10 Whole life insurance pays a pre-determined benefit either at or soon after the insured's death. x denotes force of mortality at time The actuarial symbols for accumulations and present values are modified by placing a pair of dots over the s or a. The expected value of Y is: Current payment technique (taking the total present value of the function of time representing the expected values of payments): where F(t) is the cumulative distribution function of the random variable T. The equivalence follows also from integration by parts. A life annuity is an annuity whose payments are contingent on the continuing life of the annuitant. t International Actuarial Notation125 . This tool is designed to calculate relatively simple annuity … Keeping the total payment per year equal to 1, the longer the period, the smaller the present value is due to two effects: Conversely, for contracts costing an equal lumpsum and having the same internal rate of return, the longer the period between payments, the larger the total payment per year. The actuarial present value of a life annuity of 1 per year paid continuously can be found in two ways: Aggregate payment technique (taking the expected value of the total present value): This is similar to the method for a life insurance policy. Thus: an annuity payable so long as at least one of the three lives (x), (y) and (z) is alive. {\displaystyle \,{\overline {A}}_{x}} surviving to age %%EOF The accrual formula could be based on … �h���s��:6l�4ԑ���z���zr�wY����fF{����u�% Let G>0 (the "age at death") be the random variable that models the age at which an individual, such as (x), will die. μ t t x Makeham's formula: A = K+p(I-t)(C-K) g where: A is the present value of capital and net interest payments; K is the present value of capital payments; C is the total capital to be repaid (at redemption price); g is the rate of interest expressed per unit of the redemption price; t is the rate of tax on interest. There is no proportional payment for the time in the period of death, i.e. The formulas described above make it possible—and relatively easy, if you don't mind the math—to determine the present or future value of either an ordinary annuity or an annuity due. The probability of a future payment is based on assumptions about the person's future mortality which is typically estimated using a life table. Whole life insurance pays a pre-determined benefit either at or soon after the insured's death. Since T is a function of G and x we will write T=T(G,x). {\displaystyle x} The actuarial present value of one unit of whole life insurance issued to (x) is denoted by the symbol $${\displaystyle \,A_{x}}$$ or $${\displaystyle \,{\overline {A}}_{x}}$$ in actuarial notation. in actuarial notation. ¯ To determine the actuarial present value of the benefit we need to calculate the expected value A variable annuity fluctuates with the returns on the mutual funds it is invested in. Finally, let Z be the present value random variable of a whole life insurance benefit of 1 payable at time T. Then: where i is the effective annual interest rate and δ is the equivalent force of interest. Here we present the 2017 period life table for the Social Security area population.For this table, … number appears over the bar, then unity is supposed and the meaning is at least one survivor. �'����I�! {\displaystyle \,_{t}p_{x}} 0000003675 00000 n Exam FM/2 Interest Theory Formulas . The symbol (x) is used to denote "a life aged x" where x is a non-random parameter that is assumed to be greater than zero. {\displaystyle x} A fixed annuity guarantees payment of a set amount for the term of the agreement. It can't go down (or up). + Actuarial present value factors for annuities, life insurance, life expectancy; plus commutation functions, tables, etc. . The annuity payment formula is used to calculate the periodic payment on an annuity. And let T (the future lifetime random variable) be the time elapsed between age-x and whatever age (x) is at the time the benefit is paid (even though (x) is most likely dead at that time). of this random variable Z. 254 0 obj<>stream {\displaystyle x+t} xڴV}P�����$|��͒@��.1�бK�D>�&*ڠ=�!�a�LPIEA� z��8�����Ǎp���G[:Ci;s�י����wf���}���=�����Q!�B���v(Z� is the probability that (x+t) dies within one year. {\displaystyle \,E(Z)} x Thus if the annual interest rate is 12% then $${\displaystyle \,i=0.12}$$. Value of annuity … Rate Per Period As with any financial formula that involves a rate, it is important to make sure that the rate is consistent with the other variables in the formula. ) T x In practice life annuities are not paid continuously. The APV of whole-life assurance can be derived from the APV of a whole-life annuity-due this way: In the case where the annuity and life assurance are not whole life, one should replace the assurance with an n-year endowment assurance (which can be expressed as the sum of an n-year term assurance and an n-year pure endowment), and the annuity with an n-year annuity due. For an n-year life annuity-immediate: Find expression for the present value random variable. Then T(G, x) := ceiling(G - x) is the number of "whole years" (rounded upwards) lived by (x) beyond age x, so that the actuarial present value of one unit of insurance is given by: where This is a collaboration of formulas for the interest theory section of the SOA Exam FM / CAS Exam 2. 0000000496 00000 n An annuity is a series of periodic payments that are received at a future date. The payments are made on average half a period later than in the continuous case. Z For example, a three year term life insurance of$100,000 payable at the end of year of death has actuarial present value, For example, suppose that there is a 90% chance of an individual surviving any given year (i.e. + f a series of payments which may or may not be made). A large library of mortality tables and mortality improvement scales. ; Ability to use generational mortality, and the new 2-dimensional rates in Scale BB-2D, MP-2014, MP-2015, MP-2016, MP-2017, or MP-2018. trailer t A A for a life aged 0000003070 00000 n The Society of Actuaries (SOA) developed the Annuity Factor Calculator to calculate an annuity factor using user-selected annuity forms, mortality tables and projection scales commonly used for defined benefit pension plans in the United States or Canada. This study sheet is a free non-copyrighted … This tool is designed to calculate relatively simple annuity factors for users who are accustomed to making actuarial … <]>> For example, a temporary annuity … 0000003482 00000 n x A variable annuity plan is usually a career accumulation plan in which the plan document defines the amount of benefit that accrues to a participant each year. And let T (the future lifetime random variable) be the time elapsed between age-x and whatever age (x) is at the time the benefit is paid (even though (x) is most likely dead at that time). x The last displayed integral, like all expectation formulas… an annuity … 0000002843 00000 n {\displaystyle x+t} A period life table is based on the mortality experience of a population during a relatively short period of time. You have 20 years of service left and you … Then, and at interest rate 6% the actuarial present value of one unit of the three year term insurance is. Life assurance as a function of the life annuity, https://en.wikipedia.org/w/index.php?title=Actuarial_present_value&oldid=929088712, Creative Commons Attribution-ShareAlike License. Actuarial Mathematics 1: Whole Life Premiums and Reserves: Actuarial Mathematics 1: Joint Life Annuities: Actuarial Mathematics 2: Comparing Tails via Density and Hazard Functions: Loss Models … + The actuarial present value of one unit of whole life insurance issued to (x) is denoted by the symbol A quick video to show you how to derive the formulas for an annuity due. EAC Present Value Tools is an Excel Add-in for actuaries and employee benefit professionals, containing a large collection of Excel functions for actuarial present value of annuities, life insurance, life expectancy, actuarial … and ( If the payments are made at the end of each period the actuarial present value is given by. Actuarial Mathematics (Second Edition), 1997, by Bowers, N.L., Gerber, H.U., Hickman, J.C., Jones, D.A. + The age of the annuitant is an important consideration in calculating the actuarial present value of an annuity… Annuity Formula – Example #2 Let say your age is 30 years and you want to get retired at the age of 50 years and you expect that you will live for another 25 years. The Society of Actuaries (SOA) developed the Annuity Factor Calculator to calculate an annuity factor using user-selected annuity forms, mortality tables and projection scales commonly used for defined benefit pension plans in the United States or Canada. The present value portion of the formula … premium formula, namely the pure n-year endowment. 0000002983 00000 n startxref • An annuity-due is an annuity for which the payments are made at the beginning of the payment periods • The first payment is made at time 0, and the last payment is made at time n−1. July 10, 2017 10:32 Financial Mathematics for Actuaries, 2nd Edition 9.61in x 6.69in b3009-ch02 page 42 42 CHAPTER2 Example 2.2: Calculate the present value of an annuity-immediate of amount $100 paid annually for5years attherateofinterest of9%perannum using formula 0000004196 00000 n The present value of annuity formula relies on the concept of time value of money, in that one dollar present day is worth more than that same dollar at a future date. is the probability of a life age 0 The proofs are rather similar to the annuity immediate proofs. {\displaystyle f_{T}} {\displaystyle {}_{t}p_{x}} B��屏����#�,#��������'+�8#����ad>=��:��ʦ0s��}�G�o��=x��z��L���s_6�t�]wU��F�[��,M�����52�%1����2�xQ9�)�;�VUE&�5]sg�� 245 0 obj <> endobj The actuarial present value (APV) is the expected value of the present value of a contingent cash flow stream (i.e. {\displaystyle \,A_{x}} T has a geometric distribution with parameter p = 0.9 and the set {1, 2, 3, ...} for its support). The value of an annuity at the valuation date is the single sum value at the valuation date in which one is indifferent to receiving instead of receiving the periodic payments that form the annuity. p where Express formulas for its actuarial present value or expectation. {\displaystyle \,q_{x+t}} t Actuarial present values are typically calculated for the benefit-payment or series of payments associated with life insurance and life annuities. For an n-year deferred whole life annuity … The expected present value of$1 one year in the future if the policyholder aged x is alive at that time is denoted in older books as nEx and is called the actuarial … This time the random variable Y is the total present value random variable of an annuity of 1 per year, issued to a life aged x, paid continuously as long as the person is alive, and is given by: where T=T(x) is the future lifetime random variable for a person age x. Each of the following annuities-due have an actuarial PV of 60,000: (1) life annuity-due of 7,500 on (25) (2) life annuity-due of 12,300 on (35) (3) life annuity-due of 9,400 on (25) that makes at most 10 … "j����>���gs�|��0�=P��8�"���r��p��#vp@���-x�@=@ׇ��h�,N��I��c�~˫����r� k���T��Ip�\��,���]�mƇ�FG��븅l� �*~��j����p,�H��!�벷��-�Іo�לV��u>b�dO�z ��hZn��Aq�"��Gnj��a�a�e���oܴE�:ƺ��i�k�,�SmD��n)�M������nQf��+� �cu�j6��r�k�H�Z��&s���='Ğ��v�o�.f=3���u Ciecka: The First Mathematically Correct Life Annuity Valuation Formula 63 References De Witt, Jan, Value of Life Annuities in Proportion to Redeemable Annui- ties, 1671, published in Dutch with an English translation in Hendricks (1852, 1853). is the probability that (x) survives to age x+t, and $${\displaystyle \,i}$$ is the annual effective interest rate, which is the "true" rate of interest over a year. E ���db��8��m��LO�aK��*߃��j���%�q�d ���%�rd�����]4UY�BC��K37L�ל�l�*�F0��5C'i�F�"��x�siɓ�(�@�,>R�t ����1��:HUv:�]u8�}�JK }�6�����#N�\���X�$�q��8��) �����.�m��>�:Jv�W���^��,�h��eDd��r,)��c�|x0(�u�y]#)r���_����iWZ'"Pd��� ;:?\0$Q��i�I���-��������3�4���+�ti�b�%{��W92b�"��-(1^\�lIs����Ғ��ݱ2�C�l�Lse"���?�FG#�_�����/�F��l��Z����u�_ӟ�}s�=Ik�ޮl�_�*7Q�kP?kWj�x�o]���đ�6L����� �d �2E�EOٳ�{#z���wg(U5^�]�����pp�o�4�ߍ��h�uU{iZ�JoE�/�o�8����-��-s���R�r7x2-��p�(�Ly���Ï�/���Ws��������b��M�2�2q�kU�p��3j����1��� �ZE |�IL&��������[��Eݷ�BD=S ��U���E� �T;�5w�#=��a�rP1X]�p�?9��H��N��U��4?��N9@�Z��f�"V%��٠�8�\]4LPFkE��9�ɿ4?WX?���ӾoM� Commons Attribution-ShareAlike License APV ) is the expected value of a future payment based. Present values are typically calculated for the time in the period of death about person! In which case it is called an annuity-due life annuity, https //en.wikipedia.org/w/index.php... Payments which may or may not be made ) a future payment is based on about! 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Designed to calculate relatively simple annuity … Exam FM/2 interest Theory formulas example, temporary! Relatively simple annuity factors for users who are accustomed to making actuarial International! Benefit is payable at the end of year of death its actuarial present value of the three year insurance! It ca n't go down ( or up ) the end of each period the symbols... Annuity is a series of payments associated with life insurance pays a pre-determined benefit either or. Function of G and x we will concentrate on the basic level annuity SOA Exam FM / CAS 2! Received at a future payment is based on assumptions about the person 's future mortality is. Based on assumptions about the person 's future mortality which is typically using. An annuity … Exam FM/2 interest Theory section of the SOA Exam FM / CAS Exam 2 for on half. | 6,899 | 24,806 | {"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} | 3.21875 | 3 | CC-MAIN-2021-31 | latest | en | 0.768534 |
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Syntax for Nested Functions
Posted 9 years ago
I'm hoping this is a relatively simple question, but I'm quite new to Mathematica and am struggling with exactly how to express what I'd like it to do. The following isn't what I'm trying to do, but a very simplified example with the same fundamental issue. We want to calculate the total weight of a collection of balls 1, 2,...,N. There are only a certain number of types of ball, 1, 2,...T. Ball number x is of type balltype[x]. The weight of a ball of type x is weight[x]. So the weight of ball number x is expressed by weight[balltype[x]]. I want to define the function that is the total weight of all N balls, and it should be something like: totalweight[weight_, balltype_, N_] := Sum[weight[balltype[x]], {x, 1, N}] The problem is that I'm not now sure how then to ask for the derivative of totalweight with respect to the weight of a specific type of ball, say weight[1]. The answer should be something like Sum[Boole[balltype[x]=1],{x,1,N}] i.e. the total number of balls of type 1 that are in the collection, although there are may be many ways to express this. My question is: how can I express the function totalweight properly, and how (for example) would I ask for the derivative of totalweight wrt the weight of a particular ball type weight[x]? This seems like a simple sort of problem, but I can't seem to find anything addressing it, although that could be because I don't know the right terms to search for. Very many thanks in advance.
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Posted 9 years ago
To understand your problem, I made small example and set up a function to find the total weight of all balls of a certain type: In[3]:= SeedRandom[1234](*10 ball types with a different weight each \ between 10 and 50*) ballWeightsByType = RandomReal[{10, 50}, 10] Out[4]= {45.0643, 30.8786, 13.4489, 25.1165, 10.4658, 47.0906, \ 31.7503, 29.1733, 19.814, 40.3958} In[5]:= (*random collection of 50 balls identified by their ball type*) In[6]:= ballsByType = RandomInteger[{1, 10}, 50] Out[6]= {3, 9, 9, 1, 5, 8, 4, 7, 6, 4, 8, 10, 1, 5, 5, 3, 3, 5, 8, 6, \ 9, 3, 3, 7, 8, 3, 6, 3, 1, 3, 8, 10, 6, 7, 8, 1, 10, 6, 8, 8, 8, 10, \ 8, 8, 6, 1, 6, 5, 4, 10} In[7]:= (*get weight of balltype x*) In[8]:= ballWeight[x_?IntegerQ] := First@Take[ballWeightsByType, {x}] In[9]:= ballWeight[7] Out[9]= 31.7503 In[10]:= (*list of {type,weight} of all 50 balls sorted by type*) In[11]:= lst = SortBy[{#, ballWeight[#]} & /@ ballsByType, First] Out[11]= {{1, 45.0643}, {1, 45.0643}, {1, 45.0643}, {1, 45.0643}, {1, 45.0643}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {3, 13.4489}, {4, 25.1165}, {4, 25.1165}, {4, 25.1165}, {5, 10.4658}, {5, 10.4658}, {5, 10.4658}, {5, 10.4658}, {5, 10.4658}, {6, 47.0906}, {6, 47.0906}, {6, 47.0906}, {6, 47.0906}, {6, 47.0906}, {6, 47.0906}, {6, 47.0906}, {7, 31.7503}, {7, 31.7503}, {7, 31.7503}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {8, 29.1733}, {9, 19.814}, {9, 19.814}, {9, 19.814}, {10, 40.3958}, {10, 40.3958}, {10, 40.3958}, {10, 40.3958}, {10, 40.3958}} In[12]:= (*number of balls of type 7*) In[13]:= Count[lst, {7, _}] Out[13]= 3 In[14]:= (*weight of all balls of type 6*) In[15]:= % ballWeight[7] Out[15]= 31.7503 Null In[16]:= weightTotalByType[type_?IntegerQ] := Count[lst, {type, _}]*ballWeight[type] In[17]:= weightTotalByType[7] Out[17]= 95.2508 In[18]:= (*list of {type,total weight of all balls of type} of all 50 \ balls sorted by type*) In[19]:= {#, weightTotalByType[#]} & /@ Range[10] Out[19]= {{1, 225.322}, {2, 0.}, {3, 107.591}, {4, 75.3496}, {5, 52.3289}, {6, 329.635}, {7, 95.2508}, {8, 320.906}, {9, 59.4419}, {10, 201.979}} Attachments:
Posted 9 years ago
Many thanks for this - yes, this is exactly the sort of problem I am thinking about. However, I don't want to analyse a specific set of weights and types of ball. I want to leave them as free variables - weight[1], balltype[3] etc., and then manipulate the 'totalweight' function using those variables, to (for example) calculate the derivative of totalweight with respect to the weight of a specific ball type (e.g. weight[1]). I am trying to do this using statements like D[totalweight[weight, balltype, N], weight[1]] but I'm certain this is not well-formed, and it returns the output '0'.(The problem I'm actually looking at is to derive maximisation conditions for a big PDF, which expresses the outcome of a series of experiments, where each experiment has an individual PDF depending on its 'type' - but I didn't want to complicate things when the problem I've got is basically a syntactical one.) | 1,746 | 4,791 | {"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} | 3.203125 | 3 | CC-MAIN-2024-33 | latest | en | 0.93241 |
https://whatisconvert.com/152-feet-second-in-kilometers-hour | 1,643,428,490,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00238.warc.gz | 637,729,295 | 7,585 | # What is 152 Feet/Second in Kilometers/Hour?
## Convert 152 Feet/Second to Kilometers/Hour
To calculate 152 Feet/Second to the corresponding value in Kilometers/Hour, multiply the quantity in Feet/Second by 1.0972799999991 (conversion factor). In this case we should multiply 152 Feet/Second by 1.0972799999991 to get the equivalent result in Kilometers/Hour:
152 Feet/Second x 1.0972799999991 = 166.78655999987 Kilometers/Hour
152 Feet/Second is equivalent to 166.78655999987 Kilometers/Hour.
## How to convert from Feet/Second to Kilometers/Hour
The conversion factor from Feet/Second to Kilometers/Hour is 1.0972799999991. To find out how many Feet/Second in Kilometers/Hour, multiply by the conversion factor or use the Velocity converter above. One hundred fifty-two Feet/Second is equivalent to one hundred sixty-six point seven eight seven Kilometers/Hour.
## Definition of Foot/Second
The foot per second (plural feet per second) is a unit of both speed (scalar) and velocity (vector quantity, which includes direction). It expresses the distance in feet (ft) traveled or displaced, divided by the time in seconds (s, or sec). The corresponding unit in the International System of Units (SI) is the metre per second. Abbreviations include ft/s, ft/sec and fps, and the rarely used scientific notation ft s−1.
## Definition of Kilometer/Hour
The kilometre per hour (American English: kilometer per hour) is a unit of speed, expressing the number of kilometres travelled in one hour. The unit symbol is km/h. Worldwide, it is the most commonly used unit of speed on road signs and car speedometers. Although the metre was formally defined in 1799, the term "kilometres per hour" did not come into immediate use – the myriametre (10,000 metres) and myriametre per hour were preferred to kilometres and kilometres per hour.
## Using the Feet/Second to Kilometers/Hour converter you can get answers to questions like the following:
• How many Kilometers/Hour are in 152 Feet/Second?
• 152 Feet/Second is equal to how many Kilometers/Hour?
• How to convert 152 Feet/Second to Kilometers/Hour?
• How many is 152 Feet/Second in Kilometers/Hour?
• What is 152 Feet/Second in Kilometers/Hour?
• How much is 152 Feet/Second in Kilometers/Hour?
• How many km/h are in 152 ft/s?
• 152 ft/s is equal to how many km/h?
• How to convert 152 ft/s to km/h?
• How many is 152 ft/s in km/h?
• What is 152 ft/s in km/h?
• How much is 152 ft/s in km/h? | 630 | 2,451 | {"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} | 3.421875 | 3 | CC-MAIN-2022-05 | latest | en | 0.879562 |
https://studyres.com/doc/1265520/water-potential-math | 1,620,832,621,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990929.24/warc/CC-MAIN-20210512131604-20210512161604-00021.warc.gz | 580,352,072 | 9,416 | Survey
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```AP Water Potential Math
Name:
Introduction:
Water potential is the measure of water’s potential energy or it’s ability to do work. Water potential quantifies
the tendency of water to move from one area to another due to osmosis, gravity, mechanical pressure, or matrix
effects such as capillary action (which is caused by surface tension).
Water potential integrates a variety of different potential drivers of water movement, which may operate in the
same or different directions. Within complex biological systems, it is common for many potential factors to be
important. For example, the addition of solutes to water lowers the water's potential (makes it more negative),
just as the increase in pressure increases its potential (makes it more positive). If there is no restriction on flow,
water will move from an area of higher water potential to an area that has a lower water potential.
Ψ = Ψ𝑠 + Ψ𝑝
Pressure potential is based on mechanical pressure, and is an important component of the total water potential
within plant cells. Pressure potential increases as water enters a cell. As water passes through the cell wall and
cell membrane, it increases the total amount of water present inside the cell, which exerts an outward pressure
that is opposed by the structural rigidity of the cell wall. By creating this pressure, the plant can maintain turgor,
which allows the plant to keep its rigidity. Without turgor, plants lose structure and wilt. The pressure potential in
a living plant cell is usually positive. In plasmolysed cells, pressure potential is almost zero. Negative pressure
potentials occur when water is pulled through an open system such as a plant xylem vessel. Withstanding
negative pressure potentials (frequently called tension) is an important adaptation of xylem.
Ψ𝑠 = −𝑖𝐶𝑅𝑇
Osmotic potential has important implications for many living organisms. If a living cell is surrounded by a more
concentrated solution, the cell will tend to lose water to the more negative water potential of the surrounding
environment. A soil solution also experiences osmotic potential. The osmotic potential is made possible due to the
presence of both inorganic and organic solutes in the soil solution. As water molecules increasingly clump around
solute ions or molecules, the freedom of movement, and thus the potential energy, of the water is lowered. As
the concentration of solutes is increased, the osmotic potential of the soil solution is reduced. Since water has a
tendency to move toward lower energy levels, water will want to travel toward the zone of higher solute
concentrations. Osmotic potential has an extreme influence on the rate of water uptake by plants.
a) addition of solutes on right side reduces water potential. S = -0.23. Water flows from hypotonic to
hypertonic or from high on left to low on right.
b) adding +0.23 pressure with plunger creates no net flow of water.
c) applying +0.30 pressure increases water potential solution now has of +0.07. Water moves right to left
d) negative pressure or tension using plunger decreases water potential on the left. Water moves from right
to left
Remember water always moves from [high] to [low].
Water moves from hypotonic to hypertonic.
[Solute] is related to osmotic pressure. Pressure is related to pressure potential.
Pressure raises water potential.
When working problems, use zero for pressure potential in animal cells & open beakers.
1 bar of pressure = 1 atmosphere
Directions: Answer the following questions using the equations on the front page. Keep in mind that is
measured in megapascals (MPa) and 1 Mpa = 10 atmospheres of pressure.
1. If a plant cell’s P = 2 bars and its S = -3.5 bars, what is the resulting ?
a. The plant cell from question 1 is placed in a beaker of sugar water with S = -4.0 bars. In which
direction will the net flow of water be?
b. The original cell from question 1 is placed in a beaker of sugar water with S = -0.15MPa
(megapascals). We know that 1 MPa = 10 bars. In which direction will the net flow of water be?
2. The value for in root tissue was found to be -3.3 bars. If you place the root tissue in a 0.1 M solution of
sucrose at 20°C in an open beaker, what is the of the solution, and in which direction would the net flow of
water be?
a. NaCl dissociates into 2 particles in water: Na+ and Cl-. If the solution in question 2 contained 0.1 M
NaCl instead of 0.1 M sucrose, what is the of the solution, and in which direction would the net
flow of water be?
3. A plant cell with a s of -7.5 bars keeps a constant volume when immersed in an open-beaker solution that
has a s of -4 bars. What is the cell’s P?
4. At 20°C, a plant cell containing 0.6 M glucose is in equilibrium with its surrounding solution containing 0.5 M
glucose in an open container. What is the cell’s P?
```
Related documents | 1,243 | 5,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-21 | latest | en | 0.948593 |
https://forum.math.toronto.edu/index.php?PHPSESSID=ru0n85pa62910hsu96hsjjk8i1&action=printpage;topic=1614.0 | 1,660,767,148,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573104.24/warc/CC-MAIN-20220817183340-20220817213340-00108.warc.gz | 252,229,901 | 2,980 | # Toronto Math Forum
## MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: xinran zhao on December 02, 2018, 05:46:45 PM
Title: question about a homo matrix
Post by: xinran zhao on December 02, 2018, 05:46:45 PM
Having questions, can anyone take a look.
Question: $x'=\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix} x\\$
express the general solution of the given system of equation
Title: Re: question about a homo matrix
Post by: Tianyu Guo on December 02, 2018, 06:08:46 PM
We can compute the eigenvalues first
Let $A=\begin{pmatrix}-4~~~~7\\1 ~~~-4 \end{pmatrix}$
Setting up $det\begin{pmatrix} -4-\lambda ~~~7 \\ 1 ~~~-4-\lambda \end{pmatrix} = 0$
then we find out $(-4-\lambda)^2 -7=0$
By solving the equation, we get the eigenvalue $r_1=-4+\sqrt{7}$ and $r_2=-4-\sqrt{7}$.
Now, we need to plug the eigenvalue back to the matrix.
For $r_1=-4+\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}$.
For $r_2=-4-\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
Thus, the general solution for the system of equation is
$x'= ce^{(-4+\sqrt{7})t}*\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}+ce^{(-4-\sqrt{7})t}*\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
Title: Re: question about a homo matrix
Post by: Ning Du on December 02, 2018, 06:16:04 PM
$x'=\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix}x$
$\mathrm{Eigenvalue\:}:det\begin{bmatrix}-4-r & 7 \\1 & -4-r \end{bmatrix}=0$
$(-4-r)^2-7=0, then\ r = \sqrt{7}-4,\:-4-\sqrt{7}$
$\mathrm{Eigenvectors\:for\:}λ=\sqrt{7}-4:$
$\mathrm{Solve}(A-rI):\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix}-(\sqrt{7}-4)\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=\begin{pmatrix}-\sqrt{7}&7\\ 1&-\sqrt{7}\end{pmatrix}$
$\mathrm{Reduce\:}\begin{pmatrix}-\sqrt{7}&7\\ 1&-\sqrt{7}\end{pmatrix}:\quad \begin{pmatrix}1&-\sqrt{7}\\ 0&0\end{pmatrix}$
$\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=\sqrt{7}-4$
$\left(A-\left(\sqrt{7}-4\right)I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&-\sqrt{7}\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\$
$\mathrm{This\:reduces\:to\:the\:equation}x-\sqrt{7}y=0,x=\sqrt{7}y\\$
$plug in and let y = 1$
$\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix}$
$\mathrm{Eigenvectors\:for\:}λ=-4-\sqrt{7}:$
$\mathrm{Solve\:}\:\left(A-\lambda\:I\right):\:\begin{pmatrix}-4&7\\ 1&-4\end{pmatrix}-\left(-4-\sqrt{7}\right)\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=\begin{pmatrix}\sqrt{7}&7\\ 1&\sqrt{7}\end{pmatrix}$
$\mathrm{Reduce\:}\begin{pmatrix}\sqrt{7}&7\\ 1&\sqrt{7}\end{pmatrix}:\quad \begin{pmatrix}1&\sqrt{7}\\ 0&0\end{pmatrix}\\\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=-4-\sqrt{7}$
$\left(A-\left(-4-\sqrt{7}\right)I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&\sqrt{7}\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
$\mathrm{This\:reduces\:to\:the\:equation}:x+\sqrt{7}y=0, x=-\sqrt{7}y$
$\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\end{pmatrix},\mathrm{Let\:}y=1, \begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$
$\mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}-4&7\\ 1&-4\end{pmatrix}=\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix},\:\begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$
$x={c_1e^ \sqrt{7}t-4t}\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix}+{c_2e^ \-4t-\sqrt{7}t}\begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$ | 1,418 | 3,393 | {"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} | 4.375 | 4 | CC-MAIN-2022-33 | latest | en | 0.432208 |
https://www.kodytools.com/units/radexposure/from/mrem/to/rep | 1,721,354,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00670.warc.gz | 738,137,579 | 14,075 | Millirem to Rep Converter
Convert → Rep to Millirem
1 Millirem = 0.00107185 Reps
One Millirem is Equal to How Many Reps?
The answer is one Millirem is equal to 0.00107185 Reps and that means we can also write it as 1 Millirem = 0.00107185 Reps. Feel free to use our online unit conversion calculator to convert the unit from Millirem to Rep. Just simply enter value 1 in Millirem and see the result in Rep.
Manually converting Millirem to Rep can be time-consuming,especially when you don’t have enough knowledge about Radiation Exposure units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Millirem to Rep converter tool to get the job done as soon as possible.
We have so many online tools available to convert Millirem to Rep, but not every online tool gives an accurate result and that is why we have created this online Millirem to Rep converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
How to Convert Millirem to Rep (mrem to rep)
By using our Millirem to Rep conversion tool, you know that one Millirem is equivalent to 0.00107185 Rep. Hence, to convert Millirem to Rep, we just need to multiply the number by 0.00107185. We are going to use very simple Millirem to Rep conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Millirem} = 1 \times 0.00107185 = \text{0.00107185 Reps}$$
What Unit of Measure is Millirem?
Millirem is a unit of measurement for ionizing radiation exposure. Millirem is a decimal fraction of radiation exposure unit rem. One millirem is equal to 0.001 rem.
What is the Symbol of Millirem?
The symbol of Millirem is mrem. This means you can also write one Millirem as 1 mrem.
What Unit of Measure is Rep?
Rep is a unit of measurement for ionizing radiation exposure. Rep stands for roentgen equivalent physical. Rep is defined as 83 or 93 ergs per gram of tissue or per cc of tissue.
What is the Symbol of Rep?
The symbol of Rep is rep. This means you can also write one Rep as 1 rep.
How to Use Millirem to Rep Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Millirem and in the first input field, enter a value.
• From the second dropdown, select Rep.
• Instantly, the tool will convert the value from Millirem to Rep and display the result in the second input field.
Millirem
1
Rep
0.00107185
Millirem to Rep Conversion Table
Millirem [mrem]Rep [rep]Description
1 Millirem0.00107185 Rep1 Millirem = 0.00107185 Rep
2 Millirem0.0021437 Rep2 Millirem = 0.0021437 Rep
3 Millirem0.00321555 Rep3 Millirem = 0.00321555 Rep
4 Millirem0.0042874 Rep4 Millirem = 0.0042874 Rep
5 Millirem0.00535925 Rep5 Millirem = 0.00535925 Rep
6 Millirem0.0064311 Rep6 Millirem = 0.0064311 Rep
7 Millirem0.00750295 Rep7 Millirem = 0.00750295 Rep
8 Millirem0.0085748 Rep8 Millirem = 0.0085748 Rep
9 Millirem0.00964665 Rep9 Millirem = 0.00964665 Rep
10 Millirem0.0107185 Rep10 Millirem = 0.0107185 Rep
100 Millirem0.107185 Rep100 Millirem = 0.107185 Rep
1000 Millirem1.07 Rep1000 Millirem = 1.07 Rep
Millirem to Other Units Conversion Table
ConversionDescription
1 Millirem = 2.765373e-7 Coulomb/Kilogram1 Millirem in Coulomb/Kilogram is equal to 2.765373e-7
1 Millirem = 2.765373e-10 Coulomb/Gram1 Millirem in Coulomb/Gram is equal to 2.765373e-10
1 Millirem = 2.765373e-13 Coulomb/Milligram1 Millirem in Coulomb/Milligram is equal to 2.765373e-13
1 Millirem = 2.765373e-16 Coulomb/Microgram1 Millirem in Coulomb/Microgram is equal to 2.765373e-16
1 Millirem = 0.0002765373 Millicoulomb/Kilogram1 Millirem in Millicoulomb/Kilogram is equal to 0.0002765373
1 Millirem = 2.765373e-7 Millicoulomb/Gram1 Millirem in Millicoulomb/Gram is equal to 2.765373e-7
1 Millirem = 2.765373e-10 Millicoulomb/Milligram1 Millirem in Millicoulomb/Milligram is equal to 2.765373e-10
1 Millirem = 2.765373e-13 Millicoulomb/Microgram1 Millirem in Millicoulomb/Microgram is equal to 2.765373e-13
1 Millirem = 0.2765373 Microcoulomb/Kilogram1 Millirem in Microcoulomb/Kilogram is equal to 0.2765373
1 Millirem = 0.0002765373 Microcoulomb/Gram1 Millirem in Microcoulomb/Gram is equal to 0.0002765373
1 Millirem = 2.765373e-7 Microcoulomb/Milligram1 Millirem in Microcoulomb/Milligram is equal to 2.765373e-7
1 Millirem = 2.765373e-10 Microcoulomb/Microgram1 Millirem in Microcoulomb/Microgram is equal to 2.765373e-10
1 Millirem = 0.00107185 Roentgen1 Millirem in Roentgen is equal to 0.00107185
1 Millirem = 0.00107185 Tissue Roentgen1 Millirem in Tissue Roentgen is equal to 0.00107185
1 Millirem = 1.07 Milliroentgen1 Millirem in Milliroentgen is equal to 1.07
1 Millirem = 1071.85 Microroentgen1 Millirem in Microroentgen is equal to 1071.85
1 Millirem = 0.00107185 Parker1 Millirem in Parker is equal to 0.00107185
1 Millirem = 0.00107185 Rep1 Millirem in Rep is equal to 0.00107185
1 Millirem = 0.00001 Sievert1 Millirem in Sievert is equal to 0.00001
1 Millirem = 0.01 Millisievert1 Millirem in Millisievert is equal to 0.01
1 Millirem = 10 Microsievert1 Millirem in Microsievert is equal to 10
1 Millirem = 0.001 Rem1 Millirem in Rem is equal to 0.001
1 Millirem = 1000 Microrem1 Millirem in Microrem is equal to 1000 | 1,759 | 5,273 | {"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": 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} | 2.90625 | 3 | CC-MAIN-2024-30 | latest | en | 0.92616 |
http://www.cs.virginia.edu/~robins/FPGA/Iterated_Dominance_Heuristic.html | 1,542,354,675,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742981.53/warc/CC-MAIN-20181116070420-20181116091656-00022.warc.gz | 395,604,716 | 1,861 | ## 5. Iterated Dominance Heuristic
Our second heuristic for the GSA problem greedily iterates over a given spanning arborescence construction: we repeatedly find Steiner candidates that reduce the overall spanning arborescence cost, and include them into the growing set of Steiner nodes. The heuristic which we use for producing spanning arborescences is the Dominating (DOM) heuristic, described as follows:
• DOM -- connect each sink to the closest sink or source that it dominates, and compute the shortest-paths tree over the union of these paths.
Definition 5.1 Given a set of Steiner candidate node S subset of V - N, we define the cost savings of S with respect to DOM as Delta DOM( G, N, S ) = cost(DOM( G, N)) - cost(DOM(G, N U S )).
Starting with an initially empty set of Steiner candidates S = Ø, our heuristic finds a node t in V - N which maximizes Delta DOM( G, N, S U { t } ) > 0 and repeats this procedure with S = S U { t }. The cost for DOM to span N U S will decrease with each added node t, and the construction terminates when there is no t in (V - N) - S such that Delta DOM( G, N, S U { t } ) > 0, with the final solution being DOM( G, N U S ). This Iterated Dominance (IDOM) approach is formally described in Figure 6. The IDOM heuristic can be implemented within time O(|N|*|E| + |V|*|N|^3 ).
```
Iterated Dominance (IDOM) Algorithm
Input: A weighted graph G=(V,E) and a net N subset of V
Output: Low-cost arborescence T'=(V',E') spanning N, where
N subset of V' subset of V and E' subset of E
S = Ø
Do Forever
T = { t in V - N | Delta DOM( G, N, S U {t}) > 0 }
If T = Ø Then Return DOM(G, N U S)
Find t in T with maximum Delta DOM( G, N, S U {t})
S = S U {t}
Figure 6: The Iterated Dominance algorithm.
``` | 493 | 1,742 | {"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} | 3.21875 | 3 | CC-MAIN-2018-47 | latest | en | 0.841336 |
http://mathoverflow.net/questions/73439/when-is-a-given-matrix-of-two-forms-a-curvature-form | 1,469,598,734,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00035-ip-10-185-27-174.ec2.internal.warc.gz | 153,973,216 | 15,710 | # When is a given matrix of two forms a curvature form?
Let's assume we are working over $\mathbb{R}^n$ (but feel free to change to domain to answer the question). I wish to know if the equation $F = dA + A \wedge A$ can be solved for a matrix of 1-forms $A$, given a (smooth) matrix of 2-forms $F$ which satisfies the condition $dF =B \wedge F - F \wedge B$ for some smooth matrix of 1-forms $B$ (i.e. the Bianchi identity is satisfied). Notice that this is true for line-bundles (in fact over any convex open set).
-
A necessary condition: All the coefficients of the "characteristic polynomial" of $A$ must be closed differential forms. (If we were working over $\mathbb{C}$, these would be the Chern forms. Not sure what they're called over $\mathbb{R}$, but they should be closed either way.) – David Speyer Aug 22 '11 at 21:28
The Bianchi identity guarantees this ($dtr(F^k) = ktr(dF F^{k-1})=ktr([B,F]F^{k-1})=0$) – Vamsi Aug 22 '11 at 22:03
@David: You mean $F$, not $A$, and this closure is guaranteed by the quasi-Bianchi equation $dF = B\wedge F - F\wedge B$. – Robert Bryant Aug 22 '11 at 22:08
The answer is generally 'no'; for most $F$ that satisfy your condition, there will not exist an $A$ that satisfies $F = dA + A\wedge A$.
The easiest counterexample I know of is when $n=4$ and the matrix $F$ is $2$-by-$2$. To begin, note that you can reduce to the case when both $F$ and the $A$ you seek have trace zero, i.e., they take values in ${\frak{sl}}(2,\mathbb{R})$. (The reason is that the problem breaks into the part of $F$ that is a multiple of the identity matrix and the trace-free part. I'll leave the details to you.)
One can easily check that, for the generic ${\frak{sl}}(2,\mathbb{R})$-valued matrix $F$ of $2$-forms on $\mathbb{R}^4$, the kernel of the mapping $C\mapsto F\wedge C - C\wedge F$ from ${\frak{sl}}(2,\mathbb{R})$-valued matrices $C$ of $1$-forms on $\mathbb{R}^4$ to ${\frak{sl}}(2,\mathbb{R})$-valued matrices of $3$-forms on $\mathbb{R}^4$ is zero. By dimension count, it follows that this mapping is surjective as well.
Thus, for a candidate ${\frak{sl}}(2,\mathbb{R})$-valued $2$-form $F$ that satisfies this open genericity condition, the equation $dF = F\wedge B - B\wedge F$ is always solvable for $B$, and, moreover, the solution is unique. Thus, this $B$ is the only possible candidate for $A$. Heuristically, this makes it almost immediate that, for the generic such $F$, the $B$ that you find will not satisfy $F = dB + B\wedge B$. The reason is that ${\frak{sl}}(2,\mathbb{R})$-valued $1$-forms on $\mathbb{R}^4$ depend on only $3\times 4 = 12$ arbitrary functions of $4$ variables while the generic ${\frak{sl}}(2,\mathbb{R})$-valued $2$-form $F$ depends on $3\times 6 = 18$ arbitrary functions of $4$ variables. There is no chance that you could hit each such $F$ with an $A$.
To construct an explicit example, choose an ${\frak{sl}}(2,\mathbb{R})$-valued $1$-form $A$ such that $F = dA + A\wedge A$ satisfies the genericity condition. Then, of course, $2F$ will satisfy this genericity condition as well, and, since it satisfies $d(2F) = (2F)\wedge A - A \wedge (2F)$, it follows that the only possible $1$-form whose curvature could be $2F$ is $A$. However, the curvature of $A$ is $F\not=2F$. Thus, $2F$ satisfies your condition, but it is not a curvature form. | 1,082 | 3,323 | {"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} | 3.609375 | 4 | CC-MAIN-2016-30 | latest | en | 0.882684 |
https://in.mathworks.com/matlabcentral/cody/problems/42948-find-a-specific-element-from-an-matrix/solutions/1219984 | 1,611,012,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00719.warc.gz | 389,970,567 | 16,991 | Cody
# Problem 42948. find a specific element from an matrix
Solution 1219984
Submitted on 26 Jun 2017 by Chris Cleveland
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [1 2 3; 4 5 6] y_correct = 6; assert(isequal(your_fcn_name(x),y_correct))
x = 1 2 3 4 5 6
2 Pass
x = [1 2 3; 4 5 7] y_correct = 7; assert(isequal(your_fcn_name(x),y_correct))
x = 1 2 3 4 5 7
### Community Treasure Hunt
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#### Resources tagged with Generalising similar to Playground Snapshot:
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### 2001 Spatial Oddity
##### Age 11 to 14 Challenge Level:
With one cut a piece of card 16 cm by 9 cm can be made into two pieces which can be rearranged to form a square 12 cm by 12 cm. Explain how this can be done.
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Cubes Within Cubes Revisited
##### Age 11 to 14 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
### Picturing Triangular Numbers
##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Route to Infinity
##### Age 11 to 14 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Konigsberg Plus
##### Age 11 to 14 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Tilted Squares
##### Age 11 to 14 Challenge Level:
It's easy to work out the areas of most squares that we meet, but what if they were tilted?
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Steps to the Podium
##### Age 7 to 14 Challenge Level:
It starts quite simple but great opportunities for number discoveries and patterns!
### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### Is There a Theorem?
##### Age 11 to 14 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Hidden Rectangles
##### Age 11 to 14 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
### Circles, Circles
##### Age 5 to 11 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### Dotty Triangles
##### Age 11 to 14 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Squares, Squares and More Squares
##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Enclosing Squares
##### Age 11 to 14 Challenge Level:
Can you find sets of sloping lines that enclose a square?
### Finding 3D Stacks
##### Age 7 to 11 Challenge Level:
Can you find a way of counting the spheres in these arrangements?
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Mini-max
##### Age 11 to 14 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Sum Equals Product
##### Age 11 to 14 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
### Special Sums and Products
##### Age 11 to 14 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### Snake Coils
##### Age 7 to 11 Challenge Level:
This challenge asks you to imagine a snake coiling on itself.
### Taking Steps
##### Age 7 to 11 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
##### Age 11 to 14 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Odd Squares
##### Age 7 to 11 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Magic Letters
##### Age 11 to 14 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Shear Magic
##### Age 11 to 14 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Go Forth and Generalise
##### Age 11 to 14
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Seven Squares - Group-worthy Task
##### Age 11 to 14 Challenge Level:
Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning?
### Handshakes
##### Age 11 to 14 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### More Number Pyramids
##### Age 11 to 14 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Partitioning Revisited
##### Age 11 to 14 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4
### Egyptian Fractions
##### Age 11 to 14 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written different fractions.
### Keep it Simple
##### Age 11 to 14 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions?
### Spirals, Spirals
##### Age 7 to 11 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Fault-free Rectangles
##### Age 7 to 11 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Sliding Puzzle
##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Make 37
##### Age 7 to 14 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### One, Three, Five, Seven
##### Age 11 to 16 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Cut it Out
##### Age 7 to 11 Challenge Level:
Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?
### Overlap
##### Age 11 to 14 Challenge Level:
A red square and a blue square overlap so that the corner of the red square rests on the centre of the blue square. Show that, whatever the orientation of the red square, it covers a quarter of the. . . .
### Window Frames
##### Age 5 to 14 Challenge Level:
This task encourages you to investigate the number of edging pieces and panes in different sized windows. | 2,294 | 9,322 | {"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} | 3.96875 | 4 | CC-MAIN-2019-39 | latest | en | 0.892914 |
http://www.finderchem.com/what-shape-has-1-face-and-0-edges.html | 1,500,936,971,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424931.1/warc/CC-MAIN-20170724222306-20170725002306-00067.warc.gz | 415,192,110 | 8,510 | # What shape has 1 face and 0 edges?
## We found this answers
You can manipulate and color each shape to explore the number of faces, edges, ... Colors faces, edges, ... You can download and print the Exploring Geometric Solids ... - Read more
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## What shape has 1 face and 0 edges? resources
### 3D Shapes 1 (Faces, Edges and Vertices) - Powerpoint Maths ...
3D Shapes 1 (Faces, Edges and Vertices) A Shape and Space PowerPoint Presentation: A very nice and visually appealing presentation introducing faces, ...
### This 3D shapes has no flat faces and no straight edges. It ...
this 3D shapes has no flat faces and no straight edges. ... What 3d shape has one face no edges and no vertices? ... What is 3D shape with 1 curved face and 1 edge?
### Vertices, Edges and Faces - Math is Fun - Maths Resources
Vertices, Edges and Faces. ... For many solid shapes the. Number of Faces; ... Try it on the cube: A cube has 6 Faces, 8 Vertices, and 12 Edges, so:
### What shape(3-D) has 5 vertices,8 edges,and 5 faces? - Math ...
what shape(3-D) has 5 vertices,8 edges,and 5 faces? Stacy, Here's a hint. Penny . Math Central is supported by the University of Regina and The Pacific Institute for ...
### What solid shape has two circular faces, 0 edges, 0 ...
'What solid shape has two circular faces, 0 edges, 0 vertices?' was asked by a user of Poll Everywhere to a live audience who responded via text messaging, the web ...
### Basic Haircare & Hairstyles : Hairstyles for Your Face Shape
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Automatically Create Amazing Photo Effects with our Photoshop Plugins and Filters. Simple 1 ... Edges Gen1 Photo Editor / Photoshop ... and shape, lip color and shape ...
### Platonic Solids - NLVM
Count the faces, edges or vertices of the Platonic solid; Display different Platonic solids; ... Display different Platonic solids. Click the New Shape button.
### Solid shapes and their nets - cube - Demon Internet
Game with nets of cubes. To make a net of a cube, first look at one, such as a dice. How many faces does it have? Six, so make sure that your net has six squares.
### Interactives . 3D Shapes . Polyhedron - Learner
3D Shapes There are many types ... A three-dimensional shape whose faces are polygons is known as a polyhedron. ... The other parts of a polyhedron are its edges, ...
### Platonic Solids - Why Five? - Maths is Fun
Platonic Solids - Why Five? ... A cube has 6 Faces, 8 Vertices, and 12 Edges, so: 6 + 8 - 12 = 2. ... 1/s + 1/m - 1/2 > 0. Or, more simply:
### Edges of Darkness (Video 2008) - IMDb
Edges of Darkness (2008) Video 87 min - Horror ... 2.9 . Your rating: 1 2 3 4 5 6 7 8 9 10-/ 10 X . Ratings: 2.9 / 10 from 474 users Reviews: ... 0 Check in. X Beta I ...
### Regular Polyhedra - Cut-the-Knot
Regular faces cease to be ... Since every face has p edges there would be a ... if an n-dimensional polyhedron has N 0 vertices, N 1 edges, N 2 faces ...
### Find 4 formulae that can work out the number of cubes in a ...
... painted on the outside with 0 faces painted, 1 ... of cubes in a cube that has been painted on the outside with 0 faces ... of straight lines and edges.
### What is a Mineral? Naturally Occurring- must be made in ...
1) Naturally Occurring- must be made ... Solid- has a definite volume and shape. 4) ... Crystal have flat sides called faces, that meet at sharp edges and corners ...
### Lifestyle | Shape Magazine
Get special offers from SHAPE partners and advertisers. Privacy policy. BROWSE SHAPE. Shape Home; Fitness; Healthy Eating; Weight Loss; Lifestyle; Celebrities; Sweeps ...
### Shapes at EnchantedLearning.com - ENCHANTED LEARNING HOME PAGE
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### Number of Cylinder Edges - Math Forum - Ask Dr. Math
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### Sports News & Articles – Scores, Pictures, Videos - ABC News
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### Growing and Shrinking Polygons: Round One | Hans Muller's ...
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### mask - definition of mask by The Free Dictionary
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### Platonic Solids - NLVM - Utah State University
Platonic Solids. A Platonic solid is a polyhedron whose faces are identical regular polygons. ... Click the Reset Shape button to uncolor all faces, edges, ...
### Study the Solids - NCTM Illuminations
In this interactive geometry investigation, students explore geometric solids and their properties. Specifically, students count the number of faces, edges, and ...
### Interactives . 3D Shapes . Euler's Theorem
You already know that a polyhedron has faces (F), vertices (V), and edges (E). ... they speculated that its shape was a truncated icosahedron ... (edges). From Euler ...
### How to Determine Your Body Shape | eHow
Everyone is born with a certain type of body shape that no amount of dieting and exercise is ever going to change. Use the following tips to determine your body shape ...
### IXL - Compare sides and vertices (1st grade math practice)
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### How to Choose a Haircut That Flatters Your Facial Shape: 9 ...
Three Parts: Knowing Your Face Shape Researching the Right Haircut for Your Face Making Sure You Get the Right Cut. When you're deciding which haircut will look best ...
### Which Shape Has One Vertex One Edge And Two Faces? - Blurtit
Which Shape Has One Vertex One Edge And Two Faces? ... What Shape Has 6 Faces 12 Edges And 8 Vertices? ... What solid has 1 face 0 edges and 0 vertices?
Related Questions
Recent Questions | 1,757 | 7,106 | {"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-2017-30 | latest | en | 0.893289 |
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Оценки: 236
Рецензии: 29
## О курсе
Welcome to Asymmetric Cryptography and Key Management! In asymmetric cryptography or public-key cryptography, the sender and the receiver use a pair of public-private keys, as opposed to the same symmetric key, and therefore their cryptographic operations are asymmetric. This course will first review the principles of asymmetric cryptography and describe how the use of the pair of keys can provide different security properties. Then, we will study the popular asymmetric schemes in the RSA cipher algorithm and the Diffie-Hellman Key Exchange protocol and learn how and why they work to secure communications/access. Lastly, we will discuss the key distribution and management for both symmetric keys and public keys and describe the important concepts in public-key distribution such as public-key authority, digital certificate, and public-key infrastructure. This course also describes some mathematical concepts, e.g., prime factorization and discrete logarithm, which become the bases for the security of asymmetric primitives, and working knowledge of discrete mathematics will be helpful for taking this course; the Symmetric Cryptography course (recommended to be taken before this course) also discusses modulo arithmetic. This course is cross-listed and is a part of the two specializations, the Applied Cryptography specialization and the Introduction to Applied Cryptography specialization....
## Лучшие рецензии
SC
1 июня 2021 г.
Very good intro, particularly talking about Key Management, an usually overlooked aspect of Crypto. Professor is also super clear in his explanations.\n\nRecommended :)
LS
20 мар. 2021 г.
Very understandable! Well structured lesson slides and topics. I was able to understand the content as a fresher for this topic! Thank you Mr. Chang very much!
Фильтр по:
## 1–25 из 30 отзывов о курсе Asymmetric Cryptography and Key Management
автор: Frode
14 янв. 2018 г.
Some annoying errors in quizes, and I wish there was something more about elliptic curves, but otherwise good course.
автор: Manuel A D R
27 июля 2019 г.
Excellent course Asymmetric Crytography and Key Management.
автор: CHITTILLA V V
16 янв. 2018 г.
great lectures.Thank you Sir and Thank you Coursera
автор: Anton P
25 янв. 2018 г.
автор: Dirk D S
3 авг. 2019 г.
Excellent to the point with good references
автор: Marcus D
16 дек. 2020 г.
Solid course and pretty good introduction to the topics of asymmetric cryptography. More time spent on the modular arithmetic of RSA encryption as well as the Discrete Logarithm math would have been good for strong understanding of what was going on. It was hard to understand those core topic without many external references.
автор: Sebastian C
2 июня 2021 г.
Very good intro, particularly talking about Key Management, an usually overlooked aspect of Crypto. Professor is also super clear in his explanations.
Recommended :)
автор: Shubham T
26 сент. 2020 г.
Good course to learn Asymmetric Cryptography really helped me a lot to learn , suggest all to take this specialization course to build carrier in cryptography.
автор: Liufeng S
21 мар. 2021 г.
Very understandable! Well structured lesson slides and topics. I was able to understand the content as a fresher for this topic! Thank you Mr. Chang very much!
4 дек. 2021 г.
Thank you so much Sang-Yoon Chang for the wonderful lectures that enrich and broaden my knowledge in Cryptography for my future new career I am pursuing.
автор: Hien D Q
22 авг. 2020 г.
Great course for everyone who would like to learn foundation knowledge about cryptography.
автор: Лопатин М С
26 мая 2020 г.
Курс интересный, мне понравился. Приобрел для себя новые знания. Рад такому опыту
автор: Богдан Х М
2 июня 2020 г.
I could understand and learn a lot after this course. The lecturer is so great)
17 мар. 2020 г.
Very good description of the basics and also pace of the session is good.
автор: venkat
15 сент. 2020 г.
VERY IMPRESSIVE TEACHING AND I LOVED LECTURES VERY MUCH
автор: Devarakonda S
12 мар. 2021 г.
I'm able to learn new topics that helped me in exams
автор: Aswani K C
11 авг. 2020 г.
Excellent course introducing Asymmetric crypto.
автор: Скоков Н С
26 мая 2020 г.
Very interesting and informative course!
автор: Renate K
13 февр. 2020 г.
Sang-Yoon Chang is a great teacher!
автор: Iván S M
18 окт. 2020 г.
I enjoyed this course very much.
автор: Alejandro S
13 июля 2020 г.
Nice and easy to follow course
автор: Luis V
4 дек. 2020 г.
Love the level of detail!
автор: Praneeth C
12 мая 2020 г.
Awesome course, Thanks.
автор: Mr.Arif M A
19 мая 2020 г.
Useful Concept
автор: 321810403034 g
20 авг. 2021 г.
good course | 1,166 | 4,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-27 | longest | en | 0.833727 |
https://oeis.org/A275422 | 1,582,457,084,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145767.72/warc/CC-MAIN-20200223093317-20200223123317-00204.warc.gz | 488,032,486 | 4,655 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A275422 Number A(n,k) of set partitions of [n] such that k is a multiple of each block size; square array A(n,k), n>=0, k>=0, read by antidiagonals. 10
1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 1, 2, 1, 15, 1, 1, 1, 4, 1, 52, 1, 1, 2, 2, 10, 1, 203, 1, 1, 1, 4, 5, 26, 1, 877, 1, 1, 2, 1, 11, 11, 76, 1, 4140, 1, 1, 1, 5, 1, 31, 31, 232, 1, 21147, 1, 1, 2, 1, 14, 2, 106, 106, 764, 1, 115975, 1, 1, 1, 4, 1, 46, 7, 372, 337, 2620, 1, 678570 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,6 LINKS Alois P. Heinz, Antidiagonals n = 0..200, flattened Wikipedia, Partition of a set FORMULA E.g.f. for column k>0: exp(Sum_{d|k} x^d/d!), for k=0: exp(exp(x)-1). EXAMPLE A(5,3) = 11: 123|4|5, 124|3|5, 125|3|4, 134|2|5, 135|2|4, 1|234|5, 1|235|4, 145|2|3, 1|245|3, 1|2|345, 1|2|3|4|5. A(4,4) = 11: 1234, 12|34, 12|3|4, 13|24, 13|2|4, 14|23, 1|23|4, 14|2|3, 1|24|3, 1|2|34, 1|2|3|4. A(6,5) = 7: 12345|6, 12346|5, 12356|4, 12456|3, 13456|2, 1|23456, 1|2|3|4|5|6. Square array A(n,k) begins: : 1, 1, 1, 1, 1, 1, 1, 1, 1, ... : 1, 1, 1, 1, 1, 1, 1, 1, 1, ... : 2, 1, 2, 1, 2, 1, 2, 1, 2, ... : 5, 1, 4, 2, 4, 1, 5, 1, 4, ... : 15, 1, 10, 5, 11, 1, 14, 1, 11, ... : 52, 1, 26, 11, 31, 2, 46, 1, 31, ... : 203, 1, 76, 31, 106, 7, 167, 1, 106, ... : 877, 1, 232, 106, 372, 22, 659, 2, 372, ... : 4140, 1, 764, 337, 1499, 57, 2836, 9, 1500, ... MAPLE A:= proc(n, k) option remember; `if`(n=0, 1, add( `if`(j>n, 0, A(n-j, k)*binomial(n-1, j-1)), j= `if`(k=0, 1..n, numtheory[divisors](k)))) end: seq(seq(A(n, d-n), n=0..d), d=0..14); MATHEMATICA A[n_, k_] := A[n, k] = If[n==0, 1, Sum[If[j>n, 0, A[n-j, k]*Binomial[n-1, j - 1]], {j, If[k==0, Range[n], Divisors[k]]}]]; Table[A[n, d-n], {d, 0, 14}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 08 2017, translated from Maple *) CROSSREFS Columns k=0-10 give: A000110, A000012, A000085, A190865, A190452, A275423, A275424, A275425, A275426, A275427, A275428. Main diagonal gives A275429. Sequence in context: A213945 A290771 A014651 * A169951 A174453 A082063 Adjacent sequences: A275419 A275420 A275421 * A275423 A275424 A275425 KEYWORD nonn,tabl AUTHOR Alois P. Heinz, Jul 27 2016 STATUS approved
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Last modified February 23 06:13 EST 2020. Contains 332159 sequences. (Running on oeis4.) | 1,404 | 2,796 | {"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} | 3.03125 | 3 | CC-MAIN-2020-10 | latest | en | 0.170863 |
http://mrmaisonet.com/index.php?/Number-Theory-Video/feed/rss.html | 1,371,676,736,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709224828/warc/CC-MAIN-20130516130024-00010-ip-10-60-113-184.ec2.internal.warc.gz | 175,229,388 | 4,516 | Finding The Greatest Common Factor
Watch this tutorial to see an easy method of finding the GCF of any group of numbers.
Factor Trees
Review how to break a composite number into all prime numbers by making a factor tree. At the end, express your answer using exponential notation.
Greatest Common Factor
Find the greatest common factor of a pair of numbers.
Find GCF or LCM Using Factor Trees
Find the LCM or GCF using prime factorization.
Least Common Multiple
Review how to solve a problem involving the least common multiple. | 112 | 538 | {"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-2013-20 | latest | en | 0.841798 |
https://oeis.org/A194753 | 1,618,149,510,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038062492.5/warc/CC-MAIN-20210411115126-20210411145126-00321.warc.gz | 541,387,321 | 3,520 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A194753 Number of k such that {-k*e} > {-n*e}, where { } = fractional part. 3
0, 0, 0, 3, 2, 1, 0, 6, 4, 2, 10, 7, 4, 1, 12, 8, 4, 17, 12, 7, 2, 18, 12, 6, 24, 17, 10, 3, 24, 16, 8, 31, 22, 13, 4, 30, 20, 10, 0, 29, 18, 7, 38, 26, 14, 2, 36, 23, 10, 46, 32, 18, 4, 43, 28, 13, 54, 38, 22, 6, 50, 33, 16, 62, 44, 26, 8, 57, 38, 19, 70, 50, 30, 10, 64 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,4 LINKS MATHEMATICA r = -E; p[x_] := FractionalPart[x]; u[n_, k_] := If[p[k*r] <= p[n*r], 1, 0] v[n_, k_] := If[p[k*r] > p[n*r], 1, 0] s[n_] := Sum[u[n, k], {k, 1, n}] t[n_] := Sum[v[n, k], {k, 1, n}] Table[s[n], {n, 1, 100}] (* A194752 *) Table[t[n], {n, 1, 100}] (* A194753 *) CROSSREFS Cf. A194752. Sequence in context: A051427 A194763 A194741 * A323908 A098825 A111460 Adjacent sequences: A194750 A194751 A194752 * A194754 A194755 A194756 KEYWORD nonn AUTHOR Clark Kimberling, Sep 02 2011 STATUS approved
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Last modified April 11 09:54 EDT 2021. Contains 342886 sequences. (Running on oeis4.) | 614 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-17 | latest | en | 0.695492 |
www.happynewyear2017wishesimageshd.com | 1,597,227,063,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738888.13/warc/CC-MAIN-20200812083025-20200812113025-00203.warc.gz | 142,767,308 | 9,912 | # Calculate repayment and loan interest with repayment calculator
## The low key interest rates
More and more consumers are currently being given the opportunity to carry out real estate financing. One reason is the low key interest rates, which the Meduim Centrum Bank reduced by 0.1 percentage points just a few days ago. Banks can now borrow money from the central bank at a rate of only 0.15 percent and are therefore able to provide very cheap loans. These inexpensive loans can be found particularly in the area of real estate financing, i.e. in the form of annuity or repayment loans. Interest rates on the one hand are very low, but interested loan seekers on the other hand should not forget to calculate the monthly charge.
## Low interest rates can also represent an interest trap
Ultimately, the low interest rates can also represent an interest trap, because borrowers can generally not secure the low interest rates over the entire repayment period of the loan. It can happen, for example, that the interest rate after the fixed interest rate expires in five or ten years will be significantly higher than the current one and that the monthly loan rate will therefore be difficult to bear or no longer affordable. A tool that can be used to calculate repayments and loan interest is the so-called repayment calculator. With this online calculator, the interested loan seeker can calculate the monthly loan rate at which certain repayment rates and interest rates would lead.
## Customer only has to provide some information
The operation of such computers is relatively easy, because the customer only has to provide some information. This primarily includes the desired loan amount, the term of the loan, the initial repayment and the loan interest rate that the respective bank requires. Based on this information, the calculator can be used to calculate how high the monthly rate will be. For example, if you need a loan of over 100,000 USD and decide to make an initial repayment of three percent, you will be charged an annual charge of just 5,000 USD at an interest rate of two percent.
Calculated monthly, this leads to a comparatively low credit rate of less than 430 USD. This number already shows that it has now become cheaper for many consumers to carry out real estate financing instead of paying rent. One of the advantages of the loan calculator is that the optimal repayment and interest payment can be calculated using various variables. As a result, repayment calculators can help consumers find the right amount. | 492 | 2,554 | {"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} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.957174 |
https://thoughtbot.com/blog/ruby-splat-operator | 1,722,911,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640461318.24/warc/CC-MAIN-20240806001923-20240806031923-00179.warc.gz | 445,469,864 | 14,798 | # Ruby splat operator π
Sami Birnbaum
Edited by thoughtbot
The Ruby splat operator is confusing and here is why: it does two things that are the exact opposite of each other. Let me explainβ¦
### Destructuring
It destructures an array, which looks something like this:
``````x, y, z = *[1,2,3]
puts x
# => 1
puts y
# => 2
puts z
# => 3
``````
### Constructing
But it can also be used to construct an array:
``````x = *123
# => [123]
``````
One takes an array and removes the surrounding square brackets `[]`.
`*[123] becomes 123`
The other takes a value and adds the surrounding square brackets `[]`.
`*123 becomes [123]`
### When does it destruct and when does it construct?
The only question we need to answer then is when does it destruct and when does it construct? π€
If you use it when defining a method it will take any argument you give it and construct it into an array.
``````def describe_args(*args)
puts args
puts args.class
end
describe_args(1,2,3)
# 1
# 2
# 3
# Array
``````
If you use it when passing arguments as an array to a method it will deconstruct the array into arguments.
``````def describe_first_arg(x,y,z)
puts x
puts x.class
end
describe_first_arg(*[1,2,3])
# 1
# Integer
``````
### Practical application
Hopefully, we can now understand a class method you may have come across:
``````def self.perform(*args)
new(*args).perform
end
``````
At first glance this looks strange and complex but using our newfound understanding of the splat operator we can understand what this is doing.
``````class SumTotal
def self.perform(*args) # this splat operator will construct any arguments into an array
new(*args).perform # this splat operator will deconstruct that array into arguments
end
def initialize(x,y)
@x = x
@y = y
end
def perform
puts @x + @y
end
end
SumTotal.perform(5,5)
# => 10
``````
``````SumTotal.perform(5,5)
# => 10
``````
``````sum_total = SumTotal.new(5,5)
sum_total.perform
# => 10
``````
``````class SumTotal
def self.perform(*args)
new(*args).perform
end
def initialize(x,y,z)
@x = x
@y = y
@z = z
end
def perform
puts @x + @y + @z
end
end
SumTotal.perform(5,5,5)
# => 15
``````
The `initialize` and `perform` method have changed but we can make those changes without having to worry about the `self.perform` class method.
Thank you to Starr Horne’s article Using splats to build up and tear apart arrays in Ruby, which helped to shape my understanding. | 701 | 2,437 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-33 | latest | en | 0.817174 |
https://social.msdn.microsoft.com/Forums/en-US/d9184537-00d7-481e-a587-8c9b997a0a45/how-best-to-handle-2d-graphics?forum=csharpgeneral | 1,600,539,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192783.34/warc/CC-MAIN-20200919173334-20200919203334-00074.warc.gz | 665,909,256 | 14,309 | How Best To Handle 2D Graphics
• Question
• I recently finished making a basic Tetris clone and it works fine. Only problem is there is an annoying flicker every time the screen is updated and it's given me cause to revise my paint() method which draws to the screen.
Is it perhaps purely because my paint method is too inefficient? How would others choose to output 2D graphics for simple arcade games such as Tetris or Breakout?
Currently I have a picturebox which I've set up a graphics object within to which I draw upon. I chose to do this purely 'cause this is the only form of specific graphic output I've come across in the short year that I've been learning C#. Was this wise or are there other, better methods of drawing to the screen?
Also I've been using nothing but filled rectangles to represent the squares of the game, would things maybe work more smoothly if I used small bitmaps or something instead?
Here's my current paint() method:
public void paint()
{
lock (this)
{
SolidBrush squarepen = null;
grphPicBoxGameArea.Clear(Color.White);
grphPicBoxNextBlock.Clear(Color.White);
//reads in the array of the pile and fills the corresponding squares of the grid
for (int i = 0; i < Pile.pile.GetLength(0); i++)
{
for (int j = 0; j < Pile.pile.GetLength(1); j++)
{
if (Pile.pile[i, j] != null)
{
squarepen = new SolidBrush(Pile.pile[i, j].getColour());
grphPicBoxGameArea.FillRectangle(squarepen, i * 30, (j * 30) - 120, 30, 30);
}
}
}
//reads in the array of the current block and fills the corresponding squares of the grid
Square[] currentblock = blockinuse.getBlockSquares();
for (int i = 0; i < currentblock.GetLength(0); i++)
{
int currentxpos = currentblock[i].getXpos(), currentypos = currentblock[i].getYpos();
squarepen = new SolidBrush(currentblock[i].getColour());
grphPicBoxGameArea.FillRectangle(squarepen, currentxpos * 30, (currentypos * 30) - 120, 30, 30);
}
}
}
Sunday, August 9, 2009 4:05 PM
• You should take advantage of the double buffering supported by controls. To do so, you have to implement the Paint() event of those controls to do your drawing, use e.Graphics. Force a repaint with the Invalidate() method. It is best to not use picture boxes, just the paint event of the form. Set the form's DoubleBuffered property to True.
Hans Passant.
Sunday, August 9, 2009 4:31 PM
All replies
• You should take advantage of the double buffering supported by controls. To do so, you have to implement the Paint() event of those controls to do your drawing, use e.Graphics. Force a repaint with the Invalidate() method. It is best to not use picture boxes, just the paint event of the form. Set the form's DoubleBuffered property to True.
Hans Passant.
Sunday, August 9, 2009 4:31 PM
• Works just dandy with no flicker now, thanks a bunch. =]
Sunday, August 9, 2009 7:02 PM | 736 | 2,829 | {"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-2020-40 | latest | en | 0.846152 |
http://nasawavelength.org/resource-search?facetSort=1&educationalLevel=Elementary+school%3AUpper+elementary&topicsSubjects=Physical+sciences%3AVibration+and+waves | 1,511,185,090,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806066.5/warc/CC-MAIN-20171120130647-20171120150647-00269.warc.gz | 209,512,302 | 12,370 | ## Narrow Search
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# The Electromagnetic Spectrum
In this lesson, students are introduced to the electromagnetic spectrum. They observe a demonstration of the visible light spectrum created by a flashlight and a prism, complete an activity sheet where they identify the wavelength that is involved... (View More)
# The Electromagnetic Spectrum: Wavelength and Energy
In this activity, students demonstrate the relationship between wave frequency and energy in the electromagnetic spectrum by shaking a rope to identify the relationships. This activity is part of Unit 2 in the Space Based Astronomy guide that... (View More)
# Wavelength and Energy
This is an activity about wavelength and frequency. Using a 30 to 50 foot rope and two volunteers, learners will observe as one end of the rope is shaken and wavelength patterns are created. They will estimate the wavelength, the distance between... (View More)
# The Electromagnetic Spectrum: Resonating Atmosphere
Using a paper and tape device, students experience how atoms and molecules of gas in Earth’s atmosphere absorb electromagnetic energy through resonance. This activity is part of Unit 2 in the Space Based Astronomy guide that contains background... (View More)
# The Electromagnetic Spectrum: Red Shift, Blue Shift
In this activity, a Whiffle® ball containing a battery-operated buzzer is twirled in a circle to demonstrate the Doppler effect. The demonstration is an illustration of how stellar spectra can be used to measure a star's motion relative to Earth... (View More)
# Good Vibrations
This activity is about the use of thermal emission spectroscopy in planetary research. Learners will understand how thermal spectroscopy science is used to understand the composition of the Martian surface by simulating the data collection of the... (View More)
1 | 438 | 2,165 | {"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-2017-47 | latest | en | 0.859697 |
https://www.vidyakul.com/cbse-class-12-maths-formulas-list-chapter-2-inverse-trigonometric-functions/ | 1,560,685,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00141.warc.gz | 945,616,114 | 78,092 | # Inverse Trigonometric Functions Class 12 Formulas and Notes
## Class 12 Maths Chapter 2 Inverse Trigonometric Functions Class 12 Formulas & Notes – PDF Download
Inverse function. Inverse of a function ‘f ‘ exists, if the function is one-one and onto, i.e, bijective. Since trigonometric functions are many-one over their domains, we restrict their domains and co-domains in order to make them one-one and onto and then find their inverse. Know More about these in Inverse Trigonometric Functions Class 12 Formulas and Notes List.
The topics and sub-topics covered in Inverse Trigonometric Functions Class 12 Formulas and Notes are:
2.1 Introduction
2.2 Basic Concepts
2.3 Properties of Inverse Trigonometric Functions.
Download the FREE PDF of Inverse Trigonometric Functions Class 12 Formulas and Notes and start your preparation with Vidyakul!
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http://www.caraudio.com/forums/enclosure-design-construction-help/414622-how-design-ported-box-using-pencil-paper.html | 1,394,278,751,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999654390/warc/CC-MAIN-20140305060734-00092-ip-10-183-142-35.ec2.internal.warc.gz | 281,983,118 | 16,251 | 1. ## How To: Design a ported box using pencil & paper
Things you will need:
Pencil
Paper
Calculator
I worked each equation step by step to help you guys understand. It's actually rather easy. This will be more accurate than the box calculators online. Feel free to correct any mistakes I made.
================================================== ========================================
================================================== ========================================
There are a few different equations to determine port length. They can be found on numerous sites. I use the following.
This one comes from a loudspeaker cookbook I read once upon a time. It also can be found online at numerous sites.
Lv= {(14630000 x R²) / (Fb²) x [(Vb/Np) x 1728]} - 1.463 x R
Lv= Port Length (inches)
Fb= Tune Frequency (Hertz)
R= See ******* (this will use different equations depending on round or square vents)
Vb= total internal airspace (cubic ft)
Np= Number of ports
******This is for square vents and will take place of R.
R= √[( H x W) / π]
√= square root
H= height of port (inches)
W= width of port (inches)
π = 3.141592 (pi)
******This is for round vents and will take place of R
R= (Dia / 2)
Dia= Vent Diameter
This is just an example for a 6" port.
R= (Dia / 2)
R= (6 / 2)
R= 3
================================================== ========================================
================================================== ========================================
Here we go. The box I want is 6 cubic ft. tuned to 40hz, so lets fill in what we already know. I'll be using a square vent.
Lv= {(14630000 x R²) / (Fb²) x [(Vb/Np) x 1728]} - 1.463 x R
Lv=?
Fb=40
R=?
Vb=6
Lets solve for R. This is a square so we use the asterisk for a square vent. The vent of my box is going to be 15.5" x 8", (these
numbers are up to you, multiplying these numbers together determines your port area in inches, which is a number we don't need
but will help in deciding how big to make the port).
R= √[(H x W) / π]
√= square root
H= 15.5 (inches)
W= 8 (inches)
π = 3.141592 (pi)
R= √[(15.5 x 8) / 3.141592]
R= √[124 / 3.141592]
R= √39.470434098380693610118691415053
R= 6.282549967837955334674739903437
Now let's fill in the rest of the equation and solve.
Lv= {(14630000 x R²) / (Fb²) x [(Vb/Np) x 1728]} - 1.463 x R
Lv=?
Fb=40
R=6.282549967837955334674739903437
Vb=6
Lv = {(14630000 x 6.28254996783795533467473990343700²)/ (40²) x [(6 / 1) x 1728]} - 1.463 x 6.28254996783795533467473990343700
Lv= {14630000 x 39.470434098380693610118691415053) / 1600 x [6 x 1728]} - 9.1913706029469286546291444787283
Lv= {577452450.85930954751603645540223 / 1600 x 10368} - 9.1913706029469286546291444787283
Lv= {577452450.85930954751603645540223 / 16588800} - 9.1913706029469286546291444787283
Lv= 34.809778335944103703464774751774 - 9.1913706029469286546291444787283
Lv= 25.618407732997175048835630273042
There we have it, my port will be 25.6" long.
================================================== ========================================
================================================== ========================================
Now to figure the box to acquire this port. First let's find how much air displacement this port has. We're assuming the box
has the port in the middle so we add wood thickness to the port width twice because of the walls found inside the box, the
top and bottom of the port are part of the outer walls resulting in no displacement for those. This will change depending
on how your port is. You should be able to figure out where you should and shouldn't need to add wood thickness in the
following equation. We also assume we're using 3/4" mdf for the wood.
Dp = {[Lv x (Ph + wt + wt)] x [(Pw + Wt + Wt)] / 1728} x Np
Dp= Port displacement (cubic ft)
Pw= Port width (inches)
Lv= Port length (inches)
Ph= Port height (inches)
Wt= wood thickness (inches)
Np= number of ports
Dp = {[Lv x (Ph + wt + wt)] x [(Pw + Wt + Wt)] / 1728} x Np
Dp= ? (cubic ft)
Pw= 8 (inches)
Lv= 25.6 (inches)
Ph= 15.5 (inches)
Wt= .75 (inches)
Np= 1
Dp = {[25.6 x (15.5 + 0 + 0)] x [(8 + .75 + .75)] / 1728} x 1
Dp = {[25.6 x 15.5] x [9.5] / 1728} x 1
Dp = {395.25 x 9.5 / 1728} x 1
Dp = {3754.875 / 1728} x 1
Dp = 2.1729600694444444444444444444444 x 1
Dp = 2.1729600694444444444444444444444
The equation for figuring out the displacement of round vents is as follows.
Dp = {[(R² x π) x Lv] / 1728} x Np
Dp= Port displacement (cubic ft)
π= 3.14592 (pi)
Lv= Port length (inches)
Np= Number of ports
This is just an example for 2 6" ports 17" long.
Dp = {[(R² x π) x Lv] / 1728} x Np
Dp= Port displacement (cubic ft)
R= 3 (inches)
π= 3.14592 (pi)
Lv= 17 (inches)
Np= 2
Dp = {[(3² x 3.14592) x 17] / 1728} x 2
Dp = {[(9 x 3.14592) x 17] / 1728} x 2
Dp = {[28.31328 x 17] / 1728} x 2
Dp = {481.32576 / 1728} x 2
Dp = .278545 x 2
Dp = 0.55709
================================================== ========================================
================================================== ========================================
Now lets make the box. My maximum dimensions are 35 x 30 x 17 (L x W x H) I only need to solve
for one side so pick one and go with it. I'll pick the width and use the following equation.
W = {(Dp + Vb + Sw + Ex) x 1728 / [L - (wt x 2)] x [(H - (wt x 2)]} + (wt x 2)
L= Length (inches)
W= Width (inches)
H= Height (inches)
Ex= Extra displacements: braces, dope, guns, whatever etc. (cubic ft.)
Dp= Port displacement (cubic ft)
Vb= Total internal airspace (cubic ft)
Sw= Subwoofer displacement (cubic ft)
Wt= wood thickness (inches)
We already know most of it so let's fill it in and solve it. You can find subwoofer displacement
by looking at the T/S parameters usually listed as Vd.
W = {(Dp + Vb + Sw + Ex) x 1728 / [L - (wt x 2)] x [(H - (wt x 2)]} + wt x 2
L= 35
W= ?
H= 17
Dp= 2.1729600694444444444444444444444
Vb= 6
Sw= .2
Wt= .75
Ex= 0
W = {(2.1729600694444444444444444444444 + 6 + .2 + 0) x 1728 / [35 - (.75 x2 )] x [(17 - (.75 x 2)]} + .75 x2
W = {8.372960069444444444444444444444 x 1728 / [35 - 1.5] x [17 - 1.5]} + .75 x 2
W = {14468.474999999999999999999999999 / [33.5 x 15.5]} + 1.5
w = {14468.474999999999999999999999999 / 519.25} + 1.5
w = {27.864179104477611940298507462667} + 1.5
w = 29.364179104477611940298507462667
That's it, we're done. The box is 17" x 35" x 29.3" with a square port in the center of the box 15.5" x 8" x 25.6"
================================================== ========================================
================================================== ========================================
The formulas to remember:
Vent Length
__________________________________________________ ________________________________________
Lv= {(14630000 x R²) / (Fb²) x [(Vb/Np) x 1728]} - 1.463 x R
Lv= Port Length (inches)
Fb= Tune Frequency (Hertz)
R= See ******* (this will use different equations depending on round or square vents)
Vb= total internal airspace (cubic ft)
Np= Number of ports
R value for Square vent
__________________________________________________ ________________________________________
R= √[( H x W) / π]
√= square root
H= height of port (inches)
W= width of port (inches)
π = 3.141592 (pi)
R value for Round vent
__________________________________________________ ________________________________________
R= (Dia / 2)
Dia= Vent inside Diameter
Square vent displacement
__________________________________________________ ________________________________________
Dp = {[Lv x (Ph + wt + wt)] x [(Pw + Wt + Wt)] / 1728} x Np
Dp= Port displacement (cubic ft)
Pw= Port width (inches)
Lv= Port length (inches)
Ph= Port height (inches)
Wt= wood thickness (inches)
Np= number of ports
Round vent displacement
__________________________________________________ ________________________________________
Dp = {[(R² x π) x Lv] / 1728} x Np
Dp= Port displacement (cubic ft)
π= 3.14592 (pi)
Lv= Port length (inches)
Np= Number of ports
Box Layout
__________________________________________________ ________________________________________
W = {(Dp + Vb + Sw + Ex) x 1728 / [L - (wt x 2)] x [(H - (wt x 2)]} + (wt x 2)
L= Length (inches)
W= Width (inches)
H= Height (inches)
Ex= Extra displacements: braces, dope, guns, whatever etc. (cubic ft.)
Dp= Port displacement (cubic ft)
Vb= Total internal airspace (cubic ft)
Sw= Subwoofer displacement (cubic ft)
Wt= wood thickness (inches)
2. ## Re: How To: Design a ported box using pencil & paper
I did very well in calculus thank you very much...
but I seriously just **** myself. Unless you can write a source code for that and pack it into a nice little program for people to use, then I just can't see it being sought after by very many people. I'm not doubting it's accuracy, I'm just saying it's not exactly "ca.com user friendly". People around here like instant gratification
3. ## Re: How To: Design a ported box using pencil & paper
Originally Posted by DNick454
I did very well in calculus thank you very much...
but I seriously just **** myself. Unless you can write a source code for that and pack it into a nice little program for people to use, then I just can't see it being sought after by very many people. I'm not doubting it's accuracy, I'm just saying it's not exactly "ca.com user friendly". People around here like instant gratification
I'll look into writing a program when I get home. Depends whether or not I can remember how And if I still have all my programs etc.
4. ## Re: How To: Design a ported box using pencil & paper
I could work this into an excel formula I might give it a try
5. ## Re: How To: Design a ported box using pencil & paper
That'd be insanely easy to make a little vb prog to do.
Might actually do it later tonight...
im
7. ## Re: How To: Design a ported box using pencil & paper
Originally Posted by Anniku989
That'd be insanely easy to make a little vb prog to do.
Might actually do it later tonight...
That's what I was gonna do though
8. ## Re: How To: Design a ported box using pencil & paper
OR... you could just use this...
Which is the same thing, but with much less number crunching. We use winISD to do the port length. You could also use the above method with any number of excel spreadsheets...
It's common sense, simple algebra, and has been type up nearly the same way hundreds of times...
9. ## Re: How To: Design a ported box using pencil & paper
too much math involved here. theres a reason i switched from an engineering major to a construction management major.
but... to the op, good job being able to figure that **** out
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##### Activity
Charlene Chambliss
@pzivich Thanks Paul! Looking into your sparsity suggestion I realized the DF with the interaction terms was not merging correctly with the main DF (it was dropping about 80% of the data). I fixed this issue and the model fits correctly now. Oops. Thanks for pointing me in the right direction :)
Cameron Davidson-Pilon
@CamDavidsonPilon
:raised_hands: glad you got it working!
Cameron Davidson-Pilon
@CamDavidsonPilon
I added some more helpful context for users to check when convergence fails.
Rob deCarvalho
@robdmc
Hi @CamDavidsonPilon Question about custom fitters. I was looking at this documentation https://lifelines.readthedocs.io/en/latest/jupyter_notebooks/Piecewise%20Exponential%20Models%20and%20Creating%20Custom%20Models.html . Before I put any effort into experimenting, I was wondering if it would be possible to make one of the parameters an arbitrary list. Say for example, I wanted the date associated with each element of the times parameter. If this were possible, I think it might allow me to add seasonality to a competing risk model that captures the cumulative hazard of the outcome-of-interest. So I guess my question is two-fold. a) Is that possible with lifelines, b) Does that make sense for modeling competing risk.
Rob deCarvalho
@robdmc
Here is a crude sketch of what I'd like to do.
class SeasonalHazardFitter(ParametericUnivariateFitter):
"""
The idea of this class would be to fit custom seasonality to an
exponential-like hazard model.
"""
_fitted_parameter_names = ['a_q1_', 'a_q2_', 'a_q3_', 'a_q4_' 'dates']
def _cumulative_hazard(self, params, times):
# Pull out fiscal quarters and dates corresponding to times.
# Each element of the dates array corresponds an element of the
# times array.
a_q1_, a_q2_, a_q3_, a_q4_, dates = params
# Call a function that associates fiscal quarter with date
quarters = get_fiscal_quarters(dates)
# Get the hazard for each time
q_lookup = {1: a_q1_, 2: a_q2_, 3: a_q3_, 4:a_q4}
hazards = np.array([q_lookup[quarter] for quarter in quarters])
# Return the cumulative hazard
# You'd have to be more careful to actually do the
# integration properly, but you get the idea.
return np.cumsum(hazards)
Cameron Davidson-Pilon
@CamDavidsonPilon
@robdmc, but dates isn't an unknown, is it? If not, if could be a global variable. If it is unknown, then I think you'll need to "flatten" it, i.e. one parameter for each element of the list.
Rob deCarvalho
@robdmc
@CamDavidsonPilon You are correct. dates are not an unknown. They are known constants. It makes sense that everything that goes into params should be unknown. Not sure what I was thinking there. Putting it in a global/class/instance variable makes sense. I just want to be sure I understand how _cumulative_hazard() is called.
params: get tweaked by the optimization
times: the times passed into the fitter as "durations"
return: The cumulative hazard encountered over the duration represented by each time
Is that right?
Rob deCarvalho
@robdmc
The process I am trying to model consists of two competing kinds of events. The hazard for each event is a function of date. So the cumulative hazard for each time would be the integral of the hazard from the "start_date" to the "end_date". (where these can be derived from an element of time and its corresponding date.) What I really care about is the cumulative incidence function (CIF) for each kind of event. If the idea of getting dates into the _cumulative_hazard function works, then I was hoping to use this technique to model the CIF for one of the competing event types.
Is this making sense?
Cameron Davidson-Pilon
@CamDavidsonPilon
Your explanation of _cumulative_hazard is correct. But you can also see it as simply the cumulative hazard you wish to implement (i.e., not necessary to think about "durations" or "unknowns")
I was thinking about your seasonal model, and actually tried to code something up, but there is a problem I think. The _cumulative_hazard is invoked for both the censored and uncensored data, so your code needs to handle that (and you won't know which until you see the shapes of the input data)
Cameron Davidson-Pilon
@CamDavidsonPilon
yea I don't know if this can be done... I'm playing with it locally, and having some trouble
I'll think more about it. Try to write down the hazard mathematically - I think the problem is that it is clock-time dependent.
Rob deCarvalho
@robdmc
Thank you for thinking about it. Clock-dependent hazards I think are actually pretty common
I love this interface you have for arbitrary models. If there was a way to hack that, it could be pretty useful.
Rob deCarvalho
@robdmc
maybe with (..., *args, **kwargs) to the _cumulative_hazard? I actually don't understand very well how _cumulative_hazard is used under the hood, so perhaps I'm spouting nonsense.
Cameron Davidson-Pilon
@CamDavidsonPilon
(..., *args, **kwargs) I was thinking about this, too
Clock-dependent hazards I think are actually pretty common
Agree, but I feel like the common strategy is to use a regression model or fit N univariate models (i.e. partition the data)
I think a seasonal model is a great idea, so I want this to work.
Cameron Davidson-Pilon
@CamDavidsonPilon
:wave: new lifelines release: 0.22.0. Some important API changes to take a look at, but some really powerful new regression models: https://github.com/CamDavidsonPilon/lifelines/releases/tag/v0.22.0
Julian Späth
@julianspaeth
Hi all, does lifelines somehow offer a Random Survival Forest? Or is there a specific reason why not? As there is no real python implementation of RSF and I want to implement it for my Master thesis, I was wondering if you are interested in including it into lifelines?
Cameron Davidson-Pilon
@CamDavidsonPilon
Hi Julian, lifelines does not have a RF model. Maybe scikit-learn survival does though.
lifelines has focused less on purely predictive models, and more on inference
Julian Späth
@julianspaeth
Hi Cameron, thank you for your answer. As far as I can see scikit-survival does not have a RF model. So I guess I need to implement it from scratch to use it in python. Thank you anyways 🙂
Cameron Davidson-Pilon
@CamDavidsonPilon
hm, I thought it was, okay - have fun!
Pedro Sola
@pedrosola
Hi everyone, I'm trying to fit a model onto a recurrent process. I.E: Patient returns to a doctor. Is there a way to do so using lifelines ? So far the closest that I've got was this repo: https://github.com/dunan/MultiVariatePointProcess
Cameron Davidson-Pilon
@CamDavidsonPilon
Lifelines has limited support for recurring events.
Unfortunately
mohit
@mohit-shrma
Hi, I am using CoxPHFitter with IPS weights and robust=True flag. However, the fit is taking really long time to finish. I have about million instances and 6 features in my dataset. Let me know if slower runtime is expected in weighted version and what can be done to speed it up.
mohit
@mohit-shrma
@CamDavidsonPilon Let me know if you have any suggestions for question below:
Hi, I am using CoxPHFitter with IPS weights and robust=True flag. However, the fit is taking really long time to finish. I have about million instances and 6 features in my dataset. Let me know if slower runtime is expected in weighted version and what can be done to speed it up.
Cameron Davidson-Pilon
@CamDavidsonPilon
hello! A million is a lot, much more than needed for only 6 features. I would suggest subsampling to 50k or even less, and checking the std. errors.
Cameron Davidson-Pilon
@CamDavidsonPilon
@mohit-shrma, another suggestion is to "collapse" similar rows and use weights. Ex: with only 6 variables, you likely have the same row appear twice. You can group these, assign that row an integer count, and use the weight_col argument in fit
Cameron Davidson-Pilon
@CamDavidsonPilon
:wave: minor version of lifelines released: https://github.com/CamDavidsonPilon/lifelines/releases/tag/v0.22.1
Robert Green
@rgreen13_gitlab
Hi all. I've somewhat new to using lifelines, and in using the CoxPHFitter, when I run check_assumptions, I end up with an error that reads as follows: /RuntimeWarning: overflow encountered in exp scores = weights * np.exp(np.dot(X, self.params_))
Any suggestions on dealing with this issue? I'm starting down the road of normalization, but I'm not sure if that's 100% correct.
Cameron Davidson-Pilon
@CamDavidsonPilon
@rgreen13_gitlab hi, thanks for reporting this. I'll create a bug issue around it. For now, you can try scaling and normalizing your matrix before calling .fit
Robert Green
@rgreen13_gitlab
Thanks!
sharmarishika
@sharmarishika
Hi! I am currently trying to create mixed cure models using the lifelines fitter. I saw that there is an example code in the GitHub under experiments. I was going to use this as a starting point and then adjust accordingly but I am getting an error when I run that code saying: "AttributeError: 'CureModel' object has no attribute '_primary_parameter_name'
I don't have a full understanding of the input arguments for _cumulative_hazard so I am not sure what is causing this error. Thank you!
Cameron Davidson-Pilon
@CamDavidsonPilon
@sharmarishika hi there. Are you using lifelines >= 0.22.0?
If not, try upgrading. Otherwise, if you are still getting the error, can you post the entire stack trace?
Cameron Davidson-Pilon
@CamDavidsonPilon
(Also, make sure you are subclassing ParametricRegressionFitter, and not ParametericAFTRegressionFitter)
sharmarishika
@sharmarishika
@CamDavidsonPilon ah - I think I'm using version 0.19.5! when i install through pip it says 'requirement already satisfied' - would you recommend a different way of upgrading?
in reference to the subclass my computer doesn't recognize ParametricRegressionFitter as an option but it does recognize ParametericRegressionFitter - perhaps also because of the version?
mohit
@mohit-shrma
hello! A million is a lot, much more than needed for only 6 features. I would suggest subsampling to 50k or even less, and checking the std. errors.
@CamDavidsonPilon thanks for the advice, I will try that idea.
Cameron Davidson-Pilon
@CamDavidsonPilon
@sharmarishika try pip install -U lifelines or pip install lifelines==0.21.1
sharmarishika
@sharmarishika
@CamDavidsonPilon i get an error saying that there are no matching distributions for lifelines 0.21.1 - in the list provided the most recent version is 0.19.5
Cameron Davidson-Pilon
@CamDavidsonPilon
is this conda?
sharmarishika
@sharmarishika
i was using pip, in my command line typed exactly what i mentioned above - sorry i'm quite new to this!
Cameron Davidson-Pilon
@CamDavidsonPilon
@sharmarishika ah, are you on Python2? 0.19.5 was the latest Py2 release. It's only Python3 now
sharmarishika
@sharmarishika
oh i see! ya i'm using python2.7 - i should look into doing the upgrade! thanks for the help!
Charlene Chambliss | 2,786 | 10,940 | {"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} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.878122 |
https://cboard.cprogramming.com/cplusplus-programming/141531-if-else-statement-question.html | 1,516,698,421,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891791.95/warc/CC-MAIN-20180123072105-20180123092105-00704.warc.gz | 631,878,231 | 12,978 | Thread: if and else statement question
1. if and else statement question
I was assigned to input a 4 digit by the user and output each number in every line. My question is in the code as comments that follows:
Code:
```#include <iostream>
#include <string>
using namespace std;
void a();
void b();
void c();
void d();
bool die( const string & msg );
int main(){
a();
b();
c();
d();
}
void a(){
unsigned noko1;
cin >> noko1;
if( !cin ) die( "non-numeric input" ); // if person inputs a non digit value complain and die
if( noko1 <= 0 || noko1 > 9999 ) // this states the range beetween 1 and 1000 if not it dies
die( "out of range");
if(noko1 == noko1)
cout<<(noko1%10000)/1000<<endl;
cout<<(noko1%1000)/100<<endl;
cout<<(noko1%100)/10<<endl;
cout<<(noko1%10)/1<<endl;
// else what?? if i input else <<cout"do nothing" endl;
// sends me an error
// is it ok to leave it like this without the else? thanks
} // a
void b(){
} // b
void c(){
} // c
void d(){
} // d
bool die( const string & msg ){
cerr <<endl <<"Fatal error: " <<msg <<endl;
exit( EXIT_FAILURE );
} // die```
2. This statement:
Code:
` if(noko1 == noko1)`
Will always evaluate to true. What do you even have it?
3. else isn't compulsory, if that is what you're asking.
4. if and else statement question
Originally Posted by elpedoloco
Code:
``` if(noko1 == noko1)
cout<<(noko1%10000)/1000<<endl;
cout<<(noko1%1000)/100<<endl;
cout<<(noko1%100)/10<<endl;
cout<<(noko1%10)/1<<endl;```
You may have tabbed those lines out a bit, but that doesn't mean that they are all only executed when the if-statement is true. To make that happen, you need to put curly brackets around all of those statements. The lack of these brackets is also why you are getting an error when trying to put an else in there.
You're going to need to fix that tautology though, before your program will do whatever you meant it to do.
5. thanks I am working on 5 different programs I've fixed most of my code I will post once I am done so I can get your opinion, as always, your help is appreciated. | 602 | 2,077 | {"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-2018-05 | latest | en | 0.797704 |
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# boolean operations on sets
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the
standard Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
In [7]:set('casa') & set('porca')
Out[7]:set(['a', 'c'])
In [8]:set('casa') | set('porca')
Out[8]:set(['a', 'c', 'o', 'p', 's', 'r'])
and they work correctly. Now what is confusing is that if you do:
In [5]:set('casa') and set('porca')
Out[5]:set(['a', 'p', 'c', 'r', 'o'])
In [6]:set('casa') or set('porca')
Out[6]:set(['a', 'c', 's'])
The results are not what you would expect from an AND or OR
operation, from the mathematical point of view! aparently the "and"
operation is returning the the second set, and the "or" operation is
returning the first.
If python developers wanted these operations to reflect the
traditional (Python) truth value for data structures: False for empty
data structures and True otherwise, why not return simply True or
False?
So My question is: Why has this been implemented in this way? I can
see this confusing many newbies...
Aug 6 '07 #1
7 1729
Flavio wrote:
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the
standard Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
In [7]:set('casa') & set('porca')
Out[7]:set(['a', 'c'])
In [8]:set('casa') | set('porca')
Out[8]:set(['a', 'c', 'o', 'p', 's', 'r'])
and they work correctly. Now what is confusing is that if you do:
In [5]:set('casa') and set('porca')
Out[5]:set(['a', 'p', 'c', 'r', 'o'])
In [6]:set('casa') or set('porca')
Out[6]:set(['a', 'c', 's'])
The results are not what you would expect from an AND or OR
operation, from the mathematical point of view! aparently the "and"
operation is returning the the second set, and the "or" operation is
returning the first.
If python developers wanted these operations to reflect the
traditional (Python) truth value for data structures: False for empty
data structures and True otherwise, why not return simply True or
False?
So My question is: Why has this been implemented in this way? I can
see this confusing many newbies...
it has been implemented in this way to conform with the definitions of
"and" and "or", which have never been intended to apply to set
operations. The result of these operations has always returned one of
the operands in the case where possible, and they continue to do so with
set operands.
regards
Steve
--
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----------- Thank You for Reading -------------
Aug 6 '07 #2
On Monday 06 August 2007, Flavio wrote:
So My question is: Why has this been implemented in this way? I can
see this confusing many newbies...
I did not implement this, so I cannot say, but it does have useful
side-effects, for example:
x = A or B
is equivalent to:
if A:
x = A
else:
x = B
also, in python implementations without the (y if x else z) syntax, you can
use (x and y or z) with nearly the same result*. Also, this implementation of
and/or might well be faster ;-)
*: this doesn't work the same if y is a false value; (x and [y] or [z])[0] is
less readable, but works for all y
--
Regards, Thomas Jollans
GPG key: 0xF421434B may be found on various keyservers, eg pgp.mit.edu
Hacker key <http://hackerkey.com/>:
v4sw6+8Yhw4/5ln3pr5Ock2ma2u7Lw2Nl7Di2e2t3/4TMb6HOPTen5/6g5OPa1XsMr9p-7/-6
-----BEGIN PGP SIGNATURE-----
Version: GnuPG v1.4.6 (GNU/Linux)
iD8DBQBGtz6FJpinDvQhQ0sRAsN8AJ9SsIx6gj3fG+VHtXvp1a aCJ3E2WgCfeh+y
rx90H88SVRlBZbVRXmIG9Lo=
=Qgsq
-----END PGP SIGNATURE-----
Aug 6 '07 #3
Flavio wrote:
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the
standard Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
In [7]:set('casa') & set('porca')
Out[7]:set(['a', 'c'])
In [8]:set('casa') | set('porca')
Out[8]:set(['a', 'c', 'o', 'p', 's', 'r'])
and they work correctly. Now what is confusing is that if you do:
In [5]:set('casa') and set('porca')
Out[5]:set(['a', 'p', 'c', 'r', 'o'])
In [6]:set('casa') or set('porca')
Out[6]:set(['a', 'c', 's'])
The results are not what you would expect from an AND or OR
operation, from the mathematical point of view! aparently the "and"
operation is returning the the second set, and the "or" operation is
returning the first.
If python developers wanted these operations to reflect the
traditional (Python) truth value for data structures: False for empty
data structures and True otherwise, why not return simply True or
False?
So My question is: Why has this been implemented in this way? I can
see this confusing many newbies...
It has nothing to do with sets - it stems from the fact that certain values
in python are considered false, and all others true. And these semantics
were introduced at a point where there was no explicit True/False, so the
operators were defined in exact the way you observed.
Consider this:
"foo" or "bar" -"foo"
So - nothing to do with sets.
Diez
Aug 6 '07 #4
On Mon, 06 Aug 2007 14:13:51 +0000, Flavio wrote:
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the standard
Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
In [7]:set('casa') & set('porca')
Out[7]:set(['a', 'c'])
In [8]:set('casa') | set('porca')
Out[8]:set(['a', 'c', 'o', 'p', 's', 'r'])
and they work correctly. Now what is confusing is that if you do:
In [5]:set('casa') and set('porca')
Out[5]:set(['a', 'p', 'c', 'r', 'o'])
In [6]:set('casa') or set('porca')
Out[6]:set(['a', 'c', 's'])
The results are not what you would expect from an AND or OR operation,
from the mathematical point of view! aparently the "and" operation is
returning the the second set, and the "or" operation is returning the
first.
That might be, because `and` and `or` are not mathematical in Python (at
least not as you think). All the operator bits, e.g. `|` and `&`, are
overloadable. You can just give them any meaning you want.
The `and` and `or` operator, though, are implemented in Python and there
is no way you can make them behave different from how they do it by
default. It has been discussed to remove this behaviour or make them
overloadable as well but this hasn't made it far, as far as I remember.
If python developers wanted these operations to reflect the traditional
(Python) truth value for data structures: False for empty data
structures and True otherwise, why not return simply True or False?
Because in the most cases, returning True of False simply has no
advantage. But returning the actual operands has been of fairly large
use, e.g. for replacing the if expression ("ternary operator") ``THEN if
COND else DEFAULT`` with ``COND and THEN or DEFAULT`` (which has some bad
corner cases, though).
So My question is: Why has this been implemented in this way? I can see
this confusing many newbies...
Hmm, you could be right there. But they shouldn't be biased by default
boolean behaviour, then, anyways.
Aug 6 '07 #5
In article <5h*************@mid.uni-berlin.de>,
"Diez B. Roggisch" <de***@nospam.web.dewrote:
Flavio wrote:
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the
standard Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
[...]
The results are not what you would expect from an AND or OR
operation, from the mathematical point of view! aparently the "and"
operation is returning the the second set, and the "or" operation is
returning the first.
[...]
It has nothing to do with sets - it stems from the fact that certain values
in python are considered false, and all others true. And these semantics
were introduced at a point where there was no explicit True/False, so the
operators were defined in exact the way you observed.
Consider this:
"foo" or "bar" -"foo"
So - nothing to do with sets.
In addition to what Diez wrote above, it is worth noting that the
practise of returning the value of the determining expression turns out
to be convenient for the programmer in some cases. Consider the
following example:
x = some_function(a, b, c) or another_function(d, e)
This is a rather nice shorthand notation for the following behaviour:
t = some_function(a, b, c)
if t:
x = t
else:
x = another_function(d, e)
In other words, the short-circuit behaviour of the logical operators
gives you a compact notation for evaluating certain types of conditional
expressions and capturing their values. If the "or" operator converted
the result to True or False, you could not use it this way.
Similarly,
x = some_function(a, b, c) and another_function(d, e)
.... behaves as if you had written:
x = some_function(a, b, c)
if x:
x = another_function(d, e)
Again, as above, if the results were forcibly converted to Boolean
values, you could not use the shorthand.
Now that Python provides an expression variety of "if", this is perhaps
not as useful as it once was; however, it still has a role. Suppose,
for example, that a call to some_function() is very time-consuming; you
would not want to write:
x = some_function(a, b, c) \
if some_function(a, b, c) else another_function(d, e)
.... because then some_function would get evaluated twice. Python does
not permit assignment within an expression, so you can't get rid of the
second call without changing the syntax.
Also, it is a common behaviour in many programming languages for logical
connectives to both short-circuit and yield their values, so I'd argue
that most programmers are proabably accustomed to it. The && and ||
operators of C and its descendants also behave in this manner, as do the
AND and OR of Lisp or Scheme. It is possible that beginners may find it
a little bit confusing at first, but I believe such confusion is minor
and easily remedied.
Cheers,
-M
--
Michael J. Fromberger | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA
Aug 6 '07 #6
Michael J. Fromberger <Mi******************@Clothing.Dartmouth.EDU>
wrote:
...
Also, it is a common behaviour in many programming languages for logical
connectives to both short-circuit and yield their values, so I'd argue
that most programmers are proabably accustomed to it. The && and ||
operators of C and its descendants also behave in this manner, as do the
Untrue, alas...:
brain:~ alex\$ cat a.c
#include <stdio.h>
int main()
{
printf("%d\n", 23 && 45);
return 0;
}
brain:~ alex\$ gcc a.c
brain:~ alex\$ ./a.out
1
In C, && and || _do_ "short circuit", BUT they always return 0 or 1,
*NOT* "yield their values" (interpreted as "return the false or true
value of either operand", as in Python).
Alex
Aug 7 '07 #7
Flavio a écrit :
Hi, I have been playing with set operations lately and came across a
kind of surprising result given that it is not mentioned in the
standard Python tutorial:
with python sets, intersections and unions are supposed to be done
like this:
In [7]:set('casa') & set('porca')
Out[7]:set(['a', 'c'])
In [8]:set('casa') | set('porca')
Out[8]:set(['a', 'c', 'o', 'p', 's', 'r'])
and they work correctly. Now what is confusing is that if you do:
In [5]:set('casa') and set('porca')
Out[5]:set(['a', 'p', 'c', 'r', 'o'])
In [6]:set('casa') or set('porca')
Out[6]:set(['a', 'c', 's'])
The results are not what you would expect from an AND or OR
operation, from the mathematical point of view! aparently the "and"
operation is returning the the second set, and the "or" operation is
returning the first.
the semantic of 'and' and 'or' operators in Python is well defined and
works the same for all types AFAIK.
If python developers wanted these operations to reflect the
traditional (Python) truth value for data structures: False for empty
data structures and True otherwise, why not return simply True or
False?
So My question is: Why has this been implemented in this way?
Because Python long lived without the 'bool' type - considering None,
numeric zero, empty string and empty containers as false (ie :
'nothing', and anything else as true (ie : 'something').
I can
see this confusing many newbies...
Yes, and this has been one of the arguments against the introduction of
the bool type. Changing this behaviour would have break lot of existing
code, and indeed, not changing it makes things confusing.
OTHO - and while I agree that there may be cases of useless complexities
in Python -, stripping a language from anything that might confuse a
newbie doesn't make great languages.
Aug 7 '07 #8
This thread has been closed and replies have been disabled. Please start a new discussion.
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https://people.maths.bris.ac.uk/~matyd/GroupNames/480/e10/D5xDic3byC2%5E2.html | 1,702,095,857,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100800.25/warc/CC-MAIN-20231209040008-20231209070008-00279.warc.gz | 501,175,244 | 3,219 | # Extensions 1→N→G→Q→1 with N=C22 and Q=D5×Dic3
Direct product G=N×Q with N=C22 and Q=D5×Dic3
dρLabelID
C22×D5×Dic3240C2^2xD5xDic3480,1112
Semidirect products G=N:Q with N=C22 and Q=D5×Dic3
extensionφ:Q→Aut NdρLabelID
C22⋊(D5×Dic3) = D5×A4⋊C4φ: D5×Dic3/D10S3 ⊆ Aut C22606C2^2:(D5xDic3)480,979
C222(D5×Dic3) = Dic3×C5⋊D4φ: D5×Dic3/C5×Dic3C2 ⊆ Aut C22240C2^2:2(D5xDic3)480,629
C223(D5×Dic3) = Dic1516D4φ: D5×Dic3/Dic15C2 ⊆ Aut C22240C2^2:3(D5xDic3)480,635
C224(D5×Dic3) = D5×C6.D4φ: D5×Dic3/C6×D5C2 ⊆ Aut C22120C2^2:4(D5xDic3)480,623
Non-split extensions G=N.Q with N=C22 and Q=D5×Dic3
extensionφ:Q→Aut NdρLabelID
C22.1(D5×Dic3) = D20.3Dic3φ: D5×Dic3/C5×Dic3C2 ⊆ Aut C222404C2^2.1(D5xDic3)480,359
C22.2(D5×Dic3) = D20.2Dic3φ: D5×Dic3/Dic15C2 ⊆ Aut C222404C2^2.2(D5xDic3)480,360
C22.3(D5×Dic3) = C60.28D4φ: D5×Dic3/C6×D5C2 ⊆ Aut C221204C2^2.3(D5xDic3)480,34
C22.4(D5×Dic3) = C12.6D20φ: D5×Dic3/C6×D5C2 ⊆ Aut C222404C2^2.4(D5xDic3)480,37
C22.5(D5×Dic3) = (C2×C6).D20φ: D5×Dic3/C6×D5C2 ⊆ Aut C221204C2^2.5(D5xDic3)480,71
C22.6(D5×Dic3) = D5×C4.Dic3φ: D5×Dic3/C6×D5C2 ⊆ Aut C221204C2^2.6(D5xDic3)480,358
C22.7(D5×Dic3) = (C6×Dic5)⋊7C4φ: D5×Dic3/C6×D5C2 ⊆ Aut C22240C2^2.7(D5xDic3)480,604
C22.8(D5×Dic3) = Dic5×C3⋊C8central extension (φ=1)480C2^2.8(D5xDic3)480,25
C22.9(D5×Dic3) = C30.21C42central extension (φ=1)480C2^2.9(D5xDic3)480,28
C22.10(D5×Dic3) = C60.93D4central extension (φ=1)240C2^2.10(D5xDic3)480,31
C22.11(D5×Dic3) = C60.13Q8central extension (φ=1)480C2^2.11(D5xDic3)480,58
C22.12(D5×Dic3) = C30.24C42central extension (φ=1)480C2^2.12(D5xDic3)480,70
C22.13(D5×Dic3) = C2×D5×C3⋊C8central extension (φ=1)240C2^2.13(D5xDic3)480,357
C22.14(D5×Dic3) = C2×C20.32D6central extension (φ=1)240C2^2.14(D5xDic3)480,369
C22.15(D5×Dic3) = C2×Dic3×Dic5central extension (φ=1)480C2^2.15(D5xDic3)480,603
C22.16(D5×Dic3) = C2×D10⋊Dic3central extension (φ=1)240C2^2.16(D5xDic3)480,611
C22.17(D5×Dic3) = C2×C30.Q8central extension (φ=1)480C2^2.17(D5xDic3)480,617
×
𝔽 | 1,164 | 1,960 | {"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-2023-50 | latest | en | 0.485483 |
https://www.indiabix.com/aptitude/problems-on-numbers/discussion-262-2-4 | 1,708,532,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00321.warc.gz | 881,563,775 | 9,234 | # Aptitude - Problems on Numbers - Discussion
Discussion Forum : Problems on Numbers - General Questions (Q.No. 4)
4.
The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?
4
8
16
None of these
Answer: Option
Explanation:
Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.
Let ten's and unit's digits be 2x and x respectively.
Then, (10 x 2x + x) - (10x + 2x) = 36
9x = 36
x = 4.
Required difference = (2x + x) - (2x - x) = 2x = 8.
Discussion:
44 comments Page 2 of 5.
Jonsihay said: 8 years ago
The Problem should be re-stated specially on the aspect of the ratio.
It should "NOT JUST BE" 'if the ratio between the digits of the number is 1:2. But it should be restated as 'if the ratio between the unit's and ten's digits of the number is 1:2.
Then if that's ok, I think reliably be solvable!
(1)
Leopard said: 1 decade ago
Let's the digits are x & y.
The ratio between them is x:y = 1:2.
So y=2x.
Hence y is ten digit no.
According to statement,
(10y+x)-(10x+y) = 36.
(10*2x+x)-(10x+2x) = 36.
21x-12x = 36.
9x = 36.
x = 4..
(y+x)-(y-x) = (2x+4)-(2x-4).
= (8+4)-(8-4).
= 12-4 = 8.
Maganizo Chithila said: 2 years ago
Ratio is 1:2
we will consider 1 and 2 as the numbers.
And have variable x after the subtracting.
then (21 -12)x=36.
Then 9x=36
And x=4 .
And have 2x for the second number which is 8.
the difference is 8-4 = 4.
the sum of 8+4 = 12.
Therefore the difference is 12-4 = 8.
(9)
Anju said: 6 years ago
Since they have mentioned the word "difference" in the problem so the answer will be 8 .From the 1st part of the question we can find that X-Y=4, also they have mentioned X/Y=1/2 hence Y=2X;
Therefore X~2X=4;hence X=4;Y=8;
So, the answer is :(4+8)~(4~8)=8.
(1)
Bhargavi said: 3 years ago
From 3rd question,we can say x-y=4.
We have to find diff of ( sum of digits n diff of digits).
(X+y)-(x-y)=0.
X+y=4.
From 2 equations,we get X=4, sub X value then y=0,
So 2 numbers are 40 and 04,
Sum is 44,diff is 36.
So, diff is 44 - 36 = 8.
(6)
Kcs said: 1 decade ago
Two digits x&y
x:y=1:2
x=k
y=2k
two numbers 10x+y&10y+x
10x+y=10k+2k=12k
10y+x=20k+k=21k
diff=9k=36
k=4
numbers 12k=48
21k=84
sum of dgts=12
diff of dgts=4
ans=sum-diff=8
Bhavna said: 9 years ago
Assume the 2 digit number be 84 and by interchanging the digits we get 48.
Difference 84-48 = 36.
Sum = 4+8= 12.
Difference = 8-4 = 4.
Difference between sum & diff = 12-4 = 8.
Ajeya said: 6 years ago
Lets assume n1=10a+b and n2 = a+10b where a and b are digits in a two digit number,given n1-n2 = 36,
So, 10a+b-a-10b = 36,
9a-9b = 36,
a-b = 4.
How can it be 8? Please help.
Swamy said: 7 months ago
Let's assume 2 digit number is 10.
10x - 01x = 36,
9x = 36,
x = 4.
Coming to question; 1:2;
1x:2x.
1x4: 2x4.
4:8.
Then the answer is 8.
(12)
Pavithra said: 1 decade ago
I get the answer as 16.
Diff in digits: (x+y)-(x-y)
=2y
=2(2x)
=4x.
Finding the value of x we get, x=4;
So ans = 4(4) = 16.
Post your comments here:
Your comments will be displayed after verification. | 1,201 | 3,271 | {"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} | 4.3125 | 4 | CC-MAIN-2024-10 | latest | en | 0.904647 |
https://numberworld.info/1011001100101 | 1,695,807,164,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00451.warc.gz | 476,829,931 | 3,903 | # Number 1011001100101
### Properties of number 1011001100101
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
eb645c8745
Base 32:
tdi5p1q5
sin(1011001100101)
0.49144303353175
cos(1011001100101)
-0.8709097225276
tan(1011001100101)
-0.56428699877808
ln(1011001100101)
27.641962144098
lg(1011001100101)
12.00475162816
sqrt(1011001100101)
1005485.5046698
Square(1011001100101)
1.0221232244054E+24
### Number Look Up
Look Up
1011001100101 (one trillion eleven billion one million one hundred thousand one hundred one) is a very amazing figure. The cross sum of 1011001100101 is 7. If you factorisate the number 1011001100101 you will get these result 167 * 31183 * 194141. The figure 1011001100101 has 8 divisors ( 1, 167, 31183, 194141, 5207561, 32421547, 6053898803, 1011001100101 ) whith a sum of 1017092853504. 1011001100101 is not a prime number. 1011001100101 is not a fibonacci number. The figure 1011001100101 is not a Bell Number. The figure 1011001100101 is not a Catalan Number. The convertion of 1011001100101 to base 2 (Binary) is 1110101101100100010111001000011101000101. The convertion of 1011001100101 to base 3 (Ternary) is 10120122120101221012202221. The convertion of 1011001100101 to base 4 (Quaternary) is 32231210113020131011. The convertion of 1011001100101 to base 5 (Quintal) is 113031012240200401. The convertion of 1011001100101 to base 8 (Octal) is 16554427103505. The convertion of 1011001100101 to base 16 (Hexadecimal) is eb645c8745. The convertion of 1011001100101 to base 32 is tdi5p1q5. The sine of the figure 1011001100101 is 0.49144303353175. The cosine of 1011001100101 is -0.8709097225276. The tangent of 1011001100101 is -0.56428699877808. The root of 1011001100101 is 1005485.5046698.
If you square 1011001100101 you will get the following result 1.0221232244054E+24. The natural logarithm of 1011001100101 is 27.641962144098 and the decimal logarithm is 12.00475162816. that 1011001100101 is amazing number! | 764 | 2,118 | {"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-2023-40 | latest | en | 0.702649 |
https://primes.utm.edu/curios/page.php/58.html | 1,623,619,941,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610841.7/warc/CC-MAIN-20210613192529-20210613222529-00447.warc.gz | 445,402,730 | 4,286 | # 58
This number is a composite.
58 is the sum of the first seven prime numbers.
In 1869, Landry wrote that no one of the numerous factorizations of the numbers 2n ± 1 gave as much trouble and labor as that of 258 + 1.
The smallest Smith number with a prime sum of digits. Note also that its digits and their sum are a sequence of Fibonacci numbers. [Necula]
The 58th Lucas number plus 5858 is prime. This is the largest such number less than a thousand. [Opao]
58 is a Hoax Number. 58 reversed (85) is a Hoax Number and 85 = 5 x 17. Note that 5 x 71 = 355 is also a Hoax Number. [Andrew]
The only double-digit number such that 58^n+85^n is prime for n=2, 4, 16, i.e., for three powers of 2. [Loungrides]
Exactly half of the first 58 "Lunatic numbers" are prime in the traditional sense. [Gaydos]
(There are 4 curios for this number that have not yet been approved by an editor.) | 255 | 888 | {"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-2021-25 | longest | en | 0.953492 |
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# The earth s resources are being depleted much too fast. To
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SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka
The earth s resources are being depleted much too fast. To [#permalink]
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21 Sep 2005, 10:42
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The earth’s resources are being depleted much too fast. To correct this, the United States must keep its resource consumption at present levels for many years to come.
Which of the following, if true, would most strengthen the argument above?
(A) New resource deposits are constantly being discovered.
(B) The United States consumes one-third of all resources used in the world.
(C) Other countries need economic development more than the United States does.
(D) Other countries have agreed to hold their resource consumption at present levels.
(E) The United States has been conserving resources for several years.
_________________
hey ya......
Manager
Joined: 07 Oct 2004
Posts: 99
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21 Sep 2005, 10:49
B for me. Must establish a link between what is important that the US does and the impact on Earth.
Director
Joined: 09 Jul 2005
Posts: 591
### Show Tags
21 Sep 2005, 13:09
The assumption in this passage is that the US consumption is determinant. If they don't cut rising consumption the resources will deplete sooner or later.
Therefore, the best answer is B
Director
Joined: 05 Jul 2004
Posts: 894
### Show Tags
21 Sep 2005, 16:29
B.
Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea
### Show Tags
21 Sep 2005, 16:33
One more vote for D.
The earth's resources are being depleted, and the United States must reduce the use of the resources.
But why the United States?
Because it uses MUCH of the earth's resources.
Director
Joined: 15 Aug 2005
Posts: 796
Location: Singapore
### Show Tags
21 Sep 2005, 19:43
Its a straight B!
Im sure gamjatang meant B as well!
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Cheers, Rahul.
Director
Joined: 14 Sep 2005
Posts: 985
Location: South Korea
### Show Tags
21 Sep 2005, 20:45
rahulraao wrote:
Its a straight B!
Im sure gamjatang meant B as well!
Oops!
Director
Joined: 23 Jun 2005
Posts: 841
GMAT 1: 740 Q48 V42
### Show Tags
21 Sep 2005, 20:48
B as well. But I was thinking abt D too. Since the other nations are limiting their resource consumption, USA should contain it too!
SVP
Joined: 28 May 2005
Posts: 1705
Location: Dhaka
### Show Tags
22 Sep 2005, 02:24
I picked D too...
but OA is B.
I understand now why it is B but can you please explain why it is not D.
Thanks.
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hey ya......
Senior Manager
Joined: 07 Jul 2005
Posts: 404
Location: Sunnyvale, CA
### Show Tags
22 Sep 2005, 02:48
nakib77 wrote:
The earth’s resources are being depleted much too fast. To correct this, the United States must keep its resource consumption at present levels for many years to come.
Which of the following, if true, would most strengthen the argument above?
(A) New resource deposits are constantly being discovered.
(B) The United States consumes one-third of all resources used in the world.
(C) Other countries need economic development more than the United States does.
(D) Other countries have agreed to hold their resource consumption at present levels.
(E) The United States has been conserving resources for several years.
B ...
Simple point of discussion: If the question was to tell assumption instead of "Point that streangthens" : Then what would be the anser?
ur views...
Director
Joined: 23 Jun 2005
Posts: 841
GMAT 1: 740 Q48 V42
### Show Tags
22 Sep 2005, 18:51
sgrover wrote:
Simple point of discussion: If the question was to tell assumption instead of "Point that streangthens" : Then what would be the anser?
ur views...
I'd say D
Re: CR earth's resource [#permalink] 22 Sep 2005, 18:51
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Display posts from previous: Sort by | 1,311 | 4,944 | {"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} | 3.5625 | 4 | CC-MAIN-2017-26 | latest | en | 0.934153 |
http://wiki.stat.ucla.edu/socr/index.php?title=EBook_Problems_Hypothesis_S_Mean&redirect=no | 1,596,677,631,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735990.92/warc/CC-MAIN-20200806001745-20200806031745-00211.warc.gz | 119,031,947 | 7,272 | # EBook Problems Hypothesis S Mean
## EBook Problems Set - Testing a Claim about a Mean: Small Samples
### Problem 1
To test the claim that the average home in a certain town is within 5.5 miles of the nearest fire station, and insurance company measured the distances from 25 randomly selected homes to the nearest fire station and found x-bar = 5.8 miles and sd = 2.4 miles. Determine what the insurance company found out with a test of significance. Check all that apply.
• Choose at least one answer.
(a) There is no evidence in the data to conclude that the distance is different from 5.5.
(b) The average of 5.8 miles observed is by chance.
(c) We cannot reject the null.
(d) There is evidence in the data to conclude that the distance is 5.5.
### Problem 2
For mothers who are between 20 and 40 years old and who gain between 20 and 35 during the course of their pregnancies, the average birth weight of their babies is 7.0 pounds and the standard deviation is 0.85 pounds. Researchers want to determine if mothers who are also between 20 and 40 years old but who gained less than 20 pounds during the course of their pregnancies will have babies that have different birth weights.
The average weight of babies born to a random sample of 64 mothers who had gained between 10-18 pounds during the course of their pregnancy was 6.83 pounds. Should we accept or fail to reject the null hypothesis and why?
(a) We reject the alternative hypothesis ( P = 0.0548) and conclude that the average weight on the infants born to mothers who gained 10-18 pounds is not different than mothers who gained 20-35 pounds.
(b) We reject the alternative hypothesis ( P = 0.1096) and conclude that the average weight on the infants born to mothers who gained 10-18 pounds is not different than mothers who gained 20-35 pounds.
(c) We fail to reject the null hypothesis and we are 94.52% that the average weight on the infants born to mothers who gained 10-18 pounds is not different than mothers who gained 20-35 pounds.
(d) We fail to reject the alternative hypothesis ( P = 0.1096) and conclude that the average weight on the infants born to mothers who gained 10-18 pounds is not different than mothers who gained 20-35 pounds.
### Problem 3
The average length of time required to complete a certain aptitude test is claimed to be 80 minutes. A random sample of 25 students yielded an average of 86.5 minutes and a standard deviation of 15.4 minutes. If we assume normality of the population distribution, is there evidence to reject the claim?
• Choose at least one answer.
(a) Yes, because the observed 86.5 did not happen by chance
(b) Yes, because the t-test statistic is 2.11
(c) Yes, because the observed 86.5 happened by chance
(d) No, because the probability that the null is true is > 0.05 | 664 | 2,799 | {"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} | 4.1875 | 4 | CC-MAIN-2020-34 | latest | en | 0.973694 |
https://warreninstitute.org/a-commercial-refrigerator-with-refrigerant-134a-as-the-working-fluid/ | 1,721,521,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517544.48/warc/CC-MAIN-20240720235600-20240721025600-00304.warc.gz | 529,148,270 | 23,789 | # Efficient cooling with R-134a: Commercial refrigerators at their best!
Welcome to Warren Institute! In today's article, we will delve into the fascinating world of Mathematics education. Specifically, we will explore the concept of a commercial refrigerator with refrigerant 134a as the working fluid. This refrigerant plays a crucial role in maintaining optimal temperature levels, ensuring the freshness and longevity of perishable items. Join us as we unravel the mathematical principles behind this innovative cooling technology, examining its efficiency, energy consumption, and environmental impact. Stay tuned for an informative and insightful journey into the intersection of mathematics and refrigeration!
## Understanding the Thermodynamics of a Commercial Refrigerator with Refrigerant 134a
In this section, we will delve into the mathematical concepts behind the thermodynamics of a commercial refrigerator that utilizes refrigerant 134a as its working fluid.
The discussion will cover the principles of heat transfer, the ideal gas law, and the refrigeration cycle. By understanding these concepts, educators can effectively teach students about the inner workings of commercial refrigeration systems and how they utilize mathematics to function efficiently.
## Applying Mathematical Equations in Analyzing Refrigerant 134a Performance
Here, we will explore the various mathematical equations used to analyze the performance of refrigerant 134a in a commercial refrigerator.
We will discuss equations such as the Clausius-Clapeyron equation, the specific heat capacity equation, and the entropy equation. These equations play a crucial role in determining the efficiency, cooling capacity, and overall performance of the refrigeration system. Educators can use these equations to illustrate real-world applications of mathematical concepts in the field of thermodynamics.
## Optimizing Energy Efficiency through Mathematical Modeling
This section focuses on how mathematical modeling can be employed to optimize the energy efficiency of a commercial refrigerator using refrigerant 134a.
Topics covered include modeling the refrigeration cycle, optimizing heat exchanger design, and analyzing the coefficient of performance. By utilizing mathematical models, educators can demonstrate how different variables and design parameters affect the energy efficiency of the system, leading to informed decision-making and improvements in commercial refrigeration technology.
## Enhancing Mathematics Education through Practical Applications in Refrigeration
Here, we highlight the benefits of incorporating practical applications of refrigeration systems into mathematics education.
We will discuss how teaching about commercial refrigerators and refrigerant 134a can enhance students' understanding of mathematical concepts such as thermodynamics, algebraic equations, and data analysis. By integrating real-world examples, educators can make mathematics education more engaging, relevant, and applicable to students' lives and future careers in fields related to engineering, energy efficiency, and environmental sustainability.
### How can Mathematics education be integrated into the study of commercial refrigerators with refrigerant 134a as the working fluid?
Mathematics education can be integrated into the study of commercial refrigerators with refrigerant 134a as the working fluid by incorporating mathematical concepts and principles into the analysis and calculations involved in understanding the refrigeration process. This can include using mathematical equations to determine the heat transfer rates, pressure-temperature relationships, and energy efficiency of the refrigeration system. Additionally, students can apply mathematical problem-solving skills to optimize the performance of the refrigerator, such as determining the ideal operating conditions or analyzing the impact of different refrigerant properties on system performance.
### What mathematical concepts are relevant to understanding the thermodynamic properties of refrigerant 134a in a commercial refrigerator?
Calculus is relevant to understanding the thermodynamic properties of refrigerant 134a in a commercial refrigerator. This includes concepts such as integration and differentiation to analyze temperature, pressure, and volume changes in the refrigeration cycle.
### How can data analysis and statistical methods be used to optimize the efficiency of a commercial refrigerator using refrigerant 134a?
Data analysis and statistical methods can be used to optimize the efficiency of a commercial refrigerator using refrigerant 134a by collecting and analyzing data on various factors that affect its performance, such as temperature, pressure, and energy consumption. By applying statistical techniques, such as regression analysis or hypothesis testing, we can identify the most significant variables and their relationships with the efficiency of the refrigerator. This information can then be used to make informed decisions on adjustments or improvements to the system, leading to optimal efficiency.
### Are there any mathematical models or equations that can accurately predict the performance of a commercial refrigerator with refrigerant 134a?
Yes, there are mathematical models and equations that can accurately predict the performance of a commercial refrigerator with refrigerant 134a. These models incorporate thermodynamic principles and variables such as temperature, pressure, and heat transfer coefficients to calculate the refrigeration capacity, coefficient of performance, and other performance parameters. These mathematical models are based on fundamental principles of fluid mechanics and heat transfer, and they provide valuable insights into the behavior of commercial refrigerators using refrigerant 134a.
### How can mathematical problem-solving skills be developed through the study of commercial refrigerators and their use of refrigerant 134a?
Mathematical problem-solving skills can be developed through the study of commercial refrigerators and their use of refrigerant 134a by analyzing and applying mathematical concepts such as thermodynamics, heat transfer, and fluid mechanics. Students can use mathematical equations and formulas to calculate the efficiency of refrigeration systems, understand the behavior of refrigerants under different conditions, and solve problems related to temperature, pressure, and energy consumption. This real-world application of mathematics helps students develop critical thinking, analytical skills, and problem-solving strategies that can be applied in various other contexts.
In conclusion, the use of a commercial refrigerator with refrigerant 134a as the working fluid offers a practical and tangible example for Mathematics education. By analyzing the variables involved in the refrigeration process, students can develop a deeper understanding of mathematical concepts such as pressure, temperature, and volume. This hands-on approach not only enhances their problem-solving skills but also fosters critical thinking and mathematical reasoning. Additionally, incorporating real-world applications like commercial refrigeration systems into the classroom can make Mathematics more engaging and relevant to students' lives. Thus, by leveraging the power of everyday examples, educators can effectively bridge the gap between theory and practice in Mathematics education. | 1,233 | 7,468 | {"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-2024-30 | latest | en | 0.886168 |
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Appendix 1
LESSON PLAN
SCHOOL OF EDUCATION
LESSON ORGANISATION
Year Level: 2 Time: 11am-12pm Date: 30th May 2018 Students’ Prior Knowledge:
Learning Area: Mathematics – Measurement and Geometry Students have been introduced to the order
of months and seasons in each year
Strand/Topic from the Australian Curriculum
Use a calendar to identify the date and determine the number
of days in each month (ACMMG041)
General Capabilities (that may potentially be covered in the lesson)
Literacy Numeracy ICT Critical and Ethical Personal and Intercultural
✓ ✓ competence creative thinking behaviour Social understanding
competence
Cross-curriculum priorities (may be addressed in the lesson)
Aboriginal and Torres Strait Islander Asia and Australia’s engagement with Asia Sustainability
histories and cultures
Proficiencies:(Mathematics only) Understanding, fluency
Lesson Objectives (i.e. anticipated outcomes of this lesson, in point form beginning with an action verb)
As a result of this lesson, students will be able to:
1. Date a calendar
2. Determine the number of days in each month
Teacher’s Prior Preparation/Organisation: Provision for students at educational risk:
Create class calendar Work one on one with Ellie to get her started
Poem poster on the activity
June templates
Month, season, days in month cards
Index cards for exit ticket
LESSON EVALUATION (to be completed AFTER the lesson)
Assessment of Lesson Objective and Suggestions for Improvement:
I think that my objectives were measurable however of course will be ongoing and perhaps I could have made the
objectives more specific to this lesson rather than the topic. Many students are still unsure of how many days are in
each month – though they can determine that 1) not all months have the same amount of days and 2) February is
different to the other months as it only has 29 days or 28 in a leap year.
Teacher self-reflection and self-evaluation:
After feeling quite down on myself after my Monday Maths lesson, this lesson felt like a breath of fresh air. I thought
the kids were engaged and interested in what they were doing but also clearly taking in what I was teaching them!
Even the kids who did not complete the Exit Ticket with correct answers I noticed were getting involved in the
calendar activity and asking each other questions about the different days in the month. I tried to ask some key
1
questions to get students to think about the task as not only something to learn at school but a concept they will use
throughout their whole lives. In our calendar filling activity, many of the students asked to take their calendar they
made home to put on their wall, which to me felt like an accomplishment because it shows that they were interested in
continuing to learn more about the topic. I am still trying to decide the most effective “attention grabbing” technique
and after this lesson I am thinking Stop Look and Listen gets their attention best.
LESSON DELIVERY (attach worksheets, examples, marking key, etc, as relevant)
Resources/References
Time Motivation and Introduction:
Tell students that the WALT for this lesson is to date a calendar and
11am
determine the number of days in each month.
Bring students onto the front mat and ask a series of quick questions
to revise what we did in our last lesson on months and seasons.
1. What is the first month of the year?
2. What is the order of the seasons? (Remind them of SAWS)
3. Which three months are in Summer?
4. Which three months are in Autumn?
5. Which three months are in Winter?
6. Which three months are in Spring?
Introduce the new concept by asking if anyone has a calendar they
look at every day at home (If not, suggest that that mum/dad might
have a calendar in the living room/kitchen.)
Explain that a good way to keep up with different things happening in
the month is to use a calendar.
“Can anyone explain to the class what a calendar is?”
Tell students that a calendar is a chart with twelve pages, with one
page being dedicated to each month of the year. On a calendar we
can see days, weeks and months of a particular year and we can fill it
in with different events, celebrations or seasons.
Show students the video that explains calendars, days, months and YouTube Video
(Pause at 3:25)
Present the “Class Calendar” to students and tell them that from Class Calendar
today, a new job will be added to the list of daily jobs. The job will be
to change the day, date and weather for each day of the week.
Look at the calendar with students and explain the purpose of each
box: months, days, season, weather and why it is important for us to
have a daily calendar in our classroom.
11:15am
Ask individual students to stick the correct day, date, month, weather
2
and season on the calendar for today.
Poem poster
Explain to students their task for this lesson. They will each be given
a calendar template for the month June (which begins in two days).
The first task is to think about their commitments for the month of
June and fill them in, in appropriate days. Prompt them by asking
what their daily commitments are (afterschool sport, cocurricular
activities, afterschool care, playdates) Then tell them to fill in more
11:20am formal plans (such as weekend plans, birthdays, parties). If they have
no plans ask them to make pretend plans with other students in the June calendar template
class.
Questions sheet
Give students ten minutes to do this, then hand out a sheet with
questions related to the month of June. Have them glue this sheet
into their maths book and answer the questions on the opposite page
with the June calendar as their guide.
If students finish early give them jumbled month cards, season cards
and cards signalling number of days in a month. Have them order the
months, then add the season then match the month with the number
of days in that month (refer to the poem poster if need be)
Month, seasons and
number of days cards
Call students back to the front mat to teach them the poem about
11:40am months in a year. Show them the poster and read along together.
Lesson Closure:(Review lesson objectives with students)
To wrap up the lesson and review objectives, students will complete
their exit tickets. First have them write their name on the back, then
11:45am 1. Do all months have the same amount of days?
2. Which month has 28 days?
3. How many months of the year?
4. Why do we use a calendar?
11:50am Index cards for exit
Transition: (What needs to happen prior to the next lesson?) ticket
Call students up to hand in their exit ticket then instruct them to wait
patiently on the mat for further instruction from Mrs C.
Assessment: (Were the lesson objectives met? How will these be judged?)
Collect exit tickets at the end of the lesson to see whether or not
objectives have been met. Copy answers into table below:
3
30th May 2018 Do all months Which month How many Why do we
NAMES have the has 28 days? months are use a
same amount there in a year? calendar?
of days?
Chloe ✓ ✓ ✓ ✓
Hunter X ✓ ✓ ✓
Rebecca ✓ X X ✓
Hudson ✓ X X X
Jordi ✓ ✓ X ✓
Carla ✓ ✓ ✓ ✓
Bella ✓ ✓ ✓ ✓
Shiloh X ✓ ✓ ✓
Maya ✓ X ✓ ✓
Eva ✓ ✓ ✓ ✓
Millie ✓ ✓ ✓ ½
Izzy ✓ ✓ ✓ ✓
Taylan ✓ X ✓ X
Brodie ✓ ✓ ✓ ✓
Lawrence ✓ ✓ ✓ ✓
Hayley ✓ ✓ ✓ ✓
Mia ✓ ✓ X ✓
Cohen ✓ ✓ ✓ ✓
Ellie ✓ X X X
Skye ✓ ✓ ✓ ✓
Casey ✓ ✓ ✓ ✓ | 1,766 | 7,387 | {"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-2019-35 | latest | en | 0.91685 |
https://www.stillunfold.com/science/interesting-facts-about-friction-your-teacher-won-t-tell-you | 1,611,548,869,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703564029.59/warc/CC-MAIN-20210125030118-20210125060118-00780.warc.gz | 986,530,768 | 9,694 | # 9 Interesting Facts About Friction Your Teacher Won’t Tell You
If you are a science student, then you must have studied a lot about friction. But here are some interesting friction facts that you might not have been taught in school.
3 years ago
Have you ever wondered why you don’t slide easily across the cement in your shoes and smoothly slide on a wooden floor even in socks? The answer is friction. It is a powerful force that acts between two things in contact and controls the movement of the object.
### 1. Friction Is Both Harmful And Helpful
Source = Haikudeck
Friction positively affects the car movement. For instance, without friction, your car cannot move i.e. the tyres cannot push against the surface (ground) to move the car and thereby brakes cannot stop the car.
On the other hand, it can cause problems too. Consider this example. Friction between any engine parts only increases the temperature and thus causes engine parts to break down; therefore, a coolant is added to keep the temperature normal.
And this how friction can be both harmful and helpful in practical world. (14.1)
### 2. Weight Is An Important Element In Friction
Source = Wikimedia
According to the three laws of physics, weight is an important factor in friction. The first law says that friction is directly proportional to the weight of the object. According to the second law, it is not calculated by the volume and area of an object. The last law suggests that it is independent of the speed at which an object is active.
### 3. Different Types Of Friction Reveal Different Features
Source = Myinterestingfacts
There are different types of friction you experience in your daily life. Fluid Friction depicts the friction between the layers of liquid crossing each other. Dry Friction explains the friction between the motion of two solid surfaces when comes in contact. Lubricated Friction is a fluid friction when a lubricant fluid divides into two solid surfaces.
### 4. The Laws Of Sliding Friction Date Back To 1493
Source = Internapcdn
The laws of sliding friction were introduced by Leonardo Da Vinci in 1493. Unfortunately, his laws were not documented in books and the laws remained unknown. Luckily, his laws were rediscovered by Guillaume Amontons in 1699 and later became known widely as Amonton’s 3 laws of dry friction. (14.2)
### 5. Coulomb’s Law Is Used For Calculating Force Of Dry Friction
Source = Thefamouspeople
After knowing the friction laws and understanding what friction is, Charles-Augustin de Coulomb further investigated the impact of other factors in friction. He checked the normal pressure, nature of materials, size of area and length of time the surface remained in touch with another surface.
### 6. Friction Often Heats Things Up
Source = Scienceabc
As the sliding object slows down, most of the amount of energy is transformed into heat. The friction converts ordered energy into thermal energy or heat. For example, in match lighting, the glass powder helps in creating friction that’s needed for igniting flammable compounds at the time of striking process.
### 7. There’s A Strong Connection Between Friction And Energy
Source = Mikesierra
Another interesting friction fact that you might know is that friction has a lot to do with energy. If the two objects are rubbing each other and one is sliding down, it means the objects are losing the energy. The energy released is not at all real and is rather kinetic energy.
### 8. Friction Blisters Can Cause Pain And Damage To The Tissue
Source = Thefootcaregroup
Friction blister is a tiny pocket of puffy, raised skin with fluid and is most importantly caused due to continuous pressure. It usually occurs on the feet when fitting shoes irritate toes for long period of time. The best way to prevent this is by wearing shoes that fit you well, meaning, they should not be too tight or too loose. (14.3)
### 9. Friction Is A Form Of Energy That Can’t Be Reused
Source = 2025kids
Friction is a way for a system to lose 'useful' energy. This energy doesn’t disappear and converts mostly into heat but it could also be lost to sound (for instance, things produce screeching or vibrating sound when slid across the floor) or other 'useless' forms of energy.
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By Kevin Green | 960 | 4,444 | {"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} | 3.421875 | 3 | CC-MAIN-2021-04 | longest | en | 0.933516 |
https://serverlogic3.com/what-is-xor-in-data-structure/ | 1,713,080,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00359.warc.gz | 475,630,226 | 16,126 | # What Is XOR in Data Structure?
//
Angela Bailey
What Is XOR in Data Structure?
The XOR operation, also known as the exclusive OR operation, is a fundamental bitwise operation in computer programming and data structures. XOR stands for “exclusive or” because it returns true only if exactly one of the two operands is true.
Understanding the XOR Operation
The XOR operation takes two binary values and performs a comparison bit by bit. If the corresponding bits of the operands are different, it returns 1; otherwise, it returns 0. Here’s a truth table to illustrate this:
A B A XOR B
0 0 0
0 1 1
1 0 1
1 1 0
XOR in Data Structures and Algorithms:
The XOR operation has various applications in data structures and algorithms. One common use case is linked list manipulation, specifically for implementing a doubly linked list without using extra space for storing pointers to previous and next nodes.
In a traditional doubly linked list, each node contains pointers to both the previous and next nodes. However, with XOR, we can store the XOR of the addresses of the previous and next nodes in a single pointer. This allows us to traverse the linked list in both directions without using additional memory for storing pointers.
Here’s an example to illustrate this:
• Create a structure for each node in the linked list. The structure should contain a value and a pointer.
• Define two variables: ‘prev’ and ‘curr’, initialized to NULL.
• To insert a new node at the beginning of the list, create a new node and set its value. Set its pointer as XOR of ‘prev’ and NULL.
• If ‘prev’ is not NULL, update its pointer by XORing the address of ‘curr’ with the address stored in its pointer.
• Update ‘prev’ as ‘curr’ and ‘curr’ as newly created node.
Benefits of Using XOR:
The use of XOR in data structures offers several advantages:
• Saves Memory: By storing two addresses in one pointer, we can reduce memory consumption, which is particularly useful when dealing with large data sets.
• In-Place Operations: Operations like reversing a linked list or swapping elements in an array can be performed without requiring additional space for temporary variables.
Conclusion:
XOR is an essential operation in computer programming and data structures. It allows us to perform bitwise operations efficiently while saving memory.
In data structures like linked lists, it enables us to implement efficient algorithms without using extra space for pointers. Understanding XOR opens up possibilities for optimizing code and solving complex problems more efficiently. | 523 | 2,573 | {"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} | 3.296875 | 3 | CC-MAIN-2024-18 | latest | en | 0.902129 |
https://charline-picon.com/what-is-the-least-common-multiple-of-16-and-20/ | 1,638,376,006,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360803.6/warc/CC-MAIN-20211201143545-20211201173545-00046.warc.gz | 255,997,167 | 4,540 | LCM that 16 and 20 is the the smallest number amongst all common multiples that 16 and 20. The first couple of multiples the 16 and also 20 room (16, 32, 48, 64, 80, 96, . . . ) and also (20, 40, 60, 80, 100, . . . ) respectively. There room 3 commonly used techniques to find LCM that 16 and 20 - by division method, by element factorization, and also by listing multiples.
You are watching: What is the least common multiple of 16 and 20
1 LCM the 16 and 20 2 List of Methods 3 Solved Examples 4 FAQs
Answer: LCM the 16 and 20 is 80.
Explanation:
The LCM of 2 non-zero integers, x(16) and y(20), is the smallest confident integer m(80) the is divisible through both x(16) and also y(20) without any type of remainder.
Let's look at the various methods for finding the LCM that 16 and 20.
By department MethodBy element Factorization MethodBy Listing Multiples
### LCM that 16 and 20 by department Method
To calculation the LCM of 16 and also 20 by the department method, we will divide the numbers(16, 20) by your prime determinants (preferably common). The product of these divisors provides the LCM of 16 and also 20.
Step 3: continue the actions until just 1s space left in the critical row.
The LCM of 16 and also 20 is the product of every prime numbers on the left, i.e. LCM(16, 20) by division method = 2 × 2 × 2 × 2 × 5 = 80.
### LCM the 16 and 20 by prime Factorization
Prime administrate of 16 and 20 is (2 × 2 × 2 × 2) = 24 and (2 × 2 × 5) = 22 × 51 respectively. LCM that 16 and 20 deserve to be acquired by multiplying prime components raised to their respective highest possible power, i.e. 24 × 51 = 80.Hence, the LCM of 16 and 20 by element factorization is 80.
### LCM that 16 and also 20 through Listing Multiples
To calculation the LCM that 16 and 20 by listing the end the typical multiples, we can follow the given listed below steps:
Step 1: list a few multiples the 16 (16, 32, 48, 64, 80, 96, . . . ) and 20 (20, 40, 60, 80, 100, . . . . )Step 2: The common multiples indigenous the multiples the 16 and 20 room 80, 160, . . .Step 3: The smallest typical multiple of 16 and also 20 is 80.
See more: How Far Is Cheyenne Wyoming From Denver, Co And Cheyenne, Wy
∴ The least common multiple of 16 and also 20 = 80. | 690 | 2,256 | {"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} | 4.5625 | 5 | CC-MAIN-2021-49 | longest | en | 0.925274 |
http://harrypotter.wikia.com/wiki/Wizarding_currency?oldid=814036 | 1,454,719,339,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701145578.23/warc/CC-MAIN-20160205193905-00110-ip-10-236-182-209.ec2.internal.warc.gz | 94,124,733 | 29,606 | Wizarding Money
12,603pages on
this wiki
Currency in the wizarding Britain consists of three different coins. In decreasing order of value, they are: Galleon, Sickle and Knut. They are gold, silver, and bronze, respectively. According to Rubeus Hagrid, there are 17 Sickles in a Galleon, and 29 Knuts in a Sickle, meaning there are 493 Knuts to a Galleon. Around the edge of each coin is a series of numerals which represent a serial number belonging to the Goblin that cast the coin.
Money in itself is thought to be one of the five exceptions to Gamp's Law of Elemental Transfiguration[citation needed], meaning it cannot be created from nothing. Aside from the Philosopher's Stone which can convert other metals into gold there seems to be no other method of obtaining it. Attempting to duplicate money with the Geminio spell is also ineffective, as duplicates created from Geminio are worthless.
Symbols
The three denominations of wizarding currency were sometimes represented with the following set of symbols:[1]
• — Galleon
• — Sickle
• — Knut
The symbol ʛ was also used to represent the Galleon.[2]
Exchange Rate
One Knut is One Sickle is One Galleon is 1 Knut 29 Knuts 493 Knuts 0.03448... Sickles 1 Sickle 17 Sickles 0.002028... Galleons 0.05882... Galleons 1 Galleon
Converted into other currencies
According to J. K. Rowling, the approximate value of a Galleon is "About five Great British pounds, though the exchange rate varies!".[3]
This is consistent with the "textbooks" Rowling wrote for charity (Fantastic Beasts and Where to Find Them and Quidditch Through the Ages), which states that GB£174 million/US\$250 million is equivalent to 34 million Galleons[4] (or 34,000,872 Galleons, 14 Sickles, 7 Knuts to be exact[5]) and works out as approximately £5.12/\$7.35 per Galleon, but does not match the wizarding prices printed on the backs of the books (£4.99 or 1 Galleon, 11 Sickles; and \$3.99 or 14 Sickles, 3 Knuts for the original UK and US editions respectively).
With this information, we can calculate the value of wizarding currency in Muggle money.[6] The amounts below are approximate, and were accurate as of December 5, 2010, though it is unlikely that the value of a Galleon scales with US currency. As the gold standard values of wizarding currencies are unknown, it is difficult to estimate a present day conversion rate, and one would have to rely on Rowling's older approximations.
Table: Estimation of Wizarding Currency into Muggle Money
1 Galleon 1 Sickle 1 Knut
Pound Sterling GBP £4.93 £0.29 £0.01
U.S. Dollar USD \$7.35 \$0.46 \$0.02
Euro EUR €5.90 €0.34 €0.01
Japanese Yen JPY ¥653.66 ¥37.91 ¥1.31
Swiss Franc CHF 7.75 Fr. 0.45 Fr. 0.02 Fr.
Australian Dollar AUD \$7.96 \$0.46 \$0.02
Danish Kroner DKK 43.94 kr. 2.55 kr. 0.09 kr.
South African Rand ZAR R 54.01 R 3.13 R 0.11
Indian Rupee INR Rs. 354.67 Rs. 20.57 Rs. 0.71
Hong Kong Dollar HKD \$60.64 \$3.52 \$0.12
Philippine Peso PHP ₱342.24 ₱19.85 ₱0.68
Thai Baht THB ฿245.36 ฿14.43 ฿0.49
Swedish Krona SEK 53.81 kr 3.12 kr 0.11 kr
Serbian Dinar RSD 631.25 RSD 36.61 RSD 1.26 RSD
Argentine Peso ARS \$25.34 \$1.49 \$0.03
Note that the Galleon/Pound rate cited by Rowling is probably that offered by Gringotts and bears no relation to the precious-metal value of Wizarding coins. The "gold coins the size of hubcaps" mentioned in reference to the Quidditch World Cup would be much larger than the British five-pound Quintuple Sovereign today sold for its bullion value of hundreds of pounds sterling (though this hubcap reference may have been an exaggeration). However, it is unclear whether the coins were Galleons, or the currency of some other Wizard community. Certainly, if the coins were indeed of such size, there could have been no talking about "handfuls" of them.
It should be mentioned that Rowling's exchange rates between Galleons and Muggle currency are very far off from reality, assuming that Galleons are made from gold. At the time of this entry (April 5, 2010) one gram of gold is worth €26,84 or £23,69 and using regular sized coins as a comparison (one 20 €-cent coin weighs 5.74g for example, 10p weigh 6.50g) would allow for an exchange rate in excess of 100-1 instead of the 5-1 that is accepted as canon. There are possible explanations for this, the most logical ones being that a Galleon is not made from pure gold but rather has a gold core of 0.21 grams or less, or for that matter is only gold-coloured and not made of actual gold at all. Another reason, considering the sometimes mentioned logic of the wizarding world, is the way Galleons may possibly be created and equipped with safety measures such as charms that will not allow its owners to exchange them in the Muggle world outside official goblin establishments.
Furthermore, the value of the coins presented by Rowling at the end of Fantastic Beasts and Where to Find Them is extremely unlikely as this would make the overall currency rate of all the coins extremely low indeed. For example, in the fifth book, Harry, Ron and Hermione all buy a Butterbeer each; the price for the three is six Sickles - so two Sickles each. Going by the currency approximations this would make a beer cost about 60p - British currency - which is of course ridiculously low (unless perhaps the Butterbeer is a sort of cheap non-alcoholic drink like soda pop, which would be plausible as it is regularly drunk by minors). Moreover, in the second book Molly Weasley is described to only take "one gold galleon" from her vault at Gringotts which is not much at all. Of course, it is common knowledge that the Weasleys are poor but they have never been described as starving and have always been able to buy the equipment they need with the little money they have, even if it is second hand. However, many people would just assume that everything was much cheaper in the Wizarding world as they don't need to spend money on manufacturing but most of it does seem slightly ridiculous. A more appropriate value of the Galleon would be around £24.60 as this would be a rate that would fit with Britains present climate.
Fake Galleons
Leprechaun Gold
• Galleons made of Leprechaun gold were common at Quidditch games where Leprechauns are the mascots for the Irish team. These Galleons are occasionally in temporary circulation (they vanish a few hours after appearing), but goblin experts at Gringotts can differentiate them from real ones.
Names
Galleons were Spanish treasure ships often raided by pirates. "Sickle" is the ancient Greek term for the shekel, the currency of Judea, as rendered by William Tyndale in his translation of the Greek New Testament, (later the King James Bible). Also translated as silverling. "Knut" or "Canute" is the name of an 11th-century King of England.
"Unum Galleon" written on the coins literally means "One Galleon". "Unum" may be used as Latin for "One".
Behind the scenes
• It was said by J.K. Rowling that goblins get muggle money back into circulation in the event that muggle-borns need to exchange pounds for Galleons.[citation needed]
• In the films, the Galleon seems to be represented by the heraldic "fleam" symbol.[7] In Pottermore, however, the Galleon is represented by the voiced uvular implosive, ʛ.
• In the films, Wizarding money appears as fairly average-sized round coins, not that dissimilar from that used by Muggles. In the books, though they are never particularly described beyond their metal and the Knuts being called "tiny" once, there are several references to Wizarding money and Muggle money appearing quite differently.[8][9]
Diameters of coins
Compared to existing coinage in the United States, The Galleon is the same diameter as the U.S. Silver Eagle; the Silver Sickle is slightly larger than the 1/2 Dollar and the Bronze Knut is slightly larger than the nickel.[citation needed]
Notes and references
1. Pottermore
2. http://web.archive.org/web/20091023035038/http://www.accio-quote.org/articles/2001/0301-comicrelief-staff.htm
3. "Widespread amusement is converted into large amounts of money (over 250 million dollars since they started in 1985 – which is the equivalent of over 174 million pounds or thirty-four million Galleons)." –Albus Dumbledore on Comic Relief in his Foreword to Quidditch Through the Ages
4. "Comic Relief has raised 174 million pounds since 1985 (thirty-four million, eight hundred and seventy-two Galleons, fourteen Sickles and seven Knuts)." –Albus Dumbledore on Comic Relief in his Foreword to Fantastic Beasts and Where to Find Them
5. Precise for the time of Dumbledore's in-universe writing of the Forewords, some time before his death on 30 June 1997. Rowling herself said she wrote the books immediately after finishing Harry Potter and the Goblet of Fire (released on 8 July 2000), and the Comic Relief books were released on 12 March 2001, meaning the \$250m/£174m/ʛ34m actually applies to this time.
6. http://www.accio-quote.org/articles/2000/1000-aol-chat.htm
7. "Weird!...What a shape! This is money?" -Ron on a 50 pence piece in Chapter 12, Sorcerer's Stone
8. "I had two try and pay me with great gold coins the size of hubcaps ten minutes ago." -Mr. Roberts, a Muggle, on wizards attempting to pay him in their own currency in Chapter 7, Goblet of Fire | 2,393 | 9,260 | {"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} | 2.6875 | 3 | CC-MAIN-2016-07 | longest | en | 0.924699 |
https://stacks.math.columbia.edu/tag/0EAT | 1,657,050,913,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00431.warc.gz | 581,054,431 | 6,409 | Lemma 42.5.3. Let $(A, \mathfrak m)$ be a Noetherian local ring of dimension $1$. Let $A \subset B$ be a finite ring extension with $B/A$ annihilated by a power of $\mathfrak m$ and $\mathfrak m$ not an associated prime of $B$. For $a, b \in A$ nonzerodivisors we have
$\partial _ A(a, b) = \prod \text{Norm}_{\kappa (\mathfrak m_ j)/\kappa (\mathfrak m)}(\partial _{B_ j}(a, b))$
where the product is over the maximal ideals $\mathfrak m_ j$ of $B$ and $B_ j = B_{\mathfrak m_ j}$.
Proof. Choose $B_ j \subset C_ j$ as in Lemma 42.4.4 for $a, b$. By Lemma 42.4.1 we can choose a finite ring extension $B \subset C$ with $C_ j \cong C_{\mathfrak m_ j}$ for all $j$. Let $\mathfrak m_{j, k} \subset C$ be the maximal ideals of $C$ lying over $\mathfrak m_ j$. Let
$a = u_{j, k}\pi _{j, k}^{f_{j, k}},\quad b = v_{j, k}\pi _{j, k}^{g_{j, k}}$
be the local factorizations which exist by our choice of $C_ j \cong C_{\mathfrak m_ j}$. By definition we have
$\partial _ A(a, b) = \prod \nolimits _{j, k} \text{Norm}_{\kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m)} ((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}} \bmod \mathfrak m_{j, k})^{m_{j, k}}$
and
$\partial _{B_ j}(a, b) = \prod \nolimits _ k \text{Norm}_{\kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m_ j)} ((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}} \bmod \mathfrak m_{j, k})^{m_{j, k}}$
The result follows by transitivity of norms for $\kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m_ j)/\kappa (\mathfrak m)$, see Fields, Lemma 9.20.5. $\square$
There are also:
• 2 comment(s) on Section 42.5: Tame symbols
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 711 | 1,810 | {"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": 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": 2, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-27 | latest | en | 0.632732 |
https://www.jiskha.com/11th_grade/?page=4 | 1,516,318,343,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887660.30/warc/CC-MAIN-20180118230513-20180119010513-00231.warc.gz | 940,839,814 | 12,185 | I have to determine the muzzle velocity of a toy gun by treating it as a horizontal projectile. What could be a possible hypothesis when a toy gun is shot 5 meters away from the bucket and the ball which was loaded in it is supposed to fall in?
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Suppose you surveyed 50 students and 34 said that they approve of how the president is doing. Construct 90%, 95%, and 99% confidence intervals for the true proportion of students in your school who approve of how the president is doing. Be sure to specifically calculate the ...
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This problem is related to Chapter-Sets. Please solve the question using x method. x method means x ∈ A ∩ B Q. If A-B=A then show that A∩B=Ø
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Write your own journal entry from the perspective of a member of Bradford's colony. What do you think of these unruly pirates? What shocks you the most about these surly characters? How do the manners of the pirates differ from yours?
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How fast must a 1,000 kg car be moving to have a kinetic energy of: (a) 2.0 x 10^3, (b) 2.0 x 10^5, (c) 1.0 kW x h? How high would you have to lift a 1,000 kg car to give it a potential energy of: (a) 2.0 x 10^3, (b) 2.0 x 10^5, (c) 1.0 kW x h?
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Write your own journal entry from the perspective of a member of Bradford's colony. What do you think of these unruly pirates? What shocks you the most about these surly characters? How do the manners of the pirates differ from yours?
Write your own journal entry from the perspective of a member of Bradford's colony. What do you think of these unruly pirates? What shocks you the most about these surly characters? How do the manners of the pirates differ from yours?
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What is the value of a charge of a body that carries 20 excess electrons? I think that the question asks you to pretend that something is negatively charged by 20 electrons. So I think you take the charge of an electron, which is 1.6 x 10^-19, and multiply it by 20. If you do ... | 5,258 | 20,776 | {"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-2018-05 | latest | en | 0.943236 |
https://community.qlik.com/t5/QlikView-Layout-Visualizations/Calculated-Dimension-based-on-condition/td-p/333434 | 1,547,763,497,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659340.1/warc/CC-MAIN-20190117204652-20190117230652-00104.warc.gz | 477,012,422 | 46,245 | # QlikView Layout & Visualizations
Discussion Board for collaboration on QlikView Layout & Visualizations.
Not applicable
## Calculated Dimension... based on condition...
Hi,
I have a dataset containing the ages of a bunch of people. I also have these linked to Age Groups, i.e 0-10, 11-20, 21-30 etc etc. I have multiple Age Group Types, so the age group type described previously was 10 year incremental, but i also have a 5 year incremental group (0-5, 6-10, 11-15 etc etc), plus others.
I have a chart with the dimensions AgeGroup & Gender. The expression is a simple count.
For the chart to display correctly, i need to either select a single AgeGroupType, Or, Add the equivalent set analysis into my expression. What i WANT to acheive is...
If a single AgeGroupType is NOT selected then by default use AgeGroupType X, else use whatever AgeGroupType is selected.
I've tried building an 'If' statement into my expression but it just didnt work. it was something like...
If(Count({\$} Distinct [AgeGroupTypeName])=5
//Then
,Count(DISTINCT{\$<AgeGroupTypeName={'Ten-year intevals'}>}EpisodeId)
//Else
,Count(DISTINCT{\$}EpisodeId)
Thanks,
-Mathew
Tags (5)
1 Solution
Accepted Solutions
MVP
## Re: Calculated Dimension... based on condition...
Hi
You could try this:
If(GetSelectedCount(AgeGroupTypeName) <> 1,
Count({<AgeGroupTypeName = {'Ten-year intervals'}>} DISTINCT EpisodeId),
Count({<AgeGroupTypeName = {"\$(=GetFieldSelections(AgeGroupTypeName))"}>} DISTINCT EpisodeId))
This should use 'Ten-year intervals' unless exactly one AgeGroupTypeName is selected, in whih case it should use the selecetd one.
Hope that helps
Jonathan
Logic will get you from a to b. Imagination will take you everywhere. - A Einstein
3 Replies
MVP
## Re: Calculated Dimension... based on condition...
Hi
You could try this:
If(GetSelectedCount(AgeGroupTypeName) <> 1,
Count({<AgeGroupTypeName = {'Ten-year intervals'}>} DISTINCT EpisodeId),
Count({<AgeGroupTypeName = {"\$(=GetFieldSelections(AgeGroupTypeName))"}>} DISTINCT EpisodeId))
This should use 'Ten-year intervals' unless exactly one AgeGroupTypeName is selected, in whih case it should use the selecetd one.
Hope that helps
Jonathan
Logic will get you from a to b. Imagination will take you everywhere. - A Einstein
Not applicable
## Re: Calculated Dimension... based on condition...
Awesome Thanks Jonathan. I was on the right track i just needed guiding.
Employee
## Re: Calculated Dimension... based on condition...
Hi Mathew,
I would create a table for diferent kind of age goups.
Now you can control if there is a age group type selected or not in the chart:
Finally you can use conditional view for the diferent age groups in the dimension sheet.
I used this expression.
=SubStringCount(Concat(_dimension, '|'), 'Grupo 1')
I've developed a very small example to test it. If you send me an email I can send you by email.
I found this technique in this demo:
http://eu.demo.qlik.com/detail.aspx?appName=Whats%20New%20in%20QlikView11.qvw | 752 | 3,037 | {"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} | 2.765625 | 3 | CC-MAIN-2019-04 | longest | en | 0.789408 |
https://www.tutorialspoint.com/python-get-the-index-of-first-element-greater-than-k | 1,674,949,978,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00545.warc.gz | 1,079,431,456 | 10,936 | # Python - Get the Index of first element greater than K
The values of items in a python list are not necessarily in any sorted order. More over there may be situation when we are interested only in certain values greater than a specific value. In this article we will see how we can get the
## Using Enumeration
Using enumeration we get both the index and value of the elements in the list. Then we apply the greater than condition to get only the first element where the condition is satisfied. The next function goes through each list element one by one.
## Example
Live Demo
List = [21,10,24,40.5,11]
print("Given list: " + str(List))
#Using next() + enumerate()
result = next(k for k, value in enumerate(List)
if value > 25)print("Index is: ",result)
Running the above code gives us the following result
## Output
Given list: [21, 10, 24, 40.5, 11]
Index is: 3
## Using Filter and Lambda Functions
In the next example we take a lambda function to compare the given value with value at each index and then filter out those which satisfy the required condition. From the list of elements satisfying the required condition, we choose the first element at index 0 for our answer.
## Example
Live Demo
List = [21,10,24,40.5,11]
print("Given list: " + str(List))
#Using filter() + lambda
result = list(filter(lambda k: k > 25, List))[0]
print("Index is: ",List.index(result))
Running the above code gives us the following result
## Output
Given list: [21, 10, 24, 40.5, 11]
Index is: 3
## Using map and lambda
In the next example we take a similar approach but use map instead of filter. The map function is used to loop through each of the elements. Whenever the condition becomes true that index is captured.
## Example
Live Demo
List = [21,10,24,40.5,11]
print("Given list: " + str(List))
result = list(map(lambda k: k > 25, List)).index(True)
print("Index is: ",(result))
Running the above code gives us the following result
## Output
Given list: [21, 10, 24, 40.5, 11]
Index is: 3 | 519 | 2,009 | {"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} | 2.953125 | 3 | CC-MAIN-2023-06 | latest | en | 0.758054 |
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Ex2-5210
# Ex2-5210 - EXAM INFORMATION Harmonic Oscillator Hamiltonian...
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Unformatted text preview: EXAM INFORMATION Harmonic Oscillator Hamiltonian: Energy Levels: Anharmonic Oscillator Energies: Rigid Rotor Hamiltonian: Quantum Numbers: J = 0, 1, 2, 3, ... M=0, ±1, ±2, ..., ±J Energy Levels: Moment of Inertia: Linear Polyatomic: Diatomic: Boltzmann Distribution: First Order Perturbation Theory: INTEGRALS Constants and Conversions: h = 6.63x10-34 J·s ħ = h/2 π = 1.05x10-34 J·s k = 1.38x10-23 J/K c = 3.00x10 8 m/s = 3.00x10 10 cm/s N A = 6.02x10 23 mol-1 1 Å = 10-10 m 1 amu = 1.66x10-27 kg 1 J = 1 kg·m 2 /s 2 1 N = 1 kg·m/s 2 2 2 2 2 2 1 2 kx dx d H +- = μ μ πν ϖ ϖ k n n E n = = = + = 2 , 2 , 1 , 2 1 & ∑ = 2 i i r m I 2 1 2 1 2 m m m m r I + = = μ μ ∫ = ∆ τ ψ ψ d H E ) ( ) 1 ( ) ( * kT E i i i e g N /- ∝ ∫ ∫ ∫ ∫ ∞- ∞- ∞- ∞- = = = = 3 6 2 4 2 16 15 8 3 4 1 2 1 2 2 2 2 β π β β π β β π β β π β β β β dx e x dx e x dx e x dx e x x x x ∂ ∂ + ∂ ∂ ∂...
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Ex2-5210 - EXAM INFORMATION Harmonic Oscillator Hamiltonian...
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Ask a homework question - tutors are online | 604 | 1,522 | {"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-2018-09 | latest | en | 0.635505 |
https://www.wordunscrambler.net/unscramble-horedss | 1,670,653,251,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711712.26/warc/CC-MAIN-20221210042021-20221210072021-00280.warc.gz | 1,190,611,421 | 16,183 | Unscramble HOREDSS
HOREDSS unscrambles into 158 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT!
2 letter words made by unscrambling HOREDSS
Unscrambled 17 2 Letter Words
Above are the words made by unscrambling HOREDSS (DEHORSS). To further help you, here are a few lists related to/with the letters HOREDSS
The Value of HOREDSS In Word Scramble Games
The letters HOREDSS are worth 11 points in Scrabble
The letters HOREDSS are worth 10 in points Words With Friends
• H = 3 points in WWF & 4 points in Scrabble
• O = 1 points in WWF & 1 points in Scrabble
• R = 1 points in WWF & 1 points in Scrabble
• E = 1 points in WWF & 1 points in Scrabble
• D = 2 points in WWF & 2 points in Scrabble
• S = 1 points in WWF & 1 points in Scrabble
• S = 1 points in WWF & 1 points in Scrabble
What Does HOREDSS Mean... If you Unscramble it?
Possible Definitions of HOREDSS
If we unscramble these letters, HOREDSS, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
shoders
• Sorry. I don't have the meaning of this word.
Permutations of HOREDSS
According to our other word scramble maker, HOREDSS can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
Definition of Permutation
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the letters horedss scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time HOREDSS comes up in a word scramble game.
We stopped it at 50, but there are so many ways to scramble HOREDSS!
Scramble Words
Unscramble these letters to make words...
scrambled using word scrambler... | 507 | 1,854 | {"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-2022-49 | latest | en | 0.903768 |
https://www.tutorialspoint.com/return-the-fractional-and-integral-parts-of-a-value-in-numpy | 1,680,222,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949506.62/warc/CC-MAIN-20230330225648-20230331015648-00398.warc.gz | 1,151,956,889 | 10,305 | # Return the fractional and integral parts of a value in Numpy
To return the fractional and integral parts of a value, use the numpy.modf() method in Python Numpy. The fractional and integral parts are negative if the given number is negative.
The out is a location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned. A tuple (possible only as a keyword argument) must have length equal to the number of outputs.
The condition is broadcast over the input. At locations where the condition is True, the out array will be set to the ufunc result. Elsewhere, the out array will retain its original value. Note that if an uninitialized out array is created via the default out=None, locations within it where the condition is False will remain uninitialized.
## Steps
At first, import the required library −
import numpy as np
To return the fractional and integral parts of a value, use the numpy.modf() method in Python Numpy −
print("Returning the fractional and integral parts...")
Check for float −
print("Result? ", np.modf(0.1))
print("Result? ", np.modf(-0.7))
Check for int and inf −
print("Result? ", np.modf(5))
print("Result? ", np.modf(-np.inf))
Check for nan and inf −
print("Result? ", np.modf(np.nan))
print("Result? ", np.modf(np.inf))
Check for log −
print("Result? ", np.modf(np.log(1)))
print("Result? ", np.modf(np.log(2)))
## Example
import numpy as np
# To return the fractional and integral parts of a value, use the numpy.modf() method in Python Numpy
print("Returning the fractional and integral parts...")
# Check for float
print("Result? ", np.modf(0.1))
print("Result? ", np.modf(-0.7))
# Check for int and inf
print("Result? ", np.modf(5))
print("Result? ", np.modf(-np.inf))
# Check for nan and inf
print("Result? ", np.modf(np.nan))
print("Result? ", np.modf(np.inf))
# Check for log
print("Result? ", np.modf(np.log(1)))
print("Result? ", np.modf(np.log(2)))
## Output
Returning the fractional and integral parts...
Result? (0.1, 0.0)
Result? (-0.7, -0.0)
Result? (0.0, 5.0)
Result? (-0.0, -inf)
Result? (nan, nan)
Result? (0.0, inf)
Result? (0.0, 0.0)
Result? (0.6931471805599453, 0.0) | 615 | 2,257 | {"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} | 3.40625 | 3 | CC-MAIN-2023-14 | latest | en | 0.707102 |
http://algebra-expression.com/algebra-expression-simplifying/graphing-function/algebra-1-prentice-hall.html | 1,519,054,771,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812756.57/warc/CC-MAIN-20180219151705-20180219171705-00684.warc.gz | 16,256,970 | 10,694 | Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
Ineq. #7:
Ineq. #8:
Ineq. #9:
Solve for:
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algebra 1 prentice hall textbook
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Author Message
Dreh
Registered: 14.01.2002
From:
Posted: Thursday 28th of Dec 20:13 hi People out there I really hope some math wiz reads this. I am stuck on this homework that I have to turn in in the next couple of days and I can’t seem to find a way to complete it. You see, my professor has given us this test that includes algebra 1 prentice hall textbook, reducing fractions and logarithms and I just can’t make head or tail out of it. I am thinking of going to some private tutor to help me solve it. If someone can give me some suggestions, I will very appreciative.
oc_rana
Registered: 08.03.2007
From: egypt,alexandria
Posted: Saturday 30th of Dec 18:09 How about giving a little more particulars of what exactly is your trouble with algebra 1 prentice hall textbook? This would help in finding out ways to look for a solution . Finding a teacher these days quickly enough and that too at a price tag that you can pay for can be a wearisome task. On the other hand, these days there are programs that are offered to assist you with your math problems. All you need to do is to choose the most suitable one. With just a click the right answer pops up. Not only this, it assists you to arriving at the answer. This way you also get to learn to get at the correct answer.
MoonBuggy
Registered: 23.11.2001
From: Leeds, UK
Posted: Monday 01st of Jan 09:50 Thanks for the suggestion . Algebrator is indeed a great algebra software. I was able to get answers to difficulties I had about binomials, like denominators and algebra formulas. You simply type in a problem, click on Solve and you get the all the answers you need. You can use it for any number of algebra things, like Algebra 2, Pre Algebra and Basic Math. I am really happy with Algebrator.
albochg
Registered: 09.04.2006
From:
Posted: Tuesday 02nd of Jan 18:40 I hope this software has an easy interface . Can I have a look at it?
Dolknankey
Registered: 24.10.2003
From: Where the trout streams flow and the air is nice
Posted: Wednesday 03rd of Jan 13:08 I recommend trying out Algebrator. It not only assists you with your math problems, but also provides all the required steps in detail so that you can improve the understanding of the subject.
Majnatto
Registered: 17.10.2003
From: Ontario
Posted: Thursday 04th of Jan 07:25 Cool, i got it from here : http://www.algebra-expression.com/simplifying-radical-expressions.html. Oh!, and when you look at the page check their unconditional guarantee! Good luck dude contact me if you need anything. | 899 | 3,500 | {"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-2018-09 | latest | en | 0.910413 |
https://primepaperhelp.com/2020/11/10/accounting-for-decision-makers-accounting-homework-help/ | 1,620,591,302,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989012.26/warc/CC-MAIN-20210509183309-20210509213309-00534.warc.gz | 504,805,758 | 10,345 | # Accounting for decision-makers | Accounting homework help
Compare two drug companies – Merck, a U.S. company, and Novartis, a Swiss company. The annual reports for these companies are attached. Download the two PDF files and save them to your computer. As you review these two 2016 annual reports, you will see some differences in presentation between the U.S. company and the Swiss company, because Merck uses U.S. generally accepted accounting principles (GAAP) and Novartis uses International Financial Reporting Standards (IFRS). The companies present the same information, just in different formats.
You do not need to read every word in these two annual reports, but you should scan through them to see the different things reported. Then, answer the following questions and compute the following ratio and trend analysis for the two companies.
Prepare a 3 page narrative report (including exhibits) in Microsoft Word with at least 2 scholarly sources. You will embed Microsoft Excel spreadsheets with your calculations into your Word document. Follow the attached templates and answer each required item. When narrative explanation is required, use complete sentences in your paper and citations in APA format.
1. Find the balance sheet for each company (2016 Merck; Novartis PDF files above). Enter the following numbers for each company into the Excel template provided and explain in the narrative what these numbers tell you about the companies.
• Accounting equation: Total Assets = Total Liabilities + Total Equity
2. Find the income statement for each company (2016 Merck; Novartis PDF files above). Enter in the net income for each company from the income statement into the Excel template and discuss what this number tells you about the companies in the narrative.
3. For each company, find the statement of equity and locate the Retained Earnings (2016 Merck; Novartis PDF files above). Enter the number into the Excel template and discuss what this number tells you about each company in the narrative.
4. Using the Excel template calculate the ratios below for each company for 2016 and 2015. Note, some calculations will require you to go back to the 2014 statements when you are asked for an average (those have been provided also).
• Current Ratio
• Quick Ratio
• Debt to Total Assets
• Debt to Total Equity
• Net Profit on Sales
• Return on Assets
• Return on Equity
In the narrative compare the results of the two companies and discuss what these results indicate to you as a manager and how they may be used in decision making.
5. Prepare a common size trend analysis using the Excel template provided for each company from 2015 to 2016.
In the narrative discuss what these trends might indicate to you as a manager and how they can be used in decision making.
6. After using the two sets of financial statements to prepare the items above, discuss in the narrative any differences you see between the presentation of the two sets of reports. For example, what are the differences in the balance sheet presentation? The income statement? The statement of equity? The statement of cash flows? In your opinion, which set of statements do you find easier to work with as a manager for analysis and why?
How to submit this assignment
Submit your work in a Word document using the attached templates. Your Excel tables should be embedded in the Word document.
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How has NCTM leadership shaped the evolution of teaching and learning mathematics? What are your expectations for NCTM leadership? | 497 | 2,641 | {"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} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.85254 |
https://www.mumsnet.com/Talk/pedants_corner/3583355-Ambiguous-maths-question | 1,566,496,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317339.12/warc/CC-MAIN-20190822172901-20190822194901-00503.warc.gz | 901,396,826 | 24,265 | # Ambiguous maths question
(10 Posts)
TickledOnion Sat 11-May-19 18:16:52
DD is 9. She was doing maths homework about bar charts.
18 children are asked what their favourite farm animal is. The results are: pig 3, duck 4, goat 5, cow 6.
The question is: how many children voted for duck or pig?
I know that the answer is 7 and that you add the duck and pig votes together, but DD can’t understand how that is correct. She argued that the answer is “3 or 4”.
I tried to explain it by imagining she answered the survey with duck. If I asked her “did you vote for duck or pig?” My answer is yes but she says her answer is duck.
Are we both right?
TickledOnion Sat 11-May-19 18:19:02
I think my explanation made things worse. It would make more sense if I said “put your hand up if you voted for duck or pig”.
StealthPolarBear Sat 11-May-19 18:20:28
This is where you need a Venn diagram
I agree though the or in sentences can often be confusing as I think it's used in speech to mean 'and'
TeenTimesTwo Sat 11-May-19 18:21:50
You may both be right from an English pedant point of view.
But you are right from a maths homework point of view.
She needs to get with it and learn to 'read' wordy maths questions otherwise she'll be in for years of pain and not getting marks.
PrincessTiggerlily Sat 11-May-19 18:23:13
How many children in total voted duck or pig - would have made more sense.
Can she draw circle with crosses for duck voters and pig voters - then how many voted would be them added together.
TeenTimesTwo Sat 11-May-19 18:24:31
duck 'and' pig would mean the intersection though, and as they only got one vote each it would be zero.
'or' means either A or B or both (ie the union), unless stated as 'exactly one of A or B'
StealthPolarBear Sat 11-May-19 18:24:59
But in a way it is being used in the same way as and would in the sentence. How many voted duck and (how many voted) pig. So I can see why she's annoyed.
TickledOnion Sat 11-May-19 18:27:13
Some of the later questions were similar but included “in total” at the end of the sentence. She thought that was much clearer.
NoBaggyPants Sat 11-May-19 18:29:06
You wouldn't use a Venn diagram to explain this. They're mutually exclusive events.
The question would be better phrased "how many children in total voted for duck or pig?". The putting their hands up suggestion is an excellent way of explaining it too.
TickledOnion Sat 11-May-19 18:29:07
I wonder if this is a problem in other languages or just English.
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Registering is free, quick, and means you can join in the discussion, watch threads, get discounts, win prizes and lots more. | 703 | 2,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-35 | latest | en | 0.977151 |
https://numbermatics.com/n/62549/ | 1,701,745,183,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100540.62/warc/CC-MAIN-20231205010358-20231205040358-00440.warc.gz | 516,022,981 | 6,433 | # 62549
## 62,549 is a prime number. Like all primes greater than two, it is odd and has no factors apart from itself and one.
What does the number 62549 look like?
As a prime, it is not composed of any other numbers and has no internal structure.
62549 is a prime number. Like all primes (except two), it is an odd number.
## Prime factorization of 62549:
### 62549
See below for interesting mathematical facts about the number 62549 from the Numbermatics database.
### Names of 62549
• Cardinal: 62549 can be written as Sixty-two thousand, five hundred forty-nine.
### Scientific notation
• Scientific notation: 6.2549 × 104
### Factors of 62549
• Number of distinct prime factors ω(n): 1
• Total number of prime factors Ω(n): 1
• Sum of prime factors: 62549
### Divisors of 62549
• Number of divisors d(n): 2
• Complete list of divisors:
• Sum of all divisors σ(n): 62550
• Sum of proper divisors (its aliquot sum) s(n): 1
• 62549 is a deficient number, because the sum of its proper divisors (1) is less than itself. Its deficiency is 62548
### Bases of 62549
• Binary: 11110100010101012
• Base-36: 1C9H
### Squares and roots of 62549
• 62549 squared (625492) is 3912377401
• 62549 cubed (625493) is 244715294055149
• The square root of 62549 is 250.0979807997
• The cube root of 62549 is 39.6953946103
### Scales and comparisons
How big is 62549?
• 62,549 seconds is equal to 17 hours, 22 minutes, 29 seconds.
• To count from 1 to 62,549 would take you about seventeen hours.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 62549 cubic inches would be around 3.3 feet tall.
### Recreational maths with 62549
• 62549 backwards is 94526
• The number of decimal digits it has is: 5
• The sum of 62549's digits is 26
• More coming soon!
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Keywords: Divisors of 62549, math, Factors of 62549, curriculum, school, college, exams, university, Prime factorization of 62549, STEM, science, technology, engineering, physics, economics, calculator, sixty-two thousand, five hundred forty-nine.
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http://mathhelpforum.com/math-topics/27479-physics-gravity-problem-print.html | 1,513,623,370,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948619804.88/warc/CC-MAIN-20171218180731-20171218202731-00790.warc.gz | 179,761,165 | 2,818 | Physics Gravity Problem
• Feb 4th 2008, 06:39 PM
Linnus
Physics Gravity Problem
I am having a bit of trouble with this problem...
The gravitational pull of the Moon is partially responsible for the tides of the sea. The Moon pulls on you, too, so if you are on a diet it is better to weigh yourself when this heavenly body is directly overhead! If you have a mass of 85.0 kg, how much less do you weigh if you factor in the force exerted by the Moon when it is directly overhead (compared to when it is just rising or setting)? Use the values 7.35×1022 kg for the mass of the moon, and 3.76×108 m for its distance above the surface of the Earth.
Thanks
• Feb 4th 2008, 08:16 PM
TrevorP
The gravitational force between two objects is given by the following formula:
$F = \frac{Gm_2m_2}{r^2}$
Where G is the gravitational constant, and r is the distance between the two masses.
Just google for G and you should be able to apply this to get what you want. | 256 | 958 | {"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": 1, "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-2017-51 | longest | en | 0.915926 |
http://www.upjob.in/2017/04/simplification-questions-for-bank-exams.html | 1,513,318,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567042.50/warc/CC-MAIN-20171215060102-20171215080102-00134.warc.gz | 485,597,707 | 30,515 | ### Simplification questions for bank exams with solutions
Simplification or Sarlikaran (सरलीकरण) is one of the very important topic for all banking PO, Clerk, IBPS RRB, Railway and other civil service examination. This topic usually in all Competitive Examination. Staff selection commission also included in the Syllabus of Combined Graduate level Examination, multi Tasking Examination, higher secondary level examination and other department examination. Approximately 5 to 10 questions are asked in Railway, NDA, NA, CDS and other Test. So candidates are advised to practice more and more for cracking the Numerical and problems of simplification. Upjob.in brought Simplification question and answer with complete solution. Hope following MCQ questions will be helpful in the preparation of your examination.
Q1. 16 11/13- 3 4/9 + 18 21/26 =?
(a) 11 23/234 (b) 23 11/234 (c) 32 49/234 (d) 49 31/234
Ans. (c)
Q2. 5 5/6 * 6 3/7+5 1/2 =?
(a) 40 (b) 40.5 (c) 42.5 (d) 43
Ans. (d)
Q3. (48-12*3+9) / (12-9/3) =?
(a) 4 1/3 (b) 3 (c) 2 1/3 (d) 21
Ans. (c)
Q4. (20+8*0.5) / (20-x) =12
(a) 2 (b) 8 (c) 18 (d) none of these
Ans. (c)
Q5. 4 7/8*2 4/13 =?
(a) 11 3/8 (b) 11 4/13 (c) 11 1/4 (d) 11 1/13 (e) none of these
Ans. (c)
Q6. 3/8+1/2 of 3/16=?
(a) 15/16 (b) 21/128 (c) 15/32 (d) 3/4
Ans. (c)
Q7. (7+7+7/7) / [(7+7+7)/7] =?
(a) 15/21 (b) 5 (c) 14 2/3 (d) 15 1/3
Ans. (b)
Q8. By how much does 6/ (7/8) exceed (6/7) /8 ?
(a) 6 1/8 (b) 6 3/4 (c) 7 3/4 (d) 7 5/6
Ans. (b)
Q9. 999 995/999*999=?
(a) 990809 (b) 998996 (c) 998999 (d) 999824
Ans. (b)
Q10. 9-1 2/9 of (3 3/11) / (5 1/7) of 7/9 =?
(a) 1 1/4 (b) 8 (c) 8 32/81 (d) 9
Ans. (b)
Q11. 5/8 of 24= 15/7*x, then the value of x
(a) 7/225 (b) 7 (c) 8 (d) 15
Ans. (b)
Q12. 999 1/7 + 999 2/7 + 999 3/7 + 999 4/7 + 999 5/7 + 999 6/7 = ?
(a) 2997 (b) 5979 (c) 5997 (d) 5994
Ans. (c)
Q13. (10 1/8) of (12/15) / (35/36) of (20/49) =?
(a) 17 5/12 (b) 17 8/17 (c) 20 3/25 (d) 20 103/250
Ans. (d)
Q14. 1/(1*2*3) + 1/(2*3*4) + 1/(3*4*5) + 1/(4*5*6) = ?
(a) 1/30 (b) 7/30 (c) 11/30 (d) 13/30
Ans. (b)
Q15. -7m- [3n- {8m-(4n – 10m)}] =?
(a) 11m-5n (b)11n-11m (c) 11n-7m (d) 11m-7n
Ans. (d)
Q16. (1-1/3) (1-1/4) (1-1/5) …….. (1-1/99) (1-1/100) =?
(a) 1/25 (b) 1/50 (c) 1/100 (d) 2/99
Ans. (b)
Q17. (2-1/3) (2-3/5) (2-5/7) …… (2-999/1001) =?
(a) 999/1001 (b) 1003/3 (c) 1001/3 (d) none of these
Ans. (b)
Q18. [{1+ 1/(10+ 1/10)} * {1+ 1/(10 + 1/10)}- {1-1/(10+ 1/10)} * {1- 1/(10+ 1/10)}] + [{1+ 1/ (10+ 1/10)} {1- 1/ (10 +1/10)}] simplifies to
(a) 20/101 (b) 90/101 (c) 100/101 (d) 101/100
Ans. (a)
Q19. {1+ 1/(x+1)} {1+ 1/(x+2)} {1+1/(x+3)} {1+ 1/(x+4)} =?
(a) (x+5)/(x+1) (b) {x+1/(x+5)} (c) 1/(x+5) (d) (x+6)/(x+5)
Ans. (a)
Q20. [1/ (1*2)] + [1/(2*3)] + [1/(3*4)] + ……… + [1/100*101] =?
(a) 1/100 (b) 100/101 (c) 101/102 (d) 9 9/10
Ans. (b)
Q21. (1-1/2) (1-1/3) (1-1/4) …….. (1-1/n) =?
(a) 1/n (b) 1/(n-1) (c) n/(n-1) (d) none of these
Ans. (a)
Q22. [(1/1*4) + (1/4*7) + (1/7*10) + (1/10*13) + (1/13*16)] = ?
(a) 1/3 (b) 5/16 (c) 3/8 (d) 41/7280
Ans. (b)
Q23. {1/2 + 1/6 + 1/12 + 1/20 + 1/30 + …….. + 1/n (n+1)} =?
(a) 1/n (b) 1/(n+1) (c) n/(n+1) (d) 2(n-1)/n
Ans. (c)
Q24. If x/y = 6/5, then [6/7 – {(5x-y)/(5x+y)}] =?
(a) 1/7 (b) 2/7 (c) 3/7 (d) 4/7
Ans. (a)
number series questions and answers for bank exams
Q25. If a/b = 4/3, then {(3a + 2b)/ (3a-2b)} =?
(a) -1 (b) 3 (c) 5 (d) 6
Ans. (b)
Q26. If a + 1/b =1 and b +1/c =1, then c + 1/a =?
(a) 0 (b) 1/2 (c) 1 (d) 2
Ans. (c)
Q27. If (a + 2b) =6 and ab= 4, then [2/a + 1/b] =?
(a) 1/2 (b) 1/3 (c) 3/2 (d) 2 (e) 5/2
Ans. (c)
Q28. If a/3 = b/4 = c/7, then (a + b +c)/c =?
(a) 1/7 (b) 1/2 (c) 2 (d) 7
Ans. (c)
Q29. If x/2y =3/2, then {(2x+y)/(x-2y)} =?
(a) 1/7 (b) 7 (c) 7.1 (d) none of these
Ans. (b)
Q30. If a/b = 3/4 and 8a + 5b =22, then a =?
(a) 2/3 (b) 3/2 (c) 5/8 (d) 2 (e) none of these
Ans. (b)
Q31. If P/Q=7, then (P+Q)/(P-Q) =?
(a) 7/8 (b) 1/3 (c) 2/3 (d) 4/3
Ans. (d)
Q32. If a/x + y/b =1 and b/y + z/c =1, then (x/a + c/z) =?
(a) 0 (b) b/y (c) 1 (d) y/b
Ans. (c)
Q33. If x = (1-a), y = (2a +1) and x=y, then a =?
(a) 2 (b) 1/2 (c) 0 (d) -1
Ans. (c)
Q34. If 2a +b = 5 and 3a- 4b =2, then ab =?
(a) 4 (b) 5 (c) 2 (d) 3
Ans. (c)
Q35. If a + b =5 and 3a + 2b = 20, then (3a +b) =?
(a) 10 (b) 15 (c) 20 (d) 25
Ans. (d)
• SBI Bank PO Examination Practices Set English
• Q36. (885*885*885 + 115*115*115) / (885*885 + 115*115 – 115*885) =?
(a) 115 (b) 770 (c) 885 (d) 1000
Ans. (d)
Q37. (147*147 + 147*143 +143*143) / (147*147*147 – 143*143*143) =?
(a) 1/290 (b) 290 (c) 1/4 (d) 4
Ans. (c)
Q38. {(38*38*38 + 34*34*34 + 28*28*28 – 38*34*84) / (38*38 + 34*34 +28*28 -38*34 – 34*28 -38*28)} =?
(a) 24 (b) 32 (c) 44 (d) 100
Ans. (d)
Q39. 3/48 is what part of 1/12?
(a) 3/7 (b) 1/12 (c) 4/3 (d) none of these
Ans. (d)
Q40. How many 1/8s are there in 371/2?
(a) 300 (b) 400 (c) 500 (d) cannot be determined
Ans. (a)
Q41. Two- third of 57 is how much more than one – third of 90?
(a) 8 (b) 38 (c) 30 (d) 28
Ans. (a)
Q42. If we multiply a fraction by itself and divide the product by its reciprocal, 18 26/27is obtained. The original fraction is:
(a) 8/27 (b) 2 2/3 (c) 1 1/3 (d) none of these
Ans. (b)
Q43. The smallest fraction which should be subtracted from the sum of 1 3/4, 2 1/2, 5 7/12, 3 1/3, and 2 1/4 to make the result a whole number, is:
(a) 5/12 (b) 7/12 (c) 1/2 (d) 7
Ans. (a)
Q44. Rohit was asked to find 7/9 of a fraction. But, made a mistake of dividing the given fraction by 7/9 and got an answer which exceeded the correct answer by 8/21. The correct answer is:
(a) 3/7 (b) 7/12 (c) 2/21 (d) 1/3
Ans. (b)
• Sample Question Paper SBI PO for reasoning with Answer key
• Q45. A boy was asked to multiply a number by 12. By mistake he multiplied the number by 21 and got his answer 63 more than the correct answer. The number was:
(a) 7 (b) 8 (c) 9 (d) 12
Ans. (a)
Q46. A farmer divides his herd of cows among his four sons so that first son gets one –half of the herd, the second son one –fourth, the third son one –fifth and the fourth son 7 cows. The total number of cows in the herd is:
(a) 100 (b) 140 (c) 180 (d) 240
Ans. (b)
Q47. A boy read 3/8th of a book on one day and 4/5th of the remainder on another day. If there were 30 pages unread, how many pages did the book contain?
(a) 240 (b) 300 (c) 600 (d) none of these
Ans. (a)
Q48. In an objective examination of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions, a student got a total of 387 marks. The number of questions attempted wrong were:
(a) 9 (b) 18 (c) 27 (d) 36
Ans. (a)
Q49. After measuring 120 meters of a rope, it was discovered that the meter rod was 3 cm longer. The true length of the rope measured is:
(a) 116m 40cm (b) 121m 20cm (c) 123m (d) 123m 60cm
Ans. (d)
Q50. I paid 5 of a bill. If Rs 400 of the bill is still due, what was the total amount of the bill?
(a) Rs 1000 (b) Rs 1200 (c) Rs1500 (d) Rs 1800
Ans. (a)
Q51. A man has in all Rs 640 in the denominations of one – rupee, five rupee and ten rupee notes. The number of each type of notes are equal. What is the total number of notes he has?
(a) 90 (b) 100 (c) 120 (d) 150
Ans. (c)
Q52. A bag contains three type of coins- rupee coins, 50 p coins and 25 p coins, totaling 175 coins. If the total value of the coins of each kind be the same, the total amount in the bag is
(a) Rs 75 (b) Rs 126 (c) Rs 175 (d) Rs 300
Ans. (a)
• PARTNERSHIP Mathematics Questions Answer Quantitative Aptitude
• Q53. A sum of Rs 6.25 is made up of 80 coins which are either ten- paise or five- paise. The number of five paise coins is:
(a) 25 (b) 35 (c) 40 (d) 45
Ans. (b)
Q54. A number of persons decided to raise Rs 3 lakh by equal contributions from each. Had they contributed Rs 50 each extra, the contribution would have been Rs 3.25 lakh. How many persons are there?
(a) 400 (b) 450 (c) 600 (d) can not be determined (e) none of these
Ans. (e)
Q55. Each child from a certain school can make 5 items of handicraft in a day. If 1125 handicraft items are to be displayed in an exhibition, then in how many days can 25 children make these items?
(a) 6 days (b) 7 days (c) 8 days (d) 9 days (e) none of these
Ans. (d)
Q56. On the children’s day sweets were distributed equally among 540 children. But, on that day 120 children were absent. Hence each child got 4 more sweets. How many sweets were to be distributed each child originally?
(a) 14 (b) 18 (c) 20 (d) 25 (e) none of these
Ans. (a)
• Chain Rule Unitary Method Question Answer 2017
• Q57. In an examination, a student was asked to fine 3/14 of a number. By mistake, he found 3/4 of it. His answer was 150 more than the correct answer. The given number is:
(a) 180 (b) 240 (c) 280 (d) 290
Ans. (c)
Q58. In a certain office 1/3 of the workers are women, 1/2 of the women are married and 1/3 of the married women have children. If 3/4 of the men are married and 2/3 of the married men have children. What part of workers is without children?
(a) 5/18 (b) 4/9 (c) 11/18 (d) 17/36
Ans. (c)
Q59. A tree increases annually by 1/8th of its height. What will be its height after 2 years? If it stands today 64cm high?
(a) 72cm (b) 74cm (c) 81cm (d) 85cm
Ans. (c)
Q60. A tin of oil was 4/5 full. When 6 bottles of oil were taken out and 4 bottles of oil poured into it, it was 3/4 full. How many bottles of oil can the tin contain?
(a) 10 (b) 20 (c) 30 (d) 40
Ans. (d)
Q61. How many digits are required for numbering the pages of a book having 300 pages?
(a) 299 (b) 492 (c) 789 (d) 792
Ans. (d)
Q62. A total of 324 coins of 20 paise and 25 paise make a sum of Rs71. The number of 25 paise coins is
(a) 120 (b) 124 (c) 144 (d) 200
Ans. (b)
Q63. A man spends 1/3 of the money with him on clothes, 1/5th of the remaining on food and 1/4 of the remaining on travel. Now he is left with Rs 100. How much did he have in the beginning?
(a) Rs 200 (b) Rs 250 (c) Rs 300 (d) Rs 450
Ans. (b)
Q64. Robert weight 15 kg more than Samuel. Their combined weight is 135 kg. How much does Samuel weight?
(a) 75 kg (b) 62.5 kg (c) 60 kg (d) none of these
Ans. (c)
Q65. A man purchased 147 stamps of 60 paise and 35 paise. The total amount he spent was Rs 68.20. How many 35 paise stamps were purchased by him?
(a) 60 (b) 72 (c) 76 (d) 80
Ans. (d)
Q66. If Rs 1440 is divided into two parts in which one part is 7/9 of the second, then the smaller part is
(a) Rs 405 (b) Rs 630 (c) Rs 810 (d) Rs 1035
Ans. (b)
Q67. Price of 5 chairs and 2 tables is Rs 1350. Two chairs cost as much as one table. What is the price of one chair and one table?
(a) Rs 350 (b) Rs 450 (c) Rs 500 (d) Rs 550
Ans. (b)
Q68. A class starts at 10 a.m. and last till 1.27p.m. Four periods are held during this interval. After every period, 5 minutes are given free to the students. The exact duration of each period is:
(a) 42 min (b) 48 min (c) 51min (d) 53 min
Ans. (b)
Q69. The difference of 1 3/16 and its reciprocal is:
(a) 1 1/8 (b) 1 1/3 (c) 15/16 (d) none of these
Ans. (d)
Q70.If 3x-5y =5 and x/(x+y) =5/7, then (x-y) =?
(a) 3 (b) 4 (c) 6 (d) 9
Ans. (c) | 5,310 | 13,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-51 | longest | en | 0.504698 |
http://www.varsitytutors.com/act_math-help/how-to-find-the-perimeter-of-an-acute-obtuse-isosceles-triangle | 1,484,959,643,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280891.90/warc/CC-MAIN-20170116095120-00501-ip-10-171-10-70.ec2.internal.warc.gz | 739,877,586 | 34,489 | # ACT Math : How to find the perimeter of an acute / obtuse isosceles triangle
## Example Questions
### Example Question #24 : Isosceles Triangles
The height of an isosceles triangle, dropped from the vertex to its base, is one fourth the length of the base. If the area of this triangle is , what is its perimeter?
Possible Answers:
Correct answer:
Explanation:
Based on the description of this question, you can draw your triangle as such. We can do this thanks to the nature of an isosceles triangle:
Now, you know that the area of a triangle is defined as:
So, for our data, we can say:
Solving for , we get:
Thus, .
Now, for our little triangle on the right, we can draw:
Using the Pythagorean Theorem, we know that the other side is:
This can be simplified to:
Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:
### Example Question #25 : Isosceles Triangles
The base of an isosceles triangle is five times the length of its correlative height. If the area of this triangle is , what is its perimeter?
Possible Answers:
Correct answer:
Explanation:
Based on the description of this question, you can draw your triangle as such. We can do this thanks to the nature of an isosceles triangle:
Now, you know that the area of a triangle is defined as:
So, for our data, we can say:
Solving for , we get:
Thus, .
Now, for our little triangle on the right, we can draw:
Using the Pythagorean Theorem, we know that the other side is:
This can be simplified to:
Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:
### Example Question #2 : Acute / Obtuse Isosceles Triangles
What is the area of an isosceles triangle with a vertex of degrees and two sides equal to units?
Possible Answers:
Correct answer:
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is , and the total base of the triangle is .
Now, the area of the triangle is:
or
### Example Question #3 : Acute / Obtuse Isosceles Triangles
What is the perimeter of an isosceles triangle with a vertex of degrees and two sides equal to
Possible Answers:
Correct answer:
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is and the total base of the triangle is .
Therefore, the perimeter of the triangle is: | 815 | 3,276 | {"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} | 4.625 | 5 | CC-MAIN-2017-04 | latest | en | 0.869805 |
https://www.petervaldivia.com/types-of-currents/ | 1,550,819,866,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247513661.77/warc/CC-MAIN-20190222054002-20190222080002-00130.warc.gz | 930,331,547 | 9,303 | # types of currents
## Types of currents. ac/dc
What does it mean: ac/dc? Ok, apart from a rock group?
There’re two types of current, defined based on the directions of the movement of the electrons. An electric current is called a direct current (d.c.) if electrons always flows in one direction. An electric caused by a chemical cell or a battery is a direct current.
If an electric current in a circuit reverses direction every so often, it is called an alternating current (a.c.). For example, the current that flows through an domestic light bulb when connected to the mains electricity supply reverses direction 100 times every second and is therefore an a1ternating current.
The circuit on the left is called a series circuit. There is only one path for the electrons to move. Electrons do not build up or leak away from the wire at any point, thus the same amount of charge is passing through every point at any given second.
### How can we measure the current?
To measure the size of an electric current we use an ammeter. Smaller currents are measured on milliammeters.
The current is the same at every point in the wire in a series circuit.
If a million electrons leave the positive terminal in the battery, a million electrons will arrive at the negative terminal of the battery.
Of course, if every electron makes a works to light up the bulb, its energy will be reduced every time it passes the bulb. They lose potential energy.
### The Current at a junction of conductors
We have seen that electrons do not disappear in a circuit. In the next diagram we have five conductors meeting at a point ‘O’ (in the diagram center). There are 2 wires carrying current towards the junction and 3 wires carrying current away from the junction and no electron is lost or stored up at the junction, thus:
The sum of the current flowing into a junction = The sum of the current leaving the junction.
In this case, I1 + I2 = I3 + I4 + I5
Example: If I3 = I2 = 10 Amp and I4 = I5 = 3 Amps, calculate the value of I1
Exercises:
1º Find the current flowing through wire W in the next figure:
2º Find the value of the current X in every circuit shown in the next figure
3º) What the reading on each of the ammeters in the next exercises
4º In the next circuit, A3 reads 2A and the bulbs are identical. What do A1 and A2 read?
5º A2 reads 10A. The bulbs are identical. What are the values of A1, A3 and A4
The bulbs are different. A2 reads 7A and A4 reads 20A. Calculate A1 and A3
Electricity always take the shortest path (way) to the ground.
As your body is 60% water and that makes you a good conductor of electricity, if a lightning falls on a tree and you touch the tree you become the best path ( the path with of least resistance ) to the ground and you have a good chance of getting electrocuted.
On the left, A Eucalyptus tree that was blown apart by a lightning strike
### Lightning Safety:
• If you are outdoor, avoid high ground and open spaces and all metal objects, including wires. Unsafe sites include trees, rain shelters or electricity pylons. Never touch the sides of a car.
• If you are indoors, avoid water, stay away from the doors and windows. Do not use the telephone. Unplug and stay away from electrical appliances, computers and Tv sets
Dictionary:
Appliances: machines, using electricity, that do a job for us in the house.
Pylons: An electricity pylon or transmission tower is a tall, usually steel, structure used to support overhead electricity conductors for electric power transmission. | 827 | 3,557 | {"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} | 4.1875 | 4 | CC-MAIN-2019-09 | longest | en | 0.919406 |
https://codegolf.stackexchange.com/questions/265159/convert-numbers-to-dice-patterns | 1,718,356,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00285.warc.gz | 153,399,472 | 46,555 | # Convert numbers to dice patterns
The way points are assigned on a dice follows a regular pattern, the center dot is present if and only if the number is odd. To represent the even numbers, pairs of dots on opposite sides of the center are added, which will stay present for all successive numbers.
1:
O
2: 3: 4: 5: 6:
O O O O O O O O
O O O O
O O O O O O O O
This "dice pattern" can be extended to numbers greater than 6.
• The dots are arranged in a square of size k with k being the smalest odd integer that satisfies k*k>=n
• the dots are added in layers for the center outwards
• one each layer the corners are added first
• then the midpoints of the sides, first on the vertical, then on the horizontal sides.
• for the remaining pairs of dots on each level you can choose any convenient order
Due to there already being solutions a the time it was added, the next constraint is an optional additional requirement:
• If the number of dots is divisible by 4 the pattern should be symmetric under reflections along the coordinate axes (through the center)
// one possible way to add the remaining dots
7: 8:
O O OOO
OOO O O
O O OOO
10: 12: 14: 16: 18: 20: 22: 24:
O O O O O O O O O O O O O O OOO O OOOOO
OOO OOO OOO OOO OOOO OOOOO OOOOO OOOOO
O O O O OO OO OO OO OO OO OO OO OO OO OO OO
OOO OOO OOO OOO OOOO OOOOO OOOOO OOOOO
O O O O O O O O O O O O O O O OOO OOOOO
26: 30: 32: 36: 40: 44:
O O O O O O O O O OO O OO OO O OO
OOOOO OOOOO OOOOO OOOOOOO OOOOOOO OOOOOOO
OOOOO OOOOO OOOOO OOOOO OOOOO OOOOOOO
OO OO OOO OOO OOO OOO OOO OOO OOO OOO OOO OOO
OOOOO OOOOO OOOOO OOOOO OOOOO OOOOOOO
OOOOO OOOOO OOOOO OOOOOOO OOOOOOO OOOOOOO
O O O O O O O O O OO O OO OO O OO
46: 48:
OOOO OO OOOOOOO
OOOOOOO OOOOOOO
OOOOOOO OOOOOOO
OOO OOO OOO OOO
OOOOOOO OOOOOOO
OOOOOOO OOOOOOO
OO OOOO OOOOOOO
Your task is to write a program of function, that given a positive integer as input outputs the dice pattern for that number.
## Rules:
• Trailing white-spaces are allowed
• Additional white-spaces at the start of each line are allowed, as long as all lines are indented by the same amount
• You can replace O by any non-whitespace character
• You may return a matrix of Truthy/Falsey values instead of printing the outputs
• Your program has to work for inputs at least up to 2601=51*51, but your algorithm should work for arbitrarily large numbers
• This is the shortest solution in bytes wins
related
• Are we supposed to use only odd values of k? Commented Sep 12, 2023 at 16:31
• @Arnauld yes (otherwise, there would not be a unique center point) Commented Sep 12, 2023 at 21:06
• "for the remaining pairs of dots on each level you can choose any convenient order" But surely the choice must be constrained by balance/symmetry, no? Commented Sep 12, 2023 at 23:50
• @Jonah I did not find a definition that captures what I would consider "balanced" patters, so for simplicity I only require that the mirror reflection of a point is added at that same time as the point. I added an optional additional symmetry requirement for even numbers of pairs. Commented Sep 13, 2023 at 7:46
– Neil
Commented Sep 13, 2023 at 20:50
# Python 3 (194 199210 bytes) -16 (- 5 thanks to Kevin)
My First golf! Returns a int, array, where len(arr) == side ** 2
def f(n):
s=2*round(.5*n**.5-1)+3;q=s*s;d,e=[1]*q,~-s>>1;r=*({*range(s-2,0,-1)}-{0,e}),e,0;d[s*s//2]=n&1;q&=-2;b=(q-n+3)//2
for k in*r[:b//2],*(s*~i for i in r[:~-b//2]):d[k]=d[~k]=0
return d
Inputs: n
Output: returns 1d array containing pips
Attempt it online!
## Formatting code:
print(n, sum(d), *(list((map(int, d[i:(i+s)]))) for i in l(0, s*s, s)), sep="\n")
## Explanation:
Calculates the correct odd-square size to have a proper center
s=2*round(.5*n**.5-1)+3
Sets a variable keeping track of how many pips there are currently
q=s*s
Creates a 1D pip array, all filled with 1s, and sets e to center index
d,e=[1]*q,~-s>>1
Creates a tuple of which order to delete pips. Goes from right right-left, center, corner. (-4 bytes)
r=*({*range(s-2,0,-1)}-{0,e}),e,0
Sets the center pip based on if it's odd
d[s*s//2]=n&1
Clears the last bit of q (sets it to even)
q&=-2
Stores number of pip-pairs to remove
b=(q-n+3)//2
Clears pips over the limit in the pip-array
for k in*r[:b//2],*(s*~i for i in r[:~-b//2]):d[k]=d[~k]=0
Returns d
return d
• Welcome to Code Golf! As a general policy, submitted answers need some way to receive input, whether it's from STDIN, ARGV, or as a function argument. Snippets of code that assume n has been defined before the program runs are not allowed. It's a good start, though! Commented Sep 13, 2023 at 21:07
• @ValueInk All fixed. Commented Sep 14, 2023 at 1:58
• Welcome to CGCC, and nice first answer. +1 from me. :) If you want, feel free to edit in this ATO link with test code. As for three minor golfs: (s-1)>>1 and (b-1)//2 can be ~-s>>1 and ~-b//2 respectively (see this general tip), and (s*~i)for can be s*~i for, for -5 bytes total. (PS: Your current function seems to be 200 bytes instead of the 199 you currently mention in your header?) EDIT: I only now noticed you mentioned your own formatting code, so feel free to replace it in the ATO link. :) Commented Sep 14, 2023 at 9:40
• Can't really edit ATO/TIO, as on a whitelisted network. You can add the link as an edit. Commented Sep 14, 2023 at 15:19
• I've added the ATO-link. Since your "Formatting code" uses an unknown l-function and seems to use d from within your function, I've just used the verbose pretty-printer I already had in my ATO-link. Feel free to change it as you see fit once you're on a regular network. Commented Sep 15, 2023 at 9:54
# JavaScript (ES7), 180 bytes
Returns a binary matrix.
n=>[...Array(w=n**.5-n/~n&~1),W=w/2].map((_,y,a)=>a.map((_,x)=>x-W|y-W?(g=k=>k--?g(k)|y==W+(A=[d=k<4?W:(k&4?k:-k)/8|0,-d,W,-W])[k&=3]&x==W+A[6>>k&3]:x%w&&y%w>0)(n-w*w+2*w&~1):n&1))
Try it online! (with post-processing for readability)
# Charcoal, 50 bytes
ＮθＦ﹪θ²ＰOＷ⁻θＬＫＡ«↙≔ⅉηＦ⁸«Ｐ✳＆⁶κ×O⌊⟦η÷⁺ιＩ§4062κ⁸⟧Ｍη✳＆⁶κ
Try it online! Link is to verbose version of code. Explanation:
Ｎθ
Input n.
Ｆ﹪θ²ＰO
If n is odd then print an O at the origin.
Ｗ⁻θＬＫＡ«
Repeat while more Os are required.
↙
Move to the bottom left corner of the next size of square.
≔ⅉη
Get the new offset (which is an eighth of the perimeter of this square).
Ｆ⁸«
Repeat to complete the perimeter of this square.
Ｐ✳＆⁶κ×O⌊⟦η÷⁺ιＩ§4062κ⁸⟧
Output a number of Os depending on how many more are needed and which part of the perimeter is being drawn, but no more than the current offset.
Ｍη✳＆⁶κ
Move to the next compass point without printing anything, which would throw off the count of Os written.
Example: When n is between 2 and 9, the number of Os is added to the given digit according to the following pattern:
604
2 2
406
This means that Os will only be drawn on the horizontal midpoints if n is 8 or 9, but will be drawn for lower values of n in the other positions. | 2,422 | 7,484 | {"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} | 3.265625 | 3 | CC-MAIN-2024-26 | latest | en | 0.763939 |
http://docplayer.net/21381938-Stoichiometry-exploring-a-student-friendly-method-of-problem-solving.html | 1,524,507,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946120.31/warc/CC-MAIN-20180423164921-20180423184921-00296.warc.gz | 87,220,961 | 25,243 | Stoichiometry Exploring a Student-Friendly Method of Problem Solving
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1 Stoichiometry Exploring a Student-Friendly Method of Problem Solving Stoichiometry comes in two forms: composition and reaction. If the relationship in question is between the quantities of each element with a compound it is called composition stoichiometry. If the relationship in question is between quantities of substances as they undergo a chemical reaction it is called reaction stoichiometry. In plain English, if you have to calculate just about anything to do with moles of chemical quantities, we collectively call those computations stoichiometry. THE MOLE CONCEPT To be successful in solving stoichiometry problems you must be familiar with the following terms and their meaning: mole -- the number of C atoms in exactly 12.0 grams of 12 C; "mole" is also a number, 6.02 x 10 23, just as the word "dozen" means 12 and "couple" means 2. Avogadro's number, n -- is equal to 6.02 x which is the number of items in a mole of anything. molar mass, mm -- the mass of one mole of particles in grams; the sum of the atomic masses in a chemical formula. When calculating be sure to multiply the atomic mass of each element by the subscript following that element. The molar mass for H 2 O is 2(1.01 g/mol for hydrogen) g/mol for oxygen = g/mol of water molecules. THE MOLE MAP The mole map provides a quick an easy way to calculate any quantity related to the number of moles. Learn to reproduce this map (in other words, memorize it) so that you may use it as a problem-solving tool when faced with stoichiometry problems. Notice that the number of moles is always in the center of the map which means when converting moles into any other unit you must multiply the number of moles by the conversion factor next to the arrow. The conversion factors included on this map will allow you to convert the number of moles to either the number of particles, mass of matter in grams, or liters of gas at standard temperature and pressure, STP (1 atm pressure and 273K). Simply remember to divide by the conversion factor when going against the arrows (to arrive at the number of moles) and multiply by the appropriate conversion factor when going in the direction of the arrows. Molar Mass 6.02 x
2 EXAMPLE 1: Complete the following table using the mole map and a periodic table. Your teacher has the correct answers. Round each molar mass to two decimal places and calculate all other values to three significant figures. Substance MM Number of Moles Mass in Grams Number of Particles Liter of STP carbon dioxide 3.00 oxygen 64.0 methane 6.25 nitrogen 9.50x Now that you understand how to use the mole map, you must realize that successful stoichiometry problem solving requires that you have demonstrated proficiency in the following individual skills: Correctly writing chemical formulas--this requires knowledge of your polyatomic ions and being able to use the periodic table to deduce what you have not had to memorize. Correctly calculating molar masses from a chemical formula. Correctly balancing chemical equations. SOLVING STOICHIMETRY PROBLEMS This method will use a table to help you organize your thoughts and focus on the task at hand. Each time you embark upon solving a reaction stoichiometry problem, reproduce the row headings found below and place them on the left side of your paper as shown: Molar Mass: Balanced Eq'n: mole:mole # of moles: amount: We will work through Example 2 and add the appropriate information to your table. Molar Mass: Bal Eq: mole:mole # of moles: amount:
3 EXAMPLE 2: A. What mass of oxygen will react with 96.1 grams of propane? (Notice neither the chemical formulas nor the balanced equation were provided. You will have to supply those.) 1. Write a chemical equation. Double check to be sure you have the correct chemical formulas! 2. Calculate the molar masses and write them in parentheses above the formulas--soon you will figure out you do not have to do this for every reactant and product, just those dealt with in the question. 3. Balance the chemical equations using coefficients. Do not attempt to change the subscripts! 4. Look at the coefficients in the balanced equation. These numbers ARE the mole:mole ratio. Re-write them under the balanced equation so they are easy to see. 5. Next re-read the problem and look for some information about a given amount of one of the substances-- in this example its 96.1 g of propane. 6. Use the mole map to calculate the number of moles of something. Start at 96.1 grams, divide (against the arrow) by molar mass to get the # of moles of propane. 7. USE the mole:mole ratio to find the moles of EVERYTHING! If 1 = 2.18 mol propane, then moles of oxygen equals 5(2.18) or 10.9 moles etc.(if you found the number of moles for something that has a coefficient other than "1", just divide by the coefficient it has and then multiply by the new coefficient.) Leave everything in your calculator--we only rounded to save space! 8. Re-read the problem to determine which amount you are seeking. First, we'll find the mass of oxygen since that's what the problem asked moles x g/mol = 349 g oxygen. THE BEST PART OF THIS METHOD COMPARED TO DIMENSIONAL ANALYSIS: B. What if part (b) of example 2 asked for liters of CO 2 at STP? Simply take the number of moles of CO 2 from the table and use the mole map to solve for the liters of gas at STP. Start in the middle of the mole map with 6.54 moles and multiply by 22.4 L/mol to determine that 146 L of CO 2 gas is produced at STP. C. What if part (c) asked how many water molecules are produced? Simply multiply the number of moles of water 8.72 mol x 6.02 x to determine that 5.25 x molecules of water were produced.
4 CALCULATIONS INVOLVING LIMTING REACTANTS Ever notice how hot dogs are sold in packages of 10 and the buns come in packages of 8? The bun is the limiting reactant and limits hot dog production to 8 when one package of each is purchased. The limiting reactant (or reagent) is the one that is (completely) consumed first in a chemical reaction. Plan of Attack for Limiting Reactants: 1. You have to recognize that you are dealing with a limiting reactant problem and you need a plan of attack! Your clue is that you are given TWO starting amounts of matter. 2. Calculate the number of moles. i. Set up your table like before, only now you have been given TWO initial amounts and must calculate the number of moles of BOTH substances to get started. 3. Divide each number of moles by its coefficient in the balanced equation. i. The limiting reactant is the smaller value. ii. Be sure to cross out the original amount given for the reactant in excess. iii. Also cross out the number of moles you calculated for the reactant in excess. 4. Recalculate the number of moles for the reactant in excess using the mole:mole ratio and the actual number of moles of the limiting reactant. EXAMPLE 3: The Haber process synthesizes ammonia, from the reaction of nitrogen and hydrogen gases and is often used for fertilizer production. The nitrogen gas is drawn from the atmosphere while the hydrogen gas is obtained by the reaction of methane and water vapor. It is carried out at 250 atm and 400 C in the presence of a catalyst. The process ultimately saved millions from starvation. The reaction is shown below: N 2 (g) + 3H 2 (g) 2NH 3 (g) Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. (a) Calculate the maximum mass of ammonia that can be produced. (b) Identify the limiting reactant. Show calculations to justify your answer. (c) Calculate the mass of the reactant that is in excess. (Don t look at the bottom yet, the answer is down there.) Molar Mass: Bal Eq: mole:mole # of moles: amount: Divide each number of moles by its coefficient. Compare the simplified moles values and select the smaller number of moles as the limiting amount, which in turn, determines the limiting reactant. N 2 has a coefficient of one already, so just focus on the 892 moles present. For H 2, your focus becomes 2, = 825 moles, which is smaller than the number of N 2 moles, so H 2 is your limiting reagent.
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PERIODIC TABLE OF ELEMENTS. 4/23/14 Chapter 7: Chemical Reactions 1
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More information | 9,579 | 38,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-17 | longest | en | 0.921102 |
https://www.numbersaplenty.com/161434649 | 1,716,634,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00642.warc.gz | 807,853,463 | 3,208 | Search a number
161434649 is a prime number
BaseRepresentation
bin10011001111101…
…00110000011001
3102020202201201002
421213310300121
5312311402044
624004034345
74000112645
oct1147646031
9366681632
10161434649
1183142272
12460929b5
13275a377c
1417623c25
15e28c74e
hex99f4c19
161434649 has 2 divisors, whose sum is σ = 161434650. Its totient is φ = 161434648.
The previous prime is 161434643. The next prime is 161434667. The reversal of 161434649 is 946434161.
161434649 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 115777600 + 45657049 = 10760^2 + 6757^2 .
It is a cyclic number.
It is not a de Polignac number, because 161434649 - 24 = 161434633 is a prime.
It is a super-2 number, since 2×1614346492 = 52122291795506402, which contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (161434643) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 80717324 + 80717325.
It is an arithmetic number, because the mean of its divisors is an integer number (80717325).
Almost surely, 2161434649 is an apocalyptic number.
It is an amenable number.
161434649 is a deficient number, since it is larger than the sum of its proper divisors (1).
161434649 is an equidigital number, since it uses as much as digits as its factorization.
161434649 is an evil number, because the sum of its binary digits is even.
The product of its digits is 62208, while the sum is 38.
The square root of 161434649 is about 12705.6935662718. The cubic root of 161434649 is about 544.5012961766.
It can be divided in two parts, 1614 and 34649, that added together give a palindrome (36263).
The spelling of 161434649 in words is "one hundred sixty-one million, four hundred thirty-four thousand, six hundred forty-nine". | 583 | 1,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | latest | en | 0.868199 |
https://ianmcgregorphotography.com/checker-60 | 1,675,139,534,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499842.81/warc/CC-MAIN-20230131023947-20230131053947-00365.warc.gz | 327,037,821 | 4,193 | Math homework answers app can be found online or in mathematical textbooks. We can solving math problem.
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Solving linear equations is the most computationally challenging part of first-order and second-order numerical algorithms. The existing direct and indirect methods either require a large amount of computation or compromise the accuracy of the solution. In this paper, an easy to calculate decomposition method is proposed to solve the sparse linear system in cone optimization. Its iteration is easy to handle, highly parallelizable, and has a closed form solution. The algorithm can be easily implemented on a distributed platform, such as a graphics processing unit, with an order of magnitude of time improvement.
Compared with the traditional CFD method, which needs to solve the complex second-order nonlinear N-S equation, the Boltzmann equation solved by LBM is a simple linear equation, and its mathematical processing is much simpler. In addition, LBM shows propulsion in the time domain based on the statistical distribution of particles, which is completely different from the traditional CFD Euler method based on implicit iteration. Zhang Chaoyang first reviewed the power series solution of Legendre equation with netizens, and explained under what circumstances the power series solution can bring convenience to the solution process. For this homogeneous linear equation, if the recurrence relationship of the coefficients obtained after substituting the solution in the form of power series is only related to two coefficients, it will be easy to get the growth of the coefficients, so that it is easy to analyze the boundary behavior of the solution; If the obtained coefficient recurrence relationship is related to multiple coefficients, it is difficult to know the changes of the coefficients. The exploration of quantum computing for scientific computing can start with HHL quantum algorithm.
When buying a scanner, you should pay special attention to this function. But at present, these three scanners, Only the scanners of the brand of Chengzhe support formula recognition, so in the formula recognition link, we can't compare the formula recognition effects of deli and Kemi scanners. We can only show you the formula recognition effects of this Chengzhe et25. We use the high school mathematics summary as the material. After the scanning process of Chengzhe et25, we can accurately recognize it and become editable, This effect is a little unexpected.
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https://www.flexiprep.com/NCERT-Exemplar-Solutions/Physics/Class-11/NCERT-Physics-Class-11-Exemplar-Ch-7-System-of-Particles-And-Rotational-Motion-Part-2.html | 1,579,530,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598800.30/warc/CC-MAIN-20200120135447-20200120164447-00233.warc.gz | 878,133,478 | 12,061 | # NCERT Physics Class 11 Exemplar Ch 7 System Of Particles And Rotational Motion Part 2
## MCQs Type Questions
Q. 5 A uniform square plate has a small piece Q of an irregular shape removed and glued to the center of the plate leaving a hole behind (Fig.). The moment of inertia about the z-axis is then
(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner.
Q.6 In problem 7.5, the CM of the plate is now in the following quadrant of x-y plane,
(a) I
(b) II
(c) III
(d) IV
Q.7. The density of a non-uniform rod of length 1m is given where a and b are constants and.
The centre of mass of the rod will be at
(a)
(b)
(c)
(d)
Answer: When , the density becomes uniform and hence the centre of mass is at .
Only option (a) tends to
Q. 8 A Merry-go-round, made of a ring-like platform of radius and mass , is revolving with angular speed . A person of mass is standing on it. At one instant, the person jumps off the round, radially away from the centre of the round (as seen from the round). The speed of the round afterwards is
(a)
(b)
(c)
(d) | 289 | 1,085 | {"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-2020-05 | longest | en | 0.891631 |
https://www.statology.org/one-way-anova-by-hand/ | 1,717,085,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00111.warc.gz | 877,239,703 | 22,436 | # How to Perform a One-Way ANOVA by Hand
A one-way ANOVA (“analysis of variance”) compares the means of three or more independent groups to determine if there is a statistically significant difference between the corresponding population means.
This tutorial explains how to perform a one-way ANOVA by hand.
### Example: One-Way ANOVA by Hand
Suppose we want to know whether or not three different exam prep programs lead to different mean scores on a certain exam. To test this, we recruit 30 students to participate in a study and split them into three groups.
The students in each group are randomly assigned to use one of the three exam prep programs for the next three weeks to prepare for an exam. At the end of the three weeks, all of the students take the same exam.
The exam scores for each group are shown below:
Use the following steps to perform a one-way ANOVA by hand to determine if the mean exam score is different between the three groups:
Step 1: Calculate the group means and the overall mean.
First, we will calculate the mean for all three groups along with the overall mean:
Step 2: Calculate SSR.
Next, we will calculate the regression sum of squares (SSR) using the following formula:
nΣ(XjX..)2
where:
• n: the sample size of group j
• Σ: a greek symbol that means “sum”
• Xj: the mean of group j
• X..: the overall mean
In our example, we calculate that SSR = 10(83.4-85.8)2 + 10(89.3-85.8)2 + 10(84.7-85.8)2 = 192.2
Step 3: Calculate SSE.
Next, we will calculate the error sum of squares (SSE) using the following formula:
Σ(XijXj)2
where:
• Σ: a greek symbol that means “sum”
• Xij: the ith observation in group j
• Xj: the mean of group j
In our example, we calculate SSE as follows:
Group 1: (85-83.4)2 + (86-83.4)+ (88-83.4)+ (75-83.4)+ (78-83.4)+ (94-83.4)+ (98-83.4)+ (79-83.4)+ (71-83.4)+ (80-83.4)640.4
Group 2: (91-89.3)2 + (92-89.3)+ (93-89.3)+ (85-89.3)+ (87-89.3)+ (84-89.3)+ (82-89.3)+ (88-89.3)+ (95-89.3)+ (96-89.3)208.1
Group 3: (79-84.7)2 + (78-84.7)+ (88-84.7)+ (94-84.7)+ (92-84.7)+ (85-84.7)+ (83-84.7)+ (85-84.7)+ (82-84.7)+ (81-84.7)252.1
SSE: 640.4 + 208.1 + 252.1 = 1100.6
Step 4: Calculate SST.
Next, we will calculate the total sum of squares (SST) using the following formula:
SST = SSR + SSE
In our example, SST = 192.2 + 1100.6 = 1292.8
Step 5: Fill in the ANOVA table.
Now that we have SSR, SSE, and SST, we can fill in the ANOVA table:
Source Sum of Squares (SS) df Mean Squares (MS) F
Treatment 192.2 2 96.1 2.358
Error 1100.6 27 40.8
Total 1292.8 29
Here is how we calculated the various numbers in the table:
• df treatment: k-1 = 3-1 = 2
• df error: n-k = 30-3 = 27
• df total: n-1 = 30-1 = 29
• MS treatment: SST / df treatment = 192.2 / 2 = 96.1
• MS error: SSE / df error = 1100.6 / 27 = 40.8
• F: MS treatment / MS error = 96.1 / 40.8 = 2.358
Note: n = total observations, k = number of groups
Step 6: Interpret the results.
The F test statistic for this one-way ANOVA is 2.358. To determine if this is a statistically significant result, we must compare this to the F critical value found in the F distribution table with the following values:
• α (significance level) = 0.05
• DF1 (numerator degrees of freedom) = df treatment = 2
• DF2 (denominator degrees of freedom) = df error = 27
We find that the F critical value is 3.3541.
Since the F test statistic in the ANOVA table is less than the F critical value in the F distribution table, we fail to reject the null hypothesis. This means we don’t have sufficient evidence to say that there is a statistically significant difference between the mean exam scores of the three groups.
Bonus Resource: Use this One-Way ANOVA Calculator to automatically perform a one-way ANOVA for up to five samples.
May 29, 2024
May 13, 2024
April 25, 2024
## 21 Replies to “How to Perform a One-Way ANOVA by Hand”
Thank you Zack, well done. I am going to use your example for a graduate student struggling to understanding the concept of ANOVA….
2. France Jacob Punay says:
Very helpful for me to learn as a student
thankyou for the lesson. I’ve learned a lot
4. omi says:
what is the IV and the DV in this example?
5. Manish says:
Thanks Zach. This was very helpful.
6. Edwin says:
Think there is an issue with how you are calculating the F statistic. You say that to calculate MS treatment: SST / df treatment = 192.2 / 2 = 96.1. However you are plugging in 192.2 and not the 1292.8 that you calculated for SST. This effects the final result.
7. Soenhay says:
MS treatment: SST / df treatment = 192.2 / 2 = 96.1
Should be:
MS treatment: SSR / df treatment = 192.2 / 2 = 96.1
8. Lucy says:
THANK YOU VERY MUCH. I REALLY UNDRESTOOD THIS STEPWISE EXPLANATION
9. Birhanu Takele says:
The question in my mind was solved by your explanation in this section.
I am glad to say thank you!
10. Morateng says:
Good afternoon
I’ve got an assignment regards One way analysis of variance& I’m struggling with it
11. simon says:
Wow, this is so much informative. The concepts and steps are clearly explained and it is generally easier to understand
12. giga says:
Is the formula for SSR correct? Some sources say it’s not multiplied by n.
13. giga says:
NVM, I double-checked with Walpole. It’s the same.
14. Acaye Phillips says:
This has been more than resourceful to me , I can’t express my applause by word of mouth
15. Rogelio J. Bahian says:
I like this presentation of calculation of O e Way ANOVA by hand or manually.
I can easily teach my students taking up research.
Thank you so much.
16. Susie Diaz says:
I need help with my psych2 class. ANOVA-Quiz. Can you take this 12 question quiz for a fee
17. Alfred Lopez says:
how did you come up with that f critical value of 3.3541? Where is that coming from?
18. George Mbega says:
This Write up made me remember what I had forgotten, even my lecturer, having done my coursework in 2007.
Thanks, this is good work.
19. ROBERT NELSON RWEBANGIRA says:
Well simplified for easier understanding of the complex subject.
20. Ryan says:
This was truly amazing. Super easy to follow and super easy to understand and apply to our own data. You really saved my math IA. Thank you so much!
21. osogo victor says:
clear and understandable | 1,894 | 6,259 | {"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} | 4.84375 | 5 | CC-MAIN-2024-22 | latest | en | 0.794819 |
http://www.mathisfunforum.com/viewtopic.php?pid=34276 | 1,500,633,776,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423769.10/warc/CC-MAIN-20170721102310-20170721122310-00451.warc.gz | 498,984,878 | 4,393 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °
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## #1 2006-05-02 05:31:13
RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212
### Curve Sketching
I am having a little trouble with scaling curves, for example:
y = -2x^3
I know that;
y = af(x) is scaled in the y direction by a
y = f(ax) is scaled in the x direction by 1/a
So how can I tell if y = -2x^3 is y = -2f(x) or f(-2x)???
Aloha Nui means Goodbye.
Offline
## #2 2006-05-02 08:57:39
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Curve Sketching
y = -2(x)^3 or y = (-2x)^3?
y = -2(x)^3:
Draw the sketch x^3, then flip it over the y axis. Now stretch it upwards by a factor of two. Plotting some points also helps. For example, plot x = -2, 1, 0, 1, 2, then connect the dots keeping in mind it should look something like x^3.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
## #3 2006-05-02 11:14:51
RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212
### Re: Curve Sketching
I know that the curve is exactly the same but if you are asked to describe the transformation, should I say that it's scaled in the y-direction by a factor -2 or in the x-direction by factor -1/2?
Aloha Nui means Goodbye.
Offline
## #4 2006-05-02 11:41:23
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Curve Sketching
Separate the - from the 2. You flip because of the negative, you strech/shirk because of the 2.
But remember that streching the y by a factor of 2 is exactly the same as shrinking the x by a factor of 2, which is exactly the same as streching the x by a factor of 1/2.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
## #5 2006-05-03 00:12:18
RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212
### Re: Curve Sketching
So I wont have any questions in the format y = -2x^3 asking in which axis I need to scale?
FYI. The correct term for streching or shrinking the curve is to "scale" it by a "factor"
Aloha Nui means Goodbye.
Offline
## #6 2006-05-03 00:23:12
Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791
### Re: Curve Sketching
Right, but "scale" it by a "factor" may not make sense to someone who doesn't know that's the correct term for streching/shrinking
And you probably will have questions like this, but there are just multiple correct answers. Just like if I asked you to give me two numbers when added together that give me 9.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
## #7 2006-05-03 00:43:29
RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212
### Re: Curve Sketching
Ah good point!!! I'll write both answers
Aloha Nui means Goodbye.
Offline
## #8 2006-05-03 02:58:54
George,Y
Member
Registered: 2006-03-12
Posts: 1,306
### Re: Curve Sketching
Compare y=f(ax[sub]2[/sub]) and y=f(x)
when the two y equals, usually the value in the bracket equals.
ax[sub]2[/sub]=x
thus x[sub]2[/sub]=x/a
X'(y-Xβ)=0
Offline
## #9 2006-05-03 04:14:07
RickyOswaldIOW
Member
Registered: 2005-11-18
Posts: 212
### Re: Curve Sketching
-2x² - 7x + 15
-2[x² + 7/2x] + 15
-2[(x + 7/4)² - 49/16] + 15
-2(x + 7/4)² + 18 + 1/16
Sketch the curve of -2x² - 7x + 15:
2(x + 7/4)² = 18 + 1/16
(x + 7/4)² = 9 + 1/32
x + 7/4 = ±√9 + 1/32
x = - 7/4 ±√9 + 1/32
So the curve would cross the x-axis at -4.76 and +1.26 which, as you already know, is wrong!.
Aloha Nui means Goodbye.
Offline
## #10 2006-05-04 15:52:02
George,Y
Member
Registered: 2006-03-12
Posts: 1,306
### Re: Curve Sketching
Yes, but a trick will solve the controversy
devide the original curve or function into such pieces that in each piece x and y are one by one, use my procedure seperately and then stick the pieces by original sequence.
X'(y-Xβ)=0
Offline | 1,436 | 4,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2017-30 | latest | en | 0.865066 |
https://ncatlab.org/nlab/show/Hida+theory | 1,726,370,145,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-38/segments/1725700651614.9/warc/CC-MAIN-20240915020916-20240915050916-00284.warc.gz | 384,404,497 | 6,909 | Contents
# Contents
## Idea
Hida theory is an approach to p-adic interpolation of modular forms in the ordinary case.
## Definitions
This section follows chapter 8 of DarmonRotgerAWS.
Let $\Lambda$ be the Iwasawa algebra $\mathbb{Z}_{p}[[\Gamma]]$ where $\Gamma=(1+p\mathbb{Z}_{p})^{\times}$. We have that $\Lambda\cong \mathbb{Z}_{p}[[T]]$. We can further identify $\Lambda$ as the ring of rigid analytic functions on the weight space
$\mathcal{X}:=\Hom_{cts}(\Gamma,\mathbb{C}^{\times}).$
A point $\nu\in \mathcal{X}$ is called an arithmetic point if there exists and integer $k$ (called the weight) and a Dirichlet character $\chi$ of p-power order and values in $\mathbb{C}_{p}^{\times}$ such that
$\nu(x)=\chi(x)x^{k}$
for all $x\in (1+p\mathbb{Z}_{p})$.
More generally a finite extension $\widetilde{\Lambda}$ of $\Lambda$ can be identified with the ring of rigid analytic functions on an etale cover $\widetilde{\mathcal{X}}$ of $\mathcal{X}$.
A $\Lambda$-adic modular form of tame level $N$ and tame character $\chi$ is a pair $(\widetilde{\Lambda},\underline{g})$ where $\widetilde{\Lambda}$ is a finite extension of $\Lambda$ and
$\underline{g}:=\sum \underline{a}_{n}q^{n}$
where $\underline{a}_{n}\in \widetilde{\Lambda}$, such that for almost all $\nu\in\widetilde{X}$ of weight $k\in\mathbb{Z}\geq 2$, the power series
$\underline{g}:=\sum \underline{a}_{n}(\nu)q^{n}$
is the q-expansion of a classical normalized ordinary eigenform of weight $k$ and level $N$.
## References
• Chris Williams, An Introduction to Hida Theory (pdf)
• Cameron Franc, Hida Theory (pdf)
A discussion may also be found in chapter 8 of
• Henri Darmon and Victor Rotger, Algebraic Cycles and Stark-Heegner Points, (pdf)
Last revised on December 2, 2022 at 17:00:39. See the history of this page for a list of all contributions to it. | 562 | 1,845 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "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} | 2.765625 | 3 | CC-MAIN-2024-38 | latest | en | 0.654856 |
http://www.jiskha.com/display.cgi?id=1372733194 | 1,462,534,568,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861754936.62/warc/CC-MAIN-20160428164234-00010-ip-10-239-7-51.ec2.internal.warc.gz | 592,014,405 | 3,893 | Friday
May 6, 2016
Homework Help: Algebra
Posted by Jack on Monday, July 1, 2013 at 10:46pm.
An executive flew in the corporate jet to a meeting in a city 1700 kilometers away. After traveling the same amount of time on the return flight, the pilot mentioned that they still had 300 kilometers to go. The air speed of the plane was 600 kilometers per hour. How fast was the wind blowing? (Assume that the wind direction was parallel to the flight path and constant all day.)
• Algebra - Steve, Tuesday, July 2, 2013 at 12:03am
time = distance / speed, so
1700/(600+x) = 1400/(600-x)
x = 58.065 km/hr | 170 | 604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2016-18 | longest | en | 0.968081 |
https://www.cnblogs.com/iwtwiioi/p/4097231.html | 1,620,875,059,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00489.warc.gz | 740,003,261 | 7,953 | # 【BZOJ】1044: [HAOI2008]木棍分割（二分+dp）
http://www.lydsy.com/JudgeOnline/problem.php?id=1044
orz
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mkpii make_pair<int, int>
#define pdi pair<double, int>
#define mkpdi make_pair<double, int>
#define pli pair<ll, int>
#define mkpli make_pair<ll, int>
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << '\t'; cout << endl
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=50005, MD=10007;
int n, a[N], sum[N], f[N], d[N], m, ans;
bool check(int x) {
int tot=0, s=0;
for1(i, 1, n) {
if(a[i]>x) return false;
s+=a[i];
if(s>x) { s=a[i]; ++tot; }
if(tot>m) return false;
}
return true;
}
int main() {
int l=1, r=0;
for1(i, 1, n) read(a[i]), r+=a[i], sum[i]=sum[i-1]+a[i];
while(l<=r) {
int mid=(l+r)>>1;
if(check(mid)) r=mid-1;
else l=mid+1;
}
ans=r+1;
printf("%d", ans);
for1(i, 0, n) if(sum[i]<=ans) d[i]=1; else break;
for1(j, 1, m) {
int k=0;
f[0]=d[0];
for1(i, 1, n) f[i]=(f[i-1]+d[i])%MD;
for1(i, j+1, n) {
while(k<i && sum[i]-sum[k]>ans) ++k;
d[i]=(f[i-1]-f[k-1]+MD)%MD;
}
// for1(i, j+1, n) {
// for1(k, 0, i-1) if(sum[i]-sum[k]<=ans) d[i][j]=(d[i][j]+d[k][j-1]);
// }
}
printf(" %d\n", d[n]);
return 0;
}
3 2
1
1
10
10 2
## HINT
n<=50000, 0<=m<=min(n-1,1000).
1<=Li<=1000.
## Source
posted @ 2014-11-14 14:25 iwtwiioi 阅读(247) 评论(0编辑 收藏 | 926 | 2,246 | {"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} | 2.828125 | 3 | CC-MAIN-2021-21 | latest | en | 0.108299 |
https://blogs.perl.org/users/e_choroba/2020/04/perl-weekly-challenge-054-kth-permutation-sequence-collatz-conjecture.html | 1,679,603,142,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945183.40/warc/CC-MAIN-20230323194025-20230323224025-00427.warc.gz | 168,160,786 | 6,041 | ## Kth Permutation Sequence
Write a script to accept two integers n (>=1) and k (>=1). It should print the k-th permutation of n integers.
For example, n=3 and k=4, the possible permutation sequences are listed below:
```123
132
213
231
312
321```
The script should print the 4th permutation sequence 231.
The straightforward way is to generate all the permutations in the correct order and then output the kth one. To generate them, we can use recursion: To get all the permutations of n elements, start with each element and extend it with all the permutations of the remaining elements.
``````#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
sub perm {
my (\$src) = @_;
return [\$src] if 1 == @\$src;
my @perms;
for my \$s (@\$src) {
my \$subperms = perm([grep \$s != \$_, @\$src]);
push @perms, map [\$s, @\$_], @\$subperms;
}
return \@perms
}
sub kth_perm {
my (\$n, \$k) = @_;
return @{ perm([1 .. \$n])->[\$k - 1] }
}
my (\$n, \$k) = @ARGV;
say kth_perm(\$n, \$k);``````
The problem of this solution is performance. It takes it more than 2 seconds to output the 5678th permutation of 1 .. 9. If we replace the `!=` with `ne` so we can process permutations of letters, too, it takes twice as more.
There is a way how to construct the kth permutation directly without enumerating the previous ones. Let’s have a closer look: The first character divides all the permutations into n groups.
```1234 2134 3124 4123
1243 2143 3142 4132
1324 2314 3214 4213
1342 2341 3241 4231
1423 2413 3412 4312
1432 2431 3421 4321```
Size of each group is the number of permutations of n - 1 elements (i.e. the characters following the first one). And the same applies to subgroups of each group, e.g.
```1234 1324 1423
1243 1342 1432```
Size of each group is the number of permutations of the remaining elements.
The number of all the permutations of x elements is x! or x factorial, i.e. `product(1 .. \$x)`. If the k is given, e.g. 15, we can divide it by the size of a subgroup to know what group we can find the result in (`int(15 / 6) == 3`). To repeat the step recursively, we need to know what k will be used in the subgroup. If we try searching several times, we’ll realise it’s the remainder of the division, i.e. `15 % 6 == 3` (in the code bellow, you can notice I simplified the border conditions).
``````#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
use List::Util qw{ product };
sub perm_recurse {
my (\$k, @n) = @_;
return "" unless @n;
my \$factorial = product(1 .. @n);
my \$step = \$factorial / @n;
my \$select = int(\$k / \$step);
--\$select unless \$k % \$step;
return \$n[\$select]
. perm_recurse((\$k % \$step) || \$step,
@n[ grep \$_ != \$select, 0 .. \$#n ])
}
sub kth_perm { perm_recurse(\$_[1], 1 .. \$_[0]) }
my (\$n, \$k) = @ARGV;
say kth_perm(\$n, \$k);``````
This code finds the 5678th permutations of 9 elements in less than 0.01 seconds. Even calculating `perm_recurse(10, 'a' .. 'z')` takes less than a 0.1 seconds, not talking about the memory consumption of the naive approach.
## Collatz Conjecture
It is thought that the following sequence will always reach 1:
```\$n = \$n / 2 when \$n is even
\$n = 3*\$n + 1 when \$n is odd```
For example, if we start at 23, we get the following sequence:
`23 → 70 → 35 → 106 → 53 → 160 → 80 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1`
Write a function that finds the Collatz sequence for any positive integer. Notice how the sequence itself may go far above the original starting number.
### Extra Credit
Have your script calculate the sequence length for all starting numbers up to 1000000 (1e6), and output the starting number and sequence length for the longest 20 sequences.
I started with the naive implementation again. This time, there’s not trick, including the extra credit: just enumerate all the sequences, remember them in an array, sort them by size and output the top 20.
``````#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
sub collatz {
my (\$start) = @_;
my @seq = \$start;
push @seq, (\$seq[-1] / 2, 3 * \$seq[-1] + 1)[\$seq[-1] % 2]
while \$seq[-1] != 1;
return @seq
}
my @sizes;
push @sizes, [\$_, scalar collatz(\$_)] for 1 .. 1e6;
say "@\$_" for reverse +(sort { \$b->[1] <=> \$a->[1] } @sizes)[0 .. 19];``````
I was worried about the performance again. The program took 38 seconds to compute the extra credit task which didn’t sound fast. I tried to only keep a heap of the 20 longest sequences instead of storing all of them, but it was in fact even slower. I also compared my solution to Laurent’s one, realising dynamic programming would shave off about 6 seconds (about 15%). I wasn’t sure such a small gain was worth the effort, so I stopped there. | 1,418 | 4,769 | {"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-2023-14 | latest | en | 0.710863 |
https://au.mathworks.com/matlabcentral/answers/279106-how-to-set-y-axis-as-log-scale?s_tid=faqs_link | 1,722,760,611,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00698.warc.gz | 89,118,436 | 35,618 | # how to set y-axis as log scale?
4,177 views (last 30 days)
Rohit Bhoi on 15 Apr 2016
Commented: Adam Danz on 12 Jul 2024 at 15:08
I am plotting x-y plot using plot function. I want to set only y-axis as log scale and x-axis as linear? How to do that? I used loglog function but it scales both axis but I want only Y.
Walter Roberson on 22 Sep 2023
Edited: MathWorks Support Team on 22 Sep 2023
The best way to create that type of axes is to use the semilogy function. Alternatively, you can set the ‘YScale’ property on the axes:
set(gca, 'YScale', 'log')
***Update from Mathworks Support Team - September 2023***
As of R2023b, you can also use the 'yscale ' function.
Mr Thadi on 4 Jul 2024
Thank you Mr walter,
i have attached one image,
in that my data file having values from 100 to 14000.
if i plot histogram without x axis set properties it showing like figure 'A', in that bins starts from 0-300,300-600...etc like that.
but if i use set (gca,'xscale','log') command to the x scale it showing initial bin from 300-600,600-900....etc.you can see in figure 'B'
my question is what about values in between 0-300 from figure 'B' , why these are not appearing how to solve it,if i want to get xscale in log values histogram initial bin from 0-300,300-600...etc are needed to change properties in that to command?can you please explain me
i can provide my data file also to you
Adam Danz on 12 Jul 2024 at 15:08
This issue arises because the first bin edge is at x=0 and log(0)=-inf which cannot be represented graphically.
Assuming there are no data less than or equal to 0 and no bin edges less than 0, you could set the first bin edge according to the smallest value in the data.
x = rand(1,1000)*10000;
minPositiveValue = min(x(x>0),[],'all');
minbin = 10^floor(log10(minPositiveValue));
ax = axes();
h = histogram(ax, x);
ax.XScale = 'log';
if h.BinEdges(1) == 0
h.BinEdges(1) = minbin;
end
Toshia M on 20 Sep 2023
Starting in R2023b, you can change the scale of any axis after you create the plot by calling the xscale, yscale, or zscale function.
For example, create a plot of two vectors x and y. Then set the scale of the y-axis to logarithmic.
x = 1:100;
y = x.^2;
plot(x,y)
grid on
yscale log
Rohit Sinha on 27 Apr 2022
The easiest way to do this is simply use the following command instead of plot
semilogy(x,y);
This will plot x axis on a linear scale and y axis on a log scale. Similarly, if you want to plot x axis on log scale and y axis on a linear scale, you can use
semilogx(x,y) ;
Walter Roberson on 27 Apr 2022
semilogy() is the first thing I mentioned in my answer in 2016.
Nicholas Santiago on 4 Nov 2022
yo i totally missed that I generally only read the bold stuff, thanks a ton!
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1-5.
The area of the rectangle below is $24$ square units. On graph paper, draw and label all possible rectangles with an area of $24$ square units. Use only whole numbers for the dimensions (measurements).
$\text{Area}=24\;\text{square units}$
$1$ by $24$, $2$ by $12$, $3$ by $8$, and $4$ by $6$
1. Find the perimeter of each of these rectangles. You may want to refer to the Math Notes box for this lesson for more information on the perimeter of a figure.
The perimeter is the distance around the outside of a figure.
$50$, $28$, $22$, and $20$ units
2. Of these rectangles, which has the largest perimeter? Which has the smallest perimeter? Describe these shapes. Remember to use complete sentences.
The largest perimeter is a thin rectangle, P = $50$ units;
the smallest is close to a square shape, P = $20$ units. | 236 | 883 | {"found_math": true, "script_math_tex": 17, "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} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.867433 |
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High School Math symbols
all the symbols in math that are used in 6th grade
Terms in this set (...)
plus
+
minus
times
×
divided by
÷
percent
%
equals
=
does not equal
approximately equal to
less than
<
greater than
>
less than or equal to
greater than or equal to
similar
square root
cube root
³√
positive
negative
power to
pi
π
infinity
degrees
°
plus or minus
±
Parallel to
perpendicular to
|x|
absolute value of x
∠A
angle A
∆ABC
Triangle ABC
8ⁿ
eight to the power of n | 149 | 486 | {"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} | 2.828125 | 3 | CC-MAIN-2019-18 | longest | en | 0.90666 |
http://inotivity.com/m1815b4JP/qU11465cK/ | 1,576,413,944,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00285.warc.gz | 70,863,492 | 13,316 | # Worksheet Rounding Games 3rd Grade 2nd Stages Of Mitosis Number Bonds To Simple Division Worksheets Factor Pairs Distributive Property Solver Multiplication Reading Skills Common Core Math
Published at Wednesday, May 01st, 2019 - 00:53:33 AM. Worksheet. By Geneva Pereira.
Conclusion, There are two fundamental problems with worksheets. First, young children do not learn from them what teachers and parents believe they do (Kostelnik, Soderman, & Whiren, 1993). Second, children’s time should be spent in more beneficial endeavors (Willis, 1995). The use of abstract numerals and letters, rather than concrete materials, puts too many young children at risk of school failure. This has implications for years to come. Worksheets and workbooks should be used in schools only when children are older and developmentally ready to profit from them (Bredekamp, S. & Rosegrant, T., 1992). Our challenge is to convince parents and others that in a play-based, developmentally appropriate curriculum children are learning important knowledge, skills, and attitudes that will help them be successful in school and later life.
Cognitive Development, Most preschool and kindergarten children are in what Piaget described as the preoperational stage of cognitive development. Letters and numerals typically mean little to the three- to six-year-olds in this stage. These children use concrete rather than abstract symbols to represent objects and ideas (Bodrova & Leong, 1996). Through pretending, children develop the ability mentally to represent the world (Bredekamp, 1987; Stone, 1995). Reading requires a child to look at symbols or representations (i.e., letters and words) and extract meaning from them. A play-based curriculum offers children opportunities throughout the day to develop the ability to think abstractly by experiencing real objects using their senses (Bredekamp, 1987; Kostelnik, Soderman, & Whiren, 1993). Blocks can represent an airplane or a train. High heels can transform a preschooler into a mother or princess. Blocks and high heels are three dimensional, tangible objects. Sufficient practice using concrete objects as symbols is a necessary prerequisite to the use and comprehension of print (Stone, 1995).
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May 01, 2019 | Suzette Couturier | 742 | 3,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-51 | latest | en | 0.924586 |
http://conversion.org/length/light-day/french-charriere | 1,721,099,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00409.warc.gz | 6,262,477 | 7,381 | # light-day to french; charriere conversion
Conversion number between light-day and french; charriere [F] is 7.77062051136 × 10+16. This means, that light-day is bigger unit than french; charriere.
### Contents [show][hide]
Switch to reverse conversion:
from french; charriere to light-day conversion
### Enter the number in light-day:
Decimal Fraction Exponential Expression
light-day
eg.: 10.12345 or 1.123e5
Result in french; charriere
?
precision 0 1 2 3 4 5 6 7 8 9 [info] Decimal: Exponential:
### Calculation process of conversion value
• 1 light-day = (exactly) (25902068371200) / ((1/3000)) = 7.77062051136 × 10+16 french; charriere
• 1 french; charriere = (exactly) ((1/3000)) / (25902068371200) = 1.286898515425 × 10-17 light-day
• ? light-day × (25902068371200 ("m"/"light-day")) / ((1/3000) ("m"/"french; charriere")) = ? french; charriere
### High precision conversion
If conversion between light-day to metre and metre to french; charriere is exactly definied, high precision conversion from light-day to french; charriere is enabled.
Decimal places: (0-800)
light-day
Result in french; charriere:
?
### light-day to french; charriere conversion chart
Start value: [light-day] Step size [light-day] How many lines? (max 100)
visual:
light-dayfrench; charriere
00
107.77062051136 × 10+17
201.554124102272 × 10+18
302.331186153408 × 10+18
403.108248204544 × 10+18
503.88531025568 × 10+18
604.662372306816 × 10+18
705.439434357952 × 10+18
806.216496409088 × 10+18
906.993558460224 × 10+18
1007.77062051136 × 10+18
1108.547682562496 × 10+18
Copy to Excel
## Multiple conversion
Enter numbers in light-day and click convert button.
One number per line.
Converted numbers in french; charriere:
Click to select all
## Details about light-day and french; charriere units:
Convert Light-day to other unit:
### light-day
Definition of light-day unit: ≡ 24 light-hours .
Convert French; charriere to other unit:
### french; charriere
Definition of french; charriere unit: ≡ 1⁄3 mm .
← Back to Length units | 625 | 2,042 | {"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} | 2.953125 | 3 | CC-MAIN-2024-30 | latest | en | 0.680375 |
https://www.yaclass.in/p/mathematics-cbse/class-10/circles-21891/circles-11518/re-443c8904-04b4-4474-a267-3a34247e7011 | 1,718,308,389,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00327.warc.gz | 959,864,381 | 12,728 | PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free Demo
In the earlier classes, we have learnt what circles and the terms related to the circle are. Now, let us analyse what happens if a circle and a line are on a plane.
Let us discuss some situations when a circle and a line intersect.
Situation 1: The line $$AB$$ does not touch the circle.
Here, there is no common point between the straight line $$AB$$ and the circle.
Therefore, the number of points of intersection is zero.
Situation 2: The line $$AB$$ touches the circle at one point.
Here, there is one common point $$P$$ between the straight line $$AB$$ and the circle.
The line $$AB$$ is called the tangent to the circle at $$P$$.
Therefore, the number of points of intersection is one.
Situation 3: The line $$AB$$ touches the circle at two points.
Here, there are two common points $$P$$ and $$Q$$ between the straight line $$AB$$ and the circle.
The line $$AB$$ is called the secant of the circle.
Therefore, the number of points of intersection is two.
Important!
The line segment inscribed in a circle is called the chord of the circle.
The chord is a sub-section of a secant. | 422 | 1,240 | {"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} | 3.8125 | 4 | CC-MAIN-2024-26 | longest | en | 0.697642 |
https://www.physicsforums.com/threads/why-does-the-universe-move-towards-an-observer-at-rest.648359/ | 1,603,661,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00443.warc.gz | 860,266,424 | 21,682 | # Why does the universe move towards an observer at rest?
In relativity, we do not talk about just space anymore, but space-time, with time being just another dimension.
An observer A at rest in an IRF who considers himself at rest at t=0 x=0, has a far away clock at t=100 for example moving towards him, i would like to say at 1second per second, but i guess that wouldn't make sense. It is however moving towards him at SOME pace.
Someone moving at vrel relative to observer A, sees him moving towards the clock along the x and t axis.
There has to be a reason for that, because otherwise it seems logical to assume that we would just stand at a fixed(static) coordinate without moving towards either the x or t axis.
Last edited:
## Answers and Replies
Related Special and General Relativity News on Phys.org
Nugatory
Mentor
An observer A at rest in an IRF who considers himself at rest at t=0 x=0
Wait a moment.... He can't be at rest at t=0, x=0 because he's moving in the +t direction so his t coordinate is continually increasing. In fact, if he's carrying a stopwatch that he started at t=0,x=0 his t coordinate will always be the time that the stopwatch reads: x=0,t=1 when the stopwatch reads 1 second, x=0,t=2 when the stopwatch reads 2 seconds, and so forth.
has a far away clock at t=100 for example moving towards him
That doesn't make any sense either. t=100 is 100 seconds into the future, when his stopwatch would read 100. I don't see how a clock, or anything else for that matter, could start there and move towards our observer at t=0 (and moving in the direction of increasing t) unless the clock or whatever were traveling back in time. And I kinda doubt that's what you meant... so what did you mean?
There has to be a reason for that, because otherwise it seems logical to assume that we would just stand at a fixed(static) coordinate without moving towards either the x or t axis.
We're never standing at fixed static coordinates. We're always moving forward in the t direction (and after a half-century or so of moving in that direction, we start to get old and wish that we weren't moving quite so quickly in the direction of increasing t).
So it's not that the universe is "moving towards an observer". Everything, absolutely everything, is moving in spacetime. There are inertial reference frames in which the x coordinate of an object's is constant (it's at rest in that frame) and there are inertial reference frames in which the x coordinate is changes with time and is equal to vt (the object is moving at a speed of v in that frame, or equivalently the origin of the frame is moving at a speed of -v relative to the object), but there are no frames in which the t coordinate is not increasing.
Wait a moment.... He can't be at rest at t=0, x=0 because he's moving in the +t direction so his t coordinate is continually increasing. In fact, if he's carrying a stopwatch that he started at t=0,x=0 his t coordinate will always be the time that the stopwatch reads: x=0,t=1 when the stopwatch reads 1 second, x=0,t=2 when the stopwatch reads 2 seconds, and so forth.
That doesn't make any sense either. t=100 is 100 seconds into the future, when his stopwatch would read 100. I don't see how a clock, or anything else for that matter, could start there and move towards our observer at t=0 (and moving in the direction of increasing t) unless the clock or whatever were traveling back in time. And I kinda doubt that's what you meant... so what did you mean?
We're never standing at fixed static coordinates. We're always moving forward in the t direction (and after a half-century or so of moving in that direction, we start to get old and wish that we weren't moving quite so quickly in the direction of increasing t).
So it's not that the universe is "moving towards an observer". Everything, absolutely everything, is moving in spacetime. There are inertial reference frames in which the x coordinate of an object's is constant (it's at rest in that frame) and there are inertial reference frames in which the x coordinate is changes with time and is equal to vt (the object is moving at a speed of v in that frame, or equivalently the origin of the frame is moving at a speed of -v relative to the object), but there are no frames in which the t coordinate is not increasing.
why do you think the t axis is special compared to the x axis? Why when an object on the x axis is moving towards us, an observer considers himself at rest, while when an object at a distance on the t axis moving towards us you consider the object at rest?
Even if we were to take your point of view. Why do we move at all towards an object in the future? Shouldn't we remain static at one point unless there is a reason for that motion?
Nugatory
Mentor
why do you think the t axis is special compared to the x axis? Why when an object on the x axis is moving towards us, an observer considers himself at rest, while when an object at a distance on the t axis moving towards us you consider the object at rest?
I'm sorry, but I still don't understand what you mean by "an object at a distance on the t axis moving towards us". Can you construct a thought experiment in which such motion could be considered to be happening?
Even if we were to take your point of view. Why do we move at all towards an object in the future? Shouldn't we remain static at one point unless there is a reason for that motion?
I don't know of any good answer to that "Why?" question. But no matter why it happens, it appears to be a fact that all objects are always moving in the time direction of space-time. I can always find a frame in which I am at rest in space (my x, y, and z coordinates are constant) but I cannot find a frame in which I stop aging, the t coordinate is constant, and time is not passing.
It's tempting to explain this by saying that the 4-velocity of an object is never zero. But that's not really an explanation, it's just a mathematical restatement of what we're trying to explain.
Chestermiller
Mentor
I have written up a set of notes on SR that specifically addresses the questions you are asking. The write up is in the form of a Word document which I can attach to an email. If you would like a copy of the write up, send me an email message at chestermiller@mindspring.com. The write up focuses on "what is it about the fundamental geometric structure and kinematics of 4D spacetime that can give rise to the peculiar effects that we observe?"
chet
PAllen
Science Advisor
2019 Award
In relativity, we do not talk about just space anymore, but space-time, with time being just another dimension.
An observer A at rest in an IRF who considers himself at rest at t=0 x=0, has a far away clock at t=100 for example moving towards him, i would like to say at 1second per second, but i guess that wouldn't make sense. It is however moving towards him at SOME pace.
Someone moving at vrel relative to observer A, sees him moving towards the clock along the x and t axis.
There has to be a reason for that, because otherwise it seems logical to assume that we would just stand at a fixed(static) coordinate without moving towards either the x or t axis.
If you want to take the view of time as established dimension as much like spatial dimensions as possible (you can't make it exactly the same - the metric distinguishes timelike and spacelike), then the most natural picture is the block universe. This is the image I get from your description.
Then, you cannot talk about the point (t,x)=(0,0) moving at all, it just is. Meanwhile, if you speak of the world line of a particle, e.g. the line (t,x)=(t,0), the line just is. When you see a curve drawn on a piece of paper, you don't insist there is an 'existence point' moving along it.
The line (t,x)=(t,vt) also just is, in this picture. Within this particular frame, we can say the first line represents a stationary particle because dx/dt=0; and the second represents particle moving at v in this frame.
It seems your difficulty is trying to mix an 'evolving time' picture with a block universe picture. Both are workable, but the mixture is leading to confusion.
Chestermiller
Mentor
It is not the universe that is moving toward us. 4D spacetime is stationary, and it is we who are doing the moving. Within or rest frame of reference, we are unaware of our motion because, as 3D beings with our inherent physical limitations, we cannot see into our own time direction. But we are moving into our own time direction at the speed of light.
Each inertial frame of reference has its own time direction, and the time direction for one frame is not the same as the time direction for any other frame. All objects and observers that are at rest within a given 3D frame of reference, as well as the frame of reference itself, are moving in the time direction for that frame at the speed of light. Because the time directions of objects in different frames of reference are different, so also are their velocity vectors through 4D spacetime. The differences between these 4D velocity vectors represents their relative velocities, and because the magnitude of the 4D velocities of all objects is the speed of light, even very tiny differences in their time directions corresponds to significant relative velocities. Of course, we can only observe the spatial components of these relative velocities, since, as 3D beings, we cannot see the component in the time direction.
Since 4D spacetime is stationary and it is we who are doing the moving through spacetime at the speed of light, even when we are at rest within our own inertial frame of reference, we can still accurately determine how far we are traveling through spacetime. This is because our clocks can be regarded not only as timepieces but also as odometers. For every tick of our clock, we travel 1 light-second along our own world line.
PAllen
Science Advisor
2019 Award
It is not the universe that is moving toward us. 4D spacetime is stationary, and it is we who are doing the moving. Within or rest frame of reference, we are unaware of our motion because, as 3D beings with our inherent physical limitations, we cannot see into our own time direction. But we are moving into our own time direction at the speed of light.
I know there are a number of writers who say this, but quite a number of us here at PF (and before we ever discovered PF) find this idea of moving at the speed of light through time to be bogus. For example, you will not find it in any of Einstein's writing or any historic books on SR. It comes, basically, from giving meaning to d(c $\tau$)/dt as speed through time along a world line. More traditionally, this (without the c) is simply called the coordinate time dilation factor. Further, since t is an arbitrary coordinate quantity, it is quite arbitrary. Unlike 4 velocity (D/D $\tau$), it does not generalize meaningfully to GR, where it is often useful to have unusual coordinates (4 space like, none time like; or 2 light like, two space like). Even in SR, both Bondi and Dirac were fond of coordinates where two were light like, two spacelike, none timelike. 4-velocity works equally well in all of these, but this 'speed through time' becomes undefinable.
I think it also fosters over-interpreting time dilation (versus observables: differential aging comparing two paths between evetns; Doppler rate comparing exchanged signals). Dirac coordinates show the arbitrary character of time dilation per se - it doesn't exist at all in these coordinates (there is no t coordinate), while all observables are still well defined and easily computed.
Chestermiller
Mentor
I know there are a number of writers who say this, but quite a number of us here at PF (and before we ever discovered PF) find this idea of moving at the speed of light through time to be bogus. For example, you will not find it in any of Einstein's writing or any historic books on SR. It comes, basically, from giving meaning to d(c $\tau$)/dt as speed through time along a world line. More traditionally, this (without the c) is simply called the coordinate time dilation factor. Further, since t is an arbitrary coordinate quantity, it is quite arbitrary. Unlike 4 velocity (D/D $\tau$), it does not generalize meaningfully to GR, where it is often useful to have unusual coordinates (4 space like, none time like; or 2 light like, two space like). Even in SR, both Bondi and Dirac were fond of coordinates where two were light like, two spacelike, none timelike. 4-velocity works equally well in all of these, but this 'speed through time' becomes undefinable.
I think it also fosters over-interpreting time dilation (versus observables: differential aging comparing two paths between evetns; Doppler rate comparing exchanged signals). Dirac coordinates show the arbitrary character of time dilation per se - it doesn't exist at all in these coordinates (there is no t coordinate), while all observables are still well defined and easily computed.
It was not my intention to give the impression that apparently came through to you. I was referring to an object or observer in its rest frame moving through spacetime at the speed of light. In the rest frame, the clock time is equal to the proper time, such that you would have d(cτ)/dτ = c, without a time dilation factor.
I was also thinking about 4 velocity when I was talking about relative velocities of objects (i.e., in a 4D sense), although I neglected to mention that in order to keep things simple for the OP. When considering relative velocities in 4D, the relative velocity is just the difference of the 4 velocities. In this case, the spatial components contain the relativity factor, and the time component of relative velocity is (γ - 1)c.
The viewpoint of 4 velocity being equal in magnitude to c carries directly over into GR.
PAllen
Science Advisor
2019 Award
It was not my intention to give the impression that apparently came through to you. I was referring to an object or observer in its rest frame moving through spacetime at the speed of light. In the rest frame, the clock time is equal to the proper time, such that you would have d(cτ)/dτ = c, without a time dilation factor.
I was also thinking about 4 velocity when I was talking about relative velocities of objects (i.e., in a 4D sense), although I neglected to mention that in order to keep things simple for the OP. When considering relative velocities in 4D, the relative velocity is just the difference of the 4 velocities. In this case, the spatial components contain the relativity factor, and the time component of relative velocity is (γ - 1)c.
The viewpoint of 4 velocity being equal in magnitude to c carries directly over into GR.
I actually use the metric convention where 4-velocity norm is 1 (even when not setting c=1). Has this changed physics any?
I never picture particle moving through spacetime. In the spacetime picture, a world line exists, there is no motion.
I think of the 'angle' between 4 vectors as their relative velocity, as either would measure the other in their basis.
I am not saying there is anything inherently wrong with speed through space time. Just pointing out, FYI, that it is not part of the history of SR in either Einstein's initial point of view or in the spacetime picture developed from Minkowski to Synge (who is actually the person responsible for wide acceptance of space time diagrams), and isn't necessary to understand any part of SR or GR. It is a fairly recent conceptual innovation I don't see as as adding any value. A number of us here have spent time dealing with confusion it seems to instill in some people.
So definitely go ahead and use this idea if you find it helpful. Just be aware if its (lack of) history and that a number of people here (besides myself) find it adds confusion with no value.
Chestermiller
Mentor
I actually use the metric convention where 4-velocity norm is 1 (even when not setting c=1). Has this changed physics any?
I never picture particle moving through spacetime. In the spacetime picture, a world line exists, there is no motion.
I think of the 'angle' between 4 vectors as their relative velocity, as either would measure the other in their basis.
I am not saying there is anything inherently wrong with speed through space time. Just pointing out, FYI, that it is not part of the history of SR in either Einstein's initial point of view or in the spacetime picture developed from Minkowski to Synge (who is actually the person responsible for wide acceptance of space time diagrams), and isn't necessary to understand any part of SR or GR. It is a fairly recent conceptual innovation I don't see as as adding any value. A number of us here have spent time dealing with confusion it seems to instill in some people.
So definitely go ahead and use this idea if you find it helpful. Just be aware if its (lack of) history and that a number of people here (besides myself) find it adds confusion with no value.
Thanks for the history lesson, but I'm not impressed.
Since you brought up the subject of confusion, let's talk about confusion. As a person with an engineering background coming at SR for the first time, I was quite confused by all the standard texts I read. Moreover, day after day, I see postings by new initiates to SR suffering the same confusion that I did trying to learn SR using the conceptualization that you so strongly advocate. They all ask the same questions over and over again. If this isn't confusion, I don't know what is.
In the end, I gave up on these texts, and decided to work things out on my own, starting from the Lorentz Transformation. This led to the much simpler conceptual picture that developed, and clarified for me what is really going on in terms of the fundamental geometry and kinematics of 4D spacetime. I'm sure I'm not the first one to do this. But, for me, it has simplified things tremendously, and removed all the confusion. Moreover, it has never failed me in enabling me to solve problems in SR with virtually no trouble. As I mentioned in a previous posting, I am prepared to make my notes available to whomever is interested. In addition, PAllen, I welcome any critique you might have on these notes. Just because SR has historically been conceptualized in a certain way doesn't necessarily mean that it is the best way. I might also mention that other people who have read my notes have provided very favorable feedback on its ability to reduce their confusion.
One final point: As an engineer, I am comfortable with expressing the 4 velocity in either dimensional form (with a norm of c), or, as we engineers would call it, dimensionless form, with a norm of 1. Either way is fine with me. However, working with the parameters in dimensional form makes it easier to make the connection with conventional Newtonian mechanics, in which velocity and acceleration are usually expressed dimensionally. | 4,226 | 18,958 | {"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} | 3.703125 | 4 | CC-MAIN-2020-45 | latest | en | 0.970666 |
https://kheavan.wordpress.com/2011/08/29/problem-303-van-khea/ | 1,521,872,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649931.17/warc/CC-MAIN-20180324054204-20180324074204-00300.warc.gz | 652,154,297 | 20,788 | # Problem 303 Van Khea
If $a, b, x, y, z$ are positive real numbers such that $a+b=1$. Prove that
$\displaystyle x^3+y^3+z^3+6abxyz$$\geq (2a-b^2)(x^2y+y^2z+z^2x)+(2b-a^2)(xy^2+yz^2+zx^2)$ | 95 | 189 | {"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": 4, "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-2018-13 | latest | en | 0.37768 |
http://mathhelpforum.com/differential-geometry/157160-someone-told-me-statement-true.html | 1,526,959,100,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864622.33/warc/CC-MAIN-20180522014949-20180522034949-00386.warc.gz | 191,755,927 | 9,193 | # Thread: Someone told me this statement was true...
1. ## Someone told me this statement was true...
That if D is a closed set then D = Cl(Int(D)) where Cl and Int denote closure and interior. I don't think it's true though. Counter example:
Let $\displaystyle D = {0} \cup [1,2]$.
D is closed, since it is the union of two closed sets, and Int(D) = (1,2). Therefore Cl(Int(D)) = Cl(1,2) = [1,2], which is a proper subset of D.
Is this correct?
2. Yes. Furthermore, consider any set of finitely many points in R. It will be a closed set and have empty interior, so the formula will fail . | 164 | 595 | {"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} | 3.09375 | 3 | CC-MAIN-2018-22 | latest | en | 0.967984 |
https://community.fabric.microsoft.com/t5/Desktop/ABS-by-product-Platform/m-p/1082788/highlight/true | 1,716,730,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00291.warc.gz | 159,076,819 | 117,477 | cancel
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Helper II
## ABS by product Platform
Hi,
I have the following Dataset:
Platform Product Act (Sum measure) Forecast (Sum Measure) ABS (ABS = ABS('Actual Consol'[Act]-[Forecast])) A A1 80 20 60 A A2 40 56 16 B B1 120 150 30 C C1 5 1 4 C C2 1 0 1 C C3 3 4 1 C C4 4 7 3 C C5 95 81 14
I would like to calculate the ABS of the product platform but according to each product.
I tried this
ACC ABS = CALCULATE(ABS([Contract Act]-[Contract Forecast]),ALLEXCEPT(Matrix_Lookup,Matrix_Lookup[Product])) but that gave me the total ABS for each line,
See column ABS by Product, that is the resulet I want
Platform Act Forecast ABS ACC ABS ABS by Product - want A 120 76 44 29 76 B 120 150 30 29 30 C 108 93 15 29 23 Total 348 319 29 29 129
Thanks!
6 REPLIES 6
Community Support
Hi @Chedva ,
Could you please consider sharing the relationship and fields between these tables or a sample .pbix file for further discussion? I have created the table relationship like this and create a measure to get your expected output but not certain your dataset relationship:
``````ABS by product =
CALCULATE(
SUMX(
'act table',
[ABS]
),
ALLEXCEPT(
'act table',
'act table'[Platform]
)
)``````
Best Regards,
Yingjie Li
If this post helps then please consider Accept it as the solution to help the other members find it more quickly.
Super User
Just use SUM('Table'[ABS])
If you put those items in a Table visualization.
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External Tools: MSHGQM
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The Definitive Guide to Power Query (M)
DAX is easy, CALCULATE makes DAX hard...
Helper II
@Greg_Deckler @az38
ABS is a measure, not in a specific table. the table I showed is a table in the report section.
Act - a measure - that sums the measure from the actual table
Forecast - a measure that sums the forecast from forecast table.
Matrix_Lookup is a table with relationship to both actual table and forecast table on the product. and it has the platform there as well for slicing the report.
Hope I was clear - thanks for you help!!!
Super User
Well then this looks like a measure aggregation problem. See my blog article about that here: https://community.powerbi.com/t5/Community-Blog/Design-Pattern-Groups-and-Super-Groups/ba-p/138149
Become an expert!: Enterprise DNA
External Tools: MSHGQM
Latest book!:
The Definitive Guide to Power Query (M)
DAX is easy, CALCULATE makes DAX hard...
Community Champion
this doesn't work?
``````Measure =
CALCULATE(
SUMX('Matrix_Lookup', ABS([Act]-[Forecast])),
ALLEXCEPT(Matrix_Lookup, Matrix_Lookup[Platform])
)``````
do not hesitate to give a kudo to useful posts and mark solutions as solution
Community Champion
Hi @Chedva
try this measure
``````Measure =
CALCULATE(
SUMX('Matrix_Lookup', ABS('Matrix_Lookup'[Act (Sum measure)]-'Matrix_Lookup'[Forecast (Sum Measure)])),
ALLEXCEPT(Matrix_Lookup, Matrix_Lookup[Platform])
)``````
do not hesitate to give a kudo to useful posts and mark solutions as solution
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Top Kudoed Authors | 923 | 3,484 | {"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} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.857108 |
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Assignment Help:
Derivation of Formulas
i) Future Value of an Annuity
Future value of an annuity is
FVAn = A(1 + k)n -1 + A (1 + k)n - 2 + .......A (1 + k) + A ...............(a1)
Multiplying both sides of the equation a1 by (1 + k) gives.
(FVAn) (1 + k) = A (1 + k)n +A(1 +k)n -1 +... A (1 +k)2 +A (1 +k) .......(a2)
Subtracting eq. (a1) from eq. (a2) yields
FVAnk = A[((1 + k)n - 1)/k] ......................................(a3)
Dividing both sides of eq. (a3) by k yields
FVAn = A[((1 + k)n - 1)/k]
ii) Present Value of an Annuity
The present value of an annuity as:
PVAnk = A (1 + k)-1 + A(1 + k)-2 + .... + A(1 + k)- n ............(a 4)
Multiplying both sides of Eq (a 4) by (1+ k) provides:
PVAn (1 + k) = A + A (1 + k)-1 + ...... + A (1 + k)-n +1 .....................(a5)
Subtracting eq (a4) by (a5) yields:
PVAnk = A[1 - (1 + k)-n]
= A [((1 + k)]n - 1)/(k (1 + k)n) .....................(a6)
Dividing both the sides of Eq (a6) with k outcomes in as:
PVAn = A [((1 + k)]n - 1)/(k (1 + k)n)
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inventory ratio of 4 compared to 7.1 | 929 | 3,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-49 | latest | en | 0.789865 |
https://achievetampabay.org/how-much-time-does-a-builder-potion-take-off-new-update/ | 1,709,578,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476464.74/warc/CC-MAIN-20240304165127-20240304195127-00765.warc.gz | 89,772,966 | 39,294 | Home » How Much Time Does A Builder Potion Take Off? New Update
How Much Time Does A Builder Potion Take Off? New Update
Let’s discuss the question: how much time does a builder potion take off. We summarize all relevant answers in section Q&A of website Achievetampabay.org in category: Blog Finance. See more related questions in the comments below.
How many hours does builder potion take off?
It speeds up the builder by 10X for 1 hour.
How much does a builder potion take off?
The Builder Potion accelerates all construction work to 10 times the original rate in the Home Village for one hour. Compared to the normal rate, using a Builder Potion saves nine hours of builder time per builder, as each builder will do 10 hours’ worth of work in one hour.
UNDERSTANDING HOW TO USE BUILDER POTIONS! – Fix That Rush Ep.11 – \”Clash of Clans\”
UNDERSTANDING HOW TO USE BUILDER POTIONS! – Fix That Rush Ep.11 – \”Clash of Clans\”
UNDERSTANDING HOW TO USE BUILDER POTIONS! – Fix That Rush Ep.11 – \”Clash of Clans\”
How fast is the builder potion?
The Builder Potion boosts the player’s Builders in the Home Village for 1 hour. All Builders will work ten times as fast as normal, each one gaining a total of 9 hours. This Potion applies only to the builders on your Home Village, including the Master Builder if he is gearing up a defense over there.
Can you use 2 builder potions at once?
No, magic item effects do not stack. The second one you use will just override the first one. If you use multiple resource potions, only their durations will stack, not the amount of resources generated. If you use two potions it will last twice as long not twice as fast.
How long is 24 hours at 10x speed?
10h 10x faster is one hour, so get any upgrade timer, take away a 10h turn them to 1 add them back. 1d timer for example, 24-10=14 that 10 goes faster by 10 times so its a 1h put the one bacc with the regular timer 15h total. Am not replying more screw this. The correct answer is 9 HOURS!
Are builder potions worth it COC?
If you are earning builder potion by completing any limited-time event then it is surely worth it. If you are buying builder potions for clan war league medals then also these potions are worth it. I have found builder potions from season rewards the best trade.
How fast is research potion?
Conversation. A brand new Magic Item is on the way! The Research Potion will speed up your Laboratory Research times by 10x for an hour! ⌛ Plus, some changes in the League Shop, and Magic Hammers cooldown: clashofclans.com/blog/game-upda…
WHEN TO USE A BUILDER POTION | Fix that Engineer | Clash of Clans
WHEN TO USE A BUILDER POTION | Fix that Engineer | Clash of Clans
WHEN TO USE A BUILDER POTION | Fix that Engineer | Clash of Clans
How fast is the research potion COC?
A Research potion will give a boost of 24 hours in 1 hour. In simple terms, it will reduce A DAY WORTH of upgrade in just one hour.
Does builder boost last forever?
Builder Boost:
A Gold account can complete 24 hours of building in only 19.2 hours, leaving 4.8 hours of free builder time!
Do boosts stack in COC?
They stack sequentially (one starts after the other is finished). You don’t get double or triple boost for one hour.
Does boosting builders stack?
No, they don’t stack.
How much does the resource potion boost?
How much boost does a resource potion give? A resource potion gives you a 2x boost for 24 hours.
How do you get the 6th builder in clash of clans?
Once you have upgraded the O.T.T.O hut to level 5, you are able to place the Master Builder’s Hut within the Home Village. When you switch back to the Home Village, the “Unplaced Building” button will display and you are able to place the Master Builder’s Hut within your Home Village.
BUILDER POTION – Is it worth it?
BUILDER POTION – Is it worth it?
BUILDER POTION – Is it worth it?
Does Hammer of building need resources?
Hammers, however, have the power to start and finish an upgrade, without using any resource and skipping the time. This makes Hammers by far the most powerful Magic Items on the game.
Is builder potion better than Hammer of building?
-If you can keep at least 3 heroes upgrading simultaneously, buy builder potions to get some hero progress. –If you have a building that takes more than 14 days (Clan Castle comes in mind), then a hammer would be better.
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https://answers.search.yahoo.com/search?p=SPE&ei=UTF-8&flt=cat%3AVideo%2B%2526%2BOnline%2BGames%3Bmarket%3Aca%3Branking%3Adate%3Bage%3A1y%3B&xargs=0&fr2=sortBy&b=1&context=%7C%7Cgsmcontext%3A%3Aage%3A%3A1y%7Cgsmcontext%3A%3Amarket%3A%3Aca | 1,603,690,356,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00132.warc.gz | 202,690,189 | 34,750 | # Yahoo Web Search
related to: SPE
1. ### A model car of mass 5kg slides down a frictionless ramp into a spring with constant k=4.9kN/m. ?
a) GPE became SPE mgh = ½kx² 5kg * 9.8m/s² * h = ½ * 4900N...
1 Answers · Science & Mathematics · 11/10/2020
2. ### A block with mass 2kg is held against a spring with spring constant 250N/m. The block compresses the spring 22cm from equilibrium.?
a) SPE = ½kx² = ½ * 250N/m * (0.22m)² = 6.05...
1 Answers · Science & Mathematics · 12/10/2020
3. ### Physics question?
• maximum stretch GPE is a minimum SPE is a maximum KE is zero • at equilibrium GPE has...
1 Answers · Science & Mathematics · 20/08/2020
4. ### A 700 gram block is released from a height h above an ideal spring of spring constant 400 N/m. The block sticks to the spring?
...compression, the block has fallen through h + 0.19m PE becomes SPE : 0.700kg * 9.8m/s² * (h + 0.19m) = ½ * 400N/m * (0.19m...
1 Answers · Science & Mathematics · 10/06/2020
5. ### A model car of mass 5kg slides down a frictionless ramp into spring with spring constant 4.9kN/m. Spring experiences max compression of 22cm?
... left the ramp when it contacts the spring): a) GPE become SPE : mgh = ½kx² 5kg * 9.8m/s² * h = ½ * 4900N...
1 Answers · Science & Mathematics · 20/07/2020 | 425 | 1,278 | {"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} | 3.03125 | 3 | CC-MAIN-2020-45 | latest | en | 0.732914 |
https://dalneitzel.com/2019/05/08/ffgm-treasure-hunt-9/ | 1,603,246,134,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00559.warc.gz | 292,420,638 | 29,315 | # FFGM Treasure Hunt #9
###### BY JENNY
FFGM #9 has been solved and the medal claimed on May 25th, 2019
dal is an occasional filmmaker, writer and photographer who lives on an island in Washington State's Salish Sea.
## 96 thoughts on “FFGM Treasure Hunt #9”
1. The roman numerals VI, VII, and VIII are flawed (the I’s are on the wrong side of the V’s) – don’t know if that’s a lead or an accident. The hour and minute hands are also not in sync, they’d never share that placement.
S’all I got.
2. The numbers in the upper right say, “the hands of time move forward never back” (1=a , 2=b, etc.)
3. The numbers on top are a simple letter/number substitution and spell out:
The Hands of Time Move Forward Never Back
4. There are two phrases that hint at Forrest Fenn’s poem:
“In a place of riches new and old” (“and hint of riches new and old”)
“You can discover the secret where” (“I can keep my secret where”)
Just a nod to FF?
5. This remains unclaimed…… might it break a record for staying unsolved the longest?
6. So, if the numbers convert to “The hands of time move forward never back”
Then, the 2 little white hands pointing to the clock may lead somewhere.
In clockwise order, the first little white hand is pointing to the 16th space and the second little white hand is pointing to the 53rd space.
Put them together and get, 1653. Could this mean the year 1653?
The little red birds beak is pointing to the 25th space, and another little red bird is in lower right corner where the STONE is secreted in some leaves.
Can the area where the stone is hidden have to do with a 25th something?
There is a leaf pointing to the 37th space on clock.
The stone is hidden is some leaves.
The clock hands read 3:03 roughly.
The red X’s are in spaces 1,12,22,27,36,40,47 and 51.
The mouse’s nose is pointing to the 7th space and it’s tail is pointing to the 9th space.
• Another way the clock hands can be interpreted is instead of 3:03 it could be 3 and 18 or 318.
The big hand is pointing to the 3rd space on clock and the little hand is pointing to the 18th space on clock.
7. Didn’t get anywhere with this line of thinking, but I was going with the intervals of 5 spaces on the clock being a Bacon cipher. The blank spaces are A and the xs (and possibly the other spaces with things pointing to them) are B. And since the hands of time always go forward, I, II, III, IIII, IX, X, XI, XII are to be read clockwise, but the others are counter clockwise since the numbers are backwards. To go forwards, you’d have to go in the direction of VI for instance, not IV.
Sounds good! But didn’t get anything meaningful.
8. I also tried to force the solution. I’m pretty sure it’s New England judging by the number of black oak leaves on the ground in the video. So tried to use Google Maps to check out every AT parking lot to see if it matched up. Nada.
• I’d focus on NH since it’s the ninth state of the union and this is FFGM #9. Probably a reach, but who knows.
9. In MA there is a Blackstone (the ‘stone’ in the corner is black) and a Hop Brook this and that. But they are nowhere near the AT.
• There is also a cardinal bird over the stone or (redbird). REDSTONE Granite Quarry in New Hampshire.
• This beats it…:) I think it was 3 days …. quickest was the last one at around just 5 hours….
• Wow, I forgot about that rhyme, good job.
The hickory could have something to do with barbecue, smoke, wood.
The dock could have something to do with a marina or boats.
All very interesting.
10. I’m new to this hunt forum but find it very interesting and exciting. For clarification does the hunt number correlate with the valley map ? ie. Hunt #9 and #9 on the valley map the James River valley.
• Hi Mark! And Welcome….No, the map numbers do NOT connect…. The map is only used to show where the Medallions could be hidden —in the shaded areas—..please ignore all else… best of luck!
• Thanks Jenny for the info, by chance did you see my email or comment on the MR. Beil’s website defaulting to an SQL error ? Any help on being able to register with his site would be greatly appreciated.
• I know he has been updating his site, and many people are having issues not only registering, but everything else. I will see when he feels it all might be done though….. and get back to you…
11. Every time I think I’ve figured it out, I can’t find any lot that matches the video. Nothing paved, that size, with a road near it that has no lines painted, etc.
• Hi Lisa, No…. keep looking…. It’s still out there….. Next one, and last in this first series of FFGM Treasure Hunts will be posted May 22– whether this one is found or not…but it should be… there aren’t any difficult ciphers needed to solve for the location….
• Jenny – Thanks! But that “secret where” part reminds me of Forrest’s version of this quote:
“Three may keep a secret, if two of them are dead.”
– Benjamin Franklin, Poor Richard’s Almanac
And then there is that CLOCK, like in that treasure movie with Nicholas Cage.
12. Two things:
1) 8 red x’s -> X8 -> Exit ??
2) We can get the hands to be in synch if we rotate the clock face a couple clicks clockwise. It would be as if the clock were hung on the wall crookedly but the time would be 3 o/clock. (code in upper right says hands cannot move backward, but it says nothing of moving the clock face) Also the right angle of the hands as they are, relative to each other, occurs only at 3 o’clock at the top of any hour. (that angle occurs more often, but none of these other occurrences is at the top of an hour except 3.) And I am thinking the mouse/hickory dickory dock tells us the clock is striking the top of an hour.
13. No hint yet please. I’m still making progress. Could be coincidence but i think i found the state.
14. I’m thinking White Mountain NF in New Hampshire.
Possibly Franconia Notch SP in the area of Old Man of the Mountain.
The White hands pointing to clock=White Mountain NF.
What I thought was a mouse may be a Hampster=Hampshire.
The red bird is possibly a Cardinal?
I can’t let go of idea about 2 White hands pointing to clock at space 16 and space 53. Put together is 1653. Year 1653?
Could the Red Ribbon above “In a place of riches new and old” have any meaning?
New and Old = New England or New Hampshire.
Why is there references to both a White Stone and a Black Stone?
I’m clueless on Hop, Skip, Jump.
“Clues this time of a different kind”
Jenny says upthread, “There aren’t any difficult ciphers”
So, why is this so hard?
15. Here’s a solution for your amusement. It fits perfectly.
“Hop, skip, or a jump.”
Jump Hill Preserve in CT.
“Hands of time move forward, never back.”
If you travel where time is moving forward, it means you’re going east. Jump Hill is in Easton, CT.
The word Stone in black type.
Jump HIll is on Black Rock Blvd.
“Clues this time are a different find, but lead to FFGM #9”
The Jump Hill trail system was recently connected to another series of trails, giving about 18 miles of trails in total. This clue describes Jump Hill, but that’s not where it is. You have to follow the trails to another parking lot.
The border around the clue is red.
Some of the trails are in Redding.
The clock has numbers that are opposite what they should be.
Some of the trails are in Weston, instead of Easton.
So, I say this solution is for your amusement because there’s just one problem. Easton CT is nowhere near the Appalachian Valley! I spent way too much time on this solution before actually checking it on a map.
• SD
I’d love it if this was right, I only live a hop skip jump away from there!
16. Personally I don’t like hints unless I’ve given up. I used to have a German teacher in high school that all the kids found really frustrating. She would ask you how to say a phrase, and as you were formulating it, she would blurt out the answer.
I think Jennie should wait until the next clue is released. I’d rather the race be between people, not against a clock (no pun intended). That’s just me, of course.
17. Oh I’m so sorry, I didn’t see these opinions on hints until just now— when I came here to post a hint….lol……and I have already announced on my FB page I would be giving a hint… so I feel I must— (because many over there are expecting and waiting patiently for one)…. However, please know it will only help focus on what has been seen already…….
The Hint for FFGM #9 Treasure Hunt:
You’ve got X marks the spot
And also one full minute
18. Lots of excellent solutions are popping up here. Here’s mine that turned out to be a dud.
The red-x code is 1,12,22,27,36,40,47, and 51. Taking the 1st, 12th…words in the poem above gives the phrase, “IN RESTS PAVED SEE JUMP DISCOVER CLUES”.
I did a recon today to Jenny Jump State Forest in New Jersey. The bird is pointing at the number 25, so I searched primarily around what would have been campsite 25, but it doesn’t exist. The campsite map goes 22, 23….then 26,27 with a small parking area on google satellite where 25 would have been, but it can’t be seen easily. On the ground it was just a dirt spot with no pavement, here’s a photo in case anyone’s solution pointed to Jenny Jump:
https://ibb.co/sWjzq8V
There are also roughly 37 campsites at Jenny Jump, and the “weeds” point towards the number 37. This was my perfect solution, but it’s back to the drawing board.
• The red X is X Marks the Spot. I believe the 8Xs represent the 8 that have already been found.
19. I give up!
I’m goin back to Forrest’s puzzle!
Jenny should be working for NASA or the CIA or something!
Peace!
20. I didn’t know there was a FB page. There are over 130 comments. Some theories are way out there. But some seem plausible.
So…here are a couple of things I picked up.
If you start at the top of the clock and count spaces, ABC etc., then start back at A when you get to the end, the white hands point to PA.
If you do the same thing, but start at the position of the minute hand, you get NY.
This one I figured out: If you take the clock hands and move them forward as a unit, there are two places where they are both pointing at red Xs: 33 spaces forward and 9 spaces forward. Then if you count 9 and 33 from the end of the Hands of Time sentence, you get NH.
So it could be any of three states. Personally, I’m leaning towards PA.
• SD
That’s all super interesting.
I’m thinkin it’s somewhere North of Santa Fe because, NO I’m only kidding!
I do think it’s somewhere in northern New England most likely New Hampshire cause the previous 8 hunts had a pattern leading Northward and the 8th stone was found in Vermont.
21. Oh, also from the FB page: the clock with the weird numbers is a sundial. There’s a picture there that shows a standard sundial and it has IIII for the number four, and reversed letters for the VI VII and VIII.
22. OK, I had a momentary lapse of sanity, I’m feeling better now.
So, Jenny gave us a hint upthread:
You’ve got X marks the spot
And also one full minute
Here’s what I’m thinking, the first task has something to do with all the red X’s. Starting at the first space with the red X or the “One Full Minute.”
That’s all I got, no clue what happens next. 🙁
• SD
Check out Lafayette campground at Franconia Notch, the Appalachian trail comes out right there. The parking areas across the highway look promising but can’t get a good view from google. There are so many possibilities around that area.
• I looked at it. So…see if everyone agrees, but when I look at the video of Jenny planting it, I see a few things.
1. The road next to the parking lot looks like there are no painted lines.
2. It looks like the land on the far side of the road is sloping up. It also looks like the land at the right side of the lot is sloping up as well, like the road is running in a little valley.
3. There’s a strip of grass bordering the road.
4. It looks like there’s an exit off the road that you can just see on the other side with grass in the middle.
5. Jenny is planting in the corner nearest the road. BUT the poem says it is planted in a back corner. Which would mean the exit to the parking lot is not off the road we see. There must be another road you turn on off the main road, and then into the parking lot.
Also there aren’t many trees in the corner where Jenny is planting.
So these are all things to consider when using Google Maps.
• SD
I agree. I think parking lot runs parallel with road.
Also, it may be at a rest area. Line in poem reads, “ The white stone RESTS to claim the gold.”
23. I’m thining it’s lake George area southeast corner of Adirondack park. Route 9 leads right to it and it’s a hop skip and jump away from last find. They have old battle recreations. The ribbon in the puzzle looks just like the rotary, southern part of lake George. Can’t figure out what the clock/sundial have to do with the area. Maybe someone can help.
24. The red X’s fall on the numbers 1-12-22-27-36-40-47-51. But if we don’t count the #1 red x as it is behind the second hand, and the hands of time move forward never back, the first red x is actually #61. Counting the letters in the poem you get IDHOELTR – DIRT HOLE.
Googling dirt hole does not come up with a location in NY, PA, NJ etc. and most Google results are for “dirt hole set'” a form of trapping. The mouse may have something to do with trap or there may be a quarry near the location.
Could also just be a coincidence but thought I’d share in case it jogs something loose for someone else.
• Jenny’s new clue mentions Rabbit Hole (Dirthole). I did some research and some call quarry holes rabbit holes. The bird in the corner over stone could stand for Redstone. Redstone Granite Quarry in New Hampshire. It’s off of route 302 near the Appalachian Trail.
25. I posted this on the FB page because the clock style/type was being talked about……. and thought I should share it here too—
I don’t want anyone going down to deep of rabbit holes….so let me say this—. the clock type/style is not a clue….. Great thoughts though….and I will keep all these in mind for other hunts…
• Jenny
You say the clock type/style is not a clue, but does that refer to the Roman numerals that are misspelled?
For instance:
IIiI should be IV
IV should be VI
IIV should be VII
IIIV should be VIII
26. I think its WHITE ROCK NEW MEXICO . The sign with trees around, near the secret city Los-Alamos WHERE discoveries are made
27. Hey Jenny
By any chance, does your car have a “This car climbed Mt Washington” bumper sticker on it? 🙂
• No…it doesn’t…
• Although we have been there….. a few years ago….
28. Clue #2:
A Treasure’s Trove First see
with 58712643
29. Alrighty then,
A second clue reads, “A Treasure’s Trove First see
with 58712643”
Why is Treasure’s, Trove, and First capitalized?
On first clue upthread, Jenny says, “please know it will only help focus on what has been seen already”
The word First has been used in both clues, first task and First see. (SEEN)
Could the numbers “58712643” have something to do with the Roman numerals? There is a red X near the numbers 5(V), 8(IIIV), 7(IIV),12(XII), not really 6(IV), 4(IIII), not really 3(III)
See and three(3) rhyme.
Matching the numbers 58712643 to the Words In first line of poem:
“riches old and in a new of place”
Matching the numbers 58712643 to the letters in first line of poem:
“LECINAPA” = “IN A PLACE”, That is a remarkable coincidence.
• There are 8 red X’s
58712643 = 8 digits
Matching 58712643 to red X’s by alphabet starting at first minute = JYUALNAV
30. Jericho lake state park NH looks interesting.
I didn’t see anything around Redstone.
31. So, if the previous hunts have something to do with the 8 red X’s and the code 58712643, then maybe they have something to do with the locations of the previous 8 hunts.
This is what I got from looking at the solves.
Hunt 1 – Lost mountain trail, Sky Meadows State Park, Virginia
Hunt 2 – Bear Mountain State Park, New York
Hunt 3 – Rest Area, Tennessee
Hunt 4 – Ascalon Trailhead, Cloudland Canyon State Park, Georgia
Hunt 5 – Arcadia Boat Launch, Virginia
Hunt 6 – Glen Onoko Falls Trailhead, Pennsylvania
Hunt 7 – General Store, Pine Grove Furnace State Park, Pennsylvania
Hunt 8 – Emerald Lake State Park, Vermont
It seems that State Parks are the particular area for hiding the stones and the Appalachian trail is near to some.
32. Cardinal is NC state bird.
Nantahala SP (Murphy) was where the Treasure’s Trove butterfly jewel was found.
Don’t think any of the eight coins have been found in NC yet. V
• Fantastic sleuthing Matt!
Many possibilities around that area.
There’s a Hickory as well, for mouse rhyme.
I looked at the Treasure’s Trove jewels and locations.
The Grasshopper was in James Baird SP New York.
That’s just a Hop Skip Jump away from Bear Mountain SP where FFGM #2 was found.
So many possibilities.
• The Great Smokies and the area in and around Newfound Gap looks interesting, trailheads and parking pullouts galore, and the Appalachian trail is right there.
• I thought about that too. Still wondering with Treasure’s Trove First means. The Dragonfly was the first found and was at Ricketts Glen State Park, Pennsylvania. But the Spider was the first one listed on Jenny’s write up which was in Prickett’s Fort State Park, West Virginia.
And then the Butterfly is specifically mentioned in her blog post.
None of the jewels have 8 letters, but if the Xs on the clock translate into BUTERFLY, that would be close enough. Is there anything about the butterfly that could be considered a First?
33. “In a place of riches new and old”
Could this mean, New and Old Testament, meaning-The Bible Belt?
Bible Belt states- North Carolina, Georgia, South Carolina, Tennessee, Mississippi, Alabama.
34. State Birds.
West Virginia= Northern Cardinal
Virginia= Cardinal
Kentucky= Northern Cardinal
North Carolina= Cardinal
I got excited about North Carolina until I looked up state birds for VA, KY, WV.
I think the Cardinal has something to do with one of these 4 states.
35. Jenny,
I wonder if you could double-check the clue? I got something that is just one letter off from the name of a park. I did notice a typo on two of your other clues, so I just wondered if you could check it. If there is a typo, it might explain why nobody has been able to solve it.
Thanks.
• There isn’t a typo in the solution… it’s fine…
36. FFGM #10 was posted earlier today on MW, and successfully claimed! Congrats to Amy S (and family!)….. (more coming soon)
So this means FFGM #9 is the only one left to solve/claim from the first series! Best of luck to all!
37. 58712643 Puts the lines in order to spell “Treasure.”
I.e., the put the first line of the poem 5th, put the second line 8th, put the third line 7th, put the fourth line, 1st; then, the bolded letters will be in the order to spell “Treasure.” I then shifted the lines horizontally, so that “Treasure” was spelled down in a line. It did not help. I found nothing.
Also, the x’s are on ALVAJNUY on the clock, and with the poem in its original order, you can actually spell that too. I shifted the original lines to get ALVAJNUY in a line, and, got…nothing.
I counted shifts of the first letter in each line relative to the red x at the beginning of original line 8. Still got nothing.
The other minutes marked (with the hands, clock hands, mouse and bird and plant) land on CGIPRKYA, which very nearly, but not quite anagrams to CITY PARK.
38. So here are some random things I’ve noticed on this puzzle. I must admit that I’m only casually following along and some of this may have been quite possibly picked up on and I missed it but….
First, I think that the cardinal has to do with cardinal point, as in point of prominence, rather than the bird itself. If true most likely the starting point of where our clock hands need to move forward to but there’s also the chance that they land there as an ending point as well.
Next observation is that there are 8 red xs and 8 highlighted letters, one per Iine of the poem. Jenny gives us the second clue and it’s the numbers 1-8 scrambled. The highlighted letters are S E R T R U A E or the word TREASURE itself. When you put the letters in the order Jenny gives it becomes R E A S E U T E.
I’m thinking that each one of these letters pairs up with an x on the clock. And when we figure out the correct method, it may turn into a helpful phrase, maybe the whole word that the highlighted letter is from somehow gets placed on the clock, although I tend to think it’s just letters.
My last thoughts go to the obvious, the hands must refer to movement of the hands of the clock. Perhaps a clue to a resting point or a reference to the spacing of them as they are moved. Which one corresponds with the hour and minute hand is anyone’s guess. I think that the blade of grass and the red eye of the mouse that both occupy spaces on the clock face must be specific points of reference as well.
39. Solved! Congrats to Susie D! She has successfully claimed FFGM #9! (More coming soon) | 5,201 | 21,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2020-45 | latest | en | 0.916443 |
https://www.physicsforums.com/threads/circular-motion-finding-revolutions.442117/ | 1,553,580,115,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204857.82/warc/CC-MAIN-20190326054828-20190326080828-00153.warc.gz | 872,101,105 | 17,804 | # Circular motion - finding revolutions?
#### wschmidt22
i've been struggling for hours upon hours attempting to find the answer to simple problems and i'm extremely depressed..
how do i find the amount of revolutions if i'm given the constant angular acceleration (rad/s^2) and the time (seconds)?
#### Doc Al
Mentor
Could you solve the analogous linear problem? If you were given the time and a constant linear acceleration, could you solve for the distance traveled? It's the same problem.
Angular quantities obey kinematic equations analogous to those obeyed by linear quantities. See: http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin"
You also need to be able to convert from radians to revolutions. How are they related?
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#### wschmidt22
i've spent hours crunching numbers into equations.. find total radians then divide by 2*pi? i'm burnt out.. how about you just tell me what i do
#### wschmidt22
or at least put it into english for me, to get the amount of rotations, when you have acceleration and time.. you need to do what?
#### Doc Al
Mentor
How about you post the exact problem you are trying to solve and show what you've done so far.
You didn't answer my question: Do you know how to solve the linear kinematic problem?
[I'll move this to HW: Intro Physics.]
#### wschmidt22
i'm simply asking the proper method for finding revolutions when given accel and time. i've struggled for hours and i'm pretty fed up.. if you wanna do the whole "answer my question with a question" thing, then go for it dude.. i'm just looking for a little help.
and yeah you just plug the values in the equation.. pretty straightforward.
#### Doc Al
Mentor
and yeah you just plug the values in the equation.. pretty straightforward.
What equation are you using?
#### wschmidt22
you're killin me here doc.. i've tried everything.. i know this is easy. tell me this: do i need to find the radius? if so, how do i do that using rad/s? is it vT/(2*pi)? c'mon doc.. work with me here.
#### Doc Al
Mentor
Let me get this straight. You've worked on this for hours, yet you won't take 5 minutes to state the exact problem and show what you've done?
No, you don't need the radius. (At least as far as I can tell, since you still haven't stated the full problem.) So far, it sounds like a kinematics problem. But who knows?
(FYI: The kinematics equations are all contained in that link I gave.)
#### wschmidt22
A wheel starts from rest with a constant angular acceleration of 2.50 rad/s2 and rolls for 7.72 seconds.
How many revolutions does it go through?
answer is 11.86.. how the hell do you get there?
#### Doc Al
Mentor
There are several ways to go:
Look for a kinematic equation that relates distance/displacement (or angle) to acceleration and time. Here's a list: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"
Or you can just figure out the average angular speed and use that to figure out the angle swept out during that time.
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#### wschmidt22
k. just used displacement & time:
x = x_0 + v_0 t + (1/2) a t^2
then divided by 2*pi
now how about when they give the radius and time? which equations do i need there?
ex:
While riding on a Ferris wheel at a constant speed, you measure on your stopwatch that it takes the wheel 41.87 seconds to finish one cycle. The radius of the wheel is 15.20 meters. What is the angular velocity of the wheel's rotation?
#### zgozvrm
Could you solve the analogous linear problem? If you were given the time and a constant linear acceleration, could you solve for the distance traveled? It's the same problem.
Let's say a truck is accelerating at a constant rate of 2 [itex]m/s^2[/tex] and is traveling in a straight line.
How far does the truck travel in 5 seconds?
#### Doc Al
Mentor
k. just used displacement & time:
x = x_0 + v_0 t + (1/2) a t^2
then divided by 2*pi
Good.
now how about when they give the radius and time? which equations do i need there?
ex:
While riding on a Ferris wheel at a constant speed, you measure on your stopwatch that it takes the wheel 41.87 seconds to finish one cycle. The radius of the wheel is 15.20 meters. What is the angular velocity of the wheel's rotation?
Since you're only asked about angular velocity, the radius is irrelevant. And since the velocity is constant, use the rotational analog of distance = velocity*time.
#### Doc Al
Mentor
Let's say a truck is accelerating at a constant rate of 2 [itex]m/s^2[/tex] and is traveling in a straight line.
How far does the truck travel in 5 seconds?
That depends on its initial velocity. But you'd use the same kinematic equation.
#### zgozvrm
That depends on its initial velocity. But you'd use the same kinematic equation.
That's my point ... there is no initial velocity given in the original question! (Just acceleration and time)
#### Doc Al
Mentor
That's my point ... there is no initial velocity given in the original question! (Just acceleration and time)
That's why I kept insisting that the full question be posted. And it was...
A wheel starts from rest...
#### zgozvrm
My bad! I made my post before updating my browser so didn't see that.
I was trying to help the OP see that either the question was incomplete or that the answer was that you couldn't give the displacement based on the data given.
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• Solo and co-op problem solving | 1,367 | 5,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-13 | latest | en | 0.898802 |
https://ch.mathworks.com/matlabcentral/cody/problems/43584-find-pseudo-cyclic-number/solutions/1049329 | 1,582,379,217,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145676.44/warc/CC-MAIN-20200222115524-20200222145524-00538.warc.gz | 328,775,838 | 15,674 | Cody
# Problem 43584. Find Pseudo-Cyclic Number
Solution 1049329
Submitted on 9 Nov 2016 by Gyoung Deuk Kim
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; y_correct = false; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
2 Pass
x = 230769; y_correct = true; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0
3 Pass
x = 307692; y_correct = true; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0
4 Pass
x = 307691; y_correct = false; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
5 Pass
x = 307693; y_correct = false; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
6 Pass
x = 2; y_correct = false; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
7 Pass
x = 285714; y_correct = true; assert(isequal(cyclicNumber(x),y_correct))
8 Pass
x = 55; y_correct = false; assert(isequal(cyclicNumber(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
9 Pass
x = 142857142857; y_correct = true; assert(isequal(cyclicNumber(x),y_correct))
10 Pass
x = 142857; y_correct = true; assert(isequal(cyclicNumber(x),y_correct)) | 558 | 1,691 | {"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} | 3.28125 | 3 | CC-MAIN-2020-10 | latest | en | 0.57932 |
http://www.learnersplanet.com/content/test-7-21 | 1,550,567,076,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489729.11/warc/CC-MAIN-20190219081639-20190219103639-00128.warc.gz | 377,829,173 | 9,618 | # Math Olympiad Grade 3 Addition & Subtraction Test-7
Addition and Subtraction worksheet
1. If this equals to one, then
What is
A. 24 B. 20 C. 16 D. 10
1. Madhavi wanted to buy a pen that costs Rs. 15. She had Rs. 7. How much more does she need to buy the pen?
A. Rs.22 B. Rs. 12 C. Rs.8 D. Rs.3
Click here for math practice tests
1. A number machine takes any number put into it, adds 20, then subtracts 10, and then adds 8. If the number 13 is put into the machine, what number comes out?
A. 13 B. 23 C. 31 D. 33
1. Asha had 90 days of summer vacation. Each day she did one fun exercise. She swam on 33 of the days and played basketball on 24 of the days. She rode her bike on all the other days of her vacation. Which number sentence shows one way to find the number of days that Asha rode her bike?
A. 90 – 33 + 24 B. 90 + 24 + 33
C. 90 + 33 – 24 D. 90 – 24 – 33
1. What numeral means the same as 800 – 6?
A. 860 B. 806 C. 794 D. 608
Answers
(11)–A; (12)–C; (13)–C; (14)–D; (15)–C | 386 | 1,250 | {"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} | 4.25 | 4 | CC-MAIN-2019-09 | latest | en | 0.85923 |
http://www.mywordsolution.com/question/if-typographical-errors-occur-randomly-about-how/940942 | 1,527,136,309,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865913.52/warc/CC-MAIN-20180524033910-20180524053910-00101.warc.gz | 428,111,160 | 7,997 | +1-415-315-9853
info@mywordsolution.com
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A thick book with 1200 pages has quite a few typographical errors. There are only 180 pages without typographical errors in the whole book. If typographical errors occur randomly, about how many pages in the book have three typographical errors? What is the median number of typographical errors per page?
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Create a provider database and related reports and queries to capture contact information for potential PC component pro | 1,021 | 4,749 | {"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-2018-22 | latest | en | 0.841266 |
https://cybergeeksquad.co/2021/07/longest-substring-without-two-contiguous.html | 1,675,735,017,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500368.7/warc/CC-MAIN-20230207004322-20230207034322-00292.warc.gz | 206,261,545 | 57,456 | # Longest Substring Without Two Contiguous Occurrences of Letter
Microsoft OA 2023 Question Longest Substring Without Two Contiguous Occurrences of Letter. We provide solution to Microsoft Online Assessment Questions and do check out our Microsoft OA 2023 Questions List below.
## Longest Substring Without Two Contiguous Occurrences of Letter Solution
Given a string Given a string s containing only a and b, find longest substring of s such that s does not contain more than two contiguous occurrences of a and b
Example 1:
Input: aabbaaaaabb
Output: aabbaa
Example 2:
Input: aabbaabbaabbaaa
Output:Â aabbaabbaabbaa
Also See: Amazon OA Online Assessment 2023 Questions and Answers
### SOLUTION
Program: Longest Substring Without Two Contiguous Occurrences of Letter Solution in C++
Approach:
Scan the string from left to right
1. drop all sequences of the same chars longer than 2;
2. count length of sequences of different chars;
3. save length of the longest sequence.
Algorithm:
We should have a counter of the same chars and index which points to the last two chars of this sequence.
Lets call them “count” and “start”
And we should have two variable where we will save length of the longest sequence of different chars
and index which points to the beginning of this sequence.
Lets call them “max_length” and “start_ml”
So the algorithm in general:
Scan the string from left to right:
1. Take next char of the string;
2. If next char is the same as the previous one increase counter of the same chars “count”;
1.1 If next char is different drop counter of the same chars “count” to 1;
3. If number of previous the same chars is:
2.1 More than 2. Move pointer to the beginning of the current sequence of different chars “start” to
the previous char before the current one, to keep only two the same chars at the beginning of the sequence.
drop counter of the same chars “count” to 2.
2.2. Less or equal 2. Check if current sequence of different chars is longer than current maximum.
If yes – update maximum to the current length. Save pointer to the beginning of the new longest sequence.
4. Go to phase 0;
``````#include <iostream>
#include <vector>
using namespace std;
string solution(const string &s) {
int s_size = s.size();
// start position and length of the longest sequence
// which doesn't contain 3 contiguous occurrences of a and b
int start_ml = 0, max_length = 0;
int start = 0; // start of current processing string of the same letters.
int count = 1; // length of current processing string of the same letters.
for (int i = 1; i < s_size; ++i) {
if (s[i] == s[i - 1]) {
// if we met two the same letters increase the counter of the same letters
count++;
}
else {
// if next letter is different drop the counter to 1
count = 1;
}
if (count <= 2) {
// if the sequence of different letters continuing, set it's current length as
// max_length if it is bigger than current max_length
// "i - start + 1" is length of the current processed sequence
if (i - start + 1 > max_length) {
max_length = i - start + 1;
start_ml = start;
}
}
else {
// if the sequence of the same letters continuing,
// move the pointer to points to the last two chars of this sequence
// drop the count to 2
start = i - 1;
count = 2;
}
}
return s.substr(start_ml, max_length);
}
int main() {
cout << "Result: " << solution("aabbaabbaabbaa") << " Expected: aabbaabbaabbaa" << endl;
cout << "Result: " << solution("aabbaaaaabb") << " Expected: aabbaa" << endl;
return 0;
}`````` | 879 | 3,486 | {"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-2023-06 | latest | en | 0.821017 |
https://math.stackexchange.com/questions/4344071/if-two-triangles-are-similar-are-the-part-of-each-triangle-similar-in-proportio/4344080 | 1,660,778,323,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00211.warc.gz | 363,274,050 | 66,035 | # If two triangles are similar, are the part of each triangle similar in proportion as well?
My question is raised from this problem:
It's known that the isosceles triangle $$AB$$ is similar to $$CD$$. Let's say the ratio of the area of those two triangle is $$x\,:\,y$$.
My question is, if I draw a cyan line so that I have a triangle $$E$$ (the big cyan triangle) so that it will similar to the small triangle $$D$$ (small cyan), will the ratio of $$E$$ and $$D$$ be $$x\,:\,y$$ as well? If yes, could you explain it briefly regarding the proof?
My original problem, I have to find the ratio of the triangle $$A$$ and $$D$$ which looks like they're similar as well. However, that's not what I'm going to ask here. The question has already written above. Mentioning my original problem, probably you could understand why I ask this.
From the problem statement we can say wlog the base of $$AB$$ is $$\sqrt{x}$$ and the base of $$CD$$ is $$\sqrt{y}$$.
Given that $$E$$ and $$D$$ are similar, by the SSS criterion, their ratio is also $$\sqrt{x}:\sqrt{y}$$ as in the figure below, where $$\sqrt{x}$$ is the side of E and $$\sqrt{y}$$ is the side of D
• Thank you for your comment. Could you give me a link about what LLL criterion is? I have tried to search on Google and couldn't find one. Dec 29, 2021 at 1:52
• LLL, commonly known as SSS Dec 29, 2021 at 1:55
• LOL. I haven't used this criterion since high school and was thinking "it was length-length-length?"
– Rol
Dec 29, 2021 at 1:59
• You state that there is a ratio $x:y$ between the lengths, but you haven't gone into whether it's the same ratio as exists between the larger triangles. Also, the asker used $x:y$ for the ratio of areas, not of lengths. Dec 29, 2021 at 2:00
• @TimPederick thank you! I fixed these issues
– Rol
Dec 29, 2021 at 2:13
Yes, similarity preserves the same proportions between parts as it does the whole (angles equal, lengths scaled $$1:n$$, areas scaled $$1:n^2$$).
I'm not familiar with any succinct statement of this fact, unlike CPCTC for congruence, but the principle is the same. $$AB$$ is the image of $$CD$$ under a certain combination of translation, rotation, and scaling. The same translation, rotation and scaling will produce an image of $$D$$ that is similar: $$E$$. | 625 | 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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-33 | latest | en | 0.964026 |
http://www.chegg.com/homework-help/working-papers-to-accompany-fundamentals-of-financial-accounting-3rd-edition-solutions-9780077269708 | 1,475,328,905,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662856.94/warc/CC-MAIN-20160924173742-00240-ip-10-143-35-109.ec2.internal.warc.gz | 380,712,112 | 15,869 | # Working Papers to accompany Fundamentals of Financial Accounting (3rd Edition) View more editions
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Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 4
a) Assets = Liabilities + Stockholders’ Equity
= \$13,750 + \$4,450
= \$18,200
= Assets reported on the balance sheet
• Step 2 of 4
b) Net Income = Revenue – Expenses
= \$10,500 - \$9,200
= \$1,300
= Net income reported on the income statement
• Step 3 of 4
c) Beginning Retained Earnings (R/E) + Net Income – Dividends = Ending R/E
\$3,500 + \$1,300 - \$500 = \$4,300
• Step 4 of 4
d) Beginning Cash + Cash Flows from Operating Activities + Cash Flows from Investing Activities + Cash Flows from Financing Activities = Ending Cash
\$1,000 + \$1,600 + (\$1,000) + (\$900) = \$700
Corresponding Textbook
Working Papers to accompany Fundamentals of Financial Accounting | 3rd Edition
9780077269708ISBN-13: 0077269705ISBN: Fred PhillipsAuthors: | 327 | 1,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2016-40 | latest | en | 0.716666 |
https://www.lmfdb.org/Character/Dirichlet/287/t/62?number_field= | 1,713,714,618,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817780.88/warc/CC-MAIN-20240421132819-20240421162819-00339.warc.gz | 782,350,622 | 6,519 | # Properties
Label 287.62 Modulus $287$ Conductor $287$ Order $20$ Real no Primitive yes Minimal yes Parity odd
# Related objects
Show commands: PariGP / SageMath
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(287, base_ring=CyclotomicField(20))
M = H._module
chi = DirichletCharacter(H, M([10,7]))
pari: [g,chi] = znchar(Mod(62,287))
## Basic properties
Modulus: $$287$$ Conductor: $$287$$ sage: chi.conductor() pari: znconreyconductor(g,chi) Order: $$20$$ sage: chi.multiplicative_order() pari: charorder(g,chi) Real: no Primitive: yes sage: chi.is_primitive() pari: #znconreyconductor(g,chi)==1 Minimal: yes Parity: odd sage: chi.is_odd() pari: zncharisodd(g,chi)
## Galois orbit 287.t
sage: chi.galois_orbit()
order = charorder(g,chi)
[ charpow(g,chi, k % order) | k <-[1..order-1], gcd(k,order)==1 ]
## Related number fields
Field of values: $$\Q(\zeta_{20})$$ Fixed field: 20.0.1241291203716901419760482976136252014889.1
## Values on generators
$$(206,211)$$ → $$(-1,e\left(\frac{7}{20}\right))$$
## First values
$$a$$ $$-1$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$8$$ $$9$$ $$10$$ $$11$$ $$12$$ $$\chi_{ 287 }(62, a)$$ $$-1$$ $$1$$ $$e\left(\frac{1}{10}\right)$$ $$-i$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{1}{5}\right)$$ $$e\left(\frac{17}{20}\right)$$ $$e\left(\frac{3}{10}\right)$$ $$-1$$ $$e\left(\frac{3}{10}\right)$$ $$e\left(\frac{1}{20}\right)$$ $$e\left(\frac{19}{20}\right)$$
sage: chi.jacobi_sum(n)
$$\chi_{ 287 }(62,a) \;$$ at $$\;a =$$ e.g. 2
## Gauss sum
sage: chi.gauss_sum(a)
pari: znchargauss(g,chi,a)
$$\tau_{ a }( \chi_{ 287 }(62,·) )\;$$ at $$\;a =$$ e.g. 2
## Jacobi sum
sage: chi.jacobi_sum(n)
$$J(\chi_{ 287 }(62,·),\chi_{ 287 }(n,·)) \;$$ for $$\; n =$$ e.g. 1
## Kloosterman sum
sage: chi.kloosterman_sum(a,b)
$$K(a,b,\chi_{ 287 }(62,·)) \;$$ at $$\; a,b =$$ e.g. 1,2 | 739 | 1,870 | {"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} | 3.078125 | 3 | CC-MAIN-2024-18 | latest | en | 0.279132 |
https://transwikia.com/engineering/if-an-aircrafts-wing-blows-air-downwards-how-is-the-momentum-conserved-in-level-flight/ | 1,723,297,363,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00145.warc.gz | 450,108,469 | 14,124 | # If an aircraft's wing blows air downwards, how is the momentum conserved in level flight?
Engineering Asked on January 6, 2021
An aircraft’s wing generates force of lift by directing the oncoming airflow down. How is this momentum balanced when the aircraft is not climbing, but maintains level flight?
The angle of a typical wing in flight is slightly greater than zero. So the wing/airfoil deflects the air downwards thus creating the necessary Lift for keeping the airplane up.
Regarding the momentum, in order for the air to change direction because a force is applied on it. Due to the Newton's second law of motion, an equal and opposite force is applied by the air on the foil. That force is the aerodynamic force, which is decomposed to lift and drag.
In order for the plane to maintain height and speed, the following forces are at equilibrium:
• On the horizontal axis: Thrust and Drag
• On the vertical axis: Lift and Weight
Answered by NMech on January 6, 2021
Lift and gravity are "balancing" forces, just as thrust and drag.
The engine provides energy to the system. When there is a "surplus" of energy, it is converted to lift and speed. There is a statement in the aviation world, "The throttle provides, the elevator divides." As power is increased, if one desires to maintain level flight, the elevator is deflected appropriately, and the speed will increase.
Depending on one's perspective, some of the lift is generated by a reduction of pressure above the wing surface. This also applies to the above statements. It is not necessarily all deflection downward.
Answered by fred_dot_u on January 6, 2021
TLDR: If I stand in front of two wheels and push equally so that they counterrotate, the linear momentum starts and ends at zero, and the rotational momentum starts and ends at zero (counterrotation cancels momentum of individual wheels), but there is a force exerted on me.
In the 2D flow model
When you consider a large control boundary around the wing, all the vertical moment vanishes in the flow. Behind the foil a sufficient distance, the flow is once again perfectly horizontal - it's identical to the upstream flow. This is because the flow field is modeled as the sum of two items. The first is the farfield flow - straight, uniform flow from front to back. The second is a circulation flow. From a distance, this looks like a circular vortex centered on the wing, and the velocity trends to zero as you get further away. All the vertical momentum terms in the flow are contained in this circulation flow, and there is as much going up in front of the wing as there is going down behind the wing.
The combined fields yield a force on the wing, due to the different velocities and pressures on the top and bottom surfaces of the wing. But there is no work being done because there is no translation in the direction of the force. And there is no momentum change in the fluid mass taken as a whole. The flow far downstream is identical to the flow far upstream. This is true not just in steady state, but in startup and slowdown as well.
In the 3D flow model
Things get a bit more complicated. Here, there is a change between the upstream far field velocity and the downstream far field velocity . But just as in the 2D case, there is no energy exchange between the wing and the flow. So the transverse momentum in the wake has to have come from the fluid itself. What happens is that the wake is travelling a bit slower, on average, and this momentum deficit is manifested in the transverse flow that persists downstream of the wing. But the momentum in the transverse plane of the wake still balances. There is as much going up as going down, and there is as much going right as going left.
So for real-world objects in a breeze, if there is a lift force
1. There will be a net momentum change in the downstream flow that corresponds to that force. There will be persistent downstream transverse flows.
2. The energy needed to produce the transverse flow in the wake comes from the only available energy source - the initial velocity of the upstream flow. So there is an inescapable stream-wise energy and momentum defect in the wake, and this produces drag on the object.
Momentum comes in two flavors - linear and angular momentum. Both are conserved quantities when no work is done on the system, as is the case here.
In the 3D case, where the wake flow does not return to the upstream condition, both types of momentum are none the less conserved.
Viewing the transverse velocities in a transverse plane far behind the wing, there is equal linear momentum up and down, and equal linear momentum right and left. Summed over the domain, the transverse linear momentum is unchanged. Between the wingtips, there is downwash. Outboard of the wingtips, there is upwash.
The rotational momentum starts at zero in front and must remain at zero behind the wing. The wake consists of a pair of counterrotating vortexes that perfectly cancel each other's angular momentum regardless of any asymmetry of the wing. Creating these wake features requires a force. That force is mostly in the transverse plane, and we call the in-plane component lift, but that force can't ever be entirely in the transverse plane. It always has a drag component.
Answered by Phil Sweet on January 6, 2021
An airplane in level flight, cruising, needs lift to fight the gravity. It lacks buoyancy as opposed to blimps.
It sometimes needs even more power when cruising in not the optimal configuration of payload and misplaced CG of the load.
So even when a plane is not climbing it needs lift. That lift is provided by changing the airflow momentum.
Answered by kamran on January 6, 2021 | 1,221 | 5,718 | {"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} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.945128 |
https://www.toprankers.com/rrb-group-d-exam-analysis-5th-december-2018 | 1,628,185,669,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046156141.29/warc/CC-MAIN-20210805161906-20210805191906-00308.warc.gz | 1,068,321,289 | 14,490 | ## RRB Group D Exam Analysis 5th December 2018 & Questions Asked
The 2nd Shift of RRB Group D Exam of 5th December 2018, is over. Exam Analysts at TopRankers, are here to provide a detailed analysis of the exam. This analysis shall help in evaluating your performance. You can even get an estimate of your score. All those who are yet to appear for the exam can find this analysis helpful too. The RRB Group D Exam consists of 4 sections - Mathematics, General Intelligence and Reasoning, General Science and General Awareness & Current Affairs. The overall exam timing is 90 minutes and for every wrong answer, 0.25 marks will be deducted. Explore the overall as well as sectional difficulty level, topic-wise weightage, no. of good attempts and Questions asked.
### RRB Group D Shift 2 Exam Analysis - 5th December
The table below shows the safer attempts of RRB Group D Shift 2.
RRB Group D Exam No. of Good Attempts - Shift 2
Section No. of Qs Difficulty Level No. of Good Attempts
Mathematics 25 Moderate 11-13
General Intelligence and Reasoning 30 Moderate 18-21
General Science 25 Moderate to easy 17-19
General Awareness & Current Affairs 20 Moderate 10-13
Overall 100 Moderate 63-70
#### Overview of RRB Group D Shift 2 Exam - 5th December 2018
The shift 2 of RRB Group D exam was moderate. Questions trends now are changing. In place of direct questions, now questions are coming in a twisted way. So keep practicing and keep following the recent analysis questions.
#### RRB Group D Exam Analysis 5th December Shift 2 - General Awareness & Current Affairs
• Who is the Finance and Industrial Minister of India?
• Which is the oldest monuments in India?
• Where is the head of World Health Organization situated?
• Who is the President of National Development Council in India?
• When did the battle of Haldighati take place?
• How many people have been selected for Padma Bibhushan Award in 2018?
#### RRB Group D Exam Analysis 5th December Shift 2 – Mathematics
• Time and Work
• Profit and Loss
• SI and CI
• Mensuration
• Trigonometry
• The sum of 2 numbers is 48 while their difference is 8. Find the product of the two numbers.
• Pipe A can fill a tank in 10 hrs while Pipe B can empty the tank in 15 hours. If both the pipes are opened, how long will it take to fill the tank?
• A certain amount deposited at bank got doubled in 10 years. What is the rate percent applicable at simple interest?
#### RRB Group D Exam Analysis 5th December Shift 2 - General Science
• Which microorganism helps in formation of curd?
• Where are the Red Blood Cells formed?
• What is the Chemical name of Table Salt?
#### RRB Group D Exam Analysis 5th December Shift 2 - General Intelligence and Reasoning
• Mirror Image
• Number Series
• Blood Relation
• Venn Diagram
• Calendar based problems
• Direction Problems
• Miscellaneous qts.
### RRB Group D Shift 1 Exam Analysis - 5th December
The table below shows the safer attempts of RRB Group D Shift 1.
RRB Group D Exam No. of Good Attempts - Shift 1
Section No. of Qs Difficulty Level No. of Good Attempts
Mathematics 25 Moderate 13-15
General Intelligence and Reasoning 30 Moderate 20-23
General Science 25 Moderate to easy 16-18
General Awareness & Current Affairs 20 Moderate 13-15
Overall 100 Moderate 65-73
#### Overview of RRB Group D Shift 1 Exam - 5th December 2018
The shift 1 of RRB Group D exam was moderate. Questions trends now are changing. In place of direct questions, now questions are coming in a twisted way. So keep practicing and keep following the recent analysis questions.
#### RRB Group D Exam Analysis 5th December Shift 1 - General Awareness & Current Affairs
• Where will 38th National Games be organized?
• In which state is the Sukhna Lake is situated?
• When was the first Television in Indian inaugurated?
• Where is venue of the 21st Common Wealth Games?
• When was the first World Bee Day Celebrated?
• Which is the biggest bay in the world?
#### RRB Group D Exam Analysis 5th December Shift 1 – Mathematics
• What is the difference in Simple Interest and Compound Interest on a sum of Rs. 1250 for 4% rate of interest for 2 years.
• A 300 m long traion moving at a speed of 72km/hr. How much time will it take to cross a platform of 200 metres?
• If on selling an object at Rs. 500, there is a profit of 25%. At what rate should the object be sold to make a profit of 20%?
• A and B can do a piece of work in 3 days, while A alone can do the work in 4 days. Den in how many days will B complete the work?
#### RRB Group D Exam Analysis 5th December Shift 1 - General Science
• Biology : 8-9 Qs
• Physics: 6-8 Qs
• Chemistry: 7-8
• Which planet is called the "Blue Planet of Solar system"?
• Name the instrument used to measure electric current.
• Which Vitamins are obtained from the rays of the Sun?
#### RRB Group D Exam Analysis 5th December Shift 1 - General Intelligence and Reasoning
• Mirror Image
• Number Series
• Blood Relation
• Venn Diagram
• Calendar based problems
• Miscellaneous qts.
#### Tips for upcoming Shifts
Go through the basic concepts of Science. Also, revise through the formula charts for the numerical paper. Go through on the current affairs of last 3 months. | 1,285 | 5,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.932923 |
http://stackoverflow.com/questions/11620914/removing-nan-values-from-an-array?answertab=oldest | 1,386,941,892,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164943590/warc/CC-MAIN-20131204134903-00027-ip-10-33-133-15.ec2.internal.warc.gz | 167,657,741 | 13,152 | # Removing nan values from an array
I want to figure out how to remove nan values from my array. It looks something like this:
``````x = [1400, 1500, 1600, nan, nan, nan ,1700] #Not in this exact configuration
``````
I'm relatively new to python so I'm still learning. Any tips?
-
If you're using numpy for your arrays, you can also use
``````x = x[numpy.logical_not(numpy.isnan(x))]
``````
-
Or `x = x[numpy.isfinite(x)]` – lazy1 Jul 23 '12 at 22:29
These both work! I'm so glad there are so many ways to do this. Thank you! – Dax Feliz Jul 23 '12 at 22:47
Or `x = x[~numpy.isnan(x)]`, which is equivalent to mutzmatron's original answer, but shorter. In case you want to keep your infinities around, know that `numpy.isfinite(numpy.inf) == False`, of course, but `~numpy.isnan(numpy.inf) == True`. – chbrown Nov 19 at 19:02
@dax-felizv I agree with @chbrown, NaN and Infinite are not the same in `numpy`. @chbrown - thanks for pointing out the shorthand for `logical_not`, though beware that it is considerably slower - stackoverflow.com/questions/15998188/…, stackoverflow.com/questions/13600988/… – mutzmatron Nov 20 at 19:45
Hmm, @mutzmatron -- I figured they did the same thing underneath the hood, and I'm getting very similar results with timeit (as did @unutbu at that first link): `python -m timeit -s "import numpy; bools = numpy.random.uniform(size=10000) >= 0.5" "numpy.logical_not(bools)"` vs. `python -m timeit -s "import numpy; bools = numpy.random.uniform(size=10000) >= 0.5" "~bools"` (`numpy.__version__ == '1.8.0'`) – chbrown Nov 20 at 22:41
show 1 more comment
Try this:
``````import math
print [value for value in x if not math.isnan(value)]
``````
For more, read on List Comprehensions.
-
AWESOME! That did it! Thank you so much! I wish one line of code could fix all my problems :) – Dax Feliz Jul 23 '12 at 22:45
If you're using numpy both my answer and that by @lazy1 are almost an order of magnitude faster than the list comprehension - lazy1's solution is slightly faster (though technically will also not return any infinity values). – mutzmatron Jul 24 '12 at 13:54 | 624 | 2,112 | {"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-2013-48 | latest | en | 0.837132 |
https://paylawexam.com/what-is-a-civil-infraction-2 | 1,723,240,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00139.warc.gz | 350,919,919 | 28,265 | # What is a Civil Infraction?
What is a Civil Infraction? Of these 3D-printed forms, one is a 2-panel of 4-bristle modules, and the other 3D-printing the 4-bristle modules, on the 3-layer-on-the-board. [INFO] a 2-panel of 4-bristle modules [INFO] a 2-panel of 4-bristle modules [INFO] a 2-panel of 4-bristle modules ## 5.14 About a 3D-printed block A 3D-printed block can be a real 3-dimensional surface or 2-3D shape, after 3D-printing. A 3-dimensional surface is useful or not, since it might be difficult or impossible to trace. # 5.15 A 3D-printed block, for fixing print and mount A 3-dimensional surface has its unique property that the thickness of ink is not directly proportional to the surface area. A 3-dimensional block is a 3-dimensional surface, and what makes a 3-dimensional sheet so printed is clearly a 3-dimensional sheet. 2D and 3-dimensional objects can be found on a printed object. With enough ink, the surface does not form an ancillary point, and the total number of parts of a 3-dimensional object can be calculated if the surface is provided at the top with the aesium-like pattern. ### 5.16 3-dimensional surface, (a) The shapes of the 3-dimensional forms ### 5.17 Approximation of a 3-dimensional surface, and the equations Example 5.16 #### To fit a 3-dimensional model The 5.16 solution of the above definition (in this case – 6: 6) is not given, due to numerical inconsistencies – but is allowed. We refer to section 5.11, with description context,What is a Civil Infraction? The earliest form of an indentation in a metal is a single layer of lead or a plug and insulator compound (see the page on its website). Its origin was likely a method of compressing gases, i.e of using a pressure to compress out the initial energy dissipation (temperature). Several early works have indicated how the compound fills the volume of a unit electrode—this led to the discovery of an idea which, in its later history, is largely understood as a technique for the creation of circuits. For example, such “deterministic processes” have played a part in designing electronic systems, although many models, such as those based on particle physics, have been proposed.
The former were not very interesting, as they typically do not produce more energy. The latter were rather interesting as well, and made a substantial contribution to the design of quantum computers and catwaw engines. Why the classical mechanism for constructing circuits was such a big leap toward a computer First there was both abstract concepts underlying the nature of electronic circuits and the nature of mechanical calculations, which represented a vast variety of possible processes. The content model was quite complex and had to be described, quite literally, in terms of the particle world, whereas mechanical calculations were mostly about the wave equations in the mechanics of computer simulation. These models were much more intuitive than the more complex models, and were viewed as a complete structure of a fluid, the state and distribution of which was of fluidless quality. However, the first fully “functional” models were built by means of mechanical analysis, which proved to be very scientific and accurate, although not usually built with precision; an analysis is then used to calculate mechanical parameters in an electronic circuit, and to obtain electronic properties such as the mechanical properties that are produced by the mechanical operation of the circuit. In order to have a method of construction simple, at the beginning there was no physical mechanism of manipulation beyond what was attainable by mechanical analysisWhat is a Civil Infraction? and the Nature and Function of Evolutionary Social Networks and their Application to Human Bioarchaeology? The Nature of Social Networks – Social Networks: A Theory of Connectivity & Evolutionary Bijlinger, R. S. (N/A) Abstract Social networks are formed from self-local and linear scales by interactions between persons, connected by some entities known as ‘social network’, organised for various reasons. Social networks have evolved as the result of a process of social segregation of individuals. It starts when the social network is brought together and as it grows, relationships/groups within groups can and need an up and coming evolutionary social network – this is called social segregation. Social networks are then ‘transformed’ into individual one-dimensional (1D) and ‘volatile’. There are two main processes of social mobility and separation – the distance and the mass. Eventually, we end up with a self-gravitated self-gravitating network. A First Opinion – Science: A Life-Ups in Human Social Networks. On February 10, 2016, Dr. Aruna S. Vaz, postdoctoral researcher and an associate postdoctoral researcher at the University of Texas at Austin, USA conceived the work “The Nature and Function of Evolutionary Social Networks and their Application to Human Bioarchaeology.” The idea to study this nature and process of social evolution lies beyond any current papers we’ve done so far and so far contains unique data on how individual and society life and behaviour are organised. This work was funded by the National Science Foundation through R02EB006164.
## I Need read To Do My Homework For Me
From the vantage point of classical statistical physics, a natural question is how the distribution of social networks fit with the usual structure of physics. It takes the form of three distributions and we ‘like’ this distribution if we can identify its logarithm of mass
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Celebrate success in law with our comprehensive examination services – Your path to excellence awaits! | 1,217 | 5,808 | {"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} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.923089 |
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