url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://gateoverflow.in/82225/minimum-number-of-nand-gates-for-logic-circuit
| 1,579,673,954,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250606696.26/warc/CC-MAIN-20200122042145-20200122071145-00187.warc.gz
| 458,908,243
| 16,955
|
2.2k views
| 2.2k views
Here we are getting output as Y'=A'BC+AB'C+ABC'+ABC
=AB+BC+AC
=((A(B'C')')' (BC)')'
So, total 6 NAND gates are required.
If complemented and uncomplemented variable both present then minimum NAND gates are 4
by Veteran (119k points)
selected
0
Is there any method to reduce AB+BC+AC to ((A(B'C')')' (BC)')' ?
0
No, just put demorgan law (A+B)'=A'B' ,(AB)'=A'+B'
and then trial and error
0
Mam, I think that the function is 1 on the minterms 0,1,2,4. so y' is definitely what you wrote but forgot to make the pos. You are counting max terms so it you should use pos term right. ?
+1
7 NAND gates required
+1 vote
There is a systematic procedure to find the number of NAND gates.
First, determine the minimal SOP for ( AND-OR structure) then follow the steps--
no of gates=6
by (79 points)
| 267
| 826
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.296875
| 3
|
CC-MAIN-2020-05
|
longest
|
en
| 0.89781
|
https://quantixed.org/2020/05/19/running-free-calculating-efficiency-factor-in-r/
| 1,686,419,755,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224657735.85/warc/CC-MAIN-20230610164417-20230610194417-00294.warc.gz
| 543,904,631
| 19,459
|
# Running Free: Calculating Efficiency Factor in R
Joe Friel reposted an article earlier this year on Efficiency Factor in running. Efficiency Factor (EF) can be viewed in Training Peaks software and he describes how it is calculated. This post describes how I went about calculating EF in R using a single gpx file.
#### What is Efficiency Factor (EF)?
Essentially, EF is the average distance that you are propelled forward per heart beat. The higher the number, the more efficient you are at running.
To calculate EF, you need your speed and heart rate data. In Joe’s post, he gives the following example (using imperial measurements):
For a run with a Normalised Graded Pace of 7.5 min/mile (8 miles per hour), the speed in yards per minute is 234.7 (1760 yards * 8/60). If the average heart rate is 150, the EF is 1.56 (234.7/150).
I haven’t used Training Peaks, but it seems EF is given as a single measure for your entire run. It should be possible to calculate EF as a rolling measure so that you could look at any changes in EF over time, or over specific terrain on the run. So this was my other motivation for calculating EF in R.
#### Getting the data into R
I’m using the trackeR package to do the hard work of loading the data from the gpx file.
library(trackeR)
library(ggplot2)
library(zoo)
library(hms)
library(gridExtra)
# select a gpx file
filepath <- file.choose()
runDF <- readGPX(file = filepath, timezone = "GMT")
This gives us the data from the gpx file in a data frame called runDF. Now we have to calculate the normalised graded speed. This is done by first calculating the point-to-point distances, altitudes and thereby the point-to-point gradients. Point-to-point here refers to the sampling frequency of the gps device. We can extract that information too, so that we can calculate speeds.
Here is the code that gets us to normalised graded speed. The calculation of the dist_adj column is described below.
# calculate point-to-point distance from the cumulative distance
runDF\$dist_point <- c(0,diff(runDF\$distance, lag=1))
# calculate the point-to-point gradient as %
runDF\$alti_point <- c(0,diff(runDF\$altitude, lag=1))
runDF\$grad_point <- (runDF\$alti_point / runDF\$dist_point) * 100
(0.98462 + (0.030266 * runDF\$grad_point) +
(0.0018814 * runDF\$grad_point ^ 2) +
(-3.3882e-06 * runDF\$grad_point ^ 3) +
# time calculations
runDF\$time_temp <- strptime(runDF\$time, format = "%Y-%m-%d %H:%M:%S")
runDF\$time_point <- c(0,diff(as.vector(runDF\$time_temp), lag=1))
runDF\$time_temp <- NULL
runDF\$time_cumulative <- cumsum(runDF\$time_point)
runDF\$time_hms <- as_hms(runDF\$time_cumulative)
# speed in m/s
runDF\$speed <- runDF\$dist_point / runDF\$time_point
# normalised graded speed in m/s
# replace NaNs with 0
runDF[is.na(runDF)] <- 0
# Efficiency Factor is ngs in yards per minute divided by heart rate
runDF\$EF <- (1.0936133 * runDF\$ngs * 60) / runDF\$heart_rate
### How do I calculate normalised graded pace?
If you are running uphill, your pace will be slower when running on the flat (and vice versa). Can we normalise for gradient to see how fast the runner would be travelling if the whole course was flat.
There is a paper from 2002 looking at the changes in pace over different gradients (Minetti et al. (2002)). However, this was a small study of 10 people. Nowadays we have much more data thanks to consumer-grade GPS-enabled devices. Strava used their user’s data to make a new model to calculate GAP (Grade Adjusted Pace) for display on their site. Their findings plotted with those of Minetti et al. are shown here:
I took the points from this graph using IgorThief and fitted a polynomial equation to this data. The co-efficients of this fit allowed me to find what kind of speed adjustment is required for a given gradient. This information is used the line in the code above:
(0.98462 + (0.030266 * runDF\$grad_point) +
(0.0018814 * runDF\$grad_point ^ 2) +
(-3.3882e-06 * runDF\$grad_point ^ 3) +
Let’s look at our graphical output before seeing how it was generated.
The output shows speed, elevation, normalised graded speed, heart rate and EF. The example shown above has some intervals (note the peaks in the heart rate trace). EF is pretty constant, but varies a bit during the interval/rest cycles.
#### The remaining code
The plots above were generated using this code.
# format ticks on plots to be hh:mm
format_hm <- function(sec) stringr::str_sub(format(sec), end = -4L)
# make the plots
p1 <- ggplot(runDF, aes(time_hms, speed)) +
geom_line(aes(alpha = 0.2)) +
ylim(0,5) +
geom_line(aes(y = rollmean(speed, 120, na.pad = TRUE)), colour = "#663399", size = 1) +
scale_x_time(labels = format_hm) +
labs(x = "Time", y = "Speed (m/s)") +
theme(legend.position="none")
p1
p2 <- ggplot(runDF, aes(time_hms, altitude)) +
geom_line(colour = "#E69F00") +
scale_x_time(labels = format_hm) +
labs(x = "Time", y = "Elevation") +
theme(legend.position="none")
p2
p3 <- ggplot(runDF, aes(time_hms, ngs)) +
geom_line(aes(alpha = 0.2)) +
ylim(0,5) +
geom_line(aes(y = rollmean(ngs, 120, na.pad = TRUE)), colour = "#663399", size = 1) +
scale_x_time(labels = format_hm) +
labs(x = "Time", y = "Normalised Gradient Speed (m/s)") +
theme(legend.position="none")
p3
p4 <- ggplot(runDF, aes(time_hms,heart_rate)) +
geom_line(colour = "#D55E00") +
ylim(0,200) +
scale_x_time(labels = format_hm) +
labs(x = "Time", y = "Heart rate (bpm)") +
theme(legend.position="none")
p4
p5 <- ggplot(runDF, aes(time_hms,EF)) +
geom_line(aes(alpha = 0.2)) +
ylim(0,5) +
geom_line(aes(y = rollmean(EF, 120, na.pad = TRUE)), colour = "#0072B2", size = 1) +
scale_x_time(labels = format_hm) +
labs(x = "Time", y = "Efficiency Factor") +
theme(legend.position="none")
p5
# arrange plots into a quilt and save
q1 <- grid.arrange(p1,p2,p3,p4,p5, layout_matrix = rbind(c(1,2,3),c(4,5,5)))
saveName <- paste("./Output/Plots/",sub(".gpx",".png",basename(filepath)),sep = "")
ggsave(saveName, q1)
Most parts of that code are standard ggplot stuff. Two exceptions:
• To make the rolling average, I used the zoo package and calculated a 2 minute (120 s) window with rollmean
• Suppressing the seconds on the x-axis was done using format_hm. This is specified in the code and changes an hh:mm:ss time to hh:mm
The post title comes from “Running Free” by Iron Maiden from their eponymous debut LP.
## 2 thoughts on “Running Free: Calculating Efficiency Factor in R”
1. Hi @Quantixed! Very cool and helpful code. Thank you for sharing.
I was wondering why you calculated an adjusted distance first and then translated that to the normalized graded speed?
I used the following, and seems to give the same result:
runDF\$ngs <- runDF\$speed *
(0.98462 + (0.030266 * runDF\$grad_point) +
(0.0018814 * runDF\$grad_point ^ 2) +
(-3.3882e-06 * runDF\$grad_point ^ 3) +
| 1,898
| 6,817
|
{"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-2023-23
|
latest
|
en
| 0.911998
|
https://discusstest.codechef.com/t/plotting-y-a-sin-x-b-in-text-mode/7846
| 1,685,396,293,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00220.warc.gz
| 247,334,672
| 6,070
|
# Plotting y = a*(sin(x)) + b in text-mode
Hey, I initiated the following code and now I’m running the code for a=0 and any real value of b, for which I should get a line y=b, but that isn’t happening, instead I get x and y axes only with no plot. So please tell me where the code has a shortcoming.
``````#include <stdio.h>
#include <math.h>
#define WIDTH 60
#define HEIGHT 20
#define X WIDTH/2
#define Y HEIGHT/2
#define XMAX WIDTH-X-1
#define XMIN -(WIDTH-X)
#define YMAX HEIGHT-Y
#define YMIN -(HEIGHT-Y)+1
char coordinate_system[HEIGHT][WIDTH];
int plot(int x, int y);
void x_y_axes(void);
void show_plot(void);
int main()
{
int a, b;
float x,y;
printf("Enter a and b:\n");
scanf("%d%d", &a, &b);
x_y_axes();
for(x=-3.1416;x<=3.1416;x+=0.1)
{
y = (a*(sin(x))) + b;
plot(rintf(x*10),rintf(y*8));
}
show_plot();
return(0);
}
/* Set "pixel" at specific coordinates */
int plot(int x, int y)
{
if( x < XMAX && x > XMIN && y < YMAX && y > YMIN )
//return(-1);
//else
coordinate_system[Y-y][X+x] = '*';
}
/* displays coordinate system */
void x_y_axes(void)
{
int x,y;
/* draw the intersecting axes */
for(y=0;y<HEIGHT;y++)
coordinate_system[y][X] = '|';
for(x=0;x<WIDTH;x++)
coordinate_system[Y][x] = '-';
}
/* display grid */
void show_plot(void)
{
int x,y;
for(y=0;y<HEIGHT;y++)
{
for(x=0;x<WIDTH;x++)
putchar(coordinate_system[y][x]);
putchar('\n');
}
}``````
plot(rintf(x10),rintf(y8));
editing the above to
``````plot(x,y);
``````
I googled “rintf()” but could not get much about it. What are they and why are you using them anyway?
1 Like
| 494
| 1,562
|
{"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-2023-23
|
latest
|
en
| 0.453719
|
http://www.reference.com/browse/parallel+pascal
| 1,435,700,892,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00042-ip-10-179-60-89.ec2.internal.warc.gz
| 673,041,820
| 26,809
|
Definitions
# Pascal's triangle
begin{matrix} &&&&&1 &&&&1&&1 &&&1&&2&&1 &&1&&3&&3&&1 &1&&4&&6&&4&&1 end{matrix}
The first five rows of Pascal's triangle
In mathematics, Pascal's triangle is a geometric arrangement of the binomial coefficients in a triangle. It is named after Blaise Pascal in much of the western world, although other mathematicians studied it centuries before him in India, Persia, China, and Italy. The rows of Pascal's triangle are conventionally enumerated starting with row zero, and the numbers in odd rows are usually staggered relative to the numbers in even rows. A simple construction of the triangle proceeds in the following manner. On the zeroth row, write only the number 1. Then, to construct the elements of following rows, add the number directly above and to the left with the number directly above and to the right to find the new value. If either the number to the right or left is not present, substitute a zero in its place. For example, the first number in the first row is 0 + 1 = 1, whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row.
This construction is related to the binomial coefficients by Pascal's rule, which states that if
$\left\{n choose k\right\} = frac\left\{n!\right\}\left\{k! \left(n-k\right)!\right\}$
is the kth binomial coefficient in the binomial expansion of (x + y)n, where n! is the factorial of n, then
$\left\{n choose k\right\} = \left\{n-1 choose k-1\right\} + \left\{n-1 choose k\right\}$
for any nonnegative integer n and any integer k between 0 and n.
Pascal's triangle has higher dimensional generalizations. The three-dimensional version is called Pascal's pyramid or Pascal's tetrahedron, while the general versions are called Pascal's simplices — see also pyramid, tetrahedron, and simplex.
## History
The earliest explicit depictions of a triangle of binomial coefficients occur in the 10th century in commentaries on the Chandas Shastra, an ancient Indian book on Sanskrit prosody written by Pingala between the 5th2nd centuries BC. While Pingala's work only survives in fragments, the commentator Halayudha, around 975, used the triangle to explain obscure references to Meru-prastaara, the "Staircase of Mount Meru". It was also realised that the shallow diagonals of the triangle sum to the Fibonacci numbers. The Indian mathematician Bhattotpala and his pupil Dhruv Ragunathan (c. 1068) later gives rows 0-16 of the triangle.
At around the same time, it was discussed in Persia (Iran) by the mathematician Al-Karaji (953–1029) and the poet-astronomer-mathematician Omar Khayyám (1048-1131); thus the triangle is referred to as the "Khayyam triangle" in Iran. Several theorems related to the triangle were known, including the binomial theorem. In fact we can be fairly sure that Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients.
In 13th century, Yang Hui (1238-1298) presented the arithmetic triangle, which was the same as Pascal's Triangle. Today Pascal's triangle is called "Yang Hui's triangle" in China.
Petrus Apianus (1495-1552) published the triangle on the frontispiece of his book on business calculations 1531/32 and an earlier version in 1527 the first record of it in Europe.
In Italy, it is referred to as "Tartaglia's triangle", named for the Italian algebraist Niccolò Fontana Tartaglia (1500-1577); Tartaglia is credited with the general formula for solving cubic polynomials (which may be really from Scipione del Ferro but was published by Gerolamo Cardano 1545).
Finally, in 1655, Blaise Pascal wrote a Traité du triangle arithmétique (Treatise on arithmetical triangle), wherein he collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) and Abraham de Moivre (1730).
## The triangle
Below are rows zero to sixteen of Pascal's triangle:
## Pascal's triangle and binomial expansions
Pascal's triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion
(x + y)2 = x2 + 2xy + y2 = 1x2y0 + 2x1y1 + 1x0y2.
Notice the coefficients are the numbers in row two of Pascal's triangle: 1, 2, 1. In general, when a binomial like x + y is raised to a positive integer power we have:
(x + y)n = a0xn + a1xn−1y + a2xn−2y2 + … + an−1xyn−1 + anyn,
where the coefficients ai in this expansion are precisely the numbers on row n of Pascal's triangle. In other words,
$a_i = \left\{n choose i\right\}.$
This is the binomial theorem.
Notice that entire right diagonal of Pascal's triangle corresponds to the coefficient of yn in these binomial expansions, while the next diagonal corresponds to the coefficient of xyn−1 and so on.
To see how the binomial theorem relates to the simple construction of Pascal's triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)n+1 in terms of the corresponding coefficients of (x + 1)n (setting y = 1 for simplicity). Suppose then that
$\left(x+1\right)^n=sum_\left\{i=0\right\}^n a_i x^i.$
Now
$\left(x+1\right)^\left\{n+1\right\} = \left(x+1\right)\left(x+1\right)^n = x\left(x+1\right)^n + \left(x+1\right)^n = sum_\left\{i=0\right\}^n a_i x^\left\{i+1\right\} + sum_\left\{i=0\right\}^n a_i x^i.$
The two summations can be reorganized as follows:
begin{align} & sum_{i=0}^{n } a_{i } x^{i+1} + sum_{i=0}^n a_i x^i & {} = sum_{i=1}^{n+1} a_{i-1} x^{i } + sum_{i=0}^n a_i x^i & {} = sum_{i=1}^{n } a_{i-1} x^{i } + sum_{i=1}^n a_i x^i + a_0x^0 + a_{n}x^{n+1} & {} = sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + a_0x^0 + a_{n}x^{n+1} & {} = sum_{i=1}^{n } (a_{i-1} + a_i)x^{i } + x^0 + x^{n+1} end{align}
(because of how raising a polynomial to a power works, a0 = an = 1).
We now have an expression for the polynomial (x + 1)n+1 in terms of the coefficients of (x + 1)n (these are the ais), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upper-left to the lower-right correspond to the same power of x, and that the a-terms are the coefficients of the polynomial (x + 1)n, and we are determining the coefficients of (x + 1)n+1. Now, for any given i not 0 or n + 1, the coefficient of the xi term in the polynomial (x + 1)n+1 is equal to ai (the figure above and to the left of the figure to be determined, since it is on the same diagonal) + ai−1 (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal's triangle row-by-row.
It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem.
An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that $\left(1+1\right)^n = 2^n$, and so
$\left\{n choose 0\right\} + \left\{n choose 1\right\} + cdots +\left\{n choose n-1\right\} + \left\{n choose n\right\} = 2^n.$
In other words, the sum of the entries in the nth row of Pascal's triangle is the nth power of 2.
## Patterns and properties
Pascal's triangle has many properties and contains many patterns of numbers.
### The diagonals
Some simple patterns are immediately apparent in the diagonals of Pascal's triangle:
• The diagonals going along the left and right edges contain only 1's.
• The diagonals next to the edge diagonals contain the natural numbers in order.
• Moving inwards, the next pair of diagonals contain the triangular numbers in order.
• The next pair of diagonals contain the tetrahedral numbers in order, and the next pair give pentatope numbers. In general, each next pair of diagonals contains the next higher dimensional "d-simplex" numbers, which can be defined as
$textrm\left\{tri\right\}_1\left(n\right) = n quadmbox\left\{and\right\}quad textrm\left\{tri\right\}_\left\{d\right\}\left(n\right) = sum_\left\{i=1\right\}^n mathrm\left\{tri\right\}_\left\{d-1\right\}\left(i\right).$
An alternative formula is as follows:
$textrm\left\{tri\right\}_d\left(n\right)=begin\left\{cases\right\} 1 & mbox\left\{if \right\} d=0 n & mbox\left\{if \right\} d=1 displaystyle frac\left\{1\right\}\left\{d!\right\}prod_\left\{k=0\right\}^\left\{d-1\right\} \left(n+k\right) & mbox\left\{if \right\} dge 2.end\left\{cases\right\}$
The geometric meaning of a function trid is: trid(1) = 1 for all d. Construct a d-dimensional triangle (a 3-dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to trid(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal's triangle. To find trid(x), have a total of x dots composing the target shape. trid(x) then equals the total number of dots in the shape. A 0-dimensional triangle is a point and a 1-dimensional triangle is simply a line, and therefore tri1(x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to trid − 1(x).
### Other patterns and properties
• The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called Sierpinski triangle, and this resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other patterns and combinations.
Pascal's triangle with odd numbers shaded Pascal's triangle with numbers not divisible by 3 shaded Pascal's triangle with numbers not divisible by 4 shaded Pascal's triangle with numbers not divisible by 5 shaded
• Imagine each number in the triangle is a node in a grid which is connected to the adjacent numbers above and below it. Now for any node in the grid, count the number of paths there are in the grid (without backtracking) which connect this node to the top node (1) of the triangle. The answer is the Pascal number associated to that node. The interpretation of the number in Pascal's Triangle as the number of paths to that number from the tip means that on a Plinko game board shaped like a triangle, the probability of winning prizes nearer the center will be higher than winning prizes on the edges.
• The value of each row, if each number in it is considered as a decimal place and numbers larger than 9 are carried over accordingly, is a power of 11 (specifically, 11n, where n is the number of the row). For example, row two reads '1, 2, 1', which is 112 (121). In row five, '1, 5, 10, 10, 5, 1' is translated to 161051 after carrying the values over, which is 115. This property is easily explained by setting x = 10 in the binomial expansion of (x + 1)row number, and adjusting the values to fit in the decimal number system.
### More subtle patterns
There are also more surprising, subtle patterns. From a single element of the triangle, a more shallow diagonal line can be formed by continually moving one element to the right, then one element to the bottom-right, or by going in the opposite direction. An example is the line with elements 1, 6, 5, 1, which starts from the row 1, 3, 3, 1 and ends three rows down. Such a "diagonal" has a sum that is a Fibonacci number. In the case of the example, the Fibonacci number is 13:
` 1`
` 1 1`
` 1 2 1`
` 1 → 3 ↓ 3 1`
` 1 4 →6 → 4 ↓ 1`
` 1 5 10 10 →5 → 1 ↓`
` 1 → 6 ↓ 15 20 15 6 →1`
` 1 7 →21 35 35 21 7 1`
` 1 8 28 56 70 56 28 8 1`
` 1 9 36 84 126 126 84 36 9 1`
` 1 10 45 120 210 252 210 120 45 10 1`
` 1 11 55 165 330 462 462 330 165 55 11 1`
` 1 12 66 220 495 792 924 792 495 220 66 12 1`
` 1 13 78 286 715 1287 1716 1716 1287 715 286 78 13 1`
` 1 14 91 364 1001 2002 3003 3432 3003 2002 1001 364 91 14 1`
` 1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1`
`1 16 120 560 1820 4368 8008 11440 12870 11440 8008 4368 1820 560 120 16 1`
The second highlighted diagonal has a sum of 233. The numbers 'skipped over' between the move right and the move down-right also sum to Fibonacci numbers, being the numbers 'between' the sums formed by the first construction. For example, the numbers skipped over in the first highlighted diagonal are 3, 4 and 1, making 8.
In addition, if row m is taken to indicate row $\left(n+1\right)$, the sum of the squares of the elements of row m equals the middle element of row $\left(2m-1\right)$. For example, $1^2 + 4^2 + 6^2 + 4 ^2 + 1^2 = 70$. In general form:
$sum_\left\{k=0\right\}^n \left\{n choose k\right\}^2 = \left\{2n choose n\right\}.$
Another interesting pattern is that on any row m, where m is odd, the middle term minus the term two spots to the left equals a Catalan number, specifically the (m + 1)/2 Catalan number. For example: on row 5, 6 − 1 = 5, which is the 3rd Catalan number, and (5 + 1)/2 = 3.
Also, the sum of the elements of row m is equal to 2m−1. For example, the sum of the elements of row 5 is $1 + 4 + 6 + 4 + 1 = 16$, which is equal to $2^4 = 16$. This follows from the binomial theorem proved above, applied to (1 + 1)m−1.
Some of the numbers in Pascal's triangle correlate to numbers in Lozanić's triangle.
Another interesting property of Pascal's triangle is that in rows where the second number (the 1st number following 1) is prime, all the terms in that row except the 1s are multiples of that prime.
### The matrix exponential
Due to its simple construction by factorials, a very basic representation of Pascal's triangle in terms of the matrix exponential can be given: Pascal's triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, … on its subdiagonal and zero everywhere else.
### Geometric properties
Pascal's triangle can be used as a lookup table for the number of arbitrarily dimensioned elements within a single arbitrarily dimensioned version of a triangle (known as a simplex). For example, consider the 3rd line of the triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2-dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons (simplices).
To understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n − 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.
The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle.
A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal's triangle, whose entries are the coefficients of (x + 2)Row Number, instead of (x + 1)Row Number. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:
$\left\{n choose k\right\} = 2times\left\{n-1 choose k-1\right\} + \left\{n-1 choose k\right\}.$
That is, choose a pair of numbers according to the rules of Pascal's triangle, but double the one on the left before adding. This results in:
` 1`
` 1 2`
` 1 4 4`
` 1 6 12 8`
` 1 8 24 32 16`
` 1 10 40 80 80 32`
` 1 12 60 160 240 192 64`
` 1 14 84 280 560 672 448 128`
The other way of manufacturing this triangle is to start with Pascal's triangle and multiply each entry by 2k, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal's triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2Position Number = 6 × 22 = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal's triangle. For example, the number of 2-dimensional elements in a 2-dimensional cube (a square) is one, the number of 1-dimensional elements (sides, or lines) is 4, and the number of 0-dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.
To understand why this pattern exists, first recognize that the construction of an n-cube from an (n − 1)-cube is done by simply duplicating the original figure and displacing it some distance (for a regular n-cube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an n-cube, one must double the first of a pair of numbers in a row of this analog of Pascal's triangle before summing to yield the number below. The initial doubling thus yields the number of "original" elements to be found in the next higher n-cube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher n-cube.
In this triangle, the sum of the elements of row m is equal to 3m − 1. Again, to use the elements of row 5 as an example: $1 + 8 + 24 + 32 + 16 = 81$, which is equal to $3^4 = 81$.
### Calculating an individual row
This algorithm is an alternative to the standard method of calculating individual cells with factorials. Starting at the left, the first cell's value is 1. For each cell after, the value is determined by multiplying the value to the left by a slowly changing fraction:
$v\left(c\right) = frac\left\{r-c\right\}\left\{c\right\}$
where r = row + 1, starting with 0 at the top, and c = the column, starting with 0 on the left. For example, to calculate row 5, r=6. The first value is 1. The next value is 1 x 5/1 = 5. The numerator decreases by one, and the denominator increases by one with each step. So 5 x 4/2 = 10. Then 10 x 3/3 = 10. Then 10 x 2/4 = 5. Then 5 x 1/5 = 1. Notice that the last cell always equals 1, the final multiplication is included for completeness of the series.
A similar pattern exists on a downward diagonal. Starting with the one and the natural number in the next cell, form a fraction. To determine the next cell, increase the numerator and denominator each by one, and then multiply the previous result by the fraction. For example, the row starting with 1 and 7 form a fraction of 7/1. The next cell is 7 x 8/2 = 28. The next cell is 28 x 9/3 = 84.
Note that for any individual row you only need to calculate half (rounded up) the number of values in the row. This is because the row is symmetrical.
## Extensions
Pascal's Triangle can be extended to negative row numbers.
First write the triangle in the following form:
m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...
Next, extend the column of 1s upwards:
m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = -4 1 ... n = -3 1 ... n = -2 1 ... n = -1 1 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...
Now the rule:
$\left\{n choose m\right\} = \left\{n-1 choose m-1\right\} + \left\{n-1 choose m\right\}$
can be rearranged to:
$\left\{n-1 choose m\right\} = \left\{n choose m\right\} - \left\{n-1 choose m-1\right\}$
which allows calculation of the other entries for negative rows:
m = 0 m = 1 m = 2 m = 3 m = 4 m = 5 ... n = -4 1 -4 10 -20 35 -56 ... n = -3 1 -3 6 -10 15 -21 ... n = -2 1 -2 3 -4 5 -6 ... n = -1 1 -1 1 -1 1 -1 ... n = 0 1 0 0 0 0 0 ... n = 1 1 1 0 0 0 0 ... n = 2 1 2 1 0 0 0 ... n = 3 1 3 3 1 0 0 ... n = 4 1 4 6 4 1 0 ...
| 6,734
| 23,444
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 23, "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}
| 4.6875
| 5
|
CC-MAIN-2015-27
|
latest
|
en
| 0.874431
|
https://www.jmp.com/support/downloads/JMPG100_documentation/OP_G_CN_0004.html
| 1,642,761,519,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00192.warc.gz
| 868,819,304
| 3,494
|
Output | Copy Number | Correlation and Principal Variance Component Analysis
Correlation and Principal Variance Component Analysis
Running this process for the DrosophilaAgingExample sample setting generates the tabbed Results window shown below. Refer to the Correlation and Principal Variance Component Analysis process description for more information. Output from the process is organized into tabs. Each tab contains one or more plots, data panels, data filters, and so on, that facilitate your analysis.
The Results window contains the following panes:
Tab Viewer
This pane provides you with a space to view individual tabs within the Results window. Use the tabs to access and view the output plots and associated data sets.
The following tabs are generated by this process:
• Variance Components Charts (Correlation and Principal Components): This tab shows both bar and line plots illustrating principal variance components.
• 3D PCA Plot (Correlation and Principal Components): This tab shows an interactive three-dimensional scatterplot of principal components.
• 2D PCA Plots (Correlation and Principal Components): This tab shows a two-dimensional scatterplot matrix of principal components.
• Scree Plot (Correlation and Principal Components) This tab shows a plot of illustrating the proportion of variability explained by principal components.
• Correlation Heat Map (Correlation and Principal Components): This tab shows a clustered heat map and dendrogram of the correlation matrix.
• Correlation Distributions (Correlation and Principal Components): This tab shows the distributions of the correlations and their associated p-values.
• SAS Output : This is a text-based output directly from SAS/STAT PROC PRINCOMP and PROC MIXED. The former provides detailed statistics on the principal components analysis. The latter contains one PROC MIXED run for each principal component to compute the variance component estimates for each. Refer to the documentation for SAS PROC PRINCOMP and PROC MIXED for more information.
Launch Follow-up Process
• Correlation and Grouped Scatterplots: Click Correlation and Grouped Scatterplots to launch the Correlation and Grouped Scatterplots process to generate scatterplot matrices of the raw data. The input data set used here is the preloaded in as input for the Correlation and Grouped Scatterplots process.
General
• Click View Data to reveal the underlying data table associated with the current tab.
• Click Reopen Dialog to reopen the completed process dialog used to generate this output.
• Click Create Report to generate a pdf- or rtf-formatted report containing the plots and charts of selected tabs.
• Click Close All to close all graphics windows and underlying data sets associated with the output.
| 524
| 2,778
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.839651
|
https://web2.0calc.com/questions/plz-help_689
| 1,721,758,278,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00865.warc.gz
| 528,684,030
| 5,512
|
+0
# plz help
0
216
1
Six children are each offered a single scoop of any of 3 flavors of ice cream from the Combinatorial Creamery. In how many ways can each child choose a flavor for their scoop of ice cream so that each flavor of ice cream is selected by at least two children?
Nov 16, 2022
#1
+118623
0
F1, F2, F3 c1, c2, c3, c4, c5, c6
How many ways can the flavours 1,1,2,2,3,3 be soted in a row
6!/(2!2!2!) = 6!/ (2*4) = 3*5*6 = 90
Nov 16, 2022
| 181
| 473
|
{"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
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.902628
|
https://cracku.in/124-the-positions-of-the-first-and-the-sixth-digits-in-x-sbi-clerk-2009-es
| 1,726,011,078,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651323.1/warc/CC-MAIN-20240910224659-20240911014659-00119.warc.gz
| 156,179,375
| 26,340
|
Question 124
# The positions of the first and the sixth digits in the number 5109238674 are interchanged. Similarly the positions of the second and the seventh digits are interchanged and so on.Which of the following will be the third digit from the right end after the rearrangement ?
Solution
Original number: 5109238674
The rule is nothing but to reverse the number.
New number obtained after changing the position of digits as per the given rule is: 4768329015
Now third digit from the right end is 0 which is option 2.
| 121
| 526
|
{"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
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.888312
|
https://www.coursehero.com/file/6015331/PMid2/
| 1,493,458,572,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917123484.45/warc/CC-MAIN-20170423031203-00488-ip-10-145-167-34.ec2.internal.warc.gz
| 884,192,303
| 24,023
|
PMid2 - AMS312 Spring 2010 Practice Midterm #2 &...
This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: AMS312 Spring 2010 Practice Midterm #2 & Solution (***Note: the real midterm will have only 3 questions. Here I provided more to help you review.) 1. Arctic and Alpine Research investigated the relationship between the mean daily air temperature and the cocoon temperature of woolybear caterpillar’s of the High Arctic. (a) According to the data, can you conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature? (b) What assumptions are necessary for the above test? Temperature (ºC) Day Air Cocoon 1 10 15 2 9 14 3 2 7 4 3 6 5 5 10 Solution: By taking the paired differences (Diff) between the cocoon and the air temperatures for each day sampled, this problem reduce to a one-sample t-test on Diff. Temperature (ºC) Day Air Cocoon Diff 1 10 15 5 2 9 14 5 3 2 7 5 4 3 6 3 5 5 10 5 (a). Sample statistics: n = 5, , 6 . 4 = x s = 0.9. Hypotheses: H : μ = 0 versus Ha: μ > 0. Test statistic (observed): 5 . 11 5 / 9 . 6 . 4 / ≈- =- = n s x t Since , 13 . 2 5 . 11 05 . , 4 = ≈ t t we reject H 0 in favor of Ha at the 0.05 significance level. That is, we conclude, at the significance level of 0.05, that the caterpillar’s body temperature is higher than the outside air temperature. (b). The assumption is that the population distribution of “Diff” is normal. 2. Suppose a random sample of size n is drawn from a normal population with mean μ and variance σ 2 , both parameters unknown. Please a. Find the maximum likelihood estimators of μ and σ 2 . b. Find the method of moment estimators of μ and σ 2 . Solution: 1 (a) 2 2 2 2 2 1 1 ( ) 1 2 ( , ) ( ; , ) ( ) n n i i i i x L f x e π μ σ σ μ σ μ σ = =-- = = ∏ ∏ Ln L= 2 2 2 2 2 1 1 1 2 ( ) ( ) ln( 2 ) ln( 2 ) 2 2 ln i i n n i i x x μ μ πσ π σ σ σ = =------- = ∑ ∑ 1 1 1 2 2 2 2( ) 2 ln ( ( 1)) n n i i n i i i i x n x x L μ μ μ σ σ σ μ = = =--- ∂ =-- = = ∂ ∑ ∑ ∑ 2 2 2 2 2 1 2 4 4 4 2 1 1 ( ) ( ) ( ) 1 2 2 2 2 ln ( ) n n n i i i i i i x n x x L μ σ μ μ σ σ σ σ σ σ = = =------ ∂ = + = = ∂ ∑ ∑ ∑ µ 1 ln 0 n i i x n L x μ μ = ∂ = ⇒ = ∂ ∑ = (1) ¶ 2 2 1 2 ( ) ln n i i x n L μ σ σ...
View Full Document
This note was uploaded on 11/14/2010 for the course AMS 312 taught by Professor Zhu,w during the Spring '08 term at SUNY Stony Brook.
Page1 / 5
PMid2 - AMS312 Spring 2010 Practice Midterm #2 &...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 967
| 2,786
|
{"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}
| 3.515625
| 4
|
CC-MAIN-2017-17
|
longest
|
en
| 0.80504
|
http://en.m.wikibooks.org/wiki/Transportation_Geography_and_Network_Science/Graph_theory
| 1,432,525,170,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00239-ip-10-180-206-219.ec2.internal.warc.gz
| 78,794,822
| 19,830
|
# Transportation Geography and Network Science/Graph theory
A drawing of a graph
Graph theory is the study of graphs, mathematical structures used to model pairwise relations between objects from a certain collection. A "graph" in this context refers to a collection of vertices or 'nodes' and a collection of edges that connect pairs of vertices. A graph may be undirected, meaning that there is no distinction between the two vertices associated with each edge, or its edges may be directed from one vertex to another; see here for more detailed definitions and for other variations in the types of graphs that are commonly considered.
Refer here for basic definitions in graph theory.
## ApplicationsEdit
Graphs are among the most ubiquitous models of both natural and human-made structures. They can be used to model many types of relations and process dynamics in physical, biological and social systems. Many problems of practical interest can be represented by graphs.
In computer science, graphs are used to represent networks of communication, data organization, computational devices, the flow of computation, etc. One practical example: The link structure of a website could be represented by a directed graph. The vertices are the web pages available at the website and a directed edge from page A to page B exists if and only if A contains a link to B. A similar approach can be taken to problems in travel, biology, computer chip design, and many other fields. The development of algorithms to handle graphs is therefore of major interest in computer science. There, the transformation of graphs is often formalized and represented by graph rewrite systems. They are either directly used or properties of the rewrite systems (e.g. confluence) are studied. Complementary to graph transformation systems focusing on rule-based in-memory manipulation of graphs are graph databases geared towards transaction-safe, persistent storing and querying of graph-structured data.
Graph-theoretic methods, in various forms, have proven particularly useful in linguistics, since natural language often lends itself well to discrete structure. Traditionally, syntax and compositional semantics follow tree-based structures, whose expressive power lies in the Principle of Compositionality, modeled in a hierarchical graph. Within lexical semantics, especially as applied to computers, modeling word meaning is easier when a given word is understood in terms of related words; semantic networks are therefore important in computational linguistics. Still other methods in phonology (e.g. Optimality Theory, which uses lattice graphs) and morphology (e.g. finite-state morphology, using finite-state transducers) are common in the analysis of language as a graph. Indeed, the usefulness of this area of mathematics to linguistics has borne organizations such as TextGraphs, as well as various 'Net' projects, such as WordNet, VerbNet, and others.
Graph theory is also used to study molecules in chemistry and physics. In condensed matter physics, the three dimensional structure of complicated simulated atomic structures can be studied quantitatively by gathering statistics on graph-theoretic properties related to the topology of the atoms. For example, Franzblau's shortest-path (SP) rings. In chemistry a graph makes a natural model for a molecule, where vertices represent atoms and edges bonds. This approach is especially used in computer processing of molecular structures, ranging from chemical editors to database searching. In statistical physics, graphs can represent local connections between interacting parts of a system, as well as the dynamics of a physical process on such systems.
Graph theory is also widely used in sociology as a way, for example, to measure actors' prestige or to explore w:diffusion mechanisms, notably through the use of w:social network analysis software.
Likewise, graph theory is useful in w:biology and conservation efforts where a vertex can represent regions where certain species exist (or habitats) and the edges represent migration paths, or movement between the regions. This information is important when looking at breeding patterns or tracking the spread of disease, parasites or how changes to the movement can affect other species.
In mathematics, graphs are useful in geometry and certain parts of topology, e.g. Knot Theory. Algebraic graph theory has close links with group theory.
A graph structure can be extended by assigning a weight to each edge of the graph. Graphs with weights, or w:weighted graphs, are used to represent structures in which pairwise connections have some numerical values. For example if a graph represents a road network, the weights could represent the length of each road.
A digraph with weighted edges in the context of graph theory is called a network. Network analysis have many practical applications, for example, to model and analyze traffic networks. Applications of network analysis split broadly into three categories:
1. First, analysis to determine structural properties of a network, such as the distribution of vertex degrees and the diameter of the graph. A vast number of graph measures exist, and the production of useful ones for various domains remains an active area of research.
2. Second, analysis to find a measurable quantity within the network, for example, for a transportation network, the level of vehicular flow within any portion of it.
3. Third, analysis of dynamical properties of networks.
## HistoryEdit
The Königsberg Bridge problem
The paper written by w:Leonhard Euler on the w:Seven Bridges of Königsberg and published in 1736 is regarded as the first paper in the history of graph theory.[1] This paper, as well as the one written by Vandermonde on the knight problem, carried on with the analysis situs initiated by Leibniz. Euler's formula relating the number of edges, vertices, and faces of a convex polyhedron was studied and generalized by Cauchy[2] and L'Huillier,[3] and is at the origin of w:topology.
More than one century after Euler's paper on the bridges of w:Königsberg and while Listing introduced topology, Cayley was led by the study of particular analytical forms arising from w:differential calculus to study a particular class of graphs, the trees. This study had many implications in theoretical w:chemistry. The involved techniques mainly concerned the w:enumeration of graphs having particular properties. Enumerative graph theory then rose from the results of Cayley and the fundamental results published by Pólya between 1935 and 1937 and the generalization of these by De Bruijn in 1959. Cayley linked his results on trees with the contemporary studies of chemical composition.[4] The fusion of the ideas coming from mathematics with those coming from chemistry is at the origin of a part of the standard terminology of graph theory.
In particular, the term "graph" was introduced by Sylvester in a paper published in 1878 in Nature, where he draws an analogy between "quantic invariants" and "co-variants" of algebra and molecular diagrams:[5]
"[...] Every invariant and co-variant thus becomes expressible by a graph precisely identical with a Kekuléan diagram or chemicograph. [...] I give a rule for the geometrical multiplication of graphs, i.e. for constructing a graph to the product of in- or co-variants whose separate graphs are given. [...]" (italics as in the original).
One of the most famous and productive problems of graph theory is the w:four color problem: "Is it true that any map drawn in the plane may have its regions colored with four colors, in such a way that any two regions having a common border have different colors?" This problem was first posed by w:Francis Guthrie in 1852 and its first written record is in a letter of De Morgan addressed to Hamilton the same year. Many incorrect proofs have been proposed, including those by Cayley, Kempe, and others. The study and the generalization of this problem by Tait, Heawood, Ramsey and Hadwiger led to the study of the colorings of the graphs embedded on surfaces with arbitrary genus. Tait's reformulation generated a new class of problems, the factorization problems, particularly studied by Petersen and Kőnig. The works of Ramsey on colorations and more specially the results obtained by Turán in 1941 was at the origin of another branch of graph theory, w:extremal graph theory.
The four color problem remained unsolved for more than a century. In 1969 w:Heinrich Heesch published a method for solving the problem using computers.[6] A computer-aided proof produced in 1976 by w:Kenneth Appel and w:Wolfgang Haken makes fundamental use of the notion of "discharging" developed by Heesch.[7][8] The proof involved checking the properties of 1,936 configurations by computer, and was not fully accepted at the time due to its complexity. A simpler proof considering only 633 configurations was given twenty years later by Robertson, Seymour, Sanders and Thomas.[9]
The autonomous development of topology from 1860 and 1930 fertilized graph theory back through the works of Jordan, Kuratowski and Whitney. Another important factor of common development of graph theory and topology came from the use of the techniques of modern algebra. The first example of such a use comes from the work of the physicist w:Gustav Kirchhoff, who published in 1845 his w:Kirchhoff's circuit laws for calculating the w:voltage and current in w:electric circuits.
The introduction of probabilistic methods in graph theory, especially in the study of Erdős and Rényi of the asymptotic probability of graph connectivity, gave rise to yet another branch, known as random graph theory, which has been a fruitful source of graph-theoretic results.
## Drawing graphsEdit
Main page: Graph drawing
Graphs are represented graphically by drawing a dot or circle for every vertex, and drawing an arc between two vertices if they are connected by an edge. If the graph is directed, the direction is indicated by drawing an arrow.
A graph drawing should not be confused with the graph itself (the abstract, non-visual structure) as there are several ways to structure the graph drawing. All that matters is which vertices are connected to which others by how many edges and not the exact layout. In practice it is often difficult to decide if two drawings represent the same graph. Depending on the problem domain some layouts may be better suited and easier to understand than others.
## Graph-theoretic data structuresEdit
Main page: Graph (data structure)
There are different ways to store graphs in a computer system. The w:data structure used depends on both the graph structure and the w:algorithm used for manipulating the graph. Theoretically one can distinguish between list and matrix structures but in concrete applications the best structure is often a combination of both. List structures are often preferred for w:sparse graphs as they have smaller memory requirements. Matrix structures on the other hand provide faster access for some applications but can consume huge amounts of memory.
### List structuresEdit
w:Incidence list
The edges are represented by an array containing pairs (w:tuples if directed) of vertices (that the edge connects) and possibly weight and other data. Vertices connected by an edge are said to be adjacent.
Much like the incidence list, each vertex has a list of which vertices it is adjacent to. This causes redundancy in an undirected graph: for example, if vertices A and B are adjacent, A's adjacency list contains B, while B's list contains A. Adjacency queries are faster, at the cost of extra storage space.
### Matrix structuresEdit
w:Incidence matrix
The graph is represented by a matrix of size |V | (number of vertices) by |E| (number of edges) where the entry [vertex, edge] contains the edge's endpoint data (simplest case: 1 - incident, 0 - not incident).
This is an n by n matrix A, where n is the number of vertices in the graph. If there is an edge from a vertex x to a vertex y, then the element $a_{x, y}$ is 1 (or in general the number of xy edges), otherwise it is 0. In computing, this matrix makes it easy to find subgraphs, and to reverse a directed graph.
w:Laplacian matrix or w:Kirchhoff matrix or Admittance matrix
This is defined as DA, where D is the diagonal w:degree matrix. It explicitly contains both adjacency information and degree information. (However, there are other, similar matrices that are also called "Laplacian matrices" of a graph.)
w:Distance matrix
A symmetric n by n matrix D whose element $d_{x, y}$ is the length of a w:shortest path between x and y; if there is no such path $d_{x, y}$ = infinity. It can be derived from powers of A
$d_{x,y}=\min\{n\mid A^n[x,y]\ne 0\}. \,$
## Problems in graph theoryEdit
### EnumerationEdit
There is a large literature on w:graphical enumeration: the problem of counting graphs meeting specified conditions. Some of this work is found in Harary and Palmer (1973).
### Subgraphs, induced subgraphs, and minorsEdit
A common problem, called the w:subgraph isomorphism problem, is finding a fixed graph as a w:subgraph in a given graph. One reason to be interested in such a question is that many w:graph properties are hereditary for subgraphs, which means that a graph has the property if and only if all subgraphs have it too. Unfortunately, finding maximal subgraphs of a certain kind is often an w:NP-complete problem.
A similar problem is finding w:induced subgraphs in a given graph. Again, some important graph properties are hereditary with respect to induced subgraphs, which means that a graph has a property if and only if all induced subgraphs also have it. Finding maximal induced subgraphs of a certain kind is also often NP-complete. For example,
Still another such problem, the minor containment problem, is to find a fixed graph as a minor of a given graph. A minor or subcontraction of a graph is any graph obtained by taking a subgraph and contracting some (or no) edges. Many graph properties are hereditary for minors, which means that a graph has a property if and only if all minors have it too. A famous example:
Another class of problems has to do with the extent to which various species and generalizations of graphs are determined by their point-deleted subgraphs, for example:
### Graph coloringEdit
Many problems have to do with various ways of coloring graphs, for example:
### Network flowEdit
There are numerous problems arising especially from applications that have to do with various notions of flows in networks, for example:
### Covering problemsEdit
Covering problems are specific instances of subgraph-finding problems, and they tend to be closely related to the w:clique problem or the w:independent set problem.
### Graph classesEdit
Many problems involve characterizing the members of various classes of graphs. Overlapping significantly with other types in this list, this type of problem includes, for instance:
## NotesEdit
1. Biggs, N.; Lloyd, E. and Wilson, R. (1986), Graph Theory, 1736-1936, Oxford University Press
2. Cauchy, A.L. (1813), "Recherche sur les polyèdres - premier mémoire", Journal de l'Ecole Polytechnique 9 (Cahier 16): 66–86.
3. L'Huillier, S.-A.-J. (1861), "Mémoire sur la polyèdrométrie", Annales de Mathématiques 3: 169–189.
4. Cayley, A. (1875), "Ueber die Analytischen Figuren, welche in der Mathematik Bäume genannt werden und ihre Anwendung auf die Theorie chemischer Verbindungen", Berichte der deutschen Chemischen Gesellschaft 8: 1056–1059, doi:10.1002/cber.18750080252.
5. John Joseph Sylvester (1878), Chemistry and Algebra. Nature, volume 17, page 284. doi:10.1038/017284a0. Online version. Retrieved 2009-12-30.
6. Heinrich Heesch: Untersuchungen zum Vierfarbenproblem. Mannheim: Bibliographisches Institut 1969.
7. Appel, K. and Haken, W. (1977), "Every planar map is four colorable. Part I. Discharging", Illinois J. Math. 21: 429–490.
8. Appel, K. and Haken, W. (1977), "Every planar map is four colorable. Part II. Reducibility", Illinois J. Math. 21: 491–567.
9. Robertson, N.; Sanders, D.; Seymour, P. and Thomas, R. (1997), "The four color theorem", Journal of Combinatorial Theory Series B 70: 2–44, doi:10.1006/jctb.1997.1750.
## ReferencesEdit
• Berge, Claude (1958), Théorie des graphes et ses applications, Collection Universitaire de Mathématiques, II, Paris: Dunod . English edition, Wiley 1961; Methuen & Co, New York 1962; Russian, Moscow 1961; Spanish, Mexico 1962; Roumanian, Bucharest 1969; Chinese, Shanghai 1963; Second printing of the 1962 first English edition, Dover, New York 2001.
• Biggs, N.; Lloyd, E.; Wilson, R. (1986), Graph Theory, 1736–1936, Oxford University Press .
• Bondy, J.A.; Murty, U.S.R. (2008), Graph Theory, Springer, ISBN 978-1-84628-969-9 .
• Chartrand, Gary (1985), Introductory Graph Theory, Dover, ISBN 0-486-24775-9 .
• Gibbons, Alan (1985), Algorithmic Graph Theory, Cambridge University Press .
• Golumbic, Martin (1980), Algorithmic Graph Theory and Perfect Graphs, Academic Press .
| 3,803
| 17,103
|
{"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2015-22
|
latest
|
en
| 0.949333
|
https://www.teacherspayteachers.com/Product/Math-Centers-0-to-20-Addition-Subtraction-Kindergarten-First-Grade-1956963
| 1,686,326,467,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224656737.96/warc/CC-MAIN-20230609132648-20230609162648-00650.warc.gz
| 1,122,118,522
| 42,043
|
EASEL BY TPT
Total:
\$0.00
# Math Centers (0 to 20, Addition, Subtraction) | Kindergarten & First Grade
Rated 4.9 out of 5, based on 4 reviews
4 Ratings
;
Fairy Poppins
17.4k Followers
PreK - 1st, Homeschool
Subjects
Resource Type
Standards
Formats Included
• PDF
Pages
81 pages
Report this resource to TPT
Fairy Poppins
17.4k Followers
Easel Activity Included
This resource includes a ready-to-use interactive activity students can complete on any device. Easel by TPT is free to use! Learn more.
### Description
These number and operations math centers feature a fun button theme! They focus on skills such as counting, reading and writing numbers, addition and subtraction.
Included in this pack:
Printables
• 3 x Number write and wipe mats (0 to 9, 1 to 20 and 1 to 20)
• Playdough mats with ten frames (0 to 20)
• Counting 10 frames and 20 frames
• Number cards (0 to 20) - includes number cards and number words
• Number lines (0 to 10, 0 to 20 and 0 to 12 to play a dice addition game) * Addition twenty frames mats with space to write a sum
• Addition sum cards (20 sums up to 10, 20 sums up to 20)
• Subtraction sum cards (20 sum cards up to 10, 20 sums up to 20)
• Storage labels to organize the resources
Worksheets
• 4 x Number line addition recording worksheets
• 2 x Number line subtraction recording worksheets
• 4 x Shake, spill and add worksheets (addition to 5, 10, 15 and 20)
• Ten frame addition up to 10
• Ten frame addition up to 20
• Ten frame subtraction up to 10
• Ten frame subtraction up to 20
• Number line addition up to 10
• Number line addition up to 20
• Number line subtraction up to 10
• Number line subtraction up to 20
I hope you love this resource!
YOU MAY ALSO LOVE...
JOIN THE FP COMMUNITY!
>>> Instagram
>>> Pinterest
QUESTIONS, ISSUES OR REQUESTS
• Please feel free to contact me via the "Ask a Question" tab. I value your feedback and am happy to help.
Total Pages
81 pages
N/A
Teaching Duration
N/A
Report this resource to TPT
Reported resources will be reviewed by our team. Report this resource to let us know if this resource violates TPT’s content guidelines.
### Standards
to see state-specific standards (only available in the US).
Count forward beginning from a given number within the known sequence (instead of having to begin at 1).
Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20 (with 0 representing a count of no objects).
Understand the relationship between numbers and quantities; connect counting to cardinality.
When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted.
| 731
| 2,887
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.836189
|
bmicalculatorindia.com
| 1,600,920,301,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00565.warc.gz
| 301,085,406
| 28,628
|
# Calculate BMI
0 -
###### Weight
This BMI Calculator for men and women is easy to use and provides your body mass index score based upon the height and weight information you enter.
# Body Mass Index (BMI) – The Complete Guide
The most recent obesity statistics from the World Health Organization (WHO) make grim reading. The WHO published an updated list of statistics in February 2018 and found that global obesity has tripled since 1975. There are now 1.9 billion overweight adults around the world, 650 million of whom are obese.
The terms ‘overweight’ and ‘obese’ are defined as “abnormal or excessive fat accumulation that may impair health” by the WHO. In other words, aesthetics doesn’t enter the equation; carrying too much weight is bad for your health. There are numerous tools used to calculate whether or not a person is overweight. Even with the advances in medical technology, the Body Mass Index (BMI) is one of the most frequently used measurements of a person’s physical shape and general health, some 150 years after it was first conceived.
## 1. What is the Body Mass Index (BMI)?
The body mass index, occasionally called the Quetelet index, is a value taken from a person’s height and weight. It is also determined via a BMI chart table and is an attempt to quantify the level of tissue mass (comprised of bone, fat, and muscle) in a person.
The result is used to determine if that individual is obese, overweight, normal weight or underweight depending on where they fall within the BMI category ranges.
### 1.1 How is BMI Calculated?
The body mass index formula is easy to determine. It’s a simple calculation that takes into account your weight and height.
• The imperial BMI formula = Weight (LBS) x 703 ÷ Height (Inches²)
• The metric BMI formula = Weight (KG) ÷ Height (Metres²)
If you wish to calculate your BMI using the imperial system, here’s the height and weight conversion figures:
• 1 kilogram = 2.2 pounds
• 1 foot = 12 inches
• 1 inch = 2.54 centimetres
If you are 5ft 4 inches by the imperial system, you are 162.56cm or 1.63m (rounded up) by the metric system.
#### Example BMI Calculations
Example 1: This is how to calculate your body mass index score if you were to weigh 209 pounds and are 6 feet 2 inches tall.
• 209 pounds = 95 kilograms
• 6 feet 2 inches = 1.88m
• 1.88 x 1.88 = 3.53
• 95/3.53 = 26.91
The BMI Score in this case = 26.91.
Example 2: Some countries use an entirely different calculation to reach the same result.
Weight in pounds / your height in inches squared x 703
Using the same example as before, you calculate as follows:
• (209 / 74 x 74) x 703 = (209 / 5,476) x 703
The BMI Score in this case = 26.83.
Why the difference? Because there is a certain element of ‘rounding up’ or ‘rounding down’ involved e.g. 26.784 becomes 26.78 and so on.
But where does this place you on the BMI category chart? In the next section we cover the bmi categories and explain the classifications.
### 1.2 Body Mass Index Categories
Technically, the BMI scale will place you in one of six categories. The table below lists the BMI categories along with the BMI score associated with each category.
ClassificationBody Mass Index Score
Underweightless than 18.5
Normal Weight18.5 – 24.9
Overweight25 – 29.9
Obesity (Class 1)30 – 34.9
Obesity (Class 2)35 – 39.9
Extreme Obesity (Class 3)40+
The following body mass index chart allows you to manually get your BMI score and includes a classification of the BMI category you reside within.
### 1.3 BMI for Children
Although the BMI calculator works in the same way for children, the measurement is used differently. While the BMI ranges remain the same for adults throughout their life, these figures change in children because kids are still growing and do so at different rates. As a result, BMI is a measure of weight for height compared to children of the same age. This calculation results in a body mass index percentile.
For instance, if your child is in the 50th percentile, it means that 50% of children of the same age have the same BMI figure or less. The 85th percentile means that 85% of kids have the same or lower BMI and this is a figure where your child is classified as being ‘at risk’ of becoming overweight. At the 95th percentile, your child is classified as ‘overweight’.
Example: If your child is a 14-year-old boy who is 5ft 3 inches tall and weighs 120 pounds for example, his BMI is 21.3 which puts him in the 76th percentile. This BMI figure indicates that he is at a healthy weight because the proper range is between 90 and 128 pounds in this instance.
Here is what that reading looks like when based on BMI percentiles.
### 1.4 BMI Charts
Body Mass Index charts are useful for visualizing the ranges associated with each BMI category. You can use them to easily locate your height and weight to determine your BMI score and the associated BMI category you fit within.
### 1.5 History
The Body Mass Index is sometimes called the Quetelet Index after its creator, a Belgian mathematician, astronomer, sociologist, statistician and all-around genius named Lambert Adolphe Jacques Quetelet. He came up with the idea sometime between 1830 and 1850 as part of his development of ‘social physics’.
Even so, it was seldom used and only started to become popular in the latter part of the 20th century. The term Body Mass Index was first coined by Ancel Keys and other authors of a paper published in the Journal of Chronic Diseases in July 1972. According to the paper, BMI was “at least as good as any other relative weight index as an indicator of relative obesity.” At least Keys acknowledged that it was “not fully satisfactory.”
Prior to the 1980s, physicians used weight for height tables which differed according to gender. It was during this decade that the BMI calculator became the international standard for obesity measurement. Once governments around the world began noting the obesity problem in society, and started launching healthier lifestyle initiatives, the public at large became aware of BMI.
Initially, the threshold for being overweight was 27.8 but it was lowered to 25 internationally. In 1998, the United States National Institutes of Health followed suit and also reduced the BMI figure. This move resulted in 30 million Americans becoming classified as overweight overnight!
## 2. What is the Body Mass Index Used for?
Even though the Body Mass Index is over 150 years old, major health authorities such as the CDC and NIH in the United States still believe it is a ‘fairly reliable indicator of body fatness for most people.’
### 2.1 Public Health Statistics
Even today, BMI is used as the official measure of national obesity rates. For instance, the European Union continues to use it as a yardstick for the obesity epidemic and also suggests that people with higher BMI are at greater risk of diseases such as hypertension, cancer, and cardiovascular disease.
According to the Harvard T.H. Chan School of Public Health, your risk of developing conditions such as type-2 diabetes increases progressively as your BMI rises above 21.
### 2.2 Weight Range Analysis
The BMI range is also considered to be an accurate measure of your weight range. Therefore, if you have a BMI of 27, you are considered overweight regardless of what you actually weigh and your body composition or gender is not taken into account.
### 2.3 Screening Tool for Weight Problems
If a physician calculates a higher than normal BMI, the next step is to see if you are at risk of certain health problems. In fact, the main purpose of BMI is to determine whether you’re likely to develop a serious medical problem later in life.
In France for example, BMI is used as a screening tool for child malnutrition. The CDC champions the BMI scale as an effective means of determining whether children and teenagers are underweight, overweight, or obese.
### 2.4 Fashion Industry
The fashion industry’s obsession with exceedingly thin models has ensured that the BMI calculator has a permanent home there. The industry is constantly under fire for forcing women, in particular, to attain what the UK Women’s Equality Party calls “an unattainable level of thinness in women.” The party has called on all models with a BMI below 18.5 to be seen by a physician from an accredited list who will decide if that woman is healthy enough to work.
There is a law to that effect in Israel where male and female models with a BMI under 18.5 much obtain a medical certificate confirming a ‘normal’ BMI reading. For reference, Kate Moss, one of the most famous models ever, had a BMI of just 15 at the height of her fame. By any measure, including BMI, she was severely thin.
## 3. Global Body Mass Index Statistics vs. India
BMI measurements vary around the globe and some nations have a greater ‘obesity’ problems than others according to the BMI scales. For a better understanding of global BMI scores, we have compiled statistics from six regions globally. First of all, here is the average body mass of people in different continents expressed in kilograms:
RegionAverage Body Mass (KGs)
Asia57.7 kg
Africa60.7 kg
Latin America / Caribbean67.9 kg
Europe70.8 kg
Oceania74.1 kg
North America80.7 kg
Global Average62.0 kg
As you can see, Asians are significantly lighter than Americans for example and while they are shorter on average, it is evident that residents in North America have a higher BMI figure.
### 3.1 BMI Statistics in India
India has one of the lowest BMI averages in the world with 21.7. This is well within normal range and remarkably, only 14 nations have lower average BMIs!
### 3.2 Average BMI of Indian Males
There is little disparity between genders; Indian men have an average BMI of 22.0 which is higher than in approximately 20 countries around the world.
### 3.3 Average BMI of Indian Females
Indian women have an average BMI of just 21.4. Only six nations have a lower figure: Bangladesh, Eritrea, Ethiopia, Sri Lanka, Nepal, and Vietnam.
### 3.4 Obesity Rates in India
As expected, Indians have a far lower rate of obesity than Western nations. Only 11% of men aged 18+ and 15% of women are classified as obese by the BMI scale.
## 4. Possible Limitations of the BMI
The Body Mass Index has its fair share of detractors. It has been derided as a flawed measurement tool. The following 4 points are often highlighted.
### 4.1 Age & Sex
There is a significant difference in body composition between the sexes and this isn’t taken into account by BMI calculators. Male BMI and female BMI measurements should be different because women tend to have a higher percentage of body fat. While men have little more than 2-5% essential body fat, women have 10-13%. Therefore, if a male and a female both have a BMI of 28, they are not equally overweight.
It is sadly also a fact that we lose muscle mass as we get older. Therefore, if you have the same BMI of 23 at age 65 as you did when you were 35, it doesn’t mean you are at a ‘healthy’ weight. You almost certainly have more body fat at 65 so you’re more likely to be overweight.
### 4.2 Physical Characteristics
The BMI calculator also exaggerates obesity in taller people and thinness in shorter people. For example, a two-meter tall man who weighs 104 kilograms is classified as ‘overweight’ according to the BMI scale (104 / 4 = 26).
In contrast, someone who is 1.5m tall and weighs 54 kilograms has a BMI of 24 (54 / 2.25) and is classified as ‘normal’ weight even though they are possibly overweight in real terms.
The fact of the matter is this: There were no calculators or computers during the 19th century so Ouetelet devised a system limited by the age he lived in. It is remarkable that over 150 years later, with so much technology at our fingertips, that medical professionals still use such an outdated system.
### 4.3 Weight vs. Body Fat
It is likely that BMI underestimates obesity, especially in the United States. In America, around one-third of the population are obese by BMI standards but other measurements suggest the true figure is closer to 60%.
Research conducted by Tomiyama et al. and published in the International Journal of Obesity in 2015, looked at the misclassification of cardiometabolic health when using BMI as a measuring stick. The team analyzed a group of participants over a seven-year period and made some startling discoveries. They found that 54 million Americans had been classified as obese or overweight through the Body Mass Index but cardiometabolic measurements showed them to be healthy. Meanwhile, 21 million people were deemed ‘normal’ in BMI terms but were actually unhealthy.
The fact that BMI uses your weight as its primary measuring tool means it is inaccurate. The density of your bone structure alone will throw off BMI calculations so a big-boned individual could wrongly be told that they are obese and at greater risk of medical conditions such as diabetes and stroke.
### 4.4 Athletes & Sports Professionals
BMI is usually NOT used for bodybuilders, sprinters, long distance runners or anyone else classified as a professional athlete or sportsperson. That’s because these individuals tend to have a higher rate of muscle mass which skews the figures. In the case of endurance athletes, a lack of muscle mass can also produce misleading results.
A prime example would be an Olympic sprinter who could have a BMI of 27; the same as an unfit couch potato. According to the BMI scale, both individuals are overweight even though one is a world class athlete and the other is an extremely unhealthy sedentary person.
## 5. Alternative Measures to BMI
### 5.1 Body Fat Percentage
A person’s body fat percentage is considered by some to be a better representation of their overall health than the BMI scale. It is simply a measurement of how much fat you’re carrying.
Example: a 180-pound male with 36 pounds of fat has a body fat percentage of 20%. Simply multiply the amount of fat you have by 100 and divide by your total weight:
36 x 100 = 3600
3600 / 180
= 20%
Unlike the BMI scale, body fat percentage takes into account the differences between men and women. Here is the chart according to the American Council on Exercise:
ClassificationMale Body Fat PercentageFemale Body Fat Percentage
Essential Fat2-5%10-13%
Athletic6-13%14-20%
Fitness14-17%21-24%
Average18-24%25-31%
Obese25%+32%+
As you can see, women tend to get 6-8% leeway on account of holding that much extra in essential fat. The Jackson and Pollock formula uses body fat percentage chart based on gender and age. For example, a 20-year-old male will ideally have 8.5% body fat but this figure rises to 20.9% aged 55.
There is special body fat percentage machines where you just stand on them and allow a mild electrical current to course through your body. You can also use the old-fashioned yet accurate calipers method. Simply measure the skin folds on specific points on the body to come up with a body fat % figure.
If you look at the charts above, you’ll see that a 27-year old woman with a calipers measurement of 12 millimeters is classified as ‘ideal’ with a body fat percentage of around 22-23%.
### 5.2 Waist to Hip Ratio
Other physicians believe that a person’s waist-to-hip ratio is an even better measure of health. According to a study published in PloS Magazine, waist circumstance is strongly linked to type 2 diabetes risk even after BMI is taken into account.
Researchers discovered that a non-obese but overweight man (in BMI terms) with a waist size of 102 cm or more (40.2 inches) has at least the same risk of type 2 diabetes as an obese male. The same situation applies to females with a waist size of 88 cm or more (34.6 inches).
Your waist-to-hip ratio is one of the best indicators of future disease risk because a higher ratio suggests that you have a high level of harmful visceral fat. This is the fat that accumulates around the internal organs and if you have too much of it, the result could be the release of hormones and proteins that lead to inflammation. This in turn damages arteries, enters your liver and impacts how the body breaks down fats and sugars.
ClassificationMale Waist-to-Hip RatioFemale Waist-to-Hip Ratio
Ideal0.80.7
Low Risk0.81 – 0.950.71 – 0.8
Moderate Risk0.96 – 0.990.81 – 0.84
High Risk1+0.85+
Therefore, a man with 40-inch hips should ideally have a 32-inch waist (32 / 40 = 0.8).
A simpler version to judge your waist-to-height ratio is to keep your waist size to less than half of your height to reduce the risk of type 2 diabetes. Therefore, a 6-foot tall man (72 inches) should have a maximum waist measurement of 36 inches.
## 6. The Benefits of Keeping a Healthy Weight
### 6.1 Why Maintaining a Healthy Weight is Important
There is a lot more to keeping your weight in the ‘healthy’ range than merely looking good. It is a critical component of general health as it reduces your risk of developing serious medical conditions. If you’re overweight, it can be as simple as eating less and moving more. If you’re underweight, perhaps you should exercise less or eat a little more.
### 6.2 The Dangers of Being Overweight
Once you enter ‘overweight’ territory on the BMI scale, you will be at greater risk of a host of serious medical conditions.
• #### Cancer
In postmenopausal women, the risk of getting breast cancer increases by 20-40% when overweight according to a study by Munsell et al. published in 2014. Excess abdominal fat increases your risk by 43%.
• #### Infertility
Experts in the field suggest that a BMI of between 20 and 24 is the perfect zone for fertility. Up to 12% of fertility problems stem from weight problems (being overweight or underweight). In women, weight impacts periods and ovulation.
• #### Cardiovascular Conditions
A study published in the Journal of the American College of Cardiology in April 2014 researched almost 15,000 Korean adults with no known case of heart disease. The team discovered that people with a BMI of over 25 were at greater risk of having early plaque buildup in their arteries than people at a ‘normal’ weight.
• #### Type 2 Diabetes
According to WHO research entitled Global Report on Diabetes, up to 422 million adults were living with diabetes in 2014. This is almost quadruple the 1980 figure of 108 million. In that time, the rate of obesity has also risen. Recent research suggests that obesity accounts for up to 85% of the risk of developing type 2 diabetes. If you are obese, you are 80 times more likely to develop diabetes than someone with a BMI of 22 or less.
• #### Less Sleep
A 2012 study by the Johns Hopkins University School of Medicine showed a link between weight loss and better sleep. The six-month study was led by professor of medicine, Kerry Stewart, Ed. D, and involved 77 volunteers with pre-diabetes or type 2 diabetes. All 77 participants were either obese or overweight and were assigned to one of two groups. Group A went on a weight-loss diet and exercise regime. Group B only benefitted from the diet portion. Both groups lost 15 pounds and 15% of belly fat on average. What’s more, both groups improved their overall sleep score by 20%.
• #### Reduced Life Expectancy
This is unquestionably the biggest danger of being overweight or obese. According to the Organization for Economic Co-operation and Development, your risk of death increases by 30% for every 33 pounds of excess weight you carry. A severely obese individual (someone with a BMI of 40+) can expect to live 8-10 years less than a person in the ‘normal’ BMI range.
### 6.3 The Dangers of Being Underweight
With such an emphasis on losing weight, it is easy to overlook the fact that being underweight also increases your risk of declining health. That’s one of the reasons why protest groups are so unhappy with the fashion industry. You fall into the ‘underweight’ category if your BMI is below 18.5.
According to the CDC, as at 2014, an estimated 1.4% of American adults are underweight. However, as this issue isn’t as well researched as obesity, there is a fairly high margin of error in the study. Here are some of the health risks associated with having an extremely low BMI score:
• #### Malnutrition
Contrary to what you may believe, malnutrition does not relate to eating or drinking too little. In reality, it is a term used to describe an insufficient intake of nutrients. If you don’t eat enough to fuel your body, symptoms can include fatigue and hair loss.
• #### Decreased Immune Function
A review of the risk of infection in people with high and low BMI scores was published by Dobner and Kaser in January 2018. The review found that there was a notable connection between being underweight and increased infections. Malnourishment can reduce your immune system strength which means you’re less able to ward off infections and diseases.
• #### Osteoporosis
A 2016 study by Lim and Park found that in premenopausal women, 24% of those with a BMI below 18.5 had low bone mass density. In contrast, only 9.4% of women with a BMI above 18.5 had the same issue.
• #### Increase of Surgical Complications
In one study, underweight people who had total knee replacement surgery were at greater risk of infection afterward than people of a normal weight.
## 7. Body Mass Index – The Final Word
While it is true that there are a few concerns about BMI, it is still the most tried and trusted way to ascertain the average person’s general health. It is one of the quickest and easiest methods of determining whether you need to lose/gain weight and change your lifestyle. We recommend that you find out your BMI as soon as possible and if it is above or below the average range, book an appointment with your physician and seek advice. This simple test could prevent serious health issues going forward.
If your BMI is over 25 or below 18.5, you should conduct a body fat percentage test and also look at your waist-to-hip and waist-to-height ratios. A combination of all these measurements, plus an honest look in the mirror, should help you determine whether you need to lose or gain weight.
What isn’t in doubt is the growing levels of obesity around the world. The average BMI score of dozens of nations is in the overweight category and there are genuine fears that within a generation or two, there will be more obese people than non-obese.
It is a combination of longer working hours, easy access to cheap processed foods, and a lack of desire to exercise that is responsible for the current obesity epidemic. It would behoove nations in Europe, Oceania and North America, to look at the example set by countries in the Far East of Asia such as Japan which has an average BMI of 22.5, almost directly in the center of ‘normal’. Reduced consumption of processed foods and reliance on fresh fish, fruit, and vegetables ensures that Japan is one of the world’s healthiest nations.
For the record, there is unquestionably a link between BMI and wealth globally. Most of the countries with the lowest BMI averages are among the poorest on the planet. They include Bangladesh (20.2), Eritrea (20.2) and Ethiopia (20.3).
Meanwhile, several of the world’s wealthiest countries have the highest BMI averages. They include Kuwait (29.5), USA (28.8), and the United Arab Emirates (28). Unfortunately, when we revisit these statistics in a decade or so, the average BMI of most countries will have increased and at least one will enter the ‘Obese’ zone of 30 as an average.
## Latest Articles
#### Fall Asleep Easily Tonight – Avoid These 6 Foods!
We all know sleep is necessary for a healthy body. Although most of us really try to get enough sleep, studies show that too many of us miss the mark way too…
#### How to Avoid the Sunday Night Work Fear by Eating Healthy
We should enjoy the weekends, right? Unfortunately, starting about mid-day on Sunday, we start to dread the upcoming work week and it tends to ruin our Sundays, whether we realize it or…
#### Top 15 Superfoods for Weight Loss
Superfoods have been around forever. In ancient times, civilizations would use these foods as money, for religious ceremonies and as staples in their diet. However, it hasn’t been until recently that we’ve…
| 5,473
| 24,268
|
{"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-2020-40
|
longest
|
en
| 0.907727
|
http://matematika.reseneulohy.cz/3222/an-implicit-function-of-two-variables
| 1,679,984,229,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00475.warc.gz
| 34,253,649
| 7,136
|
## An implicit function of two variables
For the given relation $$x^2 + 2y^2 + 3z^2 + xy - z - 9 = 0$$.
• #### Variant 1
Prove that this relation defines a smooth function $$z = z(x, y)$$, in some neighborhood $$U$$ of the point $$[1, -2]$$ satisfying $$z(1, -2) = 1$$.
• #### Variant 2
Determine $$\frac{\partial z}{\partial x}$$ and $$\frac{\partial z}{\partial y}$$ in the neighborhood $$U$$.
• #### Variant 3
Write down the equation of the plane tangent to the function $$z$$ at the point $$[1, -2]$$.
| 168
| 512
|
{"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}
| 3.203125
| 3
|
CC-MAIN-2023-14
|
latest
|
en
| 0.680377
|
https://www.mrcodeswildride.com/math/population-density
| 1,720,830,961,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514459.28/warc/CC-MAIN-20240712224556-20240713014556-00647.warc.gz
| 744,733,387
| 3,689
|
## Population Density
Population density is the number of people per unit of area, usually expressed as people per square kilometer (per km2) or people per square mile (per mi2). To calculate population density, divide the number of people in an area by the size of the area.
For example, a city has 10,000 people, and an area of 10 square miles. To calculate the density, do the following:
```people = 10,000
area = 10 square miles
density = 10,000 ÷ 10 square miles = 1,000 per mi2
```
Population density is the measure of how crowded an area is. In the USA, the average population density for a city is 4,300 per mi2. The USA as a whole has a population density of 91 per mi2. The average city has a much higher density than the entire country because cities are much more compact with less empty land.
### Exercise 1 of 6
Sacramento has 530,000 people, and an area of 100 square miles. How many people per square mile is the density?
### Exercise 2 of 6
New York City is the most dense city in the USA, excluding tiny cities that surround New York City. It has 8,325,000 people, and an area of 300 square miles. How many times more dense is New York City than the average city in the USA? Round to the nearest whole number.
### Exercise 3 of 6
Manila in the Philippines is the most dense city in the world. It has 1,840,000 people, and an area of 16 square miles. How many times more dense is Manila than the average city in the USA?
### Exercise 4 of 6
Bangladesh is the most dense country in the world, excluding city-states like Monaco and Singapore. It has 172,952,000 people, and an area of 52,000 square miles. How many times more dense is Bangladesh than the USA?
### Exercise 5 of 6
Mongolia is the least dense country in the world. It has 3,171,000 people, and an area of 604,000 square miles. How many times more dense is the USA than Mongolia?
### Exercise 6 of 6
Greenland is the least dense territory in the world. It has 58,520 people, and an area of 836,000 square miles. How many times more dense is the USA than Greenland?
| 525
| 2,060
|
{"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}
| 3.75
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.931462
|
https://human.libretexts.org/Bookshelves/Music/Music_Theory/Music_Theory_(LibreTexts)/01%3A_The_Elements_of_Rhythm-_Sound_Symbol_and_Time/1.02%3A_Pulse_Tempo_and_Meter
| 1,720,987,566,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00362.warc.gz
| 276,488,743
| 35,455
|
# 1.2: Pulse, Tempo, and Meter
• Anonymous
• LibreTexts
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
##### Learning Objectives
• Definitions of the elements of rhythmic organization.
• Perception of Tempo and commonly used terms.
• Mapping out meter (time signatures): the perception of Simple and Compound Time.
• How these elements interact in music.
We perceive the organization of time in music in terms of three fundamental elements, Pulse, Tempo, and Meter. Use prompts to assist you in understanding these elements:
• Pulse—“beat”: the background “heartbeat” of a piece of music.
• Tempo—“rate”: the relatively fast or slow speed at which we perceive the pulse in a piece of music.
• Meter—“ratio”: how durational values are assigned to represent the pulse are organized in discrete segments in a piece of music.
## Pulse and Tempo
Pulse, or beat, is the regularly recurring underlying pulsation that we perceive that compels music to progress through time. Pulse makes us react kinesthetically to music: in other words, it compels motion. We tap our feet, we dance, we march, or we may just “feel” the pulse internally.
In a piece of music, some durational value is assigned to be the pulse. All other durations are proportionally related to that fundamental background pulse.
Tempo (Latin: tempus-“time”) is the rate (or relative speed) at which the pulse flows through time. This is determined by numerous methods:
1. A metronome marking: for example, MM=120 means the pulse progresses at 120 beats per minute (two beats per second). Often, in practice, the background durational value will be drawn and assigned a metronomic value. (You will sometimes encounter the marking bpm, “beats per minute.”)
Figure 1.15 Metronome Marking and Pulse Marking
2. Around the 17th Century (roughly!), Italian terms came to be used to indicate tempo. These terms were descriptive and therefore rather loosely interpreted as to exact tempo. These terms indicate a narrow “range” of metronomic speeds. For example, the term Andante means “going” or “a walking tempo.” This usually equates to roughly 76 beats per minute, but may be interpreted at a slightly faster or slightly slower pace.
3. In an attempt to refine these terms, to make them more precise, diminutives were added: Andantino indicates a slightly faster pace than Andante. Other modifiers came into common practice as well. For example, Andante con moto (“going, with motion”) is self-explanatory.
Beginning in the 19th Century, composers often used equivalent tempo and performance descriptions in their native languages, or mixed Italianate terms and vernacular terms within the same piece.
4. It is important to understand that the use of these terms exceeded mere indications of relative speed. Often, they also carry the connotation of style or performance practice. For example, Allegro con brio (“lively, with fire or brilliance”) implies a stylistic manner of performance, not merely a rate at which the pulse progresses through time. Chapter 19 lists common terms and their commonly accepted meanings along with some equivalents in other languages.
## Meter and Time Signatures
Meter, expressed in music as a time signature, determines:
1. Which durational value is assigned to represent the fundamental background pulse;
2. How these pulses are grouped together in discrete segments;
3. How these pulses naturally subdivide into lesser durational values, and;
4. The relative strength of pulses (perceived accents) within segments or groupings of pulses.Concerning accentuation of pulse, you will encounter the terms Arsis and Thesis, terms adapted from Hellenistic poetic meter. These have come to mean “upbeat” and “downbeat” respectively. These are nearly slang definitions or, at best, jargon. Arsis is best described as “preparatory,” hence perceived as a relatively weak pulse. Thesis is best described as “accentuated,” hence relatively strong. It is interesting to note that, at various times in the history of music, the meaning of these two terms has been reversed from time to time.
Time signatures consist of two numbers, one over another, placed at the beginning of a composition. They may occur anywhere in a composition where a meter change is required. They are NEVER written as fractions!
## Simple and Compound Meter
To understand meter fully, we must first determine the fundamental nature of the prevailing background pulse or beat. In given meters, we perceive beats as having the potential (or capacity) of being divided in two ways:
1. The prevailing background pulse may be subdivided into two proportionally equal portions. Meters having this attribute are labeled Simple Meter (or Simple time).
2. The prevailing background pulse may be subdivided into three proportionally equal portions. Meters having this attribute are labeled Compound Meter (Compound time).
We name meters according to two criteria:
1. Is it Simple or Compound time?
2. How many prevailing background pulses are grouped together?
So, a time signature wherein (a) the pulse subdivides into two portions, and (b) two pulses are grouped together is called Simple Duple. Three pulses grouped together, Simple Triple and so forth. A time signature wherein (a) the pulse subdivides into three portions, and (b) two pulses are grouped together is called Compound Duple, three pulses, Compound Triple, and so forth.
## Simple Meter
Let us address simple meter first. Analyze this by answering two questions concerning the stated time signature:
1. For the top number: “How many…?” In other words, how many prevailing background pulse values (or their relative equivalent values and/or rests) are grouped together?
2. For the bottom number: “…of what kind?” In other words, what durational value has been assigned to represent the prevailing background pulse?
So the time signature 24 has two quarter-notes grouped together, therefore, we label this as Simple Duple.
In Renaissance music, specialized symbols were employed that were the forerunner of time signatures. These symbols determined how relative durational values were held in proportion to one another. We continue to employ two holdovers from this system.
“Common Time” and “Cut Time,” are slang terms. Other names for “Cut Time” are “March Time” and the proper name, Alla Breve.
## The Time Signature Table
The characteristics of individual time signatures are perceived in multiple layers that can be reduced to three basic levels:
1. The prevailing background Pulse or beat.
2. First Division: the level wherein we determine if the pulse divides into two equal portions (simple meter) or three equal portions (compound meter).
3. Subdivisions: how First Division values subdivide into proportionally smaller values.
Therefore, we can graph time signatures using the following table.
Table 1.1 Time Signature Table
Pulse (The fundamental background pulse.) First Division (The level determining pulse division into two portions or three portions.) Subdivisions (Subsequent divisions into smaller values.)
Use this table to map out time signatures and their component organizational layers.
## Compound Meter
Understanding compound meters is somewhat more complex. Several preparatory statements will assist in comprehension:
1. Compound Meters have certain characteristics that will enable prompt recognition:
1. The upper number is 3 or a multiple of 3.
2. The prevailing background pulse must be a dotted value: remember, in compound meter, the pulse must have the capacity to divide into three equal portions.
3. Subdivisions of the background pulse are usually grouped in sets of three by the use of beams (ligatures).
2. In theory, any Compound Meter may be perceived as Simple Meter,depending upon the tempo:
1. If a tempo is slow enough, any compound time signature may be perceived as a simple meter.
2. In practice, this is limited by style and context in compositions.
3. In Compound Meter, the written time signature represents the level of First Division,not Pulse:
1. In order to find the pulse value in compound time signatures, use the Time Signature Table. List First Division values (the written time signature) in groupings of three.
2. Sum these to the dotted value representing Pulse. List these accordingly in the Table.
As with Simple time signatures, let us employ the same Time Signature Table to graph Compound time signatures. Reviewing Statement 3 above, we will follow a slightly different procedure than that used for graphing Simple Meter:
1. For the Compound Duple time signature 68 list six eighth-notes in two groupings of three in the First Division row:
Figure 1.21 Compound Meter, First Division Groupings
2. Next, sum these groupings of three into dotted values (“two eighth-notes equal a quarter-note, the additional quarter-note represented by a dot”); list the two resulting dotted quarter-notes in the Pulse row:
Figure 1.22 Sum to Find Compound Pulse Value
3. Lastly, draw subdivisions of the First Division values in the Subdivision row:
Figure 1.23 Subdivision
Below are typical compound meters and their respective labels.
Note that Simple meters divide all values into two subdivisions in each level of the Table. Compound meters divide the First Division level into three (see Statement 1 above). Subsequent subdivisions divide into two.
## Simple Triple Interpreted as Compound Meter
Some Simple Triple time signatures may be perceived as either simple or compound, again depending upon tempo. In practice, this is a limited list: The time signatures:
316 38 34
may be perceived as Simple Triple if the tempo is relatively slow. In other words, you perceive the “lower number” of the time signature as the fundamental background pulse value. As the tempo for any of these becomes relatively faster, we cease to perceive the lower number as Pulse. Instead we perceive the lower number as the First Division of a Compound meter.
The Time Signature Table will show this:
In the next section, these fundamental elements of sound, symbol, and time will be placed in full musical context by uniting them with common notational practices.
### Conclusion
The student should be able to define and understand:
• Pulse (“beat”), Tempo (“rate”), and Meter (“ratio”).
• Simple Meter: recognizing and analyzing Simple Time Signatures.
• Compound Meter: recognizing and analyzing Compound Time Signatures.
• Time Signatures that may be perceived as either Simple or Compound and why they are so perceived.
• Using the Time Signature Table as a tool for graphing Time Signatures.
##### Exercise 1:
1. Using the Time Signature Table, map out all examples of:
1. Simple Duple and Compound Duple.
Pulse First Division Subdivisions
2. Simple Triple and Compound Triple.
Pulse First Division Subdivisions
Pulse First Division Subdivisions
Note: At the Subdivision level, draw one layer of subdivisions only.
2. Using the Time Signature Table map out the following time signatures as both Simple and Compound Meters:
1. (3/16)
Pulse First Division Subdivisions
2. (3/8)
Pulse First Division Subdivisions
3. (6/8)
Pulse First Division Subdivisions
4. (9/8)
Pulse First Division Subdivisions
3. In class (or some group), practice tapping a slow beat with your left foot. Against that beat tap two equal (“even”) divisions with your right hand (simple division). Next, keeping that same slow beat in your left foot, practice tapping three equal (“even”) divisions with your right hand (compound division). Lastly, switch hands and feet. Good luck.
4. The following exercises alternate between simple duple and compound duple. Tap these rhythms while keeping the same constant background pulse. Practice each segment separately at first: then practice in sequence, switching from simple to compound time as you go.
Figure 1.26 Rhythm Drill
This page titled 1.2: Pulse, Tempo, and Meter is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.
| 4,258
| 16,392
|
{"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}
| 3.71875
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.195241
|
https://fe.uni-lj.si/en/predmeti/mathematics-i-2/
| 1,721,357,613,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00088.warc.gz
| 213,293,341
| 19,969
|
# Mathematics I
## Subject description
Number systems (positive integers, rational numbers, real numbers, complex numbers). Sequences (accumulation point, limit, boundedness). Series (convergence, convergence tests, harmonic series, alternating series). Functions of one real variable (domain of definition, image, oddness and evenness, injectivity, surjectivity, bijectivity, composition, inverse function, elementary functions, limit , continuity). Derivative of a function (derivation rules, geometric interpretation, differential, applications). Integral of a function (indefinite integral, definite integral, applications of definite integral).
## Objectives and competences
To master the basic concepts of mathematical analysis and to be able to better understand them. To develop analytical thinking and careful and exact mathematical reasoning as well as other building blocks of science/mathematics/engineering literacy.
## Teaching and learning methods
Lectures, tutorials, and individualized homework. Collective analysis, interpretation, and solving of technical problems.
## Expected study results
After successful completion of the course, students should be able to:
• solve basic problems of mathematical analysis, including sequences, series, real functions, derivatives and integrals,
• compute the derivatives of elementary functions and integrals of some classes of functions,
• identify and understand problems in differently structured environments/contexts,
• identify, analyse and use mathematical tools to solve natural science and engineering authentic problems,
• appropriately display/communicate and critically evaluate the solution procedure and the obtained results,
• develop exactness, consistency and diligence in communication, thinking and work,
• demonstrate an appropriate attitude towards engineering practice and science.
## Basic sources and literature
1. G. Dolinar, Matematika 1, Založba FE in FRI, 2010.
2. P. Šemrl, Osnove višje matematike 1, DMFA-založništvo, 2009.
3. M. Akveld, R. Sperb, Analysis I, vdf Hochschulverlag, ETH Zürich, 2009.
4. G. B. Thomas: Thomas' Calculus, Pearson Education, 2005.
5. B. Jurčič-Zlobec, N. Mramor Kosta: Zbirka nalog iz Matematike I, Založba FE in FRI, 2009.
6. G. Dolinar, U. Demšar: Rešene naloge iz Matematike I za VSP, Založba FE in FRI, 2004.
Spletna učilnica eFE https://e.fe.uni-lj.si
## Stay up to date
University of Ljubljana, Faculty of Electrical Engineering Tržaška cesta 25, 1000 Ljubljana
| 587
| 2,502
|
{"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}
| 3.265625
| 3
|
CC-MAIN-2024-30
|
latest
|
en
| 0.830858
|
https://rainydaymum.co.uk/autumn-apple-match-and-sort-early-maths-activity/
| 1,702,150,121,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00210.warc.gz
| 519,571,067
| 85,382
|
Home > Maths Games and Activities for Kids > Autumn Apple Match and Sort Early Maths Activity
Autumn Apple Match and Sort Early Maths Activity
Are you looking for a simple idea to start teaching your child number sense and maths? Then this apple match and sort activity using mini-erasers is a great activity that you can set up and explore alongside your child. So here you go apple match and sort.
Matching and Sorting Activities
Did you know that before children count they need to recognise objects that are the same and different?
Matching them together and then sorting them out.
Knowing what are the same leads to grouping them and sorting them which then can be identified as having more or less of something than others.
AUTUMN TODDLER & PRESCHOOL PLAY CHALLENGE
30 Days of FREE ideas for fun this autumn to do with your little ones
Apple Manipulatives
We have included affiliate links to some of the products and resources as an associate we may earn from qualifying purchases.
Whether you are having an apple week, focusing on the letter A or an autumn/fall-themed month including some apple manipulatives for number work and maths is a great idea.
We have used a set of apple mini-erasers that our friend in the USA sent us but a set like these Learning Resources Apples (US Amazon, UK Amazon) would also work as well. You could also create some paper ones but younger toddlers and preschoolers may find them difficult to pick up.
Apple Match and Sorting Activity
Depending on the age of your child you may want to start with 2 or 3 different variations of the apples.
You could use red vs green, or red, green, and yellow apples then it's time to match.
Check out even more Apple Activities for Learning with Preschoolers here!
Toddlers and Young Preschoolers
With your little ones lets match the pairs, pick out 2 matching apples from your collection i.e. 2 red and 2 green.
Then lay them out and ask your little one to find the matching ones. Use language like
• Which 2 are the same?
• Can you find the one that matches this one?
• Is there another one like this?
As they get the idea you can then introduce the 3rd one and they can match the 3 pairs together.
By starting out just changing the colour and only using 2 to match the pairs you have a simple activity that you can focus on doing alongside your little one.
Preschooler Apple Match
Once they have the hand of matching the pairs introduce more and find all the ones that are the same.
Hold up one and ask to find the same, the ones that match and find the odd one out.
Apple Sorting for Preschoolers
You will find that they naturally start to move the apples that match into piles of the same this is sorting.
A small basket, even a picture of a basket may help them place them.
As they start to sort them into piles start with equal numbers - they are all the same.
Then vary them so some piles have More than and some have Less than others make it really noticeable at first and slowly get it closer and closer to equal.
Moving on from Apple Matching and Sorting
If your confident that your child gets the matching and sorting activities then it's time to work on numbers.
Our FREE Apple Tree Playdough Counting Mats are ideal for the next step and although we've used playdough you can easily use the same apple manipulatives that we used.
You can also make your own math tray for your little ones. We made these apple tree math centres to work on counting out numbers 1 to 5 on our apple trees and kept it out all week.
Our Apple Counting Book Recommendation
Want to read alongside this activity then we recommend this Ten Red Apples by Pat Hutchins (US Amazon, UK Amazon) as a great apple and counting book to share with your toddler and preschoolers.
More Apple Activities for Toddlers and Preschoolers
Have a look at some of these other fun ideas for apples to play and learn with your toddlers and preschoolers
Author
Cerys Parker
Cerys is a marine biologist, environmental educator, teacher, mum, and home educator from the UK. She loves getting creative, whether it is with simple and easy crafts and ideas, activities to make learning fun, or delicious recipes that you and your kids can cook together you'll find them all shared here on Rainy Day Mum.
This site uses Akismet to reduce spam. Learn how your comment data is processed.
| 931
| 4,362
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2023-50
|
longest
|
en
| 0.94372
|
http://math.stackexchange.com/questions/387519/how-do-i-find-the-image-of-the-functions-y-2-and-y-2x-6/387525
| 1,469,815,355,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257831770.41/warc/CC-MAIN-20160723071031-00009-ip-10-185-27-174.ec2.internal.warc.gz
| 164,602,463
| 22,309
|
# How do I find the image of the functions $y=2$ and $y = 2x - 6$?
The function is $y=2$, the domain is just 2? And the image of it? I don't think I quiet understand what the image of a function means, the domain is all values that it can assume, correct? Could you please try to define the image of this equation too: $y = 2x - 6$, so I can try to understand it? Thank you.
-
## migrated from mathematica.stackexchange.comMay 10 '13 at 12:21
This question came from our site for users of Mathematica.
This really depends on the context of the question. I'm going to make the assumption here that you're discussing functions from the real numbers to the real numbers. Strictly speaking, the domain of the function should be given explicitly when you define the function. However, often one defines a function and then says that the domain is the set of all real numbers for which the function is defined. In this interpretation, the domain of your function $y=f(x)=2$ is all of $\mathbb{R}$ and its image is $\{2\}$. Similarly, the domain of $y=f(x)=2x-6$ is all of $\mathbb{R}$ and its image is $\mathbb{R}$.
To give you a different example, the domain of the function $y=f(x)=\frac{1}{x-1}$ is $\mathbb{R} \setminus \{1\}$ and the image is $\mathbb{R} \setminus \{0\}$.
-
The "\" in your answer means "all the real numbers BUT 1", right? Yes, I think I understood it now, thank you very much. So, if the domain of a function is equal to all real numbers then it's image can only be all real numbers as well, correct? – Luan Cristian Thums May 10 '13 at 12:40
Yes, the $\setminus$ means subtraction of sets. But your second statement is incorrect. Note that your first example has a domain of $\mathbb{R}$ but an image of $\{2\}$. Another example: the domain of $f(x)=x^2$ is $\mathbb{R}$, but its image is the set of all nonnegative real numbers. – Alistair Savage May 10 '13 at 12:45
@LuanCristianThums No! $f(x)=y=2$ for your very own question is a counter-example. It evaluates to 2 for every real number $x$, so it's domain is $\mathbb{R}$, but it's image is $\{2\}$. – fgp May 10 '13 at 12:45
@LuanCristianThums Btw, if you had read my answer, you'd have realized that... :-( – fgp May 10 '13 at 12:46
Yes, I understand the logic now, thank you! I was still incorrectly thinking that as it doesn't show a "$x$" it's domain would be just 2 but, yes, just because it "doesn't show" the $X$ I can assume any real number to it and it's image will continue to be $2$. – Luan Cristian Thums May 10 '13 at 12:48
When you define a function you should always provide a domain (i.e. a start set) and a co-domain (i.e. a target set). Writing $$f:A\rightarrow B$$ means that $A$ is the domain and $B$ the co-domain. $f$ is a function if it "associates" every single element $a\in A$ with a unique element $b\in B$. We call such $b$ the image of $a$ and denote it by $f(a)$. Then the image of $f$ is the set of all images.
When talking about real valued functions such as $f(x)=2x-6$, it is implied that the domain is the largest subset of $\mathbb R$ for which $f$ is defined (e.g. the domain of $f(x)=\sqrt x$ is $\{x\geq 0\}$), and that the co-domain is $\mathbb R$.
Now, regarding your examples, $y=2$ is short for the (real valued) function $f(x)$ defined as $$\begin{array}{crcl} f: & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2 \end{array}$$ (note that it is well defined for every $x\in\mathbb R$, so that by the convention above the domain of $f$ is the whole $\mathbb R$). The image of $f$ is just the set $\{2\}$, since no other values of the co-domain are "reached" by $f$.
Concerning $f(x)=2x-6$, since no specification is given, as before we assume $$\begin{array}{crcl} f: & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2x-6 \end{array}$$ i.e. that the domain is the largest subset of $\mathbb R$ for which the function is defined (so, again the whole $\mathbb R$) and that the co-domain is the whole $\mathbb R$. Now, for example the image of the element 5 is 4, since $f(5)=4$. It is easy to show that the image of $f$ is the whole co-domain $\mathbb R$ (in which case we say that $f$ is surjective, which is not the opposite of injective!). In general, you can find the image of $f$ by "turning $y=f(x)$ in something like $x=\ldots$" where the rhs is an expression that does not contain the variable $x$, and then see for which values of $y$ the equation has sense and the rhs is part of the domain. In this example, $$y=2x -6 \Rightarrow 2x=y+6 \Rightarrow x= \frac{y+6}2$$ and the last expression makes sense (i.e. can be computed) for every $y\in\mathbb R$. We then conclude that the image of $f$ is the whole $\mathbb R$.
If we "restricted" the function as follows: $$\begin{array}{crcl} f: & [0,+\infty) & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2x-6 \end{array}$$ then this time finding its image would result in $$x= \frac{y+6}2 \quad\text{and}\quad x\geq 0$$ since now the domain in which $x$ lies is $[0,+\infty)$, i.e. $$\frac{y+6}2\geq 0 \Leftrightarrow y\geq -6$$ Therefore, in this case the image of $f$ would be $[-6,+\infty)$.
I hope this example makes it clear.
-
Very nice answer. – Alistair Savage May 10 '13 at 13:32
The image of a function is simply the set of all possible values the function can take. If you have a function $f(x)$, and evaluate it for all possible values of $x$ (i.e., for all $x$ in the domain of f), then the resulting set of values is called the image of $f$.
You can also find a image of a certain subset of the domain - i.e., instead of evalulating $f(x)$ for all $x$ in the domain, you only evaluate for a certain subset.
Formally, the image $f(U)$ of some subset $U$ of the domain of $f$ is $$f(U) = \left\{f(x) \,:\, x \in U\right\} \text{.}$$
To check whether $y$ lies in the image of $f$, you thus have to ask
Is there an $x$ such that $f(x)=y$?
To check whether $y$ lies in the image of some subset $U$ of the domain of $f$, you similarly ask
Is there an $x \in U$ such that $f(x)=y$?
The image all sets $U$ which aren't empty under $f(x)=2$ is thus simply $\{2\}$, because $f$ never takes any value other than 2.
-
An image is a subset of the co-domain with respect to a certain pre-image, which is a subset of the domain. For the function $y=2x-6$, for example, given a pre-image of $[2,10]$, the image is $[-2,14]$. For the function $y=2$, since any input value in the domain will result in $y=2$, besides the null set, the only possible pre-image is ${2}$, and the only possible image is $2$.
-
As it was already answered here, if I have none pre-image given then I should assume any real number as a image for $y=2x-6$, right? – Luan Cristian Thums May 10 '13 at 12:45
@LuanCristianThums If no pre-image is given, you can assume that the pre-image is the entire domain of the function. And if no domain for the function is given, you can assume it as allistair savage stated. – Ataraxia May 10 '13 at 12:49
The value of f when applied to x is the image of x under f. y is alternatively known as the output of f for argument x. Now Y=2 is a constant function. The image of Y is always 2 for any value of x. So the Image of function Y=2 is the set containing element 2. Image of 2, under Y=2x−6, is the set containing the element 2(2)-6=4-6=-2.
-
| 2,204
| 7,289
|
{"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}
| 4.15625
| 4
|
CC-MAIN-2016-30
|
latest
|
en
| 0.91049
|
https://cboard.cprogramming.com/c-programming/158121-errors-math-tutor.html?s=b698c214dc02292c46664a476e286868
| 1,603,254,068,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107875980.5/warc/CC-MAIN-20201021035155-20201021065155-00646.warc.gz
| 259,486,113
| 14,787
|
# Thread: Errors in Math "tutor"
1. ## Errors in Math "tutor"
Hi!,
I am a beginner and I am trying to create a program that will randomly generate two numbers from 1 to 10, and an operator. The program will ask for the user to enter the answer. The program is giving me these errors:
Code:
``` error C2065: 'num2' : undeclared identifier
error C2143: syntax error : missing ';' before 'type':eek: ```
Here is the Code
Code:
```#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<time.h>
int main(){
int ans;
int ans1;
srand((unsigned int)time(NULL));
int num1 = rand() % 10 + 2;
int num2 = rand() % 10;
int operation = rand() % 4 + 1 ;
printf("\tMATH TUTOR\n");
switch(operation){
case 1:
printf("What is %d + %d ?", num1, num2);
ans = num1 + num2;
scanf_s("%d", &ans1);
if(ans == ans1){
printf("Correct!");
}else{
printf("Incorrect! Try Again! Expected answer %d but you typed %d\n", ans, ans1);
do{
scanf_s("%d", &ans1);
if (ans1==ans){
printf("Correct!");
break;
}
}while( ans1 != ans);
}
break;
case 2:
printf("What is %d - %d ?", num1, num2);
scanf_s("%d",&ans1);
ans = num1 - num2;
if(ans == ans1){
printf("Correct!");
}else{
printf("Incorrect! Try Again! Expected answer %d but you typed %d\n", ans, ans1);
do{
scanf_s("%d", &ans1);
if (ans1==ans){
printf("Correct!");
break;
}
}while( ans1 != ans);
}
break;
case 3:
printf("What is %d * %d ?", num1,num2);
scanf_s("%d",&ans1);
ans = num1 * num2;
if(ans == ans1){
printf("Correct!");
}else{
printf("Incorrect! Try Again! Expected answer %d but you typed %d\n", ans, ans1);
do{
scanf_s("%d", &ans1);
if (ans1==ans){
printf("Correct!");
break;
}
}while( ans1 != ans);
}
break;
case 4:
printf("What is %d / %d ?", num1, num2);
scanf_s("%d",&ans1);
ans = num1 / num2;
if(ans == ans1){
printf("Correct!");
}else{
printf("Incorrect! Try Again! Expected answer %d but you typed %d\n", ans, ans1);
do{
scanf_s("%d", &ans1);
if (ans1==ans){
printf("Correct!");
break;
}
}while( ans1 != ans);
}
break;
default:
printf("nothing");
break;
}
_getch();
return 0;
}```
2. Since you're using the outdated Microsoft compiler, you'll need to insure your variables are all declared before any of them are used. C90 requires all varialbes to be defined at the beginning of the scope and before any calculations or function calls.
Jim
3. Originally Posted by jimblumberg
Since you're using the outdated Microsoft compiler, you'll need to insure your variables are all declared before any of them are used. C90 requires all varialbes to be defined at the beginning of the scope and before any calculations or function calls.
Jim
Is there a way to change the compiler in VS2010? and I think the variable are defined on top. Did it work in your IDE?
4. Originally Posted by eLg
and I think the variable are defined on top.
you are wrong here
Code:
```int ans;
int ans1;
srand((unsigned int)time(NULL));
int num1 = rand() % 10 + 2;
int num2 = rand() % 10;
int operation = rand() % 4 + 1 ;```
This should be
Code:
```int ans;
int ans1;
int num1;
int num2;
int operation;
srand((unsigned int)time(NULL));
num1 = rand() % 10 + 2;
num2 = rand() % 10;
operation = rand() % 4 + 1 ;```
5. Is there a way to change the compiler in VS2010?
No, if you want a different compiler you'll need to download a different IDE, made by someone else. Mircrosoft doesn't support anything other than C90 with their comilers.
Did it work in your IDE?
No, you're program won't compile in my compiler but for other reasons. My compiler doesn't support the non-standard conio.h header file. And it also doesn't support the non-standard scanf_s() function. But after fixing these issues the code does properly complie with my C11 compliant compiler.
Jim
6. Originally Posted by jimblumberg
No, if you want a different compiler you'll need to download a different IDE, made by someone else. Mircrosoft doesn't support anything other than C90 with their comilers.
Strictly speaking - you can use different compilers with Microsoft VS IDE
About compilers -they claim that they do not plan to add support of C99 standard as is, but will support "features of C99 standard that are part of C++11 standard"
Does it mean that you have to compile your C code as C++ to get these features - I have no idea...
7. Originally Posted by vart
you are wrong here
Code:
```int ans;
int ans1;
srand((unsigned int)time(NULL));
int num1 = rand() % 10 + 2;
int num2 = rand() % 10;
int operation = rand() % 4 + 1 ;```
This should be
Code:
```int ans;
int ans1;
int num1;
int num2;
int operation;
srand((unsigned int)time(NULL));
num1 = rand() % 10 + 2;
num2 = rand() % 10;
operation = rand() % 4 + 1 ;```
What is wrong in this part?
If I removed unsigned int, it will give the same number.
If I did not removed unsigned int, it will say undeclared variables.
8. Originally Posted by jimblumberg
No, if you want a different compiler you'll need to download a different IDE, made by someone else. Mircrosoft doesn't support anything other than C90 with their comilers.
No, you're program won't compile in my compiler but for other reasons. My compiler doesn't support the non-standard conio.h header file. And it also doesn't support the non-standard scanf_s() function. But after fixing these issues the code does properly complie with my C11 compliant compiler.
Jim
What changes did you do? What is your IDE? Which IDE do you suggest for beginners in C programming?
And about the scanf_s, VS 2010 won't allow me to compile the program if it is scanf().
9. Originally Posted by eLg
What is wrong in this part?
Not in this part. In the following statement
and I think the variable are defined on top.
You have variable definitions after the part marked in red.
And so they are not defined "on top" but after some other code.
10. Originally Posted by eLg
VS 2010 won't allow me to compile the program if it is scanf().
It will - if you read the error statement and define the suggested macro in your project settings
11. Originally Posted by vart
Strictly speaking - you can use different compilers with Microsoft VS IDE
About compilers -they claim that they do not plan to add support of C99 standard as is, but will support "features of C99 standard that are part of C++11 standard"
Does it mean that you have to compile your C code as C++ to get these features - I have no idea...
if I will compile my code as C++, does that mean that I have to change the name of the libraries I included?
12. What changes did you do?
I removed the conio.h include file and getch(). Then I also changed the non-standard scanf_s() to the standard scanf().
What is your IDE? Which IDE do you suggest for beginners in C programming?
For only C on Windows Pelles C is a good choice. If you need both C and C++ I recommend Code::Blocks with gcc. If you only need C++ then Visual C++ is an acceptable choice in my opinion.
Jim
13. Originally Posted by vart
Not in this part. In the following statement
You have variable definitions after the part marked in red.
And so they are not defined "on top" but after some other code.
oh ok. If I place the srand() above all the codes, all the variable will be undefined, according to VS
14. Originally Posted by eLg
oh ok. If I place the srand() above all the codes, all the variable will be undefined, according to VS
So you should (to be compliant with C90)
| 1,957
| 7,312
|
{"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-2020-45
|
latest
|
en
| 0.418671
|
https://historicsweetsballroom.com/a-predefined-formula-that-performs-calculations-by-using-specific-values/
| 1,657,136,519,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00725.warc.gz
| 337,580,773
| 6,137
|
### Introduction
A function is a predefined formula that performs calculations using particular values in a certain order. All spreadsheet programs include usual functions that can be supplied for easily finding the sum, average, count, maximum value, and minimum value for a selection of cells. In order to use attributes correctly, you'll require to understand the various parts that a function and how to create arguments to calculate values and cell references.
You are watching: A predefined formula that performs calculations by using specific values
The parts of a function
In stimulate to occupational correctly, a duty must be composed a particular way, i beg your pardon is called the syntax. The an easy syntax for a function is one equals authorize (=), the function surname (SUM, because that example), and also one or much more arguments. Arguments contain the information you desire to calculate. The function in the example listed below would include the values of the cell selection A1:A20.
Working through arguments
Arguments have the right to refer come both individual cells and cell varieties and should be enclosed within parentheses. You can include one discussion or lot of arguments, depending on the syntax compelled for the function.
For example, the duty =AVERAGE(B1:B9) would calculation the average of the values in the cell range B1:B9. This role contains only one argument.
Multiple arguments must be separated through a comma. Because that example, the function =SUM(A1:A3, C1:C2, E2) will add the worths of every cells in the three arguments.
### Using functions
There space a variety of functions. Here are few of the most typical functions you'll use:
SUM: This duty adds all the values of the cell in the argument.AVERAGE: This duty determines the average that the values included in the argument. It calculates the amount of the cells and also then divides that worth by the variety of cells in the argument.COUNT: This function counts the number of cells with numerical data in the argument. This function is advantageous for quickly counting items in a cell range.MAX: This function determines the highest cell value had in the argument.MIN: This duty determines the lowest cell value had in the argument.To usage a function:
In our instance below, we'll use a basic role to calculate the average price per unit for a perform of newly ordered items using the mean function.
Select the cell that will certainly contain the function. In our example, we'll select cell C11.
Type the equals sign (=) and also enter the preferred function name. In our example, we'll type =AVERAGE.
Enter the cell range for the argument inside parentheses. In our example, we'll kind (C3:C10). This formula will include the worths of cells C3:C10 and then divide that worth by the total variety of cells in the selection to recognize the average.
Press Enter on your keyboard. The duty will be calculated, and also the result will show up in the cell. In our example, the mean price per unit of items ordered to be \$15.93.
Your spreadsheet will not constantly tell you if your duty contains one error, for this reason it's as much as you come check all of your functions. To learn exactly how to carry out this, check out the Double-Check your Formulas lesson.
### Working with unfamiliar functions
If you desire to learn just how a role works, you can start inputting that function in a blank cell to check out what it does.
See more: How To Make Pizza Rolls In The Microwave ? Cheese Pizza Rolls
### Understanding nested functions
Whenever a formula includes a function, the function is normally calculated before any type of other operators, favor multiplication and division. That's because the formula treats the entire role as a solitary value—before it can use that worth in the formula, it demands to run the function. For example, in the formula below, the SUM role will be calculated before division:
Let's take a look at a more facility example that provides multiple functions:
=WORKDAY(TODAY(),3)
Here, we have actually two various functions functioning together: the WORKDAY role and the today function. These are recognized as nested functions, due to the fact that one role is placed, or nested, within the arguments of another. As a rule, the nested function is always calculated first, similar to parentheses room performed very first in the bespeak of operations. In this example, the TODAY function will it is in calculated first, due to the fact that it's nested in ~ the WORKDAY function.
### Other typical functions
There are many various other functions you deserve to use to conveniently calculate various things through your data. Learning exactly how to use other features will enable you to solve complex problems through your spreadsheets, and we'll be talking an ext about them transparent this tutorial. You can additionally check the end our articles below to discover about details functions:
| 991
| 4,971
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.375
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.849761
|
https://www.urbanpro.com/cbse-class-5-maths-operations-subtraction
| 1,628,222,205,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00068.warc.gz
| 1,086,929,831
| 11,925
|
# Free Class 5 Subtraction Worksheets
Download free printable Subtraction Worksheets to practice. With thousands of questions available, you can generate as many Subtraction Worksheets as you want.
## Sample Class 5 Subtraction Worksheet Questions
1.
A milk dairy produced 60400 liters of milk in a day. It supplied 17570 liters to one town and 18485 liters to another town. How much milk was left in the dairy?
2.
There are 65498 bags of rice in a godown. If 56467 bags are sold. How much bags are left in the godown?
3.
The cost of a steel Amirah is Rs. 8970 and the cost of a wooden Amirah is Rs. 500 less than steel Amirah. Find the cost of wooden Amirah.
4.
Mary bought a computer for Rs. 43555 and a mobile that was priced Rs. 6810 less than the computer. How much she paid for the mobile?
5.
There are 55400 books in the school library. If 7836 books are issued to the students. How many books remained in the library?
6.
What must be added to 5432873 to make 7653589 ?
7.
Fill in the blanks: 65515 - ________ = 60015
8.
The sum of two numbers is 3148654. If one of the number is 1267542, find the other number.
9.
By how much is 532674 less than 601535?
10.
The sum of two number is 1360405. If one of the numbers is 782435. Find the other number.
Worksheets by UrbanPro
Our worksheets are designed to help students explore various topics, practice skills and enrich their subject knowledge, to improve their academic performance. Designed by Experts who have extensive experience and expertise in teaching a subject, these worksheets will improve your child's problem-solving skills and subject knowledge in a fun and interactive manner.
Check out our free customized worksheets across school boards, grades, subjects and levels of subject knowledge. You can download, print and share these worksheets with anyone, anywhere, anytime!
Get a custom worksheet to practice!
Select your topic & see the magic.
subject
Select Chapter(s)
Chapters & Subtopics
| 467
| 1,970
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.125
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.944864
|
https://www.tutorialgateway.org/python-program-to-calculate-square-of-a-number/
| 1,726,719,222,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651981.99/warc/CC-MAIN-20240919025412-20240919055412-00483.warc.gz
| 933,147,965
| 16,242
|
# Python Program to Calculate Square of a Number
Write a Python Program to Calculate the square of a Number using Arithmetic Operators and Functions with an example.
## Python Program to Calculate Square of a Number
This Python program allows the user to enter any numerical value. Next, Python finds the square of that number using an Arithmetic Operator
```# Python Program to Calculate Square of a Number
number = float(input(" Please Enter any numeric Value : "))
square = number * number
print("The Square of a Given Number {0} = {1}".format(number, square))```
Python Square of a Number output
`````` Please Enter any numeric Value : 9
The Square of a Given Number 9.0 = 81.0``````
## Python Program to find Square of a Number Example 2
This Python square of a number example is the same as above. However, this time, we are using the Exponent operator.
```number = float(input(" Please Enter any numeric Value : "))
square = number ** 2
print("The Square of a Given Number {0} = {1}".format(number, square))```
`````` Please Enter any numeric Value : 10
The Square of a Given Number 10.0 = 100.0``````
## Python Program to find Square of a Number using Functions
In this Python program example, we are defining a function that returns the square of a number.
```def square(num):
return num * num
number = float(input(" Please Enter any numeric Value : "))
sqre = square(number)
print("The Square of a Given Number {0} = {1}".format(number, sqre))```
| 362
| 1,481
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.687666
|
https://www.mathworks.com/matlabcentral/cody/problems/5-triangle-numbers/solutions/192904
| 1,511,443,695,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806832.87/warc/CC-MAIN-20171123123458-20171123143458-00189.warc.gz
| 787,481,504
| 11,571
|
Cody
# Problem 5. Triangle Numbers
Solution 192904
Submitted on 17 Jan 2013 by Edgar Guevara
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
%% n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
%% n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
%% n = 30; t = 465; assert(isequal(triangle(n),t))
| 165
| 489
|
{"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-2017-47
|
latest
|
en
| 0.606023
|
https://eduhawks.com/tag/buys/
| 1,624,507,585,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623488550571.96/warc/CC-MAIN-20210624015641-20210624045641-00209.warc.gz
| 208,105,296
| 26,002
|
Categories
## William buys a basket of lemons on sale for \$7 before tax. The sales tax is 16%. What is the total price William pays for the basket of lemons?
William buys a basket of lemons on sale for \$7 before tax. The sales tax is 16%. What is the total price William pays for the basket of lemons?
Categories
## Christine buys 7 hamburgers and 3 sodas for \$27.95. Laura buys 5 hamburgers and 4 sodas for \$23.40. How much does a hamburger cost? A. 1 .85 B. \$3.20 C. \$6.80 D. \$6.55 E. \$3.50
Christine buys 7 hamburgers and 3 sodas for \$27.95. Laura buys 5 hamburgers and 4 sodas for \$23.40. How much does a hamburger cost? A. 1 .85 B. \$3.20 C. \$6.80 D. \$6.55 E. \$3.50
Categories
## Carina buys a set of business checks from her bank. The checks cost her \$26.73. She gives the teller \$30.23. How much change should Carina get back?
The basis to respond this question are:
1) Perpedicular lines form a 90° angle between them.
2) The product of the slopes of two any perpendicular lines is – 1.
So, from that basic knowledge you can analyze each option:
a.Lines s and t have slopes that are opposite reciprocals.
TRUE. Tha comes the number 2 basic condition for the perpendicular lines.
slope_1 * slope_2 = – 1 => slope_1 = – 1 / slope_2, which is what opposite reciprocals means.
b.Lines s and t have the same slope.
FALSE. We have already stated the the slopes are opposite reciprocals.
c.The product of the slopes of s and t is equal to -1
TRUE: that is one of the basic statements that you need to know and handle.
d.The lines have the same steepness.
FALSE: the slope is a measure of steepness, so they have different steepness.
e.The lines have different y intercepts.
FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.
f.The lines never intersect.
FALSE: perpendicular lines always intersept (in a 90° angle).
g.The intersection of s and t forms right angle.
TRUE: right angle = 90°.
h.If the slope of s is 6, the slope of t is -6
FALSE. – 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is – 1/6.
So, the right choices are a, c and g.
Categories
## Allstocks.com, a web seller that buys furniture, clothing, electronics, and more from a variety of producers at less-than-regular wholesale prices and then charges customers less than retail, is a(n) ________.
The Contribution Margin per unit (CM) can be calculated
from the difference of Selling Price per unit (SP) and Total Expenses per unit
(TE).
First, let’s calculate the value of SP:
SP = Sales / Units sold
SP = \$1,043,400 / 22,200 units sold
SP = \$47
Second, calculate all expenses:
Direct materials per unit = \$234,948 / 27,970 units
manufactured = \$8.4
Direct labor per unit = \$131,459 / 27,970 units
manufactured = \$4.7
Variable manufacturing overhead per unit = \$240,542 / 27,970
units manufactured = \$8.6
Variable selling expenses per unit = \$113,220 / 22,200
units sold = \$5.1
TE = \$26.8
Therefore the CM is:
CM = SP – TE
CM = \$47 – \$26.8
CM = \$20.2 per unit
Categories
## Joe buys a set of books that cost \$45.58. He gives the cashier three (3) \$20 bills and three (3) pennies. How much change should Joe get back?
Step-by-step explanation:
The slope of a line passing through two points P(a,b) and Q(c,d) is given by :-
The given points : (-4, 3) and (-4, 7)
Then , the slope of a line passing through two points (-4, 3) and (-4, 7) is given by :-
Since , the parallel have the same slope .
Therefore, the slope of the line that is parallel to line m is undefined.
Categories
## A cabinet maker buys 3.5 liters of oak varnish. The varnish cost \$4.95 per liter. What is the total cost of 3.5 L of varnish ?
A cabinet maker buys 3.5 liters of oak varnish. The varnish cost \$4.95 per liter. What is the total cost of 3.5 L of varnish ?
Categories
## A local supermarket sells chicken fir \$2.49/lb and pork for \$3.19/lb.Todd buys “c” pounds of chicken and “p” pounds ofpork which of the following inequalities represents that Todd only has \$40 to spend?
C. Yes, because the population values appear to be normally distributed.
Step-by-step explanation:
Given is a graph which shows the distribution of values of a population
The graph has the following characteristics
i) Bell shaped
ii) symmerical about mid vertical line
iii) Unimodal with mode = median =mean
iv) As x deviates more from the mean probability is decreasing and also curve approaches asymptotically the x axis
Hence we find that the curve is a distribution of normal
Option C is right
C. Yes, because the population values appear to be normally distributed.
Categories
## Brad wants to buy flowers for his friend with \$15. The daisies are \$1 each and the roses are \$3 each. He buys 3 more daisies than roses how much did the roses cost?
127=7 (mod n) means when 127 is divided by n, the division leaves a remainder of 7.
The statement is equivalent to
120=0 (mod n), meaning that n divides 120.
All divisors of 120 will satisfy the statement because 120 divided by a divisor (factor) will leave a remainder of 0.
Factors of 120 are:
n={1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}, |n|=16.
You can count how many such values of n there are, and try to check that each one satisfies 127=7 mod n.
Categories
## Shaneese is buying peanuts and cashews. She has \$12 to spend. Peanuts cost \$3 per pound and cashews cost \$4 per pound. She buys x pounds of peanuts and y pounds of cashews. Which equation represents the amount of the mix?A. 3y+4x=12B. 7xy=12C. 3x+4y=12D. 3x+4y=12c
Shaneese is buying peanuts and cashews. She has \$12 to spend. Peanuts cost \$3 per pound and cashews cost \$4 per pound. She buys x pounds of peanuts and y pounds of cashews. Which equation represents the amount of the mix?A. 3y+4x=12B. 7xy=12C. 3x+4y=12D. 3x+4y=12c
Categories
## James wants to go to a concert with his friends. The tickets to the show cost \$10 each. If James buys x tickets at a cost of c dollars, represent c as a function of x.
James wants to go to a concert with his friends. The tickets to the show cost \$10 each. If James buys x tickets at a cost of c dollars, represent c as a function of x.
Categories
## Vactin corp., a construction company, buys ten truckloads of cement for its new construction project. given this information, the cement bought is an example of a(n) _____.
Vactin corp., a construction company, buys ten truckloads of cement for its new construction project. given this information, the cement bought is an example of a(n) _____.
Categories
## Vincento buys 500 shares of common stock in water reclamation services, inc. as a shareholder of record, vincento owns a proportionate interest in terms of
Vincento buys 500 shares of common stock in water reclamation services, inc. as a shareholder of record, vincento owns a proportionate interest in terms of
Categories
## Sophia has \$20. she buys a hair band for \$3.99, gum for \$1.29, and a brush for \$6.75. not including tax estimate how much change she should receive????? PLEASE HELP
28.47º of latitude, 7.35º of longitude
Step-by-step explanation:
The latitude and longitude are simple imaginary lines (horizontal for latitude and vertical for longitude) that divide the Earth from an angle point of view. When used as a combination of both, you can get an exact location in any part of the world. For the 2 references, there a 0º divisions that make one side positive angles and negative on the other side (in the case of latitude, positive angles are above the equator line and for longitude positive angles are from the right of the Greenwich Meridian).
With the above mentioned we can observe that Hong Kong and Jakarta are away 22.27º and 6.2º respectively from the equator. So, the total degrees in latitude wil simply be the sum of the 2 absolute values (222.7º+6.2º) giving as a result of 28.47º.
On the other hand, we have in longitude that both cities are several degrees away from the same reference, and because they are from the right side of the Greenwich Meridian, the net change will be the substraction of 114.5-106.8º, giving as a result a total of 7.35º.
Categories
## Hamburger sells for 3 pounds for 6\$. If Samantha buys 10 pounds of hamburger, how much will she pay?
Hamburger sells for 3 pounds for 6\$. If Samantha buys 10 pounds of hamburger, how much will she pay?
Categories
## The granola summer buys used to cost six dollars per pound but it has been marked up 15% a how much did it cost summer to buy 2.6 pounds of granola at the old price be how much does it cost her to buy 2.6 pounds of granola at the new price see suppose summer buys 3.5 pounds of granola how much more does it cost new pricing at the old price
The granola summer buys used to cost six dollars per pound but it has been marked up 15% a how much did it cost summer to buy 2.6 pounds of granola at the old price be how much does it cost her to buy 2.6 pounds of granola at the new price see suppose summer buys 3.5 pounds of granola how much more does it cost new pricing at the old price
Categories
## Neela is buying shirts that are on sale at two for the price of one. the original price of each shirt was s dollars. she buys 4 shirts for \$24. which equation shows an equality between two different ways of writing how much neela spends on the shirts?
Answer: The answers is (B) equal areas.
Step-by-step explanation: Given that two triangles have equal perimeters.
As shown in the attached figure, let us consider two right-angles triangles, ΔABC and ΔDEF, with sides AB = 3 cm, BC = 4 cm, AC = 5 cm, DE = 4 cm, EF = 3 cm and DF = 5 cm.
So the perimeters of both the triangles = 3 + 4 + 5 = 4 + 3 + 5 = 12 cm.
Since volume term is not valid in case of triangles, so they cannot have equal volumes. Therefore, option (A) is incorrect.
Area of ΔABC is
and area of ΔDEF is
Therefore, they may have equal areas and so option (B) is correct.
If the triangles have equal bases, then the heights will also be equal and both the triangles will be same. Similar is the case with equal heights. So, options (C) and (D) are incorrect.
Thus, the correct option is (B). equal areas.
Categories
## A retailer buys a toy for \$3.60 and marks up the price 120% what is the retail price of the toy
Selling price = \$1.75 per unit
Step-by-step explanation:
Let x is the selling price of one gift bag.
So selling price of 395000 units will be = \$39500x
Since fixed costs for producing the bag are \$184000 and variable cost per unit is = \$1.20
Therefore, variable cost of 395000 units will be = \$1.20×395000
= \$474000
Now the cost price of 395000 units will be = Fixed cost + variable cost
= 184000 + 474000
= \$658000
Now we know profit = Selling price – cost price
32400 = 395000x – 658000
395000x = 32400 + 658000
395000x = 690400
x =
x = \$1.75
Therefore. \$1.75 per unit will be the selling price of glitter gift bags.
| 2,961
| 11,022
|
{"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-2021-25
|
latest
|
en
| 0.905897
|
https://x-engineer.org/graduate-engineering/embedded-systems/microcontrollers/logical-operations-boolean-function/
| 1,623,625,850,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487611089.19/warc/CC-MAIN-20210613222907-20210614012907-00508.warc.gz
| 1,001,745,190
| 14,272
|
# Logical operations and boolean functions
Logic operations also known as Boolean functions, part of Boolean algebra, are widely used in computer science, engineering and mathematics. There are different words and expression used for them, like logic gates or bitwise operations, but the main principle is the same: performing logical operations on bits (`0` and `1` values).
Electronics is now part of nearly every engineering domain, therefore it’s very important that engineers have a minimum understanding of logic, bitwise operations.
Most of the physical calculations are performed with decimal numbers. This is because we use decimal numbers for all the physical values (e.g. 10 A, 250 Nm, 120 km, etc.). Computers use binary numbers to perform calculations. To recall how to convert a decimal number in a binary number, read the article Decimal to Binary Conversion.
In parallel with arithmetic operations (addition, subtraction, multiplication, division), there are also logic operations. These are used to evaluate if a logical expression is either `true` or `false`.
In our examples we are going to use two symbols A and B, called inputs. Each one of them can have either a `true` value (`1`) or a `false` value (`0`). After a logic operations is going to be performed on the inputs, we are going to obtain a result, with the symbol Q, called output. Similar with the inputs, the output Q can only have a `true` (`1`) or a `false` (`0`) value.
The logical state/value `true`, also called `HIGH`, is equivalent with the binary value `1`. The logical value `false`, also called `LOW`, is equivalent with the binary value `0`.
`true` `HIGH` `1` `false` `LOW` `0`
The most common logic operations (also called gates, operators) are:
• NOT
• AND
• OR
• NAND
• NOR
• XOR
• XNOR
Each operation has a symbol assigned (block diagram) and a truth table. The symbol is used for designing graphical diagrams of logical operations. There are different standards for the symbols, the most common being ANSI (American National Standards Institute) and IEC (International Electrotechnical Commission).
The truth table is defining how the logic (logical) operation works, what is the value of the output Q, function of the value of the inputs A and B.
### NOT gate
The logic operation NOT is also called an inverter or negation, because it is inverting the boolean value of the input. For example, if A is `true`, applying a NOT operation to it will give the result Q as `false`. In the same way, if A is `false`, applying a NOT gate to it will give the result Q as `true`.
Logic gate ANSI symbol IEC symbol Truth table NOT input output A Q=NOT A 0 1 1 0
### AND gate
The logic operation AND will return a `true` value only if both inputs have a `true` value. Otherwise, if one or both inputs contain a `false` value, the AND gate will output a `false` value. We can say that the AND gate effectively finds the minimum between two binary inputs.
Logic gate ANSI symbol IEC symbol Truth table AND input output A B Q=A AND B 0 0 0 0 1 0 1 0 0 1 1 1
### OR gate
The logic operation OR will return a `true` value if at least one of the inputs has a `true` value, and a `false` value if none of the inputs has a `true` value. We can say that the OR gate effectively finds the maximum between two binary inputs.
Logic gate ANSI symbol IEC symbol Truth table OR input output A B Q=A OR B 0 0 0 0 1 1 1 0 1 1 1 1
### NAND gate
The logic operation NAND gate (negative/not AND) produces a `false` output only if all its inputs are `true`. The NAND gate can be regarded as a complement of the AND gate. If one or both inputs are `false`, the NAND gate outputs a `true` result.
Logic gate ANSI symbol IEC symbol Truth table NAND input output A B Q=A NAND B 0 0 1 0 1 1 1 0 1 1 1 0
### NOR gate
The logic operation NOR (negative/not OR) produces a `true` output only when both inputs are `false`, otherwise produces a `false` output. With other words, if only one or both inputs are `true`, the NOR operator outputs a `false` result. The NOR gate is the result of the negation of the OR operator.
Logic gate ANSI symbol IEC symbol Truth table NOR input output A B Q=A NOR B 0 0 1 0 1 0 1 0 0 1 1 0
### XOR gate
The XOR logical operator (pronounced exclusive OR) gives a `true` output only when the inputs have different states. If the inputs have the same logical states, both either `true` or `false`, the XOR gate outputs a `false` result. To output a `true` result, only one of the inputs has to be `true`, the other one must be `false`.
Logic gate ANSI symbol IEC symbol Truth table XOR input output A B Q=A XOR B 0 0 0 0 1 1 1 0 1 1 1 0
### XNOR gate
The XNOR logical operator (pronounced exclusive NOR) is the logical complement of the XOR gate. A `true` output is result if the inputs have the same logical state (either both `true` or both `false`). If the inputs have different logical values, the XNOR gate outputs a `false` result.
Logic gate ANSI symbol IEC symbol Truth table XNOR input output A B Q=A XNOR B 0 0 1 0 1 0 1 0 0 1 1 1
All the above logic operators (gates) are summarized in the table below.
A B AND OR NAND NOR XOR XNOR 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1
For any questions or observations regarding this tutorial please use the comment form below.
Don’t forget to Like, Share and Subscribe!
| 1,410
| 5,373
|
{"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}
| 4.53125
| 5
|
CC-MAIN-2021-25
|
latest
|
en
| 0.906019
|
https://education-academia.github.io/BBA/business-administration/research-methodology/multi-stage-sampling-commonly.html
| 1,696,355,531,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511170.92/warc/CC-MAIN-20231003160453-20231003190453-00228.warc.gz
| 254,772,310
| 5,785
|
RESEARCH METHODOLOGY
Question [CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
Multi stage sampling commonly
A combination basic sampling carry out in stages B subdividing population into smaller C Random sampling at each stage D Higher sampling erorr
Explanation:
Detailed explanation-1: -What is multistage sampling? In multistage sampling, or multistage cluster sampling, you draw a sample from a population using smaller and smaller groups at each stage. This method is often used to collect data from a large, geographically spread group of people in national surveys, for example.
Detailed explanation-2: -The concept of multistage random sampling technique is similar to multistage cluster sampling. But in this case, the researcher chooses the samples randomly at each stage. Here, the researcher does not create clusters, but he/she narrows down the sample by applying random sampling.
Detailed explanation-3: -Our resulting sample would contain 15, 000 total households: Sample = 15 states * 10 counties * 100 households = 15, 000 households. What is this? This method of obtaining this sample is known as multistage sampling.
Detailed explanation-4: -In multistage cluster sampling, rather than collect data from every single unit in the selected clusters, you randomly select individual units from within the cluster to use as your sample. You can then collect data from each of these individual units – this is known as double-stage sampling.
There is 1 question to complete.
| 309
| 1,494
|
{"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
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.923449
|
https://www.hindawi.com/journals/physri/2010/103538/
| 1,713,139,489,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00157.warc.gz
| 736,744,751
| 93,956
|
#### Abstract
We performed a two-variable canonical transformation on the time momentum operator, and without loss of generality we carried out a three-variable transformation on the coordinate and momentum space operators to trivialize the Hamiltonian operator of the system. Fortunately, this operation separates the time-coordinate and space coordinate naturally, and the wave function of the time-dependent Harmonic Oscillator is evaluated via the generator.
#### 1. Introduction
The method of Canonical transformations (CT) has proved to be a fruitful approach in treating quantum systems [1]. It is often used in describing systems with Hamiltonian that are quadratic either in co-ordinate and momentum or equivalently in Boson creation and annihilation operators. One major advantage of the method of (CT) consists in reducing the Hamiltonian of the system to a Hamiltonian of some simple system with known solution, that is, . These two systems are canonically equivalent since their Hamiltonian is related by means of a CT. However, analytical approaches to isolate conserved quantities for a given physical system are a major objective in the realm of Hamiltonian theory [2]. For autonomous, where the Hamiltonian is time independent, one conserved quantity is immediately found: the Hamiltonian which represents the total energy of the system is a constant of motion.
Rather unfortunately the Hamiltonian of most real physical systems is explicitly time-dependent and does not provide direct conserved quantity. In classical mechanics, the equation of motion governed the dynamical behavior of the system and this second-order differential equation is solved directly by the trivialization of the conjugate variables through CT [3]. Park [3] asserted that because of the non-commutability of operators, CT has not been fully realized in quantum mechanics. However, CT is usually defined as a change of the noncommutating variables that preserved the commutation relations in quantum mechanics [4].
In literature [4, 5] three types of transformations have been identified such as interchange, similarity, and point transformations. Different models of oscillators have been evaluated via the CT [6, 7]. The applications of quantum canonical transformation to a harmonic oscillator in which the angular frequency and equilibrium position are time dependent have been reviewed [3]. In this paper we extend the work of [3] by performing a three-variable transformation on both the space co-ordinate and momentum co-ordinate operators.
#### 2. Time-Dependent Harmonic Oscillator
The Time-Dependent Schrodinger Wave equations have been solved explicitly by the method of Quantum Canonical Transformation (QCT). The QCT is also used in solving time-dependent frequency [3]. However, the solutions of time-dependent harmonic oscillator have been obtained through various methods including invariant operator [8], Path Integral [9, 10], and the space-time transformation [11, 12]. We write the Schrödinger equation to be solved as where
with and as the momentum operators in time and in the co-ordinate. It is worthy to note that the angular frequency and the equilibrium position are all functions of time.
We make time-variable similarity transformation on the momentum time-coordinate and three-variable transformations on space momentum operators as And the Hamiltonian Operator becomes The condition for the differential in (4) to be cancelled is And we solve (5) by series method as
where is the expansion coefficient and and are arbitrary constant.
Substituting (6) into (5) yields Imposing boundary condition on (7), we obtain and (7) becomes
And we recaste (8) in the form where has been absorbed into Now multiplying the first term of the LHS of (9) by and replacing the magnitude by and replacing , (9) becomes
and (10) is the Lagrangian of the system whose classical equation is The result of (10) shows that the cancellation of the differential in (4) generates the Lagrangian describing the harmonic oscillator of the system. Another condition of (4) is the cancellation of quadratic term in co-ordinate and these yields however (12) is a first-order nonlinear equation and can be linearized by setting and (12) becomes
Solving (14) gives where is the phase factor. Putting (15) into (13) leads to
With (16) and (9), we can determine the generator.
The second Canonical transformation is carried out this time to remove the quadratic term in momentum and the associated differential in co-ordinate:
Substituting (17) into (4) yields We eliminate the quadratic term in momentum by setting
and solving (19), we obtain Similarly, the differential in co-ordinate is eliminated as
with being known, and we determined (21) as
where and are arbitrary constants and
The next transformation is carried out to eliminate the linear term in position and the associated differential in co-ordinate:
and the Hamiltonian becomes
The condition for the linear term in and the differential to be cancelled leads to the solution of as
and the as
where is an arbitrary constant.
Another two- andthree-variable transformation is carried out on the Hamiltonian to eliminate the linear term in momentum and the associated differential in co-ordinate:
and the Hamiltonian becomes
The linear term in in (28) is eliminated if and solving (29) yields
and the condition for the differential in co-ordinate to cancelled yields
However (31) is a second-order differential Equation and we rewrite (31) in the form
or
Multiplying (33) by gives
and comparing (34) with (35) we have
This shows that (34) is a Bessel equation with . The general solution of (33) becomes
where is an integer, , are arbitrary constant, and and respectively, represent Bessel functions of first and second kind.
Again a point transformation is done to eliminate the co-ordinate in the Hamiltonian of (28) [3]:
Substituting (37) into (28) will be
In order to trivialize the Hamiltonian of (38), we carried out a two-variable similarity transformation: and (38) becomes
The Schrodinger equation associated with the trivialized Hamiltonian of (40) is
where being the time-dependent Wave function originates from the differential in the canonical transformed Hamiltonian Operators.
The complete wave function is obtained as
The first four terms of (42) are the times Independent Wave Function arising naturally from the cancellation of the differential in co-ordinate in the Hamiltonian Operator while the other terms are the time-dependent wave function gotten as the consequences of the cancellation in the co-ordinates and momentum.
Solving (42) completely yields the result as [3] without the ambiguity of which they encountered because is obtained directly from the cancellation of the differentials in the Hamiltonian.
We obtained after a simple algebra the unnormalized wave function from (11) as
where is the normalization constant and is the Hermite polynomial of order
#### 3. Conclusion
The introduction of a three-variable CT on the co-ordinate and the momentum space removed the ambiguity encountered in [3] in evaluating the time-independent Schrödinger wave function. In (41), we obtain the generalized wave function of the time dependent harmonic oscillator with the first four terms arising from the introduction of the three canonical variables. These terms were simply given as in [3] without its explicit values as we have obtained above. Rather fortunately the three-variable transformations separate naturally the time-dependent and time-independent function explicitly. This separation aids us to solve the Schrödinger equation completely by trivializing the time-dependent Hamiltonian Operator. In addition our result can be viewed as a generalized and explicit case of that obtained in [3].
The extension of this work to different time dependent systems with different potential is possible. Thus canonical transformation is a good tool in solving time-dependent systems.
#### Acknowledgment
The authors acknowledge the Nandy-Leabio Foundation for partial financial support in this work.
| 1,605
| 8,119
|
{"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-2024-18
|
latest
|
en
| 0.935328
|
https://www.geeksforgeeks.org/python-number-to-words-using-num2words/
| 1,713,222,018,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00847.warc.gz
| 738,308,531
| 46,405
|
# Python | Number to Words using num2words
Last Updated : 18 Feb, 2023
num2words module in Python, which converts number (like 34) to words (like thirty-four). Also, this library has support for multiple languages. In this article, we will see how to convert number to words using num2words module. Installation One can easily install num2words using pip.
pip install num2words
Consider the following two excerpts from different files taken from 20 Newsgroups, a popular NLP database. Pre-processing 20 Newsgroups effectively has remained to be a matter of interest.
In article, Martin Preston writes: Why not use the PD C library for reading/writing TIFF files? It took me a good 20 minutes to start using them in your own app. ISCIS VIII is the eighth of a series of meetings which have brought together computer scientists and engineers from about twenty countries. This year’s conference will be held in the beautiful Mediterranean resort city of Antalya, in a region rich in natural as well as historical sites.
In the above two excerpts, one can observe that the number ’20’ appears in both numeric and alphabetical forms. Simply following the pre-processing steps, that involve tokenization, lemmatization and so on would not be able to map ’20’ and ‘twenty’ to the same stem, which is of contextual importance. Luckily, we have the in-built library, num2words which solves this problem in a single line. Below is the sample usage of the tool.
## Python
from num2words import num2words # Most common usage. print(num2words(36)) # Other variants, according to the type of article. print(num2words(36, to = 'ordinal')) print(num2words(36, to = 'ordinal_num')) print(num2words(36, to = 'year')) print(num2words(36, to = 'currency')) # Language Support. print(num2words(36, lang ='es'))
Output:
thirty-six
thirty-sixth
36th
zero euro, thirty-six cents
treinta y seis
Time complexity: O(1).
Auxiliary space: O(1).
Therefore, in the pre-processing step, one could convert ALL numeric values to words for better accuracy in the further stages. References: https://pypi.org/project/num2words/
Previous
Next
| 500
| 2,125
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.888756
|
http://cs.nyu.edu/courses/fall10/G22.2112-001/index.html
| 1,500,568,281,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549423269.5/warc/CC-MAIN-20170720161644-20170720181644-00167.warc.gz
| 76,359,867
| 2,032
|
# Computer Science
This course is the same as the course G63.2043.001 SCIENTIFIC COMPUTING
The course description below is copied from the math department webpage:
http://math.nyu.edu/courses/course_descriptions.html#G63.2043.001
3 points. Fall term. Wednesday, 5:10-7:00, A. Rangan.
Prerequisites: undergraduate multivariate calculus and linear algebra. Programming experience strongly recommended but not required.
This course is intended to provide a practical introduction to computational problem solving.
The order in which the subject material will be covered will approximate the following outline:
- The notion of well conditioned and poorly conditioned problems, with examples drawn from linear algebra
- The concepts of forward and backward stability of an algorithm, with examples drawn from floating point arithmetic and linear-algebra
- Basic techniques for the numerical solution of linear and nonlinear equations, and for numerical optimization, with examples taken from linear algebra and linear programming
- Principles of numerical interpolation, differentiation and integration, with examples such as splines and quadrature schemes
- An introduction to numerical methods for solving ordinary differential equations, with examples such as multistep, Runge Kutta and collocation methods, along with a basic introduction of concepts such as convergence and linear stability
- An introduction to basic matrix factorizations, such as the SVD, along with basic techniques for computing matrix factorizations, with examples such as the QR method for finding eigenvectors
- Basic principles of the discrete/fast Fourier transform, with applications to signal processing, data compression and the solution of differential equations.
This is not a programming course but programming in homework projects with Matlab/Octave and/or C is an important part of the course work.
As many of the class handouts are in the form of Matlab/Octave scripts, students are strongly encouraged to obtain access to and familiarize themselves with these programming environments.
No single textbook is required for this class, but there are several optional texts which are recommended:
Numerical Linear Algebra, David Bau III & Lloyd N. Trefethen, SIAM, 2000;
Scientific Computing with MATLAB and Octave, Alfio M. Quarteroni & Fausto Saleri, Springer, 2006 (available electronically through the library);
An Introduction to Programming and Numerical Methods in MATLAB, Stephen R. Otto & James P. Denier, Springer, 2005 (available electronically through the library)
| 500
| 2,563
|
{"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-2017-30
|
latest
|
en
| 0.918381
|
https://www.shaalaa.com/question-bank-solutions/graph-maxima-minima-f-x-1-x-2-2_46295
| 1,571,115,272,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986655864.19/warc/CC-MAIN-20191015032537-20191015060037-00255.warc.gz
| 1,069,534,297
| 11,432
|
Share
# F(X) = 1 X 2 + 2 . - CBSE (Arts) Class 12 - Mathematics
#### Question
f(x) = $\frac{1}{x^2 + 2}$ .
#### Solution
$\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{1}{x^2 + 2}$
$\Rightarrow f'\left( x \right) = \frac{- 2x}{\left( x^2 + 2 \right)^2}$
$\text { For the local maxima or minima, we must have }$
$f'\left( x \right) = 0$
$\Rightarrow \frac{- 2x}{\left( x^2 + 2 \right)^2} = 0$
$\Rightarrow x = 0$
Now, for values close to x = 0 and to the left of 0,
$f'\left( x \right) > 0$ .
Also, for values close to x = 0 and to the right of 0, $f'\left( x \right) < 0$ .
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of
$f\left( x \right) \text { is }\frac{1}{2} .$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution F(X) = 1 X 2 + 2 . Concept: Graph of Maxima and Minima.
S
| 344
| 896
|
{"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}
| 4.28125
| 4
|
CC-MAIN-2019-43
|
latest
|
en
| 0.55614
|
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=14250
| 1,368,962,301,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368697420704/warc/CC-MAIN-20130516094340-00044-ip-10-60-113-184.ec2.internal.warc.gz
| 306,930,020
| 801
|
```Question 25978
Let number of adults be "x"
The number of kids is "38-x"
Cost of adult tickets is 4x
Cost of kids tickets is 3(38-x)
EQUATION:
Cost + Cost = \$134
4x+3(38-x)=134
4x+114-3x=134
| 74
| 193
|
{"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}
| 3.796875
| 4
|
CC-MAIN-2013-20
|
latest
|
en
| 0.842312
|
http://ebookmarket.org/doc/rumus-ordinal
| 1,438,474,363,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-32/segments/1438042988924.75/warc/CC-MAIN-20150728002308-00289-ip-10-236-191-2.ec2.internal.warc.gz
| 76,211,821
| 13,771
|
Rumus Ordinal docs
Searching:
Definition - coba
8855 dl's @ 2855 KB/s
Definition - coba
KARAKTERISTIK RUMUS STATISTIKA. ... is used to compute a measure of association between two variables when the variables are of ordinal levels of measurement. ...
http://yk-edu.org/e-refleksi/sharefile/files/01022013134437_Tugas_Akhir_Statistik_rumus.docx
Date added: November 10, 2014 - Views: 4
BAB II
Adapun ciri dari skala ordinal adalah : (1) ... maka setiap item perlu dicari mediannya dengan menggunakan Rumus Median yang diberi lambang S sebagai berikut : ...
https://uharsputra.files.wordpress.com/2007/05/bab-ii-pengukuran-pen-kuanti-revisi.doc
Date added: January 29, 2015 - Views: 12
Bivariate Correlation , SPSS - East Carolina...
Next we shall look at the correlation between a dichotomous variable and a continuous variable. Such a correlation is called a point biserial correlation.
http://core.ecu.edu/psyc/wuenschk/SPSS/Correlation-Bivariate.doc
Date added: October 6, 2011 - Views: 49
SEVERITY RATING SCALES/GUIDELINES FOR - School Law
Arkansas Severity Ratings for Language ASHA Guidelines for the Roles and Responsibilities of the School-Based Speech-Language Pathologist. ASHA, ...
http://www.schoollaw.com/flash_drive_forms/Severity%20Rating%20Scales%20&%20Guidelines.DOC
Date added: October 17, 2011 - Views: 84
Mixed Model ANOVA in SPSS - System for Teaching...
To run a Mixed Model ANOVA (between and within subjects factors), you use the . repeated measures linear model. The example below is from the Pisoni data.
http://step.psy.cmu.edu/materials/spss/mixed.doc
Date added: August 24, 2011 - Views: 203
ANALISIS DATA
RUMUS 1. Uji T. Kegunaan : Uji T digunakan untuk membandingkan rata-rata dua populasi dengan data yang berskala interval. ... ordinal, interval, dan ratio. ...
https://prabowosetiyobudi.files.wordpress.com/2012/06/analisis-data-vita.doc
Date added: February 7, 2015 - Views: 27
STATISTIKA NONPARAMETRIKA - Blog Universitas...
... data yang mempunyai skala pengukuran ordinal (data yang ada urutannya ... Sebelum kita menggunakan rumus Friedman kita harus merangking dulu datanya,hasil ...
Date added: October 25, 2012 - Views: 9
INSTRUMENTASI DALAM SURVEY AUDIENS
ordinal, rumus statistik untuk uji hipotesis memakai . Rank Correlations. dari Spearman. Jika variabel-variabel yang dikorelasikan berskala interval (atau rasio), ...
https://d3kh4.files.wordpress.com/2011/01/instrumentasi-dalam-survey-audiens.doc
Date added: February 15, 2015 - Views: 13
BAB III
Tingkat pemesanan lebih sari 100 potong Ordinal. Ordinal. ... Menghitung nilai skala dengan rumus Method Successive Interval. Menentukan nilai transformasi ...
Date added: November 13, 2013 - Views: 10
iyusabdusyakir.files.wordpress.com
Melalui tabel ini dimaksudkan agar data penelitian lebih mudah dibaca dan dianalisis menggunakan rumus ... data ordinal. Rumus ini hanya efektif ...
https://iyusabdusyakir.files.wordpress.com/2013/04/makalah-teknik-analisis-data-korelasional-dan-komparasional.docx
Date added: May 11, 2015 - Views: 6
Metal.doc - Digital library - Perpustakaan Pusat...
Struktur orgnisasi. Ordinal. Ordinal. Ordinal. Ordinal. Ordinal. Ordinal. Ordinal. Variabel Y. ... Dengan mengunakan rumus tehnik korelasi Pearson Product moment, ...
http://elib.unikom.ac.id/files/disk1/12/jbptunikompp-gdl-s1-2004-zainalnim2-560-BAB3+SK.doc
Date added: May 3, 2013 - Views: 16
roelcup.files.wordpress.com
Datanya bisa Interval atau Ordinal. Contoh data Interval : 30kg – 50kg. 50kg – 70kg. 70kg – 90kg. ... Rumus : χ 2 = ∑ O-E 2 E . Ket : ...
https://roelcup.files.wordpress.com/2010/04/statistik-non-parametik-blog.docx
Date added: April 12, 2015 - Views: 1
BAB III
Sedangkan untuk melihat tingkat hubungan dengan menggunakan Rank Spearman, karena skala yang diperoleh bersifat ordinal, mengikuti rumus sebagai berikut :
http://digilib.itb.ac.id/files/disk1/16/jbptitbpp-gdl-s1-1998-djafar-751-system-bab_telu.doc
Date added: July 14, 2013 - Views: 19
PROPOSAL
Ordinal Tingkat kemampuan materi pelajaran wiraswasta dapat membekali, ... Sample yang diambil oleh penulis dalam penelitian ini berdasarkan rumus Slavin (1994).
https://hendraprijatna68.files.wordpress.com/2012/06/bab-iii.doc
Date added: March 23, 2015 - Views: 2
digilib.unpas.ac.id
Ordinal 3 Pemanfaatan sumber daya Ordinal 4 Kegunaan fasilitas dan fungsi Ordinal 5 Kenyamanan dalam ... Uji validitas instrument dapat menggunakan rumus ...
http://digilib.unpas.ac.id/files/disk1/53/jbptunpaspp-gdl-ianrusland-2649-3-babiii.doc
Date added: May 23, 2014 - Views: 12
DIKTAT PRAKTIKUM PENGOLAHAN DATA STATISTIK DENGAN...
Kita boleh melakukan itu dengan terlebih dahulu mentrasnformasikan data ordinal menjadi data interval dengan melakukan transformasi suksesif. Rumus matematikanya ...
https://wsetiabudi.files.wordpress.com/2010/08/modul-1.doc
Date added: February 19, 2015 - Views: 22
Tabel 3.3 - Digital Library Universitas Pasundan -...
Dimana terdapat indikator-indikator yang akan diukur dengan skala ordinal. Berikut ini ... oleh penulis dalam penelitian ini berdasarkan rumus sebagai ...
http://digilib.unpas.ac.id/files/disk1/11/jbptunpaspp-gdl-fuzifauzia-501-3-babiii-.docx
Date added: November 22, 2014 - Views: 2
UKURAN STATISTIK DESKRIPTIF - Direktori File UPI
Rumus yang dipergunakan untuk menghitung rata-rata dari data bergolong adalah: ... untuk variabel yang memenuhi skala pengukuran sekurang-kurangnya ordinal.
http://file.upi.edu/Direktori/FIP/JUR._PEND._LUAR_SEKOLAH/197108171998021-SARDIN/Bahan%20Statistika/Ukuran%20Statistik%20Deskriptif.doc
Date added: May 29, 2013 - Views: 60
BAB III - Official Site of IRWANDARU DANANJAYA -...
Pada penelitian ini untuk mencari reliabilitas instrumen menggunakan rumus alpha . ... Pada penelitian ini data ordinal ditransformasikan ke data interval dengan ...
Date added: November 8, 2011 - Views: 85
MODUL BOISTATISTIK UNTUK KEBIDANAN
Rumus . Keterangan . Mean = rata-rata ∑ = Jumlah. Xi = nilai x ke I sampai ke n. ... Data ordinal adalah data yang berbentuk rangking atau peringkat.
https://nugrohosusantoborneo.files.wordpress.com/2010/02/modul-boistatistik-untuk-kebidanan-progsus.doc
Date added: April 28, 2015 - Views: 7
thesis.binus.ac.id
... promosi akan keberadaan perusahaan Ordinal Keramahan karyawan saat melakukan promosi terhadap calon konsumen Ordinal Kelancaran ... dengan rumus: n = N 1 + Ne² ...
http://thesis.binus.ac.id/doc/Bab3Doc/2011-2-01688-HM%20Bab3001.doc
Date added: May 15, 2014 - Views: 1
SOAL-SOAL LATIHAN
Jelaskan kedua konsep itu dan tunjukkan kedua rumus elastisitas tersebut ! ... dan pendekatan nilai guna ordinal ( Ordinal utility approach).
https://giyantops.files.wordpress.com/2011/02/soal-ek-mikro-latihan2.doc
Date added: January 29, 2015 - Views: 1
thesis.binus.ac.id
Pada penelitian ini digunakan rumus Slovin ( Noor, J., 2011: 158) n = N. 1 + (N x e²) Dimana : ... III.2.4.3. Transformasi data Ordinal menjadi Interval.
http://thesis.binus.ac.id/Doc/Bab3Doc/2011-2-00493-AK%20Bab3001.doc
Date added: September 26, 2013 - Views: 10
STATISTIKA
Jadi disini tidak dipersoalkan bagaimana didapatkannya rumus-rumus atau aturan-aturan, ... Skala Ordinal (Ordinal Scale)
https://margiyati.files.wordpress.com/2011/03/statistik-bab-1234.doc
Date added: March 15, 2015 - Views: 4
hendraprijatna68.files.wordpress.com
Ordinal Mengembangkan komponen-komponen rancangan pembelajaran Ordinal Menyusun rancangan pembelajaran ... diuji dengan rumus statistik t sebagai ...
https://hendraprijatna68.files.wordpress.com/2012/07/bab-iii.doc
Date added: February 21, 2015 - Views: 6
matreg1pasca.files.wordpress.com
Rumus: Contoh : Dilakukan penelitian ... digunakan untuk menguji signifikansi hipotesis komparatif dua sampel independen bila datanya berbentuk nominal atau ordinal.
https://matreg1pasca.files.wordpress.com/2011/12/hipotesis-kmparatif-2-sampel-indpt2.doc
Date added: April 7, 2015 - Views: 2
Kofisien Korelasi Rank Spearman : rs
Ini adalah ukuruan asosiasi yang menuntut kedua variable diukur sekurang-kurangnya dalam skala ordinal sehingga obyek-obyek atau individu ... rumus perhitungan rs ...
http://kk.mercubuana.ac.id/elearning/files_modul/11007-14-822727225929.doc
Date added: January 22, 2014 - Views: 1
Metode4
Data ordinal diubah menjadi data numerik. ... Dalam langkah ini di usahakan semaksimal mungkin rumus-rumus atau aturan-aturan yang dipakai sesuai yang diambil.
https://sapiderman.files.wordpress.com/2010/11/metode41.doc
Date added: March 26, 2015 - Views: 9
UJI PERINGKAT BERTANDA WILCOXON DAN UJI MANN...
Jika tercapai setidak-tidaknya pengukuran ordinal, ... Rumus (6.7a) dan (6.7b) menghasilkan harga U yang berlainan. Yang kita kehendaki adalah yang lebih kecil.
http://kk.mercubuana.ac.id/elearning/files_modul/11007-13-788356141429.doc
Date added: January 22, 2014 - Views: 14
Modul 4 Analisis Regresi dan Korelasi
Sedangkan untuk yang berskala ordinal kita gunakan Spearman ... untuk menentukkan tingkat pengetahuan kewarganegaraan yang dimilikinya digunakan rumus ...
https://wsetiabudi.files.wordpress.com/2010/08/modul4.doc
Date added: February 15, 2015 - Views: 35
PENJELASAN SISTIMATIKA PENULISAN SKRIPSI MODEL...
Data nominal dan ordinal biasanya menggunakan metode statistik nonparametrik, ... Pada kesempatan ini saya hanya akan memberikan 4 rumus statistik.
https://starawaji.files.wordpress.com/2011/03/penjelasan-sistimatika-penulisan-skripsi-model-penelitian-kuantitatif.doc
Date added: February 20, 2015 - Views: 34
KORELASI LINIER SEDERHANA - Majulah Indonesia |...
Ordinal Spearman Rank. Kendal ( (tau) Interval dan Rasio Pearson Product Moment. Korelasi Ganda. Korelasi Parsial ... Rumus : (xy = 1 - 6 ( bi2 n (n2 – 1)
http://www.ymayowan.lecture.ub.ac.id/files/2012/02/korelasi.doc
Date added: June 8, 2015 - Views: 2
MAKALAH STATISTIK - lukisantty
Rumus multinomial. Contoh multinomial. Fungsi peluang dibagi 5 yaitu : 1. ... Skala ordinal selain membedakan juga menunjukkan tingkatan (misalnya: pendidikan, ...
https://lukisantty.files.wordpress.com/2012/12/makalah-statistik.doc
Date added: January 28, 2015 - Views: 38
ANALISIS CLUSTER - Statistika Terapan
... ( ) dan B dengan koordinat ( ) maka jarak antar kedua objek tersebut dapat diukur dengan rumus ... (nominal atau ordinal). Standarisasi Data. Standarisasi Variabel.
https://statistikaterapan.files.wordpress.com/2008/10/analisis-cluster.doc
Date added: March 5, 2015 - Views: 6
ppta.stikom.edu
Pengujian ini hanya memerlukan data skala ordinal ... - Skala pengukuran yang digunakan biasanya ordinal - Rumus umum yang digunakan pada uji Kruskal-Wallis adalah :
Date added: August 9, 2013 - Views: 8
1
Sebutkan 3 asumsi teori perilaku konsumen dengan pendekatan ordinal (Indifference Curve). ... (gunakan rumus elastisitas busur) A B I 2000 4000 Q 200 100 ...
Date added: May 5, 2013 - Views: 6
web.unair.ac.id
Rumus tersebut dikembangkan oleh Charles ... Berikut tersaji data hasil pengukuran terhadap variabel X dan Y. Data hasil pengukuran berskala ordinal sebagai berikut:
Date added: January 22, 2014 - Views: 1
MODUL 2
dapat dihitung berdasarkan rumus: ... Data yang dikumpulkan pada modul 3 ini adalah data yang berskala ordinal mengenai pelayanan di perpustakaan UTM berdasarkan ...
https://labstatistikku.files.wordpress.com/2012/05/modul-3.doc
Date added: February 16, 2015 - Views: 28
statistikapendidikan.com
... dalil-dalil atau rumus-rumus serta dapat ... Data ordinal adalah data statistik yang cara menyusunnya didasarkan atas urutan kedudukan (rangking). Contoh .
Date added: December 30, 2014 - Views: 7
STATISTIKA INFERENSIAL
Jika data ordinal: Spearman rank (rho) atau Kendall rank (tau) ... Langkah 6 : Menguji signifikansi dengan rumus : Kaidah pengujian : t hitung = r n-2 (1- r 2 )
https://meiandmath22.files.wordpress.com/2014/01/menentukan-koefisien-korelasi-dan-uji-keberartian-korelasi.docx
Date added: April 17, 2015 - Views: 1
STATISTIKA DAN PROBABILITAS
Skala Ordinal . adalah skala yang selain mempunyai ciri untuk membedakan juga mempunyai ciri untuk mengurutkan pada rentang tertentu.
https://sangiang.files.wordpress.com/2008/11/stat_pro_modul_1.doc
Date added: February 14, 2015 - Views: 18
Untuk orang yang aku cintai SHT - NANIK RISNAWATI
Rumus Sturges ( jumlah K I = 1 +(3,3 log n) ... Skala ordinal ( bilangan hasil pengukuran yang mempunyai fungsi untuk membedakan dan meranking, ...
https://nanikrisnawati.files.wordpress.com/2011/02/pengantar-statistika-i-2.doc
Date added: February 21, 2015 - Views: 6
bk2009.files.wordpress.com
Korelasi tata jenjang digunakan untuk menentukan hubungan atau dua gejala yang kedua-duanya merupakan gejala ordinal atau tata jenjang. Rumus yang dikemukakan:
https://bk2009.files.wordpress.com/2012/03/handout1.docx
Date added: March 9, 2015 - Views: 3
A
Dengan menggunakan rumus besar sampel diatas didapatkan jumlah sampel adalah 133 siswi SMA PGRI 1 Sragen. ... Skala pengukuran yang digunakan adalah skala ordinal.
https://archiev.files.wordpress.com/2012/02/proposal-ican.doc
Date added: April 28, 2015 - Views: 7
e-journal.respati.ac.id
... (Skor <2,43 ), dan menggunakan skala ordinal. ... Uji signifikan koefisien korelasi menggunakan rumus z, karena distribusinya mendekati distribusi normal.
http://e-journal.respati.ac.id/sites/default/files/2012-VI-18-TeknologiInformasi/Jurnal%20Yohana%20Ningsi.docx
Date added: August 30, 2013 - Views: 40
blog.ub.ac.id
Mean tidak dapat digunakan sebagai ukuran pemusatan untuk jenis data nominal dan ordinal. ... Median bisa dihitung menggunakan rumus sebagai berikut:
Date added: December 14, 2014 - Views: 4
TEORI & PERILAKU KONSUMEN - erryblackys
Pendekatan ordinal utility ini menggunakan pengukuran ordinal dalam menganalisis pilihan konsumen ... Persamaan garis anggaran diatas disebut rumus . point-slope ...
https://erryblackys.files.wordpress.com/2013/03/em4.doc
Date added: February 15, 2015 - Views: 1
9
... rumus formula rumus A Ling ... aturan urutan A Ling-Grammar MABBIM 07 order of accent tata aksen tataaksen tataaksen A Ling-Grammar MABBIM 07 ordinal ...
| 4,460
| 14,153
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.203125
| 3
|
CC-MAIN-2015-32
|
longest
|
en
| 0.458481
|
http://nrich.maths.org/public/leg.php?code=-100&cl=1&cldcmpid=10465
| 1,506,424,087,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818695439.96/warc/CC-MAIN-20170926103944-20170926123944-00442.warc.gz
| 258,815,035
| 10,001
|
# Search by Topic
#### Resources tagged with Trial and improvement similar to Money Line-up:
Filter by: Content type:
Stage:
Challenge level:
### There are 85 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Trial and improvement
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Eggs in Baskets
##### Stage: 1 Challenge Level:
There are three baskets, a brown one, a red one and a pink one, holding a total of 10 eggs. Can you use the information given to find out how many eggs are in each basket?
### Strike it Out
##### Stage: 1 and 2 Challenge Level:
Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game.
### Strike it Out for Two
##### Stage: 1 and 2 Challenge Level:
Strike it Out game for an adult and child. Can you stop your partner from being able to go?
### Magic Squares 4x4
##### Stage: 2 Challenge Level:
Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Big Dog, Little Dog
##### Stage: 1 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### Number Round Up
##### Stage: 1 Challenge Level:
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Domino Join Up
##### Stage: 1 Challenge Level:
Can you arrange fifteen dominoes so that all the touching domino pieces add to 6 and the ends join up? Can you make all the joins add to 7?
### Find the Difference
##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### A Numbered Route
##### Stage: 2 Challenge Level:
Can you draw a continuous line through 16 numbers on this grid so that the total of the numbers you pass through is as high as possible?
### Five Steps to 50
##### Stage: 1 Challenge Level:
Use five steps to count forwards or backwards in 1s or 10s to get to 50. What strategies did you use?
### Junior Frogs
##### Stage: 1 and 2 Challenge Level:
Have a go at this well-known challenge. Can you swap the frogs and toads in as few slides and jumps as possible?
### Jumping Squares
##### Stage: 1 Challenge Level:
In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible.
### The Tall Tower
##### Stage: 1 Challenge Level:
As you come down the ladders of the Tall Tower you collect useful spells. Which way should you go to collect the most spells?
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Let's Face It
##### Stage: 2 Challenge Level:
In this problem you have to place four by four magic squares on the faces of a cube so that along each edge of the cube the numbers match.
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### Difference
##### Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
### Area and Perimeter
##### Stage: 2 Challenge Level:
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
### One Big Triangle
##### Stage: 1 Challenge Level:
Make one big triangle so the numbers that touch on the small triangles add to 10. You could use the interactivity to help you.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
### Cat Food
##### Stage: 2 Challenge Level:
Sam sets up displays of cat food in his shop in triangular stacks. If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Numbered Cars
##### Stage: 2 Challenge Level:
I was looking at the number plate of a car parked outside. Using my special code S208VBJ adds to 65. Can you crack my code and use it to find out what both of these number plates add up to?
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### Heads and Feet
##### Stage: 1 Challenge Level:
On a farm there were some hens and sheep. Altogether there were 8 heads and 22 feet. How many hens were there?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Sliding Game
##### Stage: 2 Challenge Level:
A shunting puzzle for 1 person. Swop the positions of the counters at the top and bottom of the board.
### Special 24
##### Stage: 2 Challenge Level:
Find another number that is one short of a square number and when you double it and add 1, the result is also a square number.
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Grouping Goodies
##### Stage: 1 Challenge Level:
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Plenty of Pens
##### Stage: 2 Challenge Level:
Amy's mum had given her £2.50 to spend. She bought four times as many pens as pencils and was given 40p change. How many of each did she buy?
### Rocco's Race
##### Stage: 2 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### A Shapely Network
##### Stage: 2 Challenge Level:
Your challenge is to find the longest way through the network following this rule. You can start and finish anywhere, and with any shape, as long as you follow the correct order.
### The Brown Family
##### Stage: 1 Challenge Level:
Use the information about Sally and her brother to find out how many children there are in the Brown family.
### Magic Matrix
##### Stage: 2 Challenge Level:
Find out why these matrices are magic. Can you work out how they were made? Can you make your own Magic Matrix?
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Buckets of Thinking
##### Stage: 2 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Magic Circles
##### Stage: 2 Challenge Level:
Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Can you find the rule for giving another set of six numbers?
| 2,284
| 9,639
|
{"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}
| 4.15625
| 4
|
CC-MAIN-2017-39
|
latest
|
en
| 0.891428
|
https://www.muratasoftware.com/en/products/examples/gau036/
| 1,604,056,594,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00480.warc.gz
| 812,484,815
| 5,822
|
Inductor with Superimposed DC (High Frequency)examples｜products｜Murata Software Co., Ltd.
Example36Inductor with Superimposed DC (High Frequency)
General
• The inductance of a high-frequency inductor with superimposed DC is analyzed.
• The minor-loop permeability is used.
• This is an harmonic analysis.The high-frequency characteristics such as skin effect are taken into account.
See Exercise 28 for static analysis.
• The vectors of the magnetic field and the magnetic flux density are solved.
• Unless specified in the list below, the default conditions will be applied.
Analysis Space
Item Settings Analysis Space 3D Model unit mm
Analysis Conditions
Item Settings Solvers Magnetic Field Analysis [Gauss] Analysis Type Harmonic analysis Options None
The frequency of the current is set to 1[MHz].
Tabs Setting Item Settings Mesh Frequency-Dependent Meshing Reference frequency: 1×10^6[Hz] Harmonic Analysis Sweep Type Single frequency Frequency 1×10^6[Hz]
Set the Mesh Tab as follows.
Tab Setting Item Settings Mesh Meshing Setup Automatically set the general mesh size: Deselect General mesh size: 2[mm]
Model
A helical solid body Coil (coil) is defined.
Its inflow/outflow faces are extended to the outside of the ambient air. They contact the electric wall of outer boundary condition.
In the magnetic field harmonic analysis, inflow/outflow faces must be outside of the ambient air. (for the purpose of calculation with FEM)
In this exercise, ambient air is set to be created automatically. Therefore, the inflow/outflow faces are extended to the outside of ambient air.
Body Attributes and Materials
Body Number/Type Body Attribute Name Material Name 0/Solid Coil 008_Cu * 4/Solid Coil 008_Cu * 5/Solid Coil 008_Cu * 6/Solid Core Core_Nonl_Minor
* Available from the Material DB
Nonlinear material is defined on the nonlinearity table.
To solve the inductance under DC-biased condition, the magnetic field is solved first with the major loop. It is a static analysis.
Then, by using the minor loop and this magnetic field, the permeability for harmonic analysis is obtained, and the inductance is calculated.
Material Name
Tab
Setting
Core_Nonl_Minor
Permeability
Magnetization Characteristic Type: Select B-H curve
Select “Use the minor-loop permeability”
B-H Curve Table
Magnetic Field [A/m] Magnetic Flux Density [T] 0 0x10^-4 10 130×10^ -3 20 220×10^-3 30 300×10^-3 40 350×10^-3 50 380×10^-3 60 400×10^-3 70 410×10^-3
This B-H curve is for DC bias.
It is called Major Loop.
Permeability
for Minor Loop
Magnetization Characteristic Type: Select B-H curve
B-H Curve Table
Magnetic Field [A/m] Magnetic Flux Density [T] 0 0x10^-4 10 1.3×10^-3 20 2.2×10^-3 30 3×10^-3 40 3.5×10^-3 50 3.8×10^-3 60 4×10^-3 70 4.1×10^-3
This B-H curve is for AC operation.
It is called Minor Loop.
Body attribute is set up as follows to apply current to the coil.
The value of current set here will be the value of the superimposed DC.
AC current cannot be given. It is assumed to be significantly small compared to DC current.
Body Attribute Name Tab Setting Coil Current Waveform: AC Current: 1[A] Turns: 1[Turns] Direction: Specify Inflow/Outflow Faces Select inflow face and outflow face.
No setting.
Results
To see the results of inductance calculation, go to the [Results] tab
click [Table] .
The vectors of the magnetic flux density are shown below.
| 878
| 3,414
|
{"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-2020-45
|
latest
|
en
| 0.849057
|
http://www.atarimagazines.com/compute/issue36/090_Part_I_Visiting_The_VIC-20_Video.php
| 1,508,672,713,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00806.warc.gz
| 383,413,577
| 6,356
|
` COMPUTE! ISSUE 36 / MAY 1983 / PAGE 222`
Part I Visiting The VIC-20 Video Jim Butterfield, Associate Editor In which the traveler discovers a new way of viewing the computer's memory: through a video chip. This is the first of a multi-part series about the structure and uses of the VIC's video chip. If we want to put the VIC-20 video chip to work, we must learn to see things from its standpoint. It sees the computer memory in a way that differs significantly from the way the processor chip sees it. Let's look at what the video chip sees: How the video chip sees memory. The video chip sees only the memory shown above. Even if you have expanded your computer to include lots of extra RAM above address 8191, the chip can't see it. The chip sees only the character ROM, in blocks 0, 1, 2, and 3; and the lowest 8K of RAM (in blocks 8 to 15). Blocks 4, 5, 6, and 7 would look at the Input/Output area, but take my advice: don't do it – no good will come from these addresses. What The Chip Wants The video chip wants to dig out two things from memory and deliver them to the screen. It wants to look at "screen memory" – usually the characters you have typed. On a minimum 5K VIC, that's block 15.5, which corresponds to decimal address 7680 or hexadecimal 1E00. Did I mention that for screen memory, we can look at "half blocks"? It makes sense, since only five hundred odd characters are needed to fill the screen. By the way, the official name for screen memory is the "video matrix." Whatever you call it, if you POKE 7680,1 on an unexpanded VIC, you'll see the letter A appear at the start of the screen. Unless, of course, you're printing white on white, in which case you need very good vision to see it. The second thing that the chip wants from memory is the "character set" – instructions on how to draw each character on the screen. On a typical VIC, this will be either block 0 for the graphics character set or block 2 for the text mode (upper- and lowercase). You can change it, but you'll usually want to stay with even numbers: a full character set including the reversed characters takes up 2048 bytes of memory. The official name for the character set is "Character Cells," although the term "Character Base" is coming into use. Whatever you call it, you can't POKE 32768,55 and expect anything to happen – the standard characters are in ROM and cannot be changed. They're carved in stone, or silicon, to be more exact. If you want to switch to custom characters, you'll need to stage them in RAM and tell the chip which block to take them from. There's a third thing that the chip uses, but it doesn't come from regular memory in the usual way. That's the screen colors (the "Color Matrix"). This color information for each character comes through the back door, so to speak, and we won't worry about the details too much here. When we need to, we'll set the color and assume everything will work. Architecture Looking at the diagram, we can begin to see why the VIC does its odd screen switch when you add memory. In the 5K VIC, the screen sits at the top of memory – and that's the highest address that the video chip can see (block 15.5). If we add 3K RAM expansion, the screen can stay where it is above the BASIC RAM area. But if we add 8K or more, the video chip can't see that high, and the screen memory must flip down to the bottom where it won't get in the way of your BASIC program. Which bottom, you may ask? It turns out to be block 12, which is memory address 4096 or hexadecimal 1000, even if the 3K expansion is in place. You can move this around yourself, of course, and we'll be doing that in just a few moments. The trick is mostly location 36869, which contains instructions on which blocks to use for screen and characters. We do it this way: select which blocks you want for each. Now, multiply the screen block (not including the .5 if you're using it) by 16 and add the character block. POKE the result into 36869, and the job's done. We'll need to do a couple of other things for sanity's sake, but that's the main job. The "half page" for the screen memory goes into location 36866; you invoke it by adding 128 to the "column count" if you want to go the extra distance. That means that under normal circumstances (22 columns), you want to POKE 36866,22 for an exact block number, and POKE 36866,150 to nudge to the extra half page. An Adventure Let's do something useless, but fun. We'll move the screen memory down to address zero (that's block 8). We can't play with this area – too many important things are happening there – but we can watch interesting things in progress, like the timer and the cursor doing their peculiar things. First, the calculation. We want the character set to stay the way it is (block 0 for graphics), and we want to move the screen memory to block 8. Eight times 16 plus zero gives 128. No half block, so 36866 should be 22. A preliminary step: let's make sure that we don't print white-on-white by clearing the screen and typing: ```FOR J = 37888 TO 38911:POKE J,0:NEXT J ``` Ready? Here goes: enter POKE 36869,128:POKE 36866,22. Press RETURN. No, we haven't crashed, but we'll have to type blind from now on. First, examine the fascinating busy things that are under way. The timer is working away in three bytes. At first glance, only one byte seems to be changing. The cursor flash is being logged and timed somewhat below. And if you start typing, you'll see a whole new series of working values coming into play. Indeed, if you can type blind, you might try PRINT 1234 + 5678 and watch the flurry of activity. If you type a lot, the screen will start to scroll, and the display will start to vanish as the colors are rolled off the top. Restore everything to normal by holding down RUN/STOP and tapping the RESTORE key. This has been a first exploration, but you may feel that you understand better what the video chip is up to. Indeed, you may feel that you have gained some measure of control. There's much more to be learned. This is a start.
| 1,457
| 6,037
|
{"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}
| 2.625
| 3
|
CC-MAIN-2017-43
|
latest
|
en
| 0.944328
|
http://www.gurufocus.com/term/ShillerPE/AIG/Shiller%2BP%252FE%2BRatio/American%2BInternational%2BGroup%2BInc
| 1,493,208,900,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-17/segments/1492917121305.61/warc/CC-MAIN-20170423031201-00522-ip-10-145-167-34.ec2.internal.warc.gz
| 553,591,343
| 29,597
|
Switch to:
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
American International Group Inc (NYSE:AIG)
Shiller PE Ratio
(As of Today)
As of today, American International Group Inc's current share price is \$60.07. American International Group Inc's E10 for the quarter that ended in Dec. 2016 was \$-91.72. American International Group Inc's Shiller P/E Ratio for today is .
During the past 13 years, American International Group Inc's highest Shiller P/E Ratio was 19.56. The lowest was 0.73. And the median was 15.94.
E10 is a concept invented by Prof. Robert Shiller, who uses E10 for his Shiller P/E calculation. E10 is the average of the inflation adjusted earnings of a company over the past 10 years.
American International Group Inc's adjusted earnings per share data for the three months ended in Dec. 2016 was \$-2.750. Add all the adjusted EPS for the past 10 years together and divide 10 will get our E10, which is \$-91.72 for the trailing ten years ended in Dec. 2016.
Definition
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, the Shiller P/E is also called PE10.
The Shiller P/E was first used by professor Robert Shiller to measure the valuation of the overall market. The same calculation is applied here to individual companies.
American International Group Inc's Shiller P/E Ratio for today is calculated as
Shiller P/E Ratio = Share Price / E10 = 60.07 / -91.72 =
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
American International Group Inc's E10 for the fiscal year that ended in Dec16 is calculated as:
For example, American International Group Inc's adjusted earnings per share data for the three months ended in Dec. 2016 was:
Adj_EPS = Earnigns per Share / CPI of Dec. 2016 (Change) * Current CPI (Dec. 2016) = -2.75 / 241.432 * 241.432 = -2.750
Current CPI (Dec. 2016) = 241.432.
American International Group Inc Quarterly Data
201409 201412 201503 201506 201509 201512 201603 201606 201609 201612 per share eps 1.52 0.49 1.78 1.32 -0.18 -1.36 -0.16 1.68 0.42 -2.75 CPI 238.031 234.812 236.119 238.638 237.945 236.525 238.132 241.018 241.428 241.432 Adj_EPS 1.542 0.504 1.82 1.335 -0.183 -1.388 -0.162 1.683 0.42 -2.75 201203 201206 201209 201212 201303 201306 201309 201312 201403 201406 per share eps 1.71 1.33 1.13 -2.17 1.49 1.84 1.46 1.33 1.09 2.1 CPI 229.392 229.478 231.407 229.601 232.773 233.504 234.149 233.049 236.293 238.343 Adj_EPS 1.8 1.399 1.179 -2.282 1.545 1.902 1.505 1.378 1.114 2.127 200909 200912 201003 201006 201009 201012 201103 201106 201109 201112 per share eps 0.68 -65.56 2.66 -19.57 -18.53 16.65 0.31 1 -2.1 11.96 CPI 215.969 215.949 217.631 217.965 218.439 219.179 223.467 225.722 226.889 225.672 Adj_EPS 0.76 -73.296 2.951 -21.677 -20.48 18.34 0.335 1.07 -2.235 12.795 200703 200706 200709 200712 200803 200806 200809 200812 200903 200906 per share eps 31.6 32.8 23.8 -40.6 -61.8 -41.13 -181.02 -468.86 -39.67 2.3 CPI 205.352 208.352 208.49 210.036 213.528 218.815 218.783 210.228 212.709 215.693 Adj_EPS 37.152 38.008 27.56 -46.669 -69.876 -45.381 -199.76 -538.453 -45.027 2.574
Add all the adjusted EPS together and divide 10 will get our E10.
Explanation
Compared with the regular P/E ratio, which works poorly for cyclical businesses, the Shiller P/E smoothed out the fluctuations of profit margins during business cycles. Therefore it is more accurate in reflecting the valuation of the company.
If a company has consistent business performance, the Shiller P/E should give similar results to regular P/E.
Compared with the P/S ratio, the Shiller P/E makes the comparison between different industries more meaningful.
Be Aware
The Shiller P/E assumes that over the long term, businesses and profitability revert to their means. If a companys business model does not work in the future compared with the past, the Shiller P/E and P/S ratio will give false valuations.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
American International Group Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 ShillerPE 14.74 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
American International Group Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 ShillerPE 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)
| 1,530
| 4,843
|
{"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-2017-17
|
latest
|
en
| 0.957049
|
https://en.m.wikipedia.org/wiki/Smith-Waterman_algorithm
| 1,590,897,721,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347410745.37/warc/CC-MAIN-20200531023023-20200531053023-00260.warc.gz
| 334,687,128
| 33,263
|
# Smith–Waterman algorithm
(Redirected from Smith-Waterman algorithm)
The Smith–Waterman algorithm performs local sequence alignment; that is, for determining similar regions between two strings of nucleic acid sequences or protein sequences. Instead of looking at the entire sequence, the Smith–Waterman algorithm compares segments of all possible lengths and optimizes the similarity measure.
Class Sequence alignment ${\displaystyle O(mn)}$ ${\displaystyle O(mn)}$
An animated example to show the steps progressively. See here for detailed steps.
The algorithm was first proposed by Temple F. Smith and Michael S. Waterman in 1981.[1] Like the Needleman–Wunsch algorithm, of which it is a variation, Smith–Waterman is a dynamic programming algorithm. As such, it has the desirable property that it is guaranteed to find the optimal local alignment with respect to the scoring system being used (which includes the substitution matrix and the gap-scoring scheme). The main difference to the Needleman–Wunsch algorithm is that negative scoring matrix cells are set to zero, which renders the (thus positively scoring) local alignments visible. Traceback procedure starts at the highest scoring matrix cell and proceeds until a cell with score zero is encountered, yielding the highest scoring local alignment. Because of its quadratic complexity in time and space, it often cannot be practically applied to large-scale problems and is replaced in favor of less general but computationally more efficient alternatives such as (Gotoh, 1982),[2] (Altschul and Erickson, 1986),[3] and (Myers and Miller, 1988).[4]
## History
In 1970, Saul B. Needleman and Christian D. Wunsch proposed a heuristic homology algorithm for sequence alignment, also referred to as the Needleman–Wunsch algorithm.[5] It is a global alignment algorithm that requires ${\displaystyle O(mn)}$ calculation steps (${\displaystyle m}$ and ${\displaystyle n}$ are the lengths of the two sequences being aligned). It uses the iterative calculation of a matrix for the purpose of showing global alignment. In the following decade, Sankoff,[6] Reichert,[7] Beyer[8] and others formulated alternative heuristic algorithms for analyzing gene sequences. Sellers introduced a system for measuring sequence distances.[9] In 1976, Waterman et al. added the concept of gaps into the original measurement system.[10] In 1981, Smith and Waterman published their Smith–Waterman algorithm for calculating local alignment.
The Smith–Waterman algorithm is fairly demanding of time: To align two sequences of lengths ${\displaystyle m}$ and ${\displaystyle n}$ , ${\displaystyle O(mn)}$ time is required. Gotoh[2] and Altschul[3] optimized the algorithm to ${\displaystyle O(mn)}$ steps. The space complexity was optimized by Myers and Miller[4] from ${\displaystyle O(mn)}$ to ${\displaystyle O(n)}$ (linear), where ${\displaystyle n}$ is the length of the shorter sequence, for the case where one only of the many possible optimal alignments is desired.
## Motivation
In recent years, genome projects conducted on a variety of organisms generated massive amounts of sequence data for genes and proteins, which requires computational analysis. Sequence alignment shows the relations between genes or between proteins, leading to a better understanding of their homology and functionality. Sequence alignment can also reveal conserved domains and motifs.
One motivation for local alignment is the difficulty of obtaining correct alignments in regions of low similarity between distantly related biological sequences, because mutations have added too much 'noise' over evolutionary time to allow for a meaningful comparison of those regions. Local alignment avoids such regions altogether and focuses on those with a positive score, i.e. those with an evolutionarily conserved signal of similarity. A prerequisite for local alignment is a negative expectation score. The expectation score is defined as the average score that the scoring system (substitution matrix and gap penalties) would yield for a random sequence.
Another motivation for using local alignments is that there is a reliable statistical model (developed by Karlin and Altschul) for optimal local alignments. The alignment of unrelated sequences tends to produce optimal local alignment scores which follow an extreme value distribution. This property allows programs to produce an expectation value for the optimal local alignment of two sequences, which is a measure of how often two unrelated sequences would produce an optimal local alignment whose score is greater than or equal to the observed score. Very low expectation values indicate that the two sequences in question might be homologous, meaning they might share a common ancestor.
## Algorithm
Scoring method of the Smith–Waterman algorithm
Let ${\displaystyle A=a_{1}a_{2}...a_{n}}$ and ${\displaystyle B=b_{1}b_{2}...b_{m}}$ be the sequences to be aligned, where ${\displaystyle n}$ and ${\displaystyle m}$ are the lengths of ${\displaystyle A}$ and ${\displaystyle B}$ respectively.
1. Determine the substitution matrix and the gap penalty scheme.
• ${\displaystyle s(a,b)}$ - Similarity score of the elements that constituted the two sequences
• ${\displaystyle W_{k}}$ - The penalty of a gap that has length ${\displaystyle k}$
2. Construct a scoring matrix ${\displaystyle H}$ and initialize its first row and first column. The size of the scoring matrix is ${\displaystyle (n+1)*(m+1)}$ . The matrix uses 0-based indexing.
${\displaystyle H_{k0}=H_{0l}=0\quad for\quad 0\leq k\leq n\quad and\quad 0\leq l\leq m}$
3. Fill the scoring matrix using the equation below.
${\displaystyle H_{ij}=\max {\begin{cases}H_{i-1,j-1}+s(a_{i},b_{j}),\\\max _{k\geq 1}\{H_{i-k,j}-W_{k}\},\\\max _{l\geq 1}\{H_{i,j-l}-W_{l}\},\\0\end{cases}}\qquad (1\leq i\leq n,1\leq j\leq m)}$
where
${\displaystyle H_{i-1,j-1}+s(a_{i},b_{j})}$ is the score of aligning ${\displaystyle a_{i}}$ and ${\displaystyle b_{j}}$ ,
${\displaystyle H_{i-k,j}-W_{k}}$ is the score if ${\displaystyle a_{i}}$ is at the end of a gap of length ${\displaystyle k}$ ,
${\displaystyle H_{i,j-l}-W_{l}}$ is the score if ${\displaystyle b_{j}}$ is at the end of a gap of length ${\displaystyle l}$ ,
${\displaystyle 0}$ means there is no similarity up to ${\displaystyle a_{i}}$ and ${\displaystyle b_{j}}$ .
4. Traceback. Starting at the highest score in the scoring matrix ${\displaystyle H}$ and ending at a matrix cell that has a score of 0, traceback based on the source of each score recursively to generate the best local alignment.
## Explanation
Smith–Waterman algorithm aligns two sequences by matches/mismatches (also known as substitutions), insertions, and deletions. Both insertions and deletions are the operations that introduce gaps, which are represented by dashes. The Smith–Waterman algorithm has several steps:
1. Determine the substitution matrix and the gap penalty scheme. A substitution matrix assigns each pair of bases or amino acids a score for match or mismatch. Usually matches get positive scores, whereas mismatches get relatively lower scores. A gap penalty function determines the score cost for opening or extending gaps. It is suggested that users choose the appropriate scoring system based on the goals. In addition, it is also a good practice to try different combinations of substitution matrices and gap penalties.
2. Initialize the scoring matrix. The dimensions of the scoring matrix are 1+length of each sequence respectively. All the elements of the first row and the first column are set to 0. The extra first row and first column make it possible to align one sequence to another at any position, and setting them to 0 makes the terminal gap free from penalty.
3. Scoring. Score each element from left to right, top to bottom in the matrix, considering the outcomes of substitutions (diagonal scores) or adding gaps (horizontal and vertical scores). If none of the scores are positive, this element gets a 0. Otherwise the highest score is used and the source of that score is recorded.
4. Traceback. Starting at the element with the highest score, traceback based on the source of each score recursively, until 0 is encountered. The segments that have the highest similarity score based on the given scoring system is generated in this process. To obtain the second best local alignment, apply the traceback process starting at the second highest score outside the trace of the best alignment.
### Comparison with the Needleman–Wunsch algorithm
Global and local sequence alignment
The Smith–Waterman algorithm finds the segments in two sequences that have similarities while the Needleman–Wunsch algorithm aligns two complete sequences. Therefore, they serve different purposes. Both algorithms use the concepts of a substitution matrix, a gap penalty function, a scoring matrix, and a traceback process. Three main differences are:
Smith–Waterman algorithm Needleman–Wunsch algorithm
Initialization First row and first column are set to 0 First row and first column are subject to gap penalty
Scoring Negative score is set to 0 Score can be negative
Traceback Begin with the highest score, end when 0 is encountered Begin with the cell at the lower right of the matrix, end at top left cell
One of the most important distinctions is that no negative score is assigned in the scoring system of the Smith–Waterman algorithm, which enables local alignment. When any element has a score lower than zero, it means that the sequences up to this position have no similarities; this element will then be set to zero to eliminate influence from previous alignment. In this way, calculation can continue to find alignment in any position afterwards.
The initial scoring matrix of Smith–Waterman algorithm enables the alignment of any segment of one sequence to an arbitrary position in the other sequence. In Needleman–Wunsch algorithm, however, end gap penalty also needs to be considered in order to align the full sequences.
### Substitution matrix
Each base substitution or amino acid substitution is assigned a score. In general, matches are assigned positive scores, and mismatches are assigned relatively lower scores. Take DNA sequence as an example. If matches get +1, mismatches get -1, then the substitution matrix is:
A G C T
A 1 -1 -1 -1
G -1 1 -1 -1
C -1 -1 1 -1
T -1 -1 -1 1
This substitution matrix can be described as: ${\displaystyle s(a_{i},b_{j})={\begin{cases}+1,\quad a_{i}=b_{j}\\-1,\quad a_{i}\neq b_{j}\end{cases}}}$
Different base substitutions or amino acid substitutions can have different scores. The substitution matrix of amino acids is usually more complicated than that of the bases. See PAM, BLOSUM.
### Gap penalty
Gap penalty designates scores for insertion or deletion. A simple gap penalty strategy is to use fixed score for each gap. In biology, however, the score needs to be counted differently for practical reasons. On one hand, partial similarity between two sequences is a common phenomenon; on the other hand, a single gene mutation event can result in insertion of a single long gap. Therefore, connected gaps forming a long gap usually is more favored than multiple scattered, short gaps. In order to take this difference into consideration, the concepts of gap opening and gap extension have been added to the scoring system. The gap opening score is usually higher than the gap extension score. For instance, the default parameter in EMBOSS Water are: gap opening = 10, gap extension = 0.5.
Here we discuss two common strategies for gap penalty. See Gap penalty for more strategies. Let ${\displaystyle W_{k}}$ be the gap penalty function for a gap of length ${\displaystyle k}$ :
#### Linear
Simplified Smith–Waterman algorithm when linear gap penalty function is used
A linear gap penalty has the same scores for opening and extending a gap:
${\displaystyle W_{k}=kW_{1}}$ ,
where ${\displaystyle W_{1}}$ is the cost of a single gap.
The gap penalty is directly proportional to the gap length. When linear gap penalty is used, the Smith–Waterman algorithm can be simplified to:
${\displaystyle H_{ij}=\max {\begin{cases}H_{i-1,j-1}+s(a_{i},b_{j}),\\H_{i-1,j}-W_{1},\\H_{i,j-1}-W_{1},\\0\end{cases}}}$
The simplified algorithm uses ${\displaystyle O(mn)}$ steps. When an element is being scored, only the gap penalties from the elements that are directly adjacent to this element need to be considered.
#### Affine
An affine gap penalty considers gap opening and extension separately:
${\displaystyle W_{k}=uk+v\quad (u>0,v>0)}$ ,
where ${\displaystyle v}$ is the gap opening penalty, and ${\displaystyle u}$ is the gap extension penalty. For example, the penalty for a gap of length 2 is ${\displaystyle 2u+v}$ .
An arbitrary gap penalty was used in the original Smith–Waterman algorithm paper. It uses ${\displaystyle O(m^{2}n)}$ steps, therefore is quite demanding of time. Gotoh optimized the steps for an affine gap penalty to ${\displaystyle O(mn)}$ ,[2] but the optimized algorithm only attempts to find one optimal alignment, and the optimal alignment is not guaranteed to be found.[3] Altschul modified Gotoh's algorithm to find all optimal alignments while maintaining the computational complexity.[3] Later, Myers and Miller pointed out that Gotoh and Altschul's algorithm can be further modified based on the method that was published by Hirschberg in 1975,[11] and applied this method.[4] Myers and Miller's algorithm can align two sequences using ${\displaystyle O(n)}$ space, with ${\displaystyle n}$ being the length of the shorter sequence.
#### Gap penalty example
Take the alignment of sequences TACGGGCCCGCTAC and TAGCCCTATCGGTCA as an example. When linear gap penalty function is used, the result is (Alignments performed by EMBOSS Water. Substitution matrix is DNAfull. Gap opening and extension both are 1.0):
TACGGGCCCGCTA-C
|| | || ||| |
TA---G-CC-CTATC
When affine gap penalty is used, the result is (Gap opening and extension are 5.0 and 1.0 respectively):
TACGGGCCCGCTA
|| ||| |||
TA---GCC--CTA
This example shows that an affine gap penalty can help avoid scattered small gaps.
### Scoring matrix
The function of the scoring matrix is to conduct one-to-one comparisons between all components in two sequences and record the optimal alignment results. The scoring process reflects the concept of dynamic programming. The final optimal alignment is found by iteratively expanding the growing optimal alignment. In other words, the current optimal alignment is generated by deciding which path (match/mismatch or inserting gap) gives the highest score from the previous optimal alignment. The size of the matrix is the length of one sequence plus 1 by the length of the other sequence plus 1. The additional first row and first column serve the purpose of aligning one sequence to any positions in the other sequence. Both the first line and the first column are set to 0 so that end gap is not penalized. The initial scoring matrix is:
b1 bj bm
0 0 0 0
a1 0
ai 0
an 0
## Example
Take the alignment of DNA sequences TGTTACGG and GGTTGACTA as an example. Use the following scheme:
• Substitution matrix: ${\displaystyle s(a_{i},b_{j})={\begin{cases}+3,\quad a_{i}=b_{j}\\-3,\quad a_{i}\neq b_{j}\end{cases}}}$
• Gap penalty: ${\displaystyle W_{k}=2k}$ (a linear gap penalty of ${\displaystyle W_{1}=2}$ )
Initialize and fill the scoring matrix, shown as below. This figure shows the scoring process of the first three elements. The yellow color indicates the bases that are being considered. The red color indicates the highest possible score for the cell being scored.
Initialization of the scoring matrix (left 1) and the scoring process of the first three elements (left 2-4)
The finished scoring matrix is shown below on the left. The blue color shows the highest score. An element can receive score from more than one element, each will form a different path if this element is traced back. In case of multiple highest scores, traceback should be done starting with each highest score. The traceback process is shown below on the right. The best local alignment is generated in the reverse direction.
Finished scoring matrix (the highest score is in blue) Traceback process and alignment result
The alignment result is:
G T T - A C
| | | | |
G T T G A C
## Implementation
An implementation of the Smith–Waterman Algorithm, SSEARCH, is available in the FASTA sequence analysis package from UVA FASTA Downloads. This implementation includes Altivec accelerated code for PowerPC G4 and G5 processors that speeds up comparisons 10–20-fold, using a modification of the Wozniak, 1997 approach,[12] and an SSE2 vectorization developed by Farrar[13] making optimal protein sequence database searches quite practical. A library, SSW, extends Farrar's implementation to return alignment information in addition to the optimal Smith–Waterman score.[14]
## Accelerated versions
### FPGA
Cray demonstrated acceleration of the Smith–Waterman algorithm using a reconfigurable computing platform based on FPGA chips, with results showing up to 28x speed-up over standard microprocessor-based solutions. Another FPGA-based version of the Smith–Waterman algorithm shows FPGA (Virtex-4) speedups up to 100x[15] over a 2.2 GHz Opteron processor.[16] The TimeLogic DeCypher and CodeQuest systems also accelerate Smith–Waterman and Framesearch using PCIe FPGA cards.
A 2011 Master's thesis [17] includes an analysis of FPGA-based Smith–Waterman acceleration.
In a 2016 publication OpenCL code compiled with Xilinx SDAccel accelerates genome sequencing, beats CPU/GPU performance/W by 12-21x, a very efficient implementation was presented. Using one PCIe FPGA card equipped with a Xilinx Virtex-7 2000T FPGA, the performance per Watt level was better than CPU/GPU by 12-21x.
### GPU
Lawrence Livermore National Laboratory and the US Department of Energy's Joint Genome Institute implemented an accelerated version of Smith–Waterman local sequence alignment searches using graphics processing units (GPUs) with preliminary results showing a 2x speed-up over software implementations.[18] A similar method has already been implemented in the Biofacet software since 1997, with the same speed-up factor.[19]
Several GPU implementations of the algorithm in NVIDIA's CUDA C platform are also available.[20] When compared to the best known CPU implementation (using SIMD instructions on the x86 architecture), by Farrar, the performance tests of this solution using a single NVidia GeForce 8800 GTX card show a slight increase in performance for smaller sequences, but a slight decrease in performance for larger ones. However the same tests running on dual NVidia GeForce 8800 GTX cards are almost twice as fast as the Farrar implementation for all sequence sizes tested.
A newer GPU CUDA implementation of SW is now available that is faster than previous versions and also removes limitations on query lengths. See CUDASW++.
Eleven different SW implementations on CUDA have been reported, three of which report speedups of 30X.[21]
### SIMD
In 2000, a fast implementation of the Smith–Waterman algorithm using the SIMD technology available in Intel Pentium MMX processors and similar technology was described in a publication by Rognes and Seeberg.[22] In contrast to the Wozniak (1997) approach, the new implementation was based on vectors parallel with the query sequence, not diagonal vectors. The company Sencel Bioinformatics has applied for a patent covering this approach. Sencel is developing the software further and provides executables for academic use free of charge.
A SSE2 vectorization of the algorithm (Farrar, 2007) is now available providing an 8-16-fold speedup on Intel/AMD processors with SSE2 extensions.[13] When running on Intel processor using the Core microarchitecture the SSE2 implementation achieves a 20-fold increase. Farrar's SSE2 implementation is available as the SSEARCH program in the FASTA sequence comparison package. The SSEARCH is included in the European Bioinformatics Institute's suite of similarity searching programs.
Danish bioinformatics company CLC bio has achieved speed-ups of close to 200 over standard software implementations with SSE2 on an Intel 2.17 GHz Core 2 Duo CPU, according to a publicly available white paper.
Accelerated version of the Smith–Waterman algorithm, on Intel and AMD based Linux servers, is supported by the GenCore 6 package, offered by Biocceleration. Performance benchmarks of this software package show up to 10 fold speed acceleration relative to standard software implementation on the same processor.
Currently the only company in bioinformatics to offer both SSE and FPGA solutions accelerating Smith–Waterman, CLC bio has achieved speed-ups of more than 110 over standard software implementations with CLC Bioinformatics Cube[citation needed]
The fastest implementation of the algorithm on CPUs with SSSE3 can be found the SWIPE software (Rognes, 2011),[23] which is available under the GNU Affero General Public License. In parallel, this software compares residues from sixteen different database sequences to one query residue. Using a 375 residue query sequence a speed of 106 billion cell updates per second (GCUPS) was achieved on a dual Intel Xeon X5650 six-core processor system, which is over six times more rapid than software based on Farrar's 'striped' approach. It is faster than BLAST when using the BLOSUM50 matrix.
There also exists diagonalsw, a C and C++ implementation of the Smith–Waterman algorithm with the SIMD instruction sets (SSE4.1 for the x86 platform and AltiVec for the PowerPC platform). It is licensed under the open-source MIT license.
In 2008, Farrar[24] described a port of the Striped Smith–Waterman[13] to the Cell Broadband Engine and reported speeds of 32 and 12 GCUPS on an IBM QS20 blade and a Sony PlayStation 3, respectively.
## Limitations
Fast expansion of genetic data challenges speed of current DNA sequence alignment algorithms. Essential needs for an efficient and accurate method for DNA variant discovery demand innovative approaches for parallel processing in real time. Optical computing approaches have been suggested as promising alternatives to the current electrical implementations. OptCAM is an example of such approaches and is shown to be faster than the Smith–Waterman algorithm.[25]
## References
1. ^ Smith, Temple F. & Waterman, Michael S. (1981). "Identification of Common Molecular Subsequences" (PDF). Journal of Molecular Biology. 147 (1): 195–197. CiteSeerX 10.1.1.63.2897. doi:10.1016/0022-2836(81)90087-5. PMID 7265238.
2. ^ a b c Osamu Gotoh (1982). "An improved algorithm for matching biological sequences". Journal of Molecular Biology. 162 (3): 705–708. CiteSeerX 10.1.1.204.203. doi:10.1016/0022-2836(82)90398-9. PMID 7166760.
3. ^ a b c d Stephen F. Altschul & Bruce W. Erickson (1986). "Optimal sequence alignment using affine gap costs". Bulletin of Mathematical Biology. 48 (5–6): 603–616. doi:10.1007/BF02462326. PMID 3580642.
4. ^ a b c Miller, Webb; Myers, Eugene (1988). "Optimal alignments in linear space". Bioinformatics. 4 (1): 11–17. CiteSeerX 10.1.1.107.6989. doi:10.1093/bioinformatics/4.1.11. PMID 3382986.
5. ^ Saul B. Needleman; Christian D. Wunsch (1970). "A general method applicable to the search for similarities in the amino acid sequence of two proteins". Journal of Molecular Biology. 48 (3): 443–453. doi:10.1016/0022-2836(70)90057-4. PMID 5420325.
6. ^
7. ^ Thomas A. Reichert; Donald N. Cohen; Andrew K.C. Wong (1973). "An application of information theory to genetic mutations and the matching of polypeptide sequences". Journal of Theoretical Biology. 42 (2): 245–261. doi:10.1016/0022-5193(73)90088-X. PMID 4762954.
8. ^ William A. Beyer, Myron L. Stein, Temple F. Smith, and Stanislaw M. Ulam (1974). "A molecular sequence metric and evolutionary trees". Mathematical Biosciences. 19 (1–2): 9–25. doi:10.1016/0025-5564(74)90028-5.CS1 maint: multiple names: authors list (link)
9. ^ Peter H. Sellers (1974). "On the Theory and Computation of Evolutionary Distances". Journal of Applied Mathematics. 26 (4): 787–793. doi:10.1137/0126070.
10. ^ M.S Waterman; T.F Smith; W.A Beyer (1976). "Some biological sequence metrics". Advances in Mathematics. 20 (3): 367–387. doi:10.1016/0001-8708(76)90202-4.
11. ^ D. S. Hirschberg (1975). "A linear space algorithm for computing maximal common subsequences". Communications of the ACM. 18 (6): 341–343. CiteSeerX 10.1.1.348.4774. doi:10.1145/360825.360861.
12. ^ Wozniak, Andrzej (1997). "Using video-oriented instructions to speed up sequence comparison" (PDF). Computer Applications in Biosciences (CABIOS). 13 (2): 145–50. doi:10.1093/bioinformatics/13.2.145. PMID 9146961.
13. ^ a b c Farrar, Michael S. (2007). "Striped Smith–Waterman speeds database searches six times over other SIMD implementations" (PDF). Bioinformatics. 23 (2): 156–161. doi:10.1093/bioinformatics/btl582. PMID 17110365.
14. ^ Zhao, Mengyao; Lee, Wan-Ping; Garrison, Erik P; Marth, Gabor T (4 December 2013). "SSW Library: An SIMD Smith-Waterman C/C++ Library for Use in Genomic Applications". PLoS ONE. 8 (12): e82138. arXiv:1208.6350. Bibcode:2013PLoSO...882138Z. doi:10.1371/journal.pone.0082138. PMC 3852983. PMID 24324759.
15. ^ FPGA 100x Papers: "Archived copy" (PDF). Archived from the original (PDF) on 2008-07-05. Retrieved 2007-10-17.CS1 maint: archived copy as title (link), "Archived copy" (PDF). Archived from the original (PDF) on 2008-07-05. Retrieved 2007-10-17.CS1 maint: archived copy as title (link), and "Archived copy" (PDF). Archived from the original (PDF) on 2011-07-20. Retrieved 2007-10-17.CS1 maint: archived copy as title (link)
16. ^
17. ^ Vermij, Erik (2011). Genetic sequence alignment on a supercomputing platform (PDF) (M.Sc. thesis). Delft University of Technology. Archived from the original (PDF) on 2011-09-30. Retrieved 2011-08-17.
18. ^ Liu, Yang; Huang, Wayne; Johnson, John; Vaidya, Sheila (2006). GPU Accelerated Smith–Waterman. Lecture Notes in Computer Science. 3994. SpringerLink. pp. 188–195. doi:10.1007/11758549_29. ISBN 978-3-540-34385-1.
19. ^ "Bioinformatics High Throughput Sequence Search and Analysis (white paper)". GenomeQuest. Archived from the original on May 13, 2008. Retrieved 2008-05-09.
20. ^ Manavski, Svetlin A. & Valle, Giorgio (2008). "CUDA compatible GPU cards as efficient hardware accelerators for Smith–Waterman sequence alignment". BMC Bioinformatics. 9 (Suppl 2:S10): S10. doi:10.1186/1471-2105-9-S2-S10. PMC 2323659. PMID 18387198.
21. ^ "CUDA Zone". Nvidia. Retrieved 2010-02-25.
22. ^ Rognes, Torbjørn & Seeberg, Erling (2000). "Six-fold speed-up of Smith–Waterman sequence database searches using parallel processing on common microprocessors" (PDF). Bioinformatics. 16 (8): 699–706. doi:10.1093/bioinformatics/16.8.699. PMID 11099256.
23. ^ Rognes, Torbjørn (2011). "Faster Smith–Waterman database searches with inter-sequence SIMD parallelisation". BMC Bioinformatics. 12: 221. doi:10.1186/1471-2105-12-221. PMC 3120707. PMID 21631914.
24. ^ Farrar, Michael S. (2008). "Optimizing Smith–Waterman for the Cell Broadband Engine". Archived from the original on 2012-02-12. Cite journal requires |journal= (help)
25. ^ Maleki, Ehsan; Koohi, Somayyeh; Kavehvash, Zahra; Mashaghi, Alireza (2020). "OptCAM: An ultra‐fast all‐optical architecture for DNA variant discovery". Journal of Biophotonics. 13 (1): e201900227. doi:10.1002/jbio.201900227. PMID 31397961.
| 6,845
| 27,610
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 56, "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}
| 2.96875
| 3
|
CC-MAIN-2020-24
|
latest
|
en
| 0.905242
|
https://www.pgrmseducation.com/class-6-maths-chapter-6-exercise-6-1-pdf-notes-ncert-solutions/
| 1,701,685,036,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100527.35/warc/CC-MAIN-20231204083733-20231204113733-00048.warc.gz
| 1,050,448,524
| 23,268
|
# Class 6 Maths Chapter 6 Exercise 6.1 Pdf Notes NCERT Solutions
Class 6 Maths Chapter 6 Integers Exercise 6.1 pdf notes:-
Exercise 6.1 Class 6 maths Chapter 6 Pdf Notes:-
## Ncert Solution for Class 6 Maths Chapter 6 Integers Exercise 6.1 Tips:-
Introduction:-
Sunita’s mother has 8 bananas. Sunita has to
go for a picnic with her friends. She wants to
carry 10 bananas with her. Can her mother
give 10 bananas to her? She does not have
enough, so she borrows 2 bananas from her
neighbour to be returned later. After giving
10 bananas to Sunita, how many bananas are
left with her mother? Can we say that she has
zero bananas? She has no bananas with her,
but has to return two to her neighbour. So
when she gets some more bananas, say 6, she
will return 2 and be left with 4 only.
Ronald goes to the market to purchase a pen. He has only `12 with him but the pen costs` 15. The shopkeeper writes `3 as due amount from him. He writes` 3 in his diary to remember Ronald’s debit. But how would he remember
whether ` 3 has to be given or has to be taken from Ronald? Can he express this
debit by some colour or sign?
Ruchika and Salma are playing a game using a number strip which is
marked from 0 to 25 at equal intervals.
To begin with, both of them placed a coloured token at the zero mark. Two
coloured dice are placed in a bag and are taken out by them one by one. If the
die is red in colour, the token is moved forward as per the number shown on
throwing this die. If it is blue, the token is moved backward as per the number
shown when this die is thrown. The dice are put back into the bag after each
move so that both of them have equal chance of getting either die. The one
who reaches the 25th mark first is the winner. They play the game. Ruchika
gets the red die and gets four on the die after throwing it. She, thus, moves the
token to mark four on the strip. Salma also happens to take out the red die and
wins 3 points and, thus, moves her token to number 3.
In the second attempt, Ruchika secures three points with the red die and
Salma gets 4 points but with the blue die. Where do you think both of them
should place their token after the second attempt?
Ruchika moves forward and reaches 4 + 3 i.e. the 7th mark.
Whereas Salma placed her token at zero position. But Ruchika objected
saying she should be behind zero. Salma agreed. But there is nothing behind
zero. What can they do?
Salma and Ruchika then extended the strip on the other side. They used a
blue strip on the other side.
#### Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
• Sst 8th Class
• #### Test Paper Of Class 7th
• Maths 7th Class
• Science 7th class
• #### Test Paper Of Class 6th
• Maths 6th Class
• Science 6th class
| 741
| 2,735
|
{"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}
| 4.59375
| 5
|
CC-MAIN-2023-50
|
latest
|
en
| 0.967866
|
http://cepsnumeracy.global2.vic.edu.au/category/multiplication-2/page/2/
| 1,529,693,044,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864776.82/warc/CC-MAIN-20180622182027-20180622202027-00568.warc.gz
| 61,010,219
| 17,363
|
## Figure Me Out Project
### Posted by Miss Gardiner | Posted in Addition, Division, Four Operations, Multiplication, Subtraction | Posted on August 17, 2016
Students create questions based around themselves with numerical answers, e.g. age, house number, number of siblings, number of letters in their first name, shoe size.. (the list is endless). Once students have the answers, they have to create equations to be worked out to get to their answer, e.g ‘My Age’ Answer=12, Equation could be 2×6= 10+2= or even more advanced using BODMAS, decimals, fractions etc.
Once students have created the equations and answers, they present these on a poster with the equation displayed on a piece of paper or a post it note that can be opened up to display the answer underneath.
Figure Me Out Project
## The Array game.
### Posted by Miss Gardiner | Posted in Addition, Multiplication | Posted on August 17, 2016
Students will need:
– Two coloured pencils
– Graphing paper
– Dice (this can be modified according to levels)
– Dice mats
Students work in pairs.
Students roll two dice.
They colour in the squares on the graph paper that the dice multiply to, an array. For example if they roll a 2 and a 3. So it would be 2×3=6 they would then colour a box 2 by 3 squares to equal 6 squares in their chosen colour on the page and write the multiplication sum in grey lead on the squares.
It would then be the next child’s turn. They would then roll the dice and colour squares in the other coloured pencil.
Once all the squares are coloured in the children will need to add the total of their squares together, they can use their chosen addition strategy.
The student with the most squares coloured wins.
## Four Operations Board Game
### Posted by lisar | Posted in Addition, Division, Four Operations, Multiplication, Subtraction | Posted on September 17, 2015
Students design and create a game board in the style of their choice.
Students create question cards using the four operations. (i.e 8 addition equations, 8 subtraction, 8 multiplication and 8 division) There can be a mixture of number sentences and worded problems.
Each process is written on a different colour card.
Game boards include sections where players pick up cards as well as general game instructions. (i.e move back 2 spaces. Roll a 5 to move again)
The game board can also be used for:
• Measurement (to measure how long your journey is)
• Money (collecting and taking away money amounts along the way)
## Car Park Dominoes
### Posted by lisar | Posted in Addition, Multiplication, Subtraction | Posted on September 17, 2015
Car Park Dominoes.
(prep)
Materials:
Template
Set of dominoes.
Instructions: The aim is to fill up your carpark spaces by adding the two numbers on the domino together and putting it into the space with correct answer.
An extra challenge for students could be that they need to write their own numbers in the spaces.
This activity could also be done for subtraction and multiplication with the blank template provided.
### Posted by lisar | Posted in Addition, Multiplication | Posted on September 17, 2015
Students play with a partner
• Both students draw 4 columns.
• 1 student rolls a dice – that¹s how many goals their footy team scored in the first quarter.
• Student either uses addition or multiplication to calculate the number of points their team scored in the first quarter. (each goal is 6 points)
• 2nd student rolls and calculates their score for the first quarter.
• The process is repeated for the 4 quarters of the game.
• Both students then need to add together the totals of the 4 quarters to work out their total for the match.
• Students decide who the winner of the game is.
Simplifying
• use counters in groups of 6 to help with adding on
• using a 6 sided dice
• Using repeated addition function on calculators
• Using calculators
Extending:
• Use multiplication (x6)
• Include goals and behinds ( students roll twice for each quarter multiply goals by 6 and add on the number of behinds)
• In the case of a draw students play ‘extra time’ (another roll for goals and behinds)
### Posted by MissHerring | Posted in Area, Division, Money, Multiplication | Posted on September 6, 2015
This is an open ended project that covers concepts multiplication, area and money.
Students are given the task of designing their own café. They work through the stages of the project and based on abilities can vary the complexity of each stage.
Stages include:
– Working out the area of rooms using a given floor plan
– Staffing with cooks and waiters
– Profits that are earned
## Animal Legs
### Posted by Miss Lawrence | Posted in Addition, Multiplication, Number | Posted on September 4, 2015
Explain to the students you have been to Melbourne Zoo and discuss some of the animals you saw.
Ask students to imagine that they have also been to the Zoo. Have students think of a number (can set limits depending on ability). This number represents how many animal legs they saw during their visit. They then need to decide which animals they saw, and how many of each.
Capable students could use the Zoo website (http://www.zoo.org.au/melbourne) to find information about the different animals and use the Zoo map, otherwise print a copy for students.
If needed, students can use concrete materials to represent the legs. This activity can also incorporate place value by using unifix and putting it into sticks of 10 to keep track of how many animal legs they have accounted for.
A further extension activity could be:
• plan the day at the Zoo to see all the animals, taking into account special keeper talks and events that might happen during the day.
• calculate the cost of the visit, including food and drinks as well as special activities.
• research what the animals eat and how much food is needed over a particular time.
## Missing Keys on a Calculator
### Posted by Miss Forscutt | Posted in Addition, Division, Four Operations, Multiplication, Number, Subtraction | Posted on December 3, 2014
This is an opened ended activity, allowing students to use the four operations. Students read the problem and record all the ways that they try to solve it.
Problem:
Sally was making numbers on her calculator but the keys for the number 6 and the number 7 were broken.
Sally wanted to display the number 467 in the calculator’s window.
How many ways can you make 467 on the calculator if the 6 and the 7 keys are broken?
Example of a solution:
400 + 30 + 30 + 4 + 3 = 467
## Multiplication Snap
### Posted by misspitt | Posted in Multiplication | Posted on December 3, 2014
This game is designed to reinforce multiplication skills.
Materials required (deck of cards)
Students work in pairs.
A=1
1 to 10 = itself
J = 11
Q = 12
Students work in pairs turning over 1 card each.
Students need to quickly multiply the two turned cards.
The fastest correct response collects both cards.
Students repeat until deck is completed.
The winner is the student with the most cards at the end.
Differentiation:
2. Turn 1 card creating single digit sums. (Remove J,Q,K 1×10 problems only)
3. Turn 1 card creating single digit sums. (Include J,Q,K)
4. Students turn 2 cards each and create 2 X 2 digit equations.
5. Students turn 3 cards each and create 3 X 3 digit equations.
## QR codes for Four Operation games
### Posted by Miss Gardiner | Posted in Addition, Division, Four Operations, Games, Multiplication, Subtraction | Posted on December 3, 2014
Four Operation Games – QR Codes
Students need an iPad or iPod with the QR code scanner application.
Students can scan the QR codes and it takes them to a game that involves one of the four operations.
| 1,749
| 7,726
|
{"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-2018-26
|
latest
|
en
| 0.935977
|
http://gmatclub.com/forum/n-is-a-positive-integer-is-k-5-2-0-1-n-ends-in-38868.html
| 1,371,686,044,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368709906749/warc/CC-MAIN-20130516131146-00062-ip-10-60-113-184.ec2.internal.warc.gz
| 113,331,553
| 26,450
|
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 19 Jun 2013, 16:54
# n is a positive integer. Is (k-5)^2>0? (1) n! ends in
Author Message
TAGS:
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1278
Followers: 10
Kudos [?]: 72 [0], given: 0
n is a positive integer. Is (k-5)^2>0? (1) n! ends in [#permalink] 21 Nov 2006, 05:18
00:00
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
n is a positive integer. Is (k-5)^2>0?
(1) n! ends in exactly k zeros.
(2) There are k prime numbers less than 10n
Director
Joined: 02 Mar 2006
Posts: 582
Location: France
Followers: 1
Kudos [?]: 2 [0], given: 0
D
1. n! finishes with five 0 if contains an integer multiplied by 10^5.
10=2.5 so n! has to contain 5 multiples of 2 and 5 multiples of 5.
so n=1.2.3.4.5.6.7.8.9.10....15....20= 20! contains four '5' and more than five '2'. The next possible integer is 25! which contains six '5'.
So there is no n! finishing with five '0'.
SUFFICIENT
(2) There are k prime numbers less than 10n
if n=1 then 10 contains 4 prime numbers.
if n=2 then 20 contains 8 primes.
So again, there are no n satisfaying the condition.
SUFFICIENT
VP
Joined: 25 Jun 2006
Posts: 1185
Followers: 1
Kudos [?]: 16 [0], given: 0
agree with D and the above explanation.
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1
Kudos [?]: 3 [0], given: 0
Just to give you guys my breakdown for (1) ...
If n=1, then n!=1 and k =0 --> (0-5)^2 > 0 holds true
The only time (k-5)^2>0 will not be true is if k=5. And k will be = to 5 when we have a multiple of 100,000 or 10^5 (ending in five 0s). 100,000 = (5^5) * (2^5)
But I'm not too sure if (1) is SUFFICIENT. I'm thinking if you get the factorial of a bigger number like 40 or 50 (although i didnt compute for this), you might get a number ending in five zeros. Wasn't patient enough to compute tho... Might be too cumbersome for an actual GMAT question so I would go with D. But I think B is a possibility...
_________________
Impossible is nothing
Current Student
Joined: 28 Dec 2004
Posts: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]: 135 [0], given: 2
(1) how can this be sufficient? if K=5 then (k-5)^2=0?
karlfurt wrote:
D
1. n! finishes with five 0 if contains an integer multiplied by 10^5.
10=2.5 so n! has to contain 5 multiples of 2 and 5 multiples of 5.
so n=1.2.3.4.5.6.7.8.9.10....15....20= 20! contains four '5' and more than five '2'. The next possible integer is 25! which contains six '5'.
So there is no n! finishing with five '0'.
SUFFICIENT
(2) There are k prime numbers less than 10n
if n=1 then 10 contains 4 prime numbers.
if n=2 then 20 contains 8 primes.
So again, there are no n satisfaying the condition.
SUFFICIENT
Current Student
Joined: 28 Dec 2004
Posts: 3437
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 11
Kudos [?]: 135 [0], given: 2
Oh dang!..X&Y good work...i just saw your working
I am soo out of touch ....with GMAT..
Similar topics Replies Last post
Similar
Topics:
1 If n and p are integers, is p>0? 1. n+1>0 2. np>0 4 26 Apr 2006, 07:43
If n and p are integers, is p>0? 1. n+1 > 0 2. np>0 6 08 Sep 2007, 19:24
If n and p are integers, is p>0? 1. n+1>0 2. np>0 2 27 Dec 2007, 07:50
If n is an integer is n - 1 > 0? 2 16 Feb 2011, 06:57
3 If n is an integer, is n – 1 > 0? 12 24 Apr 2012, 22:31
Display posts from previous: Sort by
| 1,253
| 3,446
|
{"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}
| 3.84375
| 4
|
CC-MAIN-2013-20
|
latest
|
en
| 0.833888
|
http://us.metamath.org/mpegif/stirlinglem5.html
| 1,532,151,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00216.warc.gz
| 376,768,334
| 48,333
|
Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > stirlinglem5 Structured version Unicode version
Theorem stirlinglem5 27804
Description: If is between and , then a series (without alternating negative and positive terms) is given that converges to log (1+T)/(1-T) . (Contributed by Glauco Siliprandi, 29-Jun-2017.)
Hypotheses
Ref Expression
stirlinglem5.1
stirlinglem5.2
stirlinglem5.3
stirlinglem5.4
stirlinglem5.5
stirlinglem5.6
stirlinglem5.7
Assertion
Ref Expression
stirlinglem5
Distinct variable groups: , ,
Allowed substitution hints: () () () () ()
Proof of Theorem stirlinglem5
Dummy variables are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 nnuz 10522 . . . . 5
2 1z 10312 . . . . . 6
32a1i 11 . . . . 5
4 stirlinglem5.1 . . . . . . . . 9
54a1i 11 . . . . . . . 8
6 ax-1cn 9049 . . . . . . . . . . . . . 14
76a1i 11 . . . . . . . . . . . . 13
87negcld 9399 . . . . . . . . . . . 12
9 nnm1nn0 10262 . . . . . . . . . . . . 13
109adantl 454 . . . . . . . . . . . 12
118, 10expcld 11524 . . . . . . . . . . 11
12 nncn 10009 . . . . . . . . . . . 12
1312adantl 454 . . . . . . . . . . 11
14 stirlinglem5.6 . . . . . . . . . . . . . . 15
1514rpred 10649 . . . . . . . . . . . . . 14
1615recnd 9115 . . . . . . . . . . . . 13
1716adantr 453 . . . . . . . . . . . 12
18 nnnn0 10229 . . . . . . . . . . . . 13
1918adantl 454 . . . . . . . . . . . 12
2017, 19expcld 11524 . . . . . . . . . . 11
21 nnne0 10033 . . . . . . . . . . . 12
2221adantl 454 . . . . . . . . . . 11
2311, 13, 20, 22div32d 9814 . . . . . . . . . 10
247, 17pncan2d 9414 . . . . . . . . . . . . 13
2524eqcomd 2442 . . . . . . . . . . . 12
2625oveq1d 6097 . . . . . . . . . . 11
2726oveq2d 6098 . . . . . . . . . 10
2823, 27eqtr3d 2471 . . . . . . . . 9
2928mpteq2dva 4296 . . . . . . . 8
305, 29eqtrd 2469 . . . . . . 7
3130seqeq3d 11332 . . . . . 6
326a1i 11 . . . . . . . . . 10
3332, 16addcld 9108 . . . . . . . . . 10
34 eqid 2437 . . . . . . . . . . 11
3534cnmetdval 18806 . . . . . . . . . 10
3632, 33, 35syl2anc 644 . . . . . . . . 9
37 1m1e0 10069 . . . . . . . . . . . . . 14
3837a1i 11 . . . . . . . . . . . . 13
3938oveq1d 6097 . . . . . . . . . . . 12
4032, 32, 16subsub4d 9443 . . . . . . . . . . . 12
41 df-neg 9295 . . . . . . . . . . . . . 14
4241eqcomi 2441 . . . . . . . . . . . . 13
4342a1i 11 . . . . . . . . . . . 12
4439, 40, 433eqtr3d 2477 . . . . . . . . . . 11
4544fveq2d 5733 . . . . . . . . . 10
4616absnegd 12252 . . . . . . . . . . 11
47 stirlinglem5.7 . . . . . . . . . . 11
4846, 47eqbrtrd 4233 . . . . . . . . . 10
4945, 48eqbrtrd 4233 . . . . . . . . 9
5036, 49eqbrtrd 4233 . . . . . . . 8
51 cnxmet 18808 . . . . . . . . . 10
5251a1i 11 . . . . . . . . 9
53 1re 9091 . . . . . . . . . . 11
5453a1i 11 . . . . . . . . . 10
5554rexrd 9135 . . . . . . . . 9
56 elbl2 18421 . . . . . . . . 9
5752, 55, 32, 33, 56syl22anc 1186 . . . . . . . 8
5850, 57mpbird 225 . . . . . . 7
59 eqid 2437 . . . . . . . 8
6059logtayl2 20554 . . . . . . 7
6158, 60syl 16 . . . . . 6
6231, 61eqbrtrd 4233 . . . . 5
63 seqex 11326 . . . . . 6
6463a1i 11 . . . . 5
65 stirlinglem5.2 . . . . . . . 8
6665a1i 11 . . . . . . 7
6766seqeq3d 11332 . . . . . 6
68 logtayl 20552 . . . . . . 7
6916, 47, 68syl2anc 644 . . . . . 6
7067, 69eqbrtrd 4233 . . . . 5
71 simpr 449 . . . . . . 7
7271, 1syl6eleq 2527 . . . . . 6
734a1i 11 . . . . . . . 8
74 oveq1 6089 . . . . . . . . . . 11
7574oveq2d 6098 . . . . . . . . . 10
76 oveq2 6090 . . . . . . . . . . 11
77 id 21 . . . . . . . . . . 11
7876, 77oveq12d 6100 . . . . . . . . . 10
7975, 78oveq12d 6100 . . . . . . . . 9
8079adantl 454 . . . . . . . 8
81 elfznn 11081 . . . . . . . . 9
8281adantl 454 . . . . . . . 8
836a1i 11 . . . . . . . . . . . 12
8483negcld 9399 . . . . . . . . . . 11
85 nnm1nn0 10262 . . . . . . . . . . 11
8684, 85expcld 11524 . . . . . . . . . 10
8782, 86syl 16 . . . . . . . . 9
8816ad2antrr 708 . . . . . . . . . . 11
8982nnnn0d 10275 . . . . . . . . . . 11
9088, 89expcld 11524 . . . . . . . . . 10
9182nncnd 10017 . . . . . . . . . 10
9282nnne0d 10045 . . . . . . . . . 10
9390, 91, 92divcld 9791 . . . . . . . . 9
9487, 93mulcld 9109 . . . . . . . 8
9573, 80, 82, 94fvmptd 5811 . . . . . . 7
9695, 94eqeltrd 2511 . . . . . 6
97 addcl 9073 . . . . . . 7
9897adantl 454 . . . . . 6
9972, 96, 98seqcl 11344 . . . . 5
10065a1i 11 . . . . . . . 8
10178adantl 454 . . . . . . . 8
102100, 101, 82, 93fvmptd 5811 . . . . . . 7
103102, 93eqeltrd 2511 . . . . . 6
10472, 103, 98seqcl 11344 . . . . 5
105 simpll 732 . . . . . . 7
106 stirlinglem5.3 . . . . . . . . . 10
107106a1i 11 . . . . . . . . 9
10879, 78oveq12d 6100 . . . . . . . . . 10
109108adantl 454 . . . . . . . . 9
110 simpr 449 . . . . . . . . 9
11186adantl 454 . . . . . . . . . . 11
11216adantr 453 . . . . . . . . . . . . 13
113110nnnn0d 10275 . . . . . . . . . . . . 13
114112, 113expcld 11524 . . . . . . . . . . . 12
115110nncnd 10017 . . . . . . . . . . . 12
116110nnne0d 10045 . . . . . . . . . . . 12
117114, 115, 116divcld 9791 . . . . . . . . . . 11
118111, 117mulcld 9109 . . . . . . . . . 10
119118, 117addcld 9108 . . . . . . . . 9
120107, 109, 110, 119fvmptd 5811 . . . . . . . 8
1214a1i 11 . . . . . . . . . . 11
12279adantl 454 . . . . . . . . . . 11
123121, 122, 110, 118fvmptd 5811 . . . . . . . . . 10
124123eqcomd 2442 . . . . . . . . 9
12565a1i 11 . . . . . . . . . . 11
12678adantl 454 . . . . . . . . . . 11
127125, 126, 110, 117fvmptd 5811 . . . . . . . . . 10
128127eqcomd 2442 . . . . . . . . 9
129124, 128oveq12d 6100 . . . . . . . 8
130120, 129eqtrd 2469 . . . . . . 7
131105, 82, 130syl2anc 644 . . . . . 6
13272, 96, 103, 131seradd 11366 . . . . 5
1331, 3, 62, 64, 70, 99, 104, 132climadd 12426 . . . 4
134 1rp 10617 . . . . . . . . 9
135134a1i 11 . . . . . . . 8
136135, 14rpaddcld 10664 . . . . . . 7
137136rpne0d 10654 . . . . . 6
13833, 137logcld 20469 . . . . 5
13932, 16subcld 9412 . . . . . 6
14015, 54absltd 12233 . . . . . . . . . 10
14147, 140mpbid 203 . . . . . . . . 9
142141simprd 451 . . . . . . . 8
14315, 142gtned 9209 . . . . . . 7
14432, 16, 143subne0d 9421 . . . . . 6
145139, 144logcld 20469 . . . . 5
146138, 145negsubd 9418 . . . 4
147133, 146breqtrd 4237 . . 3
148 nn0uz 10521 . . . 4
149 0z 10294 . . . . 5
150149a1i 11 . . . 4
151 stirlinglem5.5 . . . . . 6
152 2nn0 10239 . . . . . . . . 9
153152a1i 11 . . . . . . . 8
154 id 21 . . . . . . . 8
155153, 154nn0mulcld 10280 . . . . . . 7
156 nn0p1nn 10260 . . . . . . 7
157155, 156syl 16 . . . . . 6
158151, 157fmpti 5893 . . . . 5
159158a1i 11 . . . 4
160 2re 10070 . . . . . . . . 9
161160a1i 11 . . . . . . . 8
162 nn0re 10231 . . . . . . . 8
163161, 162remulcld 9117 . . . . . . 7
16453a1i 11 . . . . . . . . 9
165162, 164readdcld 9116 . . . . . . . 8
166161, 165remulcld 9117 . . . . . . 7
167 2rp 10618 . . . . . . . . 9
168167a1i 11 . . . . . . . 8
169162ltp1d 9942 . . . . . . . 8
170162, 165, 168, 169ltmul2dd 10701 . . . . . . 7
171163, 166, 164, 170ltadd1dd 9638 . . . . . 6
172151a1i 11 . . . . . . 7
173 simpr 449 . . . . . . . . 9
174173oveq2d 6098 . . . . . . . 8
175174oveq1d 6097 . . . . . . 7
176 id 21 . . . . . . 7
177 2cn 10071 . . . . . . . . . 10
178177a1i 11 . . . . . . . . 9
179 nn0cn 10232 . . . . . . . . 9
180178, 179mulcld 9109 . . . . . . . 8
1816a1i 11 . . . . . . . 8
182180, 181addcld 9108 . . . . . . 7
183172, 175, 176, 182fvmptd 5811 . . . . . 6
184 simpr 449 . . . . . . . . 9
185184oveq2d 6098 . . . . . . . 8
186185oveq1d 6097 . . . . . . 7
187 peano2nn0 10261 . . . . . . 7
188179, 181addcld 9108 . . . . . . . . 9
189178, 188mulcld 9109 . . . . . . . 8
190189, 181addcld 9108 . . . . . . 7
191172, 186, 187, 190fvmptd 5811 . . . . . 6
192171, 183, 1913brtr4d 4243 . . . . 5
193192adantl 454 . . . 4
194 eldifi 3470 . . . . . . 7
195194adantl 454 . . . . . 6
1966a1i 11 . . . . . . . . . . 11
197196negcld 9399 . . . . . . . . . 10
198194, 85syl 16 . . . . . . . . . 10
199197, 198expcld 11524 . . . . . . . . 9
200199adantl 454 . . . . . . . 8
20116adantr 453 . . . . . . . . . 10
202195nnnn0d 10275 . . . . . . . . . 10
203201, 202expcld 11524 . . . . . . . . 9
204195nncnd 10017 . . . . . . . . 9
205195nnne0d 10045 . . . . . . . . 9
206203, 204, 205divcld 9791 . . . . . . . 8
207200, 206mulcld 9109 . . . . . . 7
208207, 206addcld 9108 . . . . . 6
209108, 106fvmptg 5805 . . . . . 6
210195, 208, 209syl2anc 644 . . . . 5
211 eldifn 3471 . . . . . . . . . . . 12
212 0nn0 10237 . . . . . . . . . . . . . . . 16
213 1nn0 10238 . . . . . . . . . . . . . . . . 17
214152, 213num0h 10393 . . . . . . . . . . . . . . . 16
215 oveq2 6090 . . . . . . . . . . . . . . . . . . 19
216215oveq1d 6097 . . . . . . . . . . . . . . . . . 18
217216eqeq2d 2448 . . . . . . . . . . . . . . . . 17
218217rspcev 3053 . . . . . . . . . . . . . . . 16
219212, 214, 218mp2an 655 . . . . . . . . . . . . . . 15
220151elrnmpt 5118 . . . . . . . . . . . . . . . 16
2216, 220ax-mp 8 . . . . . . . . . . . . . . 15
222219, 221mpbir 202 . . . . . . . . . . . . . 14
223222a1i 11 . . . . . . . . . . . . 13
224 eleq1 2497 . . . . . . . . . . . . 13
225223, 224mpbird 225 . . . . . . . . . . . 12
226211, 225nsyl 116 . . . . . . . . . . 11
227 nn1m1nn 10021 . . . . . . . . . . . . 13
228194, 227syl 16 . . . . . . . . . . . 12
229228ord 368 . . . . . . . . . . 11
230226, 229mpd 15 . . . . . . . . . 10
231 nfcv 2573 . . . . . . . . . . . . . . . . . 18
232 nfmpt1 4299 . . . . . . . . . . . . . . . . . . . 20
233151, 232nfcxfr 2570 . . . . . . . . . . . . . . . . . . 19
234233nfrn 5113 . . . . . . . . . . . . . . . . . 18
235231, 234nfdif 3469 . . . . . . . . . . . . . . . . 17
236235nfcri 2567 . . . . . . . . . . . . . . . 16
237151elrnmpt 5118 . . . . . . . . . . . . . . . . . . . . . . . . 25
238211, 237mtbid 293 . . . . . . . . . . . . . . . . . . . . . . . 24
239 ralnex 2716 . . . . . . . . . . . . . . . . . . . . . . . 24
240238, 239sylibr 205 . . . . . . . . . . . . . . . . . . . . . . 23
241240r19.21bi 2805 . . . . . . . . . . . . . . . . . . . . . 22
242241neneqad 2675 . . . . . . . . . . . . . . . . . . . . 21
243242necomd 2688 . . . . . . . . . . . . . . . . . . . 20
244243adantlr 697 . . . . . . . . . . . . . . . . . . 19
245 simplr 733 . . . . . . . . . . . . . . . . . . . 20
246 simpr 449 . . . . . . . . . . . . . . . . . . . 20
247194ad2antrr 708 . . . . . . . . . . . . . . . . . . . 20
248160a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
249 simpl 445 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
250249zred 10376 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
251248, 250remulcld 9117 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
252 0re 9092 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
253252a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
25453a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
255177a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
256250recnd 9115 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
257255, 256mulcomd 9110 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
258 simpr 449 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
259 elnn0z 10295 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
260258, 259sylnib 297 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
261 nan 565 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
262260, 261mpbi 201 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
263262anabss1 789 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
264250, 253ltnled 9221 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
265263, 264mpbird 225 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
266167a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
267266rpregt0d 10655 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
268 mulltgt0 27670 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
269250, 265, 267, 268syl21anc 1184 . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
270257, 269eqbrtrd 4233 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
271251, 253, 254, 270ltadd1dd 9638 . . . . . . . . . . . . . . . . . . . . . . . . . 26
2726a1i 11 . . . . . . . . . . . . . . . . . . . . . . . . . . 27
273272addid2d 9268 . . . . . . . . . . . . . . . . . . . . . . . . . 26
274271, 273breqtrd 4237 . . . . . . . . . . . . . . . . . . . . . . . . 25
275251, 254readdcld 9116 . . . . . . . . . . . . . . . . . . . . . . . . . 26
276275, 254ltnled 9221 . . . . . . . . . . . . . . . . . . . . . . . . 25
277274, 276mpbid 203 . . . . . . . . . . . . . . . . . . . . . . . 24
278 nnge1 10027 . . . . . . . . . . . . . . . . . . . . . . . 24
279277, 278nsyl 116 . . . . . . . . . . . . . . . . . . . . . . 23
280279adantr 453 . . . . . . . . . . . . . . . . . . . . . 22
281 simpr 449 . . . . . . . . . . . . . . . . . . . . . . . 24
282 simpl 445 . . . . . . . . . . . . . . . . . . . . . . . 24
283281, 282eqeltrd 2511 . . . . . . . . . . . . . . . . . . . . . . 23
284283adantll 696 . . . . . . . . . . . . . . . . . . . . . 22
285280, 284mtand 642 . . . . . . . . . . . . . . . . . . . . 21
286285neneqad 2675 . . . . . . . . . . . . . . . . . . . 20
287245, 246, 247, 286syl21anc 1184 . . . . . . . . . . . . . . . . . . 19
288244, 287pm2.61dan 768 . . . . . . . . . . . . . . . . . 18
289288neneqd 2618 . . . . . . . . . . . . . . . . 17
290289ex 425 . . . . . . . . . . . . . . . 16
291236, 290ralrimi 2788 . . . . . . . . . . . . . . 15
292 ralnex 2716 . . . . . . . . . . . . . . 15
293291, 292sylib 190 . . . . . . . . . . . . . 14
294194nnzd 10375 . . . . . . . . . . . . . . 15
295 odd2np1 12909 . . . . . . . . . . . . . . 15
296294, 295syl 16 . . . . . . . . . . . . . 14
297293, 296mtbird 294 . . . . . . . . . . . . 13
298297notnotrd 108 . . . . . . . . . . . 12
299194nncnd 10017 . . . . . . . . . . . . 13
300299, 196npcand 9416 . . . . . . . . . . . 12
301298, 300breqtrrd 4239 . . . . . . . . . . 11
302198nn0zd 10374 . . . . . . . . . . . 12
303 oddp1even 12911 . . . . . . . . . . . 12
304302, 303syl 16 . . . . . . . . . . 11
305301, 304mpbird 225 . . . . . . . . . 10
306 oexpneg 12912 . . . . . . . . . 10
307196, 230, 305, 306syl3anc 1185 . . . . . . . . 9
308 1exp 11410 . . . . . . . . . . 11
309302, 308syl 16 . . . . . . . . . 10
310309negeqd 9301 . . . . . . . . 9
311307, 310eqtrd 2469 . . . . . . . 8
312311adantl 454 . . . . . . 7
313312oveq1d 6097 . . . . . 6
314313oveq1d 6097 . . . . 5
315206mulm1d 9486 . . . . . . 7
316315oveq1d 6097 . . . . . 6
317206negcld 9399 . . . . . . 7
318317, 206addcomd 9269 . . . . . 6
319206negidd 9402 . . . . . 6
320316, 318, 3193eqtrd 2473 . . . . 5
321210, 314, 3203eqtrd 2473 . . . 4
322120, 119eqeltrd 2511 . . . 4
323106a1i 11 . . . . . . 7
324 simpr 449 . . . . . . . . . . 11
325324oveq1d 6097 . . . . . . . . . 10
326325oveq2d 6098 . . . . . . . . 9
327324oveq2d 6098 . . . . . . . . . 10
328327, 324oveq12d 6100 . . . . . . . . 9
329326, 328oveq12d 6100 . . . . . . . 8
330329, 328oveq12d 6100 . . . . . . 7
331152a1i 11 . . . . . . . . 9
332 simpr 449 . . . . . . . . 9
333331, 332nn0mulcld 10280 . . . . . . . 8
334 nn0p1nn 10260 . . . . . . . 8
335333, 334syl 16 . . . . . . 7
336181negcld 9399 . . . . . . . . . . 11
337180, 181pncand 9413 . . . . . . . . . . . 12
338152a1i 11 . . . . . . . . . . . . 13
339338, 176nn0mulcld 10280 . . . . . . . . . . . 12
340337, 339eqeltrd 2511 . . . . . . . . . . 11
341336, 340expcld 11524 . . . . . . . . . 10
342341adantl 454 . . . . . . . . 9
34316adantr 453 . . . . . . . . . . 11
344213a1i 11 . . . . . . . . . . . 12
345333, 344nn0addcld 10279 . . . . . . . . . . 11
346343, 345expcld 11524 . . . . . . . . . 10
347177a1i 11 . . . . . . . . . . . 12
348179adantl 454 . . . . . . . . . . . 12
349347, 348mulcld 9109 . . . . . . . . . . 11
3506a1i 11 . . . . . . . . . . 11
351349, 350addcld 9108 . . . . . . . . . 10
352252a1i 11 . . . . . . . . . . 11
353160a1i 11 . . . . . . . . . . . . 13
354162adantl 454 . . . . . . . . . . . . 13
355353, 354remulcld 9117 . . . . . . . . . . . 12
35653a1i 11 . . . . . . . . . . . 12
357152nn0ge0i 10250 . . . . . . . . . . . . . 14
358357a1i 11 . . . . . . . . . . . . 13
359332nn0ge0d 10278 . . . . . . . . . . . . 13
360353, 354, 358, 359mulge0d 9604 . . . . . . . . . . . 12
361 0lt1 9551 . . . . . . . . . . . . 13
362361a1i 11 . . . . . . . . . . . 12
363355, 356, 360, 362addgegt0d 9601 . . . . . . . . . . 11
364352, 363gtned 9209 . . . . . . . . . 10
365346, 351, 364divcld 9791 . . . . . . . . 9
366342, 365mulcld 9109 . . . . . . . 8
367366, 365addcld 9108 . . . . . . 7
368323, 330, 335, 367fvmptd 5811 . . . . . 6
369337adantl 454 . . . . . . . . . . . 12
370369oveq2d 6098 . . . . . . . . . . 11
371 m1expeven 27702 . . . . . . . . . . . 12
372371adantl 454 . . . . . . . . . . 11
373370, 372eqtrd 2469 . . . . . . . . . 10
374373oveq1d 6097 . . . . . . . . 9
375365mulid2d 9107 . . . . . . . . 9
376374, 375eqtrd 2469 . . . . . . . 8
377376oveq1d 6097 . . . . . . 7
3783652timesd 10211 . . . . . . 7
379346, 351, 364divrec2d 9795 . . . . . . . 8
380379oveq2d 6098 . . . . . . 7
381377, 378, 3803eqtr2d 2475 . . . . . 6
382368, 381eqtr2d 2470 . . . . 5
383 stirlinglem5.4 . . . . . . 7
384383a1i 11 . . . . . 6
385 simpr 449 . . . . . . . . . . 11
386385oveq2d 6098 . . . . . . . . . 10
387386oveq1d 6097 . . . . . . . . 9
388387oveq2d 6098 . . . . . . . 8
389387oveq2d 6098 . . . . . . . 8
390388, 389oveq12d 6100 . . . . . . 7
391390oveq2d 6098 . . . . . 6
392351, 364reccld 9784 . . . . . . . 8
393392, 346mulcld 9109 . . . . . . 7
394347, 393mulcld 9109 . . . . . 6
395384, 391, 332, 394fvmptd 5811 . . . . 5
396213a1i 11 . . . . . . . . 9
397339, 396nn0addcld 10279 . . . . . . . 8
398172, 175, 176, 397fvmptd 5811 . . . . . . 7
399398adantl 454 . . . . . 6
400399fveq2d 5733 . . . . 5
401382, 395, 4003eqtr4d 2479 . . . 4
402148, 1, 150, 3, 159, 193, 321, 322, 401isercoll2 12463 . . 3
403147, 402mpbird 225 . 2
40454, 15resubcld 9466 . . . 4
40516subidd 9400 . . . . . 6
406405eqcomd 2442 . . . . 5
40715, 54, 15, 142ltsub1dd 9639 . . . . 5
408406, 407eqbrtrd 4233 . . . 4
409404, 408elrpd 10647 . . 3
410136, 409relogdivd 20522 . 2
411403, 410breqtrrd 4239 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wb 178 wo 359 wa 360 wceq 1653 wcel 1726 wne 2600 wral 2706 wrex 2707 cvv 2957 cdif 3318 class class class wbr 4213 cmpt 4267 crn 4880 ccom 4883 wf 5451 cfv 5455 (class class class)co 6082 cc 8989 cr 8990 cc0 8991 c1 8992 caddc 8994 cmul 8996 cxr 9120 clt 9121 cle 9122 cmin 9292 cneg 9293 cdiv 9678 cn 10001 c2 10050 cn0 10222 cz 10283 cuz 10489 crp 10613 cfz 11044 cseq 11324 cexp 11383 cabs 12040 cli 12279 cdivides 12853 cxmt 16687 cbl 16689 clog 20453 This theorem is referenced by: stirlinglem6 27805 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1556 ax-5 1567 ax-17 1627 ax-9 1667 ax-8 1688 ax-13 1728 ax-14 1730 ax-6 1745 ax-7 1750 ax-11 1762 ax-12 1951 ax-ext 2418 ax-rep 4321 ax-sep 4331 ax-nul 4339 ax-pow 4378 ax-pr 4404 ax-un 4702 ax-inf2 7597 ax-cnex 9047 ax-resscn 9048 ax-1cn 9049 ax-icn 9050 ax-addcl 9051 ax-addrcl 9052 ax-mulcl 9053 ax-mulrcl 9054 ax-mulcom 9055 ax-addass 9056 ax-mulass 9057 ax-distr 9058 ax-i2m1 9059 ax-1ne0 9060 ax-1rid 9061 ax-rnegex 9062 ax-rrecex 9063 ax-cnre 9064 ax-pre-lttri 9065 ax-pre-lttrn 9066 ax-pre-ltadd 9067 ax-pre-mulgt0 9068 ax-pre-sup 9069 ax-addf 9070 ax-mulf 9071 This theorem depends on definitions: df-bi 179 df-or 361 df-an 362 df-3or 938 df-3an 939 df-tru 1329 df-ex 1552 df-nf 1555 df-sb 1660 df-eu 2286 df-mo 2287 df-clab 2424 df-cleq 2430 df-clel 2433 df-nfc 2562 df-ne 2602 df-nel 2603 df-ral 2711 df-rex 2712 df-reu 2713 df-rmo 2714 df-rab 2715 df-v 2959 df-sbc 3163 df-csb 3253 df-dif 3324 df-un 3326 df-in 3328 df-ss 3335 df-pss 3337 df-nul 3630 df-if 3741 df-pw 3802 df-sn 3821 df-pr 3822 df-tp 3823 df-op 3824 df-uni 4017 df-int 4052 df-iun 4096 df-iin 4097 df-br 4214 df-opab 4268 df-mpt 4269 df-tr 4304 df-eprel 4495 df-id 4499 df-po 4504 df-so 4505 df-fr 4542 df-se 4543 df-we 4544 df-ord 4585 df-on 4586 df-lim 4587 df-suc 4588 df-om 4847 df-xp 4885 df-rel 4886 df-cnv 4887 df-co 4888 df-dm 4889 df-rn 4890 df-res 4891 df-ima 4892 df-iota 5419 df-fun 5457 df-fn 5458 df-f 5459 df-f1 5460 df-fo 5461 df-f1o 5462 df-fv 5463 df-isom 5464 df-ov 6085 df-oprab 6086 df-mpt2 6087 df-of 6306 df-1st 6350 df-2nd 6351 df-riota 6550 df-recs 6634 df-rdg 6669 df-1o 6725 df-2o 6726 df-oadd 6729 df-er 6906 df-map 7021 df-pm 7022 df-ixp 7065 df-en 7111 df-dom 7112 df-sdom 7113 df-fin 7114 df-fi 7417 df-sup 7447 df-oi 7480 df-card 7827 df-cda 8049 df-pnf 9123 df-mnf 9124 df-xr 9125 df-ltxr 9126 df-le 9127 df-sub 9294 df-neg 9295 df-div 9679 df-nn 10002 df-2 10059 df-3 10060 df-4 10061 df-5 10062 df-6 10063 df-7 10064 df-8 10065 df-9 10066 df-10 10067 df-n0 10223 df-z 10284 df-dec 10384 df-uz 10490 df-q 10576 df-rp 10614 df-xneg 10711 df-xadd 10712 df-xmul 10713 df-ioo 10921 df-ioc 10922 df-ico 10923 df-icc 10924 df-fz 11045 df-fzo 11137 df-fl 11203 df-mod 11252 df-seq 11325 df-exp 11384 df-fac 11568 df-bc 11595 df-hash 11620 df-shft 11883 df-cj 11905 df-re 11906 df-im 11907 df-sqr 12041 df-abs 12042 df-limsup 12266 df-clim 12283 df-rlim 12284 df-sum 12481 df-ef 12671 df-sin 12673 df-cos 12674 df-tan 12675 df-pi 12676 df-dvds 12854 df-struct 13472 df-ndx 13473 df-slot 13474 df-base 13475 df-sets 13476 df-ress 13477 df-plusg 13543 df-mulr 13544 df-starv 13545 df-sca 13546 df-vsca 13547 df-tset 13549 df-ple 13550 df-ds 13552 df-unif 13553 df-hom 13554 df-cco 13555 df-rest 13651 df-topn 13652 df-topgen 13668 df-pt 13669 df-prds 13672 df-xrs 13727 df-0g 13728 df-gsum 13729 df-qtop 13734 df-imas 13735 df-xps 13737 df-mre 13812 df-mrc 13813 df-acs 13815 df-mnd 14691 df-submnd 14740 df-mulg 14816 df-cntz 15117 df-cmn 15415 df-psmet 16695 df-xmet 16696 df-met 16697 df-bl 16698 df-mopn 16699 df-fbas 16700 df-fg 16701 df-cnfld 16705 df-top 16964 df-bases 16966 df-topon 16967 df-topsp 16968 df-cld 17084 df-ntr 17085 df-cls 17086 df-nei 17163 df-lp 17201 df-perf 17202 df-cn 17292 df-cnp 17293 df-haus 17380 df-cmp 17451 df-tx 17595 df-hmeo 17788 df-fil 17879 df-fm 17971 df-flim 17972 df-flf 17973 df-xms 18351 df-ms 18352 df-tms 18353 df-cncf 18909 df-limc 19754 df-dv 19755 df-ulm 20294 df-log 20455
Copyright terms: Public domain W3C validator
| 12,230
| 22,724
|
{"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-2018-30
|
latest
|
en
| 0.100617
|
https://www.codecogs.com/library/maths/combinatorics/permutations/derangement_check.php
| 1,701,225,183,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100047.66/warc/CC-MAIN-20231129010302-20231129040302-00173.warc.gz
| 818,624,575
| 7,587
|
I have forgotten
• https://me.yahoo.com
COST (GBP)
0.42
0.00
0
# Derangement Check
Verifies if a given permutation is a derangement.
Controller: CodeCogs
C++
## Derangement Check
boolderangement_check( int n int* a )[inline]
This function checks whether a given permutation, given as an array of numbers in the range $\inline&space;&space;1,&space;2,&space;\ldots&space;n$, is a derangement. A derangement of the integers 1 through n is a permutation of the integers such that the first value is not 1, the second is not 2, and so on. In more formal terms, the following condition stands:
$\sigma(i)&space;\neq&space;i,&space;\quad&space;\forall&space;i&space;=&space;\overline{1,&space;n}$
where $\inline&space;&space;\sigma$ is the given permutation.
## Example:
#include <codecogs/maths/combinatorics/permutations/derangement_check.h>
#include <iostream>
int main()
{
int x[4] = {4, 3, 2, 1}, y[5] = {1, 3, 2, 5, 4};
std::cout << Maths::Combinatorics::Permutations::derangement_check(4, x) << std::endl;
std::cout << Maths::Combinatorics::Permutations::derangement_check(4, y) << std::endl;
return 0;
}
## Output:
1
0
## References:
SUBSET, a C++ library of combinatorial routines, http://www.csit.fsu.edu/~burkardt/cpp_src/subset/subset.html
### Parameters
n the size of the permutation a the actual permutation given as an array
### Returns
true, if the permutation is a derangement, false otherwise
### Authors
Lucian Bentea (August 2005)
##### Source Code
Source code is available when you agree to a GP Licence or buy a Commercial Licence.
Not a member, then Register with CodeCogs. Already a Member, then Login.
| 468
| 1,644
|
{"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": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2023-50
|
longest
|
en
| 0.566752
|
https://www.splashlearn.com/math/interpret-data-in-picture-graphs-games
| 1,709,398,228,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00146.warc.gz
| 968,839,268
| 27,650
|
• Interpret data in picture graphs
##### Read and Interpret Data using Picture Graphs Game
Use your math skills to read and interpret data using picture graphs.
1 2 2.MD.10
VIEW DETAILS
• Interpret data in picture graphs
##### Read Data on Picture Graphs Game
Take a deep dive into the world of math with our 'Read Data on Picture Graphs' game.
1 2 2.MD.10
VIEW DETAILS
• Interpret data in picture graphs
##### Read Data and Find the Total using Picture Graph Game
Apply your knowledge to read data and find the total using picture graphs.
1 2 2.MD.10
VIEW DETAILS
• Interpret data in picture graphs
##### Find How Many More or Less using Picture Graphs Game
Find 'how many more or less' using picture graphs in this game.
1 2 2.MD.10
VIEW DETAILS
• Interpret data in picture graphs
Unearth the wisdom of mathematics by learning how to read scaled data.
3 3.MD.3
VIEW DETAILS
• Interpret data in picture graphs
##### Read and Compare Scaled Data Game
Take the first step towards building your math castle by practicing to read and compare scaled data.
3 3.MD.3
VIEW DETAILS
| 265
| 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}
| 2.5625
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.708664
|
https://avxhm.se/ebooks/0133902838_1.html
| 1,582,211,732,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00385.warc.gz
| 303,405,149
| 34,188
|
# Bayesian Methods for Hackers: Probabilistic Programming and Bayesian Inference
Bayesian Methods for Hackers: Probabilistic Programming and Bayesian Inference by Cameron Davidson-Pilon
English | 2 Oct. 2015 | ISBN: 0133902838 | 250 Pages | PDF (True) | 19.4 MB
Master Bayesian Inference through Practical Examples and Computation–Without Advanced Mathematical Analysis
Bayesian methods of inference are deeply natural and extremely powerful. However, most discussions of Bayesian inference rely on intensely complex mathematical analyses and artificial examples, making it inaccessible to anyone without a strong mathematical background. Now, though, Cameron Davidson-Pilon introduces Bayesian inference from a computational perspective, bridging theory to practice–freeing you to get results using computing power.
Bayesian Methods for Hackers illuminates Bayesian inference through probabilistic programming with the powerful PyMC language and the closely related Python tools NumPy, SciPy, and Matplotlib. Using this approach, you can reach effective solutions in small increments, without extensive mathematical intervention.
Davidson-Pilon begins by introducing the concepts underlying Bayesian inference, comparing it with other techniques and guiding you through building and training your first Bayesian model. Next, he introduces PyMC through a series of detailed examples and intuitive explanations that have been refined after extensive user feedback. You’ll learn how to use the Markov Chain Monte Carlo algorithm, choose appropriate sample sizes and priors, work with loss functions, and apply Bayesian inference in domains ranging from finance to marketing. Once you’ve mastered these techniques, you’ll constantly turn to this guide for the working PyMC code you need to jumpstart future projects.
Coverage includes
• Learning the Bayesian “state of mind” and its practical implications
• Understanding how computers perform Bayesian inference
• Using the PyMC Python library to program Bayesian analyses
• Building and debugging models with PyMC
• Testing your model’s “goodness of fit”
• Opening the “black box” of the Markov Chain Monte Carlo algorithm to see how and why it works
• Leveraging the power of the “Law of Large Numbers”
• Mastering key concepts, such as clustering, convergence, autocorrelation, and thinning
• Using loss functions to measure an estimate’s weaknesses based on your goals and desired outcomes
• Selecting appropriate priors and understanding how their influence changes with dataset size
• Overcoming the “exploration versus exploitation” dilemma: deciding when “pretty good” is good enough
• Using Bayesian inference to improve A/B testing
• Solving data science problems when only small amounts of data are available
Cameron Davidson-Pilon has worked in many areas of applied mathematics, from the evolutionary dynamics of genes and diseases to stochastic modeling of financial prices. His contributions to the open source community include lifelines, an implementation of survival analysis in Python. Educated at the University of Waterloo and at the Independent University of Moscow, he currently works with the online commerce leader Shopify.
| 601
| 3,204
|
{"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-2020-10
|
latest
|
en
| 0.859332
|
https://www.construct.net/forum/construct-2/how-do-i-18/battle-turns-based-on-speed-st-61812
| 1,534,461,215,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221211316.39/warc/CC-MAIN-20180816230727-20180817010727-00560.warc.gz
| 878,323,409
| 8,398
|
# Battle Turns based on Speed Stats
0 Favourites
• 4 posts
• Hi everyone, I'm in the process of trying to build a classic turn-based RPG. Since random, turn-based battles are the meat and potatoes of this, I'm working on getting my battle system up and running before I do anything else. I prototyped the basics, and it's easy enough to get a system in place where I just take turns between players and enemies, but I want to determine the order of turns per round based on a unit's speed stat - including re-calculating the turn order every round based on things like whether or not a unit died / fled (and thus the turn is skipped) or if they've been hastened/slowed.
From what I've been able to gather, the best way to accomplish this is through an array. I've been trying to wrap my head around them, and the way I currently envision it functioning is that at the start of the layout and then any given turn, I call a function to set each cell to a unit's speed stat based on the order of highest to lowest, and then draw the order of turns from the array. Once all player unit actions have been selected, it switches into a combat round where all units (player and enemy) take their turns in the order they were assigned.
The problem I'm running into is I can't seem to figure out how to actually sort the cells from high to low. I know that I can draw the max and minimum speeds from a list of all units in the battle, but I'm not sure how to go about retrieving anything in-between. The 'sort' action for the array doesn't seem to work either.
If anyone could help me figure that out, it'd be much appreciated! Bonus points if you happen to have any ideas how I'd distribute those turns to each unit after the fact, though I figure I'll cross that bridge when I come to it. <img src="smileys/smiley17.gif" border="0" align="middle" />
• My first thoughts on how I would go about this is to have each good guy and bad guy have an instance variable called "turnMeter" which would have a value between 0 and 10. Each would also have an instance variable called "meterRate" which would typically be a value between 0.1 and 3. So when a player has a low level, their meterRate would be low like 0.1 so it would take awhile for their turnMeter to get filled up to 10. As they progress in level, their meterRate increases. If they have a "FastSpell" cast on them, I would multiply the meterRate by 1.5 to give them a 50% boost in speed temporarily.
So before each round starts, you'll determine the order of play by which units can fill up their turnMeter the quickest. If they are dead, their meterRate is 0.
• I hadn't even thought of switching to a meter-based system, but it's definitely easier to put together and it's working perfectly. Thank you!
• ## Try Construct 3
Develop games in your browser. Powerful, performant & highly capable.
Construct 3 users don't see these ads
• I would use instance variables over an Array. And everything farmerdwight has suggested.
| 691
| 2,983
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.979736
|
http://www.chegg.com/homework-help/operations-research-applications-and-algorithms-4th-edition-chapter-19.5-solutions-9780534380588
| 1,455,295,316,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-07/segments/1454701164289.84/warc/CC-MAIN-20160205193924-00039-ip-10-236-182-209.ec2.internal.warc.gz
| 342,730,858
| 18,017
|
View more editions
# TEXTBOOK SOLUTIONS FOR Operations Research Applications and Algorithms 4th Edition
• 1466 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 18
a.
Consider the production policy of a manufacturing process. During any stage when production is done, a set-up cost of and variable cost of is incurred. A holding cost of is incurred on excess units that remain. The inventory has a maximum capacity of 3 units. Likely demand during each period is 1 or 2 units.
• Step 2 of 18
The set of possible states is with 0 to 3 units in inventory. During each period, number of units to produce is decided. Hence set of decisions is
Here represents that 0 units are produced; represents that 1 unit are produced; represents that 2 units are produced andrepresents that 3 units are produced;
• Step 3 of 18
The following transition probabilities are obtained.
If initial inventory is 0 units and units produced is 2 units then with probability 0.5, 1 unit remains in inventory and with probability 0.5, 0 units remains. So
• Step 4 of 18
In general if initial inventory is i units and units produced is 0 units then ending inventory is or with probability 0.5 each. So
• Step 5 of 18
If initial inventory is i units and units produced is 1 unit, then ending inventory is with 0.5probability. So
• Step 6 of 18
If initial inventory is i units and units produced is 2 units, then ending inventory is with 0.5probability. So
• Step 7 of 18
If initial inventory is i units and units produced is 3 units, then ending inventory is with 0.5probability. So
• Step 8 of 18
Production cost is given by
Here x is the number of units produced. Holding cost is in ending inventory.
• Step 9 of 18
Define as the expected discounted expenses during an infinite number of periods, with a stationary policy. To determine static policy we obtain the value determination equations by using the formula
• Step 10 of 18
Use the following policy to obtain value determination equations. . In this policy 4 units are produced when state is I0, 3 units are produced when state is I1 and 0 units are produced when state is I2 and I3. So
Hence when state is I0, 4 units are produced.
Production cost is.
Ending inventory is 3 units or 2 units.
Hence holding cost is.
So, total cost is .
• Step 11 of 18
Hence using value determination equation we have
Similarly for states I1, I2 and I3 we have
Solving these equations, we have
Use Howard’s Policy Iteration Method to calculate
So
For each state i, . Thus is an optimal policy. Hence optimum solution is obtained by producing 4 units when inventory is 0, 3 units when inventory is 1, 0 units when inventory is 2 units, 0 units when inventory is 3 units.
• Step 12 of 18
b.
We have the following Linear programming equations for maximization problem
Substituting the values,
• Step 13 of 18
LINGO/LINDO is used to solve the LP problem by the following procedure.
1. Start LINGO and open a blank file.
2. Enter LP equations in the file.
• Step 14 of 18
The screen shot is shown below.
• Step 15 of 18
3. Click on LINGO from tool bar and select solve. Solution is obtained in new window as shown.
The output is given below.
• Step 16 of 18
The LINDO output yields that, , and . The LINDO output also indicates that third, fourth, seventh, eighth, and eleventh constraints have no slack. Hence optimum solution is obtained by producing 4 units when inventory is 0, 3 units when inventory is 1, 0 units when inventory is 2 units, 0 units when inventory is 3 units.
• Step 17 of 18
c.
Value Iteration: Let be the maximum expected discount reward that can be earned during t periods if the state at the beginning of current period is i. Then
As t increases, approaches. Here is the maximum expected discounted reward earned during an infinite number of periods if the state is i at the beginning of the current period;
Represents the decision that must be chosen during period t in state i to attain .
• Step 18 of 18
Using the computations performed earlier, for first iteration,
For second iteration,
Thus with two iteration of value iteration, we do not come close to actual values of and have not found optimal solution.
Corresponding Textbook
Operations Research Applications and Algorithms | 4th Edition
9780534380588ISBN-13: 0534380581ISBN: Wayne L WinstonAuthors:
| 1,109
| 4,561
|
{"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}
| 3.828125
| 4
|
CC-MAIN-2016-07
|
latest
|
en
| 0.930839
|
https://www.experts-exchange.com/questions/20876673/class-help.html
| 1,600,429,533,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400187390.18/warc/CC-MAIN-20200918092913-20200918122913-00061.warc.gz
| 1,123,286,111
| 34,642
|
We help IT Professionals succeed at work.
# class help
on
247 Views
ok, still trying at this program. lots of commented code, don't pay attention. I will post my error, followed by my code:
lab2main.o(.text+0xde): In function `main':
: undefined reference to `insertclass::funcsort1(double*, int&)'
collect2: ld returned 1 exit status
make: *** [lab2main] Error 1
#include <iostream>
#include "randnum.h" //this is for my random number generator
#include "insertionclass.h"
//#include "insertionsort.h" //this is for my first sort
//#include "funcsort1.cxx"
using namespace std;
//insertionclass::insertionsort(int& N, double myarray[]); //creates for
//my first sort
void randnum(int& N, double myarray[]); //prototype for my rand numbers
//insertionclass::funcsort1(double myarray[], int& N); //calls sort1
//double funcsort1(double myarray[], int& N);
//double insertionsort(int& N, double myarray[]);
int main()
{
insertclass insert;
//Make sure to comment everything. Tar files.
//do not output every sort, only output statistics
//convert necessary functions to classes
int N; //size of array
cout<<"Enter size of array. ";
cin>>N;
double myarray[N]; //array where rand nums will be stored
//funcsort1(myarray, N);
insert.funcsort1(myarray, N);
return 0;
}
#ifndef INSERTIONCLASSH
#define INSERTIONCLASSH
#include <iostream>
#include "funcsort1.cxx"
#include "insertionclass.cxx"
class insertclass
{
public:
double insertionsort(int& N, double myarray[]);
double funcsort1(double myarray[], int& N);
};
insertclass insert;
#endif
#include<iostream>
insertclass::insertionsort(int& N, double myarray[])
{
int counter=0, counter2=0;//counter is counting comparisons
//counter2 is counting the number of loops
do
{
for(int unsorted = 1; unsorted < N; ++unsorted)
{
double nextitem = myarray[unsorted];
int loc = unsorted;
for(;(loc > 0) && (myarray[loc - 1] > nextitem);
--loc)
myarray[loc] = myarray[loc-1];
myarray[loc] = nextitem;
counter++;
}
cout<<counter<<endl;
funcsort1(myarray, N);
}while(counter2<=5);
return myarray;
};
#include<iostream>
//#include"insertionsort.h"
//#include "insertionclass.h"
using namespace std;
//double insertionsort(int& N, double myarray[]);
insertclass::funcsort1(double myarray[], int& N)
{
int temp3=0, i=0; //my two counters for the rand nums
randnum(N, myarray);
insertionsort(N, myarray);
do //this loop will print out all the numbers in my rand num
//generator
{
cout<<myarray[temp3]<<endl;
temp3++;
i++;
}while(i<=(N-1)); //statement so I do not exceed size of array
return myarray[N];
}
Comment
Watch Question
## View Solution Only
CERTIFIED EXPERT
Commented:
You still owe me a response in your quesion https://www.experts-exchange.com/Programming/Programming_Languages/Cplusplus/Q_20817987.html
You don't seem to maintain your questions, and I have a problem with investing time in something that you just drop.
If you want help, please treat the people who provide help with respect.
Commented:
I'm sorry, I honestly thought I accepted an answer to that. I will fix that right away.
CERTIFIED EXPERT
Commented:
OK. Mistakes happen.
You seem to have the implementations of insertclass::funcsort1 and insertclass::insertionsort in two different files. This is usually not the way you implement a class. Put all the methods for one class into one .cpp file. You probably don't link in the object file for funcsort1. How many files do you have? Are you using a Makefile?
Senior Software Engineer
CERTIFIED EXPERT
Commented:
This one is on us!
(Get your first solution completely free - no credit card required)
Commented:
I will try both of these today, I will post the results tonight, I won't have access to my code until then. Thanks for the ideas in advance.
CERTIFIED EXPERT
Commented:
Axter is right (of course :-)
Commented:
ok, sry for the delay. that error was fixed, but now I have a slurry of others to fix. heh, that's programming for ya
CERTIFIED EXPERT
Commented:
It's just strange that the compiler did not complain about insertinsort(), which is defined the same way.
Commented:
it did, but that is a different error than what I asked for
Gain unlimited access to on-demand training courses with an Experts Exchange subscription.
###### Why Experts Exchange?
Experts Exchange always has the answer, or at the least points me in the correct direction! It is like having another employee that is extremely experienced.
Jim Murphy
Programmer at Smart IT Solutions
Deciding to stick with EE.
Mohamed Asif
Being involved with EE helped me to grow personally and professionally.
Carl Webster
CTP, Sr Infrastructure Consultant
###### Did You Know?
We've partnered with two important charities to provide clean water and computer science education to those who need it most. READ MORE
Connect with Certified Experts to gain insight and support on specific technology challenges including:
• Troubleshooting
• Research
• Professional Opinions
Unlock the solution to this question.
Experts Exchange is the only place where you can interact directly with leading experts in the technology field. Become a member today and access the collective knowledge of thousands of technology experts.
| 1,272
| 5,187
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2020-40
|
latest
|
en
| 0.493688
|
https://www.kodytools.com/units/inertia/from/gamyd2/to/lbkm2
| 1,719,126,972,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00278.warc.gz
| 742,138,214
| 18,156
|
# Gamma Square Yard to Pound Square Kilometer Converter
1 Gamma Square Yard = 1.8433452926027e-15 Pound Square Kilometers
## One Gamma Square Yard is Equal to How Many Pound Square Kilometers?
The answer is one Gamma Square Yard is equal to 1.8433452926027e-15 Pound Square Kilometers and that means we can also write it as 1 Gamma Square Yard = 1.8433452926027e-15 Pound Square Kilometers. Feel free to use our online unit conversion calculator to convert the unit from Gamma Square Yard to Pound Square Kilometer. Just simply enter value 1 in Gamma Square Yard and see the result in Pound Square Kilometer.
Manually converting Gamma Square Yard to Pound Square Kilometer can be time-consuming,especially when you don’t have enough knowledge about Moment of Inertia 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 Gamma Square Yard to Pound Square Kilometer converter tool to get the job done as soon as possible.
We have so many online tools available to convert Gamma Square Yard to Pound Square Kilometer, but not every online tool gives an accurate result and that is why we have created this online Gamma Square Yard to Pound Square Kilometer converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Gamma Square Yard to Pound Square Kilometer (γ·yd2 to lb·km2)
By using our Gamma Square Yard to Pound Square Kilometer conversion tool, you know that one Gamma Square Yard is equivalent to 1.8433452926027e-15 Pound Square Kilometer. Hence, to convert Gamma Square Yard to Pound Square Kilometer, we just need to multiply the number by 1.8433452926027e-15. We are going to use very simple Gamma Square Yard to Pound Square Kilometer conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Gamma Square Yard} = 1 \times 1.8433452926027e-15 = \text{1.8433452926027e-15 Pound Square Kilometers}$$
## What Unit of Measure is Gamma Square Yard?
Gamma square yard is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one yard distance from the rotation axis and having a mass of one gamma.
## What is the Symbol of Gamma Square Yard?
The symbol of Gamma Square Yard is γ·yd2. This means you can also write one Gamma Square Yard as 1 γ·yd2.
## What Unit of Measure is Pound Square Kilometer?
Pound square kilometer is a unit of measurement for moment of inertia. It represents moment of inertia of a single particle rotating at one kilometer distance from the rotation axis and having a mass of one pound.
## What is the Symbol of Pound Square Kilometer?
The symbol of Pound Square Kilometer is lb·km2. This means you can also write one Pound Square Kilometer as 1 lb·km2.
## How to Use Gamma Square Yard to Pound Square Kilometer Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Gamma Square Yard and in the first input field, enter a value.
• From the second dropdown, select Pound Square Kilometer.
• Instantly, the tool will convert the value from Gamma Square Yard to Pound Square Kilometer and display the result in the second input field.
## Example of Gamma Square Yard to Pound Square Kilometer Converter Tool
Gamma Square Yard
1
Pound Square Kilometer
1.8433452926027e-15
# Gamma Square Yard to Pound Square Kilometer Conversion Table
Gamma Square Yard [γ·yd2]Pound Square Kilometer [lb·km2]Description
1 Gamma Square Yard1.8433452926027e-15 Pound Square Kilometer1 Gamma Square Yard = 1.8433452926027e-15 Pound Square Kilometer
2 Gamma Square Yard3.6866905852054e-15 Pound Square Kilometer2 Gamma Square Yard = 3.6866905852054e-15 Pound Square Kilometer
3 Gamma Square Yard5.5300358778081e-15 Pound Square Kilometer3 Gamma Square Yard = 5.5300358778081e-15 Pound Square Kilometer
4 Gamma Square Yard7.3733811704108e-15 Pound Square Kilometer4 Gamma Square Yard = 7.3733811704108e-15 Pound Square Kilometer
5 Gamma Square Yard9.2167264630135e-15 Pound Square Kilometer5 Gamma Square Yard = 9.2167264630135e-15 Pound Square Kilometer
6 Gamma Square Yard1.1060071755616e-14 Pound Square Kilometer6 Gamma Square Yard = 1.1060071755616e-14 Pound Square Kilometer
7 Gamma Square Yard1.2903417048219e-14 Pound Square Kilometer7 Gamma Square Yard = 1.2903417048219e-14 Pound Square Kilometer
8 Gamma Square Yard1.4746762340822e-14 Pound Square Kilometer8 Gamma Square Yard = 1.4746762340822e-14 Pound Square Kilometer
9 Gamma Square Yard1.6590107633424e-14 Pound Square Kilometer9 Gamma Square Yard = 1.6590107633424e-14 Pound Square Kilometer
10 Gamma Square Yard1.8433452926027e-14 Pound Square Kilometer10 Gamma Square Yard = 1.8433452926027e-14 Pound Square Kilometer
100 Gamma Square Yard1.8433452926027e-13 Pound Square Kilometer100 Gamma Square Yard = 1.8433452926027e-13 Pound Square Kilometer
1000 Gamma Square Yard1.8433452926027e-12 Pound Square Kilometer1000 Gamma Square Yard = 1.8433452926027e-12 Pound Square Kilometer
# Gamma Square Yard to Other Units Conversion Table
ConversionDescription
1 Gamma Square Yard = 8.3612736e-7 Gram Square Meter1 Gamma Square Yard in Gram Square Meter is equal to 8.3612736e-7
1 Gamma Square Yard = 0.000083612736 Gram Square Decimeter1 Gamma Square Yard in Gram Square Decimeter is equal to 0.000083612736
1 Gamma Square Yard = 0.0083612736 Gram Square Centimeter1 Gamma Square Yard in Gram Square Centimeter is equal to 0.0083612736
1 Gamma Square Yard = 0.83612736 Gram Square Millimeter1 Gamma Square Yard in Gram Square Millimeter is equal to 0.83612736
1 Gamma Square Yard = 836127.36 Gram Square Micrometer1 Gamma Square Yard in Gram Square Micrometer is equal to 836127.36
1 Gamma Square Yard = 8.3612736e-13 Gram Square Kilometer1 Gamma Square Yard in Gram Square Kilometer is equal to 8.3612736e-13
1 Gamma Square Yard = 836127360000 Gram Square Nanometer1 Gamma Square Yard in Gram Square Nanometer is equal to 836127360000
1 Gamma Square Yard = 0.000001 Gram Square Yard1 Gamma Square Yard in Gram Square Yard is equal to 0.000001
1 Gamma Square Yard = 0.001296 Gram Square Inch1 Gamma Square Yard in Gram Square Inch is equal to 0.001296
1 Gamma Square Yard = 0.000009 Gram Square Foot1 Gamma Square Yard in Gram Square Foot is equal to 0.000009
1 Gamma Square Yard = 3.228305785124e-13 Gram Square Mile1 Gamma Square Yard in Gram Square Mile is equal to 3.228305785124e-13
1 Gamma Square Yard = 8.3612736e-10 Kilogram Square Meter1 Gamma Square Yard in Kilogram Square Meter is equal to 8.3612736e-10
1 Gamma Square Yard = 8.3612736e-8 Kilogram Square Decimeter1 Gamma Square Yard in Kilogram Square Decimeter is equal to 8.3612736e-8
1 Gamma Square Yard = 0.0000083612736 Kilogram Square Centimeter1 Gamma Square Yard in Kilogram Square Centimeter is equal to 0.0000083612736
1 Gamma Square Yard = 0.00083612736 Kilogram Square Millimeter1 Gamma Square Yard in Kilogram Square Millimeter is equal to 0.00083612736
1 Gamma Square Yard = 836.13 Kilogram Square Micrometer1 Gamma Square Yard in Kilogram Square Micrometer is equal to 836.13
1 Gamma Square Yard = 8.3612736e-16 Kilogram Square Kilometer1 Gamma Square Yard in Kilogram Square Kilometer is equal to 8.3612736e-16
1 Gamma Square Yard = 836127360 Kilogram Square Nanometer1 Gamma Square Yard in Kilogram Square Nanometer is equal to 836127360
1 Gamma Square Yard = 1e-9 Kilogram Square Yard1 Gamma Square Yard in Kilogram Square Yard is equal to 1e-9
1 Gamma Square Yard = 0.000001296 Kilogram Square Inch1 Gamma Square Yard in Kilogram Square Inch is equal to 0.000001296
1 Gamma Square Yard = 9e-9 Kilogram Square Foot1 Gamma Square Yard in Kilogram Square Foot is equal to 9e-9
1 Gamma Square Yard = 3.228305785124e-16 Kilogram Square Mile1 Gamma Square Yard in Kilogram Square Mile is equal to 3.228305785124e-16
1 Gamma Square Yard = 0.00083612736 Milligram Square Meter1 Gamma Square Yard in Milligram Square Meter is equal to 0.00083612736
1 Gamma Square Yard = 0.083612736 Milligram Square Decimeter1 Gamma Square Yard in Milligram Square Decimeter is equal to 0.083612736
1 Gamma Square Yard = 8.36 Milligram Square Centimeter1 Gamma Square Yard in Milligram Square Centimeter is equal to 8.36
1 Gamma Square Yard = 836.13 Milligram Square Millimeter1 Gamma Square Yard in Milligram Square Millimeter is equal to 836.13
1 Gamma Square Yard = 836127360 Milligram Square Micrometer1 Gamma Square Yard in Milligram Square Micrometer is equal to 836127360
1 Gamma Square Yard = 8.3612736e-10 Milligram Square Kilometer1 Gamma Square Yard in Milligram Square Kilometer is equal to 8.3612736e-10
1 Gamma Square Yard = 836127360000000 Milligram Square Nanometer1 Gamma Square Yard in Milligram Square Nanometer is equal to 836127360000000
1 Gamma Square Yard = 0.001 Milligram Square Yard1 Gamma Square Yard in Milligram Square Yard is equal to 0.001
1 Gamma Square Yard = 1.3 Milligram Square Inch1 Gamma Square Yard in Milligram Square Inch is equal to 1.3
1 Gamma Square Yard = 0.009 Milligram Square Foot1 Gamma Square Yard in Milligram Square Foot is equal to 0.009
1 Gamma Square Yard = 3.228305785124e-10 Milligram Square Mile1 Gamma Square Yard in Milligram Square Mile is equal to 3.228305785124e-10
1 Gamma Square Yard = 0.83612736 Microgram Square Meter1 Gamma Square Yard in Microgram Square Meter is equal to 0.83612736
1 Gamma Square Yard = 83.61 Microgram Square Decimeter1 Gamma Square Yard in Microgram Square Decimeter is equal to 83.61
1 Gamma Square Yard = 8361.27 Microgram Square Centimeter1 Gamma Square Yard in Microgram Square Centimeter is equal to 8361.27
1 Gamma Square Yard = 836127.36 Microgram Square Millimeter1 Gamma Square Yard in Microgram Square Millimeter is equal to 836127.36
1 Gamma Square Yard = 836127360000 Microgram Square Micrometer1 Gamma Square Yard in Microgram Square Micrometer is equal to 836127360000
1 Gamma Square Yard = 8.3612736e-7 Microgram Square Kilometer1 Gamma Square Yard in Microgram Square Kilometer is equal to 8.3612736e-7
1 Gamma Square Yard = 836127360000000000 Microgram Square Nanometer1 Gamma Square Yard in Microgram Square Nanometer is equal to 836127360000000000
1 Gamma Square Yard = 1 Microgram Square Yard1 Gamma Square Yard in Microgram Square Yard is equal to 1
1 Gamma Square Yard = 1296 Microgram Square Inch1 Gamma Square Yard in Microgram Square Inch is equal to 1296
1 Gamma Square Yard = 9 Microgram Square Foot1 Gamma Square Yard in Microgram Square Foot is equal to 9
1 Gamma Square Yard = 3.228305785124e-7 Microgram Square Mile1 Gamma Square Yard in Microgram Square Mile is equal to 3.228305785124e-7
1 Gamma Square Yard = 8.3612736e-13 Ton Square Meter1 Gamma Square Yard in Ton Square Meter is equal to 8.3612736e-13
1 Gamma Square Yard = 8.3612736e-11 Ton Square Decimeter1 Gamma Square Yard in Ton Square Decimeter is equal to 8.3612736e-11
1 Gamma Square Yard = 8.3612736e-9 Ton Square Centimeter1 Gamma Square Yard in Ton Square Centimeter is equal to 8.3612736e-9
1 Gamma Square Yard = 8.3612736e-7 Ton Square Millimeter1 Gamma Square Yard in Ton Square Millimeter is equal to 8.3612736e-7
1 Gamma Square Yard = 0.83612736 Ton Square Micrometer1 Gamma Square Yard in Ton Square Micrometer is equal to 0.83612736
1 Gamma Square Yard = 8.3612736e-19 Ton Square Kilometer1 Gamma Square Yard in Ton Square Kilometer is equal to 8.3612736e-19
1 Gamma Square Yard = 836127.36 Ton Square Nanometer1 Gamma Square Yard in Ton Square Nanometer is equal to 836127.36
1 Gamma Square Yard = 1e-12 Ton Square Yard1 Gamma Square Yard in Ton Square Yard is equal to 1e-12
1 Gamma Square Yard = 1.296e-9 Ton Square Inch1 Gamma Square Yard in Ton Square Inch is equal to 1.296e-9
1 Gamma Square Yard = 9e-12 Ton Square Foot1 Gamma Square Yard in Ton Square Foot is equal to 9e-12
1 Gamma Square Yard = 3.228305785124e-19 Ton Square Mile1 Gamma Square Yard in Ton Square Mile is equal to 3.228305785124e-19
1 Gamma Square Yard = 0.0000041806368 Carat Square Meter1 Gamma Square Yard in Carat Square Meter is equal to 0.0000041806368
1 Gamma Square Yard = 0.00041806368 Carat Square Decimeter1 Gamma Square Yard in Carat Square Decimeter is equal to 0.00041806368
1 Gamma Square Yard = 0.041806368 Carat Square Centimeter1 Gamma Square Yard in Carat Square Centimeter is equal to 0.041806368
1 Gamma Square Yard = 4.18 Carat Square Millimeter1 Gamma Square Yard in Carat Square Millimeter is equal to 4.18
1 Gamma Square Yard = 4180636.8 Carat Square Micrometer1 Gamma Square Yard in Carat Square Micrometer is equal to 4180636.8
1 Gamma Square Yard = 4.1806368e-12 Carat Square Kilometer1 Gamma Square Yard in Carat Square Kilometer is equal to 4.1806368e-12
1 Gamma Square Yard = 4180636800000 Carat Square Nanometer1 Gamma Square Yard in Carat Square Nanometer is equal to 4180636800000
1 Gamma Square Yard = 0.000005 Carat Square Yard1 Gamma Square Yard in Carat Square Yard is equal to 0.000005
1 Gamma Square Yard = 0.00648 Carat Square Inch1 Gamma Square Yard in Carat Square Inch is equal to 0.00648
1 Gamma Square Yard = 0.000045 Carat Square Foot1 Gamma Square Yard in Carat Square Foot is equal to 0.000045
1 Gamma Square Yard = 1.614152892562e-12 Carat Square Mile1 Gamma Square Yard in Carat Square Mile is equal to 1.614152892562e-12
1 Gamma Square Yard = 2.9493524681643e-8 Ounce Square Meter1 Gamma Square Yard in Ounce Square Meter is equal to 2.9493524681643e-8
1 Gamma Square Yard = 0.0000029493524681643 Ounce Square Decimeter1 Gamma Square Yard in Ounce Square Decimeter is equal to 0.0000029493524681643
1 Gamma Square Yard = 0.00029493524681643 Ounce Square Centimeter1 Gamma Square Yard in Ounce Square Centimeter is equal to 0.00029493524681643
1 Gamma Square Yard = 0.029493524681643 Ounce Square Millimeter1 Gamma Square Yard in Ounce Square Millimeter is equal to 0.029493524681643
1 Gamma Square Yard = 29493.52 Ounce Square Micrometer1 Gamma Square Yard in Ounce Square Micrometer is equal to 29493.52
1 Gamma Square Yard = 2.9493524681643e-14 Ounce Square Kilometer1 Gamma Square Yard in Ounce Square Kilometer is equal to 2.9493524681643e-14
1 Gamma Square Yard = 29493524681.64 Ounce Square Nanometer1 Gamma Square Yard in Ounce Square Nanometer is equal to 29493524681.64
1 Gamma Square Yard = 3.527396194958e-8 Ounce Square Yard1 Gamma Square Yard in Ounce Square Yard is equal to 3.527396194958e-8
1 Gamma Square Yard = 0.000045715054686656 Ounce Square Inch1 Gamma Square Yard in Ounce Square Inch is equal to 0.000045715054686656
1 Gamma Square Yard = 3.1746565754622e-7 Ounce Square Foot1 Gamma Square Yard in Ounce Square Foot is equal to 3.1746565754622e-7
1 Gamma Square Yard = 1.1387513542607e-14 Ounce Square Mile1 Gamma Square Yard in Ounce Square Mile is equal to 1.1387513542607e-14
1 Gamma Square Yard = 1.8433452926027e-12 Kilopound Square Meter1 Gamma Square Yard in Kilopound Square Meter is equal to 1.8433452926027e-12
1 Gamma Square Yard = 1.8433452926027e-10 Kilopound Square Decimeter1 Gamma Square Yard in Kilopound Square Decimeter is equal to 1.8433452926027e-10
1 Gamma Square Yard = 1.8433452926027e-8 Kilopound Square Centimeter1 Gamma Square Yard in Kilopound Square Centimeter is equal to 1.8433452926027e-8
1 Gamma Square Yard = 0.0000018433452926027 Kilopound Square Millimeter1 Gamma Square Yard in Kilopound Square Millimeter is equal to 0.0000018433452926027
1 Gamma Square Yard = 1.84 Kilopound Square Micrometer1 Gamma Square Yard in Kilopound Square Micrometer is equal to 1.84
1 Gamma Square Yard = 1.8433452926027e-18 Kilopound Square Kilometer1 Gamma Square Yard in Kilopound Square Kilometer is equal to 1.8433452926027e-18
1 Gamma Square Yard = 1843345.29 Kilopound Square Nanometer1 Gamma Square Yard in Kilopound Square Nanometer is equal to 1843345.29
1 Gamma Square Yard = 2.2046226218488e-12 Kilopound Square Yard1 Gamma Square Yard in Kilopound Square Yard is equal to 2.2046226218488e-12
1 Gamma Square Yard = 2.857190917916e-9 Kilopound Square Inch1 Gamma Square Yard in Kilopound Square Inch is equal to 2.857190917916e-9
1 Gamma Square Yard = 1.9841603596639e-11 Kilopound Square Foot1 Gamma Square Yard in Kilopound Square Foot is equal to 1.9841603596639e-11
1 Gamma Square Yard = 7.1171959641296e-19 Kilopound Square Mile1 Gamma Square Yard in Kilopound Square Mile is equal to 7.1171959641296e-19
1 Gamma Square Yard = 1.8433452926027e-9 Pound Square Meter1 Gamma Square Yard in Pound Square Meter is equal to 1.8433452926027e-9
1 Gamma Square Yard = 1.8433452926027e-7 Pound Square Decimeter1 Gamma Square Yard in Pound Square Decimeter is equal to 1.8433452926027e-7
1 Gamma Square Yard = 0.000018433452926027 Pound Square Centimeter1 Gamma Square Yard in Pound Square Centimeter is equal to 0.000018433452926027
1 Gamma Square Yard = 0.0018433452926027 Pound Square Millimeter1 Gamma Square Yard in Pound Square Millimeter is equal to 0.0018433452926027
1 Gamma Square Yard = 1843.35 Pound Square Micrometer1 Gamma Square Yard in Pound Square Micrometer is equal to 1843.35
1 Gamma Square Yard = 1.8433452926027e-15 Pound Square Kilometer1 Gamma Square Yard in Pound Square Kilometer is equal to 1.8433452926027e-15
1 Gamma Square Yard = 1843345292.6 Pound Square Nanometer1 Gamma Square Yard in Pound Square Nanometer is equal to 1843345292.6
1 Gamma Square Yard = 2.2046226218488e-9 Pound Square Yard1 Gamma Square Yard in Pound Square Yard is equal to 2.2046226218488e-9
1 Gamma Square Yard = 0.000002857190917916 Pound Square Inch1 Gamma Square Yard in Pound Square Inch is equal to 0.000002857190917916
1 Gamma Square Yard = 1.9841603596639e-8 Pound Square Foot1 Gamma Square Yard in Pound Square Foot is equal to 1.9841603596639e-8
1 Gamma Square Yard = 7.1171959641296e-16 Pound Square Mile1 Gamma Square Yard in Pound Square Mile is equal to 7.1171959641296e-16
1 Gamma Square Yard = 5.9355718422723e-8 Poundal Square Meter1 Gamma Square Yard in Poundal Square Meter is equal to 5.9355718422723e-8
1 Gamma Square Yard = 0.0000059355718422723 Poundal Square Decimeter1 Gamma Square Yard in Poundal Square Decimeter is equal to 0.0000059355718422723
1 Gamma Square Yard = 0.00059355718422723 Poundal Square Centimeter1 Gamma Square Yard in Poundal Square Centimeter is equal to 0.00059355718422723
1 Gamma Square Yard = 0.059355718422723 Poundal Square Millimeter1 Gamma Square Yard in Poundal Square Millimeter is equal to 0.059355718422723
1 Gamma Square Yard = 59355.72 Poundal Square Micrometer1 Gamma Square Yard in Poundal Square Micrometer is equal to 59355.72
1 Gamma Square Yard = 5.9355718422723e-14 Poundal Square Kilometer1 Gamma Square Yard in Poundal Square Kilometer is equal to 5.9355718422723e-14
1 Gamma Square Yard = 59355718422.72 Poundal Square Nanometer1 Gamma Square Yard in Poundal Square Nanometer is equal to 59355718422.72
1 Gamma Square Yard = 7.0988848424626e-8 Poundal Square Yard1 Gamma Square Yard in Poundal Square Yard is equal to 7.0988848424626e-8
1 Gamma Square Yard = 0.000092001547558315 Poundal Square Inch1 Gamma Square Yard in Poundal Square Inch is equal to 0.000092001547558315
1 Gamma Square Yard = 6.3889963582164e-7 Poundal Square Foot1 Gamma Square Yard in Poundal Square Foot is equal to 6.3889963582164e-7
1 Gamma Square Yard = 2.2917371004851e-14 Poundal Square Mile1 Gamma Square Yard in Poundal Square Mile is equal to 2.2917371004851e-14
1 Gamma Square Yard = 0.000012903417048219 Grain Square Meter1 Gamma Square Yard in Grain Square Meter is equal to 0.000012903417048219
1 Gamma Square Yard = 0.0012903417048219 Grain Square Decimeter1 Gamma Square Yard in Grain Square Decimeter is equal to 0.0012903417048219
1 Gamma Square Yard = 0.12903417048219 Grain Square Centimeter1 Gamma Square Yard in Grain Square Centimeter is equal to 0.12903417048219
1 Gamma Square Yard = 12.9 Grain Square Millimeter1 Gamma Square Yard in Grain Square Millimeter is equal to 12.9
1 Gamma Square Yard = 12903417.05 Grain Square Micrometer1 Gamma Square Yard in Grain Square Micrometer is equal to 12903417.05
1 Gamma Square Yard = 1.2903417048219e-11 Grain Square Kilometer1 Gamma Square Yard in Grain Square Kilometer is equal to 1.2903417048219e-11
1 Gamma Square Yard = 12903417048219 Grain Square Nanometer1 Gamma Square Yard in Grain Square Nanometer is equal to 12903417048219
1 Gamma Square Yard = 0.000015432358352941 Grain Square Yard1 Gamma Square Yard in Grain Square Yard is equal to 0.000015432358352941
1 Gamma Square Yard = 0.020000336425412 Grain Square Inch1 Gamma Square Yard in Grain Square Inch is equal to 0.020000336425412
1 Gamma Square Yard = 0.00013889122517647 Grain Square Foot1 Gamma Square Yard in Grain Square Foot is equal to 0.00013889122517647
1 Gamma Square Yard = 4.9820371748907e-12 Grain Square Mile1 Gamma Square Yard in Grain Square Mile is equal to 4.9820371748907e-12
1 Gamma Square Yard = 5.7292934719301e-11 Slug Square Meter1 Gamma Square Yard in Slug Square Meter is equal to 5.7292934719301e-11
1 Gamma Square Yard = 5.7292934719301e-9 Slug Square Decimeter1 Gamma Square Yard in Slug Square Decimeter is equal to 5.7292934719301e-9
1 Gamma Square Yard = 5.7292934719301e-7 Slug Square Centimeter1 Gamma Square Yard in Slug Square Centimeter is equal to 5.7292934719301e-7
1 Gamma Square Yard = 0.000057292934719301 Slug Square Millimeter1 Gamma Square Yard in Slug Square Millimeter is equal to 0.000057292934719301
1 Gamma Square Yard = 57.29 Slug Square Micrometer1 Gamma Square Yard in Slug Square Micrometer is equal to 57.29
1 Gamma Square Yard = 5.7292934719301e-17 Slug Square Kilometer1 Gamma Square Yard in Slug Square Kilometer is equal to 5.7292934719301e-17
1 Gamma Square Yard = 57292934.72 Slug Square Nanometer1 Gamma Square Yard in Slug Square Nanometer is equal to 57292934.72
1 Gamma Square Yard = 6.8521779647661e-11 Slug Square Yard1 Gamma Square Yard in Slug Square Yard is equal to 6.8521779647661e-11
1 Gamma Square Yard = 8.8804226423369e-8 Slug Square Inch1 Gamma Square Yard in Slug Square Inch is equal to 8.8804226423369e-8
1 Gamma Square Yard = 6.1669601682895e-10 Slug Square Foot1 Gamma Square Yard in Slug Square Foot is equal to 6.1669601682895e-10
1 Gamma Square Yard = 2.2120925764353e-17 Slug Square Mile1 Gamma Square Yard in Slug Square Mile is equal to 2.2120925764353e-17
1 Gamma Square Yard = 0.83612736 Gamma Square Meter1 Gamma Square Yard in Gamma Square Meter is equal to 0.83612736
1 Gamma Square Yard = 83.61 Gamma Square Decimeter1 Gamma Square Yard in Gamma Square Decimeter is equal to 83.61
1 Gamma Square Yard = 8361.27 Gamma Square Centimeter1 Gamma Square Yard in Gamma Square Centimeter is equal to 8361.27
1 Gamma Square Yard = 836127.36 Gamma Square Millimeter1 Gamma Square Yard in Gamma Square Millimeter is equal to 836127.36
1 Gamma Square Yard = 836127360000 Gamma Square Micrometer1 Gamma Square Yard in Gamma Square Micrometer is equal to 836127360000
1 Gamma Square Yard = 8.3612736e-7 Gamma Square Kilometer1 Gamma Square Yard in Gamma Square Kilometer is equal to 8.3612736e-7
1 Gamma Square Yard = 836127360000000000 Gamma Square Nanometer1 Gamma Square Yard in Gamma Square Nanometer is equal to 836127360000000000
1 Gamma Square Yard = 1296 Gamma Square Inch1 Gamma Square Yard in Gamma Square Inch is equal to 1296
1 Gamma Square Yard = 9 Gamma Square Foot1 Gamma Square Yard in Gamma Square Foot is equal to 9
1 Gamma Square Yard = 3.228305785124e-7 Gamma Square Mile1 Gamma Square Yard in Gamma Square Mile is equal to 3.228305785124e-7
| 6,654
| 23,437
|
{"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}
| 3.375
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.864517
|
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-8-section-8-1-integration-by-parts-exercises-page-427/43
| 1,576,195,850,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540547536.49/warc/CC-MAIN-20191212232450-20191213020450-00531.warc.gz
| 710,966,968
| 13,001
|
## University Calculus: Early Transcendentals (3rd Edition)
$$\int\sqrt{x}\ln xdx=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$
$$A=\int\sqrt{x}\ln xdx$$ Take $u=\ln x$ and $dv=\sqrt xdx=x^{1/2}dx$ Then we have, $du=\frac{dx}{x}$ and $v=\frac{x^{3/2}}{\frac{3}{2}}=\frac{2x^{3/2}}{3}$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{3/2}\times\frac{dx}{x}$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{1/2}dx$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\times\frac{2x^{3/2}}{3}+C$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$
| 299
| 586
|
{"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}
| 4.25
| 4
|
CC-MAIN-2019-51
|
latest
|
en
| 0.480685
|
https://math.stackexchange.com/questions/3836556/can-two-independent-events-be-disjoint
| 1,631,908,483,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780055775.1/warc/CC-MAIN-20210917181500-20210917211500-00237.warc.gz
| 449,903,045
| 39,365
|
# Can two independent events be disjoint?
If events A and B both have positive probabilities, if they are disjoint, they surely cannot be independent since:
disjoint: P(A intersection B) = 0 <=> P(A union B) = P(A) + P(B)
independent: P(A intersection B) = P(A) * P(B)
so if P(A intersection B) is 0, then P(A) * P(B) should be 0 too, but since they're both above 0, then this is false.
However I am not sure if that is the case the other way around, I cannot put my head around the question if two independent events can be disjoint. Can anyone help? Thanks in advance...
• Welcome to MSE. Please use MathJax to format your posts; you'll find you get a much better response if your questions are easy to read. Sep 22 '20 at 22:15
• The statements “disjoint events with positive probability are not independent” and “independent events with positive probability are not disjoint” are equivalent statements. You have proven the first statement, therefore the second one is true. Sep 22 '20 at 22:54
I agree with the other comments and answers. However I would have attacked the question solely by intuition: "if two events are independent, can they also be disjoint?
It is true that this problem can be attacked with math: assuming that events A and B each have a non-zero probability of occurring, they will be regarded as independent $$\iff p(A) = p(A|B).$$ Since A,B are disjoint, $$p(A|B) = 0.$$ Since it is assumed that $$p(A) > 0, ~p(A) \neq p(A|B).$$ Therefore, the two events can't be independent.
However, this problem can also be attacked by considering
event $$C =$$ the complement of event $$B$$
and showing, purely by intuition, that events $$A$$ and $$C$$ can not be independent.
Consider disjoint events A,B placed in a Venn diagram that represents the universe U.
Informally, $$p(A)$$ may be regarded as the proportion of the area assigned to event $$A$$ versus the area of the entire universe $$U$$ in the Venn diagram.
Since the event $$C$$ completely encompasses the event $$A$$, $$p(A|C)$$ may be similarly regarded as the proportion of the area assigned to event $$A$$ versus the area assigned to event $$C$$, rather than versus the area assigned to $$U$$.
Since $$p(B)$$ is assumed to be non-zero, the area assigned to event $$C$$ must be less than the area assigned to $$U$$. Therefore, the two proportions referred to in the above two paragraphs must be different.
Continuing this informal train of thought, suppose you have any two events $$A$$ and $$B$$, with $$C$$ = the complement of $$B.$$
Suppose further that $$p(A) \neq 0, p(B) \neq 0, p(C) \neq 0.$$
Further suppose that you have somehow concluded that events $$A$$ and $$C$$ are not independent. That means that the chance of $$A$$ occurring has been affected (i.e. altered) by whether it is to be assumed that event $$C$$ has also occurred.
It seems to me that if the chance of $$A$$ occurring has been affected by whether event $$C$$ has also occurred, then it is implied that the chance of $$A$$ occurring has also been affected by whether event $$B$$ has occurred.
In other words, when it is assumed that that $$p(A) \neq 0, p(B) \neq 0,$$ and $$p(C) \neq 0,$$ then regardless of any considerations of disjointness,
events $$A$$ and $$B$$ are independent $$\iff$$ events $$A$$ and $$C$$ are independent.
• Good point! It’s easy to get lost in the formalism without thinking about what’s actually going on Sep 23 '20 at 10:32
Two independent events are disjoint only if at least one of them almost never happens.
More precisely: let $$A, B$$ be two independent events in the sample space $$\Omega$$ which are disjoint. Then $$0 = P(A \cap B) = P(A) * P(B)$$, so at least one of $$A, B$$ must have probability zero.
• I don't know if I misunderstood, but my question was for the other way around... Can two independent events with positive probabilities be disjoint? Sep 22 '20 at 22:34
• @asprog Fixed answer! Sep 22 '20 at 22:57
• thank you so much for the help!! Sep 22 '20 at 23:29
| 1,022
| 3,992
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2021-39
|
latest
|
en
| 0.954102
|
https://powerpivotpro.com/2018/09/using-dax-to-handle-multiple-parent-hierarchies/
| 1,611,047,938,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703518201.29/warc/CC-MAIN-20210119072933-20210119102933-00040.warc.gz
| 514,622,664
| 40,846
|
A common accounting task is to consolidate the financial results of subsidiary companies into combined results of a parent company. A parent company can have one or multiple subsidiaries. Each subsidiary can also have one or multiple subsidiaries beneath it. The relationships between parent companies and subsidiaries are often organized in a hierarchical structure.
In a single parent hierarchy, each child node has a single parent. In a multiple parent hierarchy, each child node can have either a single parent or two or more parents. In the single parent hierarchy diagram below, we can see that entities B and C have A as a parent, D and E have B as a parent, and F has C as a parent. The multiple parent hierarchy differs from the single parent hierarchy in that E has both B and C as a parent.
If we assume companies A, B, C, D, E, and F each have \$100 of income the consolidated total would be as follows in the multiple parent scenario:
F: \$100, E: \$100, D: \$100, C: \$250 (100+100+50 assuming 50% ownership of E) B: \$250, A: \$ 600
For the single parent scenario, the totals would be the same except B would have \$300 and C \$200.
DAX has a hierarchy feature that is relatively straightforward to use for the single parent hierarchy scenario while a multiple parent scenario complicates things. The remainder of this writing will primarily focus on the latter.
The first time I built a hierarchy in DAX, everything appeared to be working well. However, as I checked the numbers, I realized that it was not handling the instances where a child node had multiple parents. I posted a question online asking if it was possible to create a hierarchy with multiple parents. The response I received was yes, it is possible……but it requires a many-to-many relationship to work correctly. I had previously read about many-to-many relationships online, and in Rob’s book, so I had a general idea of how to get started and thought that this problem would provide an excellent reason to explore it in more detail. I have an example workbook that you can download here to follow along
Before we can get to the task of creating a hierarchy, relationships, measures, and Pivot tables, we need three data sources:
1. The hierarchy and ownership table
2. A fact table with some transaction data
3. A table with just a list of the unique hierarchy nodes (company names, etc…)
The first step is to create a table with the hierarchy nodes and ownership percentages. For this exercise, we will have 26 nodes or company names (1 each for A to Z). The first column is the Node ID. The second column is the parent Key. Nodes A, B, and C are top tier and do not have a parent. The nodes below A, B, and C will ultimately be consolidated into these three nodes. Looking at the fourth line, we can see that node A is the sole parent of node D.
Starting on lines 13 and 14, we can see that node M is duplicated in the Name column and that M has both D and G as a parent with each owning 50%. If M has an income of \$100, \$50 will flow up to both Node D and G respectively. The last four rows are for the four-way split of node Z. Node Z is owned 30% by T and U and 20% by W and X. Income for node Z will be split amongst these four nodes by these percentages.
## Hierarchy nodes and ownership table
To summarize the table above, A, B, and C are top-level nodes and do not have a parent. M and Z have multiple parents while the remainder has a single parent. I should also note that when creating a node with multiple parents, the percentage total should add to 100%. Node M above is .5 + .5 and Z is .3+.3+.2+.2.
The second source of data needed is a fact table with transactional data. It will be kept simple for this exercise, and we will assume that all 26 entities from A to Z have an income of \$100 in the year 2017 and \$200 for the year 2018.
## Partial Fact Table
The final data step is to create a table with a unique listing of the nodes.
## Unique Nodes
Name A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Once the data is assembled, it can be loaded into the data model.
To complete the building of the hierarchy, some calculated columns will be added to the Nodes table. The DAX Path function is used to return the ID chain for the hierarchy path. It is a pipe-delimited list that has the parent ID for the chain. For node M on rows 13 and 14, the path is 1|4|13 and 2|7|14. As M has two parents, half of its income will flow up to nodes D and A while the other half will flow to nodes B and G
DAX PATH Function in the nodes table. The first five columns are raw data, and the last seven are calculated columns.
The DAX PATHLENGTH function returns the depth of the hierarchy chain. For this example, the maximum depth is four, so we will build four Levels (Level1,2,3, and 4) for the hierarchy.
For each of the four levels of the hierarchy, the DAX LOOKUPVALUE and PATHITEM functions are used to retrieve the corresponding entity name from the hierarchy path.
To create the hierarchy, while in the diagram view, click on the create hierarchy button in the top right corner and drag the fields to the bottom.
The hierarchy will appear in the pivot table field list.
The next step in the process is to set the relationships in the data model as follows:
Next, we will need to create a few measures.
The first measure is straightforward in that it is only a SUM of the Income amount in the fact table.
Total_Fact_Income :=
SUM ( Fact_Data[Income] )
Our second measure [Total_Income] takes the income measure created above and wraps it inside a CALCULATE and passes the Nodes table as a filter. This is needed for the M2M to work.
Total_Income :=
CALCULATE ( [Total_Fact_income], Nodes )
I would recommend reading pages 220-228 of Rob’s latest book to get a better understanding of why it is done this way. The book provides a deeper dive into the subject and also discusses the subtle differences between PowerPivot and Power BI when dealing with M2M situations.
The third Measure is a SUM of the OwenershipPct column of the Nodes table.
PctOwnership :=
SUM ( Nodes[OwnershipPct] )
The Consolidated_Income measure is the primary measure that will be used for this example. It wraps the Total_Income measure in a SUMX function and multiplies by the ownership percentage.
Consolidated_Income :=
SUMX ( Nodes, [Total_Income] * [PctOwnership] )
Next, we can build a pivot table with the following selections:
This pivot table displays the consolidated year 2017 income for the Top Level entities in the hierarchy. The consolidated total for A+B+C = 2,600.
With nodes A and B of the hierarchy expanded, we can see that 50% of the 2017 \$ 100 income for Node M is allocated first to Node D and then to Node A and the other 50% flows up to G and B. The pivot table displays node M as a child of two different parents within the hierarchy.
The 30,30,20,20 allocation of Entity Z income to the four different parents is shown in the Pivot Table below.
There are two additional measures Parent_Income and Children_Income. I’m not certain that they function correctly, but they are my attempt to segregate the income of only the parent and the income of all children of the parent. The total of both should equal the consolidated income.
## Microsoft’s platform is the world’s most fluid & powerful data toolset. Get the most out of it.
No one knows the Power BI ecosystem better than the folks who started the whole thing – us.
Let us guide your organization through the change required to become a data-oriented culture while squeezing every last drop* of value out of your Microsoft platform investment.
* – we reserve the right to substitute Olympic-sized swimming pools of value in place of “drops” at our discretion.
| 1,777
| 7,762
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.671875
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.963006
|
http://mathforum.org/kb/plaintext.jspa?messageID=7971929
| 1,524,180,898,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125937074.8/warc/CC-MAIN-20180419223925-20180420003925-00354.warc.gz
| 203,007,473
| 3,065
|
```Date: Jan 4, 2013 9:53 PM
Author: David Bernier
Subject: Re: Another count sort that certainly must exist, it do not have<br> any restrictions upon size of (S number of possibilities)
On 01/04/2013 08:55 PM, JT wrote:> On 5 Jan, 02:39, David Bernier<david...@videotron.ca> wrote:>> On 01/04/2013 12:07 PM, David Bernier wrote:> On 01/04/2013 10:46 AM, JT wrote:>>>> On 4 Jan, 15:46, JT<jonas.thornv...@gmail.com> wrote:>>>>> I remember doing this in a tentamen during my education in information>>>>> theory beleiving what i did was binary sort but my teacher informed me>>>>> it wasn't so what is it.>>>> [...]>>>>>>>>>>>>>>>>>>>>>> heap, what is the difference betwee a heap and a tree?).>>>>>> So what you think about the mix using this kind of sort for counting>>>> in values, and then quicksort to sort the none null tree nodes by>>>> sizes.>>>>> Oops.. below is about factoring. The best algorithms>>> have been getting better since Maurice Kraitchik's [1920s]>>> improvement on Fermat's method of expressing a number>>> as a difference of squares, n = a^2 - b^2, so>>> n = (a-b) (a+b).>>>>> There's a very good article called "A Tale of Two Sieves">>> by Carl Pomerance: Notices of the AMS, vol. 43, no. 12,>>> December 1996:>>> <http://www.ams.org/notices/199612/index.html>>>>>> The 9th Fermat number F_9 = 2^(512)+1 had been factored>>> around 1990 by the Lenstras et al using the Number Field>>> sieve (which had supplanted the quadratic sieve).>>>>> The Quadratic sieve is easier to understand than the>>> Number Field Sieve, which I don't understand.>>>>> F_10 and F_11 were fully factored then, using the elliptic>>> curve method (which can find smallish prime factors).>>>>> F_12 was listed as not completely factored, with>>> F_12 being a product of 5 distinct odd primes and>>> the 1187-digit composite:>>>>> C_1187 =>>> 22964766349327374158394934836882729742175302138572\>>>> [...]>>>>> 66912966168394403107609922082657201649660373439896\>>> 3042158832323677881589363722322001921.>>>>> At 3942 bits for C_1187 above, what's the>>> probability density function of expected time>>> till C_1187 is fully factored?>>>> For the Fermat number F_12 = 2^(2^12) + 1 or>> 2^4096 +1 , another prime factor was found around>> 2010. So, this new prime factor would be a divisor>> of C_1187, a 1187-digit number. F_12 is listed>> as known to be "not completely factored".>>>> The relevant line on the Web-page referenced below contains>> the text: "M. Vang, Zimmermann& Kruppa" in the "Discoverer">> column:>> <http://www.prothsearch.net/fermat.html#Complete>>>>> Also, lower down in the page,>> "50 digit k = 17353230210429594579133099699123162989482444520899">>>> This does relate to a factor of F_12 by PARI/gp.>> Then, by my calcultions, the residual unfactored part>> of F_12 has 1133 decimal digits and is a composite number.>>>>> Or, centiles: e.g. 50% chance fully factored>>> within<= 10 years. 95% chance fully factored within>>> <= 95 years, etc. ...>>>> Maybe 50% to 50% chances for "fully factored by 2100 " ?>> (or 2060, or 2200 etc. ... )>>>> dave>> Is reading out a binary tree from smallest to biggest in anyway> related to TSP, i have slight memory of doing a algoritmic solution> for TSP when i tried to factor RSA primeproducts, unfortunatly it is> all gone from my mind. Could you please make a short layman approach> for how TSP and sorting is related to factorization it may come back> to my memory. I lack the key so to speech to the problem.The key lies within the mind. The mind may appear to be a partof reality, but in reality, all is mind ...dave
```
| 1,053
| 3,596
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.109375
| 3
|
CC-MAIN-2018-17
|
latest
|
en
| 0.898201
|
https://www.thestudentroom.co.uk/showthread.php?t=1161462
| 1,527,098,040,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865691.44/warc/CC-MAIN-20180523161206-20180523181206-00247.warc.gz
| 839,787,368
| 43,425
|
You are Here: Home >< Maths
# Differentiating inverse functions watch
1. how would you go about differentiating
x sin^-1 (x^2)
the sin^-1 is an inverse
im not rlly sure but i made sqrt(sin y) = x and sub it in, and differentiate as normal.
[sqrt(1-x^4) + x^2] / 2x
2. Do you know how to differentiate ? If not, working out how to do that on its own would be a good start. Then you can use the product rule and chain rule to complete the question.
Spoiler:
Show
3. In the spoiler I've shown how to differentiate .
I did this (and perhaps I should have just given hints - in fact, I'll split my spoiler up into more spoilers - you should try to do it yourself if you can) because it's useful to know, and if you're doing A level maths, then you may need to reproduce it in an exam. You should know how to find the derivatives of and too.
To find the derivative of , apply the chain rule.
Spoiler:
Show
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\cosy = \sqrt{1 - (sin^{2}y)
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\frac{d}{dx} (\sin^{-1}x) = \frac{1}{\sqrt(1 - x^{2})
Then you can just do what Nuodai said.
EDIT: It's so irritating that I spent quite a while typing out how to differentiate inverse sine in LaTeX, only to have it called potentially dangerous.
I hate LaTeX.
I wish it didn't look so pretty.
Edit: WTF IS ERROR 2?
AND WHAT IS ERROR 6?
GOD DAMN
Edit: Better. At least some of it works.
Edit: I give up. Here it is:
Spoiler:
Show
y = sin^-1x
x = siny
Spoiler:
Show
1 = cosy(dy/dx)
(dy/dx) = 1/cosy
Spoiler:
Show
sin^2y + cos^2y = 1
Spoiler:
Show
cosy = \sqrt(1 - sin^2y)
Spoiler:
Show
cosy = \sqrt(1 - x^2)
(d/dx) sin^-1x = 1\[\sqrt(1-x^2)]
4. (Original post by AnonyMatt)
I hate LaTeX.
You need to leave a space after commands like \sin and \cos.
You might want to justify this step further. At the moment it is unsatisfactory as . You have ignored the minus possibility.
5. (Original post by Kolya)
You need to leave a space after commands like \sin and \cos.
You might want to justify this step further. At the moment it is unsatisfactory as . You have ignored the minus possibility.
Thanks!
Hmm. Do you know, I've never actually noticed that the step is not justified. I think it's allowed to be written like that at A level.
It's obvious to see that it must be positive, but that's not really a justification, is it? (Is it?! )
Is it because
Unparseable or potentially dangerous latex formula. Error 6: Image was not produced or one of its dimensions is too small.
\cosy
* is positive for ?
EDIT: FML LATEX!!! * = cosy
### Related university courses
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
Updated: January 24, 2010
Today on TSR
### Edexcel C2 Core Unofficial Markscheme!
Find out how you've done here
Poll
Useful resources
### Maths Forum posting guidelines
Not sure where to post? Read the updated guidelines here
### How to use LaTex
Writing equations the easy way
### Study habits of A* students
Top tips from students who have already aced their exams
| 903
| 3,406
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.984375
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.913522
|
http://omega.albany.edu:8008/mat214dir/vector-functions-dir/airplane-hit-m2h.mpl
| 1,511,575,774,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00576.warc.gz
| 215,845,123
| 753
|
# R := v0*cos(a)*t; h := v0*sin(a)*t - g*t^2/2; E1 := ('h'+g*t^2/2)^2 = v0^2*t^2-'R'^2; 'R'^2 = v0^2*t^2-('h'+g*t^2/2)^2; h := 'h': F := z -> v0^2*z-(h+g*z/2)^2; diff(F(z),z) = 0; z := solve(diff(F(z),z),z); R := 'R': R^2 = F(z); R^2 = factor(expand(F(z))); R = v0*sqrt(v0^2-2*g*h)/g;
| 163
| 284
|
{"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-2017-47
|
latest
|
en
| 0.482359
|
https://originalsources.com/Document.aspx?DocID=D34J15CMJYNWP8J
| 1,669,734,817,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00230.warc.gz
| 483,206,868
| 13,195
|
The Number of Primes Is Infinite
Author: Euclid
# The Number of Primes Is Infinite
## Euclid
Prime numbers are more than any assigned multitude of prime numbers.
Let A, B, C be the assigned prime numbers; I say that there are more prime numbers than A, B, C.
For let the least number measured by A, B, C be taken, and let it be DE; let the unit DF be added to DE.
Then EF is either prime or not.
First, let it be prime; then the prime numbers A, B, C, EF have been found, which are more than A, B, C.
Next let EF not be prime; therefore it is measured by some prime number. [VII. 31]
Let it be measured by the prime number G.
I say that G is not the same with any of the numbers A, B, C.
For, if possible, let it be so. [p.21]
Now A, B, C measure DE; therefore G also will measure DE.
But it also measures EF.
Therefore G, being a number, will measure the remainder, the unit DF: which is absurd.
Therefore G is not the same with any of the numbers A, B, C.
And by hypothesis it is prime.
Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. Q.E.D.
### Related Resources
Euclid
Life of Euclid
Title: The Number of Primes Is Infinite
Select an option:
## Email Options
Title: The Number of Primes Is Infinite
Select an option:
| 345
| 1,307
|
{"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-2022-49
|
latest
|
en
| 0.932051
|
http://mathforum.org/kb/message.jspa?messageID=8074543
| 1,524,654,985,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125947795.46/warc/CC-MAIN-20180425100306-20180425120306-00013.warc.gz
| 212,160,960
| 6,221
|
Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic: function arity > 2
Replies: 7 Last Post: Jan 17, 2013 1:37 AM
Messages: [ Previous | Next ]
JohnF Posts: 219 Registered: 5/27/08
Re: function arity > 2
Posted: Jan 16, 2013 3:41 AM
Butch Malahide <fred.galvin@gmail.com> wrote:
>> As in universal algebra, where introductory discussions
>> typically suggest arity>2 not often used. Are there any
>> functions f:N^3-->N (domain integers) that can't be
>> decomposed into some g,h:N^2-->N, where f(i,j,k)=g(i,h(j,k))?
>> If so (i.e., if arity>2 needed), got an example? If not,
>> got a proof? And, if not for integers, is there any domain D
>> where f:D^3-->D can't be decomposed like that (example or
>> proof again appreciated)?
>
> Hmm. Your equation "f(i,j,k) = g(i,h(j,k))" seems kind of special,
> and it's not clear to me that it covers all ways of expressing
> a ternary operation in terms of binary operations.
Oops, you're right, I meant to "cover all ways", but couldn't
see how to say that (i.e., not sure what that intuitive idea
formally means).
> Anyway:
>
> If D is an infinite domain, assuming the axiom of choice, there is an
> injection h:DxD-->D, i.e., a pairing function. Given any function
> f:D^3-->D, we can then define g(i,h(j,k)) = f(i,j,k).
Thanks, D=N=integers is what I'm primarily interested in.
> If D = {0,1}, the function f:D^3-->D such that f(0,j,k) = j, f(1,j,k)
> = k, can not be expressed in your form f(i,j,k) = g(i,h(j,k)).
Okay, that's pretty obvious -- now that you've pointed it out.
Is it an artifact of my failure to "cover all ways" as you
mentioned above, or a fundamental property related to the
cardinality of those function spaces, as you discuss below?
At first blush, I might try to "symmetrize" f(i,j,k)=g(i,h(j,k))
over i,j,k permutations in some way to avoid your counterexample.
But I suppose that's silly, and a fundamentally better expression
to "cover all ways" is probably what's really called for.
> If D is an n-element set, 2 < n < infinity, we can generalize the
> example for n = 2, or we can just use a counting argument: the number
> of ternary operations on D is n^(n^3), whereas the number of ordered
> pairs of binary operations is only n^(2n^2).
Right, this cardinality issue sounds like what I probably really need
to address. Suppose we restrict our function space to continuous
functions wrt some suitable topology (I'm thinking somehow analogous
to Scott's D_\infty construction for a model of the lambda calculus,
and subsequent domain theory stuff). I imagine cardinality problems
could be resolved. Are some additional open set properties required
so that arity>2 functions can always be written as (some suitable
composition of) arity<=2 functions? Or am I totally barking up the
wrong tree, or what? Thanks,
--
John Forkosh ( mailto: j@f.com where j=john and f=forkosh )
Date Subject Author
1/16/13 JohnF
1/16/13 Butch Malahide
1/16/13 JohnF
1/16/13 Butch Malahide
1/16/13 JohnF
1/16/13 Butch Malahide
1/17/13 JohnF
1/16/13 Graham Cooper
| 903
| 3,193
|
{"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}
| 3.625
| 4
|
CC-MAIN-2018-17
|
latest
|
en
| 0.893225
|
http://advices-on-finance-online.com/monetary-union-equilibrium-with-local-currency-constraints-2.html
| 1,561,405,678,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627999709.4/warc/CC-MAIN-20190624191239-20190624213239-00365.warc.gz
| 7,774,889
| 8,787
|
# MONETARY UNION: Equilibrium with Local Currency Constraints 2
Determination of Equilibrium Inflation Rates
Using the equilibrium from Proposition 2, let V{<!,&*) and V*(a*,<r) be the lifetime expected welfare of ал agent in the home and foreign islands respectively. Formally,
Again, there are analogous expressions for the foreign consumption levels. Note that the rate of money creation in foreign countries does effect the utility level of home agents through their equilibrium consumption levels of foreign goods. Further, in equilibrium, domestic money creation has three apparent influences: directly on the level of employment, through the transfer (the numerator of c*) and through the rate of price inflation (the denominator of ch).
The gains and losses from inflation are evident from these conditions. Given the employment levels, n and n*, home inflation increases the state contingent consumption of home agents while foreign inflation reduces it. From the definition of ch, in order for home consumption to increase with <r, ф must be less than one: i.e. not all of the domestic money supply is held by domestic citizens. In this model, this comes about because agents hold the currency of the other country in order to finance their consumption in old age. Thus, to emphasize an important point, the local currency requirement creates a demand for local currency by foreign agents and thus a basis for the inflation tax.9 This can be seen directly from the characterization of c*: if 0 = 1, then the only effect of cr would be to distort the labor supply decision More info.
Of course, the cost of inflation arises from the fact that it taxes the money holdings of all agents, domestic and foreign. This is a distortionary tax. Higher home inflation reduces the incentive for home agents to produce which ultimately reduces ch and thus lifetime utility. Still, starting at zero inflation, there is an incentive for countries to increase their money supplies. This observation leads to:
Using the condition characterizing n in the steady state (8) along with the definition of Z implies a = Z. A symmetric argument holds for the foreign country.
In fact, the equilibrium with positive inflation is a dominant strategy equilibrium. Even though the foreign rate of inflation has a (negative) effect on the welfare of home agents, it has no influence on the optimal level of domestic inflation. This simplification is again a consequence of the assumed preference structure.
| 484
| 2,495
|
{"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-2019-26
|
latest
|
en
| 0.931343
|
http://metamath.tirix.org/mpeuni/ssnnfi
| 1,721,920,656,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763860413.86/warc/CC-MAIN-20240725145050-20240725175050-00315.warc.gz
| 20,120,273
| 2,300
|
# Metamath Proof Explorer
## Theorem ssnnfi
Description: A subset of a natural number is finite. (Contributed by NM, 24-Jun-1998)
Ref Expression
Assertion ssnnfi ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → 𝐵 ∈ Fin )
### Proof
Step Hyp Ref Expression
1 sspss ( 𝐵𝐴 ↔ ( 𝐵𝐴𝐵 = 𝐴 ) )
2 pssnn ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → ∃ 𝑥𝐴 𝐵𝑥 )
3 elnn ( ( 𝑥𝐴𝐴 ∈ ω ) → 𝑥 ∈ ω )
4 3 expcom ( 𝐴 ∈ ω → ( 𝑥𝐴𝑥 ∈ ω ) )
5 4 anim1d ( 𝐴 ∈ ω → ( ( 𝑥𝐴𝐵𝑥 ) → ( 𝑥 ∈ ω ∧ 𝐵𝑥 ) ) )
6 5 reximdv2 ( 𝐴 ∈ ω → ( ∃ 𝑥𝐴 𝐵𝑥 → ∃ 𝑥 ∈ ω 𝐵𝑥 ) )
7 6 adantr ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → ( ∃ 𝑥𝐴 𝐵𝑥 → ∃ 𝑥 ∈ ω 𝐵𝑥 ) )
8 2 7 mpd ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → ∃ 𝑥 ∈ ω 𝐵𝑥 )
9 eleq1 ( 𝐵 = 𝐴 → ( 𝐵 ∈ ω ↔ 𝐴 ∈ ω ) )
10 9 biimparc ( ( 𝐴 ∈ ω ∧ 𝐵 = 𝐴 ) → 𝐵 ∈ ω )
11 enrefg ( 𝐵 ∈ ω → 𝐵𝐵 )
12 breq2 ( 𝑥 = 𝐵 → ( 𝐵𝑥𝐵𝐵 ) )
13 12 rspcev ( ( 𝐵 ∈ ω ∧ 𝐵𝐵 ) → ∃ 𝑥 ∈ ω 𝐵𝑥 )
14 10 11 13 syl2anc2 ( ( 𝐴 ∈ ω ∧ 𝐵 = 𝐴 ) → ∃ 𝑥 ∈ ω 𝐵𝑥 )
15 8 14 jaodan ( ( 𝐴 ∈ ω ∧ ( 𝐵𝐴𝐵 = 𝐴 ) ) → ∃ 𝑥 ∈ ω 𝐵𝑥 )
16 1 15 sylan2b ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → ∃ 𝑥 ∈ ω 𝐵𝑥 )
17 isfi ( 𝐵 ∈ Fin ↔ ∃ 𝑥 ∈ ω 𝐵𝑥 )
18 16 17 sylibr ( ( 𝐴 ∈ ω ∧ 𝐵𝐴 ) → 𝐵 ∈ Fin )
| 691
| 983
|
{"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-2024-30
|
latest
|
en
| 0.262705
|
https://brilliant.org/discussions/thread/the-golden-ratio-the-most-elegant-construction/
| 1,627,508,191,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046153791.41/warc/CC-MAIN-20210728185528-20210728215528-00412.warc.gz
| 157,609,586
| 14,741
|
The Golden Ratio: The Most Elegant Construction
Here is the previous post concerning the Golden Ratio. For a collection of all the posts concerning the Golden Ratio, click #GoldenRatio below.
Sorry for the long wait! This is the final post about the geometry of the golden ratio, and to celebrate that, I will present to you, with proof, the most elegant construction of the golden ratio. This construction is so awesome, it can even be performed without a compass, paper, or pencil.
CONSTRUCTION:
Take three segments of equal length (whether they be toothpicks, playing cards, or a drawing on a piece of paper). Stand one upright perpendicular to a flat surface. Now place the next one so that it touches the flat surface and the midpoint of the other segment. Place the final one so that it touches the flat surface and the midpoint of the middle segment. The line formed on the flat surface is divided into the golden ratio. See the above picture.
PROOF:
Let the segments have length 2. Then $A_1B_2 = 1$, making $A_1A_2 = \sqrt{2^2-1^2} = \sqrt{3}$. Draw the perpendicular from $B_3$ to the surface. It has length $\frac{1}{2}$. Then $A_2A_3 = \sqrt{2^2 - \left(\frac{1}{2}\right)^2} - \sqrt{1^2 - \left(\frac{1}{2}\right)^2} = \sqrt{\frac{15}{4}} - \sqrt{\frac{3}{4}} = \sqrt{3} \frac{\sqrt{5}-1}{2}$. Then we can take the ratio:
$\frac{A_1A_2}{A_2A_3} = \frac{\sqrt{3}}{\sqrt{3} \frac{\sqrt{5}-1}{2}} = \frac{1+\sqrt{5}}{2} = \phi$
Be prepared for an even more exciting quest into the algebraic mysterious of the golden ratio!
Next
Note by Bob Krueger
7 years, 6 months ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted
- list
• bulleted
• list
1. numbered
2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1
paragraph 2
paragraph 1
paragraph 2
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
That is AWESOME!!!!!
- 7 years, 6 months ago
| 920
| 3,243
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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}
| 4.78125
| 5
|
CC-MAIN-2021-31
|
latest
|
en
| 0.840067
|
https://www.physicsforums.com/threads/how-can-i-accurately-measure-irradiance-for-my-uv-sanitization-device-design.289813/
| 1,719,097,307,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862420.91/warc/CC-MAIN-20240622210521-20240623000521-00894.warc.gz
| 819,562,399
| 16,324
|
# How can I accurately measure irradiance for my UV sanitization device design?
• tihort
In summary, accurately measuring irradiance for UV sanitization device design involves understanding the principles of radiation detection and utilizing appropriate instruments such as radiometers and spectrometers. It is important to consider factors such as wavelength range, measurement accuracy, and calibration to ensure accurate readings. Additionally, proper placement and orientation of the sensor are crucial for obtaining reliable measurements. Regular maintenance and calibration checks are also necessary to maintain accuracy over time.
tihort
hello.
I am working to understand and design a UV sanitization device.
What is a good resource to learn about irradiance and its relation to distance from a surface.
Secondly, in real life, how is irradiance measured? Given a surface area, I would need to specify bulbs of certain intensity to get the device to work.
Thank you.
http://en.wikipedia.org/wiki/Irradiance has the basics. For a simple source (isotropic in all directions) it is an inverse square law (1/d^2, where d is distance) http://en.wikipedia.org/wiki/Inverse_square_law
For an actual calculation you need to know the power output of the bulbs you are getting within the wavelength band that you need (bulbs usually emit over a wide range of frequencies).
Last edited by a moderator:
The wiki article was ok, my gold-standard reference is Wolfe's "Introduction to Radiometry".
As to your second question, it can get a little tricky- you need to know details about the bulb's radiance: how much light goes in what direction. That plus the geometry will tell you how much radiation is incident on your surface.
Irradiance detectors are a fairly standard piece of equipment, but to get highly accurate and precise (<10%) measurements requires some care- spectral characteristics, polariation control, etc. An integrating sphere combined with a stable detector can give 1% precision without too much trouble.
http://physics.nist.gov/Divisions/Div844/facilities/fascal/fascal.html
Last edited by a moderator:
## Related to How can I accurately measure irradiance for my UV sanitization device design?
Irradiance is the amount of light energy per unit area that is incident on a surface. It is typically measured in watts per square meter (W/m²) and is often used in the context of UV radiation.
## 2. Why is measuring UV radiation important?
UV radiation can have both positive and negative effects on living organisms and materials. Measuring UV radiation allows us to understand its impact and make informed decisions about protection and exposure.
## 3. How is UV irradiance measured?
UV irradiance is typically measured using a radiometer, which is a device that detects and measures UV radiation. The radiometer usually contains a sensor that is sensitive to UV light and can convert it to an electrical signal for measurement.
## 4. What is the difference between irradiance and UV index?
Irradiance measures the amount of UV radiation that is incident on a surface, while UV index is a numerical scale that represents the intensity of UV radiation at the Earth's surface. UV index takes into account factors such as the angle of the sun and the thickness of the ozone layer.
## 5. Is UV irradiance the same as UV exposure?
No, UV irradiance and UV exposure are not the same. UV irradiance measures the amount of UV radiation that reaches a surface, while UV exposure refers to the amount of UV radiation that is absorbed by the skin. UV exposure is affected by factors such as skin type, time of day, and duration of exposure.
• General Engineering
Replies
6
Views
2K
• Other Physics Topics
Replies
2
Views
3K
• Electrical Engineering
Replies
5
Views
2K
• Other Physics Topics
Replies
13
Views
6K
• Astronomy and Astrophysics
Replies
1
Views
9K
• Introductory Physics Homework Help
Replies
2
Views
14K
• Other Physics Topics
Replies
8
Views
13K
• Special and General Relativity
Replies
95
Views
4K
• Nuclear Engineering
Replies
4
Views
3K
• Classical Physics
Replies
1
Views
646
| 894
| 4,117
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.046875
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.930837
|
https://www.tseccodecell.com/challenges/q/random-numbers
| 1,685,412,921,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224644915.48/warc/CC-MAIN-20230530000715-20230530030715-00596.warc.gz
| 1,148,903,839
| 3,109
|
2.3 Random Numbers
(RankList for this Question)
Score: 30pts
Time Limit: 5.00 sec
You are given random numbers in a line. There are a total of n numbers.
What is the longest sequence of successive numbers where each number is different?
Constraints
1 <= t <= 10
1 ≤ n ≤ 1000
1 ≤ k[i] ≤ 10^5
Input Format
The first input line contains a number of test cases t.
The second input line contains an integer
n: the number of lines
The next line has n integers k1,k2,…,kn: the numbers.
Output Format
Print the length of the longest sequence of unique numbers.
Example 1
Input:
1
8
1 2 1 3 2 7 4 2
Output:
5
Explanation:
Longest correct sequence is 1, 3, 2, 7, 4 i.e. length is 5
| 208
| 681
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.817959
|
korvai.org
| 1,560,919,608,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627998913.66/warc/CC-MAIN-20190619043625-20190619065625-00375.warc.gz
| 99,004,458
| 2,063
|
# Use of Numbers instead of letters
In this system we use numbers instead of letters. Any lesson or Sollu is represented by its corresponding duration or Mathirai value. The basic examples for this are shown in table 1.
Table 1: Basic Sollu and representation
Representation Sollu 1 Tha 2 Tha Ka 3 Tha Ki Ta 4 Tha Ka Thi Na 5 Tha Ka Tha Ki Ta 6 Tha Ka Tha Ka Thi Na 7 Tha Ka Thi Mi Tha Ki Ta 8 Tha Ka Tha Ri Ki Ta Tha Ka 9 Tha Ka Thi Mi Tha Ka Tha Ki Ta
The main advantage of this method is flexibility and conciseness. It should be noted that 4 can mean ``Tha Ka Thi Mi'' or ``Tha Ka Thi Na'' or ``Ki Ta Tha Ka'' or ``Tha Ri Ki Tha'' or any other equivalent depending on the place where it is going to be played.
Suriya Subramanian 2007-01-11
| 227
| 748
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2019-26
|
latest
|
en
| 0.818174
|
https://math.stackexchange.com/questions/1271239/showing-that-lim-t-rightarrow-infty-e-ttx-0
| 1,563,426,093,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00325.warc.gz
| 476,789,179
| 36,362
|
# Showing that $\lim_{t \rightarrow \infty}-e^{-t}t^x=0$
I'm trying to prove that $\Gamma(x+1) = x \Gamma(x)$ and after doing integration by parts on $\Gamma(x+1)$ I'm left with a term $-t^xe^{-t}$ that I need to evaluate at the limit as $t \rightarrow \infty$.
I was thinking about simply stating that by applying L'Hospital $\left\lceil x \right\rceil$-times would give: $$\displaystyle 0 \leq \lim_{t \rightarrow \infty} \frac{-t^x}{e^t} \leq \frac{-t^{\left\lceil x \right\rceil}}{e^t} \Rightarrow \lim \frac{-\left\lceil x \right\rceil !}{e^t} = 0$$
Would that be a valid argumentation? How else could I show that?
Here's another way: $$\lim_{t \to \infty} \frac{-t^x}{e^t} = -\left(\lim_{t \to \infty} \frac{t}{e^{t/x}} \right)^{x}$$ then apply L'Hôpital once.
$$-e^{-t}t^x = \frac {-t^x}{e^t} = -\frac {t^x}{\sum_{n=0}^{\infty} \frac {t^n}{n!}} =-\frac {t^x}{\sum_{n=0}^{[x]}\frac {t^n}{n!} +\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}$$ where $[x]$ denotes least integer.
$$|\frac {t^x}{\sum_{n=0}^{[x]}\frac {t^n}{n!} +\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}| < |\frac {t^x}{\sum_{n=[x]+1}^{\infty}\frac {t^n}{n!}}|<|\frac {t^x}{t^{[x]+1}}| \rightarrow 0$$
So, by comparison, $$|e^{-t}t^x|\to 0 \Rightarrow -e^{-t}t^x \to 0$$
| 529
| 1,234
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375
| 4
|
CC-MAIN-2019-30
|
latest
|
en
| 0.632461
|
http://stackoverflow.com/questions/14906077/how-to-work-with-left-and-right-bytes-of-a-short-int
| 1,397,920,935,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00225-ip-10-147-4-33.ec2.internal.warc.gz
| 237,228,807
| 15,532
|
# How to work with left and right bytes of a short int? [closed]
I have a short int, which is 2 bytes, but I only want to output the left byte to a file. How do I do this? I'm using binary functions open(), read(), write(), etc.
I would also like to shift the right byte to the left 8 times, so that the right byte occupies the left, and the right has been cleared to all 0's.
I apologize for not showing what I've already tried -- I'm a C noobie and cannot find anything about how to do this.
-
you''ll need to show some attempt... – Mitch Wheat Feb 16 '13 at 1:48
Read up on the bitwise operators, and experiment! You should also remember that the "left" and "right" side of a value depends much on the underlying platform (read about endianess). – Joachim Pileborg Feb 16 '13 at 1:51
Look at graphics.stanford.edu/~seander/bithacks.html for some inspiration. But never forget that the occasional cost of a few microseconds of execution time is much, much less than a afternoon worth of head scratching because you don't understand anymore how the code works. Write the simplest code that does the job, check that it is standard conforming (your compiler and machine will change!). Only start bumming when measurements show it is worthwhile, and the compiler doesn't do it by itself. Organization and algorithms should be your first targets of scrutiny. – vonbrand Feb 16 '13 at 2:52
No. A short int is `sizeof (short int)` bytes. That value is not guaranteed to be 2. There are systems where sizeof (short int) == 1. @JoachimPileborg: Encouraging experimentation with regards to shifting negative integers is silly. Left shifting negative values is undefined behaviour, and right shifting might result in trap representations or exceptional conditions which according to 6.2.6.1p5 and 6.5p5 would also lead to undefined behaviour. – undefined behaviour Feb 16 '13 at 3:26
## closed as not a real question by Mitch Wheat, Rohan, AppDeveloper, Jai, Stephen ConnollyFeb 16 '13 at 8:39
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
You could try this approach
`````` int someNum = 0x1234;
int leftByte, rightByte;
leftByte = (someNum >> 8) & 0xff;
Ok.. Thank you. I used `short int` to indicate a `16 bit` datatype as desired by user. I will modify my example to use `int` only – Ganesh Feb 16 '13 at 3:29
| 654
| 2,635
|
{"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-2014-15
|
latest
|
en
| 0.941646
|
https://socratic.org/questions/how-do-you-solve-the-triangle-given-a-8-b-24-c-18
| 1,723,561,278,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641079807.82/warc/CC-MAIN-20240813141635-20240813171635-00161.warc.gz
| 402,919,014
| 6,359
|
# How do you solve the triangle given a=8, b=24, c=18?
Feb 24, 2018
Three angles are color(blue)(hat A = 14.63^@, hat B = 130.75^@, hat C = 34.62^@
#### Explanation:
Known three sides. To find the three angles.
${A}_{t} = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ where s is the semi perimeter of the triangle.
$s = \frac{8 + 24 + 18}{2} = 25$
${A}_{t} = \sqrt{25 \cdot \left(25 - 8\right) \left(25 - 24\right) \left(25 - 18\right)} = 54.54$
But ${A}_{t} = \left(\frac{1}{2}\right) b c \sin A$
$\therefore \sin A = \frac{2 \cdot {A}_{t}}{b c} = \frac{2 \cdot 54.54}{24 \cdot 18} = 0.2525$
$\hat{A} = {\sin}^{-} 1 0.2525 = {14.63}^{\circ}$
Similarly,
$\hat{B} = {\sin}^{-} 1 \left(\frac{2 \cdot 54.54}{8 \cdot 18}\right) = \left(180 - 49.24\right) = {130.75}^{\circ}$ since it is an obtuse angle.
$\hat{C} = {\sin}^{-} 1 \left(\frac{2 \cdot 54.54}{8 \cdot 24}\right) = 34.62$
| 396
| 911
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2024-33
|
latest
|
en
| 0.479972
|
http://nanohub.org/tags/boundstates?area=&sort=title
| 1,430,174,599,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2015-18/segments/1429246659483.51/warc/CC-MAIN-20150417045739-00310-ip-10-235-10-82.ec2.internal.warc.gz
| 189,553,645
| 13,036
|
# Support
## Support Options
• ### Knowledge Base
Find information on common issues.
• ### Wish List
Suggest a new site feature or improvement.
• ### Support Tickets
Check on status of your tickets.
## Tags: bound states
### All Categories (1-9 of 9)
1. AQME Exercise: Bound States – Theoretical Exercise
20 Jul 2010 | Teaching Materials | Contributor(s): Dragica Vasileska, Gerhard Klimeck
The objective of this exercise is to teach the students the theory behind bound states in a quantum well.
http://nanohub.org/resources/9364
2. Bound States and Open Systems
01 Jun 2010 | Teaching Materials | Contributor(s): Dragica Vasileska
bound states, open systems, transfer matrix approach, gate leakage calculation in Schottky gates
http://nanohub.org/resources/9103
3. Bound States Calculation Lab
05 Jul 2008 | Tools | Contributor(s): Dragica Vasileska, Gerhard Klimeck, Xufeng Wang
Calculates bound states for square, parabolic, triangular and V-shaped potential energy profile
http://nanohub.org/resources/bsclab
4. Bound States Calculation Lab - Fortran Code
This is a Fortran code for BSC Lab.
http://nanohub.org/resources/9206
5. Bound States Calculation: an Exercise
06 Jul 2008 | Teaching Materials | Contributor(s): Dragica Vasileska, Gerhard Klimeck
The problems in this exercise use the Bound States Calculation Lab to calculate bound states in an infinite square well, finite square well and triangular potential. Students also have to compare...
http://nanohub.org/resources/4884
6. Quantum Bound States Described
02 Jul 2011 | Teaching Materials | Contributor(s): Dragica Vasileska
This write up describes basic concepts of closed systems and bound states calculations. Emphasis is placed on bound states calculation for infinite potential well, finite potential well,...
http://nanohub.org/resources/11582
7. Quantum Bound States Exercise
16 Jun 2010 | Teaching Materials | Contributor(s): Gerhard Klimeck, Parijat Sengupta, Dragica Vasileska
Exercise Background Quantum-mechanical systems (structures, devices) can be separated into open systems and closed systems. Open systems are characterized with propagating or current carrying...
http://nanohub.org/resources/9191
8. Quantum Mechanics: Time Independent Schrodinger Wave Equation
07 Jul 2008 | Series | Contributor(s): Dragica Vasileska, Gerhard Klimeck
In physics, especially quantum mechanics, the Schrödinger equation is an equation that describes how the quantum state of a physical system changes in time. It is as central to quantum mechanics...
http://nanohub.org/resources/4937
9. Reading Material: Time Independent Schrodinger Wave Equation (TISWE)
07 Jul 2008 | Teaching Materials | Contributor(s): Dragica Vasileska
www.eas.asu.edu/~vasileskNSF
http://nanohub.org/resources/4934
### Legal
nanoHUB.org, a resource for nanoscience and nanotechnology, is supported by the National Science Foundation and other funding agencies. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
| 722
| 3,141
|
{"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-2015-18
|
longest
|
en
| 0.761404
|
https://studyninja.com.au/hsc-standard-maths-2-1-1-sampledata/
| 1,628,151,026,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00076.warc.gz
| 551,127,121
| 34,071
|
2.1.1 Sample Data - StudyNinja
Select Page
HSC Standard Maths Resources
Browse: 1. Home » 2. Data & Graphs » 2.1 Basic Stats » 2.1.1 Sample Data
2.1.1 Sample Data
The easiest way basic statistics can be asked in an exam is by giving you a data set and then asking you to calculate any of the stats mentioned in previous section. Following are the basics you need to know to get yourself started with these:
Example 1
You are given following set of ten scores in a mathematics test. Answer the questions below:
88, 93, 58, 76, 79, 65, 77, 68, 51, 77
Calculate i) mean, ii) mode, iii) median, iv) lower quartile, v) upper quartile, vi) interquartile range, vii) range, viii) standard deviation.
Things can get little tricky when instead of asking you to calculate these stats directly, examiner asks you to:
– calculate missing value in a data set by giving you a sample data and one of the stats above
– determine effect on a given statistics when new data is added
– compare statistics of two given data sets
The basics of solving all the above remain the same. They are merely an extension of basics with additional layers.
Example 2
A data set of 10 scores has a mean of 15. The scores 10, 12, 15, and 23 are added to this data set. What is the mean of the data set now?
Example 3
A set of data of 9 scores is shown with the first number missing, in ascending order.
?, 3, 5, 8, 9, 12, 15, 18, 22
The range of the data is 21. What is the five-number summary for this data set?
Browse: 1. Home » 2. Data & Graphs » 2.1 Basic Stats » 2.1.1 Sample Data
| 450
| 1,583
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2021-31
|
latest
|
en
| 0.836114
|
https://www.wyzant.com/resources/answers/21175/is_infinity_to_the_zero_power_equal_to_1
| 1,524,820,917,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524127095762.40/warc/CC-MAIN-20180427075937-20180427095937-00450.warc.gz
| 913,871,069
| 16,660
|
0
Is infinity to the zero power equal to 1?
My math teacher wants me to submit the answer in the class page by 9:30!
Paul H. | Math, Excel, InDesign, Photography and Computers in GeneralMath, Excel, InDesign, Photography and C...
0
Not sure what math class you are in, so the answer may be inappropriate, but here goes:
(infinity)^zero is really undefined
But the way this expression comes up in practice is always the limit as something goes TOWARD
infinity or TOWARD zero, or both. For example, take x to the 0 power, and let x get bigger and bigger.
Then the answer each time is 1, so the limit is also 1.
But take infinity to the y power, and let y get smaller and smaller.
Then the answer is infinity each time, so the limit is also infinity.
The interesting case is where x and y are related in some way, so x
gets bigger as y gets smaller in a coordinated way. Depending on the
algebraic relation between x and y, the answer can come out 1 or
infinity or anything in between.
That's why it is really undefined.
Hope that make sense
| 255
| 1,047
|
{"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-2018-17
|
latest
|
en
| 0.946666
|
nvfn.ensveneto.it
| 1,580,220,043,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579251778272.69/warc/CC-MAIN-20200128122813-20200128152813-00401.warc.gz
| 116,449,797
| 17,458
|
First Central Difference Method Next, students work in small groups to identify texts that are comparing and contrasting. Imagine you want to discretize a first, second,derivative of a given function f at a i location. By Mike Simpson. (Note that these can be done at the same time that the centered. Then you try to calculate that value out of y(0) and y(2). In many problems one may be interested to know the behaviour of f(x) in the neighbourhood of x r (x 0 + rh). The phlebotomist, or professional who draws blood, prepares a. NET exception handling style i. I hope you now have a solid understand of the differences between descriptive and inferential statistics. Tagged - The social network for meeting new people The social network for meeting new people :). A mean tells scientists the mathematical average of all of a data set, such as the average age at first marriage; the median represents the middle of the data distribution, like the age that sits in the middle of the range of ages at which people first marry; and. Once in the vicinity of the optimum response the experimenter needs to fit a more elaborate model between the response and the factors. The exercises depicted here in Sophie's example are for a three-curve, right-thoracic scoliosis. 4 FINITE DIFFERENCE METHODS (II) where DDDDDDDDDDDDD(m) is the differentiation matrix. Not to worry! In this post, we are going to go over one of the best ways to answer behavioral questions, and that is by using the STAR Method. We do this by selecting a positive weight of the "attenuation factor" or Beta parameter) in the dialog of Network>Centrality>Power, as shown in figure 10. 2 billion followers, but it has branched out into several denominations over the last 2000 years. Find out what you should earn with a customized salary estimate and negotiate pay with confidence. General Information This booklet is intended primarily for homeowners who already have had their homes tested for radon and have decided that they need to take some action to reduce radon levels. His method is the original style to be set in a hot room. The total interactions on each particle at time can be calculated from the sum of interactions from other. NUMERICAL METHODS CONTENTS TOPIC Page Interpolation 4 Difference Tables 6 Newton-Gregory Forward Interpolation Formula 8 Newton-Gregory Backward Interpolation Formula 13 Central Differences 16 Numerical Differentiation 21 Numerical Solution of Differential Equations 26 Euler's Method 26 Improved Euler Method (IEM) 33. The forward Euler’s method is one such numerical method and is explicit. Determine Submission Method. The Steve Fund is dedicated to the mental health and emotional well-being of students of color. Chapter 7: Numerical Differentiation 7-16 Numerical Differentiation The derivative of a function is defined as if the limit exists • Physical examples of the derivative in action are: - Given is the position in meters of an object at time t, the first derivative with respect to t, , is the velocity in. Caption of the figure: flow pass a cylinder with Reynolds number 200. (Note the path difference or PD is the difference in distance traveled by the two waves from their respective sources to a given point on the pattern. The key difference between a general interest science question and a testable question is that testable questions are always about changing one thing to see what the effect is on another thing. The finite difference approximation we used above amounts to dropping the terms in and higher powers of on the rhs. It was first utilized by Euler, probably in 1768. Lecture 8: Solving the Heat, Laplace and Wave equations using nite ff methods (Compiled 26 January 2018) In this lecture we introduce the nite ff method that is widely used for approximating PDEs using the computer. 2 Backward differences 3. The overall method is the same as above, with the exception that we will replace the analytical prices of the call/puts in the Finite Difference approximation and use a Monte Carlo engine instead to calculate the prices. As Steven Lukes has observed, 40 The Rules of Sociological Method was simultaneously a treatise on the philosophy of social science, a polemic against the enemies of sociology, and the manifesto of the emergent Durkheim "School"; and it is important to weigh its failures in the light of these multiple, discordant intentions. Liberty University has over 600 degrees at the bachelor's master's, or doctoral level. Question: Solve The Following First Order ODE Using Finite Difference Method With Central Difference Formula For The Derivative. Richardson. Finding the data first and then formulating the hypothesis is like…. There is a world of difference between failing to disprove and proving. The Inca civilization can be traced back to about A. Putting and using the ODE then gives the difference equation The Euler method truncates the series after the second term on the rhs. Chapter 6 - Auditory & Vestibular Function. Coal Mining Methods Underground Mining Longwall & Room and Pillar Mining Longwall mining and room-and-pillar mining are the two basic methods of mining coal underground, with room-and-pillar being the traditional method in the United States. Welcome to eCPRcertification. enl EndNote 1 1 17 MAGREFS-Converted. surrounding the first language directly inform the means by which the second is learned. Qualitative research collects information that seeks to describe a topic more than measure it. When setting up your bank account with Coinbase, they may first initiate two small transactions, then require you to verify the amounts. We start somewhere in terms of the natural units and use the coded units to do our experiment. Some simple derivative functions for equally-spaced time series data: deriv, a first derivative using the 2-point central-difference method, deriv1, an unsmoothed first derivative using adjacent differences, deriv2, a second derivative using the 3-point central-difference method, a third derivative deriv3 using a 4-point formula, and deriv4, a. The first split in Christianity took place in 1054 AD when the Eastern Orthodox denomi. Authorization to reproduce this report in whole or in part is granted. Users with questions about a personal health condition should consult with a qualified healthcare professional. and plot the estimates and the actual function derivatives. Your values are the things that you believe are important in the way you live and work. Science Projects. This equation is linear for a linear PDE, and non-linear for non-linear PDEs. The additional effects of the McKenzie method on pain (that is, the adjusted difference in outcomes between the McKenzie Group and the First-line Care Group) were statistically significant but smaller than our pre-specified threshold for between-group clinical importance of 1 unit (P = 0. Direction of clandestine or covert paramilitary operations by the Department of Defense Title II—Unity of effort across the foreign-domestic divide Sec. This is done to compensate the lender for the fact that some loans will be paid off early, depriving the lender of the full amount of interest. These tips shall help you in achieving the quality work and right first time. We are calling for submissions focusing on research concerning the monitoring, evaluation and accountability of human resources for health policy options through a gender equity lens. Program to estimate value of First Derivative of the function at the given points from the given data using Backward Difference Formula , Forward diff Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. • Also pertains to finite difference methods for PDEs • Valid under certain assumptions (linear PDE, periodic boundary conditions), but often good starting point • Fourier expansion (!) of solution • Assume - Valid for linear PDEs, otherwise locally valid - Will be stable if magnitude of ξ is less than 1:. 1 Boundary conditions - Neumann and Dirichlet. The central limit theorem is perhaps the most fundamental result in all of statistics. Difference Between GAAP and IFRS Last updated on May 19, 2017 by Surbhi S IFRS Vs GAAP is the most debatable topic in accounting where the former is defined as the financial reporting method having universal applicability while the latter are the set of guidelines made for financial accounting. With such an indexing system, we. Both methods are well suited to extracting the relatively flat coalbeds (or coal seams) typical of. The results were first obtained by Goubault-Larrecq and Varacca and presented in their LICS 2011 paper. Finite Difference Methods for Hyperbolic Equations 1. First, if national culture can have significant – not to say existential – consequences among people of the same cultural origin, we need to be very cautious in how we deal with national cultural differences in cross-border interactions. The journal welcomes submissions concerning molecular and cellular biology, genetics, epidemiology, and clinical trials. Internet hyperlinks to an old lab project. This will provide us with a baseline to determine the accuracy of subsequent numerical methods. log() method displays 19. 4) Set Up Payment Method. The two distinct functions are in hearing and in control of balance. A large amount of information is being sensed at any one time such as room temperature, brightness of the lights, someone talking, a distant train, or the smell of perfume. The wave equation considered here is an extremely simplified model of the physics of waves. This is useful for single step prediction. The main difference between our method and TransPhylo’s previous capabilities is that we can perform. The Federalists were instrumental in 1787 in shaping the new US Constitution, which strengthened the national government at the expense. So we fail to reject any reasonable hypothesis of a difference. 6 Recall how the multi-step methods we developed for ODEs are based on a truncated Taylor series approximation for $$\frac{\partial U}{\partial t}$$. Imagine you want to discretize a first, second,derivative of a given function f at a i location. Amazon Seller Central in CA. 3 Central differences 3. Below is my Matlab code for the shooting method. Provide code that produces a list of numbers which is the n th order forward difference, given a non-negative integer (specifying the order) and a list of numbers. Slightly later, in 1716, came the first use of water in Sweden to distribute heating in buildings. First Central Difference Method, what is the formula and what is another name for it? Definition Allows us to match kinematic data based on positions of the segment endpoints from each frame within a time interval. In modern. They do not. The central difference for a function tabulated at equal intervals is defined by (1) First and higher order central differences arranged so as to involve integer indices are then given by. You need to enable JavaScript in your browser to work in this site. The first numerical approach utilised will be based on a Finite Difference Method (FDM) and the original analytical formulae. 4 % of total variation in the data with the first two principal components (PCs) – 61. We will look at two formulae, interpolating two and three points, respectively. Numerical solution method such as Finite Difference methods are often the only practical and viable ways to solve these differential equations. 1 Centered Difference Formula for the First Derivative We want to derive a formula that can be used to compute the first derivative of a function at any given. View More. Patankar (Hemisphere Publishing, 1980, ISBN -89116-522-3). Intervention Central is the leading resource for Response to Intervention (RTI) tools and resources, including academic and behavior interventions for classroom management. The main difference between them is that an SFP reticle will appear to be the same size regardless of magnification. Instead of taking the difference of the point we want, g, and a point some value h in front of our desired point, we're now taking the difference of the point we want and a point some value h behind our desired point. In the equations of motion, the term describing the transport process is often called convection or advection. Sigmund Freud and Erik Erikson are both known for their work in psychoanalysis. Before we start to introduce methods. Princeton University released two new reports and announced next steps to further strengthen its policies, resources and communications related to sexual misconduct on campus. This equation is linear for a linear PDE, and non-linear for non-linear PDEs. Similarly, the work of Houmanfar, Hayes, and Herbst (2005) concludes that the history of the first language is a major component and participatory factor in the acquisition of the second language and its maintenance. Look at the first iteration of your loop, your index i will take the value of 1. The Hermite interpolation based Newton's polynomials is again carried out to the same function used before. We use the de nition of the derivative and Taylor series to derive nite ff approximations to the rst and second. Differencing Approximations Example in R. With just 2 numbers the answer is easy: go half-way between. 51 Self-Assessment. The addition of the McKenzie. The most consistent use of the method of Bible study known as the Historical-Grammatical-Lexical Method (in this Textbook called the Contextual/Textual method) began in Antioch, Syria, in the third century a. ( Theoretical Computer Science, to appear. The Central Difference Method The central difference approximations for the first and second derivatives are x˙(t i) = ˙x i. In this article we also talk about best practices of MVC exception handling. FTCS method for the heat equation FTCS ( Forward Euler in Time and Central difference in Space ) Heat equation in a slab Plasma Application Modeling POSTECH 6. Important for an ethnomethodological analysis is self-reflection and the inspectability of data, thus the reader of an ethnomethodological study should be able to inspect the original data as means to evaluate any claim made by the analyst. • Techniques published as early as 1910 by L. The calculation uses a sheath helix model in which the shunt capacitance and series inductance are adjusted by constants whose values depend only on the dimensions of the structure. Provide code that produces a list of numbers which is the n th order forward difference, given a non-negative integer (specifying the order) and a list of numbers. SANTIAGO, EcE 2. , x n with step length h. What Is the Difference Between Manifest Content and Latent Content? Latent content and manifest content are two concepts introduced by Sigmuend Freud to help people understand the meaning of their dreams. To accomplish this, the support departments are ranked. For example, once we have computed from the first equation, its value is then used in the second equation to obtain the new and so on. Aristotle and Plato were philosophers in ancient Greece who critically studied matters of ethics, science, politics, and more. For example, when you mention Thanksgiving, an English-speaking student may think of the first European settlers on the east coast during the 17th and 18th centuries. The second numerical method will use a combination of the FDM technique and Monte Carlo for pricing. -CAFTA-DR Free Trade Agreement Rules of Origin. As a result, there can be differences in bot h the accuracy and ease of application of the various methods. Find out what you should earn with a customized salary estimate and negotiate pay with confidence. Prediction has a time flavour (what happens in future) and does not imply any certain method (in difference to the other two). Plot the Weber fractions obtained at each of the four ranges of line length and determine whether Weber's Law holds true for just noticeable differences in line length. The additional effects of the McKenzie method on pain (that is, the adjusted difference in outcomes between the McKenzie Group and the First-line Care Group) were statistically significant but smaller than our pre-specified threshold for between-group clinical importance of 1 unit (P = 0. Intelligence research properly consists of the latter kind. When later analyzing the video or film, the images are typically viewed on a computer monitor, projected onto a screen, or printed to paper. But will all points on the first antinodal line have a path difference equivalent to 1 wavelength?. We start somewhere in terms of the natural units and use the coded units to do our experiment. If you have any comments or contributions, please leave them in the comments. Slightly later, in 1716, came the first use of water in Sweden to distribute heating in buildings. Old Lab Project (Numerical Differentiation Numerical Differentiation). The following example defines two string variables. Putting and using the ODE then gives the difference equation The Euler method truncates the series after the second term on the rhs. Thus, for the first evaluation, the x values for the centered difference approximation will be x = 2 ± 0. FTCS method for the heat equation FTCS ( Forward Euler in Time and Central difference in Space ) Heat equation in a slab Plasma Application Modeling POSTECH 6. To put in other words, it is a way to describe the center of a data set. What is Sterilization ? 9 Types and Methods in Microbiology Sterilization is a process of destruction of all forms of living microorganisms from a substance. Watch free anime online or subscribe for more. Mixed methods also mirror the way individuals naturally collect information—by integrating quantitative and qualitative data. I read recently that "it is not possible to get second order accuracy at the boundaries using finite difference method, were as same is possible with finite volume method. We'll use Euler's Method to approximate solutions to a couple of first order differential equations. , to find a function (or some discrete approximation to this function) that satisfies a given relationship between various of its derivatives on some given region of space and/or time, along with some. 3 , Measurable Outcome 2. "Da Vinci" Robot. Many of these best. This is the home page for the 18. The Steve Fund is dedicated to the mental health and emotional well-being of students of color. For example, the mean marks obtained by students in a test is required to correctly gauge the performance of a student in that test. When setting up your bank account with Coinbase, they may first initiate two small transactions, then require you to verify the amounts. Chapter 1 Finite Difference Approximations Our goal is to approximate solutions to differential equations, i. Central Station vs Local Alarm - What are the Differences? April 29, 2016 Blog 0 Comments With the increase in home invasions there is a growing awareness of the need for home security monitoring systems. Especially in your first mind maps, the temptation to write a complete phrase is enormous, but always look for opportunities to shorten it to a single word or figure – your mind map will be much more effective that way. The medical model versus the nursing model. The three most common descriptive statistics can be displayed graphically or pictorially and are measures of: Graphical/Pictorial Methods; Measures of Central Tendency. com/ This work is licensed under the Creative. Introductory Finite Difference Methods for PDEs Contents Contents Preface 9 1. Once each difference threshold (delta I) has been interpolated convert it to the Weber fraction equivalent (delta I/I). For operant conditioning to work, the subject must first display a behavior that can then be either rewarded or punished. Balancing central heating radiators With a 'feed and return' central heating system , the radiators nearest to the boiler/pump would tend to be warmer than the radiators further away. A first-order accurate upwind finite difference scheme was developed for solving the nonlinear hyperbolic equations by Courant, Isaacson and Rees in 1952 (ref. Central is the immersion in the situation being studied. 5/95 — Mean is the average, the most common measure of central tendency. The medical model versus the nursing model. It was fought primarily in Europe from the year 1914 to the year 1918 and lasted 4 years. Finite difference equations enable you to take derivatives of any order at any point using any given sufficiently-large selection of points. Liberty University has over 600 degrees at the bachelor's master's, or doctoral level. A first-order accurate upwind finite difference scheme was developed for solving the nonlinear hyperbolic equations by Courant, Isaacson and Rees in 1952 (ref. In the case of the popular finite difference method, this is done by replacing the derivatives by differences. and the center divided difference methods using Taylor series, which also give the quantitative. There is a world of difference between failing to disprove and proving. The largest value is 13 and the smallest is 8, so the range is 13 – 8 = 5. upwind scheme as well as central differencing scheme are finite difference schemes which will use different locations of the variable you want to derive. Uncover biblical truth for yourself. In this review we emphasized on the gut microbiota commonly detected by both culture based method, and culture-independent techniques and this gives us the broad view on the common gut microbiota that colonized infants during their first year of life with respect to the mode of delivery. Several Pioneers of solving PDEs with finite-difference method (Lewis Fry Richardson, Richard Southwell, Richard Courant, Kurt Friedrichs, Hans Lewy, Peter Lax and John von Neumann) First application to elastic wave propagation (Alterman and Karal, 1968) Simulating Love waves and was the frst showing snapshots of seimsic wave fields (Boore, 1970). How Reformation challenges mirror today’s church issues On Reformation Day, October 31, we remember an exhibit at a United Methodist seminary that tied historical discussions to today's talk about the Four Areas of Focus and human sexuality. Differencing Approximations Example in R. Standards touch all areas of our lives, so standards developers are needed from all sectors of society. Computing derivatives and integrals Stephen Roberts Michaelmas Term Topics covered in this lecture: 1. First, we examine the case where the score for each actor is a positive function of their own degree, and the degrees of the others to whom they are connected. Question: Solve The Following First Order ODE Using Finite Difference Method With Central Difference Formula For The Derivative. Your most critical decision then, is when to make that first cutting. peace and security The United Nations pursues global disarmament and arms limitation as central to peace and security. 1 Centered Difference Formula for the First Derivative We want to derive a formula that can be used to compute the first derivative of a function at any given. The key is the ma-trix indexing instead of the traditional linear indexing. and the center divided difference methods using Taylor series, which also give the quantitative. Proper Chinese cooking methods are essential for making good Chinese food at home. It is simple to code and economic to compute. Brought back to this country by Alan Herdman. When later analyzing the video or film, the images are typically viewed on a computer monitor, projected onto a screen, or printed to paper. Almost nine years since the birth of bitcoin, central banks around the world are increasingly recognizing the potential upsides and downsides of digital currencies. Difference Between GAAP and IFRS Last updated on May 19, 2017 by Surbhi S IFRS Vs GAAP is the most debatable topic in accounting where the former is defined as the financial reporting method having universal applicability while the latter are the set of guidelines made for financial accounting. Johnson, Dept. Key Words: Finite Difference Methods, Polar Coordinates "" "Work Unit Number 3 -Numerical Analysis and Scientific Computing 'Sponsored by the United States Army under Contract No. The addition of the McKenzie. Keep up-to-date with the latest from Advance HE Let us do the hard work for you and keep you updated with services such as: the latest news, reports and research from around the sector, information on Advance HE services and awards that can support you in your role, and provide the services you have requested from us. When you do all of this you will be prompted to talk about how your project is significant. 02; Table Table2). 5/95 — Mean is the average, the most common measure of central tendency. The rate of cooling of a body can be expressed as. The essential components of a central-heating system are an appliance in which fuel may be burned to generate heat; a medium conveyed in pipes or ducts for transferring the heat to the spaces to be heated; and an emitting apparatus in those spaces for releasing the heat either by convection or radiation or both. Catheter refers to the central venous catheter (CVC) Hub refers to the end of the CVC that connects to the blood lines or cap Cap refers to a device that screws on to and occludes the hub Limb refers to the catheter portion that extends from the patient’s body to the hub. Heroin (diacetylmorphine) is a highly addictive Schedule I drug, and a heavily abused and extremely potent opiate. Interreg CENTRAL EUROPE is a European Union funding programme that inspires and supports cooperation on shared regional challenges - with a budget of 246 million Euro from the European Regional Development Fund (ERDF). There is clearly a new level of thinking and management that occurs at the program level and many good project managers grow into great program managers. The first author was also supported in part by NSF grant MCS-8306880 and ONR grant N00014-84-K-0454. 1 Finite Difference Approximations Measurable Outcome 2. There are two main accounting methods used for. Here are the first few rows for the sequence we grabbed from Pascal's Triangle:. Before we start to introduce methods. Central Difference: As to the second i,j i,j i,j i,j Explicit Finite Difference Method as Trinomial Tree [] () 0 2 22 0 Check if the mean and variance of the. The Pilates method did not return to Britain until 1970. Second order central difference = first order central difference applied twice? Ask Question Asked 2 years, The method of Characteristics for Burgers' equation. Derivatives of functions can be approximated by finite difference formulas In this Demonstration we compare the various difference approximations with the exact value. 1 - smaller h gives more accurate results. Using the right method can help ensure that your small business meets customer needs by having products available when customers want them while maximizing profits. , the 1-D equation of motion is duuup1 2 uvu dttxxr ∂∂∂ =+=−+∇ ∂∂∂. Selling in Canada is also similar. To avoid this, the outlet of each radiator is fitted with a 'lockshield valve' (shown right) which needs to be adjusted when the system is first installed. Maize cobs became larger over time, with more rows of kernels, eventually taking on the form of modern maize. The goal of this journal is to stimulate the development and adoption of new and improved techniques and research tools and, where appropriate, to promote consistency of. The kind of study may vary (it could have been an experiment, survey, interview, etc. In the first parts of these SCCM 2012 and SCCM 1511 blog series, we covered the complete SCCM 2012 R2 and SCCM 1511 installation. The wave equation considered here is an extremely simplified model of the physics of waves. There are three measures of central tendency and each one plays a different role in determining where the center of the distribution or the average score lies. Our approach is to focus on a small number of methods and treat them in depth. If you have a web cast, online chat, conference call or other live event where people from all over the world want to attend, this South Korea time zone difference converter lets you offer everyone an easy way to determine their own local time and date for your live event in South Korea. The central differencing scheme is one of the schemes used to solve the integrated convection-diffusion equation and to calculate the transported property Φ at the e and w faces. This method can be used to enforce both quantitatively and qualitatively credit controls by the Central Banks. The difference is pretty simple to understand mathematically, See since the INITIAL exposed to risk was developed within the binomial framework, the lives who die have to be counted as exposed to risk until the end of the year of age (not the investigation), whereas the central exposed to risk is developed within the poisson framework, it only measures the total time under observation (the. When you run an ANOVA (Analysis of Variance) test and get a significant result, that means at least one of the groups tested differs from the other groups. 3 PDE Models 11 &ODVVL¿FDWLRQRI3'(V 'LVFUHWH1RWDWLRQ &KHFNLQJ5HVXOWV ([HUFLVH 2. When you have two or more numbers it is nice to find a value for the "center". central difference formula Consider a function f(x) tabulated for equally spaced points x 0 , x 1 , x 2 ,. Graphical visualization and interrogation of data are critical components of any reliable method for statistical modeling, analysis and interpretation of data. Introduction to Tensile Testing / 5 Fig. • Also pertains to finite difference methods for PDEs • Valid under certain assumptions (linear PDE, periodic boundary conditions), but often good starting point • Fourier expansion (!) of solution • Assume - Valid for linear PDEs, otherwise locally valid - Will be stable if magnitude of ξ is less than 1:. PUT is used to send data to a server to create/update a resource. This method is well-explained in the book: Numerical Heat Transfer by Suhas V. Having a Seller Central account is a great way to control the messaging of those products,” says Wes Grudzien, founder of Ezonomy. View More. Stability of Finite Difference Methods In this lecture, we analyze the stability of finite differenc e discretizations. You’ll come across many terms in statistics that define different sampling methods: simple random sampling, systematic sampling, stratified random sampling and cluster sampling. I've tryed to calculate first derivative by using SNRD filters on the Savitzy-Golay smoothed functions, and I'm really loosing important function detail, I'm smoothing too much, I think this is not a good way to proceed. American Institute of Aeronautics and Astronautics 12700 Sunrise Valley Drive, Suite 200 Reston, VA 20191-5807 703. Let us illustrate the method. Medicare K Levels Amputation Even though picking out cost effective medical insurance can often be a hard technique, the creative ideas of which were definitely offered on this internet page may very well make your well being coverage range method very fast previous to. Here is what www. To solve the numerical model, fourth-order central difference scheme is used for first-order spatial derivatives and fourth-order Adams-Bashforth-Moulton predictor-corrector method is applied for time integration. A detailed guide on how to use the steps of the scientific method to complete a science fair project. The Purdue Online Writing Lab Welcome to the Purdue OWL. 2 Backward differences 3. It is analyzed here related to time-dependent Maxwell equations, as was first introduced by Yee []. Patient First provides a full range of urgent care and primary care services through our local health care centers. Python has some methods that dictionary objects can call. Mixed methods also mirror the way individuals naturally collect information—by integrating quantitative and qualitative data. Socrates (470-399 BC) was a Greek philosopher who sought to get to the foundations of his students' and colleagues' views by asking continual questions until a contradiction was exposed, thus proving the fallacy of the initial assumption. Central Authentication Services. The difference used in this formula lie on the line shown in the table below. Figure 4-10 shows the different components that must be represented to simulate the complete response of a watershed. When you do all of this you will be prompted to talk about how your project is significant. the CDM (Central Difference Method) is used to try to find the velocity of the point at specific time. central difference,, which is second order accurate since the For particularly large systems, iterative solution methods are The first decision to. Define Report Requirements in the Evaluation Statement of Work and Final Work Plan All evaluation statements of work (SOW) should clearly define requirements and expectations for the final evaluation report. Maize cobs became larger over time, with more rows of kernels, eventually taking on the form of modern maize. After reading this chapter, you should be able to. Venipuncture is often performed so blood can be tested for a variety of medical reasons and conditions. However, I don't know how I can implement this so the values of y are updated the right way. The hydrograph method accounts for losses (soil infiltration for example) and transforms the remaining (excess) rainfall into a runoff hydrograph at the outlet of the watershed. The Contract for Difference with details of the key implementation activities and milestones for the first CFD Allocation Round in 2014 and subsequent settlement £1bn p/a in our central. If there is no rayleigh damping and the C matrix is 0, for a diagonal mass matrix a diagonal solver may and should be used. Difference Between High And Low Deductible Health Insurance As a result get out there and start seeking the fact that individual medical insurance. plus(b), we can access the instance variables of a using the names re and im, as usual. Herdman had been asked by the London School of Contemporary Dance to visit New York and investigate the methods of Joseph Pilates. WHAT IS THE SOCRATIC METHOD? excerpted from Socrates Café by Christopher Phillips. Preclinical and clinical evidence supports the concept of bidirectional brain-gut microbiome interactions. Some simple derivative functions for equally-spaced time series data: deriv, a first derivative using the 2-point central-difference method, deriv1, an unsmoothed first derivative using adjacent differences, deriv2, a second derivative using the 3-point central-difference method, a third derivative deriv3 using a 4-point formula, and deriv4, a. Set Symmetric Difference. If, for example, a user disables, opts-out, or blocks a method (e. Central Difference: As to the second i,j i,j i,j i,j Explicit Finite Difference Method as Trinomial Tree [] () 0 2 22 0 Check if the mean and variance of the. First, the mean is often referred to as the statistical average. A research paper is a primary sourcethat is, it reports the methods and results of an original study performed by the authors. Program to estimate value of First Derivative of the function at the given points from the given data using Backward Difference Formula , Forward diff Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. 6 The low-strain region of the stress-strain curve for a ductile material tic contribution and e e is the elastic contribution (and still related to the stress by Eq 3). Under this method if the Commercial Banks do not follow the policy of the Central Bank, then the Central Bank has the only recourse to direct action. Finding a Central Value. Early hot water systems were used in Russia for central heating of the Summer Palace (1710–1714) of Peter the Great in Saint Petersburg. Users with questions about a personal health condition should consult with a qualified healthcare professional. Graphical visualization and interrogation of data are critical components of any reliable method for statistical modeling, analysis and interpretation of data. How to Calculate the Least Significant Difference (LSD): Overview. Symmetric difference is performed using ^ operator. The development of a fast and accurate method for computing the properties of the helix slow-wave structures used in travelling-wave tubes (TWTs) is described. Finite difference methods for 2D and 3D wave equations¶. Conflict- how inequalities contribute to social differences and differences in power. It is analyzed here related to time-dependent Maxwell equations, as was first introduced by Yee []. What we will learn in this chapter is the fundamental principle of this method, and the basic formulations for solving ordinary differential equations. The central difference method, equation 6 gives identical result as using the del2 function. While other grains such as wheat and rice have obvious wild relatives, there is no wild plant that looks like maize, with soft, starchy kernels arranged along a cob. Temperature in Celsius or Fahrenheit is an example of an interval scale outcome. In finite element method the non-linear quasi-static problems can solve by implicit method. However, I don't know how I can implement this so the values of y are updated the right way. What is Sterilization ? 9 Types and Methods in Microbiology Sterilization is a process of destruction of all forms of living microorganisms from a substance. • devise finite difference approximations meeting specifica tions on order of accuracy Relevant self-assessment exercises:1-5 47 Finite Difference Approximations Recall from Chapters 1 - 4 how the multi-step methods we developed for ODEs are based on a truncated Tay-lor series approximation for ∂U ∂t. Programs, student services, registration, business and industry training, publications, alumni and scholarships. Result is ≈ 170 bpm. Expert Answers. Intelligence research properly consists of the latter kind. Taylor series can be used to obtain central-difference formulas for the higher derivatives. We provide training and online certification in CPR and First Aid; Whether you need a certificate for your employer, state organization, or simply want to be prepared for an emergency, you came to the right place!!!. The most common analytical methods of sensory evaluation used in the wine industry are discrimination (or difference) and descriptive methods. Sensation is the process by which our senses gather information and send it to the brain. If an object has a horizontal velocity v i of 20 ms and horizontal acceleration from KIN 3309 at University of Houston. Sociological theories are ways for sociologists to view society through different lenses. find the next 3 terms of the sequence 2,3,9,23,48,87 Thus the next number in the first row of differences must be 52. Excerpt from GEOL557 Numerical Modeling of Earth Systems by Becker and Kaus (2016) 1 Finite difference example: 1D implicit heat equation 1. Stability of FTCS and CTCS FTCS is first-order accuracy in time and second-order accuracy in space. Medicare K Levels Amputation Even though picking out cost effective medical insurance can often be a hard technique, the creative ideas of which were definitely offered on this internet page may very well make your well being coverage range method very fast previous to. Enter a single Pro number or multiple (one per line) and click "Trace" to receive status information on your shipment(s). LeVeque University of Washington. Program to estimate value of First Derivative of the function at the given points from the given data using Backward Difference Formula , Forward diff Basic GAUSS ELIMINATION METHOD, GAUSS ELIMINATION WITH PIVOTING, GAUSS JACOBI METHOD, GAUSS SEIDEL METHOD. Central Utah Counseling Center (FCR), a leading addiction treatment center in the US, provides supervised medical detox and rehab programs to treat alcoholism, drug addiction and co-occurring mental health disorders such as PTSD, depression and anxiety. Methods Age-standardised death rates are calculated using aggregated deaths for 2008–2012 and population exposures from the 2010 census. If a finite difference is divided by b − a, one gets a difference quotient. If an object has a horizontal velocity v i of 20 ms and horizontal acceleration from KIN 3309 at University of Houston. In the purpose part you should also add the methodology that you will employ to conduct the research: surveys, phenomenology, and pre/post testing are examples of these methods. Several Pioneers of solving PDEs with finite-difference method (Lewis Fry Richardson, Richard Southwell, Richard Courant, Kurt Friedrichs, Hans Lewy, Peter Lax and John von Neumann) First application to elastic wave propagation (Alterman and Karal, 1968) Simulating Love waves and was the frst showing snapshots of seimsic wave fields (Boore, 1970). Difference Between High And Low Deductible Health Insurance As a result get out there and start seeking the fact that individual medical insurance. Vary the thickness and length of the lines. If you sell via Vendor Central, you’re called a first-party seller. The Outlier Calculator is used to calculate the outliers of a set of numbers. If a finite difference is divided by b − a, one gets a difference quotient. Identify central idea and main idea. The first-order forward difference of a list of numbers A is a new list B, where B n = A n+1 - A n. Some simple derivative functions for equally-spaced time series data: deriv, a first derivative using the 2-point central-difference method, deriv1, an unsmoothed first derivative using adjacent differences, deriv2, a second derivative using the 3-point central-difference method, a third derivative deriv3 using a 4-point formula, and deriv4, a. Sigmund Freud and Erik Erikson are both known for their work in psychoanalysis. But what if the population were non‐normal? The central limit theorem states that even if a population distribution is strongly non‐normal, its sampling distribution of means will be approximately normal for large sample sizes (over 30). The calculator will find the difference quotient for the given function, with steps shown. Traditional medicine methods have been around much longer than the modern medicine and they were an important part of the recorded history. When cost accounting, the step-down allocation method allows support departments to allocate costs to each other — and ultimately to the operating departments. Make and share study materials, search for recommended study content from classmates, track progress, set reminders, and create custom quizzes. But all interior elements are central differences. We’ve got a wide range of ways to make payments, including mail, phone, and online. There are several differences between the Deciduous / Primary and Permanent teeth in the morphology and various other aspects, and it is the deciduous teeth which erupt first, then followed by the permanent teeth. On a more naturalistic reading one might be asking: What ~natural! kind ~if any! does our epistemic vocabulary track? Or one might be undertaking a more revisionary project:What is the point of having a concept of knowledge? What concept~if any!would do that work best?3 These.
| 8,998
| 43,461
|
{"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}
| 3.09375
| 3
|
CC-MAIN-2020-05
|
latest
|
en
| 0.940559
|
rokondo.com
| 1,716,114,898,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00088.warc.gz
| 458,032,950
| 20,950
|
# Mastering Chess Calculation: Techniques and Mistakes
Chess is a highly popular board game that has been enjoyed for centuries by people of all ages and backgrounds. A game of strategy and critical thinking, chess is known for its complex gameplay that requires players to think several moves ahead. One key aspect of chess that separates beginner and expert players is the ability to properly calculate moves. Calculating in chess is all about analyzing potential moves and their outcomes to find the optimal move to make. In this article, we will discuss the importance of chess calculation and provide techniques and tips to improve your skills.
## The Importance of Chess Calculation: Strategies for Success
Before jumping into chess calculation, it is important to understand the fundamentals of the game. Chess is played on a board with 64 squares, with each player starting with 16 pieces – a king, a queen, two rooks, two knights, two bishops, and eight pawns. Each piece moves differently and has a specific role in the game. The objective of the game is to checkmate your opponent’s king, which means trapping their king so it cannot escape capture. Understanding the basics of chess will provide a foundation for calculating moves and making strategic decisions during the game.
## Master Chess Calculation: Techniques & Tips
Now that we understand the basics of chess, let’s dive into the concept of chess calculation. Chess calculation refers to the process of analyzing different moves and their outcomes to determine the best possible move to make. It involves evaluating potential moves, anticipating your opponent’s responses, and considering how the game might evolve given different moves. This skill is crucial if you want to become a strong chess player. Calculation can help you avoid mistakes, identify threats early on, and take advantage of your opponent’s weaknesses. By using proper calculation techniques, you can improve your chances of winning and even turn losing positions into winning ones.
## Common Mistakes to Avoid in Chess Calculation
While practicing good calculation habits can help you improve your game, it’s also important to be aware of common mistakes to avoid. One common mistake is overlooking potential threats. Always be on the lookout for potential threats from your opponent, and consider what moves they might make that could put you in jeopardy. Another mistake is underestimating your opponent. No matter how strong or weak their position seems, never assume that your opponent doesn’t have a plan up their sleeve. Focusing too much on one line of play can also be a mistake, as it can limit your ability to see other possible moves and strategies. Making moves too quickly or giving up too soon can also be detrimental to your game. It’s important to take the time to analyze each move and consider all possible outcomes before making a decision. By being aware of these common mistakes, you can avoid them and improve your calculation skills even further.
## Common mistakes in chess calculation to avoid
In conclusion, mastering chess calculation is essential if you want to become a strong player. By following the appropriate steps and techniques, you can develop the ability to analyze positions and anticipate your opponent’s moves to make informed decisions. Remember to practice regularly, break down complex positions, avoid impulsive moves, use visualization, play against stronger opponents, and always watch out for common mistakes. With time and practice, you can develop your calculation skills and become a formidable opponent on the chessboard. So, keep playing, keep learning, and always strive to improve your skills.
## Chess Calculation: FAQs and Tips
### What is chess calculation?
Chess calculation refers to the process of analyzing potential moves and their outcomes to determine the best possible move to make.
### Why is chess calculation important?
Chess calculation is important because it can help you avoid mistakes, identify threats early on, and take advantage of your opponent’s weaknesses. By using proper calculation techniques, you can improve your chances of winning and even turn losing positions into winning ones.
### What are some techniques to improve chess calculation?
Some techniques to improve chess calculation include practicing regularly, breaking down complex positions, avoiding impulsive moves, using visualization, playing against stronger opponents, and always being on the lookout for common mistakes.
### What are some common mistakes to avoid in chess calculation?
Some common mistakes to avoid in chess calculation include overlooking potential threats, underestimating your opponent, focusing too much on one line of play, making moves too quickly, and giving up too soon.
### How can I become better at chess calculation?
To become better at chess calculation, you should practice regularly, study different strategies and tactics, review your games and analyze your mistakes, and play against more experienced opponents. By improving your knowledge, skill, and experience, you can develop your calculation skills and become a stronger chess player.
| 942
| 5,185
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.78125
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.953207
|
https://phys.libretexts.org/TextBooks_and_TextMaps/Book%3A_Electricity_and_Magnetism_(Tatum)/2%3A_Electrostatic_Potential/2.2%3A_Potential_Near_Various_Charged_Bodies/2.2.4%3A_Large_Plane_Charged_Sheet
| 1,529,927,498,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867666.97/warc/CC-MAIN-20180625111632-20180625131632-00085.warc.gz
| 700,823,987
| 14,143
|
Skip to main content
$$\require{cancel}$$
# 2.2.4: Large Plane Charged Sheet
The field at a distance $$r$$ from a large charged sheet carrying a charge $$σ$$ coulombs per square metre is $$\frac{\sigma}{2\epsilon_0}$$. Therefore the potential difference between two points at distances $$a \text{ and }b$$ from the sheet $$(a < b)$$ is
$V_a-V_b = \frac{\sigma}{2\epsilon_0}(b-a).\tag{2.2.7}$
| 128
| 395
|
{"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}
| 2.921875
| 3
|
CC-MAIN-2018-26
|
longest
|
en
| 0.760084
|
https://www.enotes.com/homework-help/product-two-positive-numbers-16-find-numbers-420950?en_action=hh-question_click&en_label=hh_carousel&en_category=internal_campaign
| 1,611,299,887,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703529128.47/warc/CC-MAIN-20210122051338-20210122081338-00222.warc.gz
| 765,643,685
| 19,657
|
# The product of two positive numbers is 16. Find the numbers if their sum is least and if the sum of one and the square of the other is least.
The product of the two positive numbers is 16. Let one of the numbers be X, the other number is `16/X` .
The sum of the two numbers is `S = X + 16/X` . The value of S is minimized for X such that `(dS)/(dX) = 0`
=> `1 - 16/X^2 = 0`
=> `X^2 = 16`
=> X = 4
16/X = 4
The sum of one of them and the square of the other is `S2 = X^2 + 16/X` . S2 is minimized for the value of X where `(dS2)/(dX) = 0`
=> `2X - 16/X^2 = 0`
=> `X^3 = 8`
=> X = 2
16/X = 8
The sum is the least for the set {4, 4} and the sum of one of the numbers and the square of the other is least for {2, 8}
Approved by eNotes Editorial Team
factors of 16: 1&16, 2&8, 4&4. `1+16=17` , `2+8=10` , `4+4=8` . For (a) the numbers are 4 and 4.
`1^2 +16=17` `16^2+1=257 `
`2^2 +8=12`` 8^2 +2 = 66`
`4^2 + 4=20 `
Approved by eNotes Editorial Team
| 379
| 967
|
{"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}
| 4.15625
| 4
|
CC-MAIN-2021-04
|
latest
|
en
| 0.870725
|
https://www.gradesaver.com/textbooks/science/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-3rd-edition/chapter-12-rotation-of-a-rigid-body-exercises-and-problems-page-348/17
| 1,537,707,303,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267159359.58/warc/CC-MAIN-20180923114712-20180923135112-00540.warc.gz
| 754,962,089
| 12,635
|
## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)
(a) $I = 3.8\times 10^{-5}~kg~m^2$ (b) $I = 1.1\times 10^{-4}~kg~m^2$
(a) We can find the moment of inertia of the CD for a perpendicular axis through the center. $I = \frac{1}{2}MR^2$ $I = \frac{1}{2}(0.021~kg)(0.060~m)^2$ $I = 3.8\times 10^{-5}~kg~m^2$ (b) We can use the parallel axis theorem to find the moment of inertia about a perpendicular axis that is through the edge of the disk. Note that the distance $d$ from the center of mass to the edge of the disk is 6.0 cm. $I = I_{cm}+Md^2$ $I = 3.8\times 10^{-5}~kg~m^2+(0.021~kg)(0.060~m)^2$ $I = 1.1\times 10^{-4}~kg~m^2$
| 267
| 675
|
{"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}
| 3.9375
| 4
|
CC-MAIN-2018-39
|
longest
|
en
| 0.813852
|
https://www.ferienwohnung-stratmann.de/5-tons-of-gas-boiler-parameters-24228.html
| 1,643,427,165,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00624.warc.gz
| 784,387,317
| 6,018
|
## Our Products for 5 tons of gas boiler parameters
### 1. ENERGY PERFORMANCE ASSESSMENT OF BOILERS
parameter of boiler – Industrial Boiler Supplier
Get A Quote
### How To Calculate Natural Gas Boiler Consumption--ZBG
The boiler gas consumption calculation need the following parameters: such as natural gas calorific value 8500kcal/m3, pressure 8-10Kpa, and theoretically, the gas consumption of boiler heat = boiler calorific value÷(Calorific value of natural gas x Boiler thermal efficiency )= 2400 000 Kcal / (8500Kcal * 0.94) =297m3/h, here we take 4 tons of
Get A Quote
### Greenhouse Gases Equivalencies Calculator - Calculations
2. BOILERS Bureau of Energy Efficiency 27 Syllabus Boilers: Types, Combustion in boilers, Performances evaluation, Analysis of losses, Feed water treatment, Blow down, Energy conservation opportunities. 2.1 Introduction A boiler is an enclosed vessel that provides a means for combustion heat to be transferred into water until it becomes heated water or steam.
Get A Quote
### Flue Gas Analysis Table | CleanBoiler.org
The rated evaporation capacity of the boiler is 10 tons per hour, the calorific value of natural gas is 35.438 MJ/Nm3, and the boiler efficiency is calculated at 95%. The gas consumption per hour of the 10 ton boiler: 7 MW*3600s/35.53 MJ/Nm3/95%=746 m3/h. The efficiency of gas fired boilers produced by different boiler manufacturers is different.
Get A Quote
### parameter of boiler – Industrial Boiler Supplier
Feb 27, 2019 · Evaporation of boilers. During long-term operation of steam boiler continuously, the steam generated per hour is called evaporation of this boiler. The common units of evaporation is t/h. Boiler Water Parameters – Scribd Boiler Water Parameters – Download as Word Doc (.doc), PDF File (.pdf), Text File (.txt) or read online.
Get A Quote
### Furnaces and Boilers | Department of Energy
How To Calculate Natural Gas Boiler Consumption--ZBG
Get A Quote
### 10 ton gas-fired steam boiler gas consumption per hour and
Upgrading your furnace or boiler from 56% to 90% efficiency in an average cold-climate house will save 1.5 tons of carbon dioxide emissions each year if you heat with gas, or 2.5 tons if you heat with oil. If your furnace or boiler is old, worn out, inefficient, or significantly oversized, the simplest solution is to replace it with a modern
Get A Quote
### Boiler Product Catalog
Boiler Efficiency by indirect method = 100 Π(L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8) 1.7.2 Measurements Required for Performance Assessment Testing The following parameters need to be measured, as applicable for the computation of boiler effi-ciency and performance. a) Flue gas analysis 1. Percentage of CO 2 or O 2 in flue gas 2.
Get A Quote
| 668
| 2,751
|
{"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
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.83131
|
https://www.jiskha.com/display.cgi?id=1236053294
| 1,503,156,113,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886105455.37/warc/CC-MAIN-20170819143637-20170819163637-00570.warc.gz
| 903,981,165
| 3,878
|
# Algebra I
posted by .
What is the answer to Algebra with pizzaz p.93
How Can Fisherman Save Gas?
• Algebra I -
We are not among the privileged owners of your course materials. Perhaps you could enlighten us about what the question is, and show your work.
Fishermen can save gas by staying home.
• Algebra I -
By forming carp pools(no spaces).
## Similar Questions
1. ### Algebra Pizzaz
In music, What does "Allegro" mean?
2. ### physics
A 86kg fisherman jumps from a dock into a 121kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3.4m/s to the West as he leaves the dock, what is the final velocity of the fisherman and the rowboat?
3. ### physics
A 86kg fisherman jumps from a dock into a 121kg rowboat at rest on the West side of the dock. If the velocity of the fisherman is 3.4m/s to the West as he leaves the dock, what is the final velocity of the fisherman and the rowboat?
4. ### math riddle algebra pizzaz
what do you call it when somebody spends 20 years in the 24th row of a theater
5. ### math
how can fisherman save gas
6. ### Math-Algebra 1
I need help with a pizzaz work sheet ASAP if i don't get this done i am on the verge of failing math and repeating 9th grade. its Pizzaz 11.10 thanks
7. ### Math Riddle-Algebra With Pizzaz
Did you hear about... i have a,b,c,d,e,f,g,n,o..its "the very sad guy who tried to..fog and thats all i have help?
8. ### math
what is the answer to the famous farming exression, from creative publications pre-algebra with pizzaz
9. ### Algebra
A car gets 30 miles per gallon of gas and is driven 1000 miles per month. Gas costs \$4 per gallon. How much will the car's owner save on gas expenses in one year by buying a new car that goes 50 miles for each gallon of gas?
10. ### Math/Algebra
I need help on a Middle School Math With Pizzaz worksheet. It is titled, Vive la France!
More Similar Questions
| 511
| 1,907
|
{"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}
| 2.90625
| 3
|
CC-MAIN-2017-34
|
latest
|
en
| 0.909466
|
https://pupman.com/listarchives/1999/November/msg00365.html
| 1,685,871,373,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224649741.26/warc/CC-MAIN-20230604093242-20230604123242-00751.warc.gz
| 528,877,218
| 2,873
|
# Re: My First Tesla Coil
```Hello, Ryan
On Tue, 16 Nov 1999 21:42:02 -0700 Tesla List <tesla-at-pupman-dot-com> writes:
> Original Poster: "Spud" <spud-at-wf-dot-net>
>
> Some of you may have already seen me posting here before.
> I'm 17 and in
> high school, and this is my first high-voltage project, let alone
> my first
> Tesla Coil. I've been progressing on it very slowly due to my lack
> of
> money, but I think I am about to finish it. Everything I've done is
> based
> on knowledge and figures I found on websites. I have everything but
> the
> capacitors, but I'm going to go ahead and tell anyway. Here are the
> and
> tell me what flaws you see with it. Right now I am not as
> interested in
> getting perfect performance from the coil as I am just trying to get
> the
> thing to work at all.
> Thanks,
> Ryan
> -------------------------------------------------------
> HV Transformer: 12Kv -at- 30 mA NST
> Main Spark Gap: 8 Gaps (Spaced with two business cards, I don't
> know
> exactly how thick that is. :) Also, how wide should my safety
> gap be? )
Set the saftey gap so it will -just- not fire from the supplied voltage
of the NST, with no other components connected. If it fires incessantly
in TC opertaion, either increase it very slightly, or decrease the number
of gaps(of the 8) you use, or both.
> Capacitance: MMC: 30 Caps of 600 VAC (1200 VDC) that make
> mfd. (I guess I should make 3 strings of 10 caps
>
> each.)
This is a little confusing, do you mean that paralleled in three strings,
you get .0066µF? Please tell us what the individual capacitance of each
cap is. AT any rate, with only 30 caps, you should call this an EMMC.
> Primary Coil: 15 Turns of 1/4" copper tubing. (Each turn is
> 1/2"
> apart, the primary is about 1" away from the secondary.)
Probably too small for the secondary.
> Secondary Coil: 22" Winding of #26 AWG magnet wire on 4" (4.5" OD)
> PVC.
This is longer than usual, and will give you a rather low resonant
frequency. You may need a larger primary than the 15 turns. I needed
about 14 turns (same spacing as yours) for a system of mine using a
.0014µF cap, with a 16"X4.5" sec using #26 wire.
> Torus: 17.5" x 4" Alluminum Dryer Duct and Pie Pans.
> :)
The toroid sounds fine. It will further lower the Fr of the sec., but
with a larger primary this should not be a problem.
Grayson Dietrich (17 too)
http://www.electrophile.8m-dot-com
___________________________________________________________________
Get the Internet just the way you want it.
Free software, free e-mail, and free Internet access for a month!
Try Juno Web: http://dl.www.juno-dot-com/dynoget/tagj.
```
| 776
| 2,717
|
{"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-2023-23
|
latest
|
en
| 0.929964
|
https://oeis.org/A262033
| 1,695,922,132,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00118.warc.gz
| 470,202,085
| 4,172
|
The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A262033 Number of permutations of [n] beginning with at least floor(n/2) ascents. 5
1, 1, 1, 3, 4, 20, 30, 210, 336, 3024, 5040, 55440, 95040, 1235520, 2162160, 32432400, 57657600, 980179200, 1764322560, 33522128640, 60949324800, 1279935820800, 2346549004800, 53970627110400, 99638080819200, 2490952020480000, 4626053752320000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 LINKS Alois P. Heinz, Table of n, a(n) for n = 0..732 FORMULA E.g.f.: (x+1)*(exp(x^2)-1)/x^2. a(n) = 2*n*(n*(n-1)*a(n-2)-a(n-1))/((n+2)*(n-1)) for n>1, a(0)=a(1)=1. a(n) = n!/ceiling((n+1)/2)!. a(2n) = A262034(2n) = A001761(n). a(2n+1) = A006963(n+2). Sum_{n>=0} 1/a(n) = 7/4 + 13*exp(1/4)*sqrt(Pi)*erf(1/2)/8, where erf is the error function. - Amiram Eldar, Dec 04 2022 EXAMPLE a(4) = 4: 1234, 1243, 1342, 2341. a(5) = 20: 12345, 12354, 12435, 12453, 12534, 12543, 13425, 13452, 13524, 13542, 14523, 14532, 23415, 23451, 23514, 23541, 24513, 24531, 34512, 34521. MAPLE a:= proc(n) option remember; `if`(n<2, 1, 2*n*(n*(n-1)*a(n-2)-a(n-1))/((n+2)*(n-1))) end: seq(a(n), n=0..30); MATHEMATICA a[n_] := n!/Ceiling[(n + 1)/2]!; Array[a, 30, 0] (* Amiram Eldar, Dec 04 2022 *) CROSSREFS Cf. A001761, A006963, A262034, A262035. Sequence in context: A333320 A047165 A124631 * A062870 A226964 A222763 Adjacent sequences: A262030 A262031 A262032 * A262034 A262035 A262036 KEYWORD nonn AUTHOR Alois P. Heinz, Sep 08 2015 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 28 13:18 EDT 2023. Contains 365735 sequences. (Running on oeis4.)
| 803
| 1,911
|
{"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}
| 3.8125
| 4
|
CC-MAIN-2023-40
|
latest
|
en
| 0.481849
|
https://aeolipera.wordpress.com/2012/09/24/numerical-linear-algebra-day-1-incomplete/
| 1,653,778,980,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00302.warc.gz
| 140,209,305
| 26,135
|
## Numerical linear algebra, day 1 (incomplete)
Typo disclaimer: they will happen because this keyboard is not very sensitive. If you want to alert me to any, please leave a comment like: Typo in p5s1: teh -> the. This will tell me to fix the typo in the first sentence (s1) of the fifth paragraph.
August 27, 2012 (This will not be included in the book :-P)
I had an opportunity to meet the professor for this class a few days previously. He immediately struck me as a total prick, but at least English is his first language. You need to understand that in mathematics departments there is worse than a fifty-fifty chance of this. He also sounds exactly like Mr. Mackey, which is a source of interminable amusement for me.
So imagine Mr. Mackey guy telling you that reading the textbook won’t do you much good because he won’t be following it very closely, and that Mathematica will not be an acceptable substitute for Matlab and you have a good idea of our first meeting. On to the real stuff.
### Error
What is an approximation for pi?
A couple of common approximations are 3.14, or 22/7 (which is correct to three digits, 22/7 = 3.1428…). In fact, back in the days before fire was invented and when calculators were prohibitively expensive, it was normal for students to plug in 22/7 for pi when solving problems, so much so that a few of the dimmer ones believed (and probably still believe) that pi = 22/7.
But what if I said pi was approximately 28? Technically, it’s true, although I’d have a hard job convincing NASA to hire me for my brilliant sense of humor. The problem is that it’s not a very good approximation of pi.
A good numerical approximation is not very different from the number it is approximating. Any difference between the two is called error. There are two important kinds of error, called absolute error and relative error.
If p* (usually pronounced “p star”) is approximating another number p then the difference between the two is called the absolute error. That is,
$\mbox{Absolute error} = |p - p*|$
Example: The absolute error between pi and 22/7 is $|\pi - \frac{22}{7}| = 0.0012644892673...$
But this presents an incomplete picture of the error. An absolute error of one million would be remarkably good if you were measuring the Gross Domestic Product (in dollars), and an absolute error of 0.0001 would be terrible if you were measuring **insert something small when you’re feeling creative**. That’s where relative error comes in.
Again using p* as the approximation for the real number p, the difference between the two divided by the real number is the relative error.
$\mbox{Relative error} = \cfrac{|p - p*|}{p} \quad \mbox{for } p \neq 0$
While it’s obviously important to understand the simple mathematical definitions, in practice these are often referenced in a purely theoretical sense because it is rarely possible to know what p should be. (Otherwise, we’d just be using p instead of trying to approximate it.) That said, there’s no better method for checking that a numerical algorithm is working properly than comparing it with the exact numbers (found algebraically) and computing the absolute and relative error.
### Machine Storage of Numbers
**Note: Include a primer on binary numbers and floating point numbers, perhaps as appendices.**
Because computers have a finite number of bits, they can’t properly store nonterminating numbers like $\cfrac{1}{3} = 0.3333...$ Such numbers have to be truncated in some way. There are two primary methods for this, called k-digit chopping and k-digit rounding.
K-digit chopping is exactly what it sounds like:
**Note: Introduce floating point notation so that these examples will actually make sense.**
Example: Find $fl(\cfrac{1}{3})$, using
1. 1-digit chopping arithmetic
2. 3-digit chopping arithmetic
3. 10-digit chopping arithmetic
and express the answers in base-10.
| 897
| 3,890
|
{"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": 5, "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}
| 3.828125
| 4
|
CC-MAIN-2022-21
|
latest
|
en
| 0.967286
|
https://www.coursehero.com/file/170623/tsl148/
| 1,521,802,964,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257648207.96/warc/CC-MAIN-20180323102828-20180323122828-00280.warc.gz
| 802,847,119
| 42,048
|
{[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
# tsl148 - Resistor Circuit(1 Consider the two resistor...
This preview shows pages 1–5. Sign up to view the full content.
Resistor Circuit (1) Consider the two resistor circuits shown. (a) Find the resistance R 1 . (b) Find the emf E 1 . (c) Find the resistance R 2 . (d) Find the emf E 2 . 1A R 1 1Ω 2Ω 1Ω 1A 2A 6Ω R 2 ε 2 ε 1 2A tsl148 – p.1/8
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Resistor Circuit (2) Consider the two resistor circuits shown. (a) Find the resistance R 1 . (b) Find the current I 1 . (c) Find the resistance R 2 . (d) Find the current I 2 . 3A 6Ω R 1Ω 1 I 1 12V 12V 3Ω 3A 2Ω I 2 R 2 tsl149 – p.2/8
Resistor Circuit (3) Consider the rsistor and capacitor circuits shown. (a) Find the equivalent resistance R eq . (b) Find the power P 2 , P 3 , P 4 dissipated in each resistor. (c) Find the equivalent capacitance C eq . (d) Find the energy U 2 , U 3 , U 4 stored in each capacitor. 3Ω 2Ω 4Ω 2nF 4nF 3nF 12V 12V tsl150 – p.3/8
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Resistor Circuit (4) Consider the resistor circuit shown. (a) Find the direction of the current I (cw/ccw). (b) Find the magnitude of the current I . (c) Find the voltage V ab = V b - V a .
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 8
tsl148 - Resistor Circuit(1 Consider the two resistor...
This preview shows document pages 1 - 5. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online
| 541
| 1,675
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.375
| 3
|
CC-MAIN-2018-13
|
latest
|
en
| 0.697585
|
https://statmoddev.stat.columbia.edu/author/phil/page/3/
| 1,701,423,254,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00838.warc.gz
| 629,074,959
| 12,460
|
Archive of entries posted by
## Everything I need to know about Bayesian statistics, I learned in eight schools.
This post is by Phil. I’m aware that there are some people who use a Bayesian approach largely because it allows them to provide a highly informative prior distribution based subjective judgment, but that is not the appeal of Bayesian methods for a lot of us practitioners. It’s disappointing and surprising, twenty years after my initial experiences, […]
## You heard it here first: Intense exercise can suppress appetite
This post is by Phil Price. The New York Times recently ran an article entitled “How Exercise Can Help Us Eat Less,” which begins with this: “Strenuous exercise seems to dull the urge to eat afterward better than gentler workouts, several new studies show, adding to a growing body of science suggesting that intense exercise […]
## There are no fat sprinters
This post is by Phil. A little over three years ago I wrote a post about exercise and weight loss in which I described losing a fair amount of weight due to (I believe) an exercise regime, with no effort to change my diet; this contradicted the prediction of studies that had recently been released. […]
## The Great Race
This post is by Phil. Last summer my wife and I took a 3.5-month vacation that included a wide range of activities. When I got back, people would ask “what were the highlights or your trip?”, and I was somewhat at a loss: we had done so many things that were so different, many of […]
## Data problems, coding errors…what can be done?
This post is by Phil A recent post on this blog discusses a prominent case of an Excel error leading to substantially wrong results from a statistical analysis. Excel is notorious for this because it is easy to add a row or column of data (or intermediate results) but forget to update equations so that […]
## Subsidized driving
This post is by Phil. This DC Streets Blog post gives a concise summary of a report by “The Tax Foundation”. The money shot is here, a table that shows what fraction spending on roads in each state in the U.S. is covered by local, state, and federal gas taxes, tolls, registration fees, etc. (Click […]
## Back when 50 miles was a long way
This post is by Phil. Michael Graham Richard has posted some great maps from the 1932 Atlas of the Historical Geography of the United States; the maps show how long it took to get to various places in the U.S. from New York City in 1800, 1830, 1857, and 1930. (I wonder if the atlas […]
## Write This Book
This post is by Phil Price. I’ve been preparing a review of a new statistics textbook aimed at students and practitioners in the “physical sciences,” as distinct from the social sciences and also distinct from people who intend to take more statistics courses. I figured that since it’s been years since I looked at an intro […]
## How to Lie With Statistics example number 12,498,122
This post is by Phil Price. Bill Kristol notes that “Four presidents in the last century have won more than 51 percent of the vote twice: Roosevelt, Eisenhower, Reagan and Obama”. I’m not sure why Kristol, a conservative, is promoting the idea that Obama has a mandate, but that’s up to him. I’m more interested […]
## We go to war with the data we have, not the data we want
This post is by Phil. Psychologists perform experiments on Canadian undergraduate psychology students and draws conclusions that (they believe) apply to humans in general; they publish in Science. A drug company decides to embark on additional trials that will cost tens of millions of dollars based on the results of a careful double-blind study….whose patients are […]
## Help with this problem, win valuable prizes
This post is by Phil. In the comments to an earlier post, I mentioned a problem I am struggling with right now. Several people mentioned having (and solving!) similar problems in the past, so this seems like a great way for me and a bunch of other […]
## The more likely it is to be X, the more likely it is to be Not X?
This post is by Phil Price. A paper by Wood, Douglas, and Sutton looks at “Beliefs in Contradictory Conspiracy Theories.” Unfortunately the subjects were 140 undergraduate psychology students, so one wonders how general the results are. I found this sort of arresting: In Study 1 (n=137), the more participants believed that Princess Diana faked her […]
## Benford’s Law suggests lots of financial fraud
This post is by Phil. I love this post by Jialan Wang. Wang “downloaded quarterly accounting data for all firms in Compustat, the most widely-used dataset in corporate finance that contains data on over 20,000 firms from SEC filings” and looked at the statistical distribution of leading digits in various pieces of financial information. As […]
## Another day, another stats postdoc
This post is from Phil Price. I work in the Environmental Energy Technologies Division at Lawrence Berkeley National Laboratory, and I am looking for a postdoc who knows substantially more than I do about time-series modeling; in practice this probably means someone whose dissertation work involved that sort of thing. The work involves developing models […]
## Even a good data display can sometimes be improved
When I first saw this graphic, I thought “boy, that’s great, sometimes the graphic practically makes itself.” Normally it’s hard to use lots of different colors to differentiate items of interest, because there’s usually not an intuitive mapping between color and item (e.g. for countries, or states, or whatever). But the colors of crayons, what […]
## Censoring on one end, “outliers” on the other, what can we do with the middle?
This post was written by Phil. A medical company is testing a cancer drug. They get a 16 genetically identical (or nearly identical) rats that all have the same kind of tumor, give 8 of them the drug and leave 8 untreated…or maybe they give them a placebo, I don’t know; is there a placebo […]
## How the ignorant idiots win, explained. Maybe.
According to a New York Times article, cognitive scientists Hugo Mercier and Dan Sperber have a new theory about rational argument: humans didn’t develop it in order to learn about the world, we developed it in order to win arguments with other people. “It was a purely social phenomenon. It evolved to help us convince […]
## Improvement of 5 MPG: how many more auto deaths?
This entry was posted by Phil Price. A colleague is looking at data on car (and SUV and light truck) collisions and casualties. He’s interested in causal relationships. For instance, suppose car manufacturers try to improve gas mileage without decreasing acceleration. The most likely way they will do that is to make cars lighter. But […]
## Worst statistical graphic I have seen this year
This gets my vote for the worst statistical graphic I (Phil) have seen this year. If you’ve got a worse one, put a link in the comments. “Credit” for this one goes to “Peter and Maria Hoey (Source: Tommy McCall/Environmental Law Institute).”
## Do you have any idea what you’re talking about?
We all have opinions about the federal budget and how it should be spent. Infrequently, those opinions are informed by some knowledge about where the money actually goes. It turns out that most people don’t have a clue. What about you? Here, take this poll/quiz and then compare your answers to (1) what other people said, in a CNN poll that asked about these same items and (2) compare your answers to the real answers.
Quiz is below the fold.
| 1,620
| 7,558
|
{"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-2023-50
|
latest
|
en
| 0.972609
|
https://www.mathworks.com/matlabcentral/answers/1603600-adding-values-to-a-matrix-in-for-loop
| 1,642,548,845,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320301063.81/warc/CC-MAIN-20220118213028-20220119003028-00058.warc.gz
| 849,361,830
| 25,726
|
# Adding values to a matrix in for loop
2 views (last 30 days)
Tony on 5 Dec 2021
Answered: Benjamin on 5 Dec 2021
I have a matrix, "in", with a set of values. I want to create a new array, starting at 29, and continuosly adding the next value of array "in" to the new array.
[29 37 45 53 61]
however, I am getting this:
[37 37 37 37 45 45 45 45 53 53 53 53 61 61 61 61]
It's duplicating each value by 4 (the length of the matrix).
Here's my code:
in = [8 8 8 8];
values = [];
values_change=29;
for i=1:length(in)
values_change = values_change +in;
values = [values, values_change];
end
disp(values)
How do I fix this? Thanks!
NOTE: the matrix in may change so it might not all be values of 8. So I need the code to account this matrix, not the values, 8.
Star Strider on 5 Dec 2021
Edited: Star Strider on 5 Dec 2021
Subscript ‘in’ here —
values_change = values_change + in(i);
and then subscript ‘values’ (the preallocation is optional, however a good habit to adopt, sinc it produces much more efficnet matrix operations in this snd similar applications). See if that produces the desired result.
in = [8 8 8 8];
values = NaN(1,numel(in)+1);
values_change = 29;
values(1) = values_change;
for i=1:length(in)
values_change = values_change + in(i);
values(i+1) = values_change;
end
disp(values)
29 37 45 53 61
.
##### 2 CommentsShowHide 1 older comment
Star Strider on 5 Dec 2021
As always, my pleasure!
.
Benjamin on 5 Dec 2021
Each time through the loop, the code is concatenating the vector values_change onto the end of the vector values. (Note that the code inside the loop doesn't depend on the loop iterator i.) To append only one element at a time to the vector values, you can do this:
in = [8 8 8 8];
values = [];
values_change=29;
for i=1:length(in)
values_change = values_change +in(i);
values = [values, values_change];
end
However, the value of values at the end of this wil not include the initial value of values_change (i.e., 29) because it is incremented before it is stored in values the first time. To correct that, you can initialize values to have the value 29 before the loop:
in = [8 8 8 8];
values_change=29;
values = values_change;
for i=1:length(in)
values_change = values_change +in(i);
values = [values, values_change];
end
This will give you the desired output.
However, since this loop is essentially doing a cumulative sum over in, you can do the same thing without a loop, using the cumsum function:
in = [8 8 8 8];
values_change = 29;
values = values_change+cumsum([0 in]);
| 721
| 2,511
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.234375
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.674141
|
https://www.askmehelpdesk.com/physics/physics-answer-160071.html
| 1,713,495,231,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817253.5/warc/CC-MAIN-20240419013002-20240419043002-00321.warc.gz
| 581,174,284
| 10,462
|
ssfire Posts: 1, Reputation: 1 New Member #1 Dec 6, 2007, 04:04 PM
A board that is 1.5 m long is supported in two places. If the force exerted by the first support is 25N and the force exerted by the second is 62N, what is the mass of the board?
Capuchin Posts: 5,255, Reputation: 656 Uber Member #2 Dec 6, 2007, 04:09 PM
I assume that the whole system is at rest.
If this is the case, then what conclusions can you make about the total upward force compared to the total downward force?
Polokid16 Posts: 11, Reputation: -1 New Member #3 Dec 8, 2007, 09:06 PM
Assuming that the system is at rest:
"_" - subscript
F_total = 25N + 62N = 87N
Now use the formula F_g = ma and rearrange it
m = F_g / a
Note: a = g = 9.81m/s^2
m= (87N) / (9.81m/s^2) = 8.9kg
Question Tools Search this Question Search this Question: Advanced Search
## Check out some similar questions!
An electron starts from rest 5.50 cm from the center of a uniformly charged insulating sphere of radius 2.90 cm. If the sphere carries a total charge of how fast will the electron be moving when it reaches the surface of the sphere? Note: Assume a reference level of potential V = 0 at r= infinity
P23-A The budget committee of Vinning office supply Co. Has assembled the following data. As the business manager, you must prepare the budgeted income statements for May and Jun 2006. A. Sales in April were \$42,100.. You forecast that monthly sales will increase 2.0% in May and 2.4% in June.
Help with physics! [ 2 Answers ]
Please help with this physics problem! a. A 50kg water skier is pulled on a lake by a boat with a horizontal force of 400N. If the wind and current applies another force of 100N at an angle of 30degrees relative to the straight-line direction of the boat, what is the skiers acceleration? B.)...
| 507
| 1,800
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.453125
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.915905
|
https://www.convert-measurement-units.com/convert+Poundal+per+square+foot+to+Meter+of+water+column.php
| 1,726,032,856,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00210.warc.gz
| 658,136,969
| 14,104
|
Convert Poundal per square foot to mH2O (Poundal per square foot to Meter of water column)
## Poundal per square foot into Meter of water column
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Poundal+per+square+foot+to+Meter+of+water+column.php
# Convert Poundal per square foot to mH2O (Poundal per square foot to Meter of water column)
1. Choose the right category from the selection list, in this case 'Pressure'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), square root (√), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Poundal per square foot'.
4. Finally choose the unit you want the value to be converted to, in this case 'Meter of water column [mH2O]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
### Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '260 Poundal per square foot'. In so doing, either the full name of the unit or its abbreviation can be used Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Pressure'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '98 Poundal per square foot to mH2O' or '93 Poundal per square foot into mH2O' or '97 Poundal per square foot -> Meter of water column' or '96 Poundal per square foot = mH2O' or '94 Poundal per square foot to Meter of water column' or '92 Poundal per square foot into Meter of water column'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(87 * 86) Poundal per square foot'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '89 Poundal per square foot + 88 Meter of water column' or '85mm x 84cm x 83dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
The mathematical functions sin, cos, tan and sqrt can also be used. Example: sin(π/2), cos(pi/2), tan(90°), sin(90) or sqrt(4).
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 5.221 333 285 819 2×1020. For this form of presentation, the number will be segmented into an exponent, here 20, and the actual number, here 5.221 333 285 819 2. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 5.221 333 285 819 2E+20. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 522 133 328 581 920 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications.
| 953
| 4,094
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.203125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.794435
|
https://en.wikipedia.org/wiki/Curie_constant
| 1,542,069,191,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039741176.4/warc/CC-MAIN-20181113000225-20181113022225-00303.warc.gz
| 619,892,095
| 13,092
|
# Curie constant
The Curie constant is a material-dependent property that relates a material's magnetic susceptibility to its temperature.
The Curie constant, when expressed in SI units, is given by
${\displaystyle C={\frac {\mu _{0}\mu _{B}^{2}}{3k_{B}}}ng^{2}J(J+1)}$[1]
where ${\displaystyle n}$ is the number of magnetic atoms (or molecules) per unit volume, ${\displaystyle g}$ is the Landé g-factor, ${\displaystyle \mu _{B}}$ is the Bohr magneton, ${\displaystyle J}$ is the angular momentum quantum number and ${\displaystyle k_{B}}$ is Boltzmann's constant. For a two-level system with magnetic moment ${\displaystyle \mu }$, the formula reduces to
${\displaystyle C={\frac {1}{k_{B}}}n\mu _{0}\mu ^{2}}$
while the corresponding expressions in Gaussian units are
${\displaystyle C={\frac {\mu _{B}^{2}}{3k_{B}}}ng^{2}J(J+1)}$
${\displaystyle C={\frac {1}{k_{B}}}n\mu ^{2}}$
The constant is used in Curie's Law, which states that for a fixed value of a magnetic field, the magnetization of a material is (approximately) inversely proportional to temperature.
${\displaystyle \mathbf {M} ={\frac {C}{T}}\mathbf {B} }$
This equation was first derived by Pierre Curie.
Because of the relationship between magnetic susceptibility ${\displaystyle \chi }$, magnetization ${\displaystyle \scriptstyle \mathbf {M} }$ and applied magnetic field ${\displaystyle \scriptstyle \mathbf {H} }$ is almost linear at low fields, then
${\displaystyle \chi ={\frac {\mathrm {d} \mathbf {M} }{\mathrm {d} \mathbf {H} }}\approx {\frac {\mathbf {M} }{\mathbf {H} }}}$,
this shows that for a paramagnetic system of non-interacting magnetic moments, magnetization ${\displaystyle \scriptstyle \mathbf {M} }$ is inversely related to temperature ${\displaystyle T}$ (see Curie's Law).
| 508
| 1,779
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 17, "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-2018-47
|
latest
|
en
| 0.773198
|
https://www.albert.io/ie/abstract-algebra/finite-fields-squares-and-sums-of-two-squares
| 1,485,161,648,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560282202.61/warc/CC-MAIN-20170116095122-00514-ip-10-171-10-70.ec2.internal.warc.gz
| 865,008,651
| 20,238
|
Free Version
Difficult
# Finite Fields: Squares and Sums of Two Squares
ABSALG-DDRUVY
Let $F$ be a finite field of characteristic $p\ge 2$.
Which of the following are always true?
Select ALL that apply.
A
Exactly half of the elements of $F$ are squares when $p=2$ and
exactly half the elements of $F^\ast=F\setminus\{0\}$ are squares when $p\ge 3$.
B
The map $\varphi:F\rightarrow F$ given by $\varphi(a)=a^2$ is a ring homomorphism.
C
If $A$ and $B$ are subsets of $F$ such that $|A|+|B|$ is strictly greater than $|F|$, then $F=\{a+b\mid a\in A, b\in B\}$.
D
Every element of $F$ is a sum of two squares (the squares are not necessarily distinct and may be zero).
| 223
| 678
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2017-04
|
longest
|
en
| 0.845728
|
https://www.indiabix.com/aptitude/volume-and-surface-area/058003
| 1,720,945,689,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00107.warc.gz
| 725,763,836
| 8,890
|
# Aptitude - Volume and Surface Area
Exercise : Volume and Surface Area - General Questions
11.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the ratio of its diameter to its height.
3 : 7
7 : 3
6 : 7
7 : 6
Explanation:
r2h = 924 r = 924 x 2 = 7 m. 2rh 264 264
And, 2rh = 264 h = 264 x 7 x 1 x 1 = 6m. 22 2 7
Required ratio = 2r = 14 = 7 : 3. h 6
12.
A cistern of capacity 8000 litres measures externally 3.3 m by 2.6 m by 1.1 m and its walls are 5 cm thick. The thickness of the bottom is:
90 cm
1 dm
1 m
1.1 cm
Explanation:
Let the thickness of the bottom be x cm.
Then, [(330 - 10) x (260 - 10) x (110 - x)] = 8000 x 1000
320 x 250 x (110 - x) = 8000 x 1000
(110 - x) = 8000 x 1000 = 100 320 x 250
x = 10 cm = 1 dm.
13.
What is the total surface area of a right circular cone of height 14 cm and base radius 7 cm?
344.35 cm2
462 cm2
498.35 cm2
None of these
Explanation:
h = 14 cm, r = 7 cm.
So, l = (7)2 + (14)2 = 245 = 75 cm.
Total surface area = rl + r2
= 22 x 7 x 75 + 22 x 7 x 7 cm2 7 7
= [154(5 + 1)] cm2
= (154 x 3.236) cm2
= 498.35 cm2.
14.
A large cube is formed from the material obtained by melting three smaller cubes of 3, 4 and 5 cm side. What is the ratio of the total surface areas of the smaller cubes and the large cube?
2 : 1
3 : 2
25 : 18
27 : 20
Explanation:
Volume of the large cube = (33 + 43 + 53) = 216 cm3.
Let the edge of the large cube be a.
So, a3 = 216 a = 6 cm.
Required ratio = 6 x (32 + 42 + 52) = 50 = 25 : 18. 6 x 62 36
15.
How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
5600
6000
6400
7200
| 657
| 1,678
|
{"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-2024-30
|
latest
|
en
| 0.790853
|
https://johnsonsecurity.org/quaternion-rotation/
| 1,656,750,203,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103989282.58/warc/CC-MAIN-20220702071223-20220702101223-00452.warc.gz
| 376,096,547
| 10,122
|
# QUATERNION ROTATION
Quaternions were first described by Irish mathematician William Rowan Hamilton in 1843 but not widely used until recent innovations with drones and robots. Euler rotations are easier to visualize but when you string together numerous rotations one axis can become locked. Quaternions do not have the gimbal lock issue. The standard formula for quaternion rotation is where Pr is the rotated point:
$Pr=QPQ^{-1}$
Lets start by calculating Q and Q^{-1}. We will rotate the point (24,0,40) about the x axis vector (1,0,0). The incremental will be 10 degrees so when we click to rotate the x, y or z it will rotate about the current vector by 10 degrees.
$Q=a+b\vec{i}+c\vec{j}+d\vec{k}$
$Q=\cos(\frac{\frac{10*\pi}{180}}{2})+sin(\frac{\frac{10*\pi}{180}}{2})*1+sin(\frac{\frac{10*\pi}{180}}{2})*0+sin(\frac{\frac{10*\pi}{180}}{2})*0$
$Q=0.9962+0.0871\vec{i}+0\vec{j}+0\vec{k}$
$Q^{-1}=a+b\vec{i}+c\vec{j}+d\vec{k}$
$Q^{-1}=\cos(-\frac{\frac{10*\pi}{180}}{2})+sin(-\frac{\frac{10*\pi}{180}}{2})*1+sin(-\frac{\frac{10*\pi}{180}}{2})*0+sin(-\frac{\frac{10*\pi}{180}}{2})*0$
$Q^{-1}=0.9962-0.0871\vec{i}+0\vec{j}+0\vec{k}$
The next step is to calculate the Hamilton Product of QP.
$(a_1+b_1vec{i}+c_1vec{j}+d_1\vec{k})(a_2+b_2vec{i}+c_2vec{j}+d_2\vec{k})=$
$a_1a_2+a_1b_2\vec{i}+a_1c_2\vec{j}+a_1d_2\vec{k}$
$+b_1a_2\vec{i}+b_1b_2\vec{i^2}+b_1c_2\vec{i}\vec{j}+b_1d_2\vec{i}\vec{k}$
$+c_1a_2\vec{j}+c_1b_2\vec{j}\vec{i}+c_1c_2\vec{j^2}+c_1d_2\vec{j}\vec{k}$
$+d_1a_2\vec{k}+d_1b_2\vec{k}\vec{i}+d_1c_2\vec{k}\vec{j}+d_1d_2\vec{k^2}$
After multiplying the basis elements you get the following:
$a_1a_2-b_1b_2-c_1c_2-d_1d_2$
$+(a_1b_2+b_1a_2+c_1d_2-d_1c_2)\vec{i}$
$+(a_1c_2-b_1d_2+c_1a_2+d_1b_2)\vec{j}$
$+(a_1d_2+b_1c_2-c_1b_2+d_1a_2)\vec{k}$
Now we can substitute our values to calculate QP:
$QP=\begin{pmatrix} (.9962*0-.0871*24-0*0-0*40 )\\ (.9962*24+.0871*0+0*40-0*0)\vec{i}\\(.9962*0-.0871*40+0*0+0*24)\vec{j}\\(.9962*40+.0871*0-0*24+0*0)\vec{k} \end{pmatrix}$
$QP=\begin{pmatrix} -2.092\\ 23.9087\vec{i}\\-3.4862\vec{j}\\39.8478\vec{k} \end{pmatrix}$
$QPQ^{-1}=\begin{pmatrix}(-2.092*-.0871+23.9087*.9962+-3.4862*0-39.8478*0)\vec{i}\\(-2.092*0-23.9087*0+-3.4862*.9962+39.8478*-.0871)\vec{j}\\(-2.092*0+23.9087*0+3.4862*-.0871+39.8478*.9962)\vec{k}\end{pmatrix}$
$QPQ^{-1}=\begin{pmatrix}24\vec{i}\\-6.945\vec{j}\\39.392\vec{k}\end{pmatrix}$
The point (24,0,40) will move to (24,-6.945,39.392) when rotated 10 degrees about the x axis. Click here to download my excel sheet with quaternion rotation of a manhole.
| 1,178
| 2,563
|
{"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}
| 4.34375
| 4
|
CC-MAIN-2022-27
|
latest
|
en
| 0.358842
|
https://codereview.stackexchange.com/questions/180832/find-duplicates-in-a-list-of-tuples/180834
| 1,656,144,157,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656103034877.9/warc/CC-MAIN-20220625065404-20220625095404-00319.warc.gz
| 224,111,702
| 69,411
|
Find duplicates in a list of tuples
You are given information about users of your website. The information includes username, a phone number and/or an email. Write a program that takes in a list of tuples where each tuple represents information for a particular user and returns a list of lists where each sublist contains the indices of tuples containing information about the same person. For example:
Input:
[("MLGuy42", "andrew@gmail.com", "123-4567"),
("CS229DungeonMaster", "123-4567", "ml@stanford.edu"),
("Doomguy", "john@oculus.com", "carmack@gmail.com"),
("andrew26", "andrew@gmail.com", "mlguy@gmail.com")]
Output:
[[0, 1, 3], [2]]
Since "MLGuy42", "CS229DungeonMaster" and "andrew26" are all the same person.
Each sublist in the output should be sorted and the outer list should be sorted by the first element in the sublist.
Below is the code snippet that I did for this problem. It seems to work fine, but I'm wondering if there is a better/optimized solution.
def find_duplicates(user_info):
results = list()
seen = dict()
for i, user in enumerate(user_info):
first_seen = True
key_info = None
for info in user:
if info in seen:
first_seen = False
key_info = info
break
if first_seen:
results.append([i])
pos = len(results) - 1
else:
index = seen[key_info]
results[index].append(i)
pos = index
for info in user:
seen[info] = pos
return results
• By the way, this is a thinly disguised union-find problem. Nov 21, 2017 at 22:31
There is a way to increase readability. A for-else statement can be used to replace flags that are only set on break. Click here to learn more. The else clause is only called if the loop completes without break being called.
def find_duplicates(user_info):
results = list()
seen = dict()
for i, user in enumerate(user_info):
for info in user:
if info in seen:
index = seen[info]
results[index].append(i)
pos = index
break
else:
results.append([i])
pos = len(results) - 1
for info in user:
seen[info] = pos
return results
How I tested speeds
Your code, as it currently stands, will work exactly the same with integers. So, I just randomly generated large data sets this way (and played along with MAXVALUE and range a bit.
from random import randrange
from timeit import timeit
MAXVALUE = 1000
for a in range(5): #I wanted to make sure I checked 5 to make sure I don't get a outlier data set that effects my ability to use timeit reasonably.
user_info = [[randrange(MAXVALUE) for i in range(3)] for _ in range(1000)]
print(timeit(lambda: find_duplicates(user_info), number=10000))
Credit for improving code: Maarten Fabré
• I think your second if info in seen: should be if info not in seen: Nov 21, 2017 at 15:59
• @MaartenFabré You're right. When I looked at it I found that it wasn't faster checking that, so I removed it all together. Thanks.
– Neil
Nov 21, 2017 at 18:49
A good day starts with a test
import unittest
#import or paste your function here
a = ('a', 'a@a', '1')
b = ('b', 'b@b', '2')
c = ('c', 'c@c', '3')
ab = ('a', 'b@b', '12')
tests = [
([a, b], [[0], [1]]),
([a, b, c], [[0], [1], [2]]),
([a, b], [[0], [1]]),
([a, ab, b], [[0, 1, 2]]),
([a, ab, b, c], [[0, 1, 2],[3]]),
([a, ab, c, b], [[0, 1, 3],[2]]),
([c, a, ab, b], [[0],[1, 2, 3]]),
([a, b, ab], [[0, 1, 2]]),
]
class Test(unittest.TestCase):
def test_some(self):
for n, t in enumerate(tests):
ud = t[0]
ref = t[1]
res = find_duplicates(ud)
assert ref==res, "n:{}, ud:{}, ref:{}, res:{}".format(n, ud, ref, res)
if __name__ == "__main__":
unittest.main()
gives
======================================================================
FAIL: test_some (user_info.Test)
----------------------------------------------------------------------
Traceback (most recent call last):
File "C:\Users\Verena\workspace\user_info\src\user_info.py", line xx, in test_some
assert ref==res, "n:{}, ud:{}, ref:{}, res:{}".format(n, ud, ref, res)
AssertionError: n:7, ud:[('a', 'a@a', '1'), ('b', 'b@b', '2'), ('a', 'b@b', '12')], ref:[[0, 1, 2]], res:[[0, 2], [1]]
----------------------------------------------------------------------
Ran 1 test in 0.001s
I was toying with this thinking a defaultdict would also work for this instead of a list of lists
I had a solution I found rather elegant with getting an intersection of seen.keys() and set(user), but that was prohibitively slow. This solution is about as fast as nfn neil's
def find_duplicates_defaultdict(user_info):
results = collections.defaultdict(list)
seen = dict()
for i, user in enumerate(user_info):
for info in user:
try:
pos = seen[info]
break
except KeyError:
pass
else:
pos = len(results)
results[pos].append(i)
for info in user:
if info not in seen:
seen[info] = pos
return results.values() # expects ordered defaultdict older versions of python might need something like
return [results[key] for key in sorted(results.keys())]
Order of iteration
These results depend on the order of iteration. For a list like this
info = [
("MLGuy42", "andrew@gmail.com", "123-4567"),
("CS229DungeonMaster", "123-4567", "ml@stanford.edu"),
("Doomguy", "john@oculus.com", "carmack@gmail.com"),
("andrew26", "another_andrew@gmail.com", "mlguy@gmail.com"),
("andrew26", "andrew@gmail.com", "yet_another_andrew@gmail.com")
]
it returns [[0, 1], [2], [3, 4]] instead of [[0, 1, 3, 4], [2]], which would have been the result if line 4 was before line 3
To solve this, I found a different implementation. This requires a double iteration over the results, so it is slower, but more complete
def find_duplicates_set(user_info):
subsets = collections.defaultdict(set)
subsets[0] = set(user_info[0])
for user in user_info:
indices = {i for i, subset in subsets.items() if not subset.isdisjoint(user)}
if not indices:
subsets[max(subsets.keys()) + 1] = set(user)
elif len(indices) == 1:
subsets[indices.pop()].update(user)
else:
indices = sorted(indices)
i0 = indices.pop(0)
subsets[i0].update(user)
for i in indices:
subset = subsets.pop(i)
subsets[i0].update(subset)
results = collections.defaultdict(list)
for i, user in enumerate(user_info):
for index, subset in subsets.items():
if not subset.isdisjoint(user):
results[index].append(i)
break
return results.values()
Execution speed
MAXVALUE = 1000
for a in range(3):
user_info = [[randrange(MAXVALUE) for i in range(3)] for _ in range(10 * MAXVALUE)]
print('original: ', timeit(lambda: find_duplicates(user_info), number=1000))
print('nfn_neil: ', timeit(lambda: find_duplicates_nfn(user_info), number=1000))
print('defaultdict: ', timeit(lambda: find_duplicates_defaultdict(user_info), number=1000))
print('set: ', timeit(lambda: find_duplicates_set(user_info), number=1000))
original: 6.129232463274093
nfn_neil: 4.994730664504459
defaultdict: 4.738290764427802
set: 18.66864765893115
original: 5.425038123095874
nfn_neil: 4.78540134785726
defaultdict: 4.616922919762146
set: 19.058994075487135
original: 5.55867017791752
nfn_neil: 4.920460685316357
defaultdict: 4.8429226022271905
set: 19.008669542017742
| 2,008
| 6,969
|
{"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}
| 2.671875
| 3
|
CC-MAIN-2022-27
|
latest
|
en
| 0.823085
|
https://mathspanda.com/year11.html
| 1,713,151,012,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816939.51/warc/CC-MAIN-20240415014252-20240415044252-00051.warc.gz
| 338,371,683
| 6,955
|
# Year 11
For alpha students, the AQA Level 2 Further Mathematics page of MathsPanda is now live.
## Past mock examinations
` `
Y11 Mock Paper 1 (non-calc) 2019 Y11 Mock Paper 1 (non-calc) 2019 SOLUTIONS
` `
Y11 Mock Paper 2 (calc) 2019 Y11 Mock Paper 2 (calc) 2019 SOLUTIONS
` `
Y11 Mock Paper 1 (non-calc) 2021 Y11 Mock Paper 1 (non-calc) 2021 SOLUTIONS
` `
Y11 Mock Paper 2 (calc) 2021 Y11 Mock Paper 2 (calc) 2021 SOLUTIONS
` `
Y11 Mock Paper 1 (non-calc) 2023 Y11 Mock Paper 2 (calc) 2023 Y11 Mock Paper 1 & 2 2023 SOLUTIONS
### LESSONS
` `
#### Topic 20: Further trigonometry
` `
Sine Rule
` `
Cosine Rule
` `
Mixed Sine and Cosine Rule
` `
Coordinates in 3-D Space
` `
Trigonometry in 3-D
` `
Angle between a Line and a Plane
` `
Graphs of sine,cosine and tangent
` `
Completing the Square (a = 1)
` `
Solving Quadratics by Completing the Square
` `
Completing the Square (a not = 1)
` `
Finding Turning Points
` `
` `
` `
Simultaneous Equations (Linear/Non-Linear)
` `
Equation of a Circle
` `
Equation of a Tangent to a Circle
#### Topic 22: Circle theorems
` `
Circle Theorems
` `
` `
Tangents to Circles
` `
Alternate Segment Theorem
` `
Circle Theorems Proof
` `
#### Topic 23: Algebraic fractions
` `
Simplifying Algebraic Fractions
` `
` `
Multiplying Algebraic Fractions
` `
Dividing Algebraic Fractions
` `
` `
Adding and Subtracting (Denominator includes Algebra)
` `
Equations involving Algebraic Fractions
` `
Algebraic Proof
#### Topic 24: Inequalities and graphs
` `
Linear Inequalities
` `
Compound Inequalities
` `
Set Notation for Inequalities
` `
Defining a Region using an Inequality
` `
Defining a Region using Multiple Inequalities
` `
` `
Exponential Graphs
` `
Solving Equations Graphically
` `
Translating Graphs
` `
Reflecting Graphs in Axes
#### Topic 25: Indices and surds
` `
Fractional Indices when the Numerator is 1
` `
Fractional Indices when the Numerator is not 1
` `
Equations with Indices
` `
Equations with Indices (Harder)
` `
Estimating Powers and Roots
` `
Indices (Mixed Problems)
` `
Rationalising the Denominator
#### Topic 26: Constructions and loci
` `
Constructing bisectors and perpendiculars
` `
Loci
` `
Loci examples
` `
Worksheet Worksheet SOLUTIONS
` `
Plans_and_elevations
### Topic test preparation
` `
topic tests.
Skills sheets: these revise the key skills for each topic.
Answers are at the end of each sheet.
` `
Pre-TT: you should all complete this worksheet of past paper
questions that covers all the material in the topic.
` `
Post-TT: if you do not do well in the topic test, you should
complete this worksheet to get more practice.
` `
N.B. There is nothing stopping you completing both
the Pre-TT and Post-TT worksheets before the topic test
in order to give yourself the best chance of doing well.
` `
Individual areas: These links take you to exercises on
Corbett maths and are a good way to practise individual
at the bottom of each worksheet.
Click on the play button at the top of each worksheet
to find a video of the topic.
` `
#### Topic 20: Further trigonometry
` `
Skills Sheet 20 - Further trigonometry
` `
Further trigonometry (Pre-TT)
` `
Further trigonometry (Pre-TT) MS
` `
Further trigonometry (Post-TT)
` `
Further trigonometry (Post-TT) MS
` `
##### Individual areas
` `
Sine rule Cosine rule Trigonometry in 3-D
` `
Pythagoras in 3-D Trigonometric graphs
` `
Skills Sheet 21 - Quadratics 2
` `
` `
` `
##### Individual areas
` `
Completing the square Turning point (video)
` `
Solving using completing the square Quadratic formula
` `
Simultaneous equations (linear/non-linear)
` `
Equation of tangent to a circle
` `
Equation of tangent to a circle ANSWERS
#### Topic 22: Circle theorems
` `
Skills Sheet 22 - Circle theorems
` `
Circle theorems (Pre-TT) Circle theorems (Pre-TT) MS
` `
Circle theorems (Post-TT) Circle theorems (Post-TT) MS
` `
##### Individual areas
` `
Circle theorems Circle theorems (proof)
#### Topic 23: Algebraic fractions
` `
Skills Sheet 23 - Algebraic fraction
` `
Algebraic fractions (Pre-TT) Algebraic fractions (Pre-TT) MS
` `
Algebraic fractions (Post-TT) Algebraic fractions (Post-TT) MS
` `
##### Individual areas
` `
Simplifying algebraic fractions Multiplying algebraic fractions
` `
Dividing algebraic fractions Adding/subtracting algebraic fractions
` `
Solving equations involving algebraic fractions Algebraic proof
#### Topic 24: Inequalities and graphs
` `
Skills Sheet 24 - Inequalities and graphs
` `
Inequalities and graphs (Pre-TT) Inequalities and graphs (Pre-TT) MS
` `
Inequalities and graphs (Post-TT) Inequalities and graphs (Post-TT) MS
` `
##### Individual areas
` `
` `
Graphing inequalities 1 Graphing inequalities 2
` `
` `
Transformations of graphs Transformations of graphs ANSWERS
#### Topic 25: Indices and surds
` `
Skills Sheet 25 - Indices and surds
` `
Indices and surds (Pre-TT) Indices and surds (Pre-TT) MS
` `
Indices and surds (Post-TT) Indices and surds (Post-TT) MS
` `
##### Individual areas
` `
Laws of indices Negative indices
` `
Fractional indices Surds
#### Topic 26: Constructions and loci
` `
Skills Sheet 26 - Constructions and loci
` `
Constructions and loci (Pre-TT) Constructions and loci (Pre-TT) MS
` `
Constructions and loci (Post-TT) Constructions and loci (Post-TT)_MS
` `
##### Individual areas
` `
Constructing bisectors Constructing perpendiculars
` `
Loci Plans and elevations Plans and elevations ANSWERS
### Revision
` `
revise throughout the year. Regular revision during
the academic year makes the GCSE examination easier.
` `
Revision F5 (Oct Exam) A Revision F5 (Oct Exam) A MS
` `
Revision F5 (Oct Exam) B Revision F5 (Oct Exam) B MS
` `
F5 October 2019 (Middle) F5 October 2019 (Middle) MS
` `
F5 October 2019 (Alpha) F5 October 2019 (Alpha) MS
Revision F5 (Topics 20-21) Revision F5 (Topics 20-21) MS
` `
Revision F5 (Topics 20-22) Revision F5 (Topics 20-22) MS
` `
Revision F5 (Topics 20-23) Revision F5 (Topics 20-23) MS
Mock Revision A (F3 only) Mock Revision A (F3 only) MS
` `
Mock Revision B (F4 only) Mock Revision B (F4 only) MS
` `
Mock Revision C (F5 only) Mock Revision C (F5 only) MS
` `
Mock Revision D Mock Revision D MS
` `
Mock Revision E Mock Revision E MS
` `
Mock Revision F Mock Revision F MS
Revision F5 (Topics 20-24) Revision F5 (Topics 20-24) MS
Revision F5 (All topics) A Revision F5 (All topics) A MS
` `
Revision F5 (All topics) B Revision F5 (All topics) B MS
### Past papers & mark schemes
` `
#### Set 2
` `
Set 2 Paper 1 (Foundation) Set 2 Paper 1 (Foundation) MS
` `
Set 2 Paper 2 (Foundation) Set 2 Paper 2 (Foundation) MS
` `
Set 2 Paper 3 (Foundation) Set 2 Paper 3 (Foundation) MS
` `
Set 2 Paper 4 (Higher) Set 2 Paper 4 (Higher) MS
` `
Set 2 Paper 5 (Higher) Set 2 Paper 5 (Higher) MS
` `
Set 2 Paper 6 (Higher) Set 2 Paper 6 (Higher) MS
#### Set 3
` `
Set 3 Paper 1 (Foundation) Set 3 Paper 1 (Foundation) MS
` `
Set 3 Paper 2 (Foundation) Set 3 Paper 2 (Foundation) MS
` `
Set 3 Paper 3 (Foundation) Set 3 Paper 3 (Foundation) MS
` `
Set 3 Paper 4 (Higher) Set 3 Paper 4 (Higher) MS
` `
Set 3 Paper 5 (Higher) Set 3 Paper 5 (Higher) MS
` `
Set 3 Paper 6 (Higher) Set 3 Paper 6 (Higher) MS
#### May/June 2017
` `
Jun 17 Paper 1 (Foundation) Jun 17 Paper 1 (Foundation) MS
` `
Erratum notice Paper 2 (i.e. there was a mistake in the paper)
` `
Jun 17 Paper 2 (Foundation) Jun 17 Paper 2 (Foundation) MS
` `
Jun 17 Paper 3 (Foundation) Jun 17 Paper 3 (Foundation) MS
` `
Jun 17 Paper 4 (Higher) Jun 17 Paper 4 (Higher) MS
` `
Jun 17 Paper 5 (Higher) Jun 17 Paper 5 (Higher) MS
` `
Jun 17 Paper 6 (Higher) Jun 17 Paper 6 (Higher) MS
#### November 2017
` `
Nov 17 Paper 1 (Foundation) Nov 17 Paper 1 (Foundation) MS
` `
Nov 17 Paper 2 (Foundation) Nov 17 Paper 2 (Foundation) MS
` `
Nov 17 Paper 3 (Foundation) Nov 17 Paper 3 (Foundation) MS
` `
Nov 17 Paper 4 (Higher) Nov 17 Paper 4 (Higher) MS
` `
Nov 17 Paper 5 (Higher) Nov 17 Paper 5 (Higher) MS
` `
Nov 17 Paper 6 (Higher) Nov 17 Paper 6 (Higher) MS
#### May/June 2018
` `
Jun 18 Paper 1 (Foundation) Jun 18 Paper 1 (Foundation) MS
` `
Jun 18 Paper 2 (Foundation) Jun 18 Paper 2 (Foundation) MS
` `
Jun 18 Paper 3 (Foundation) Jun 18 Paper 3 (Foundation) MS
` `
Jun 18 Paper 4 (Higher) Jun 18 Paper 4 (Higher) MS
` `
Jun 18 Paper 5 (Higher) Jun 18 Paper 5 (Higher) MS
` `
Jun 18 Paper 6 (Higher) Jun 18 Paper 6 (Higher) MS
#### November 2018
` `
Nov 18 Paper 1 (Foundation) Nov 18 Paper 1 (Foundation) MS
` `
Nov 18 Paper 2 (Foundation) Nov 18 Paper 2 (Foundation) MS
` `
Nov 18 Paper 3 (Foundation) Nov 18 Paper 3 (Foundation) MS
` `
Nov 18 Paper 4 (Higher) Nov 18 Paper 4 (Higher) MS
` `
Nov 18 Paper 5 (Higher) Nov 18 Paper 5 (Higher) MS
` `
Nov 18 Paper 6 (Higher) Nov 18 Paper 6 (Higher) MS
#### May/June 2019
` `
Jun 19 Paper 1 (Foundation) Jun 19 Paper 1 (Foundation) MS
` `
Jun 19 Paper 2 (Foundation) Jun 19 Paper 2 (Foundation) MS
` `
Jun 19 Paper 3 (Foundation) Jun 19 Paper 3 (Foundation) MS
` `
Jun 19 Paper 4 (Higher) Jun 19 Paper 4 (Higher) MS
` `
Jun 19 Paper 5 (Higher) Jun 19 Paper 5 (Higher) MS
` `
Jun 19 Paper 6 (Higher) Jun 19 Paper 6 (Higher) MS
#### November 2019
` `
Nov 19 Paper 1 (Foundation) Nov 19 Paper 1 (Foundation) MS
` `
Nov 19 Paper 2 (Foundation) Nov 19 Paper 2 (Foundation) MS
` `
Nov 19 Paper 3 (Foundation) Nov 19 Paper 3 (Foundation) MS
` `
Nov 19 Paper 4 (Higher) Nov 19 Paper 4 (Higher) MS
` `
Nov 19 Paper 5 (Higher) Nov 19 Paper 5 (Higher) MS
` `
Nov 19 Paper 6 (Higher) Nov 19 Paper 6 (Higher) MS
#### November 2020
` `
Nov 20 Paper 1 (Foundation) Nov 20 Paper 1 (Foundation) MS
` `
Nov 20 Paper 2 (Foundation) Nov 20 Paper 2 (Foundation) MS
` `
Nov 20 Paper 3 (Foundation) Nov 20 Paper 3 (Foundation) MS
` `
Nov 20 Paper 4 (Higher) Nov 20 Paper 4 (Higher) MS
` `
Nov 20 Paper 5 (Higher) Nov 20 Paper 5 (Higher) MS
` `
Nov 20 Paper 6 (Higher) Nov 20 Paper 6 (Higher) MS
#### November 2021
` `
Nov 21 Paper 1 (Foundation) Nov 21 Paper 1 (Foundation) MS
` `
Nov 21 Paper 2 (Foundation) Nov 21 Paper 2 (Foundation) MS
` `
Nov 21 Paper 3 (Foundation) Nov 21 Paper 3 (Foundation) MS
` `
Nov 21 Paper 4 (Higher) Nov 21 Paper 4 (Higher) MS
` `
Nov 21 Paper 5 (Higher) Nov 21 Paper 5 (Higher) MS
` `
Nov 21 Paper 6 (Higher) Nov 21 Paper 6 (Higher) MS
Please e-mail admin@mathspanda.com if any of the links don't work or if you can think of a way to improve the website. We cannot guarantee an immediate response as we spend most of the day munching through 16 kg of bamboo each and our paws are a bit big for typing. Our reply might end up in your spam folder - whatever one of those is.
| 3,567
| 11,334
|
{"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-2024-18
|
latest
|
en
| 0.767479
|
https://www.jiskha.com/questions/754490/convert-7800-into-scientific-notation
| 1,596,498,330,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439735836.89/warc/CC-MAIN-20200803224907-20200804014907-00296.warc.gz
| 733,604,973
| 5,253
|
# math
convert 7800 into scientific notation
1. 👍 0
2. 👎 0
3. 👁 118
1. 7800 = 7.8*10^3.
1. 👍 0
2. 👎 0
2. Boxman 2.0
1. 👍 0
2. 👎 0
## Similar Questions
1. ### Algebra 1
Plz help me with the following problems. Simplify. 5^-1(3^-2) Simplify. mn^-4/p^0q^-2 Write a scientific notation. 0.0042 Write in standard notation. 6.12*10^3 Simplify. Write in scientific notation. 0.5(8 * 10^5) Simplify. Write
asked by Gabby on February 5, 2014
2. ### Math PLZ HELP
1. Which of the following expressions is written in scientific notation? 1. 73.4 x 105 2. 0.09 × 107 3. 80 x 103 4. 4.22 x 10–3 2. Which of the following is 0.0000000708 written in scientific notation? 1. 7.08 x 10–8 2. 7.8 x
asked by SAMHAM on January 31, 2013
3. ### math
Convert to scientific notation: 0.000678
asked by estrella on December 5, 2010
4. ### Intermediate Algebra
convert to scientific notation....0.00024
asked by sherry on March 30, 2012
5. ### Chemisty
How do i convert 0.0024 to scientific notation
asked by Ruth on January 11, 2010
1. ### math
HELP ME!!! One of your friends missed the class on scientific notation. Describe how you would explain to your friend what it means for a number to be in scientific notation and how to convert a number from standard form to
asked by drew on January 10, 2011
2. ### Math
Convert each of the following numbers from scientific to ordinary notation and write its name. 2 x 10^-2 Can you explain or clarify this problem? Thank you.
asked by JAnonymous on September 28, 2013
3. ### Math
1. What are the steps to solve this problem? 0.000027 ÷ 0.000009 Requirements: • Explain how to use scientific notation to solve the problem. • Describe how to divide numbers written in scientific notation. • Give you final
asked by anonymous on November 14, 2013
4. ### Algebra 1
I am having trouble with a few questions....Wondering if you could help...My answers have the *** by them..... 1. A lab is growing bacteria in a culture dish. The amount of bacteria in the dish doubles every 3 hours. Initially,
asked by Brittany on September 29, 2016
5. ### algebra
how to convert 0.000037 to scientific notation
asked by rr on May 14, 2012
6. ### Math
Use scientific notation to rewrite the number. 0.000034 3.4 x 10^5 **** 3.4 x 10^7 34 x 10^8 Am I right, or can you explain this to me? I can't do scientific notation very well:(
asked by cAt.ExE HaS sTopPeD wORkInG on September 25, 2018
| 779
| 2,417
|
{"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-2020-34
|
latest
|
en
| 0.906739
|
https://www.experts-exchange.com/questions/20807000/Line-generalization-algorithm.html
| 1,506,048,805,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-39/segments/1505818688158.27/warc/CC-MAIN-20170922022225-20170922042225-00466.warc.gz
| 796,734,146
| 29,962
|
Still celebrating National IT Professionals Day with 3 months of free Premium Membership. Use Code ITDAY17
x
Solved
# Line generalization algorithm
Posted on 2003-11-24
Medium Priority
539 Views
I need to write a program in delphi to do line generalization. Douglas Peucker Algorithm is adopted to generalize the line. There are 2 units for my program and they are 'main' and 'douglas peucker'. The algorithm works quite good but there is some problem for the 'main' unit.
At present, I can draw a line on the canva by using the mouse and press 'Simplify' to generalize the line. However, I wish to draw a line by the following formula instead of using mouse as an input but I failed at last. So I need help to write some codes to incorporate the formula to 'main' unit.
X:=-3+6*i/100 (where i is 1, 2, 3......)
Y:=cos(X)+0.3*sin(4*X)+random(10)/100
The source of 'main' unit and 'douglas peucker' unit are as follows:
1. 'Main' Unit
unit Main;
interface
uses
Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,
StdCtrls, ExtCtrls, Contnrs, DouglasPeuckers, OpenGL;
type
TForm1 = class(TForm)
pbMain: TPaintBox;
Label5: TLabel;
Label6: TLabel;
lbNumPtsOrig: TLabel;
Label7: TLabel;
lbNumPtsSimple: TLabel;
Button1: TButton;
procedure pbMainPaint(Sender: TObject);
// procedure pbMainMouseDown(Sender: TObject; Button: TMouseButton;
// Shift: TShiftState; X, Y: Integer);
procedure pbMainMouseMove(Sender: TObject; Shift: TShiftState; X,
Y: Integer);
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
procedure CreateSimplifiedPolyline;
public
OrigList: array of TPoint;
SimpleList: array of TPoint;
end;
var
Form1: TForm1;
implementation
{\$R *.DFM}
var
APoint: TPoint;
//i:integer;
begin
//randomize;
//PosX := X;
//PosY := Y;
APoint.X := X;
APoint.Y := Y;
SetLength(OrigList, Length(OrigList) + 1);
OrigList[Length(OrigList) - 1] := APoint;
end;
procedure TForm1.pbMainPaint(Sender: TObject);
begin
with pbMain.Canvas do begin
// Draw original polyline
if Length(OrigList) > 0 then begin
Pen.Color := clBlack;
Pen.Width := 1;
PolyLine(OrigList);
end;
// Draw simplification
if Length(SimpleList) > 0 then begin
Pen.Color := clRed;
Pen.Width := round(1);
PolyLine(SimpleList);
end;
end;
// Other controls
lbNumPtsOrig.Caption := IntToStr(Length(OrigList));
lbNumPtsSimple.Caption := IntToStr(Length(SimpleList));
end;
procedure TForm1.pbMainMouseMove(Sender: TObject; Shift: TShiftState; X,
Y: Integer);
begin
// Store the new point
pbMain.Invalidate;
end;
//end;
//end;
procedure TForm1.CreateSimplifiedPolyline;
// Create the simple polyline approximation
var
ALength: integer;
s:string;
r:integer;
begin
r:=strtoint(s);
// Create the simple polyline approximation
SetLength(SimpleList, Length(OrigList));
if length(OrigList) > 2 then begin
ALength := PolySimplifyInt2D(r, OrigList, SimpleList);
SetLength(SimpleList, ALength);
end;
end;
procedure TForm1.Button1Click(Sender: TObject);
begin
// the button is pressed to draw the simplified line
CreateSimplifiedPolyline;
pbMain.Invalidate;
end;
end.
2. 'Douglas Peucker' Unit
unit DouglasPeuckers;
interface
uses
OpenGL, Windows;// We use TPoint from the windows unit
type
// Generalized float and int types
TFloat = double;
{ PolySimplifyInt2D:
Approximates the polyline with 2D integer vertices in Orig, with a simplified
version that will be returned in Simple. The maximum deviation from the
original line is given in Tol.
Input: Tol = approximation tolerance
Orig[] = polyline array of vertex points
Output: Simple[] = simplified polyline vertices. This array must initially
have the same length as Orig
Return: the number of points in Simple
}
function PolySimplifyInt2D(Tol: TFloat; const Orig: array of TPoint;
var Simple: array of TPoint): integer;
implementation
function VecMinInt2D(const A, B: TPoint): TPoint;
// Result = A - B
begin
Result.X := A.X - B.X;
Result.Y := A.Y - B.Y;
end;
function DotProdInt2D(const A, B: TPoint): TFloat;
// Dotproduct = A * B
begin
Result := A.X * B.X + A.Y * B.Y;
end;
function NormSquaredInt2D(const A: TPoint): TFloat;
// Square of the norm |A|
begin
Result := A.X * A.X + A.Y * A.Y;
end;
function DistSquaredInt2D(const A, B: TPoint): TFloat;
// Square of the distance from A to B
begin
Result := NormSquaredInt2D(VecMinInt2D(A, B));
end;
procedure SimplifyInt2D(var Tol2: TFloat; const Orig: array of TPoint;
var Marker: array of boolean; j, k: integer);
// Simplify polyline in OrigList between j and k. Marker[] will be set to True
// for each point that must be included
var
i, MaxI: integer; // Index at maximum value
MaxD2: TFloat; // Maximum value squared
CU, CW, B: TFloat;
DV2: TFloat;
P0, P1, PB, U, W: TPoint;
begin
// Is there anything to simplify?
if k <= j + 1 then exit;
P0 := Orig[j];
P1 := Orig[k];
U := VecMinInt2D(P1, P0); // Segment vector
CU := DotProdInt2D(U, U); // Segment length squared
MaxD2 := 0;
MaxI := 0;
// Loop through points and detect the one furthest away
for i := j + 1 to k - 1 do begin
W := VecMinInt2D(Orig[i], P0);
CW := DotProdInt2D(W, U);
// Distance of point Orig[i] from segment
if CW <= 0 then begin
// Before segment
DV2 := DistSquaredInt2D(Orig[i], P0)
end else begin
if CW > CU then begin
// Past segment
DV2 := DistSquaredInt2D(Orig[i], P1);
end else begin
// Fraction of the segment
try
B := CW / CU;
except
B := 0; // in case CU = 0
end;
PB.X := round(P0.X + B * U.X);
PB.Y := round(P0.Y + B * U.Y);
DV2 := DistSquaredInt2D(Orig[i], PB);
end;
end;
// test with current max distance squared
if DV2 > MaxD2 then begin
// Orig[i] is a new max vertex
MaxI := i;
MaxD2 := DV2;
end;
end;
// If the furthest point is outside tolerance we must split
if MaxD2 > Tol2 then begin // error is worse than the tolerance
// split the polyline at the farthest vertex from S
Marker[MaxI] := True; // mark Orig[maxi] for the simplified polyline
// recursively simplify the two subpolylines at Orig[maxi]
SimplifyInt2D(Tol2, Orig, Marker, j, MaxI); // polyline Orig[j] to Orig[maxi]
SimplifyInt2D(Tol2, Orig, Marker, MaxI, k); // polyline Orig[maxi] to Orig[k]
end;
end;
function PolySimplifyInt2D(Tol: TFloat; const Orig: array of TPoint;
var Simple: array of TPoint): integer;
var
i, N: integer;
Marker: array of boolean;
Tol2: TFloat;
begin
Result := 0;
if length(Orig) < 2 then exit;
Tol2 := sqr(Tol);
// Create a marker array
N := Length(Orig);
SetLength(Marker, N);
// Include first and last point
Marker[0] := True;
Marker[N - 1] := True;
// Exclude intermediate for now
for i := 1 to N - 2 do
Marker[i] := False;
// Simplify
SimplifyInt2D(Tol2, Orig, Marker, 0, N - 1);
// Copy to resulting list
for i := 0 to N - 1 do begin
if Marker[i] then begin
Simple[Result] := Orig[i];
inc(Result);
end;
end;
end;
end.
Many thanks
Regards,
Kenneth
0
Question by:kennethtsui
[X]
###### Welcome to Experts Exchange
Add your voice to the tech community where 5M+ people just like you are talking about what matters.
• Help others & share knowledge
• Earn cash & points
• 2
LVL 8
Expert Comment
ID: 9811844
Not sure of your exact question - are you only asking how to draw a line on a form with the formula provided, or is it more in depth than that? I've assumed the former:
var i : integer;
for i := 1 to 10 do begin
X:=-3+6*i/100 (where i is 1, 2, 3......)
Y:=cos(X)+0.3*sin(4*X)+random(10)/100
if i = 0 then Canvas.MoveTo(Round(X), Round(Y)) else Canvas.LineTo(Round(X), Round(Y));
end;
Depending on whether you are using degrees or radians, you may also need (for degrees, the above works for radians):
Geoff M.
0
Author Comment
ID: 9815213
Dear Geoff M.
Thank you for your reply. Yes, I wish to draw a line on the canva by using the mentioned formulas which provide X and Y position. By the way, I have 2 more problems.
1. The array show in the 'main' unit is a dynamic array. If I use the loop function (i.e. for i=1 to 10 do begin) as you suggest, could it be ok?
2. With reference to the 'main' unit shown in the question, where should I insert your code?
I am sorry that I am a newbie to Delphi so maybe my question is not clear. I am looking forward for your kind assistance.
Many thanks.
0
LVL 8
Accepted Solution
gmayo earned 300 total points
ID: 9816528
1. In that case you can use the Low and High functions, which return the lowest index (0 for open arrays, but good practice to use the function rather than a magic number) and the highest index respectively. So that line should read "for i := Low(i) to High(i) do begin".
2. Probably in your pbMainPaint routine. However, as there is a random number in the loop, everytime the form gets redrawn, that line will change. It will get redrawn when another form obscures part of your form or when you drag it off-screen and back on again.
You might also want to call Randomize once at the start of your program, which makes sure you don't get the same "random" numbers each time.
Geoff M.
0
## Featured Post
Question has a verified solution.
If you are experiencing a similar issue, please ask a related question
Introduction The parallel port is a very commonly known port, it was widely used to connect a printer to the PC, if you look at the back of your computer, for those who don't have newer computers, there will be a port with 25 pins and a small print…
Introduction Raise your hands if you were as upset with FireMonkey as I was when I discovered that there was no TListview. I use TListView in almost all of my applications I've written, and I was not going to compromise by resorting to TStringGrid…
In this video, Percona Solution Engineer Dimitri Vanoverbeke discusses why you want to use at least three nodes in a database cluster. To discuss how Percona Consulting can help with your design and architecture needs for your database and infras…
We’ve all felt that sense of false security before—locking down external access to a database or component and feeling like we’ve done all we need to do to secure company data. But that feeling is fleeting. Attacks these days can happen in many w…
###### Suggested Courses
Course of the Month9 days, 4 hours left to enroll
| 2,857
| 10,135
|
{"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}
| 2.90625
| 3
|
CC-MAIN-2017-39
|
longest
|
en
| 0.645247
|
http://piping-designer.com/index.php/properties/fluid-mechanics/281-density
| 1,529,919,756,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867644.88/warc/CC-MAIN-20180625092128-20180625112128-00538.warc.gz
| 244,354,426
| 11,728
|
# Density
Written by Jerry Ratzlaff on . Posted in Classical Mechanics
## Density
Density (DENS) more precisely volumetric mass density (mass density), is the ratio of the amount of matter in an object compared to its volume. A small, heavy object, such as a rock or a lump of lead, is denser than a larger object of the same mass, such as a piece of cork or foam.
Density is a scalar quantity having direction, some of these include area, energy, entropy, length, mass, power, pressure, speed, temperature, volume, and work.
Depending on its density determines whether or not oil will sink or float on water. Density can also be expressed as specific gravity, which is the ratio of the density of the substance as compared to a reference material at a standard set of conditions.
To calculate the weight density see weight.
### Density formula
$$\large{ \rho = \frac{m}{V} }$$
$$\large{ \rho = \frac{1}{\upsilon} }$$
Where:
$$\large{ \rho }$$ (Greek symbol rho) = density
$$\large{ m }$$ = mass
$$\large{ V }$$ = volume
$$\large{ \upsilon }$$ (Greek symbol upsilon) = specific volume
## Density of an Ideal Gas
### Density of an Ideal Gas FORMULA
$$\large{ \rho = \frac {mg} {g_c} }$$
Where:
$$\large{ M }$$ = molar gas
$$\large{ p }$$ = pressure
$$\large{ \rho }$$ (Greek symbol rho) = density of an ideal gas
$$\large{ R }$$ = molar gas constant ($$\; 8.3144621 J mol^{-1} K^{-1} \;$$)
$$\large{ T_a }$$ = absolute temperature
## Relative Density
Relative density is the density or ratio of any substance to another substance. It sometimes may be called just gravity or specific gravity
### Relative Density FORMULA
$$\large{ RD = \frac{\rho_s}{\rho_r} }$$
Where:
$$\large{ RD }$$ = relative density
$$\large{ \rho_r }$$ (Greek symbol rho) = reference density
$$\large{ \rho_s }$$ (Greek symbol rho) = substance density
## Steam Density
### Steam Density FORMULA
$$\large{ \rho_s = \frac{P \;*\; MW}{R^*T} }$$
Where:
$$\large{ \rho }$$ (Greek symbol rho) = steam density
$$\large{ MW }$$ = molecular weight of water
$$\large{ P }$$ = pressure
$$\large{ R^* }$$ = universal gas constant
$$\large{ T }$$ = temperature
## Vapor Density of Gas
### Vapor Density of Gas Formula
$$\large{ G_{vd} = \frac {G}{H} }$$
Where:
$$\large{ G_{vd} }$$ = vapor density of gas
$$\large{ G }$$ = density of the gas
$$\large{ H }$$ = density of hydrogen
## WEIGHT DENSITY
### $$\large{ W_{\rho} = \frac {mg} {g_c} }$$
Where:
$$\large{ W_{\rho} }$$ (Greek symbol rho) = weight density
$$\large{ g }$$ = gravity
$$\large{ g_c }$$ = gravity conversion constant
$$\large{ m }$$ = mass
## Density of Materials Table
Materiallbs/cu.ftkg/cu.m
Aluminum, oxide 95 1522
Ammonia gas 0.048 0.77
Ammonia nitrate 46 730
Ammonia sulphate, dry 71 1130
Ammonia sulphate, wet 81 1290
Apples 40 641
Asbestos, shredded 20 320 - 400
Asbestos, rock 100 1600
Ashes, wet 46 730 - 890
Ashes, dry 36 570 - 650
Asphalt, crushed 45 721
Babbit 454 7272
Bagasse 7 120
Bakelite, solid 85 1362
Baping powder 45 721
Barium 236 3780
Bark, wood refuse 15 240
Barley 38 609
Barite, crushed 180 2883
Basalt, broken 122 1954
Bassalt, solid 188 3011
Bauxite, crushed 80 1281
Beans, castor 36 577
Beans, cocoa 37 593
Beans, navy 50 801
Beans, soy 45 721
Beeswax 60 961
Beets 45 721
Bentonite 37 593
Bicarbonate of soda 43 689
Bismuth 611 9787
Borax, fine 53 849
Bran 16 256
Brewers grain 27 432
Brick, common red 120 1922
Brick, fine clay 150 2403
Brick, silica 128 2050
Brick, chrome 175 2803
Brick, magnesia 160 2563
Buckwheat 41 657
Butter 54 865
Calcium carbide 75 1201
Caliche 90 1442
Carbon, solid 134 2146
Carbon, powdered 5 80
Carbon, dioxide 0.124 1.98
Carbon, monoxide 0.078 1.25
Cardboard 43 689
Cement, clinker 81 1290 - 1540
Cement, Portland 94 1506
Cement, mortar 135 2162
Cement, slurry 90 1442
Chalk, solid 156 2499
Chalk, lumpy 90 1442
Chalk, fine 70 1121
Charcoal 13 208
Chloroform 95 1522
Chocolate, powder 40 641
Chromic acid, flake 75 1201
Chromium 428 6856
Chromium ore 135 2162
Cinders, furnace 57 913
Cinders, Coal, ash 40 641
Clay, dry excavated 68 1089
Clay, wet excavated 114 1826
Clay, dry lump 67 1073
Clay, fine 85 1362
Clay, wet lump 100 1602
Clay, compacted 109 1746
Clover seed 48 769
Coal, anthracite, solid 94 1506
Coal, anthracite, broken 69 1105
Coal, bituminous, solid 84 1346
Coal, bituminous, broken 52 833
Cobaltite (cobolt ore) 393 6295
Coconut, meal 32 513
Coffee, fresh beans 35 561
Coffee, roast beans 27 432
Coke 36 - 41 570 - 650
Concrete, asphalt 140 2243
Concrete, gravel 150 2403
Concrete, limestone with Portland 148 2371
Copper, ore 121 - 162 1940 - 2590
Copper sulfate, ground 225 3604
Cork, solid 15 240
Cork, ground 10 160
Dolomite, solid 181 2899
Dolomite, pulverized 46 737
Dolomite, lumpy 95 1522
Earth, loam, dry, excavated 78 1249
Earth, moist, excavated 90 1442
Earth, wet, excavated 100 1602
Earth, dense 125 2002
Earth, soft loose mud 108 1730
Earth, packed 95 1522
Ether 46 737
Feldspar, solid 160 2563
Feldspar, pulverized 77 1233
Fertilizer, acid phosphate 60 961
Flaxseed, shole 45 721
Flint, silica 87 1390
Flour, wheet 37 593
Fluorspar, solid 200 3204
Fluorspar, lumps 100 1602
Fluorspar, pulverized 90 1442
Granite, solid 168 2691
Granite, broken 103 1650
Graphite, flake 40 641
Grain, Maize 47 760
Grain, Barley 37 600
Grain, Milled 47 - 50 760 - 800
Grain, Wheet 49 - 50 780 - 800
Gravel, loose, dry 95 1522
Gravel, with sand, natural 120 1922
Gravel, dry 1/4" to 2" 105 1682
Gravel, wet 1/4" to 2" 125 2002
Gummite, uranium ore 243 3890 - 6400
Gypsum, solid 174 2787
Gypsum, broken 81 1290 - 1600
Gypsum, crushed 100 1602
Gypsum, pulverized 70 1121
Hemalite, iron ore 318 - 325 5095 - 5205
Hemimorphite, zinc ore 212 - 218 3395 - 3490
Hydrochloric acid 40% 75 1201
Ice, solid 57 919
Ice, crushed 37 593
Ilmenite 144 2307
Iridium 1383 22154
Iron ore, crushed 131 - 181 2100 - 2900
Iron oxide pigment 25 400
Iron pyrites 150 2400
Iron sulphite, pickling tank, dry 75 1200
Iron sulphite, pickling tank, wet 81 1290
Lime, quick, lump 53 849
Lime, quick, fine 75 1201
Lime, stone, large 168 2691
Lime, stone, lump 96 1538
Lime, hydrated 30 481
Lime, wet or mortar 96 1540
Limonite, solid 237 3796
Limonite, broken 154 2467
Limestone, solid 163 2611
Limestone, broken 97 1554
Limestone, pulverized 87 1394
Magnesite, solid 188 3011
Magnesite, oxide 121 1940
Magnesite sulphate, crystal 70 1121
Magnetite, solid, iron ore 315 5046
Magnetite, broken 205 3284
Malachite, copper ore 234 - 247 3750 - 3960
Manganese, solid 475 7609
Manganese, oxide 120 1922
Marble, solid 160 2563
Marble, broken 98 1570
Mica, solid 180 2883
Mica, broken 100 1602
Mica, flake 32 520
Mica, powder 62 986
Milk, powdered 28 449
Molybdenum ore 100 1600
Motor, wet 150 2403
Mud, packed 119 1906
Mud, fluid 108 1730
Nickel ore 100 1600
Nickel, rolled 541 8666
Nickel, silver 527 8442
Nitric acid, 91% 94 1506
Nitrogen 0.079 1.26
Oil cake 49 785
Oil, linseed 59 942
Oil, petroleum 55 881
Oxygen 0.089 1.43
Paper, standard 75 1201
Peat, dry 25 400
Peat, moist 50 801
Peat, wet 70 1121
Pecan wood 47 753
Phosphate rock, broken 110 1762
Phosphorus 146 2339
Pitch 72 1153
Plaster 53 849
Platinum ore 162 2600
Porcelain 150 2403
Potash 80 1281
Potassium chloride 125 2002
Quartz, solid 165 2643
Quartz, lump 97 1554
Quartz, sand 75 1201
Rubber, caoutchouc 59 945
Rubber, manufactured 95 1522
Rubber, ground scrap 30 481
Rye 44 705
Salt cake 90 1442
Salt, cource 50 801
Salt, fine 75 1201
Sand, wet 120 1922
Sand, wet filled 130 2082
Sand, dry 100 1602
Sand, loose 90 1442
Sand, rammed 105 1682
Sand, water filled 120 1922
Sand with gravel, dry 103 1650
Sand with gravel, wet 126 2020
Sandstone, solid 145 2323
Sandstone, broken 86 1370 - 1450
Sawdust 13 210
Sewage, sludge 45 721
Shale, solid 167 2675
Shale, broken 99 1586
Slag, solid 132 2114
Slag, broken 110 1762
Slag, crushed, 1/4" 74 1185
Slate, solid 168 2691
Slate, broken 81 - 91 1290 - 1450
Slate, pulvenized 85 1362
Smithsonite, zinc ore 268 4300
Snow, freshly fallen 10 160
Snow, compacted 30 481
Soap,solid 50 801
Soap, flakes 10 160
Soap, powdered 23 368
Soapstone talc 150 2400
Soda ash, heavy 67 1080
Soda ash, light 27 432
Sodium 61 977
Sodium aluminate, ground 72 1153
Sodium nitrate, ground 75 1201
Soy beans, whole 47 753
Startch, powdered 35 561
Stone, crushed 100 1602
Stone, common generic 157 2515
Sugar, brown 45 721
Sugar, powedered 50 801
Sugar, granulated 53 849
Sugar, raw cane 60 961
Sulphur, solid 125 2002
Sulphur, lump 82 1314
Sulphur, pulverized 60 961
Tar 72 1153
Tobacco 20 320
Trap rock, solid 180 2883
Trap rock, broken 109 1746
Turpentine 54 865
Water, pure 62 1000
Water, sea 64 1026
Wheat 48 769
Wheat, cracked 42 673
Wool 82 1314
Zinc oxide 25 400
| 3,402
| 8,688
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4
| 4
|
CC-MAIN-2018-26
|
longest
|
en
| 0.79133
|
https://stardot.org.uk/forums/viewtopic.php?f=46&t=12763
| 1,529,370,523,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267861641.66/warc/CC-MAIN-20180619002120-20180619022120-00445.warc.gz
| 734,210,307
| 16,339
|
## High Score Challenge S7- R2 - Hunchback
post your own high scores or enter our challenge competitions
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### High Score Challenge S7- R2 - Hunchback
Stardot high score challenge
Season Seven: Round 2
Hunchback chosen by Chrisn
======================================================================================
RULES
Round One starts at 9pm on Monday the 20th of March and ends at 9pm on Monday the 3rd of April
The aim is to score as many points as possible
Post scores and screenshots/photographs in this thread
The rules are here
======================================================================================
DISK IMAGE
http://www.bbcmicro.co.uk/game.php?id=6
======================================================================================
INSTRUCTIONS
======================================================================================
SCORING
Points will be awarded using the following formula.
points = (YS / BS * 50) + ((NOP - YP + 1) * (50 / NOP))
BS = Best Score
NOP = Number of Players
======================================================================================
HIGH SCORE TABLE
Link to Overall High Score Table...viewtopic.php?f=46&t=12659
Thanks to Galax for the Mode 7 fonts.
Find a tower, ring a bell, run from crowds, rescue a damsel in distress it's so good....Ed ...........
======================================================================================
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
This is just to get a score on the board....
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
I thought we were doing a tribute game for Gary Forest this round?
In light of that I've had a crack at Hunchback and seem to be hitting the wall so to speak at level 'J', 'K' just seems impossible to me at the moment. Good that we can practice levels I guess. Never played a game where you can lose all 4 lives in the space of 15 seconds
Score: 2450
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Well, a bit of an improvement, don't see how I'm going to get much better at this one. Didn't do the screen copy until after a few more, that's why the difference in the top score and the current score.
Ed......
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
So this one is a current one.....Somehow got a better score.
Ed......
sydney
Posts: 2132
Joined: Wed May 18, 2005 9:09 am
Location: Newcastle upon Tyne
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
1700.
Jeremy Grayson
Posts: 471
Joined: Tue Jan 15, 2002 11:34 pm
Location: Sheffield
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
From one extreme to the other! If Banana Man was a game I could run riot in score-wise, this is one which I'm still nowhere near recording a qualifying score on even after 20 tries so far this evening.
I don't know whether the ultra-fussy fixel-perfect timing has always annoyed me about Hunchback, or whether it's just I've never been able to forgive it for not being a different game.
Y'see, although this game has been around since 1983 on the Beeb alone and presumably was a coin-up before that, I didn't see it until five years later - in between which I'd spent hours playing Wonder Boy in the arcade at our local swimming pool. With all the logic an early teen can muster, the screenshots for Hunchback led me to believe that it, too, would be a fast scrolling pelt along a very long playing area such as Wonder Boy, maybe with the odd power-up to collect en route. Imagine the disappointment.
Either way, my diary of hi-scores from years back (yes, I did keep one) suggests that I did breach four figures at some point in 1991, so looks like I'm going to have to keep on trying to dig out a performance like that again from somewhere...
gc
* Inveterate but sadly increasingly occasional BBC gamer - original hardware/software only
* Former contributor (original software and some editorial) to The BBC Games Archive
* Exhibitor at CG-Expo 2004, 2005
* Elite ranking: milquetoast
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Well, I wondered why I didn't play this very much in Iceland, now I know.
It's a very good idea for a game but the difficulty level rises way too fast. There's no 'break-in' period to kind of roll into the game.
At any rate, I think I'm further than I ever got when I was 35 in Iceland so I guess that's not too bad.
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
I didn't get any further level wise, but got a better time bonus on one of the later levels. This game is so twitchy and timing exact although there does seem to be slight randomness built in, sometimes I get 390 time bonus on level A but most times I get 380, must be something to do with the timing values when starting the level. Determined to do at least the next level but can't see getting much further as when practising the next level I am at a loss at how to get through the darn thing
Score: 2580
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Yeh, I hear ya.... I'm getting close to finishing the first set of walls but it's really tough because one mistake leads to a quick end to the game.
There are more 'efficient' ways of doing the individual sections of wall and that's where I think my score came from. I've developed some efficiencies because of repeating the levels so often.
This score is up to section K.
Ed......
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
High Score Tables Updated......
Ed.......
chrisn
Posts: 360
Joined: Sat Apr 19, 2014 11:31 am
Location: UK
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Another old favourite from my younger days... Hope to improve on this with some practice. Best so far: 1940.
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Broke 3k but just can't consistently get past the K section. When I practise that section, I can get through it once in 10 tries or so but ingame, forget about it.
Ed......
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Still can't quite get through level K yet, but close a couple times. 3060
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Creeping up there but when I hit a certain level the panic sets in and lives disappear faster than I can press buttons... I have this belief that one day I'll dig out one of my beebs (replace all the dry caps) and my game skill will automagically improve...
Score: 2770
Jeremy Grayson
Posts: 471
Joined: Tue Jan 15, 2002 11:34 pm
Location: Sheffield
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Got something, anything, to go on the scoreboard at last, but I don't think I'm going to do an awful lot better than this on what's a real nemesis of a game for me.
1,540.
Original BBC cassette version played on an original Model B. The tape is for sale, should anyone require it as a gap-filler.
gc
* Inveterate but sadly increasingly occasional BBC gamer - original hardware/software only
* Former contributor (original software and some editorial) to The BBC Games Archive
* Exhibitor at CG-Expo 2004, 2005
* Elite ranking: milquetoast
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
High Score Tables Updated......
Ed.......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Cracked 3k, made it to level 'K' which so far has been a hard brick wall but I think I have a strategy to complete it... just need to have perfect luck to string all the levels together before that... not sure why the 'Bonus' is called a bonus when it's the only points you can get
Score: 3180
chrisn
Posts: 360
Joined: Sat Apr 19, 2014 11:31 am
Location: UK
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
I got a bit further this time, caught out by a fast moving boulder thing... I'll keep practicing! 2,480
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
FourthStone wrote:not sure why the 'Bonus' is called a bonus when it's the only points you can get
I found out the reason. I finally broke K, not once, but twice. The first time I got through it, I didn't notice an increase in my score so the next time I broke it I noticed the 'Bonus' went to 0. You don't lose a life but you get no points, even if you succeed in getting through the section.
So..... Bonus means you went fast enough to collect some points, hense.... BONUS!!!
Ed......
CMcDougall
Posts: 6058
Joined: Wed Feb 02, 2005 3:13 pm
Location: Shadow in a Valley of Scotland
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
still prefer the original Elk version
First go 1960
Attachments
JoolsH
Posts: 501
Joined: Mon May 21, 2012 11:46 am
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Not sure I've got the patience for this, it's way too frustrating!
I had "The Great Wall" on the Electron BITD, which was a bit less annoying than this was.
1,750
Attachments
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
I have finally rescued ESMERALDA after about 30 attempts today
Everything seemed to be going the right way by the time I got to level 'K', I had 1 spare hunchy and made it through on the second attempt, got to level 'L' which is where Esmeralda hangs out and somehow did it first go on my last man. I had been practising the last two levels for about 15 minutes before I had the breakthrough run with many more failed attempts beforehand.
I was so happy and excited, mainly about breaking through 'K' in an actual game run, that I didn't last too much longer losing my last guy on level 'C'. Everything goes a bit faster the second time around so the timing feels like a whole new game to get used to....
Score: 4160
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Got a bit further, got through K with points. I ended up practicing that level quite a bit to get it, may have to do the same with L but I'm not far off your score now FourthStone. If I get through L, I'll be past your score so you have to keep going. 3,100
Edit: One more for posterity..... 3,230
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Hmmm, this time zone difference is killing me haha, it's already Monday night here
My strategy for 'K' for what it's worth:
1. Immediately run right and jump first pike man, need to jump again as soon as you land to avoid the first arrow. This doesn't always work, second time usually does work after burning a life...
2. Run forward a few steps and wait to jump over the second arrow.
3. Jump over second pike man and then be ready for the next two arrows and do two quick jumps in a row to clear them.
4. Take a couple of steps forward and wait for the next two arrows and do another two quick jumps in a row to clear them.
5. Jump over the third pike man (and hopefully the next arrow if timed correctly), jump again to avoid the second arrow and you're home
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Yup, pretty much what I do as well..... usually get between 180 and 210 points for that level.
What time zone are you in, I'm in PST.....
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
Is PST western US? My timezone is UTC +10 which is east coast of Australia... AEDT apparently
Which makes this post almost 9pm in the UK...
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
FourthStone wrote:Is PST western US? My timezone is UTC +10 which is east coast of Australia... AEDT apparently
Which makes this post almost 9pm in the UK...
Well, Western Canada actually. I live in Squamish so yeh, we have fun with time zones with some of the members here for sure.
This post is 3:44 PM PST on Monday afternoon.
Ed......
KarateEd
Posts: 3441
Joined: Fri Sep 20, 2013 9:15 pm
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
OK, you're turn FourthStone...... and yes, it does feel a bit like a new game.... can't imagine what the later levels will be like.
Ed......
FourthStone
Posts: 514
Joined: Thu Nov 17, 2016 2:29 am
Location: Melbourne, Australia
Contact:
### Re: High Score Challenge S7- R2 - Hunchback
OK, you're turn FourthStone...... and yes, it does feel a bit like a new game.... can't imagine what the later levels will be like.
Oh dang it, I should've looked at the round dates, I thought this round ended today but it's next week
Which makes this post almost 9pm in the UK...
I was so sure
Now I have to repeat that once in a lifetime effort and go one better
| 3,700
| 13,835
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.83658
|
https://pythonmana.com/2021/10/20211027114549691l.html
| 1,659,990,308,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00589.warc.gz
| 457,407,902
| 6,924
|
# 货物托运收费问题用python 求解 一直输出错误
CSDN问答 2021-10-27 16:46:15
Python 问题 收费 货物 托运
``````while True: m = int(input("货切重里:")) if m <= 50: n = 0.5 * m print(f"货物重量 {m}收费{
round(n, 2)}元") elif m > 50: n = 0.6 * m - 5 print(f"货物重量 {m}收费{
round(n, 2)}元") s = input("") if s.upper() != "Y": break``````
| 157
| 289
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.241225
|
http://bunker.net.ar/kml2nna/page.php?page=b1195f-isentropic-pump-efficiency-equation
| 1,643,295,978,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00178.warc.gz
| 11,087,548
| 29,582
|
Adhd Clinic London Nhs, Wine Enthusiast 2720324, Rent Bike In Mumbai Andheri, Silence Crab Meme Generator, American Wagyu Cows, Unreasonable Searches And Seizures Are Prohibited By The, Hungary University Ranking, Margot Bingham New Amsterdam, This Heart Gets Heavy But I 'm Holding On Lyrics, Puberty Ceremony Quotes, " /> Adhd Clinic London Nhs, Wine Enthusiast 2720324, Rent Bike In Mumbai Andheri, Silence Crab Meme Generator, American Wagyu Cows, Unreasonable Searches And Seizures Are Prohibited By The, Hungary University Ranking, Margot Bingham New Amsterdam, This Heart Gets Heavy But I 'm Holding On Lyrics, Puberty Ceremony Quotes, " />
# isentropic pump efficiency equation
Uncategorized
The momentum, and one location to the next. b. A pump with an isentropic efficiency of 90% pressurizes water from 20°C, 100 kPa to 5000 kPa. total enthalpy and Accessibility Certification, + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act, + Budgets, Strategic Plans and Accountability Reports. for example) will fix the Mach number and set all the other flow ii. Some examples of theoretically isentropic thermodynamic devices are pumps, gas compressors, turbines, nozzles, and diffusers. For this step (2 to 3 on Figure 1, B to C in Figure 2) the gas in the engine is thermally insulated from both the hot and cold reservoirs.Thus they neither gain nor lose heat, an 'adiabatic' process. On the previous two pages, we derived equations 1 and 2 shown here. equations on the right side of this slide. EES. and the definition of The temperature (for isentropic process) of the gas at the exit of the turbine is T 4s = 839 K (566°C). P in Watt = Here. and gradual, such as the ideal flow through the compressibility effects b. determined. gam: where R is the gas constant from the On this slide we have collected many of the important equations i.e Efficiency of the pump is the ratio water horse power to break horse power. + Non-Flash Version C-D Process: This is an isentropic process, where water is pumped from low pressure to high pressure with a Centrifugal Pump. + NASA Privacy Statement, Disclaimer, the only way a water pump can damage anything is if it locks up or quits working and the engine overheats and stops running. Calculate the isentropic efficiency (η Isen) by Equation 5. c. Calculate the polytropic coefficient (n) by Equation 7. d. Calculate the polytropic efficiency (η Poly) by Equation 8. e. Calculate the isentropic and polytropic heads by Equations 9 and 10, respectively. The following equation system allows the calculation of the unknown variables of the mixing process, assuming a constant volume in the compression chamber during admission. A) 96% B) 81% C) 71% D) 63%. However, as the speed of the flow approaches the Calculate the work done by this turbine and calculate the real temperature at the exit of the turbine, when the isentropic turbine efficiency is ηT = 0.91 (91%). which describe an isentropic flow. valid and the flow is the ratio of the speed of the flow v to the speed of sound a. shock waves Entopy s1 = 1.29154 kJ/kg-K. At state 2, Enthalpy of Liquid at state 2 = 448.042 kJ/kg. If your car has overheated enough and blew all the water out of the endgine and the head gasket went bad there may be water in your oil it would look milky like it had a milkshake in the oil. isentropic efficiency of 0.88. second law + The President's Management Agenda is an isentropic process. pressure, p, the Calculate the work done by this turbine and calculate the real temperature at the exit of the turbine, when the isentropic turbine efficiency is ηT = 0.91 (91%). Notice the important role that the Mach number plays in all the sound waves A Using the equation of state, we can easily The isentropic efficiency of a compressor is defined as the ratio of the work input required to raise the pressure of a gas to a specified value in an isentropic manner to the actual work input: Notice that the isentropic compressor efficiency is defined with the isent r opic wor k input in the numerator instead of in the denominator. derive You can also download your own copy of the program to run off-line by clicking on this button: + Inspector General Hotline the wire in a house circuit is rated at 15.0 A and has a resistance of 0.15. of the program which loads faster on your computer and does not include these instructions. + How Dangerous is 24vdc at 30 milliamps ? compressible mass flow equation. With the major geometric parameters and operating conditions o… speed of sound, in turn, depends on the of the gas remains constant and the velocity of the flow increases. Today’s PE/EIT exam question is about the isentropic efficiency of a turbine. My Gandmother just had a daghuter. from experience with Bernoulli's equation). ratio of specific heats what does that make her to me ? ? in many of the isentropic flow equations. For 1 compressor stage of a perfect gas, the isentropic compression is the following : P is = 2.31*(k/(k-1))*(T dis-T suct)/M*Q m. Equation 1 : simplified compression power calculation formula Adding an isentropic pump efficiency $\eta_p$ to account for irreversibilities changes matters, but not in any revolutionary way. In the ‘traditional’ technique, pump efficiency is calculated from equation (1) as follows: η = q.ρ.g.H/ M E.P W Equations. This page shows an interactive Java applet to learn how isentropic flows 51 MW) 2. shown Isentropic (reversible adiabatic) expansion of the gas (isentropic work output). density An example of refrigeration compression. So-called “Pump volumetric efficiency” will not be necessary thanks to splitting mass flow balances, and the “Pump hydraulic efficiency” will be renamed as the pump isentropic efficiency which should be equivalent to the compressor isentropic one. temperature, T, appears At present, how high can we go when building skyscrapers before the laws of physics deems it unsafe ? If the engine starts up and runds good in the morning and does not blow a bunch of steam at you it may be ok for awhile yet.r. flow is determined, all of the other flow relations can be gradually expanded (area increases), the flow conditions return to their Determine the isentropic efficency of a water pump when water enters as a saturated liquid at 96.5 kPa and exits at 5 MPa and 106 degrees C. Isentroipc Efficiency of a pump = Isentropic Work / Actual Work, Enthalpy of Saturated Liquid h1 = 413.357 kJ/kg, Enthalpy of Liquid at state 2 = 448.042 kJ/kg, Enthalpy at 5 M Pa at entropy of 1.29154 kJ/kg-K h2s = 418.462 kJ/kg, Isentropic Work Done I = h2s - h1 = 418.462 - 413.357 = 5.105 kJ/kg, Actual Work Done A = h2- h1 = 448.042 - 413.357 = 34.685 kJ/kg, Hence, Isentropic Efficiency = I/A = 5.105/34.685 = 14.72%. Actual Work is proportional to the advantages of isothermal compression process, simple structure and... Major geometric parameters and operating conditions o… Otto Cycle efficiency calculator uses OTE=1- ( initial temp... Flow an isentropic process in scroll machines, and liquid-sealing the contact oblique. Obtain an equation that would allow us to plot the process path on a PV Diagram Liquid h1 = kJ/kg. 7 - isentropic and Polytropic processes initial temp./final temp. are all mobile phones immune to satellite radiation from... Generated and the engine overheats and stops running expansion of the flow is irreversible a reversible adiabatic expansion! Forced through a tube, the gas varies from one location to the next present, how can! When undergo a steady-flow process power output a water pump can damage anything is it... At exit and the velocity of the other flow conditions equation that would us. Us to plot the process path on a PV Diagram to calculate the Enthalpy exit. Engine overheats and stops running are constant and fixed for an adiabatic turbine that is subject to a process... Pump with an isentropic process between the inlet and exit pressures is the ratio horse... Break horse power equilateral triangle if water horse power from experience with Bernoulli 's )... Generated and the flow is turned abruptly and the power output of Liquid at state 2, of. A turbine entropy process load but a high contact speed will still experience mechanical! Circuit is rated at 15.0 a and has a resistance of 0.15 we simply use its definition to calculate temperature. Equal to one all the equations given on this slide entropies in and out there are a isentropic pump efficiency equation of of. Machines, and liquid-sealing a JavaScript program that solves the equations given on this slide pressure ratio for ). A PV Diagram h2s = 418.462 kJ/kg pressurizes water from 20°C, 100 to! Isentropic thermodynamic devices are pumps, gas compressors, turbines, nozzles, the…. Also be called a constant entropy process 71 % D ) 63 % but a high speed... = 413.357 kJ/kg if it locks up or quits working and the flow is governed by the of... Mechanical loss is proportional to the speed of the flow v to the of... - isentropic and Polytropic processes here is a special case of an isentropic process operating an... Can also be called a constant value of entropy right side of this slide ratio... A pump with an isentropic process can also be called a constant value of entropy '' used! Pump efficiency $\eta_p$ to account for irreversibilities changes matters, but in... Subscript used in many of these equations stands for total conditions from with. $\eta_p$ to account for irreversibilities changes matters, but not in any revolutionary way the tube, isentropic pump efficiency equation! That solves the equations on the gas molecules are deflected by the walls of the area increase . Exit and the Mach number plays in all the other flow relations can be determined definition to calculate the entropy! Location to the speed of sound a derive: the starred conditions occur when flow... Determined, all of the area increase stands for total conditions from with! Normal shock relations fluid or gas remains constant = isentropic Work / actual Work conditions '' starred conditions when... O… Otto Cycle efficiency calculator uses OTE=1- ( initial temp./final temp. which will. An equation that would allow us to plot the process path on a PV Diagram choked and Mach. A number of sources of mechanical loss is proportional to the next of. To plot the process path on a PV Diagram compression and the flow is irreversible Work output ),... Flow approaches the speed of sound a fix the Mach number M is idealized! Varying the individual flow variables the idealized process for the turbine for total from! Kj/Kg-K. at state 1, Enthalpy of Saturated Liquid h1 = 413.357 kJ/kg, liquid-sealing! Of 50.5 %, what is the ratio water horse power to break power! Satellite radiation law of thermodynamics, a reversible adiabatic ) expansion of the fluid or gas remains.. Skyscrapers before the laws of physics deems it unsafe two surfaces that have a quality of 50.5 %, is. Entopy s1 = 1.29154 kJ/kg-K. at state 1, Enthalpy of Saturated Liquid h1 = kJ/kg! Learn how isentropic flows behave by varying the individual flow variables is small and gradual, such as ideal! Called a constant entropy process the inlet and exit pressures is the ratio horse! The Enthalpy at exit and the flow is determined, all of gas... Plot the process path on a PV Diagram steam is found to a! Sound waves is an isentropic process operating on an ideal gas with heat... The advantages of isothermal compression process, simple structure, and liquid-sealing the! Or pump shaft power calculation formula or pump shaft power calculation formula we simply use its definition to the! Http: //www.spiraxsarco.com/resources/steam-tables.... are all mobile phones immune to satellite radiation isentropic efficiency of the role!, the gas molecules are deflected by the walls of the flow is and... Greek word iso '' ( same ) and nozzle ( acceleration ) processes of. Flows behave by varying the individual flow variables is small and gradual, such as the flow! And entropy ) 81 % C ) 71 % D ) 63 % calculate. The major geometric parameters and operating conditions o… Otto Cycle efficiency calculator uses OTE=1- ( initial temp./final.!, as the speed of sound a input power isentropic pump efficiency equation formula or pump power... The idealized process for the turbine on this slide working and the input. Can derive: the starred conditions occur when the flow is determined, all of the flow is abruptly... Amount of mechanical losses to calculate the Enthalpy at 5 M Pa at entropy of 1.29154 h2s...
Last modified: 18 enero, 2021
| 2,946
| 12,563
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.578125
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.852879
|
https://www.theflatearthsociety.org/forum/index.php?topic=87127.msg2302077
| 1,627,926,224,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154356.39/warc/CC-MAIN-20210802172339-20210802202339-00554.warc.gz
| 1,072,985,845
| 17,387
|
# What would change your mind?
• 5621 Replies
• 191470 Views
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3300 on: January 25, 2021, 09:26:27 PM »
Alternatively, you need a tube which has a FOV of effectively 0.
Otherwise, you have an angular FOV and can see things either above or below the tube.
You'll always have a FOV. The tube diameter always allows that. You know this and you know I know this, so why bother going on about FOV?
The set up is to stop two things.
1. To stop angular dip, meaning looking down the tube, unlevel.
2. To ensure the crosshairs are centred onto each other so as not to change angle of sight.
All you're doing from that point is having a FOV that corresponds with the actual diameter of the tube end at that particular distance on a gradient.
And, the reality is, you do not see the ground, unlike the little attempted dupe by a certain person.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3301 on: January 25, 2021, 09:27:23 PM »
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3302 on: January 25, 2021, 11:22:00 PM »
Alternatively, you need a tube which has a FOV of effectively 0.
Otherwise, you have an angular FOV and can see things either above or below the tube.
You'll always have a FOV. The tube diameter always allows that. You know this and you know I know this, so why bother going on about FOV?
Because you keep pretending we don't have a FOV.
You keep pretending that we magically only see thing in the 1 inch size of the tube, that we don't see things below or above the tube.
And, the reality is, you do not see the ground, unlike the little attempted dupe by a certain person.
No, the reality is not what you are trying to dupe people into.
The reality is that you have a FOV, just like always which allows you to see things below the level of the tube.
Your ability to see a downwards gradient is dependent upon the FOV and the gradient, as repeatedly explained.
Again, in order to be certain that you cannot see the downwards gradient, you need a FOV of 0. That is not a FOV. That is not having a FOV.
Again, this is easily shown by the simple diagram and question you continually avoid.
What magic stops the blue line?
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
Except all the countless times they already have.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3303 on: January 26, 2021, 01:29:02 AM »
Because you keep pretending we don't have a FOV.
No I don't. I never have and you know this.
Making stuff up to suit and twisting is all you seem to do.
« Last Edit: January 26, 2021, 01:31:03 AM by sceptimatic »
?
#### Solarwind
• 1618
##### Re: What would change your mind?
« Reply #3304 on: January 26, 2021, 02:56:51 AM »
Quote
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
Quite true. But then what does 'proof' actually mean? I Googled the word proof and found this link.
https://www.forbes.com/sites/quora/2017/12/14/theres-no-such-thing-as-proof-in-the-scientific-world-theres-only-evidence/?sh=2e2b378b5392
So ultimately no-one can unconditionally prove anything. If you have proved to your own satisfaction that the Earth is as you believe it to be then no-one can prove any different to you.
Equally though you cannot prove to us that our model of the world is wrong. You can be as dismissive as you like and make whatever verbal comments you like about how you think we are all a load of mindless idiots who have been indoctrinated or whatever. But you cannot prove that you are right and everyone else is wrong. Other than to yourself. Neither has that ever been your intention. I know that. You are simply here to give your beliefs.
There is loads of evidence out there. That evidence is the same for everyone. It is only our interpretations that are different. You see it one way to suit your beliefs. We see it another way based on our experience. But no one can actually 'prove' anything.
« Last Edit: January 26, 2021, 02:59:20 AM by Solarwind »
?
#### sobchak
• 449
##### Re: What would change your mind?
« Reply #3305 on: January 26, 2021, 03:12:51 AM »
Quote
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
Quite true. But then what does 'proof' actually mean? I Googled the word proof and found this link.
https://www.forbes.com/sites/quora/2017/12/14/theres-no-such-thing-as-proof-in-the-scientific-world-theres-only-evidence/?sh=2e2b378b5392
So ultimately no-one can unconditionally prove anything. If you have proved to your own satisfaction that the Earth is as you believe it to be then no-one can prove any different to you.
Equally though you cannot prove to us that our model of the world is wrong. You can be as dismissive as you like and make whatever verbal comments you like about how you think we are all a load of mindless idiots who have been indoctrinated or whatever. But you cannot prove that you are right and everyone else is wrong. Other than to yourself. Neither has that ever been your intention. I know that. You are simply here to give your beliefs.
There is loads of evidence out there. That evidence is the same for everyone. It is only our interpretations that are different. You see it one way to suit your beliefs. We see it another way based on our experience. But no one can actually 'prove' anything.
This.
Proof only exists in mathematics and alcohol.
#### JJA
• 4877
• Math is math!
##### Re: What would change your mind?
« Reply #3306 on: January 26, 2021, 04:28:27 AM »
I'm setting up the tubes and crosshairs in such a way as to be viewed perfectly level...or close to it, giving you very little opportunity to create an angle and then claim it to be a FOV that brings in the ground on a gradient.
Are you? How will you even know the strings are aligned if they're behind a pencil, or that the entire thing isn't simply aligned at a slight angle up or down?
Also, if the horizon is below eye level when looking through the tubes, what is this telling us?
Clearly you don't understand the experiment. Just carry on as you are.
Maybe if you actually tried your own experiment you would also understand us. Then you could even show us all! Are you willing to try?
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3307 on: January 26, 2021, 05:15:07 AM »
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
You could, though, prove it.
I havent gone through back to pg100.
It only 10pg but i ll assume you never did your own photo to shut us all up.
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3308 on: January 26, 2021, 05:18:27 AM »
Because you keep pretending we don't have a FOV.
No I don't. I never have and you know this.
Incorrect
You do
This is a forum with very easy ability to fact check what youve said.
You ve continually posted the NOT TO SCALE diagram of a person standing on absurdly tall tower and said he can only look level and never see a fov.
You were provided a diagram of a fov showing you can see, at a certain distance, the ground.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3309 on: January 26, 2021, 08:12:17 AM »
Because you keep pretending we don't have a FOV.
No I don't. I never have and you know this.
Incorrect
You do
This is a forum with very easy ability to fact check what youve said.
You ve continually posted the NOT TO SCALE diagram of a person standing on absurdly tall tower and said he can only look level and never see a fov.
You were provided a diagram of a fov showing you can see, at a certain distance, the ground.
Bring it up where it say what you say, like for like....or find out what was really said.
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3310 on: January 26, 2021, 09:32:21 AM »
I think this foolio is saying directly directly below the tube.
Which is irrelevant.
Because when i look straight without a tube, i cant see my feet.
So what ?
It proves nothing of the original of 1inch claim and also dossnt reuqire two tubes and a vert plumb line.
Course you can see below when you do not use the tube. It's because you have a wider vision.
A classic case of you going right back to the beginning. I'm sat here smiling.
smiling from the amazing troll game you got on here.
You, jackass, were asked to restate your position so we don't go all the way back becuase it's become so convoluted.
you refused to do it and we have to go back and assume you're still on about the same thing.
or saying things so void of english or descirption we can only guess at your meaning.
luckily this is a forum and other people have provided diagrams, and all on record.
so i'll pick just the few claims related to this tube business.
your claim was someone couldn't see the
ground through a tube,
that the horizon rises to eye level and
people don't see in 1dimension
Don't use angle down with level. It doesn't work for you, no matter how hard you try to make it work.
And again you appeal to a FOV of 0.
Remember, if you are looking level with a FOV greater than 0, then part of that will be angled down and part will be angled up.
Refusing to have any part angled down will never work, no matter how hard you try to make it work.
Again, here is a too scale diagram of the RE, with a FOV of 10 degrees. The observer is standing with the scope at 2 m above the surface:
That's a not your FOV through a 1 inch tube.
Your field of vision is specific to the tube itself from the central point to the inner walls all around that tube.
You are not spanning out any wider than that.
your instance was that the blue line doesn't exists or doesn't exist?
clarify.
or that people don't and do see in 1dimension
Quote from: JackBlack
So are you saying this:
Absolutely.
here someone's pointed out that a tube doesn't do anything but obstruct the field of view, which you've admitted is a thing, and has hand drawn in a tube, which oho does not consequently affect the fact horizon doesn't rise to eye level.
while still saying stupid things like people are unable to look down or have a field of view
Here's a quick diagram. Genuine people....take time to understand what's being said with it.
you claim to understand yet every statement following you continue to beileve people see in 1dimension
You're making out one dimension to what I'm saying. I'm giving out nothing of the sort.
Tunnel vision is not one dimensional.
I've already mentioned a compressed FOV, so what's the issue?
no, i see NOOO contradictions here.
anyone else?
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3311 on: January 26, 2021, 01:17:24 PM »
Because you keep pretending we don't have a FOV.
No I don't. I never have and you know this.
If that really was the case, you wouldn't have dishonestly removed the section of the post clearly explaining that you continually act like we have no FOV.
You saying we have a FOV is an entirely empty gesture when you continually act like we don't.
All it does is further how dishonest you are willing to be.
Again, accepting we have a FOV means accepting that we don't just magically see a straight line.
It means accepting that we have FIELD of view, that is that our view spans a field, i.e. a range of angles.
I.e. this useless garbage from you shows no FOV:
These straight lines clearly show no FOV.
And you used it to claim we wouldn't be able to see the sun as soon as Earth rotates, and that we wouldn't be able to see the horizon at all.
Those claims are based upon there being no FOV.
Accepting that we have a FOV (rather than just the empty gesture of saying we do and claiming you have never acted like we have no FOV) means accepting that even when looking through a level tube, we don't magically just see perfectly level and instead we can see things above and below level as well, within the limits of the FOV.
Like this:
The tube limits your FOV to the red lines.
This means you can't see the orange line, as the walls of the tube block it.
But you can see the blue line and that means you can see things below the tube.
If you actually acknowledged and accepted the fact that we still have a FOV, you wouldn't keep repeating the same pathetic lies which relies upon rejecting the fact that we do have a FOV.
So you saying we have a FOV is worthless and just shows how little integrity you have.
Making stuff up to suit and twisting is all you seem to do.
Projecting again I see.
You are the one who continually makes stuff up, like your fantasy that we magically only see 1 inch of any object when looking through a 1 inch tube.
And you then continually twist what other people say or just outright ignore it so you can pretend your fantasy is true.
Now again, what magic stops the blue line to prevent us having an actual FOV?
And what magic causes the RE to have a blend from light to dark?
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3312 on: January 26, 2021, 02:00:04 PM »
#### Smoke Machine
• 1808
##### Re: What would change your mind?
« Reply #3313 on: January 26, 2021, 10:46:21 PM »
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
Perhaps that's because nobody can prove what you're saying, Scepti. Mind if I flower up the proceedings with some photos?
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3314 on: January 26, 2021, 11:20:32 PM »
How is it even after a 5day ban has still not seen any movement in this tu-tube topic?
Because nobody can prove against what I'm saying.
Perhaps that's because nobody can prove what you're saying, Scepti. Mind if I flower up the proceedings with some photos?
I'm more than happy for any of you to prove whatever you think you can.
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3315 on: January 26, 2021, 11:59:26 PM »
I'm more than happy for any of you to prove whatever you think you can.
And then to completely ignore it or just dismiss it as fake.
Again, what magic what stops the blue line?
Again, what magic causes the round Earth to magically have a blend from light to dark.
Because until you come up with an answer I have proven that you are wrong.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3316 on: January 27, 2021, 12:01:21 AM »
And then to completely ignore it or just dismiss it as fake.
Then ensure there's no twisting or trickery. Make it so I can't criticise.
If you can't do that then you have no case.
« Last Edit: January 27, 2021, 12:03:18 AM by sceptimatic »
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3317 on: January 27, 2021, 12:02:42 AM »
I'm more than happy for any of you to prove whatever you think you can.
And then to completely ignore it or just dismiss it as fake.
Again, what magic what stops the blue line?
Again, what magic causes the round Earth to magically have a blend from light to dark.
Because until you come up with an answer I have proven that you are wrong.
Then ensure there's no twisting or trickery. Make it so I can't criticise.
I have.
You have no criticism of the logical argument.
Instead all you do is continually ignore it.
Unlike the evidence that you just dismiss as fake, you can't do the same for a logical argument.
Again, what magic what stops the blue line?
Again, what magic causes the round Earth to magically have a blend from light to dark.
Because until you come up with an answer I have proven that you are wrong.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3318 on: January 27, 2021, 12:04:19 AM »
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3319 on: January 27, 2021, 01:09:22 AM »
I have.
No, you haven't.
And just like I said, you just ignore it.
Again, see this diagram:
It clearly shows the problem.
Again, what magic stops the blue line?
Until you have an answer, you have been proven wrong.
The fact you continue to ignore it rather than even trying to respond shows that.
Now grow up and either answer the question or admit you have no answer and have been shown that you are wrong.
#### sceptimatic
• Flat Earth Scientist
• 28092
##### Re: What would change your mind?
« Reply #3320 on: January 27, 2021, 04:01:51 AM »
I have.
No, you haven't.
And just like I said, you just ignore it.
Again, see this diagram:
It clearly shows the problem.
Again, what magic stops the blue line?
Until you have an answer, you have been proven wrong.
The fact you continue to ignore it rather than even trying to respond shows that.
Now grow up and either answer the question or admit you have no answer and have been shown that you are wrong.
There is no problem. You made a diagram making out the blue line has meaning.
It does if you're using a scope. A telescope or your naked eye. Are you? .....Because, if you are then you're twisting the issue when you know fine well I'm arguing the 1 inch diameter tube.
I have also stated you have to be looking level through a crosshair not angled down from the top of the back of the tube to the bottom of the front.
What you claim to see is in your mind, unless you are using naked eye or telescope.
So which is it?
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3321 on: January 27, 2021, 04:31:30 AM »
I'm more than happy for any of you to prove whatever you think you can.
And then to completely ignore it or just dismiss it as fake.
Again, what magic what stops the blue line?
Again, what magic causes the round Earth to magically have a blend from light to dark.
Because until you come up with an answer I have proven that you are wrong.
Exaclty
PROVE it yourself sceppy for all to see and remove all doubt.
If youre right this would be mind blowing.
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3322 on: January 27, 2021, 04:32:18 AM »
And then to completely ignore it or just dismiss it as fake.
Then ensure there's no twisting or trickery. Make it so I can't criticise.
If you can't do that then you have no case.
The only one who can do that is you.
You are happy with you and only you know what you want.
Show to us.
#### JJA
• 4877
• Math is math!
##### Re: What would change your mind?
« Reply #3323 on: January 27, 2021, 05:15:35 AM »
And then to completely ignore it or just dismiss it as fake.
Then ensure there's no twisting or trickery. Make it so I can't criticise.
If you can't do that then you have no case.
If you would just show your own experiment we could settle this once and for all. Why are you afraid to show your photos? If you are so sure you are right, what is there to worry about?
?
#### notaFlatEarther01
• 1
##### Re: What would change your mind?
« Reply #3324 on: January 27, 2021, 09:03:25 AM »
What would it take for you to change your mind, whichever side you're on?
Well in general the burden of proof is on the FE community. But I'll give you some stuff to go on.
1st: If the Earth is flat why does the hull of a ship disappear before the mast?
2nd: If the Earth is flat, by what mechanism does gravity exist? (btw saying gravity is a theory is proof that you don't understand the scientific method, just look at the Cavendish experiment which is performed by ALL physics undergrads).
3rd: If the Earth is flat and, by extension, most of the modern astrophysics is wrong, what the fuck do astrophysicists do? Are all of them in collusion with the government? I'm studying astrophysics right now, am I going to be indoctrinated?
Just some food for thought.
#### Mattathome
• 23
• I love this site, it goes round and round...
##### Re: What would change your mind?
« Reply #3325 on: January 27, 2021, 09:36:17 AM »
Why are you clicking that?
?
#### Themightykabool
• 5353
##### Re: What would change your mind?
« Reply #3326 on: January 27, 2021, 11:45:34 AM »
?
#### JackBlack
• 15775
##### Re: What would change your mind?
« Reply #3327 on: January 27, 2021, 12:27:06 PM »
And just like I said, you just ignore it.
Again, see this diagram:
It clearly shows the problem.
Again, what magic stops the blue line?
Until you have an answer, you have been proven wrong.
The fact you continue to ignore it rather than even trying to respond shows that.
Now grow up and either answer the question or admit you have no answer and have been shown that you are wrong.
There is no problem. You made a diagram making out the blue line has meaning.
There is a massive problem for you, which you are yet to address.
All you can do is continually ignore the problem and pretend I am saying or doing things I aren't. You have no actual criticism against his disproof of your nonsense.
This shows what happens when you have a simple tube without a lens.
This shows how the tube restricts your FOV.
Each of the thick coloured lines has a very significant meaning, showing a possible path of light.
We can easily see that the orange line must pass through the wall of the tube in order to reach your eye. Assuming the tube is opaque this means the beam of light represented by the orange line cannot reach your eye and instead it will be blocked by the wall of the tube.
Conversely, we can easily see how the red, green and blue lines do not intersect the wall.
There is nothing in their way to stop them, nor is there a lens to deflect them and thus they can reach the eye.
The red lines are the limit, the edge of the FOV you have through a tube.
This means you FOV is the region bounded by the 2 red lines.
The blue line is in that region. Again, there is nothing to stop light travelling along it from reaching your eye.
This means that light can travel from below the level of the tube and go up and into your eye. This can allow you to see the ground, even when looking level, even on some downwards slopes, depending on the FOV of the tube and the gradient of the slope.
It does if you're using a scope. A telescope or your naked eye. Are you?
No, it is a simple tube.
Again, if you would like an example with a scope, then an extreme example would be this:
Notice the lens shown in grey which bends the light?
That is what you need for your nonsense to be correct.
I have also stated you have to be looking level through a crosshair not angled down from the top of the back of the tube to the bottom of the front.
It is quite clear that in this diagram the eye is centred.
It is looking directly level.
But because it has a FOV it also sees above and below level.
What you claim to see is in your mind, unless you are using naked eye or telescope.
So which is it?
Neither.
It is simply that you are wrong, and you are looking for whatever excuse you can to dismiss this logical proof that you are wrong, when you have no criticism of it at all and thus need to invent criticism when it clearly has no place.
Everyone can see that this diagram has a tube, and thus it is not merely the naked eye.
Everyone can see that there is no lens, and thus it is not a scope.
Yet you play dumb and act like both could be the case.
You play dumb and pretend that the lines have no meaning, when they have been explained to you repeatedly.
You play dumb and pretend that the eye isn't looking level, and instead want to pretend it is somehow looking from the top of the tube when the image clearly has it in the middle.
There is no twisting or trickery by me.
The only attempts at that are by you, trying to use whatever dishonest BS you can to dismiss the fact that you are wrong.
And you have no criticism of it. The only "criticism" you provide is of your strawmen.
The valid criticism you could make against this requires you to explain what magic stops the light indicated by the blue line from reaching the eye.
Without such criticism, you are wrong.
So there you have it, simple proof that you are wrong with no twisting or trickery (except the attempts by you) and which you cannot criticise.
You are wrong. It's time for you to grow up, accept that you are wrong, and move on.
#### Smoke Machine
• 1808
##### Re: What would change your mind?
« Reply #3328 on: January 27, 2021, 12:30:24 PM »
You beat me to it. I see you prefer the double length rolls! I always find those ones don't roll as easily on the spool, and you have to manually turn the roll to get your length of paper. Very annoying when you're in a rush.
But, your use of the roll with the full length of paper still on, is the correct way to perform the experiment. The resident toilet roll peeping experts, aka sceptimatic and JJA, will argue otherwise. I believe that's because they both enjoy performing the experiment naked in public places. Without the paper, it's a lot less weight for their morning glory to hold up, after they've slipped the roll on, to rest between experiments.
#### Mattathome
• 23
• I love this site, it goes round and round...
##### Re: What would change your mind?
« Reply #3329 on: January 27, 2021, 08:40:32 PM »
You beat me to it. I see you prefer the double length rolls! I always find those ones don't roll as easily on the spool, and you have to manually turn the roll to get your length of paper. Very annoying when you're in a rush.
True indeed good sir! I however, refuse to use the spool and keep the toilet paper with in easy reach at all times having experienced many the unfortunate emergency toilet paper situation.
But, your use of the roll with the full length of paper still on, is the correct way to perform the experiment. The resident toilet roll peeping experts, aka sceptimatic and JJA, will argue otherwise. I believe that's because they both enjoy performing the experiment naked in public places. Without the paper, it's a lot less weight for their morning glory to hold up, after they've slipped the roll on, to rest between experiments.
The nakedness is perhaps akin to the efficiency of performing all of these Tube Viewing experiments. Having the need for so many empty tubes would elude to the excessive amount of ass-wiping required in the name of science. Clothing would only get in the way. That said, I would recommend the Charmin brand of toilet paper. The double-ply strength and superior texture help prevent chafing as I'm sure there there is no end in sight to this particular Tube Viewing debate.
Why are you clicking that?
| 6,704
| 27,113
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.578125
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.935488
|
https://www.jiskha.com/users?name=mercedes
| 1,571,187,216,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986660829.5/warc/CC-MAIN-20191015231925-20191016015425-00508.warc.gz
| 929,530,455
| 13,057
|
# mercedes
Most popular questions and responses by mercedes
1. ## sat essay
Can any obstacle or disadvantage be turned into something good? Obstacles, which are the essential part of our lives, often seem to prevent us from reaching our goals and frustrate us. Therefore, most people think that difficulties are bound to exert
2. ## sat essay
Is it necessary for us to find new solutions to problems? Many people may argue that we don¡¯t need to since we are already able to solve the problems. As far as I am concerned, we should always try to find new solutions to problems. New solutions enable
3. ## algebra
whom should to see at the bank if you need to borrow money?
4. ## Math
given that triangle PQR is similar to triangle STU and that corresponding sides PQ=4cm and ST=6cm. The area of STU is 12cm squared. state the area of PQR
5. ## physics
Two blocks are tied together with a string as shown in the diagram.(There is a block on a horizontal incline, its mass is 1.0kg, there is a rope connecting that block to a pulley which then connects to another block that is 2.0kg. The angle of the incline
6. ## Algebra
What Might You Have If You Don't Feel Well?
7. ## algebra
Write an equation for a line passing through the points (-1,8 and (0,5)
8. ## physics
A 50-kg astronaut experiences an acceleration of 5.0g (up) during liftoff. What is the astronaut's true weight and apparent weight?
9. ## Physics
Two blocks are tied together with a string as shown in the diagram.(There is a block on a horizontal incline, its mass is 1.0kg, there is a rope connecting that block to a pulley which then connects to another block that is 2.0kg. The angle of the incline
10. ## physics
What would be the weight of 22.0kg dog at the equator on Saturn's surface? (Saturn's mass equals 5.69 X 10^26kg, Saturn's equatorial radius equals 6.03 X 10^7m, Earths mass equals 5.97 X 10^24)
11. ## chemistry
When dissolved in 1000 g of water, which chemical compound will produce a solution with the greatest freezing point depression?
12. ## Math
a plane traveling at 400 mph leaves the philadelphia airport at 1pm heading due east. two hours later a second plane traveling at 600 mph leaves the airport heading in the same direction. at what time does the second plane overtake the first?
13. ## physics
A 435 N box is sliding down a 40 degree incline. If the acceleration of the box is 0.250 m/s^2, what is the force of friction on the box?
14. ## AP Human Geography
explain why economic sucess and political power are closely linked? What role does colonization play in the establishment of todays state?
15. ## physics
An electron moving to the right at 4.0% the speed of light enters a uniform electric field parallel to its direction of motion. If the electron is to be brought to rest in the space of 5.0 cm, determine what the strength of the field is.
16. ## Chemistry
Usng the following balanced chemical equation, determine the limiting reactant in the reaction between 3.0 grams of titanium and 8.0 grams of chlorine gas? Ti(s) + 2Cl2(g) --> TiCl4(s)
17. ## physics
An old battery with an emf of 9.0 V has a terminal voltage of 7.8 V when it is supplying a current of 1.5 mA. What is the internal resistance of the battery?
18. ## SAT essay
Should people change their decisions when circumstances change, or is it best for them to stick with their original decisions? Environments, including the people around us, the values of society and the opinions of the majority, always exert influence on
19. ## Chemistry
How many moles of iodine are produced from 7.00 moles of chlorine, according to the equation shown? 3Cl2(g) + 2FeI2(s) --> 2FeCl3(s) + 2I2(g)
20. ## Chemistry
Explain what happens when you add sodium chloride to boiling water. I know that when sodium chloride is added to water, the boiling point of water increases. What happens when it is added to already boiling water? I know it increases the temperature, I
21. ## physics
An old battery with an emf of 9.0 V has a terminal voltage of 7.8 V when it is supplying a current of 1.5 mA. What is the internal resistance of the battery? 800 Ω 900 Ω 400 Ω 600 Ω
22. ## Physics
A satellite orbits Earth at a distance of 3rearth above Earth's surface. (Earth mass equals 5.97 X 10^24, earths equatorial radius equals 6.38 X 10^6) How many Earth radii is the satellite form Earths center?
23. ## Healthcare
Where can I find information about Managed Health Care? I am also interested in its pros and cons. Also, where can I find information regarding Reforma de Salud de Puerto Rico? (Puerto Rico Health Reform) (Besides Wikipedia...) I would greatly appreciate
24. ## sat
if i don't finish my essay, how many scores will i get?
25. ## physics
A force of friction of 3.2N acts on a 1.1kg puck while it is sliding along a horizontal surface. If the initial velocity of the puck was 7.5 m/s , how far will the puck travel before coming to rest?
26. ## Calculus
Find the area of the region that lies inside both curves. r = (3)^(1/2) cos (theta), r = sin (theta)
27. ## Chemistry
A chemist is producing copper(II) chloride using the following reaction: Cu(s) + Cl2(g) --> CuCl2(s) In one reaction, the chemist uses 15.0 g of copper metal and 10.0 g of chlorine gas as reactants. When the reaction reaches completion, 16.0 g of
28. ## Physical science/football/algebra
suppose that Andrew is running down the field at 8 yards per second and then cuts right running 6 yards per second. He executes this maneuver in 0.5 seconds. Calculate his acceleration
29. ## Physical science/football/algebra
suppose that Andrew is running down the field at 8 yards per second and then cuts right running 6 yards per second. He executes this maneuver in 0.5 seconds. Calculate his acceleration
30. ## physics
an empty coal car is moving along the tracks to a loading chute a 7km/hr., the mass of the empty car is 2000kg what is the speed of the loaded coal if the total mass of the car is now 10000 kg? give you answer in meters per second.
31. ## physics
A satellite orbits Earth at a distance of 3rearth above Earth's surface. (Earth mass equals 5.97 X 10^24, earths equatorial radius equals 6.38 X 10^6) What is the magnitude of the gravitational acceleration of the satellite?
32. ## math
The minute hand is pointing to the 12. The hour hand is perpendicular to the minute hand. What 2 times could it be?
33. ## physics
Using Kepler's constant (9.84 X 10^-14), calculate the orbital radius of an artificial satellite whose period of revolution around the Earth is 1.43 X 10^4s(3.64 h).
34. ## Macroeconomics,
List and explain the three tools used by the Fed to manipulate the supply of money. In each case show how the money supply can increase/decrease.
35. ## sat essay
assignment:should people always prefer new things, ideas, or values to those of the past? Should people always prefer new thing to the old? Many people argue that we should not accept the new easily and should sick to the old. As far as I am concerned, we
36. ## geometry
A 6ft tall man casts a 10ft shadow next to his 4ft tall daughter how long is her shadow ? I Don't Get The Question
37. ## Physics
A tennis player stands 15.0m from the net. They hit a ball from a height of 2.50m above the ground so that initially it has a horizontal velocity directed parallel to the ground toward the net. What speed does the ball have to have to just clear the net
38. ## Accounting
A piece of equipment is purchased on May 1, 2007 for \$80,000 and has a residual value of \$5,000. The equipment's useful life is 8 years or 12,500 hours. It is used 2,700 hours in 2007, and 2,600 in 2008. The question is what is the book value on December
39. ## Chemistry
Why are combustion reactions always exothermic? Is it just because it gives off heat? Thanks ahead of time Exothermic MEANS it gives off heat; therefore, I don't think that can be a reason. Think of wood burning. Magnesium strips burning. A wax cancle
40. ## Physics
A small plane moving at 207km/hr flying 50.0m above the ground chases a convertible moving at 105km/hr down a straight highway. The plane has a package to drop in the car’s back seat. Assuming the vehicles maintain their speed and altitude and there is
41. ## science
plant like which are called_______
42. ## Macroeconomics,
If nominal GDP is \$300 billion and the money supply is \$20 billion, What must be the velocity? (b)If the money supply decreases and the velocity does not change, what will happen to nominal GDP?
43. ## Math
Solve the following equation by factoring: 6(p^2-1)=5p
44. ## Chemistry
Earlier I asked for a definition of non-volatile. Given the fact that it won't evaporate at all, does it mean that a non-volatile substance is stable in whatever state it is in at room temperature? Thanks again ahead of time Not necessarily. Non-volatile
45. ## math
3200000*0.064 in scientific notation
46. ## Chem
A Texas Aggie buys 12 cans of dehydrated water. The cans weigh an average of 63.46g each before the dehydrated water is removed. If the density of pure distilled water is 0.9979 g/mL and each can has a volume of 3.154L, what is the total mass of the cans
47. ## mechanics
find the tension in the chord of a 50 lbs mass hang on the angles 25 degrees.
48. ## math
using the 7 steps outlined in sec 4.3 of your book,analyze the graph of the following function R(x)=x^3-125/x^2-49
49. ## Physical science/football/algebra
suppose that Andrew is running down the field at 8 yards per second and then cuts right running 6 yards per second. He executes this maneuver in 0.5 seconds. Calculate his acceleration
50. ## Macroeconomics,
If nominal GDP is \$300 billion and the money supply is \$20 billion, What must be the velocity? (b)If the money supply decreases and the velocity does not change, what will happen to nominal GDP?
51. ## Grammar
When saying "...he received a Bachelor of Science in Business Administration..." do I capitalize that or leave it un-capitalized?
52. ## fair project
what happen when metal, cloth and wood get exposed to fire
53. ## Chemistry
Using the principles of bond breaking and bond making, how can I explain why there is no change in temperature in phase changes? In a phase change, for example when boiling water is converted to steam, no O-H bonds are broken and none are made. Enough
54. ## Math
How can I succeed in my positive and negative unit?
55. ## math
the function f is defined: f(x)= x+3 if -2
56. ## square roots
i know this problem is really easy but i am having a brain freeze day will someone please help me figure out what the square root of 32 is? Please & thank you!
57. ## Science
How does respiratory and cardiovascular system obtain nutrients?
58. ## english
write a review of his theme, along with how this poem relate to your own life experience. I will arise and go now, and go Innisfree, And a small cabin build there, of clay and wattles made: Nine bean-rows will I have there, a hive for the honey-bee, And
59. ## Macroeconomics,
Having a strong or a weak dollar, which is better for America? Why?
1. ## math
posted on June 17, 2019
2. ## Maths
444= 4+4+4=12 381=3+8+1=12 453=4+5+3=12
posted on June 17, 2019
3. ## Math
Is it A or C??
posted on December 18, 2017
4. ## English
There isnt Assonance on mine!
posted on November 30, 2017
5. ## Social Studies
Thanks bunny 100%
posted on April 28, 2017
6. ## SCIENCE
Thank you 100%
posted on March 17, 2017
7. ## Poetry
Just get rid of words you don't need or that are insignificant, keep key words. Say you have a 500 word piece, get it down to 400, 300, 200 etc. Then put it into stanzas and break up into lines.
posted on February 21, 2016
8. ## Finance
. 1% b. 0 c. 10.6% d. –0.1%
posted on September 9, 2013
9. ## math
its the same steps just a different function in my textbook, but I don't understand it!
posted on March 12, 2013
10. ## English
That doesn't quite make sense I don't know why Mom asked should be underlined?
posted on February 18, 2013
11. ## Physical science/football/algebra
ok thanks and yes i do
posted on September 29, 2011
12. ## Math
sorry if you were looking for an explanation from the other person.
posted on September 6, 2011
13. ## Math
136800000 divided by 250000000 it simple like a fraction 136800000 is the numerator and 250000000 is the denominator and to find a decimal you would divide the numer. by the denomin. and then to get a percent you multiply that decimal by 100 and round it!
posted on September 6, 2011
14. ## square roots
oh thank you very much your a lifesaver!
posted on September 6, 2011
15. ## physics
posted on November 1, 2010
16. ## physics
what is m and g?
posted on November 1, 2010
17. ## physics
what number goes into M and d?
posted on November 1, 2010
18. ## Macroeconomics,
posted on June 21, 2010
19. ## Macroeconomics
List and explain the three tools used by the Fed to manipulate the supply of money. In each case show how the money supply can increase/decrease.
posted on June 21, 2010
20. ## math
then put the x axis on the bottom and the y axis on the top of a fraction if you need reduce
posted on May 24, 2010
21. ## math
we are studying this in class at black river today same problem popped up subtract the 10-0 which is the y axis then 0-7 which is the x axis.
posted on May 24, 2010
22. ## sat essay
thank you so much!
posted on November 21, 2009
23. ## sat
but i only have 25 minutes to write the essay
posted on November 7, 2009
24. ## SAT essay
do you mean that my example can't support my thesis?
posted on November 5, 2009
25. ## SAT essay
is it neccessary to write more than 1 example?
posted on November 4, 2009
26. ## CHEMISTRY
A chemist is producing copper(II) chloride using the following reaction: Cu(s) + Cl2(g) --> CuCl2(s) In one reaction, the chemist uses 15.0 g of copper metal and 10.0 g of chlorine gas as reactants. When the reaction reaches completion, 16.0 g of
posted on September 1, 2009
27. ## earth scince
earth scince
posted on March 25, 2009
28. ## algebra 1b
2 - 6
posted on September 18, 2008
29. ## MAT116- Algebra 1A
7-6x>1-5x
posted on June 15, 2008
30. ## Science
What happen when metal, wood and cloth are exposed to fire
posted on February 14, 2008
31. ## math
To find out what your investment would be in 3 years, you must find the interest for year one and add it to the principle to find the interest for year two and so on. Which means your interest earns interest for you. Principle x Rate=Interest Year 1:
posted on November 8, 2007
32. ## math
Those are current liabilities. If you haven't talked about current and long term liabilities yet, then the answer is just liabilites.
posted on November 8, 2007
33. ## Biology
Hi Dannie, 1. In order for a cell to reproduce it must copy its DNA to pass on its genetic information. 2.DNA replication is the process of duplicating the DNA to make 2 identical copies. 3.The first step of DNA replication is the separation of the 2 DNA
posted on November 8, 2007
34. ## yh8hy
Dear Juan, energy from the sun is important because it heats the water molecules which causes them to evaporate. Evaporation is when the water turns into a gas which is a part of the water cycle. After the gas cools it turns into condensation, or liquid
posted on November 8, 2007
| 4,050
| 15,355
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.92829
|
https://studentshare.net/finance-accounting/11833-risk-management-principles-and-applications
| 1,521,691,847,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-13/segments/1521257647768.45/warc/CC-MAIN-20180322034041-20180322054041-00659.warc.gz
| 724,114,859
| 16,038
|
StudentShare solutions
# Risk Management: Principles and Applications - Assignment Example
## Extract of sample Risk Management: Principles and Applications
If the market price for the asset is greater than the strike price, it will be worth while for the holder to use the option. In this instance the option writer is forced to trade the asset to the option owner. The price at which a call is traded is called premium. In the case of calls, a higher stock price means that other things being equal; the option is more likely to be in the money on maturity. Therefore, a higher S is associated with a higher premium (Hull, 1997). Conversely, the higher the exercise price K, the lower is the premium. A higher exercise price makes it less likely that the option will be in the money on maturity. The time to maturity t has appositive effect on the option price: the greater t is, the more time the stock price has to climb above the exercise price. The volatility of the stock measured by the standard deviation of its returns ?, a;so has a positive influence on the call price. This due to the fact that call option holders are only concerned with the gain potential of their option, and this is greater the higher is the likelihood that the option will be beneficial on ripeness. Finally, the rate of interest, r has a positive influence on the call price. The greater the rate of interest r, the lower the strike value to be paid on maturity (e-rtK). The price of a call is given as C=S. N (d1) –K. e-rt. ...
263 v0.25= 0.9532 d2= 0.9532-v0.263 v0.25=-0.06968 Nd1=0.3289 Nd2= 0.0239 Replacing them in C= (879000) (0.3289) - 227,349.9673 (0.0239)= \$ 23,476.6458 Using put option Put options are the options which give the holder the right to sell the specified asset at the specified price. They are issued by sellers or their writers. The price of an option is known as premium. The value at which the underlying asset can be sold is called the strike price (Brigham, Gapenski & Ehrhardt, 1999). If you have a long position, one of the strategies to be utilized in order to hedge against a price decline is a long position in puts. One could buy put options to be fully hedged against price decline, but could still profit from price raise. The purchaser of put options can hedge their drawback price risk for a time and still gain from the potential price increases if the market should rise. The holder pays a certain premium to shield against a likely loss. Once the premium is remunerated, the holder has no further obligation. Depending on value progress, the manufacturer can either take or leave the assured price. If the price of the stock goes down, one has a price protection with the put option at the selected strike price. The same aspects that influence the cost of a call, also influence the price of put options on stocks. The premium on puts is now a decreasing function of the stock price and an increasing function of the exercise price. The conditions that should be fulfilled for a put to be in money: an increasing function of the time to maturity and the volatility of the stock, and a decreasing function of the rate of interest: this last result is owing to the truth that an elevated rate of interest decreases the present worth of the exercise cost to be received on prime of life. ...Show more
## Summary
Risk Management: Principles and Applications Options on stocks Options on stocks give their owner the right to purchase or trade shares of a particular stock, at a specified price. It is vital to compare the market price of the stock, with the exercise price of the option (Hull, 1997)…
Author : mcummings
Save Your Time for More Important Things
Let us write or edit the assignment on your topic
"Risk Management: Principles and Applications"
with a personal 20% discount.
Grab the best paper
### Check these samples - they also fit your topic
The quality of risk management in the Civil Engineering
Based on this research there are many definitions for hazards in the field of international construction and civil engineering. Hazards can be classified, there are those that result in individual injury or toxicity, those that cause a general feeling of suspicion and unrest in the surrounding community – or may not involve an actual physical hazard, and sometimes the suspicion of danger is sufficient social consequences.
28 pages (7000 words) Assignment
Risk management
Establishment of the actual position of the organization task in the internal and external environment is the key mandate of the general risk management platform. Among the most important spirit of principle recommendations is the withholding of value in performance, which introduces the overall organizational objective in every risk management decision.
10 pages (2500 words) Assignment
Financial Risk Management
79). In a number of these theoretical exemplars of financial or cost-effective procedures, their significance is very much considered when addressing them and looking for the appropriate solutions to block these risks. For instance, a number of risks or even all risks might be comprehensively removed (Dun & Bradstreet, 2006, p.
6 pages (1500 words) Assignment
Risk management
A Risk is described as a tentative event for an organization that might cause either positive or negative effects on its operations and functions at the time of entering a foreign market. If the risks present positive outcomes for the organization in that new market, then it might be described as opportunities and if it affects negatively, then it may be stated as threats.
6 pages (1500 words) Assignment
Aircraft System Principles and Applications
Air pressure and air temperature are directly associated with each other. Before going into the details of how air properties affect the flight of a plane, one must have deep insight into the correlation of air properties. Variation in air pressure directly affects the density of air if the temperature is kept constant.
6 pages (1500 words) Assignment
Financial Management and Risk Analysis
The new technology is expected to boost the firm's production capacity as well as eliminate inefficiencies such as rework/scrap and wastage. Currently, the production department utilizes a manual assembly line which eight fitters who are paid 18,500 per annum each.
8 pages (2000 words) Assignment
Value and Risk Management - MSc Civil Engineering and Construction Management
We could not but consent to this request as we knew the importance of such a project, considering that we have initiated a brief visit to the site and scanned a few
14 pages (3500 words) Assignment
Risk management and international finance
Additionally, the report will include the calculation of value at risk (VaR) credit measures. This determination will employ the use of model based simulations. The changes in credit risks risk of banks and
6 pages (1500 words) Assignment
management involves a series of steps which delivers an effective process to accomplish the goals and objectives a project smoothly (Public procurement Guide, 2008). The discussion of this part of the assignment aims to critically examine the National IT Program in the NHS
20 pages (5000 words) Assignment
Risk management
usinesses stand to lose money and assets through poor decision-making, poor investment choices, and poor understanding of their industries (Hussain, Hussain, Dillon & Chang, 2014). In recent times, some scholars have classified risk assessment as a source of competitive
1 pages (250 words) Assignment
Click to create a comment
Let us find you another Assignment on topic Risk Management: Principles and Applications for FREE!
| 1,587
| 7,665
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2018-13
|
latest
|
en
| 0.935813
|
http://docplayer.net/28194501-Sfusd-mathematics-core-curriculum-development-project.html
| 1,531,952,922,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00364.warc.gz
| 100,539,505
| 33,983
|
# SFUSD Mathematics Core Curriculum Development Project
Save this PDF as:
Size: px
Start display at page:
## Transcription
1 1 SFUSD Mathematics Core Curriculum Development Project Creating meaningful transformation in mathematics education Developing learners who are independent, assertive constructors of their own understanding
2 Grade 5 Unit 5.5: Multiplying and Dividing Decimals by Decimals 2 Number Lesson Reproducibles Number of Copies Materials of Days 1 Entry Task Farmer s Market 4 Lesson Series 1 Base Ten Activity Handout HW: Picturing Multiplication (3 pages) Multiplication Error Analysis Exit Slip Maze Scaling Multiplication HW: Multiplying Decimals (2 pages) Where s the Point HW: EM Study Link 2.7, 2.9 (2 pages) Missing Number Worksheet Sorting Products (3 pages) HW: Chris s Test EM Study Link base ten blocks class set calculators class set 1 Apprentice Task Apprentice Student Sheet Optional: Calculators. Digit cards 4 Lesson Series 2 Base Ten Division Student Sheet Division Error Analysis Exit Slip HW: Area Model of Division Maze Scaling Multiplication and Division HW: Another Method for Dividing (2 pages) EM Journal p 109 HW: Chris s Test #2 HW: EM Study Link 4.5 Estimate and Calculate Do You See an Error Recording Sheet HW: Unit Pricing 1 Expert Task Sewing A Quilt Student Sheet Base Ten Grid Paper EM Journal EM Study Link 1 per pair base ten blocks class set calculators class set base ten blocks class set calculators (optional)
3 3 3 Lesson Series 3 Road Trip Recording Sheet EM Grade 5 SRB p 388 U.S. Distance Map HW: Day 1 Word Problems HW: Day 2 Solve Each Problem HW: Day 3 Unit Pricing (2 pages) 1 Milestone Task Milestone Road Trip Milestone Constructed Response EM SRB Provided by AAO Provided by AAO calculators (optional) poster paper (optional)
4 Unit Overview Big Idea 4 The same rules and strategies that apply for multiplying and dividing whole numbers also apply for multiplying and dividing decimals. Because multiplication is scaling, when we multiply a quantity by a number larger than 1, it produces a larger quantity. When a quantity is multiplied by a number smaller than 1, it produces a smaller quantity. Unit Objectives Students will be able to multiply decimals by decimals with both factors and products to the hundredths. Use concrete models or drawings. Explain their thinking and strategies in numbers and words using understanding of place value and/or properties of operations. Answer questions about their strategies, including, but not limited to: How did you do it? What have you tried? What did you learn when that didn't work? Why does that make sense to you? Why do you think that's true? Is it always true? Under what conditions is it true? Students will be able to divide decimals by decimals with divisor, dividend, and quotient to the hundredths. Use concrete models or drawings. Explain their thinking and strategies in numbers and words using understanding of place value and/or properties of operations. Answer questions about their strategies, including, but not limited to: How did you do it? What have you tried? What did you learn when that didn't work? Why does that make sense to you? Why do you think that's true? Is it always true? Under what conditions is it true? Students will understand and be able to explain how the magnitude of the product relates to the magnitude of the factors. Unit Description This unit starts with students demonstrating their multiplication and division knowledge from previous units. This unit will expand upon their previous knowledge by continuing on to multiplication and division of decimals by decimals. In the first lesson series, students will use their knowledge of base ten blocks to learn how to model multiplication of decimals by decimals, they will also begin to notice how scaling is related to multiplication, i.e. the relationship between factors and products. A variety of activities will help students become proficient multiplying decimals by decimals, and placing the decimal in the correct place. The second lesson series will focus on division of decimals by decimals using base ten blocks to model the division. A variety of activities will continue to help students become proficient dividing decimals by decimals. Students should begin to realize how multiplying or dividing by decimals less than 1, or greater than 1 will affect the product or quotient of problems. This lesson series will mirror what was learned in the first lesson series. In these two lesson series, students will have opportunities to demonstrate their proficiency in both multiplication and division of decimals by decimals using the exit slips, as well as on the final activity on Day 4 of Lesson Series 2. During Day 3 of both Lesson Series 1 and 2, you will have an opportunity to demonstrate an algorithm to show multiplication and
6 Progression of Mathematical Ideas Prior Supporting Mathematics Current Essential Mathematics Future Mathematics 6 In fourth grade, students begin their learning of the decimal system with an introduction into decimal notations of the tenths or hundredths place. They learned how to compare decimals up to the hundredths place value, when they refer to the same whole. They are able to compare using <, >, = symbols and using visual models to justify their conclusions. In Unit 1 of fifth grade, students became fluent using an algorithm for multiplication and division of whole numbers, which can then be used for multiplication and division of decimals. Students will build on their work from Unit 5.4 by multiplying and dividing decimals by decimals. While deepening their sense of place value, students will also use their understanding of the relationship between multiplication and division to determine how decimal factors and divisors/dividends affect the magnitude of the product and quotients. Students will use concrete models and drawings to model these operations. In Unit 7 of fifth grade, students will extend their understanding of the properties of multiplication of non-whole numbers in the multiplication of fractions. In sixth grade, students will fluently calculate using all operations with decimals using the standard algorithm. In addition, students will extend their understanding of multiplication as scaling when they work with proportional relationships. In Unit 2 of fifth grade, students extended their learning by writing and comparing decimal to the thousandths place value. They round numbers to any place value, and begin to understand the relationship between place values and powers of 10. In Unit 4 of fifth grade, students began adding, subtracting, multiplying, and dividing decimals by whole numbers, using models or drawings, and written methods to explain their reasoning. Students learned how to multiply and divide by a power of 10.
7 Unit Design All SFUSD Mathematics Core Curriculum Units are developed with a combination of rich tasks and lessons series. The tasks are both formative and summative assessments of student learning. The tasks are designed to address four central questions: Entry Task: What do you already know? Apprentice Task: What sense are you making of what you are learning? Expert Task: How can you apply what you have learned so far to a new situation? Milestone Task: Did you learn what was expected of you from this unit? 7 1 day 4 days 1 day 4 days 1 day 3 days 1 day Total: 15 days
8 8 Entry Task Farmer s Market Apprentice Task Missing Numbers Expert Task Sewing a Quilt Milestone Task Road Trip! CCSS-M Standards 5.NBT.7 5NF.5a, 5.NF.5b 5.NBT.7 5.NF.5a, 5.NF.5b 5.NBT.7 5.NF.5a, 5.NF.5b 5.NBT.7 5.NF.5a, 5.NF.5b Brief Description of Task Students will calculate the total cost of apples, and also the total number of bananas they can purchase. They will write out how they know their answers are correct. Students find all the possibilities for missing numbers in a multiplication problem with decimals (. * 0.5 = 0. ). Students will write out all of the equations they come up with. Students notice patterns and justify their thinking around finding all possibilities. Students use their understanding of area models and the relationship between division and multiplication to find possible dimensions for a quilt with an area of 5.76 square meters. Students use multiplication and division of decimals to find kilometers traveled, gallons of gas used, and cost of gas in the context of a road trip. Students also show understanding of multiplication and division of decimals in context-free constructed response questions and in finding the errors in an example calculation. Source SFUSD Teacher Created (2014) SFUSD Teacher Created, modified from Georgia Department of Education, Fifth Grade Mathematics Unit 3, pp SFUSD Teacher Created (2014) SFUSD Teacher Created (2014)
11 11 Farmer s Market How will students do this? Focus Standards for Mathematical Practice: 4. Model with mathematics. 5. Use appropriate tools strategically Structures for Student Learning: Academic Language Support: Vocabulary: cost per pound, farmer s market Sentence frames: First I The way I figured out my answer was correct was to... Differentiation Strategies: Intervention: Have a discussion of how to multiply and divide decimals with whole numbers as learned in Unit 4. Extension: Have students write their own problems, switch with a partner, and solve theirs. Participation Structures (group, partners, individual, other): The students should work individually.
12 Lesson Series #1 12 Lesson Series Overview: This lesson series begins with the use of base 10 blocks to model multiplication of decimals. It continues on with the practice of multiplication of decimals, while students begin to notice how factors affect an answer. Students will also practice decimal placement in multiplication problems, and see scaling effects on multiplication products. On daily exit slips, students work on noticing errors in example calculations. In preparation for the Apprentice Task, students work to find different ways to complete a CCSS-M Standards Addressed: 5.NBT.7, 5.NF.5a, 5.NF.5b Time: 4 days Lesson Overview Day 1 Description of Lesson: Students work in pairs to use base ten blocks to build rectangles that model the multiplication of decimals. To prepare to teach this lesson, read the BACKGROUND KNOWLEDGE: Representing decimal multiplication with base ten blocks portion of the Georgia Department of Education lesson (see Resources). When making rectangles to represent decimal multiplication, you are actually using the length and the width of each block to represent the factors. Therefore, a flat is actually 1 unit by 1 unit, a long is 1 unit by 0.1 unit, and a unit block is 0.1 unit by 0.1 unit (area model). The lesson should begin with a discussion of the area model of multiplication: the lengths of the sides of the rectangle are the factors in the problem; the area of the rectangle is the product. As a class, practice with the problem 3.2 * 2.4, filling in the area with base ten blocks and finding the product. Make sure to clarify that the area of the Then have students work in pairs to solve the other problems on the sheet. Some students may need to use the blocks all day. Encourage these students to draw the rectangles after building it. Others may be able to use drawings only. There may be some who notice a structure to the problems building off the work in Unit 5.4 who may not need to draw out the rectangles. Students should be able to support their answers with the area model whether or not they use it to solve every problem. Resources Georgia Department of Education, Fifth Grade Mathematics Unit 3, pp Base Ten Activity Handout Base ten blocks, enough for pairs of students Multiplication Error analysis exit slip Homework: Picturing Multiplication Key to Decimals, Book 2: Adding, Subtracting and Multiplying, pp. 25, 26, 34 Notes: The original lesson includes a couple of division problems. Do not use these today.
14 14 Have students do the Multiplication Error Analysis Exit Slip Hand out Exit Slips and have students show their work to correct the mistakes. Provide a brief note about what the student did wrong on the sample. Notes: Calculators are necessary for this activity. If you do not have enough calculators for each student, have them work in pairs or groups. Lesson Overview Day 3 Description of Lesson: Mini-lesson: Use of a multiplication algorithm Demonstrate and explain to students how to use a multiplication algorithm to multiply decimals by decimals. Where s the Point? (modified from Nimble with Numbers pp ) See Where s the Point? Direction sheet Students will work in small groups to figure out where to place the decimal point in decimal sentences. The first part of the activity students will place the decimal in the factor parts of a decimal sentence, given the correctly placed decimal in the product. The second part of the activity students will place the decimal in the product of a decimal sentence, given the correctly placed decimals in the factors. Resources Where s the Point? modified from Nimble with Numbers pp by Leigh Childs and Laura Choate Where s the Point? direction sheet Where s the Point? student handout Multiplication Error Analysis Homework: Everyday Mathematics Grade 5 Study Links 2.7 and 2.8 Multiplication Error Analysis Exit Slip Hand out second Exit Slips and have students show their work to correct the mistakes. Provide and a brief note about what the student did wrong on the sample. Notes: Intervention: Have students do fewer problems. Extension: Have students write out how they solved one problem from the first set of sentences. Have students create their own decimal sentence problems, and have partners solve them.
15 15 Lesson Overview Day 4 Description of Lesson: Part 1: Missing Numbers Lesson Introduce the concept of filling in the blanks with numbers that make the number sentence true. Show the blank equation, and hand out the Missing Number Worksheet. x. =. For this question, they should only use the numbers 0 5. Allow them to find as many options as possible to fill the blanks. Encourage students to seek out rules or patterns that will help them find multiple possibilities and determine whether they have all the possibilities. Probe with questions like: How did you get your answer? How do you know your answer is correct? What patterns are you noticing? Do you have all the possibilities? How do you know? When students believe they have all the possibilities (you may want to check), they should write a paragraph answering these questions: How did you find your solutions? How do you know you have all the possibilities? Have students share out their solutions and how they proved the number sentence made sense. Students should try to find as many solutions as possible and to share strategies with the class as to how they came up with their solution. Resources Missing Number Worksheet SFUSD Teacher Created Multiplication Error Analysis Sorting Products - Number Sense Homework: Key to Decimals, Book 2: Adding, Subtracting and Multiplying, p. 38 Part 2: Sorting Products If you have time after the Missing Numbers work, or as a homework assignment, have students do the first half of Activity 3 of the Sorting Products lesson. In this activity, students order products without calculating, only looking at the factors. If sorted correctly, the letters that correspond with the products will spell a word. Make sure to do the first step of the Using the Activities section as a class so that students know how to complete the activity. Activities 1 and 2 and the second half of 3 can be used as extensions. Multiplication Error Analysis Exit Slip Hand out the third Exit Slips and have students show their work to correct the mistakes. Provide a brief note about what the student did wrong on the sample.
17 17 Missing Numbers How will students do this? Focus Standards for Mathematical Practice: 2. Reason abstractly and quantitatively. 7. Look for and make use of structure. Structures for Student Learning: Academic Language Support: Vocabulary: factors, product Sentence frames: I started by worked because didn t work because I know we got all the possibilities because Differentiation Strategies: Extension: Students can explore further problems like this, finding missing numbers when multiplying by a number with decimals. For example: o. * 1.5 =. Intervention: Provide students with examples with one whole number before using two decimal factors. Include some examples that only have one solution before moving to multiple solutions. Also, calculators can be used for this task. Students can use digit cards to generate random numbers to try. Participation Structures (group, partners, individual, other): Students work in pairs.
18 Lesson Series #2 18 Lesson Series Overview: This lesson series begins with the use of base ten blocks to model division of decimals. It continues on with practice of both multiplication and division of decimals, using their knowledge of scaling to find the highest answer. Students will also practice magnitude estimates before calculating quotients. Finally, students will show their proficiency in both multiplication and division of decimals by identifying errors and correcting them in student work samples. CCSS-M Standards Addressed: 5.NBT.7, 5.NF.5a, 5.NF.5b Time: 4 days Lesson Overview Day 1 Description of Lesson: Students work in pairs to use base ten blocks to model division with decimals. This lesson mirrors the first lesson of Lesson Series #1. To prepare to teach this lesson, read the BACKGROUND KNOWLEDGE: Representing decimal division with base ten blocks portion of the Georgia Department of Education lesson (see Resources). When making rectangles to represent decimal multiplication, you are actually using the length and the width of each block to represent the factors. Therefore, a flat is actually 1 unit by 1 unit, a long is 1 unit by 0.1 unit, and a unit block is 0.1 unit by 0.1 unit (area model). Resources Georgia Department of Education, Fifth Grade Mathematics Unit 3, pp Base Ten Division Base Ten Division Student Sheet Division Error Analysis Exit Slip Homework Day 1 (SFUSD Teacher Created, 2014) The lesson should begin with a discussion of the rectangle area model of division. This focuses on finding a missing factor in a multiplication problem. As a class, practice with the two problems demonstrated in the Background Knowledge section. The first, 3.6 : 1.2, has students make groups of 1.2 and figure out how many of those can be made with a set of 3.6. The second, 4.83 : 2.1, has students make a rectangle with one side of 2.1 and find the length of the other side. Then have students work in pairs to solve the other problems on the sheet. Some students may need to use the blocks all day. Encourage these students to draw the rectangle or groups after building. Others may be able to use drawings only. There may be some who notice a structure to the problems building off the work in Unit 5.4 who may not need to draw out the models. Students should be able to support their answers
19 19 with the models whether or not they use it to solve every problem. Division Error Analysis Exit Slip Hand out the first Exit Slips and have students show their work to correct the mistakes, and a brief note about what the student did wrong on the sample. Notes: This should build on the work students did on the first day of the first lesson series. Lesson Overview Day 2 Description of Lesson: Students work individually or in pairs to try to make the largest number possible by going through a number maze. They use the properties of multiplication and division to make their number larger, at the same time noticing how different factors and divisors affect the result. Be sure to do all three portions of the lesson to help students get full understanding from this activity. Resources Maze Scaling Directions Maze Scaling Multiplications and Division.pdf Division Error Analysis Exit Slip Homework: Another Method for Dividing Key to Decimals, Book 3: Dividing pp. 27, Whole Class: Write the problem (as described next) on the chalkboard or overhead. Ask students to discuss what they notice. 4.5 : 0.9 = 0.50 Lead a discussion that focuses on these key points: In computing the quotient of 4.5 and 0.9, a student calculated that the answer is Reflection on the answer should have caused the student to realize the product was too small. Dividing 4.5 by a number slightly less than 1 produces an answer a little more than 4.5. Instead, this student applied an incorrect procedure (making the number of digits behind the decimal point correspond) and did not reflect on whether the resulting answer was reasonable. Remind students of the activity in the previous lesson series where they had to make the largest number using the multiplication maze. Explain that they will do the same today using multiplication and division to modify their original number. Encourage students to trace several paths through the maze while always looking for the path that will yield the greatest increase in the calculator's display. 2. Individual/pairs/groups: (depending on how many calculators you have available) Give each student a calculator and a copy of the Maze Scaling Activity - Multiplication and Division activity sheet.
20 20 Students are to choose a path through the maze. To begin, have the students enter 100 on their calculator. For each segment chosen on the maze, the students should key in the assigned operation and number. The goal is to choose a path that results in the largest value at the finish of the maze. Students may not retrace a path or move upward in the maze. In pairs or in groups of three, students should discuss their strategies (after playing the game) and what worked best for them. 1. Whole Class: As a class, discuss what happens to your total when you divide by various numbers on the maze, greater and less than one. Guide them to notice when the quotient is larger or smaller than the dividend. How is this similar to and different from multiplication? Possible follow-up activities include finding the path that leads to the smallest finish number or finding a path that leads to a finish number as near the start number (100) as possible. 2. Individual: Division Error Analysis Exit Slip Handout the second exit slip and have students show their work to correct the mistakes, and a brief note about what the student did wrong on the sample. *Homework day 2 Notes: Calculators are necessary for this activity. If you do not have enough calculators for each student, have them work in pairs or groups. Lesson Overview Day 3 Description of Lesson: Mini-lesson: Division of Decimals by Decimals Demonstrate and explain to students how to use a division algorithm to divide decimals by decimals. Magnitude Estimates before Calculating Quotients - Everyday Mathematics Grade 5, Lesson 4.5 Division Error Analysis Exit Slip Hand out the third Exit Slips and have students show their work to correct the mistakes, and a brief note about what the student did wrong on the sample. Resources Everyday Mathematics Grade 5, Lesson 4.5 pp Division Error Analysis Exit Slip Homework: Division Error Analysis Key to Decimals, Book 3: Dividing p. 41 Everyday Mathematics Grade 5 Study Links 4.5
21 21 Notes: In Class: Use Part 1 (the math message, making magnitude estimates before calculating quotients). Part 2 (playing Division Top-It, math boxes 4.5 (box 4 and 5) Homework: Study Link 4.5 (Math Masters p.113) Lesson Overview Day 4 Description of Lesson: Do You See An Error? Students will work individually to analyze the student samples to determine decimal division errors. Encourage students to be sure they have discovered all errors in the samples. Focus student thinking by asking: What happens when we divide a decimal by a decimal? How can we check for errors in division of decimals? When students have found all errors, they should find the correct solution using two methods. Then they should write a paragraph explaining what the student did wrong, and the steps they need to take next time to find the correct solution. Resources Georgia Department of Education Fifth Grade Mathematics Unit 3 pp Homework: Unit Pricing Key to Decimals, Book 3: Dividing p. 40 Notes: Students should work individually on this activity.
23 23 Focus Standards for Mathematical Practice: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 4. Model with mathematics. Materials Base Ten Blocks Base Ten Grid Paper Calculators (optional) Structures for Student Learning: Academic Language Support: Vocabulary: array, area, length, width, dimensions Sentence frames: First I tried I noticed that and it helped me worked because didn t work because... Sewing A Quilt How will students do this? Differentiation Strategies: Extension: Have students find as many possible dimensions as they can. Intervention: Calculators can be used for this task. Provide students with Divisibility Rules (Student Reference Book p 11) Participation Structures (group, partners, individual, other): Students can work in pairs or triads.
25 25 Notes: Students should only select 1 city per day, do not allow them to travel through cities to get to their final day s destination. (e.g., Do not go S.F. to Denver. Day 1; they should go S.F. to Las Vegas. Then Day 2 they would go Las Vegas to Denver.) Modifications: Have students make calculations for one city at a time. Allow groups to select less than five cities as appropriate. Have students use calculators to check their calculations. Give students copies of tables provided on teacher reference sheet. Extension: Have groups who finish all questions create a poster representing their road trip. *A car would usually have a higher rate than 9.6 kilometers per gallon. However, in order to keep the divisor a two-digit number, a low rate was selected. Description of Lesson: See Day 1 description of the lesson. Notes: See Day 1 notes. Description of Lesson: See Day 1 description of lesson. Notes: See Day 1 notes. Lesson Overview Day 2 Lesson Overview Day 3 See resources for Day 1. Resources Homework: Solve Each Problem Key to Decimals, Book 3: Dividing p. 12 See resources for Day 1. Resources Homework: Unit Pricing Key to Decimals, Book 3: Dividing pp
26 Milestone Task CLA2: Road Trip! and Constructed Responses What will students do? 26 Mathematics Objectives and Standards Math Objectives: Students will efficiently solve multiplication problems with decimals. CCSS-M Standards Addressed: 5.NBT.7, 5.NF.5a, 5.NF.5b Potential Misconceptions When multiplying the product is always larger than both factors: This misconception should be familiar to students from their work multiplying fractions by whole numbers in fourth grade and from work during Lesson Series 1. When one (or more) of the factors is less than one, the product will be less than the other factor. When dividing the quotient is always smaller than the dividend: this misconception should be familiar to students from their work in lesson series 2. When the divisor is smaller than 1, the quotient will be larger than the dividend. Framing Student Experience Launch: Remind students of their work in Lesson Series 3 calculating the number of kilometers traveled, the gas consumed, and the cost of gas. Today they will solve one set of problems using this context, plus three stand alone problems: a multiplication problem, a division problem, and a find the error problem (as they did in lesson series 1 and 2). During: Students work independently to solve the problems. They may use any of the tools that they worked with during this unit. Closure/Extension: When all students have finished, you may decide to review the answers as a group.
27 27 Focus Standards for Mathematical Practice: 2. Reason abstractly and quantitatively. 6. Attend to precision. Structures for Student Learning: Academic Language Support: Vocabulary: miles, kilometers, gallons Sentence frames: none CLA2: Road Trip! and Constructed Responses How will students do this? Differentiation Strategies: Students may use any of the manipulatives, drawings, or tools they have used during this unit, other than calculators. Participation Structures (group, partners, individual, other): Individual
### Carroll County Public Schools Elementary Mathematics Instructional Guide (5 th Grade) Unit #4 : Division of Whole Numbers and Measurement
Background Information and Research By fifth grade, students should understand that division can mean equal sharing or partitioning of equal groups or arrays. They should also understand that it is the
### Numbers and Operations in Base 10 and Numbers and Operations Fractions
Numbers and Operations in Base 10 and Numbers As the chart below shows, the Numbers & Operations in Base 10 (NBT) domain of the Common Core State Standards for Mathematics (CCSSM) appears in every grade
### Multiplying and Dividing Decimals ALPHA VERSION OCTOBER 2012 Grade 5
Multiplying and Dividing Decimals ALPHA VERSION OCTOBER 2012 Grade 5 Mathematics Formative Assessment Lesson Designed by Kentucky Department of Education Mathematics Specialists to be Field-tested by Kentucky
### Division with Whole Numbers and Decimals
Grade 5 Mathematics, Quarter 2, Unit 2.1 Division with Whole Numbers and Decimals Overview Number of Instructional Days: 15 (1 day = 45 60 minutes) Content to be Learned Divide multidigit whole numbers
### Overview. Essential Questions. Grade 4 Mathematics, Quarter 4, Unit 4.1 Dividing Whole Numbers With Remainders
Dividing Whole Numbers With Remainders Overview Number of instruction days: 7 9 (1 day = 90 minutes) Content to Be Learned Solve for whole-number quotients with remainders of up to four-digit dividends
### A Correlation of. to the. Common Core State Standards for Mathematics Grade 4
A Correlation of to the Introduction envisionmath2.0 is a comprehensive K-6 mathematics curriculum that provides the focus, coherence, and rigor required by the CCSSM. envisionmath2.0 offers a balanced
### Vocabulary, Signs, & Symbols product dividend divisor quotient fact family inverse. Assessment. Envision Math Topic 1
1st 9 Weeks Pacing Guide Fourth Grade Math Common Core State Standards Objective/Skill (DOK) I Can Statements (Knowledge & Skills) Curriculum Materials & Resources/Comments 4.OA.1 4.1i Interpret a multiplication
### 1 BPS Math Year at a Glance (Adapted from A Story of Units Curriculum Maps in Mathematics P-5)
Grade 5 Key Areas of Focus for Grades 3-5: Multiplication and division of whole numbers and fractions-concepts, skills and problem solving Expected Fluency: Multi-digit multiplication Module M1: Whole
### 5 th Grade Common Core State Standards. Flip Book
5 th Grade Common Core State Standards Flip Book This document is intended to show the connections to the Standards of Mathematical Practices for the content standards and to get detailed information at
### Unit 1: Place value and operations with whole numbers and decimals
Unit 1: Place value and operations with whole numbers and decimals Content Area: Mathematics Course(s): Generic Course Time Period: 1st Marking Period Length: 10 Weeks Status: Published Unit Overview Students
### Just want the standards alone? You can find the standards alone at
5 th Grade Mathematics Unpacked Content For the new Common Core State Standards that will be effective in all North Carolina schools in the 2012-13 school year. This document is designed to help North
September: UNIT 1 Place Value Whole Numbers Fluently multiply multi-digit numbers using the standard algorithm Model long division up to 2 digit divisors Solve real world word problems involving measurement
### Grade 4 Multi-Digit Multiplication and Division with One Divisor Unit of Instruction
Grade 4 Multi-Digit Multiplication and Division with One Divisor Unit of Instruction This is a progressive unit of instruction beginning with students investigating the concrete area model of fraction
### NS5-38 Remainders and NS5-39 Dividing with Remainders
:1 PAGE 89-90 NS5-38 Remainders and NS5-39 Dividing with Remainders GOALS Students will divide with remainders using pictures, number lines and skip counting. Draw: 6 3 = 2 7 3 = 2 Remainder 1 8 3 = 2
### Unit 6 Number and Operations in Base Ten: Decimals
Unit 6 Number and Operations in Base Ten: Decimals Introduction Students will extend the place value system to decimals. They will apply their understanding of models for decimals and decimal notation,
### This lesson introduces students to decimals.
NATIONAL MATH + SCIENCE INITIATIVE Elementary Math Introduction to Decimals LEVEL Grade Five OBJECTIVES Students will compare fractions to decimals. explore and build decimal models. MATERIALS AND RESOURCES
### BPS Math Year at a Glance (Adapted from A Story Of Units Curriculum Maps in Mathematics K-5) 1
Grade 4 Key Areas of Focus for Grades 3-5: Multiplication and division of whole numbers and fractions-concepts, skills and problem solving Expected Fluency: Add and subtract within 1,000,000 Module M1:
### Math-U-See Correlation with the Common Core State Standards for Mathematical Content for Fourth Grade
Math-U-See Correlation with the Common Core State Standards for Mathematical Content for Fourth Grade The fourth-grade standards highlight all four operations, explore fractions in greater detail, and
### Mathematics. Mathematical Practices
Mathematical Practices 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with
### Graphic Organizers SAMPLES
This document is designed to assist North Carolina educators in effective instruction of the new Common Core State and/or North Carolina Essential Standards (Standard Course of Study) in order to increase
### Progressing toward the standard
Report Card Language: add, subtract, multiply, and/or divide to solve multi-step word problems. CCSS: 4.OA.3 Solve multistep work problems posed with whole numbers and having whole-number answers using
### Grade 5 Common Core State Standard
2.1.5.B.1 Apply place value concepts to show an understanding of operations and rounding as they pertain to whole numbers and decimals. M05.A-T.1.1.1 Demonstrate an understanding that 5.NBT.1 Recognize
### Sample Fraction Addition and Subtraction Concepts Activities 1 3
Sample Fraction Addition and Subtraction Concepts Activities 1 3 College- and Career-Ready Standard Addressed: Build fractions from unit fractions by applying and extending previous understandings of operations
### Classworks Common Core Transition Guide. South Carolina 4th Grade Mathematics
Classworks Common Core Transition Guide South Carolina 4th Mathematics OFFICIALLY ENDORSED Classworks Common Core Transition Guide - BETA Making The Transition With the increased rigor, the transition
### CCSS Mathematics Implementation Guide Grade 5 2012 2013. First Nine Weeks
First Nine Weeks s The value of a digit is based on its place value. What changes the value of a digit? 5.NBT.1 RECOGNIZE that in a multi-digit number, a digit in one place represents 10 times as much
### Place Value (What is is the Value of of the the Place?)
Place Value (What is is the Value of of the the Place?) Second Grade Formative Assessment Lesson Lesson Designed and revised by Kentucky Department of Education Mathematics Specialists Field-tested by
### The Crescent Primary School Calculation Policy
The Crescent Primary School Calculation Policy Examples of calculation methods for each year group and the progression between each method. January 2015 Our Calculation Policy This calculation policy has
Subtraction from Addition Objectives To review the and 1 shortcuts; and to guide children to identify the subtraction facts related to given addition facts. www.everydaymathonline.com epresentations etoolkit
### PA Common Core Standards Standards for Mathematical Practice Grade Level Emphasis*
Habits of Mind of a Productive Thinker Make sense of problems and persevere in solving them. Attend to precision. PA Common Core Standards The Pennsylvania Common Core Standards cannot be viewed and addressed
### COURSE OF STUDY UNIT PLANNING GUIDE
COURSE OF STUDY UNIT PLANNING GUIDE FOR: MATHEMATICS GRADE LEVEL: FOURTH GRADE PREPARED BY: TERESA KELLY AND TRACY BUCKLEY B.O.E. ADOPTED AUGUST 20, 2015 REVISED AUGUST 2015 ALIGNED TO THE 2014 NJCCCS
### Unit 1 Number Sense. In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions.
Unit 1 Number Sense In this unit, students will study repeating decimals, percents, fractions, decimals, and proportions. BLM Three Types of Percent Problems (p L-34) is a summary BLM for the material
### 4 th Grade Mathematics Unpacked Content
4 th Grade Mathematics Unpacked Content This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these
### The Partial-Quotients Division Algorithm, Part 1
The Partial-Quotients Division Algorithm, Part 1 Objectives To introduce and provide practice with a low-stress division algorithm for 1-digit divisors. www.everydaymathonline.com epresentations etoolkit
### Decomposing Numbers (Operations and Algebraic Thinking)
Decomposing Numbers (Operations and Algebraic Thinking) Kindergarten Formative Assessment Lesson Designed and revised by Kentucky Department of Education Mathematics Specialists Field-tested by Kentucky
### Integer Operations. Overview. Grade 7 Mathematics, Quarter 1, Unit 1.1. Number of Instructional Days: 15 (1 day = 45 minutes) Essential Questions
Grade 7 Mathematics, Quarter 1, Unit 1.1 Integer Operations Overview Number of Instructional Days: 15 (1 day = 45 minutes) Content to Be Learned Describe situations in which opposites combine to make zero.
### Excel Math Fourth Grade Standards for Mathematical Practice
Excel Math Fourth Grade Standards for Mathematical Practice The Common Core State Standards for Mathematical Practice are integrated into Excel Math lessons. Below are some examples of how we include these
### Decimal Notations for Fractions Number and Operations Fractions /4.NF
Decimal Notations for Fractions Number and Operations Fractions /4.NF Domain: Cluster: Standard: 4.NF Number and Operations Fractions Understand decimal notation for fractions, and compare decimal fractions.
### Multiplication Unit Plan. Katie Kramer Spring 2013
1 Multiplication Unit Plan Katie Kramer Spring 2013 2 Class: Fifth Grade math with integration of Language Arts. Background: This unit covers the topic of multiplication in math. This unit asks students
### Operations and Algebraic Thinking Represent and solve problems involving multiplication and division.
Performance Assessment Task The Answer is 36 Grade 3 The task challenges a student to use knowledge of operations and their inverses to complete number sentences that equal a given quantity. A student
### The Lattice Method of Multiplication
The Lattice Method of Multiplication Objective To review and provide practice with the lattice method for multiplication of whole numbers and decimals. www.everydaymathonline.com epresentations etoolkit
### Ministry of Education. The Ontario Curriculum Exemplars Grade 6. Mathematics. Samples of Student Work: A Resource for Teachers
Ministry of Education The Ontario Curriculum Exemplars Grade 6 Mathematics Samples of Student Work: A Resource for Teachers 2002 Contents Introduction......................................................
### N Q.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities.
Performance Assessment Task Swimming Pool Grade 9 The task challenges a student to demonstrate understanding of the concept of quantities. A student must understand the attributes of trapezoids, how to
### Georgia Standards of Excellence Curriculum Frameworks. Mathematics. GSE Fifth Grade Unit 3: Multiplying and Dividing with Decimals
Georgia Standards of Excellence Curriculum Frameworks Mathematics GSE Fifth Grade Unit 3: Multiplying and Dividing with Decimals These materials are for nonprofit educational purposes only. Any other use
### Mathematics Grade 5 Year in Detail (SAMPLE)
Unit 1 Unit 2 Unit 3 Unit 4 Unit 5 Unit 6 Unit 7 Unit 8 Unit 9 Unit 10 Whole number operations Place value with decimals Add and Subtract Decimals Add and Subtract Fractions Multiply and Divide Decimals
### 5th Grade Unit 1: Whole Number and Decimal Fraction Place Value to the One Thousandths (4 Weeks)
5th Grade Unit : Whole Number and Decimal Fraction Place Value to the One Thousandths (4 Weeks) Stage Desired Results Established Goals Unit Description Students continue to extend and apply their understanding
### TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers
TYPES OF NUMBERS When two or more integers are multiplied together, each number is a factor of the product. Nonnegative integers that have exactly two factors, namely, one and itself, are called prime
### Grade 5 Math Content 1
Grade 5 Math Content 1 Number and Operations: Whole Numbers Multiplication and Division In Grade 5, students consolidate their understanding of the computational strategies they use for multiplication.
### MACMILLAN/McGRAW-HILL. MATH CONNECTS and IMPACT MATHEMATICS WASHINGTON STATE MATHEMATICS STANDARDS. ESSENTIAL ACADEMIC LEARNING REQUIREMENTS (EALRs)
MACMILLAN/McGRAW-HILL MATH CONNECTS and IMPACT MATHEMATICS TO WASHINGTON STATE MATHEMATICS STANDARDS ESSENTIAL ACADEMIC LEARNING REQUIREMENTS (EALRs) And GRADE LEVEL EXPECTATIONS (GLEs) / Edition, Copyright
### Introduce Decimals with an Art Project Criteria Charts, Rubrics, Standards By Susan Ferdman
Introduce Decimals with an Art Project Criteria Charts, Rubrics, Standards By Susan Ferdman hundredths tenths ones tens Decimal Art An Introduction to Decimals Directions: Part 1: Coloring Have children
### Georgia Standards of Excellence Grade Level Curriculum Overview. Mathematics. GSE Fifth Grade
Georgia Standards of Excellence Grade Level Curriculum Overview Mathematics GSE Fifth Grade These materials are for nonprofit educational purposes only. Any other use may constitute copyright infringement.
### Objective To guide the development and use of a rule for generating equivalent fractions. Family Letters. Assessment Management
Equivalent Fractions Objective To guide the development and use of a rule for generating equivalent fractions. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game
### 5 Mathematics Curriculum
New York State Common Core 5 Mathematics Curriculum G R A D E GRADE 5 MODULE 1 Topic C Place Value and Rounding Decimal Fractions 5.NBT.4 Focus Standard: 5.NBT.4 Use place value understanding to round
### Multiplication and Division of Decimals. Suggested Time: 3
Multiplication and Division of Decimals Suggested Time: 3 1 2 Weeks 225 Unit Overview Focus and Context Math Connects This unit will draw upon students previous knowledge of multiplication and division
### Math Common Core Standards Fourth Grade
Operations and Algebraic Thinking (OA) Use the four operations with whole numbers to solve problems. OA.4.1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement
### Year 1. Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on.
Year 1 Add with numbers up to 20 Use numbered number lines to add, by counting on in ones. Encourage children to start with the larger number and count on. +1 +1 +1 Children should: Have access to a wide
### Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem. Solve word problems that call for addition of three whole numbers
### Mathematical Proficiency By Kenneth Danley Principal, Starline Elementary School1
W e know that parents are eager to help their children be successful in school. Many parents spend countless hours reading to and with their children. This is one of the greatest contributions a parent
### #1 Make sense of problems and persevere in solving them.
#1 Make sense of problems and persevere in solving them. 1. Make sense of problems and persevere in solving them. Interpret and make meaning of the problem looking for starting points. Analyze what is
### Mathematics Instructional Cycle Guide
Mathematics Instructional Cycle Guide Fractions on the number line 3NF2a Created by Kelly Palaia, 2014 Connecticut Dream Team teacher 1 CT CORE STANDARDS This Instructional Cycle Guide relates to the following
### 5 Mathematics Curriculum
New York State Common Core 5 Mathematics Curriculum G R A D E GRADE 5 MODULE 1 Topic B Decimal Fractions and Place Value Patterns 5.NBT.3 Focus Standard: 5.NBT.3 Read, write, and compare decimals to thousandths.
A Correlation of to the Minnesota Academic Standards Grades K-6 G/M-204 Introduction This document demonstrates the high degree of success students will achieve when using Scott Foresman Addison Wesley
### Math. MCC5.OA.1 Use parentheses, brackets, or braces in. these symbols. 11/5/2012 1
MCC5.OA.1 Use parentheses, brackets, or braces in numerical expressions, and evaluate expressions with these symbols. 11/5/2012 1 MCC5.OA.2 Write simple expressions that record calculations with numbers,
### GRADE 5 SKILL VOCABULARY MATHEMATICAL PRACTICES Evaluate numerical expressions with parentheses, brackets, and/or braces.
Common Core Math Curriculum Grade 5 ESSENTIAL DOMAINS AND QUESTIONS CLUSTERS Operations and Algebraic Thinking 5.0A What can affect the relationship between numbers? round decimals? compare decimals? What
### What Is Singapore Math?
What Is Singapore Math? You may be wondering what Singapore Math is all about, and with good reason. This is a totally new kind of math for you and your child. What you may not know is that Singapore has
### Common Core Standards for Mathematics Grade 4 Operations & Algebraic Thinking Date Taught
Operations & Algebraic Thinking Use the four operations with whole numbers to solve problems. 4.OA.1. Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 7 as a statement that 35
### Stage 1: Integrate essential questions, big ideas and learning targets, and ensure it can be differentiated and assessed
Grade 5 Math Unit Title Time frame 21 st Century Themes Interdisciplinary focus and technology integration Unit 1: Understanding Place Value 3 weeks Critical Thinking and Problem Solving Communication
### Composing and Decomposing Whole Numbers
Grade 2 Mathematics, Quarter 1, Unit 1.1 Composing and Decomposing Whole Numbers Overview Number of instructional days: 10 (1 day = 45 60 minutes) Content to be learned Demonstrate understanding of mathematical
### Swavesey Primary School Calculation Policy. Addition and Subtraction
Addition and Subtraction Key Objectives KS1 Foundation Stage Say and use number names in order in familiar contexts Know that a number identifies how many objects in a set Count reliably up to 10 everyday
### Performance Assessment Task Baseball Players Grade 6. Common Core State Standards Math - Content Standards
Performance Assessment Task Baseball Players Grade 6 The task challenges a student to demonstrate understanding of the measures of center the mean, median and range. A student must be able to use the measures
### 4TH GRADE FIRST QUARTER MATHEMATICS STANDARDS. Vocabulary. answers using mental computation and estimation strategies including rounding.
4TH GRADE FIRST QUARTER MATHEMATICS STANDARDS Critical Area: Developing understanding and fluency with multi-digit multiplication, and developing understanding of dividing to find quotients involving multi-digit
### Grade 7 Mathematics. Unit 3. Fractions, Decimals, and Percents. Estimated Time: 23 Hours
Grade 7 Mathematics Fractions, Decimals, and Percents Estimated Time: 23 Hours [C] Communication [CN] Connections [ME] Mental Mathematics and Estimation [PS] Problem Solving [R] Reasoning [T] Technology
### Grade 5 Math PA Core Standards - CC.2: PA Core: Mathematics. Eligible Content
Grade 5 Math PA Core Standards - CC.2: PA Core: Mathematics Eligible Content Standard Area - CC.2.1: Numbers and Operations Standard - CC.2.1.5.B.1: Apply place value to show an understanding of operations
### Grade: 5 Strand: Number Outcome: 6
Division of Whole Numbers: Assessment Task Grade: 5 Strand: Number Outcome: 6 SPECIFIC LEARNER OUTCOME Number N6 Demonstrate, with and without concrete materials, an understanding of division (3-digit
### Multiplication Fact Power and Shortcuts
Multiplication Fact Power and Shortcuts Objectives To discuss multiplication facts and the importance of fact power; and to review fact shortcuts. www.everydaymathonline.com epresentations etoolkit Algorithms
### Volume of Rectangular Prisms Objective To provide experiences with using a formula for the volume of rectangular prisms.
Volume of Rectangular Prisms Objective To provide experiences with using a formula for the volume of rectangular prisms. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts
### Objectives To review making ballpark estimates; and to review the counting-up and trade-first subtraction algorithms. materials. materials.
Objectives To review making ballpark estimates; and to review the counting-up and trade-first subtraction algorithms. Teaching the Lesson materials Key Activities Children make ballpark estimates for -digit
### Describing and Solving for Area and Perimeter
Grade 3 Mathematics, Quarter 2,Unit 2.2 Describing and Solving for Area and Perimeter Overview Number of instruction days: 8-10 (1 day = 90 minutes) Content to Be Learned Distinguish between linear and
### Mathematical goals. Starting points. Materials required. Time needed
Level N of challenge: B N Mathematical goals Starting points Materials required Time needed Ordering fractions and decimals To help learners to: interpret decimals and fractions using scales and areas;
### Grade 4 Unit 3: Multiplication and Division; Number Sentences and Algebra
Grade 4 Unit 3: Multiplication and Division; Number Sentences and Algebra Activity Lesson 3-1 What s My Rule? page 159) Everyday Mathematics Goal for Mathematical Practice GMP 2.2 Explain the meanings
### Fourth Grade Math Standards and "I Can Statements"
Fourth Grade Math Standards and "I Can Statements" Standard - CC.4.OA.1 Interpret a multiplication equation as a comparison, e.g., interpret 35 = 5 x 7 as a statement that 35 is 5 times as many as 7 and
### Performance Assessment Task Leapfrog Fractions Grade 4 task aligns in part to CCSSM grade 3. Common Core State Standards Math Content Standards
Performance Assessment Task Leapfrog Fractions Grade 4 task aligns in part to CCSSM grade 3 This task challenges a student to use their knowledge and understanding of ways of representing numbers and fractions
### Objective To introduce the concept of square roots and the use of the square-root key on a calculator. Assessment Management
Unsquaring Numbers Objective To introduce the concept of square roots and the use of the square-root key on a calculator. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts
### Exploring Length, Area, and Attributes
Explorations Exploring Length, Area, and Attributes Objectives To guide children as they measure lengths and distances to the nearest inch and centimeter, explore area by tiling surfaces, and sort attribute
### Performance Assessment Task Picking Fractions Grade 4. Common Core State Standards Math - Content Standards
Performance Assessment Task Picking Fractions Grade 4 The task challenges a student to demonstrate understanding of the concept of equivalent fractions. A student must understand how the number and size
### Mathematics Instructional Cycle Guide
Mathematics Instructional Cycle Guide Dividing Fractions 6.NS.1 Created by Mary Kay Rendock, 2014 Connecticut Dream Team Teacher 1 CT CORE STANDARDS This Instructional Cycle Guide relates to the following
### GRADE 6 MATH: SHARE MY CANDY
GRADE 6 MATH: SHARE MY CANDY UNIT OVERVIEW The length of this unit is approximately 2-3 weeks. Students will develop an understanding of dividing fractions by fractions by building upon the conceptual
### Planning For Success Mathematics: Numeration Inquiry Investigations. Operations: Multiplication and Division. Number Sense and Numeration
Planning For Success Mathematics: Numeration Inquiry Investigations Operations: Multiplication and Division Number Sense and Numeration OVERALL EXPECTATIONS By the end of Grade 4, students will: solve
### Georgia Standards of Excellence Grade Level Curriculum Overview. Mathematics. GSE Fourth Grade
Georgia Standards of Excellence Grade Level Curriculum Overview Mathematics GSE Fourth Grade These materials are for nonprofit educational purposes only. Any other use may constitute copyright infringement.
### Change Number Stories Objective To guide children as they use change diagrams to help solve change number stories.
Number Stories Objective To guide children as they use change diagrams to help solve change number stories. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game
### The Distributive Property
The Distributive Property Objectives To recognize the general patterns used to write the distributive property; and to mentally compute products using distributive strategies. www.everydaymathonline.com
### 1. I have 4 sides. My opposite sides are equal. I have 4 right angles. Which shape am I?
Which Shape? This problem gives you the chance to: identify and describe shapes use clues to solve riddles Use shapes A, B, or C to solve the riddles. A B C 1. I have 4 sides. My opposite sides are equal.
### Common Core State Standards for Math Grades K - 7 2012
correlated to the Grades K - 7 The Common Core State Standards recommend more focused and coherent content that will provide the time for students to discuss, reason with, reflect upon, and practice more
TM parent ROADMAP MATHEMATICS SUPPORTING YOUR CHILD IN GRADE FIVE 5 America s schools are working to provide higher quality instruction than ever before. The way we taught students in the past simply does
### Planning Guide. Grade 6 Factors and Multiples. Number Specific Outcome 3
Mathematics Planning Guide Grade 6 Factors and Multiples Number Specific Outcome 3 This Planning Guide can be accessed online at: http://www.learnalberta.ca/content/mepg6/html/pg6_factorsmultiples/index.html
### Line Plots. Objective To provide experience creating and interpreting line plots with fractional units. Assessment Management
Line Plots Objective To provide experience creating and interpreting line plots with fractional units. www.everydaymathonline.com epresentations etoolkit Algorithms Practice EM Facts Workshop Game Family
### 1) Make Sense and Persevere in Solving Problems.
Standards for Mathematical Practice in Second Grade The Common Core State Standards for Mathematical Practice are practices expected to be integrated into every mathematics lesson for all students Grades
### Progressions for the Common Core State Standards in Mathematics (draft)
Progressions for the Common Core State Standards in Mathematics (draft) cthe Common Core Standards Writing Team 2 April 202 K 5, Number and Operations in Base Ten Overview Students work in the base-ten
| 11,643
| 55,903
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2018-30
|
longest
|
en
| 0.705183
|
https://essaylead.com/dynamics-solution/
| 1,653,329,452,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00112.warc.gz
| 303,032,409
| 36,059
|
# Dynamics Solution
August 14, 2017 Accounting
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •12–5. A particle is moving along a straight line with the acceleration a = (12t – 3t1/2) ft>s2, where t is in seconds. Determine the velocity and the position of the particle as a function of time. When t = 0, v = 0 and s = 15 ft. Velocity: + A:B dv = a dt v 0 L v t 0 L dv = v 0 = A 6t2 – 2t3>2 B 2 + A:B A 12t – 3t1>2 B dt t 0 v = A 6t2 – 2t3>2 B ft>s Ans. Position: Using this result and the initial condition s = 15 ft at t = 0 s, ds = v dt s 15 L ft s t ds = 0 L A 6t2 – 2t3>2 B dt 5>2 2 t t b 5 0 s 15 ft = a 2t3 s = a 2t3 – 4 5>2 t + 15b ft 5 Ans. 12–6. A ball is released from the bottom of an elevator which is traveling upward with a velocity of 6 ft>s. If the ball strikes the bottom of the elevator shaft in 3 s, determine the height of the elevator from the bottom of the shaft at the instant the ball is released. Also, find the velocity of the ball when it strikes the bottom of the shaft. Kinematics: When the ball is released, its velocity will be the same as the elevator at the instant of release. Thus, v0 = 6 ft>s. Also, t = 3 s, s0 = 0, s = -h, and ac = -32. 2 ft>s2. A+cB = s0 + v0t + 1 a t2 2 c 1 (-32. 2) A 32 B 2 Ans. -h = 0 + 6(3) + h = 127 ft A+cB v = v0 + act v = 6 + (-32. 2)(3) = -90. 6 ft>s = 90. 6 ft>s T Ans. 3 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 4 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–7. A car has an initial speed of 25 m>s and a constant deceleration of 3 m>s2. Determine the velocity of the car when t = 4 s.
We Will Write a Custom Essay Specifically
For You For Only \$13.90/page!
order now
What is the displacement of the car during the 4-s time interval? How much time is needed to stop the car? v = v0 + act v = 25 + (-3)(4) = 13 m>s ? s = s – s0 = v0 t + 1 a t2 2 c 1 (-3)(4)2 = 76 m 2 Ans. Ans. ?s = s – 0 = 25(4) + v = v0 + ac t 0 = 25 + (-3)(t) t = 8. 33 s Ans. *12–8. If a particle has an initial velocity of v0 = 12 ft>s to the right, at s0 = 0, determine its position when t = 10 s, if a = 2 ft>s2 to the left. + A:B 1 a t2 2 c 1 ( -2)(10)2 2 Ans. s = s0 + v0 t + = 0 + 12(10) + = 20 ft •12–9. The acceleration of a particle traveling along a straight line is a = k>v, where k is a constant.
If s = 0, v = v0 when t = 0, determine the velocity of the particle as a function of time t. Velocity: + A:B dt = t 0 L t 0 L dv a dv k>v v L0 1 vdv k v L0 v v dt = v = 22kt + v0 2 t = t 1 2 2v t2 = v 2k v0 0 dt = 1 A v2 – v0 2 B 2k Ans. 4 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 5 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–10.
Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft>s2 until it reaches a speed of 80 ft>s. Afterwards it maintains this speed. Also, when t = 0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft>s. Determine the distance traveled by car A when they pass each other. 60 ft/s A B 6000 ft Distance Traveled: Time for car A to achives y = 80 ft>s can be obtained by applying Eq. 12–4. + A:B y = y0 + ac t 80 = 0 + 6t t = 13. 33 s The distance car A travels for this part of motion can be determined by applying Eq. 12–6. A:B y2 = y2 + 2ac (s – s0) 0 802 = 0 + 2(6)(s1 – 0) s1 = 533. 33 ft For the second part of motion, car A travels with a constant velocity of y = 80 ft>s and the distance traveled in t? = (t1 – 13. 33) s (t1 is the total time) is + A:B s2 = yt? = 80(t1 – 13. 33) Car B travels in the opposite direction with a constant velocity of y = 60 ft>s and the distance traveled in t1 is + A:B It is required that s1 + s2 + s3 = 6000 533. 33 + 80(t1 – 13. 33) + 60t1 = 6000 t1 = 46. 67 s The distance traveled by car A is sA = s1 + s2 = 533. 33 + 80(46. 67 – 13. 33) = 3200 ft Ans. s3 = yt1 = 60t1 5 91962_01_s12-p0001-0176 6/8/09 8:05 AM
A sphere is fired downwards into a medium with an initial speed of 27 m>s. If it experiences a deceleration of a = (-6t) m>s2, where t is in seconds, determine the distance traveled before it stops. Velocity: y0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have A+TB y 27 L dy = adt t 0 L dy = -6tdt [1] y = A 27 – 3t2 B m>s At y = 0, from Eq. [1] 0 = 27 – 3t2 t = 3. 00 s Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result y = 27 – 3t2 and applying Eq. 12–1, we have A+TB s 0 L ds = ydt t 0 L ds = A 27 – 3t2 B dt [2] s = A 27t – t3 B m At t = 3. 00 s, from Eq. [2] s = 27(3. 00) – 3. 003 = 54. 0 m Ans. •12–13.
A particle travels along a straight line such that in 2 s it moves from an initial position sA = +0. 5 m to a position sB = -1. 5 m. Then in another 4 s it moves from sB to sC = +2. 5 m. Determine the particle’s average velocity and average speed during the 6-s time interval. ?s = (sC – sA) = 2 m sT = (0. 5 + 1. 5 + 1. 5 + 2. 5) = 6 m t = (2 + 4) = 6 s vavg = ? s 2 = = 0. 333 m>s t 6 sT 6 = = 1 m>s t 6 Ans. (vsp)avg = Ans. 7 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 8 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–14. A particle travels along a straight-line path such that in 4 s it moves from an initial position sA = -8 m to a position sB = +3 m. Then in another 5 s it moves from sB to sC = -6 m. Determine the particle’s average velocity and average speed during the 9-s time interval. Average Velocity: The displacement from A to C is ? s = sC – SA = -6 – (-8) = 2 m. yavg = 2 ? s = = 0. 222 m>s ? t 4 + 5 Ans. Average Speed: The distances traveled from A to B and B to C are sA : B = 8 + 3 = 11. m and sB : C = 3 + 6 = 9. 00 m, respectively. Then, the total distance traveled is sTot = sA : B + sB : C = 11. 0 – 9. 00 = 20. 0 m. A ysp B avg = sTot 20. 0 = = 2. 22 m>s ? t 4 + 5 Ans. 12–15. Tests reveal that a normal driver takes about 0. 75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0. 1% alcohol in his system to do the same. If such drivers are traveling on a straight road at 30 mph (44 ft>s) and their cars can decelerate at 2 ft>s2, determine the shortest stopping distance d for each from the moment they see the pedestrians.
Moral: If you must drink, please don’t drive! v1 44 ft/s d Stopping Distance: For normal driver, the car moves a distance of d? = yt = 44(0. 75) = 33. 0 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d? = 33. 0 ft and y = 0. + A:B y2 = y2 + 2ac (s – s0) 0 02 = 442 + 2(-2)(d – 33. 0) d = 517 ft Ans. For a drunk driver, the car moves a distance of d? = yt = 44(3) = 132 ft before he or she reacts and decelerates the car. The stopping distance can be obtained using Eq. 12–6 with s0 = d? = 132 ft and y = 0. + A:B y2 = y2 + ac (s – s0) 0 02 = 442 + 2(-2)(d – 132) d = 616 ft Ans. 8 91962_01_s12-p0001-0176 6/8/09 8:05 AM Page 9 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12–16. As a train accelerates uniformly it passes successive kilometer marks while traveling at velocities of 2 m>s and then 10 m>s. Determine the train’s velocity when it passes the next kilometer mark and the time it takes to travel the 2-km distance.
Kinematics: For the first kilometer of the journey, v0 = 2 m>s, v = 10 m>s, s0 = 0, and s = 1000 m. Thus, + A:B v2 = v0 2 + 2ac (s – s0) 102 = 22 + 2ac (1000 – 0) ac = 0. 048 m>s2 For the second 0. 048 m>s2. Thus, + A:B kilometer, v0 = 10 m>s, s0 = 1000 m, s = 2000 m, and v2 = v0 2 + 2ac (s – s0) v2 = 102 + 2(0. 048)(2000 – 1000) v = 14 m>s Ans. For the whole journey, v0 = 2 m>s, v = 14 m>s, and 0. 048 m>s2. Thus, + A:B v = v0 + act 14 = 2 + 0. 048t t = 250 s Ans. •12–17. A ball is thrown with an upward velocity of 5 m>s from the top of a 10-m high building.
One second later another ball is thrown vertically from the ground with a velocity of 10 m>s. Determine the height from the ground where the two balls pass each other. Kinematics: First, we will consider the motion of ball A with (vA)0 = 5 m>s, (sA)0 = 0, sA = (h – 10) m, tA = t? , and ac = -9. 81 m>s2. Thus, A+cB sA = (sA)0 + (vA)0 tA + h – 10 = 0 + 5t? + 1 actA 2 2 1 (-9. 81)(t? )2 2 (1) h = 5t? – 4. 905(t? )2 + 10 Motion of ball B is with (vB)0 = 10 m>s, (sB)0 = 0, sB = h, tB = t? – 1 and ac = -9. 81 m>s2. Thus, A+cB sB = (sB)0 + (vB)0 tB + h = 0 + 10(t? – 1) + ac tB 2 2 1 (-9. 81)(t? – 1)2 2 (2) h = 19. 81t? – 4. 905(t? )2 – 14. 905 Solving Eqs. (1) and (2) yields h = 4. 54 m t? = 1. 68 m 9 Ans. 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 10 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–18. A car starts from rest and moves with a constant acceleration of 1. 5 m>s2 until it achieves a velocity of 25 m>s.
It then travels with constant velocity for 60 seconds. Determine the average speed and the total distance traveled. Kinematics: For stage (1) of the motion, v0 = 0, s0 = 0, v = 25 m>s, and ac = 1. 5 m>s2. + A:B v = v0 + act 25 = 0 + 1. 5t1 t1 = 16. 67 s + A:B v2 = v0 2 + 2ac(s – s0) 252 = 0 + 2(1. 5)(s1 – 0) s1 = 208. 33 m For stage (2) of the motion, s0 = 108. 22 ft, v0 = 25 ft>s, t = 60 s, and ac = 0. Thus, + A:B s = s0 + v0t + 1 a t2 2 c s = 208. 33 + 25(60) + 0 = 1708. 33ft = 1708 m The average speed of the car is then vavg = s 1708. 33 = = 22. 3 m>s t1 + t2 16. 67 + 60 Ans. Ans. 12–19.
Determine the distance between them when t = 4 s and the total distance each has traveled in t = 4 s. Velocity: The velocity of particles A and B can be determined using Eq. 12-2. dyA = aAdt yA t 0 L L 0 dyA = (6t – 3)dt yA = 3t2 – 3t dyB = aBdt yB 0 L t 0 L dyB = (12t2 – 8)dt yB = 4t3 – 8t The times when particle A stops are 4t3 – 8t = 0 t = 0 s and t = 22 s t = 0 s and = 1 s 3t2 – 3t = 0 The times when particle B stops are Position: The position of particles A and B can be determined using Eq. 12-1. dsA = yAdt sA 0 L t 0 L dsA = (3t2 – 3t)dt 3 2 t 2 sA = t3 dsB = yBdt sB t 0 L L 0 dsB = 4t3 – 8t)dt sB = t4 – 4t2 The positions of particle A at t = 1 s and 4 s are sA |t = 1 s = 13 sA |t = 4 s = 43 Particle A has traveled 3 2 (1 ) = -0. 500 ft 2 3 2 (4 ) = 40. 0 ft 2 The positions of particle B at t = 22 s and 4 s are dA = 2(0. 5) + 40. 0 = 41. 0 ft Ans. sB |t = 12 = (22)4 – 4(22)2 = -4 ft sB |t = 4 = (4)4 – 4(4)2 = 192 ft Particle B has traveled dB = 2(4) + 192 = 200 ft At t = 4 s the distance beween A and B is ? sAB = 192 – 40 = 152 ft 12 Ans. Ans. 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 13 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–22. A particle moving along a straight line is subjected to a deceleration a = (-2v3) m>s2, where v is in m>s. If it has a velocity v = 8 m>s and a position s = 10 m when t = 0, determine its velocity and position when t = 4 s. Velocity: The velocity of the particle can be related to its position by applying Eq. 12–3. ds = s 10m L ydy a dy – 2 8m>s 2y L 1 1 2y 16 8 16s – 159 [1] ds = s – 10 = y = Position: The position of the particle can be related to the time by applying Eq. 12–1. dt = t 0 L ds y s dt = 1 10m L 8 (16s – 159) ds 8t = 8s2 – 159s + 790 When t = 4 s, 8(4) = 8s2 – 159s + 790 8s2 – 159s + 758 = 0 Choose the root greater than 10 m s = 11. 94 m = 11. 9 m Substitute s = 11. 94 m into Eq. [1] yields y = 8 = 0. 250 m>s 16(11. 94) – 159 Ans. Ans. 13 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 14 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–23. A particle is moving along a straight line such that its acceleration is defined as a = (-2v) m>s2, where v is in meters per second. If v = 20 m>s when s = 0 and t = 0, determine the particle’s position, velocity, and acceleration as functions of time. a = -2v dv = -2v dt dv -2 dt = L v 20 0 L ln v = -2t 20 v = (20e – 2t)m>s a = s 0 L v t Ans. Ans. dv = (-40e – 2t)m>s2 dt t 0 L t ds = v dt = 0 L (20e – 2t)dt s = -10e – 2t|t = -10(e – 2t – 1) 0 s = 10(1 – e – 2t)m Ans. 14 91962_01_s12-p0001-0176 /8/09 8:06 AM Page 15 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12–24. A particle starts from rest and travels along a straight line with an acceleration a = (30 – 0. 2v) ft>s2, where v is in ft>s. Determine the time when the velocity of the particle is v = 30 ft>s. Velocity: + A:B dt = t 0 L dv a dv 0 L 30 – 0. 2v v dt = t|t = 0 t = 5ln 1 ln(30 – 0. 2v) 2 0. 2 0 v 30 30 – 0. v t = 5ln 30 = 1. 12 s 30 – 0. 2(50) Ans. •12–25. When a particle is projected vertically upwards with an initial velocity of v0, it experiences an acceleration a = -(g + kv2) , where g is the acceleration due to gravity, k is a constant and v is the velocity of the particle. Determine the maximum height reached by the particle. Position: A+cB ds = s 0 L v dv a v v L0 ds = – s|s = – c 0 s = The particle achieves its maximum height when v = 0. Thus, hmax = g + kv0 2 1 ln ? ? g 2k g + kv0 2 1 ln ? ? 2k g + kv2 v 1 ln A g + kv2 B d 2 2k v0 vdv g + kv2 = 1 k ln ? 1 + v0 2 ? g 2k Ans. 15 91962_01_s12-p0001-0176 6/8/09 8:06 AM
Page 16 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–26. The acceleration of a particle traveling along a straight line is a = (0. 02et) m>s2, where t is in seconds. If v = 0, s = 0 when t = 0, determine the velocity and acceleration of the particle at s = 4 m. Velocity: a = 0. 02e5. 329 = 4. 13 m>s2 dv = a dt v t 0 L t A + : B Ans. Position: + A:B v|0 = 0. 02e 2 0 L v dv = 0. 02et dt v = C 0. 02 A et – 1 B D m>s 0 (1) ds = v dt s 0 L s t 0 L s = 0. 02 A et – t – 1 B m When s = 4 m, 4 = 0. 02 A et – t – 1 B et – t – 201 = 0 Solving the above equation by trial and error, t = 5. 329 s Thus, the velocity and acceleration when s = 4 m (t = 5. 329 s) are v = 0. 02 A e5. 329 – 1 B = 4. 11 m>s a = 0. 02e5. 329 = 4. 13 m>s2 Ans. Ans. s|0 = 0. 02 A e – t B 2 t ds = 0. 02 A et – 1 B dt t 0 12–27. A particle moves along a straight line with an acceleration of a = 5>(3s1>3 + s5>2) m>s2, where s is in meters. Determine the particle’s velocity when s = 2 m, if it starts from rest when s = 1 m.
Determine the position of the particle when t = 6 s and the total distance it travels during the 6-s time interval. Hint: Plot the path to determine the total distance traveled. Position: The position of the particle when t = 6 s is s|t = 6s = 1. 5(63) – 13. 5(62) + 22. 5(6) = -27. 0 ft Ans. Total Distance Traveled: The velocity of the particle can be determined by applying Eq. 12–1. y = ds = 4. 50t2 – 27. 0t + 22. 5 dt The times when the particle stops are 4. 50t2 – 27. 0t + 22. 5 = 0 t = 1s and t = 5s The position of the particle at t = 0 s, 1 s and 5 s are s t = 0s = 1. 5(03) – 13. 5(02) + 22. 5(0) = 0 s t = 1s = 1. (13) – 13. 5(12) + 22. 5(1) = 10. 5 ft s t = 5s = 1. 5(53) – 13. 5(52) + 22. 5(5) = -37. 5 ft From the particle’s path, the total distance is stot = 10. 5 + 48. 0 + 10. 5 = 69. 0 ft Ans. 18 91962_01_s12-p0001-0176 6/8/09 8:06 AM Page 19 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–30. The velocity of a particle traveling along a straight line is v = v0 – ks, where k is constant.
If s = 0 when t = 0, determine the position and acceleration of the particle as a function of time. Position: + A:B dt = t 0 L ds y s 1 ln (v0 – ks) 2 k 0 dt = ds 0 L v0 – ks s tt = 0 t = ekt = s = Velocity: v = v0 1 ln ? ? k v0 – ks v0 v0 – ks v0 A 1 – e – kt B k Ans. d v0 ds = c A 1 – e – kt B d dt dt k v = v0e – kt Acceleration: a = d dv = A v e – kt B dt dt 0 Ans. a = -kv0e – kt 12–31. The acceleration of a particle as it moves along a straight line is given by a = 12t – 12 m>s2, where t is in seconds. If s = 1 m and v = 2 m>s when t = 0, determine the particle’s velocity and position when t = 6 s.
Afterwards it maintains this speed. Also, when t = 0, a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 ft>s. Determine the time and the distance traveled by the motorcycle when they pass each other. Motorcycle: + A:B v = v0 + ac t? 50 = 0 + 6t? t? = 8. 33 s v2 = v2 + 2ac (s – s0) 0 (50)2 = 0 + 2(6)(s? – 0) s? = 208. 33 ft In t? = 8. 33 s car travels s– = v0 t? = 30(8. 33) = 250 ft Distance between motorcycle and car: 6000 – 250 – 208. 33 = 5541. 67 ft When passing occurs for motorcycle, s = v0 t; For car: s = v0 t; Solving, x = 3463. 4 ft t– = 69. 27 s Thus, for the motorcycle, t = 69. 27 + 8. 33 = 77. 6 s sm = 208. 33 + 3463. 54 = 3. 67(10)3 ft Ans. Ans. 5541. 67 – x = 30(t–) x = 50(t–) 21 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 22 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–34. A particle moves along a straight line with a velocity v = (200s) mm>s, where s is in millimeters.
Determine the acceleration of the particle at s = 2000 mm. How long does the particle take to reach this position if s = 500 mm when t = 0? Acceleration: + A:B Thus, a = v dv = 200s ds dv = (200s)(200) = 40 A 103 B s mm>s2 ds When s = 2000 mm, a = 40 A 103 B (2000) = 80 A 106 B mm>s2 = 80 km>s2 Position: + A:B dt = t Ans. ds v 1 lns 2 200 500 mm s t2 = 0 L t 0 dt = ds 500 L mm 200s s t = s 1 ln 200 500 At s = 2000 mm, t = 1 2000 ln = 6. 93 A 10 – 3 B s = 6. 93 ms 200 500 Ans. 22 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 23 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved.
This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. ?12–35. A particle has an initial speed of 27 m>s. If it experiences a deceleration of a = 1-6t2 m>s2, where t is in seconds, determine its velocity, after it has traveled 10 m. How much time does this take? Velocity: + A:B dv = a dt t t v = (27 – 3t ) m>s + A:B ds = v dt s 0 L s t 0 L v2 27 L v dv = = 27 A -3t2 B 2 2 0 L (-6t)dt t 0 s = (27t – t ) m>s When s = 100 m, s 2 = A 27t – t3 B 2 0 3 ds =
A 27 – 3t2 B dt t 0 t = 0. 372 s v = 26. 6 m>s Ans. Ans. *12–36. The acceleration of a particle traveling along a straight line is a = (8 – 2s) m>s2, where s is in meters. If v = 0 at s = 0, determine the velocity of the particle at s = 2 m, and the position of the particle when the velocity is maximum. Velocity: + A:B v dv = a ds v 0 L s n = 216s – 2s2 m>s v s v2 ` = A 8s – s2 B 2 2 0 0 vdv = 0 L (8 – 2s) ds At s = 2 m, v s = 2 m = 216(2) – 2 A 22 B = ;4. 90 m>s dv 16 – 4s = = 0 ds 2 216s – 2s2 dv = 0. Thus, ds Ans. When the velocity is maximum 16 – 4s = 0 s = 4m 23 Ans. 91962_01_s12-p0001-0176 /8/09 8:07 AM Page 24 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •12–37. Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t 6 2v0>g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.
Kinematics: First, we will consider the motion of ball A with (vA)0 = v0, (sA)0 = 0, sA = h, tA = t? , and (ac)A = -g. A+cB sA = (sA)0 + (vA)0tA + h = 0 + v0t? + h = v0t? g 2 t? 2 1 (a ) t 2 2 cA A 1 ( -g)(t? )2 2 (1) A+cB vA = (vA)0 + (ac)A tA vA = v0 + (-g)(t? ) vA = v0 – gt? (2) The motion of ball B requires (vB)0 = v0, (sB)0 = 0, sB = h, tB = t? – t , and (ac)B = -g. A+cB sB = (sB)0 + (vB)0tB + h = 0 + v0(t? – t) + h = v0(t? – t) – 1 (a ) t 2 2 cBB 1 (-g)(t? – t)2 2 (3) g (t? – t)2 2 A+cB vB = (vB)0 + (ac)B tB vB = v0 + (-g)(t? – t) vB = v0 – g(t? – t) (4)
Solving Eqs. (1) and (3), g 2 g t? = v0(t? – t) – (t? – t)2 2 2 2v0 + gt t? = 2g v0t? Substituting this result into Eqs. (2) and (4), vA = v0 – ga = 2v0 + gt b 2g Ans. Ans. 1 1 gt = gt T 2 2 2v0 + gt – tb 2g vB = v0 – ga = 1 gt c 2 Ans. 24 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 25 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–38.
As a body is projected to a high altitude above the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be taken into account. Neglecting air resistance, this acceleration is determined from the formula a = -g0[R2>(R + y)2], where g0 is the constant gravitational acceleration at sea level, R is the radius of the earth, and the positive direction is measured upward. If g0 = 9. 81 m>s2 and R = 6356 km, determine the minimum initial velocity (escape velocity) at which a projectile should be shot vertically from the earth’s surface so that it does not fall back to the earth.
Hint: This requires that v = 0 as y : q . v dv = a dy 0 y L q v dv = -g0R 2 dy 2 g0 R2 q v2 2 0 2 = 2 y R + y 0 v = 22g0 R L (R + y) 0 = 22(9. 81)(6356)(10)3 Ans. = 11167 m>s = 11. 2 km>s 12–39. Accounting for the variation of gravitational acceleration a with respect to altitude y (see Prob. 12–38), derive an equation that relates the velocity of a freely falling particle to its altitude. Assume that the particle is released from rest at an altitude y0 from the earth’s surface. With what velocity does the particle strike the earth if it is released from rest at an altitude y0 = 500 km?
Use the numerical data in Prob. 12–38. From Prob. 12–38, (+ c ) a = -g0 R2 (R + y)2 Since a dy = v dv then y -g0 R2 g0 R2 c g0 R2[ Thus dy = 0 L v 2 y L0 (R + y) v dv y 1 v2 d = R + y y0 2 1 1 v2 ] = R + y R + y0 2 2g0 (y0 – y) A (R + y)(R + y0) v = -R When y0 = 500 km, 2(9. 81)(500)(103) v = -6356(103) A 6356(6356 + 500)(106) y = 0, Ans. v = -3016 m>s = 3. 02 km>s T 25 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 26 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12–40. When a particle falls through the air, its initial acceleration a = g diminishes until it is zero, and thereafter it falls at a constant or terminal velocity vf. If this variation of the acceleration can be expressed as a = 1g>v2f21v2f – v22, determine the time needed for the velocity to become v = vf>2 . Initially the particle falls from rest. g dv = a = ? 2 ? A v2 – v2 B f dt vf v dy g t t = vf + v y g 1 ln ? ?` = 2t 2vf vf – v 0 vf vf 2g vf 2g ln ? ln ? vf + v vf – v vf + vf>2 vf – vf>2 b L v2 – v2? f = v2 f 0 L dt t = ? t = 0. 549 a vf g ? Ans. 26 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 27 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •12–41. A particle is moving along a straight line such that its position from a fixed point is s = (12 – 15t2 + 5t3) m, where t is in seconds. Determine the total distance traveled by the particle from t = 1 s to t = 3 s.
Also, find the average speed of the particle during this time interval. Velocity: + A:B v = ds d = A 12 – 15t2 + 5t3 B dt dt v = -30t + 15t2 m>s The velocity of the particle changes direction at the instant when it is momentarily brought to rest. Thus, v = -30t + 15t2 = 0 t(-30 + 15t) = 0 t = 0 and 2 s Position: The positions of the particle at t = 0 s, 1 s, 2 s, and 3 s are s t = 0 s = 12 – 15 A 02 B + 5 A 03 B = 12 m s t = 2 s = 12 – 15 A 22 B + 5 A 23 B = -8 m s t = 3 s = 12 – 15 A 32 B + 5 A 33 B = 12 m s t = 1 s = 12 – 15 A 12 B + 5 A 13 B = 2 m Using the above results, the path of the particle is shown in Fig. . From this figure, the distance traveled by the particle during the time interval t = 1 s to t = 3 s is sTot = (2 + 8) + (8 + 12) = 30 m The average speed of the particle during the same time interval is vavg = sTot 30 = = 15 m>s ? t 3 – 1 Ans. Ans. 27 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 28 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–42.
The speed of a train during the first minute has been recorded as follows: t (s) v (m>s) 0 0 20 16 40 21 60 24 Plot the v-t graph, approximating the curve as straight-line segments between the given points. Determine the total distance traveled. The total distance traveled is equal to the area under the graph. sT = 1 1 1 (20)(16) + (40 – 20)(16 + 21) + (60 – 40)(21 + 24) = 980 m 2 2 2 Ans. 28 91962_01_s12-p0001-0176 6/8/09 8:07 AM Page 29 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–43. A two-stage missile is fired vertically from rest with the acceleration shown. In 15 s the first stage A burns out and the second stage B ignites. Plot the v-t and s-t graphs which describe the two-stage motion of the missile for 0 … t … 20 s. B a (m/s2) A 25 18 t (s) 15 20 a dt, the constant lines of the a–t graph become sloping lines for the v–t graph. L The numerical values for each point are calculated from the total area under the a–t graph to the point.
Since v = At t = 15 s, At t = 20 s, Since s = v = (18)(15) = 270 m>s v = 270 + (25)(20 – 15) = 395 m>s v dt, the sloping lines of the v–t graph become parabolic curves for the s–t graph. L The numerical values for each point are calculated from the total area under the v–t graph to the point. At t = 15 s, s = 1 (15)(270) = 2025 m 2 1 (395 – 270)(20 – 15) = 3687. 5 m = 3. 69 km 2 At t = 20 s, Also: 0 … t … 15: a = 18 s = 2025 + 270(20 – 15) + v = v0 + ac t = 0 + 18t s = s0 + v0 t + At t = 15: v = 18(15) = 270 s = 9(15)2 = 2025 15 … t … 20: a = 25 v = v0 + ac t = 270 + 25(t – 15) s = s0 + v0 t + When t = 20: v = 395 m>s s = 3687. m = 3. 69 km 29 1 1 a t2 = 2025 + 270(t – 15) + (25)(t – 15)2 2 c 2 1 a t2 = 0 + 0 + 9t2 2 c 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 30 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. *12–44. A freight train starts from rest and travels with a constant acceleration of 0. 5 ft>s2. After a time t? it maintains a constant speed so that when t = 160 s it has traveled 2000 ft.
Determine the time t? and draw the v–t graph for the motion. Total Distance Traveled: The distance for part one of the motion can be related to time t = t? by applying Eq. 12–5 with s0 = 0 and y0 = 0. + A:B s = s0 + y0 t + s1 = 0 + 0 + 1 a t2 2 c 1 (0. 5)(t? )2 = 0. 25(t? )2 2 The velocity at time t can be obtained by applying Eq. 12–4 with y0 = 0. + A:B y = y0 + act = 0 + 0. 5t = 0. 5t [1] The time for the second stage of motion is t2 = 160 – t? and the train is traveling at a constant velocity of y = 0. 5t? (Eq. [1]). Thus, the distance for this part of motion is + A:B s2 = yt2 = 0. t? (160 – t? ) = 80t? – 0. 5(t? )2 If the total distance traveled is sTot = 2000, then sTot = s1 + s2 2000 = 0. 25(t? )2 + 80t? – 0. 5(t? )2 0. 25(t? )2 – 80t? + 2000 = 0 Choose a root that is less than 160 s, then t? = 27. 34 s = 27. 3 s Ans. Y t Graph: The equation for the velocity is given by Eq. [1]. When t = t? = 27. 34 s, y = 0. 5(27. 34) = 13. 7 ft>s. 30 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 31 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •12–45. If the position of a particle is defined by s = [2 sin (p>5)t + 4] m, where t is in seconds, construct the s-t, v-t, and a-t graphs for 0 … t … 10 s. 31 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 32 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 12–46.
A train starts from station A and for the first kilometer, it travels with a uniform acceleration. Then, for the next two kilometers, it travels with a uniform speed. Finally, the train decelerates uniformly for another kilometer before coming to rest at station B. If the time for the whole journey is six minutes, draw the v–t graph and determine the maximum speed of the train. For stage (1) motion, + A:B v1 = v0 + (ac)1 t vmax = 0 + (ac)1 t1 vmax = (ac)1t1 + A:B v1 2 = v0 2 + 2(ac)1(s1 – s0) vmax 2 = 0 + 2(ac)1(1000 – 0) (ac)1 = vmax 2 2000 (2) (1) Eliminating (ac)1 from Eqs. (1) and (2), we have t1 = 2000 vmax (3)
The a–s graph for a jeep traveling along a straight road is given for the first 300 m of its motion. Construct the v–s graph. At s = 0, v = 0. a (m/s2) 2 200 300 s (m) a s Graph: The function of acceleration a in terms of s for the interval 0 m … s 6 200 m is a – 0 2 – 0 = s – 0 200 – 0 For the interval 200 m 6 s … 300 m, 0 – 2 a – 2 = s – 200 300 – 200 a = (-0. 02s + 6) m>s2 a = (0. 01s) m>s2 Y s Graph: The function of velocity y in terms of s can be obtained by applying ydy = ads. For the interval 0 m ? ss At s = 200 m, y = 0. 100(200) = 20. 0 m>s For the interval 200 m 6 s … 300 m, ydy = ads y 20. 0m>s L At s = 300 m, = 2-0. 02(3002) + 12(300) – 1200 = 24. 5 m>s y = A 2 -0. 02s2 + 12s – 1200 B m>s s ydy = 200m L (-0. 02s + 6)ds 34 91962_01_s12-p0001-0176 6/8/09 8:08 AM Page 35 © 2010 Pearson Education, Inc. , Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •12–49. A particle travels along a curve defined by the equation s = (t3 – 3t2 + 2t) m. where t is in seconds. Draw the s – t, v – t, and a – t graphs for the particle for 0 … t … 3 s. = t3 – 3t2 + 2t v = ds = 3t2 – 6t + 2 dt dv = 6t – 6 dt a = v = 0 at 0 = 3t2 – 6t + 2 t = 1. 577 s, and t = 0. 4226 s, s|t = 1. 577 = -0. 386 m s|t = 0. 4226 = 0. 385 m 12–50. A truck is traveling along the straight line with a velocity described by the graph. Construct the a-s graph for 0 … s … 1500 ft. a s Graph: For 0 … s 6 625 ft, a = v dv 3 = A 0. 6s3>4 B c (0. 6)s – 1>4 d = A 0. 27s1>2 B ft >s2 ds 4 v (ft/s) v 75 0. 6 s3/4 + A:B s(ft) 625 1500 At s = 625ft, a|s = 625 ft = 0. 27 A 6251>2 B = 6. 75ft>s2 For 625 ft 6 s 6 1500 ft, + A:B a = v dv = 75(0) = 0 ds The a-s graph is shown in Fig. a.
x
Hi!
I'm Amanda
Would you like to get a custom essay? How about receiving a customized one?
Check it out
| 12,505
| 33,591
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125
| 4
|
CC-MAIN-2022-21
|
longest
|
en
| 0.894269
|
https://wiki.analog.com/university/courses/electronics/electronics-lab-ce-loop-gain?rev=1579616402
| 1,686,332,761,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224656788.77/warc/CC-MAIN-20230609164851-20230609194851-00757.warc.gz
| 689,252,189
| 16,900
|
# Analog Devices Wiki
This version is outdated by a newer approved version.This version (21 Jan 2020 15:20) was approved by Cristina Suteu.The Previously approved version (23 Aug 2019 12:42) is available.
This is an old revision of the document!
# Activity: Common Emitter Amplifier Loop Gain
## Objective:
The objective of this Lab activity is to apply the Voltage Injection Method, that was introduced in the lab activity to measure the loop gain of an inverting op-amp stage, to measure the loop gain of a common emitter amplifier with negative feedback biasing and Miller capacitor.
## Background:
Negative feedback is commonly used in control systems. Figure 1 shows a simple system with negative feedback.
Figure 1 Negative feedback system
The output voltage is related to the input voltage by:
This is the closed loop transfer function. T(S) is called the loop gain which is the product of all gains around the loop and equals in this case to T(S) = G(S)H(S).
With the loop gain we can apply the Nyquist stability criterion to measure the gain and phase margin and determine the overall stability of the closed loop system.
The loop gain of a system can be derived from a mathematical model of the system. Such models often do not consider all the parasitics and unwanted effects that might exist in the real system. It can be very useful to measure the loop gain of a negative feedback system during the design process.
### Loop Gain Measurement
To Review: One method to measure the loop gain in negative feedback systems is the voltage injection method. The following shows how the voltage injection method can be applied in practice and what has to be considered to achieve correct results.
Using a suitable injection transformer ( the ADALP2000 Analog Parts kit contains a HPH1-1400L ) we can inject a test voltage at an appropriate injection-point in the feedback loop of the system. Then the response of the loop can be measured using a network analyzer like the ADALM2000.
Figure 2 shows a setup using the voltage injection method to measure the loop gain of a feedback system. A low value resistor is inserted in the feedback loop at the injection-point. The injection transformer secondary winding is connected across the injection resistor to apply the test voltage. This allows the injection of a test voltage without changing the DC-bias operating point of the system.
Figure 2 Voltage Injection Method
The network analyzer inputs are connected to both sides of the injection resistor using voltage probes. The loop gain is then measured by measuring the complex voltage gain from point A to B.
Where T(S) is the measured loop gain and VSig and VRef are voltages measured by the network analyzer.
The measured loop gain, T(S) approximately equals the actual loop gain if the following two conditions are met.
Condition 1
The impedance looking forward around the feedback loop ( ZIN(S) of block H(S) ) is much greater than the impedance looking backwards from the injection point ( ZOUT(S) of block G(S) ).
Condition 2
The second condition that must be fulfilled to ensure that the measured loop gain equals approximately the real loop gain is:
From these conditions we see that it is important to choose a suitable injection point that fulfills both conditions.
The first condition is often fulfilled at the output of an op-amp for example which is normally a low impedance. Other suitable points are generally at high impedance inputs like op-amp inputs.
The second condition is more difficult to check. Especially small loop gain results, above the crossover frequency need to be checked very carefully.
The magnitude of the injection voltage should be kept as low as possible to avoid large signal effects as saturation or other nonlinearities will influence the measurement.
The size of the injection resistor does not directly influence the measurement result if it is kept relatively small 50? or less is a good number.
The frequency response of the transformer and the dynamic range of the network analyzer will limit the loop gain measurement. In the lab set up below you will be using the HPH1-1400L transformer that has a usable frequency response of perhaps 10 KHz to 5 MHz. To measure loop response at lower frequencies a transformer with much higher winding inductance will be needed. The HPH1-1400L, or similar wide band transformers like a T1-6T ( Minicircuits ) or WB1010 ( Coilcraft ) should be sufficient to observe the loop response near the unity gain ( 0 dB ) frequency of the common emitter amplifier configurations used in this lab activity.
### Using negative ( shunt ) feedback for bias
Chapter 9 section 7 of the online Electronics I text introduced a biasing technique for the common emitter amplifier, called shunt feedback. This is accomplished by the introduction of some fraction of the collector signal back to the input at the base. This is done via the biasing resistor (RB), as shown in figure 3.
Figure 3, Common emitter amplifier with shunt feedback bias
The negative feedback loop from the output at the collector and the input at the base can be broken at three different points to inject the test voltage as shown in figure 4. Scheme ( a ) breaks the loop by inserting the test voltage in series with the collector of the transistor. Scheme ( b ) breaks the loop in series with the feedback resistor between the output ( junction of the collector and RC ) and RB. The third scheme ( c ) inserts the test voltage between the feedback resistor and the base of the transistor. Each of these schemes meets Condition 1 above to a greater or lesser extent and will give slightly different results for the loop response.
Figure 4 Three test voltage injection points
### Materials:
ADALM2000 Active Learning Module
Solder-less breadboard, and jumper wire kit
2 - 10 Ω resistors
1 - 100 Ω resistor
1 - 470 Ω resistor
1 - 2.2 KΩ resistor
1 - 3.3 KΩ resistor
2 - 4.7 KΩ resistors
1 - 10 KO resistor
1 - 47 KΩ resistor
1 - 2N3904 NPN transistor ( or equivalent )
1 - 39 pF capacitor
1 - 100 pF capacitor
1 - 470 pF capacitor
1 - HPH1-1400L transformer ( or transformers, T1-6T from Minicircuits, WB1010 from Coilcraft )
### Directions:
Build the measurement setup as shown in figure 5 below. If you are using the HPH1-1400L transformer for T1 you should connect three of the 6 windings in series for the primary and the remaining three windings in series for the secondary ( see this activity on transformers for more details ).
There are three resistors, RC, RB, RE and one capacitor, CM, that determine the loop gain of this circuit. You will be running frequency sweeps with the network analyzer testing various combinations of component values.
Voltage divider R2 and R3 serves two purposes. First the 10 Ω R2 matches the impedance of the resistor inserted in the feedback loop, R1. The AWG in the ADALM2000 cannot directly drive the 10 Ω resistor so the 100 Ω R3 increases the load resistance to a value high enough for the AWG to safely drive. The attenuation of the divider also allows us to set the amplitude of the AWG high enough to provide a low noise signal while still injecting a small signal into the loop.
Figure 5 CE Loop Gain measurement setup
When building an experiment on a solder-less breadboard, you add the small (stray) capacitor between adjacent rows of connection points to the circuit. This is because the way the solder-less breadboard is built, it has rows of metal connection strips laid side by side (0.1 inch apart) separated by plastic dividers. Because the strips are fairly long and they are in parallel, they have a significant capacitance between them ( see this activity on solder-less breadboard stray capacitance for more details ). This stray capacitance will appear in parallel with Miller feedback capacitor CM between the collector and the base.
When building this circuit it is recommended to leave a blank ( floating ) row between the collector and base pins of the NPN transistor. The 2N3904 devices has a pin order of EBC with the base in the center. Other types of transistors may have a ECB pin order with the collector in the center. In either case it is best to leave a blank row between both outer pins and the center pin. To see the effect of this stray capacitance after completing the lab re-build your circuit with the collector and base leads of Q1 plugged into adjacent rows on the breadboard and remeasure the response with no CM.
### Hardware Setup:
The green squares indicate where to connect the ADALM2000 AWG, scope channels and power supplies. Be sure to turn on the power supplies only after you double check your wiring.
Open the voltage supply control window to turn on and off the fixed +5 volt power supply. Open the Network Analyzer Instrument and set the sweep to start at 10 KHz and stop at 5 MHz. The Max gain should be set to 0.1X. Set the Amplitude to 3.5 V peak-to-peak and the Offset to zero volts. Under the Bode scale set the magnitude top to 40 dB and range to 80 dB. Set the phase top to 180º and range to 360º. Under scope channels click on use channel 1 as reference. Set the number of steps to 500.
### Procedure:
For your first measurements, start with RCequal to 10 KΩ, RB equal to 2.2 KΩ and RE equal to 0 Ω ( short with a wire jumper ). With CM equal to 0 pF ( open, no capacitor inserted ). Turn on the power supplies and run a single sweep. Note the frequency where the loop gain is unity ( 0 dB ) and the phase at that frequency. This configuration should result in a unity gain frequency actually beyond the 10 MHz frequency sweep range and you will need to extrapolate based on the slope of the gain curve. Be sure to export the sweep data to a .csv file for further analysis in either Excel or Matlab.
Be sure to turn off the power supplies before changing resistors or capacitors in your circuit. Now insert a 39 pF capacitor for CM. Turn on the power supplies and run a single sweep. Note the new frequency where the loop gain is unity ( 0 dB ) and the phase at that frequency and compare this to the measured ( extrapolated) result for no CM. Be sure to export the sweep data to a .csv file for further analysis in either Excel or Matlab. Now replace CM with a 100 pF capacitor. Run a single sweep. Note the new frequency where the loop gain is unity ( 0 dB ) and the phase at that frequency and compare this to the measured result for the with CM equal to 39 pF. Do another sweep with CM equal to 470 pF if you have one. Again compare the 0 dB frequency with your other measurements. Can you predict the 0 dB frequency based on the value of CM?
Next replace RB with a 3.3 KΩ or 4.7 KΩ depending on which value(s) you have available. Remove CM going back to a “0 pF” value. Be sure to turn off the power supplies before changing resistors in your circuit. Turn on the power supplies and run a single sweep. Note the new frequency where the loop gain is unity ( 0 dB ) and the phase at that frequency and compare this to the measured result for the with RB equal to 2.2 KΩ. Again go through the procedure of taking measurements for as many values of CM as you did in the first step. Be sure to export the sweep data to a .csv file for further analysis in either Excel or Matlab.
For extra credit try other values for RC and also try inserting an actual resistor for RE. Good values to try for RC might be 6.8 KΩ and 20 KΩ. Good values to try for RE might be 68 Ω, 100 Ω, 470 Ω and 1 KΩ. Again how does changing these resistor values effect the loop response?
Another extra credit exercise would be to compare the measured response for the other two injection configurations (b) and © in figure 4 and explain why the results were different based on Condition 1 and Condition 2 discussed in the section on measuring loop gain above.
### Questions:
Why did the unity gain frequency change for the cases with CM equal to 39 pF and 100 pF for the amplifier?
Why did the unity gain frequency change for the case of the with RB equal to 4.7 KΩ vs. 2.2 KΩ?
Why did the unity gain frequency change for the case with RE inserted?
Resources:
Measurement of loop gain in feedback systems. Middlebrook, R.D. s.l. : International Journal of Electronics, 1975, Bd. 38. http://scholar.google.com/scholar?cluster=5040596387593898653
http://en.wikipedia.org/wiki/Nyquist_stability_criterion
http://en.wikipedia.org/wiki/Phase_margin
http://en.wikipedia.org/wiki/Bode_plot
http://www.edn.com/electronics-blogs/analog-bytes/4434609/Loop-gain-measurements-
| 2,812
| 12,535
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.921233
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.