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://www.assignmentexpert.com/homework-answers/mathematics/algebra/question-8178 | 1,585,853,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370507738.45/warc/CC-MAIN-20200402173940-20200402203940-00392.warc.gz | 778,602,659 | 85,062 | 84 564
Assignments Done
98,9%
Successfully Done
In March 2020
Answer to Question #8178 in Algebra for Beverly Beauchamp
Question #8178
Maximizing Revenue The regular price for a round-trip ticket to Las Vegas, Nevada, charged by an airline charter company is $300. For a group rate the company will reduce the price of each ticket by$1.50 for every passenger on the flight. (a) Write a formula f(x) that gives the revenue from selling x tickets. (b) Determine how many tickets should be sold to maximize the revenue. What is the maximum revenue?
1
2012-04-10T07:31:29-0400
(a) The revenue of selling x tickets is given by formula: f(x)=( 300 - 1.50x
)x
(b)f"(x)= 300 - 3x . Maximum revenue will be if f'(x)=0 :
300-3x=0
x=100
The maximum revenue is \$15000 if should be sold 100
tickets.
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | 265 | 948 | {"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} | 3.453125 | 3 | CC-MAIN-2020-16 | latest | en | 0.864244 |
https://www.learnpythonwithrune.org/category/images/ | 1,627,833,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00246.warc.gz | 891,896,367 | 41,465 | ## What will we cover in this tutorial?
In this tutorial you will learn what the Julia set is and understand how it is calculated. Also, how it translates into colorful images. In the process, we will learn how to utilize vectorization with NumPy arrays to achieve it.
## Step 1: Understand the Julia set
Juila set are closely connect to the Mandelbrot set. If you are new to the Mandelbrot set, we recommend you read this tutorial before you proceed, as it will make it easier to understand.
Julia sets can be calculated for a function f. If we consider the function f_c(z) = z^2 + c, for a complex number c, then this function is used in the Mandelbrot set.
Recall the Mandelbrot set is calculated by identifying for a point c whether the function f_c(z) = z^2 + c , for which the sequence f_c(0), f_c(f_c(0)), f_c(f_c(f_c(0))), …., does not diverge.
Said differently, for each point c on the complex plane, if the sequence does not diverge, then that point is in the Mandelbrot set.
The Julia set has c fixed and and calculates the same sequence for z in the complex plane. That is, for each point z in the complex plane if the sequence f_c(0), f_c(f_c(0)), f_c(f_c(f_c(0))), …., does not diverge it is part of the Julia set.
## Step 2: Pseudo code for Julia set of non-vectorization computation
The best way to understand is often to see the non-vectorization method to compute the Julia set.
As we consider the function f_c(z) = z^2 + c for our Julia set, we need to choose a complex number for c. Note, that complex number c can be set differently to get another Julia set.
Then each we can iterate over each point z in the complex plane.
```c = -0.8 + i*0.34
for x in [-1, 1] do:
for y in [-1, 1] do:
z = x + i*y
N = 0
while absolute(z) < 2 and N < MAX_ITERATIONS:
z = z^2 + c
set color for x,y to N
```
This provides beautiful color images of the Julia set.
## Step 3: The vectorization computation using NumPy arrays
How does that translate into code using NumPy?
```import numpy as np
import matplotlib.pyplot as plt
def julia_set(c=-0.4 + 0.6j, height=800, width=1000, x=0, y=0, zoom=1, max_iterations=100):
# To make navigation easier we calculate these values
x_width = 1.5
y_height = 1.5*height/width
x_from = x - x_width/zoom
x_to = x + x_width/zoom
y_from = y - y_height/zoom
y_to = y + y_height/zoom
# Here the actual algorithm starts
x = np.linspace(x_from, x_to, width).reshape((1, width))
y = np.linspace(y_from, y_to, height).reshape((height, 1))
z = x + 1j * y
# Initialize z to all zero
c = np.full(z.shape, c)
# To keep track in which iteration the point diverged
div_time = np.zeros(z.shape, dtype=int)
# To keep track on which points did not converge so far
m = np.full(c.shape, True, dtype=bool)
for i in range(max_iterations):
z[m] = z[m]**2 + c[m]
m[np.abs(z) > 2] = False
div_time[m] = i
return div_time
plt.imshow(julia_set(), cmap='magma')
# plt.imshow(julia_set(x=0.125, y=0.125, zoom=10), cmap='magma')
# plt.imshow(julia_set(c=-0.8j), cmap='magma')
# plt.imshow(julia_set(c=-0.8+0.156j, max_iterations=512), cmap='magma')
# plt.imshow(julia_set(c=-0.7269 + 0.1889j, max_iterations=256), cmap='magma')
plt.show()
```
## What will we cover in this tutorial?
• Understand what the Mandelbrot set it and why it is so fascinating.
• Master how to make images in multiple colors of the Mandelbrot set.
• How to implement it using NumPy vectorization.
## Step 1: What is Mandelbrot?
Mandelbrot is a set of complex numbers for which the function f(z) = z^2 + c does not converge when iterated from z=0 (from wikipedia).
Take a complex number, c, then you calculate the sequence for N iterations:
z_(n+1) = z_n + c for n = 0, 1, …, N-1
If absolute(z_(N-1)) < 2, then it is said not to diverge and is part of the Mandelbrot set.
The Mandelbrot set is part of the complex plane, which is colored by numbers part of the Mandelbrot set and not.
This only gives a block and white colored image of the complex plane, hence often the images are made more colorful by giving it colors by the iteration number it diverged. That is if z_4 diverged for a point in the complex plane, then it will be given the color 4. That is how you end up with colorful maps like this.
## Step 2: Understand the code of the non-vectorized approach to compute the Mandelbrot set
To better understand the images from the Mandelbrot set, think of the complex numbers as a diagram, where the real part of the complex number is x-axis and the imaginary part is y-axis (also called the Argand diagram).
Then each point is a complex number c. That complex number will be given a color depending on which iteration it diverges (if it is not part of the Mandelbrot set).
Now the pseudocode for that should be easy to digest.
```for x in [-2, 2] do:
for y in [-1.5, 1.5] do:
c = x + i*y
z = 0
N = 0
while absolute(z) < 2 and N < MAX_ITERATIONS:
z = z^2 + c
set color for x,y to N
```
Simple enough to understand. That is some of the beauty of it. The simplicity.
## Step 3: Make a vectorized version of the computations
Now we understand the concepts behind we should translate that into to a vectorized version. If you are new to vectorization we can recommend you read this tutorial first.
What do we achieve with vectorization? That we compute all the complex numbers simultaneously. To understand that inspect the initialization of all the points here.
```import numpy as np
def mandelbrot(height, width, x_from=-2, x_to=1, y_from=-1.5, y_to=1.5, max_iterations=100):
x = np.linspace(x_from, x_to, width).reshape((1, width))
y = np.linspace(y_from, y_to, height).reshape((height, 1))
c = x + 1j * y
```
You see that we initialize all the x-coordinates at once using the linespace. It will create an array with numbers from x_from to x_to in width points. The reshape is to fit the plane.
The same happens for y.
Then all the complex numbers are created in c = x + 1j*y, where 1j is the imaginary part of the complex number.
This leaves us to the full implementation.
There are two things we need to keep track of in order to make a colorful Mandelbrot set. First, in which iteration the point diverged. Second, to achieve that, we need to remember when a point diverged.
```import numpy as np
import matplotlib.pyplot as plt
def mandelbrot(height, width, x=-0.5, y=0, zoom=1, max_iterations=100):
# To make navigation easier we calculate these values
x_width = 1.5
y_height = 1.5*height/width
x_from = x - x_width/zoom
x_to = x + x_width/zoom
y_from = y - y_height/zoom
y_to = y + y_height/zoom
# Here the actual algorithm starts
x = np.linspace(x_from, x_to, width).reshape((1, width))
y = np.linspace(y_from, y_to, height).reshape((height, 1))
c = x + 1j * y
# Initialize z to all zero
z = np.zeros(c.shape, dtype=np.complex128)
# To keep track in which iteration the point diverged
div_time = np.zeros(z.shape, dtype=int)
# To keep track on which points did not converge so far
m = np.full(c.shape, True, dtype=bool)
for i in range(max_iterations):
z[m] = z[m]**2 + c[m]
diverged = np.greater(np.abs(z), 2, out=np.full(c.shape, False), where=m) # Find diverging
div_time[diverged] = i # set the value of the diverged iteration number
m[np.abs(z) > 2] = False # to remember which have diverged
return div_time
# Default image of Mandelbrot set
plt.imshow(mandelbrot(800, 1000), cmap='magma')
# The image below of Mandelbrot set
# plt.imshow(mandelbrot(800, 1000, -0.75, 0.0, 2, 200), cmap='magma')
# The image below of below of Mandelbrot set
# plt.imshow(mandelbrot(800, 1000, -1, 0.3, 20, 500), cmap='magma')
plt.show()
```
Notice that z[m] = z[m]**2 + c[m] only computes updates on values that are still not diverged.
I have added the following two images from above (the one not commented out is above in previous step.
## What will you learn in this tutorial?
• Where to find interesting data contained in CSV files.
• How to extract a map to plot the data on.
• Use Python to easily plot the data from the CSV file no the map.
## Step 1: Collect the data in CSV format
You can find various interesting data in CSV format on data.world that you can play around with in Python.
In this tutorial we will focus on Shooting Incidents in NYPD from the last year. You can find the data on data.world.
Looking at the data you see that each incident has latitude and longitude coordinates.
```{'INCIDENT_KEY': '184659172', 'OCCUR_DATE': '06/30/2018 12:00:00 AM', 'OCCUR_TIME': '23:41:00', 'BORO': 'BROOKLYN', 'PRECINCT': '75', 'JURISDICTION_CODE': '0', 'LOCATION_DESC': 'PVT HOUSE ', 'STATISTICAL_MURDER_FLAG': 'false', 'PERP_AGE_GROUP': '', 'PERP_SEX': '', 'PERP_RACE': '', 'VIC_AGE_GROUP': '25-44', 'VIC_SEX': 'M', 'VIC_RACE': 'BLACK', 'X_COORD_CD': '1020263', 'Y_COORD_CD': '184219', 'Latitude': '40.672250312', 'Longitude': '-73.870176252'}
```
That means we can plot on a map. Let us try to do that.
## Step 2: Export a map to plot the data
We want to plot all the shooting incidents on a map. You can use OpenStreetMap to get an image of a map.
We want a map of New York, which you can find by locating it on OpenStreetMap or pressing the link.
You should press the blue Download in the low right corner of the picture.
Also, remember to get the coordinates of the image in the left side bar, we will need them for the plot.
```map_box = [-74.4461, -73.5123, 40.4166, 41.0359]
```
## Step 3: Writing the Python code that adds data to the map
Importing data from a CVS file is easy and can be done through the standard library csv. Making plot on a graph can be done in matplotlib. If you do not have it installed already, you can do that by typing the following in a command line (or see here).
```pip install matplotlib
```
First you need to transform the CVS data of the longitude and latitude to floats.
```import csv
# The name of the input file might need to be adjusted, or the location needs to be added if it is not located in the same folder as this file.
csv_file = open('nypd-shooting-incident-data-year-to-date-1.csv')
longitude = []
latitude = []
longitude.append(float(row['Longitude']))
latitude.append(float(row['Latitude']))
```
Now you have two lists (longitude and latitude), which contains the coordinates to plot.
Then for the actual plotting into the image.
```import matplotlib.pyplot as plt
# The boundaries of the image map
map_box = [-74.4461, -73.5123, 40.4166, 41.0359]
# The name of the image of the New York map might be different.
fig, ax = plt.subplots()
ax.scatter(longitude, latitude)
ax.set_ylim(map_box[2], map_box[3])
ax.set_xlim(map_box[0], map_box[1])
ax.imshow(map_img, extent=map_box, alpha=0.9)
plt.show()
```
This will result in the following beautiful map of New York, which highlights where the shooting in the last year has occurred.
Now that is awesome. If you want to learn more, this and more is covered in my online course. Check it out.
You can also read about how to plot the mood of tweets on a leaflet map.
## The challenge
You have a quote and an image.
``` quote = "Mostly, when you see programmers, they aren’t doing anything. One of the attractive things about programmers is that you cannot tell whether or not they are working simply by looking at them. Very often they’re sitting there seemingly drinking coffee and gossiping, or just staring into space. What the programmer is trying to do is get a handle on all the individual and unrelated ideas that are scampering around in his head."
quote_by = "Charles M. Strauss"
len(quote) = 430
```
The quote is long (430 chars) and the picture might be too bright to write in white text. Also, it will take time to find the right font size.
Can you automate that getting a result that fits the Twitter recommended picture size?
Notice the following.
• The picture is dimmed to make the white text easy readable.
• The font size is automatically adjusted to fit the picture.
• The picture is cropped to fit recommended Twitter size (1024 x 512).
If this is what you are looking for, then this will automate that.
## Step 1: Find your background picture and quote
In this tutorial I will use the background image and quote given above. But you can change that to your needs.
If you chose a shorter quote the font size will adjust accordingly.
Example given here.
Notice that the following.
• The font size of the “quoted by” (here Learn Python With Rune) is adjusted to not fill more that half of the picture width.
• You can modify the margins as you like – say you want more space between the text and the logo.
## Step 2: Install the PILLOW library
The what? Install the pillow library. You can find installation documentation in their docs.
Or just type
```pip install pillow
```
The pillow library is the PIL fork, and PIL is the Python Imaging Library. Basically, you need it for processing images.
You can find various fonts in font.google.com.
For this purpose, I used the Balsamiq Sans font.
I have located the bold and the italic version in a font folder.
## Step 4: The actual Python code
This is the fun part. The actual code.
```from PIL import Image, ImageDraw, ImageFont, ImageEnhance
# picture setup - it is set up for Twitter recommendations
WIDTH = 1024
HEIGHT = 512
# the margin are set by my preferences
MARGIN = 50
MARGIN_TOP = 50
MARGIN_BOTTOM = 150
LOGO_MARGIN = 25
# font variables
FONT_SIZES = [110, 100, 90, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20]
FONT_QUOTE = 'font-text'
FONT_QUOTED_BY = 'font-quoted-by'
FONT_SIZE = 'font-size'
FONT_QUOTED_BY_SIZE = 'font-quoted-by-size'
# Font colors
WHITE = 'rgb(255, 255, 255)'
GREY = 'rgb(200, 200, 200)'
# output text
OUTPUT_QUOTE = 'quote'
OUTPUT_QUOTED_BY = 'quoted-by'
OUTPUT_LINES = 'lines'
def text_wrap_and_font_size(output, font_style, max_width, max_height):
for font_size in FONT_SIZES:
output[OUTPUT_LINES] = []
font = ImageFont.truetype(font_style[FONT_QUOTE], size=font_size, encoding="unic")
output[OUTPUT_QUOTE] = " ".join(output[OUTPUT_QUOTE].split())
if font.getsize(output[OUTPUT_QUOTE])[0] <= max_width:
output[OUTPUT_LINES].append(output[OUTPUT_QUOTE])
else:
words = output[OUTPUT_QUOTE].split()
line = ""
for word in words:
if font.getsize(line + " " + word)[0] <= max_width:
line += " " + word
else:
output[OUTPUT_LINES].append(line)
line = word
output[OUTPUT_LINES].append(line)
line_height = font.getsize('lp')[1]
quoted_by_font_size = font_size
quoted_by_font = ImageFont.truetype(font_style[FONT_QUOTED_BY], size=quoted_by_font_size, encoding="unic")
while quoted_by_font.getsize(output[OUTPUT_QUOTED_BY])[0] > max_width//2:
quoted_by_font_size -= 1
quoted_by_font = ImageFont.truetype(font_style[FONT_QUOTED_BY], size=quoted_by_font_size, encoding="unic")
if line_height*len(output[OUTPUT_LINES]) + quoted_by_font.getsize('lp')[1] < max_height:
font_style[FONT_SIZE] = font_size
font_style[FONT_QUOTED_BY_SIZE] = quoted_by_font_size
return True
# we didn't succeed find a font size that would match within the block of text
return False
def draw_text(image, output, font_style):
draw = ImageDraw.Draw(image)
lines = output[OUTPUT_LINES]
font = ImageFont.truetype(font_style[FONT_QUOTE], size=font_style[FONT_SIZE], encoding="unic")
line_height = font.getsize('lp')[1]
y = MARGIN_TOP
for line in lines:
x = (WIDTH - font.getsize(line)[0]) // 2
draw.text((x, y), line, fill=WHITE, font=font)
y = y + line_height
quoted_by = output[OUTPUT_QUOTED_BY]
quoted_by_font = ImageFont.truetype(font_style[FONT_QUOTED_BY], size=font_style[FONT_QUOTED_BY_SIZE], encoding="unic")
# position the quoted_by in the far right, but within margin
x = WIDTH - quoted_by_font.getsize(quoted_by)[0] - MARGIN
draw.text((x, y), quoted_by, fill=GREY, font=quoted_by_font)
return image
def generate_image_with_quote(input_image, quote, quote_by, font_style, output_image):
image = Image.open(input_image)
# darken the image to make output more visible
enhancer = ImageEnhance.Brightness(image)
image = enhancer.enhance(0.5)
# resize the image to fit Twitter
image = image.resize((WIDTH, HEIGHT))
# set logo on image
logo_im = Image.open("pics/logo.png")
l_width, l_height = logo_im.size
image.paste(logo_im, (WIDTH - l_width - LOGO_MARGIN, HEIGHT - l_height - LOGO_MARGIN), logo_im)
output = {OUTPUT_QUOTE: quote, OUTPUT_QUOTED_BY: quote_by}
# we should check if it returns true, but it is ignorred here
text_wrap_and_font_size(output, font_style, WIDTH - 2*MARGIN, HEIGHT - MARGIN_TOP - MARGIN_BOTTOM)
# now it is time to draw the quote on our image and save it
image = draw_text(image, output, font_style)
image.save(output_image)
def main():
# setup input and output image
input_image = "pics/background.jpg"
output_image = "quote_of_the_day.png"
# setup font type
font_style = {FONT_QUOTE: "font/BalsamiqSans-Bold.ttf", FONT_QUOTED_BY: "font/BalsamiqSans-Italic.ttf"}
quote = "Mostly, when you see programmers, they aren’t doing anything. One of the attractive things about programmers is that you cannot tell whether or not they are working simply by looking at them. Very often they’re sitting there seemingly drinking coffee and gossiping, or just staring into space. What the programmer is trying to do is get a handle on all the individual and unrelated ideas that are scampering around in his head."
# quote = "YOU ARE AWESOME!"
quote_by = "Charles M. Strauss"
# quote_by = "Learn Python With Rune"
# generates the quote image
generate_image_with_quote(input_image, quote, quote_by, font_style, output_image)
if __name__ == "__main__":
main()
```
Notice that you can change the input and output image names and locations in the main() function. Also, there you can setup the font for the quote and the quoted-by.
Finally, and obviously, you can change the quote and quote-by in the main() function.
## Step 1: Install the Pillow library
We need the Pillow library in order to manipulate images. It can be installed by using pip.
```pip install Pillow
```
This enables us to use PIL (Python Imaging Library) library.
You can browse fonts free to download from Google Font Library. In this example I use the Balsamic Sans.
Place the download in the same folder as you want to work from.
## Step 3: Find an awesome picture
I use Pexels to find free awesome pictures to use.
And place that image in your same location.
## Step 4: Create you Python program to add your awesome text
```from PIL import Image, ImageDraw, ImageFont
# Open the image (change name and location if needed)
image = Image.open('pics/background.jpg')
draw = ImageDraw.Draw(image)
# Import the font(change name and location if needed, also change font size if needed)
font = ImageFont.truetype('font/BalsamiqSans-Bold.ttf', size=200)
black = 'rgb(0, 0, 0)' # black color
# Draw the text - change position if needed
message = "This is great!"
draw.text((1000, 200), message, fill=black, font=font)
# Draw the text - change position if needed
name = 'You are AWESOME!'
draw.text((1100, 500), name, fill=black, font=font)
image.save('greeting_card.png')
``` | 5,131 | 19,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.84761 |
https://amp.doubtnut.com/question-answer/find-the-mirror-image-of-the-point-a-32-in-x-axis-31337202 | 1,611,127,637,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00616.warc.gz | 209,404,426 | 22,578 | Doubtnut is better on App
Paiye sabhi sawalon ka Video solution sirf photo khinch kar
Open App
IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
Question From class 9 Chapter CO-ORDINATE GEOMETRY
# Find the mirror image of the point in x-axis.
the dot Find the mirror image in the x-axis.
2:26
A line passing through the points then the image of the line in x-axis is
2:39
In the figureshownm the image of a real object is formed at point I. AB is the principal axis of the mirror. The mirror must be <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/MPP_PHY_C14_E01_016_Q01.png" width="80%">
2:42
Mirror image of a point in plane
8:48
A plane mirror is placed along the y-axis such that x-axis is normal to the plane of the mirror. The reflecting surface of the mirror is towards negative x-axis. The mirror moves in positive x-direction with uniform speed of 5 m/s and a point object P is moving with constant speed 3m/s in negative x-direction. find the speed of image with respect to mirror in m/s.
The line of the point (3, 8) assuming the line to be a mirror for a point Find the image in
6:25
Imagining that simple lines act like a mirror to a point, the line of point (1, 2) Find the image in
7:28
Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line .
7:20
Assertion : In case of a concave mirror if a point object is moving towards the mirror along its principal axis, then its image will move away from the mirror. <br> Reason : In case of reflection (along the principal axis of mirror) object and image always travel in opposite directions.
1:36
A concave mirror forms the real image of a point source lying on the oprical axis at a distance of from the mirror. The focal length of the mirror is . The mirror is cut into two halves and its halves are drawn a distance of apart (from each other) in a direction perpendicular to the optical axis. Find the distace (in ) between the two image formed by the two halves of the mirror.
0:53
A point object is placed 60 cm from the pole of a concave mirror of focal length 10 cm on the principal axis. <br> Find: <br> a. the position of image. <br> b. If the object is shifted 1 mm towards the mirror along principal axis, find the shift in image. Explain the result.
2:01
Point A (0, 1 cm) and B (12 cm, 5 cm) are the coordinates of object and image x-axis is the principal axis of the mirror. The, this object image pair is
2:40
A point object is kept in front of a plane mirror. The plane mirror is doing SHM of ampliture 2cm. The plane mirror moves along the x-axis which is normal to the mirror. The amplitude of the mirror is such that the object is always in front of the mirror. The amplitude of SHM of the image is
1:19
Figure shows a shperical concave mirror of focal length f=5cm with its pole at (0,0) and principal axis along x-axis. There is a point object at (-40cm, 1cm), find the postion of image. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V04_C01_S01_021_Q01.png" width="80%">
2:19
Find a point on y-axis which is equidistant from the points and .
1:46
Latest Blog Post
CBSE to Introduce Two-levels of English and Sanskrit Exam, Details Here
Two-levels of English and Sanskrit exam to be introduced in CBSE 2021-22 session. CBSE 2021 exam datesheet is expected to be released soon. Check details here.
JEE Main 2021: NTA Extends Last Date of Registration till January 23rd
JEE Main 2021 registration date extended till January 23rd. Know JEE Main important dates and other key details related to the exam!
Punjab Board Date Sheet 2021 for Classes 5, 8, 10, 12 Released, Check here
Punjab Board (PSEB) datesheet of 2021 for classes 5, 8, 10, 12 has been released. Know how to download PSEB date sheet 2021 & details related to Punjab board exam.
Delhi Schools to Reopen for Classes 10 & 12 from Jan 18, 2021
Delhi Schools to reopen for classes 10 & 12 from Jan 18, 2021. Know complete details related to Delhi school reopening and upcoming board exams.
UP School Reopen for Classes 9-12, with Fresh COVID-19 Guidelines
UP School reopen for classes 9-12, with fresh COVID-19 guidelines. Know complete details of UP School reopening and UP board exam 2021!
Goa Board Exam for Class 9 and 11 will be Held Offline by April 24
Goa board exam for class 9 and 11 will begin offline from April 24. Know complete details of Goa board exam 2021! | 1,207 | 4,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.895589 |
https://tw511.com/3/34/1222.html | 1,723,726,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00864.warc.gz | 466,054,603 | 5,394 | # NumPy - 高階索引
## 整數索引
### 範例 1
``````import numpy as np
x = np.array([[1, 2], [3, 4], [5, 6]])
y = x[[0,1,2], [0,1,0]]
print y
``````
``````[1 4 5]
``````
### 範例 2
``````import numpy as np
x = np.array([[ 0, 1, 2],[ 3, 4, 5],[ 6, 7, 8],[ 9, 10, 11]])
print '我們的陣列是：'
print x
print '\n'
rows = np.array([[0,0],[3,3]])
cols = np.array([[0,2],[0,2]])
y = x[rows,cols]
print '這個陣列的每個角處的元素是：'
print y
``````
``````我們的陣列是：
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]]
[[ 0 2]
[ 9 11]]
``````
### 範例 3
``````import numpy as np
x = np.array([[ 0, 1, 2],[ 3, 4, 5],[ 6, 7, 8],[ 9, 10, 11]])
print '我們的陣列是：'
print x
print '\n'
# 切片
z = x[1:4,1:3]
print '切片之後，我們的陣列變為：'
print z
print '\n'
# 對列使用高階索引
y = x[1:4,[1,2]]
print '對列使用高階索引來切片：'
print y
``````
``````我們的陣列是：
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]]
[[ 4 5]
[ 7 8]
[10 11]]
[[ 4 5]
[ 7 8]
[10 11]]
``````
## 布林索引
### 範例 1
``````import numpy as np
x = np.array([[ 0, 1, 2],[ 3, 4, 5],[ 6, 7, 8],[ 9, 10, 11]])
print '我們的陣列是：'
print x
print '\n'
# 現在我們會列印出大於 5 的元素
print '大於 5 的元素是：'
print x[x > 5]
``````
``````我們的陣列是：
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]]
[ 6 7 8 9 10 11]
``````
### 範例 2
``````import numpy as np
a = np.array([np.nan, 1,2,np.nan,3,4,5])
print a[~np.isnan(a)]
``````
``````[ 1. 2. 3. 4. 5.]
``````
### 範例 3
``````import numpy as np
a = np.array([1, 2+6j, 5, 3.5+5j])
print a[np.iscomplex(a)]
``````
``````[2.0+6.j 3.5+5.j]
`````` | 919 | 1,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-33 | latest | en | 0.397422 |
https://www.scribd.com/doc/263805976/casestudy2 | 1,571,375,980,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00401.warc.gz | 1,089,831,338 | 68,824 | You are on page 1of 12
# Case Study #2
## By: Nick Jackson, Ui Jeong Lee, and Makayla Neal
Overview of Lesson
## This lesson will be comprised of three stations that will
educate students on the uses and interpretations of bar
graphs, pie charts, and line graphs. Each station will be
taught by a different instructor in which the students will be
educated in an interactive learning experience that will allow
them to collect, interpret, and integrate information that is
gathered into the different graphs and charts. After the three
stations are completed, students will complete an
assessment that encompasses the information that they
learned.
Description of Learners,
Learning Envrioment,
Intended Learning Goals,
and Lesson Content.
The learners for this lesson are fourth graders that are
participating in College Mentors for Kids (CMFK) here at
Purdue University. There are twenty students in the group
and buddies who are college mentors. We are going to pair
up a student and a college mentor as one. We assumed that
there are no special needs students and exceptional child.
The learning environment will take place in the classroom. It
needs to be big enought to set a three different stations
since we are doing station teaching. The students will be
working with Ipads to collect, assemble, and integrate the
information that is provided to them in the beginning of the
class into the specified charts and graphs. The class will take
place in a tyipical classroom environment in which students
will be able access the Internet to complete the tasks
## involved with their assignment and assessment. We are
encouring buddies(college mentors) not to help kids, since
our objectives are to see if students can complete tasks by
themselves without any assistance. We encourage buddies
to just particiapte with kids. After completing the lesson, the
students will be able to construct a bar graph, pie chart, or
line graph with given data and be able to choose which data
would best fit these graphs and charts.
Standards
4.DA.3: Interpret data displayed in a circle graph.
4.DA.1: Formulate questions that can be addressed with
data. Use observations, surveys, and experiments to collect,
represent, and interpret the data using tables (including
frequency tables), line plots, and bar graphs.
Learning Objectives
Given a set of data, students will be able to
construct and interpret a bar graph without error.
Without assistance, students will be able to
construct and interpret a pie chart without error.
Given a website on iPads (technology),
students will be able to create a line graph with
collected data without error.
Given a set of data, students will be able to
determine which type of graph will best fit the data
provided without error.
Given a worksheet, students will be able to
answer questions based on the shown graph and get
90% out of 100% on the worksheet.
Materials
2. Computer with internet access
3. Poster board
4. Stickers (enough for three per student)
5. Printer paper
6. Rulers
7. Crayons
8. Example pie graph, bar graph, and line graph.
9. Discussion question worksheet(given to college
mentor)
10. Projector at the front of class
Procedure
## 1. Introduce the bell ringer, the data collection poster, when
all of the students are in the classroom.
2.Give each student three stickers.
3.Tell the students to place one sticker under each
HOUSEHOLD (not farm animals) pet that their family owns.
If a student does not have a pet, have them place one sticker
under the None Category. Their options will be Dog, Cat,
Fish, hamster, other, or none.
4. After all of the students have placed at least one of their
stickers in a category, the teacher will enter the data into a
5. Share the spreadsheet with all of the students so that they
can access the data on their Ipads. Make sure that every
6.Split the students up into three groups(mixed gender) and
assign each group to one of the three teachers. Each group
will rotate each station( each station about 10-15minutes)
and learn different types of graphs. Creating a Station
Teaching:In station teaching, the classroom is divided into
various teaching centers. The teacher and student teacher
are at particular stations; the other stations are run
independently by the students or by a teachers aide.
(Credit:http://www.asdk12.org/depts/hr/student_teaching/PD
F/The_Power_of_2.pdf)
7.Group number one will start at Makaylas station, group
two will start at Ui Jeongs station, and group 3 will start at
Nicks station. Each group will rotate stations after
completing the activity at the station.
8. At Makaylas station the topic is pie graphs. Show the
students the example of a pie graph and explain to them that
the whole circle, or pie, represents 100%. Explain to the
students that each of the parts of the graph are a
representation of data that has been collected by the teacher
from the class. Also explain to the students that the data is
not related and does not affect each other. For example, if
the number of reading books went up, the number of
## playing games would not be affected. Add up all of the given
percentages with the students to show them that they equal
100%.
9.After giving the students an example of a pie graph, have
the students access the household pet data collected at the
beginning of class on their ipads. Give the students the
graph website,
http://nces.ed.gov/nceskids/createagraph/default.aspx, and
10. Once the students are on the site, have them select pie
graph. Tell them to select the data tab and enter in an
appropriate title that fits the data, the names of the pets
where it says item label, and the value for each category.
11. After the students enter in all of the information needed,
have them select the preview tab to view the graph that
they made. Have them take a screenshot of their graph on
their ipad so they can refer to the graph in the next stations.
12. After they are done with the pie graph station, have the
students move on to Ui Jeongs station, the bar graph
station.
13. Show the students an example of a bar graph and teach
them that the data represented in the bar graph is not related
in any way. The numbers for one category of data do not
increase nor decrease any of the other categories.
14. After the students have an understanding of what a bar
graph is and how it can be used, have the students pull up
the data on their ipad that was collected at the beginning of
the class. After they have the data open, pass out pieces of
paper, crayons, and rulers.
15. With the pieces of paper, crayons, and rulers, have the
students make their own bar graphs out of the data on their
Ipad. Make sure that they include an appropriate title for the
data (could use the same one from the pie graph), a key for
the different colors to represent the different categories, and
labeled x and y-axis.
16. After the students have completed their bar graph, have
them compare the bar graph and the pie graph. Ask them
what similarities they find and what differences they notice.
17. Have the students hold onto their bar graphs and rotate
to the next station, Nicholas station.
18. At Nicholas station, they will be learning about line
graphs. Explain to the students what a line graph is and why
the data that we collected at the beginning of class cannot
be put into a line graph. Explain that the data all has to be
related and affect each other.
19. After explaining what line graphs are, pull up the
temperatures for last week on the weather app on the ipad.
Have the students record all of the temperatures for each
day.
20. After the temperatures are recorded, pull up the website
that was used to make pie graphs,
http://nces.ed.gov/nceskids/createagraph/default.aspx , on
the projector in the front of the class.
21. Once the website is opened, select line graph. After
selecting line graph, go to the the data tab. Have the
describe the weekly temperature for last week.
22. Ask them what they think the x-axis should be labeled
and what they think the y-axis should be labeled. If they
answer it correctly or incorrectly, explain to them that the xaxis is always the category that you are measuring and the
y-axis is always the values of the categories that you are
measuring. Put the days of the week in the x-axis and the
temperatures (in farenheit) in the y-axis.
23. After the axis are labeled, enter in the data set. Have the
students read aloud the temperatures and the days of the
weeks while you enter them in the computer.
24. After all of the information is filled out, select the preview
tab. Point out how the data creates a trend and allows you to
calcula te an average weekly temperature because the data
is all related.
25. After the students learn about each one of the graphs,
have them pair up with their College Mentors For Kids buddy
to disscuss the activities that they did with the graphs.
26. Pass out the discussion question worksheet and have
them fill out the questions.
27. Give them 5 minutes to complete the worksheet. After
the worksheet is completed, have the students talk about
what they wrote for the discussion questions with their
College Mentors For Kids buddy.
28. Make sure the buddies(the college mentor) ask the
students why they were not able to use the data that was
collected at the beginning of class to make a line graph. Also
have them ask the students how pie graphs and line graphs
are different. They should ask them which graph they think is
the easiest to read/make and why. The teacher will prepare a
## Questions to share with you and your
buddy:
1. What did you enjoy the
most? What graph did you liked
most?
2. How are pie charts and
line graphs different?
3. Which graph do you think
is easiest to read and make?
Why do you think so?
4. Why cant you use the
data that was collected at the
beginning of class to make a line
graph?
29. Have the students turn in the paper graph that they made
and the discussion questions at the end of the day.
Assessment
## end of class after completing all of three stations. The
assignment will have three sections that include pie charts,
bar graphs, and line graphs. The students will encounter
questions that ask them to interpret the three types of graphs
have to apply the knowledge that they learned at each of the
(See the worksheet attached in the back)
References
## Heigh Record Line Graph. (n.d.). Retrieved
March 30, 2015, from
http://www.superteacherworksheets.com/graphing/linegraph-simple-1_TWNQD.pdf
Favorite Pet Worksheet. (n.d.). Retrieved
March 30, 2015, from
http://www.theteachersguide.com/bargraphs/favoritepe
tsgraph.pdf
Summer Camp Pie Graph. (n.d.). Retrieved
March 30, 2015, from
http://www.superteacherworksheets.com/graphing/piegraph-simple-2_TWNWR.pdf
Reading a Bar Graph. (n.d.). Retrieved March
30, 2015, from
http://www.commoncoresheets.com/Math/Bar
Graphs/4 Bars/English/6.pdf
## Description of the lesson plan
In order to create a lesson plan, our group first looked at Indiana state standards. We
determined to focus on 4th grade since our group memebers had experience of teaching 4th
grade as a TIP expereience. Also, all of our group memebers liked the subject Math, therefore,
we decided to focus on Math. When we looked at 4th grade Math Indiana State Standards, two
standards caught our eyes. 4.DA.3: Interpret data displayed in a circle graph and 4.DA.1:
Formulate questions that can be addressed with data. Use observations, surveys, and
experiments to collect, represent, and interpret the data using tables (including frequency
tables), line plots, and bar graphs. After looking at standards, we figured out that there are three
different graphs we each need: Circle graph, Line graph and Bar graph. However, we figured
out that one teacher teaching all three different graphs at one time would be very difficult and
would complicate things for students because they might get confused when they learn three
different graphs at one time. Therefore, we decided to do Station Teaching(small group of
students rotate to different stations for instruction). Ui Jeong Lee, one of our team members,
thought of this teaching method since she had an experience of doing Station Teaching at her
TIP experience when teaching three different topics. She shared her experience about how
Station Teaching is a very effective teaching method. It is effective because station teaching
allows students to visit different stations during allotted time for a specific subject. It also allows
teachers to work with small groups of studetns and give more individual instructions and also
allows teachers more time to get personal connection with different learning levels of individual
students. Our lesson content is something that our group members came up with, and we did
not use any other websites to look for created lesson plans. Instead of getting ideas by looking
at created lesson plan, we tried to use our experiences from TIP experience and cooperate our
own understadnings into our lesson plan. Therefore, we feel how our lesson plan is very unique.
At the end of the procedure of our lesson plan, there is a time for assessment. Students will
have to complete a worksheet, which deals with three different graphs they have learned from
three different sections. In order to create this lesson plan, we had to look for websites and
found a professional worksheet that was created based on standards. Other than getting the
worksheet from the website, everything on the lesson plan is something we created. For
technology, we decided to incorporate iPads into the lesson plan. In order to create different
graphs, we had to make students collect some kind of data. Therefore, we decided to put the
collected data into iPads and let students carry the iPads to three different stations to create a
graph. Also, one of the stations: line graph station, we decided to let students make line graphs
on the computer. There is a computer simulation which allows students to create a line graph.
Therefore, students had to look up the collected data from the iPad and create line graphs on
the computer.
Journal Articles
1.
Brown, J. (2011, December 1). Science and Technology Educators' Enacted Curriculum: Areas
of Possible Collaboration for an Integrative STEM Approach in Public Schools. Technology and
Engineering Teacher, 30-34.
science and technology/engineering teachers and how they can partner with each other to form
stronger content. The article also offers suggestions on how these different teachers can create
and enhance partnerships by better utilizing the strengths of their disciplines by coming
together. Integrating between the STEM disciplines is an excellent way to strengthen and build
better content for a lesson plan. For this reason we were able to integrate some of the ideas of
this article into our lesson plan by collaborating technology and mathematics to create a more
interactive learning experience. The article describes effective ways of integrating between
STEM disciplines, and we were able to apply these and use technology effectively to better
engage our students in the standards for mathematics.
2.
Breiner, J. (2012, January 2). What Is STEM? A Discussion About Conceptions of STEM in
Education and Partnerships. School Science and Mathematics, 3-11.
The article What Is Stem? A Discussion About Conceptions of STEM in Education and
Partnerships discusses the main ideas behind what STEM teaching entails and attempts to
accomplish. The article mentions that conceptions of STEM often vary among person to
person, and for this reason attempts to develop a common ground among those involved with
STEM. The article aims to clearly define STEM and its goals in order to show how STEM will
influence and impact the lives of those that experience it. This journal article shaped our lesson
plan by helping us understand what STEM is and what it aims to achieve. The journal clearly
outlined what its pourposes were, and we were able to incorporate these strategies into our
lesson plan by integrating the main ideas and philosophies so that we could positively portray
STEM teaching.
Assessment Worksheet | 3,590 | 16,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2019-43 | latest | en | 0.952899 |
https://quizlet.com/explanations/questions/find-each-product-2x2-3x-3x2-4x-2-371ae987-2eea-45e9-8d56-b2128da42d02 | 1,701,852,021,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.31/warc/CC-MAIN-20231206063543-20231206093543-00252.warc.gz | 523,670,807 | 72,283 | Try the fastest way to create flashcards
Question
# Find each product. (2x² - 3x)(-3x² + 4x - 2)
Solutions
Verified
Step 1
1 of 2
Given expression is $(2x^2-3x)(-3x^2-4x-2 )$
\begin{align*} 2x^2(-3x^2-4x-2)-3x(3x^2-4x-2)\\ &=-6x^4+8x^3-4x^2 -9x^3+12x^2+6x\\ &=-6x^4-x^3+8x^2+6x \end{align*}
Use the distrubutive property by multiplying each term with $2x^2$ and $-3x$ .Then add the like terms
## Recommended textbook solutions
#### enVision Algebra 1
1st EditionISBN: 9780328931576Al Cuoco, Christine D. Thomas, Danielle Kennedy, Eric Milou, Rose Mary Zbiek
3,653 solutions
#### SpringBoard Algebra 1
1st EditionISBN: 9781457301513SpringBoard
3,022 solutions
#### Algebra 1
4th EditionISBN: 9781602773011Saxon
5,377 solutions
#### Big Ideas Math Integrated Mathematics II
1st EditionISBN: 9781680330687Boswell, Larson
4,539 solutions | 324 | 848 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 9, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.524209 |
https://mail.python.org/pipermail/tutor/2010-March/075234.html | 1,516,497,398,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00667.warc.gz | 741,645,862 | 2,987 | # [Tutor] First program
Ray Parrish crp at cmc.net
Sat Mar 13 19:04:07 CET 2010
```Luke Paireepinart wrote:
> Ray, please reply on-list in the future in case someone else has input.
>
> On Fri, Mar 12, 2010 at 8:01 PM, Ray Parrish <crp at cmc.net
> <mailto:crp at cmc.net>> wrote:
>
> Luke Paireepinart wrote:
>
> print "A %s with dimensions %sx%s has an area of %s." %
> (choice, height, width, width*height)
>
>
> Isn't it a little more understandable to use a construct like the
> following?
>
> >>> print "The area of a " + Choice + "is " str(Width) + " x " +
> str(Height) + " equals " + str(Width * Height) + " square feet"
>
> The area of a rectangle is 12 x 10 equals 120 square feet.
>
> I find that putting the variables on the end like that, when
> you're not actually applying any special formatting to them makes
> it less readable when I'm debugging my stuff, or when someone else
> is reading my code, and trying to understand it.
>
>
> Your version creates at least 10 intermediate strings before outputting.
> Remember strings are immutable in Python.
> So you're constructing strings
> The area of a
> The area of a rectangle
> The area of a rectangle is
> 12
> The area of a rectangle is 12
> The area of a rectangle is 12 x
> 10
> The area of a rectangle is 12 x 10
> The area of a rectangle is 12 x 10 equals
> 120
> The area of a rectangle is 12 x 10 equals 120
> The area of a rectangle is 12 x 10 equals 120 square feet
>
> With string formatting you avoid all of these intermediate strings, so
> it's arguably more efficient.
> Other than just viewing from a performance standpoint though, I find
> it much easier to read my version, because any computation required
> takes place at the end of the line.
> whole line to see it.
>
> It's really a personal thing, it's easier for me to read the
> formatting version than the string concatenation version, in most cases.
> Now if you had used the comma convention I would have seen your
> point. This is, I think, the easiest to read of all 3
> area = width * height
> print "The area of a", choice, "is", width, "x", height, ", which
> equals", area, "square feet."
>
> Also, why are you capitalizing variable names? That's a pretty
> unusual convention.
>
> -Luke
Thanks for letting me know how inefficient my method is. I'll remember
that, and apply your suggestions to my code from now on. So, you're
saying that the commas method also does not suffer from the overhead of
creating a bunch of individual strings?
As far as the capitalizations, it's just a habit I've held over from my
Visual Basic days, and earlier programming. It's a little easier for me
to pick out the individual words in a variable like ThisPerson as
opposed to thisperson. I have been made aware that Python standard
coding practice requires lower case variable names, and as soon as I can
force myself to do it that way, or can remember to post that way I will
be adopting the in place standards and conventions.
Is it actually supposed to be this_person? I read part of the standards
document, but can not remember right off the top of my head if
underlines between words is required.
Thanks, Ray Parrish
--
Linux dpkg Software Report script set.. | 858 | 3,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-05 | latest | en | 0.897314 |
https://goprep.co/ex-10.e-q30-300-apples-are-distributed-equally-among-a-i-1nk1cm | 1,603,499,753,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881551.11/warc/CC-MAIN-20201023234043-20201024024043-00161.warc.gz | 352,278,963 | 59,694 | # 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.
Let the total number of students be x
According to the question
taking LCM
cross multiplying
Using the splitting middle term – the middle term of the general equation is divided in two such values that:
Product = a.c
For the given equation a = 1 b = 10 c = – 3000
= 1. – 3000 = – 3000
And either of their sum or difference = b
= 10
Thus the two terms are 60 and – 50
Difference = 60 – 50 = 10
Product = 60. – 50 = – 3000
x2 + 60x – 50x – 3000 = 0
x(x + 60) – 50(x + 60) = 0
(x + 60) (x – 50) = 0
(x – 50) = 0 or (x + 60) = 0
x = 50 or x = – 60
x cannot be negative thus total number of students = 50
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Knowing the Nature of Roots44 mins
Take a Dip Into Quadratic graphs32 mins
Foundation | Practice Important Questions for Foundation54 mins
Nature of Roots of Quadratic EquationsFREE Class
Getting Familiar with Nature of Roots of Quadratic Equations51 mins
Quadratic Equation: Previous Year NTSE Questions32 mins
Champ Quiz | Quadratic Equation33 mins
Balance the Chemical Equations49 mins
Champ Quiz | Quadratic Equation48 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses | 440 | 1,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-45 | longest | en | 0.883271 |
http://www.docstoc.com/docs/107509805/Extrinsic | 1,432,528,450,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00202-ip-10-180-206-219.ec2.internal.warc.gz | 433,695,038 | 39,260 | # Extrinsic by y12b8L8
VIEWS: 7 PAGES: 20
• pg 1
``` Problem: Perform a two-tailed hypothesis test on both the intrinsic and extrinsic variables data, using a .05 significance le
5.5 Data: n μ s
3.6 25 5 1.02
6.7
5.6 (1) Formulate the hypotheses:
5.5
4.4 H0: that is μ = 5
6.5 Ha: that is μ ≠ 5
6.9
4.6 (2) Decide the test statistic and the level of significance:
6.8
5.5 t (Two-tailed), α = 0.05 Lower Critical t- score = -2.0639
3.8
6.8 (3) State the decision Rule:
4.6
6.5 Reject H0 if |t| > 2.0639
6.1
3.9 (4) Calculate the value of test statistic:
6.2
5.5 SE = s/n = 0.2040
4.4 t = (x-bar - μ)/SE = 2.0098
4.7
6.2 (5) Compare with the critical value and make a decision:
4.7
5.5 Since 2.0098 < 2.0639
4.8
Decision: There is no sufficient evidence that the mean extrinsic satisfaction is different from 5.
x-bar = 5.41
s= 1.02 (6) Calculate the p- value for the test and interpret it:
p- value = 0.0558 This is the probability of rejecting a true null hypothesis.
data, using a .05 significance level. Begin by creating a null and alternate statement. Using Microsoft Excel to process your data. Copy a
x-bar
5.41
Upper Crtical t- score = 2.0639
we fail to reject H0
faction is different from 5.
ejecting a true null hypothesis.
to process your data. Copy and paste the results in Microsoft Word. Identify the significance level, the test statistic and the critical valu
test statistic and the critical value. State whether you are rejecting or failing to reject the null hypoothesis statement. In a separate par
esis statement. In a separate paragraph provide some information on when to use a t-test and when to use a z-test and why. Also prov
to use a z-test and why. Also provide some information about why samples are used instead of populations. This will be a T-test. There
tions. This will be a T-test. There are 25 intrinsic variables as follows: 5.5, 4.2, 6.2, 4.3, 3.2, 4.5, 4.7, 5.3, 6.2, 3.4, 5.5, 5.8, 6.2, 4.2, 5
5.3, 6.2, 3.4, 5.5, 5.8, 6.2, 4.2, 5.7, 6.4, 3.7, 6.2, 6.5, 5.1, 4.5, 6.3, 6.3, 5.2, and 6.6 There are 25 extrinsic variables as follows 5.5, 3
xtrinsic variables as follows 5.5, 3.6, 6.7, 5.6, 5.5, 4.4, 6.5, 6.9, 4.6, 6.8, 5.5, 3.8, 6.8, 4.6, 6.5, 6.1, 3.9, 6.2, 5.5, 4.4, 4.7, 6.2, 4,7, 5
3.9, 6.2, 5.5, 4.4, 4.7, 6.2, 4,7, 5.5, and 4.8
Problem: Perform a two-tailed hypothesis test on both the intrinsic and extrinsic variables data, using a .05 significance le
5.5 Data: n μ s x-bar
4.2 25 5 1.04 5.27
6.2
4.3 (1) Formulate the hypotheses:
3.2
4.5 H0: that is μ = 5
4.7 Ha: that is μ ≠ 5
5.3
6.2 (2) Decide the test statistic and the level of significance:
3.4
5.5 t (Two-tailed), α = 0.05 Critical t- score = Upper Crtical t- score = 2.0639
Lower -2.0639
5.8
6.2 (3) State the decision Rule:
4.2
5.7 Reject H0 if |t| > 2.0639
6.4
3.7 (4) Calculate the value of test statistic:
6.2
6.5 SE = s/n = 0.2080
5.1 t = (x-bar - μ)/SE = 1.2981
4.5
6.3 (5) Compare with the critical value and make a decision:
6.3
5.2 Since 1.2981 < 2.0639 we fail to reject H0
6.6
Decision: There is no sufficient evidence that the mean intrinsic satisfaction is different from 5.
x-bar = 5.27
s= 1.04 (6) Calculate the p- value for the test and interpret it:
p- value = 0.2066 This is the probability of rejecting a true null hypothesis.
s data, using a .05 significance level. Begin by creating a null and alternate statement. Using Microsoft Excel to process your data. Cop
isfaction is different from 5.
t Excel to process your data. Copy and paste the results in Microsoft Word. Identify the significance level, the test statistic and the critic
vel, the test statistic and the critical value. State whether you are rejecting or failing to reject the null hypoothesis statement. In a sepa
hypoothesis statement. In a separate paragraph provide some information on when to use a t-test and when to use a z-test and why. A
d when to use a z-test and why. Also provide some information about why samples are used instead of populations. This will be a T-test
populations. This will be a T-test. There are 25 intrinsic variables as follows: 5.5, 4.2, 6.2, 4.3, 3.2, 4.5, 4.7, 5.3, 6.2, 3.4, 5.5, 5.8, 6.
.5, 4.7, 5.3, 6.2, 3.4, 5.5, 5.8, 6.2, 4.2, 5.7, 6.4, 3.7, 6.2, 6.5, 5.1, 4.5, 6.3, 6.3, 5.2, and 6.6 There are 35 extrinsic variables as follow
re 35 extrinsic variables as follows 5.5, 3.6, 6.7, 5.6, 5.5, 4.4, 6.5, 6.9, 4.6, 6.8, 5.5, 3.8, 6.8, 4.6, 6.5, 6.1, 3.9, 6.2, 5.5, 4.4, 4.7, 6.2
5, 6.1, 3.9, 6.2, 5.5, 4.4, 4.7, 6.2, 4,7, 5.5, and 4.8
```
To top | 1,880 | 5,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2015-22 | longest | en | 0.761792 |
http://perplexus.info/show.php?pid=2100&cid=18539 | 1,537,602,786,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158205.30/warc/CC-MAIN-20180922064457-20180922084857-00396.warc.gz | 184,223,637 | 4,341 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Nice Weather (Posted on 2004-11-02)
During a certain period of days in Cucumberland recently it was observed that when it rained in the afternoon, it had been clear in the morning, and when it rained in the morning, it was clear in the afternoon. (In a given morning or afternoon, it is either raining or it is clear.) It rained on 100 days, and was clear on 19 afternoons and 95 mornings. How many days were there altogether?
No Solution Yet Submitted by SilverKnight Rating: 4.0000 (2 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Solution by Equation | Comment 7 of 11 |
We know there are 19 clear afternoons and 95 clear mornings. We also know there are 100 days on which it rained. The following equation should show if there are any days that are clear both in the morning and in the afternoon.
1/2 x (19 + 95 - 100) = 1/2 x 14 = 7.
Thus, there are 7 days that are clear in the morning and in the afternoon.
Proof: With 19 - 7 days of clear afternoon, we have 12 days of rain in the morning and none in the afternoon. With 95 - 7 we have 88 days of rain in the afternoon and clear in the morning. Therefore, we have 12 + 88 = 100 days in which it rained, as the problem requires.
Gordon S.
Posted by Gordon Steel on 2004-11-02 22:05:10
Search: Search body:
Forums (0) | 386 | 1,431 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.946711 |
https://www.jiskha.com/questions/337914/ch4-2-cl2-ch2cl2-2hcl-cl-cl-bond-energy-242-kj-mole-1-calculate-the-amount-of | 1,607,103,338,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141740670.93/warc/CC-MAIN-20201204162500-20201204192500-00084.warc.gz | 703,189,615 | 5,684 | # college chemistry
CH4 + 2 Cl2--> CH2Cl2 + 2HCl
Cl-Cl bond energy: 242 kj/mole
1. Calculate the amount of energy (Joules) needed to break a single Cl-Cl bond.
2. Calculate the longest wavelength of light (meters)that can supply the energy per photon necessary to break the Cl-Cl bond.
1. 👍 0
2. 👎 3
3. 👁 1,371
1. 242 kJ/mol = 242,000 J/mol
242,000 J/mol x (1 mol/6.022 x 10^23 molecules) = E for one molecule in joules.
E = hc/wavelength.
1. 👍 1
2. 👎 0
2. 24
1. 👍 0
2. 👎 0
3. Calculate the total energy required to break bonds of the reactants if c-h is 414 kj mol
O-h 460 kj mol
o=o 499
c=o 799 kj mol
1. 👍 0
2. 👎 0
4. Ch4+2h2o ›2h2oo+co2+energy c-h 414 kj mol
o-h 460 kj mol
o=o 499 kj mol
c=h 799 kj mol
1. 👍 0
2. 👎 0
## Similar Questions
1. ### AP Chemistry
Calculate the delta H rxn for the following reaction: CH4(g)+4Cl2(g)-->CCl4(g)+4HCl Use the following reactions and given delta H's: 1) C(s)+2H2(g)-->CH4(g) delta H= -74.6 kJ 2) C(s)+2Cl2(g)-->CCl4(g) delta H= -95.7 kJ 3)
2. ### Chemistry
Use an enthalpy diagram to calculate the lattice energy of CaCl2 from the following information. Energy needed to vaporize one mole of Ca(s) is 192 kJ. For calcium, the first IE = 589.5 kJ mol-1, the second IE = 1146 kJ mol-1. The
3. ### chemistry:)
The bond enthalpy of the Br–Cl bond is equal to DH° for the reaction BrCl(g)-> Br(g) + Cl(g). Use the following data to find the bond enthalpy of the Br–Cl bond. Br2(l)--->Br2(g) ÄH=30.91 KJ/mol Br2(g)--->2Br2(g) ÄH=192.9
4. ### Chemistry
Use a Born Haber cycle to calculate the lattice energy (in kJ/mol) of Potassium Chloride (KCl) from the following data: Ionization Energy of K(g) = 431kJ/mol Electron affinity of Cl(g) = -346kJ/mol Energy to sublime K(s) =
1. ### Chemistry
CH4 + 2 O2 --> CO2 + 2 H2O + heat This reaction is exothermic because more energy is released when the products bond than is required to break the bonds of the reactants. The potential energy of the products (not including the
2. ### Chemistry
How much energy, in joules, is required to break the bond in ONE chlorine molecule? The bond energy of Cl2 is 242.6kJ/mol?
3. ### Science
When methane (CH4) combines with carbon tetrachloride , dichloromethane (CH2Cl2) is formed.
4. ### chemestry
What is occurring in the following reaction? H2 + Cl2 → 2HCl A. H2 is being reduced. B. Cl2 is being oxidized. C. H2 is gaining two electrons. D. Cl2 is acting as an oxidizing agent. is the answer C? please explain
1. ### chemistry
1)how many molecules of water are needed to produce 6 molecules of C6H12O6 according to the equation 6CO2(G)+6H2O(L)+energy TO C6H12O6(S)+6O2(G)? 2)how many mole of oxygen gas are produce when 6 mol of CO2 are consumed in the
2. ### chemistry
What is the mass of 4.80 moles of MgCl2?My answer is: Mg=24.3 g/mole Cl2=35.5g/mole*2=71g/mole so, Mg 24.3+ Cl2 71= 95.3g/mole therefore 4.80 moles * 95.3 g/mole=471.84g Am i right pls correct me if i'm wrong thanks
3. ### CHEM
a mixture of 10cm3 of methane and 10cm3 of enthane was sparked with an excess of oxygen. after coolin to room temp, the residual gas was passed through aq KOH. what volume of gas was absorbed by the alkali ? Write the equations
4. ### Chemistry
Calculate the lattice energy in kJ/mol for CaCl2 from the following information: Energy needed to vaporize one mole of Ca(s) is 192kJ. For calcium the first ionization energy is 589.5kJ/mol and the second ionization energy is | 1,148 | 3,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.667663 |
https://numbermatics.com/n/851086/ | 1,653,799,528,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00234.warc.gz | 501,736,728 | 6,125 | # 851086
## 851,086 is an even composite number composed of three prime numbers multiplied together.
What does the number 851086 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 8 divisors.
851086 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of eight divisors.
## Prime factorization of 851086:
### 2 × 19 × 22397
See below for interesting mathematical facts about the number 851086 from the Numbermatics database.
### Names of 851086
• Cardinal: 851086 can be written as Eight hundred fifty-one thousand and eighty-six.
### Scientific notation
• Scientific notation: 8.51086 × 105
### Factors of 851086
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 3
• Sum of prime factors: 22418
### Divisors of 851086
• Number of divisors d(n): 8
• Complete list of divisors:
• Sum of all divisors σ(n): 1343880
• Sum of proper divisors (its aliquot sum) s(n): 492794
• 851086 is a deficient number, because the sum of its proper divisors (492794) is less than itself. Its deficiency is 358292
### Bases of 851086
• Binary: 110011111100100011102
• Base-36: I8PA
### Squares and roots of 851086
• 851086 squared (8510862) is 724347379396
• 851086 cubed (8510863) is 616481913740624056
• The square root of 851086 is 922.5432239195
• The cube root of 851086 is 94.7671490321
### Scales and comparisons
How big is 851086?
• 851,086 seconds is equal to 1 week, 2 days, 20 hours, 24 minutes, 46 seconds.
• To count from 1 to 851,086 would take you about two days!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 851086 cubic inches would be around 7.9 feet tall.
### Recreational maths with 851086
• 851086 backwards is 680158
• The number of decimal digits it has is: 6
• The sum of 851086's digits is 28
• More coming soon!
MLA style:
"Number 851086 - Facts about the integer". Numbermatics.com. 2022. Web. 29 May 2022.
APA style:
Numbermatics. (2022). Number 851086 - Facts about the integer. Retrieved 29 May 2022, from https://numbermatics.com/n/851086/
Chicago style:
Numbermatics. 2022. "Number 851086 - Facts about the integer". https://numbermatics.com/n/851086/
The information we have on file for 851086 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 851086, math, Factors of 851086, curriculum, school, college, exams, university, Prime factorization of 851086, STEM, science, technology, engineering, physics, economics, calculator, eight hundred fifty-one thousand and eighty-six.
Oh no. Javascript is switched off in your browser.
Some bits of this website may not work unless you switch it on. | 876 | 3,329 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | latest | en | 0.861635 |
https://www.esaral.com/q/let-x-and-y-be-rational-and-irrational-numbers-21106 | 1,726,250,902,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00560.warc.gz | 682,995,942 | 11,542 | # Let x and y be rational and irrational numbers,
Question:
Let $x$ and $y$ be rational and irrational numbers, respectively. Is $x+y$ necessarily an irrational number? Give an example in support of your answer.
Solution:
Yes, $(x+y)$ is necessarily an irrational number.
e.g., Let $\quad x=2, y=\sqrt{3}$
Then, $x+y=2+\sqrt{3}$
If possible, let $x+y=2+\sqrt{3}$ be a rational number.
Consider $a=2+\sqrt{3}$
On squaring both sides, we get
$a^{2}=(2+\sqrt{3})^{2}$ [using identity $(a+b)^{2}=a^{2}+b^{2}+2 a b$ ]
$\Rightarrow$ $a^{2}=2^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})$
$\Rightarrow$ $a^{2}=4+3+4 \sqrt{3} \Rightarrow \frac{a^{2}-7}{4}=\sqrt{3}$
So, $a$ is rational $\Rightarrow \frac{a^{2}-7}{4}$ is rational $\Rightarrow \sqrt{3}$ is rational. | 281 | 791 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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-2024-38 | latest | en | 0.526695 |
https://myassignmenthelp.com/free-samples/medi11002-physics-for-health-sciences/atmospheric-pressure.html | 1,669,491,163,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00478.warc.gz | 468,115,365 | 73,578 | # MEDI11002 Physics For Health Sciences
## Questions
### Part 1
A glossary of all of the keywords you have used in your answers to Part 2 questions.
• Each keyword must be defined in your glossary. Include any symbols and unit of measure associated with the keyword provided.
• Once you have defined a keyword in your glossary, you can use that term throughout your responses to the questions in Part 2 without the need to define the term again.
• If you have used any equations to answer Part 2, include them in your glossary as well.
### Part 2
1. For each of the statements below, indicate whether the statement is true or false and explain your reasoning.
a. As a medical supply trolley is wheeled around a curved pathway, its velocity and acceleration remains unchanged.
b. When an elastic band is stretched around a patient’s arm to act as a tourniquet, there is no work done and the band has no stored energy.
2. A student healthcare professional has been asked to move a box of mass 12 kg onto a shelf that is at a height 0.3 m above the ground level and 5.5 m from where the box was originally kept. The student is holding the box and moving with the box in a straight line at a steady speed of 0.15 m/s. As the student approaches the shelf, he notices another student is opening the door adjacent to the shelf area. If the student is able to apply sufficient force to stop in 2. s, will the student avoid a collision if the opening door is 1.0 m ahead of the box? (Assume that the student is applying the same magnitude of force throughout that time).
3. Whist waiting for your next lecture session, you purchase a single-serve soft drink in a standard plastic bottle with screw-on lid. You open the lid to take a drink, and notice fluid flowing from a small hole in the bottom of the bottle. You put the lid back on the bottle to return it to the store as defective. However, once you put the lid back on, the flow from the hole stops!
i. Differentiate between the properties of the soft drink and the plastic bottle.
ii. Explain why flow did not start until the lid was off the bottle.
### Cite This Work
My Assignment Help (2021) Physics For Health Sciences [Online]. Available from: https://myassignmenthelp.com/free-samples/medi11002-physics-for-health-sciences/atmospheric-pressure.html
[Accessed 26 November 2022].
My Assignment Help. 'Physics For Health Sciences' (My Assignment Help, 2021) <https://myassignmenthelp.com/free-samples/medi11002-physics-for-health-sciences/atmospheric-pressure.html> accessed 26 November 2022.
My Assignment Help. Physics For Health Sciences [Internet]. My Assignment Help. 2021 [cited 26 November 2022]. Available from: https://myassignmenthelp.com/free-samples/medi11002-physics-for-health-sciences/atmospheric-pressure.html.
## Stuck on Any Question
250 words
### We Can Help!
Get top notch assistance from our best tutors !
### Content Removal Request
If you are the original writer of this content and no longer wish to have your work published on Myassignmenthelp.com then please raise the content removal request.
## 5% Cashback
On APP - grab it while it lasts! | 723 | 3,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-49 | latest | en | 0.940394 |
https://www.clutchprep.com/physics/practice-problems/39316/a-4-15-m-by-4-15-m-square-base-pyramid-with-height-of-2-53-m-is-placed-in-a-vert | 1,618,264,663,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00317.warc.gz | 800,863,992 | 31,058 | Gauss' Law Video Lessons
Concept
# Problem: A (4.15 m by 4.15 m) square base pyramid with height of 2.53 m is placed in a vertical electric field of 50.1 N/C. Calculate the total electric flux which goes out through the pyramid’s four slanted surfaces.
###### FREE Expert Solution
100% (420 ratings)
###### Problem Details
A (4.15 m by 4.15 m) square base pyramid with height of 2.53 m is placed in a vertical electric field of 50.1 N/C. Calculate the total electric flux which goes out through the pyramid’s four slanted surfaces.
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Gauss' Law concept. You can view video lessons to learn Gauss' Law. Or if you need more Gauss' Law practice, you can also practice Gauss' Law practice problems.
What is the difficulty of this problem?
Our tutors rated the difficulty ofA (4.15 m by 4.15 m) square base pyramid with height of 2.53...as high difficulty.
How long does this problem take to solve?
Our expert Physics tutor, Douglas took 9 minutes and 3 seconds to solve this problem. You can follow their steps in the video explanation above.
What professor is this problem relevant for?
Based on our data, we think this problem is relevant for Professor Li's class at TEXAS. | 315 | 1,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-17 | latest | en | 0.92982 |
http://mathhelpforum.com/algebra/140592-reverse-foil-factoring.html | 1,511,550,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00684.warc.gz | 187,362,756 | 10,430 | # Thread: Reverse FOIL Factoring
1. ## Reverse FOIL Factoring
My question is what to do if there is a random variable in the middle number/variable in a trinomial when factoring the trinomial using reverse FOIL. The problem itself was n^2-26mn+25n^2. When I factored it using the outer numbers/variables, I ended up getting (n-25n)(n-n), and when I redistribute everything, it becomes n2-26n^2+25n^2, and the middle number/variable should be -26mn. How do I get the m and get rid of the extra n?
2. Originally Posted by Vinefeather
My question is what to do if there is a random variable in the middle number/variable in a trinomial when factoring the trinomial using reverse FOIL. The problem itself was n^2-26mn+25n^2. When I factored it using the outer numbers/variables, I ended up getting (n-25n)(n-n), and when I redistribute everything, it becomes n2-26n^2+25n^2, and the middle number/variable should be -26mn. How do I get the m and get rid of the extra n?
Assuming there is no typo in n^2-26mn+25n^2, just combine terms and factor out an n. We have
n^2-26mn+25n^2
-26mn+26n^2
26n(-m+n)
3. Thank you sooooo much!!~ <3 Can't believe I didn't think of that sooner, lol.
4. Another question, how would I solve this: 5r^2+29rt-5t^2? I can't seem to get the 29rt no matter how hard I try... D:
5. Originally Posted by Vinefeather
Another question, how would I solve this: 5r^2+29rt-5t^2? I can't seem to get the 29rt no matter how hard I try... D:
This expression does not factor nicely, as far as I can tell. Possibly your teacher/book just intends you to leave it as it is.
I'm curious to know if there is a nice solution here, please report back if nobody answers and if you get such an answer from your teacher/book. I might just be blind.
(Edited) | 519 | 1,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-47 | longest | en | 0.927177 |
https://blogs.wsj.com/puzzle/2017/07/06/varsity-math-week-95/ | 1,553,053,965,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202199.51/warc/CC-MAIN-20190320024206-20190320050206-00062.warc.gz | 422,627,775 | 104,143 | DOW JONES, A NEWS CORP COMPANY
News Corp is a network of leading companies in the worlds of diversified media, news, education, and information services.
DOW JONES
NEWS CORP
# Varsity Math Week 95
Photo: Luci Gutierrez
This week’s problems are based on a 2005 Math Horizons article written by the great mathematics and science writer Martin Gardner (1914-2010).
Proper Place
Consider all of the counting numbers squashed into one long string of digits, like so: 12345678910111213141516… The “proper place” of a number is the position that the first digit of its occurrence in counting order occupies in this string of digits. For example, the proper place of 13 is 16, because the “1” of the “13” occurs in the 16th position in this sequence
of digits.
What is the proper place of 2017?
Earliest Bird
Martin Gardner called a number an “early bird” if it appears in the above sequence before its proper place. For example, 12 is an early bird because “12” appears as the first two digits of the sequence, even though its proper place is 14. In fact, it is exactly 13 places early; that’s its proper place minus the position of the first digit of its actual first occurrence in the sequence. On the other hand, some numbers like 11 are not early birds at all; their first occurrences are exactly in their proper places. As a warm-up for this problem, you can figure out whether or not 2017 is an early bird.
What early bird less than or equal to 1,000 is early by the most places?
*****
See answers to last week’s puzzles below. Come back next week for answers and more puzzles. | 377 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-13 | latest | en | 0.934895 |
https://gmatclub.com/forum/at-a-certain-riding-school-students-are-randomly-assigned-10839.html?fl=similar | 1,498,207,899,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320040.36/warc/CC-MAIN-20170623082050-20170623102050-00640.warc.gz | 748,705,471 | 47,945 | It is currently 23 Jun 2017, 01:51
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# At a certain riding school, students are randomly assigned
Author Message
Manager
Joined: 02 Jul 2004
Posts: 51
At a certain riding school, students are randomly assigned [#permalink]
### Show Tags
13 Oct 2004, 23:13
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
At a certain riding school, students are randomly assigned to the horses they will use during any given lesson. The school has exactly 5 male horses and 6 female horses. If there are exactly 3 women and 8 men in the class, what is the probability that a student will be assigned a horse of the same sex?
I'll post the answer but this problem threw me off even though it's not that hard.
Manager
Joined: 02 Jul 2004
Posts: 51
### Show Tags
18 Oct 2004, 16:44
OA is 58/121.
Possible assignment of 11 horses to 11 people = 121.
Possible ways to assign 6 female horses to 3 women = 18.
Possible ways to assign 5 male horses to 8 men = 40.
Possible ways to assign horse to someone of same gender = 18 + 48 = 58
Probability of being assigned to a horse of the same gender 58/121
Manager
Joined: 13 Oct 2004
Posts: 50
### Show Tags
20 Oct 2004, 22:20
I got the same answer, slightly different way to look at it
P(Man) = 8/11 , P (Woman) = 3/11
P(Male Horse) = 5/11, P (Fem Horse) = 6/11
so the required prob = P(Man)*P(Male Horse) + P(Woman)*P(Fem Horse)
= 8/11*5/11 + 3/11*6/11 = (40 + 18)/121 = 58/121
Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
### Show Tags
24 Oct 2004, 02:45
PracticeMore wrote:
I got the same answer, slightly different way to look at it
P(Man) = 8/11 , P (Woman) = 3/11
P(Male Horse) = 5/11, P (Fem Horse) = 6/11
so the required prob = P(Man)*P(Male Horse) + P(Woman)*P(Fem Horse)
= 8/11*5/11 + 3/11*6/11 = (40 + 18)/121 = 58/121
wonderfull method buddy
thanks,
Dharmin
_________________
Perseverance, Hard Work and self confidence
Manager
Joined: 19 Jun 2004
Posts: 79
### Show Tags
24 Oct 2004, 18:13
nice... good method, i prefer in this way
==> # of possibles outcomes:
we need a couple of one horse (11C1) and one student (11C1), so we obtain: 11C1 * 11C1 = 11*11 = 121 couples
==> # of favorable outcomes:
A couple of one male horse (5C1) and one man (8C1), so we obtain:
5C1 * 8C1 = 5*8 = 40 couples
A couple of one fem horse (6C1) and one woman (3C1), so we obtain:
6C1 * 3C1 = 6*3 = 18 couples
==> Therefore, the probability will be: (40+18)/121 = 58/121
hope it helps
Intern
Joined: 26 Aug 2004
Posts: 12
### Show Tags
24 Oct 2004, 21:48
I too used the method used by "PracticeMore". I guess, if one has a consistent method that works for all problems, then he/she is ok
24 Oct 2004, 21:48
Display posts from previous: Sort by | 1,067 | 3,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | longest | en | 0.927208 |
http://clay6.com/qa/74852/in-p-n-junction-diode-the-current-i-can-be-expressed-i-i-0-exp-bigg-large-f | 1,601,163,368,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00037.warc.gz | 23,593,245 | 6,374 | Comment
Share
Q)
# In p-n junction diode , the current I , can be expressed $I= I_0 exp \bigg[ \large\frac{eV}{K_{B}T}-1 \bigg]$ Where $I_0$ is called the satiation current , V is the voltage across the dicode and in positive for forward bias and negative for reverse bias and I is the current through the dicode , $K_B$ is the Boltzmann constant $(8.6 \times 10^{-5}e^v/k)$ and T is the absolute temperature . If for a given diode $I_0 = 5 \times 10^{-12}A$ and $T=300\;K$ , then what will be the forward current at a forward voltage of $0.6\;V$?
$\begin{array}{1 1} 0.63\;A \\ 0.063\;A \\ 6.3\;A \\ 0.0063\;A \end{array}$ | 226 | 625 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-40 | latest | en | 0.728689 |
http://www.mathsisfun.com/puzzles/who-weighs-what-solution.html | 1,440,928,411,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065241.14/warc/CC-MAIN-20150827025425-00191-ip-10-171-96-226.ec2.internal.warc.gz | 566,269,221 | 3,042 | # Who Weighs What Puzzle - Solution
### The Puzzle:
Hungry Horace likes to save money whenever he can (so that he's got plenty left to buy more food) so when he went swimming with some of his friends he had a clever idea to use the weighing machine to weigh him and his two friends for only one 10c coin!
Once the weighing machine has shown a reading the dial can only go down to a lower weight. So this is what Horace did. He and his two friends sorted themselves out in order of weight (they knew that Horace was the heaviest and that Tiny Tim was the lightest), and then followed this plan:
Hungry Horace and Curly Kate put the 10c in and got on the scales.
The dial showed 85 kg
Tiny Tim got on and Curly Kate got off.
The dial went down to 75 kg
Curly Kate got back on and Hungry Horace got off.
The dial went down to 60 kg.
Find the correct individual weights of Hungry Horace, Curly Kate and Tiny Tim.
### Our Solution:
From the question we know:
1) Horace + Kate = 85
2) Horace + Tim = 75
3) Kate + Tim = 60
Steps 1 and 2 tell us that Curly Kate is 10 kg heavier than Tiny Tim.
Step 3 therefore tells us that Tim is 25 kg and Kate is 35 kg.
Similarly we can work out that Hungry Horace is 15 kg heavier than Curly Kate. So Horace is 50 kg.
Puzzle Author: Stephen Froggatt
See this puzzle without solution
Discuss this puzzle at the Math is Fun Forum | 357 | 1,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2015-35 | longest | en | 0.962031 |
https://outer-court.com/basic/echo/T1630.HTM | 1,726,170,877,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00351.warc.gz | 408,368,424 | 2,701 | # Formula Solver 1.4 6/
``` BBS: Inland Empire Archive
Date: 03-19-93 (21:00) Number: 373
From: QUINN TYLER JACKSON Refer#: NONE
To: ALL Recvd: NO
Subj: Formula Solver 1.4 6/ Conf: (2) Quik_Bas```
```>>> Continued from previous message
SELECT CASE Phase%
CASE 1
' See if something of a higher precedence should be done first.
CALL sqjDesParse(2, x)
Op\$ = TOKEN\$(LvlPtr)
' The lowest level of precedence is handled by this Level.
DO WHILE fqjInList(LOGICAL, Op\$)
CALL sqjGetToken
CALL sqjDesParse(2, y)
CALL sqjApplyOp(Op\$, x, y)
Op\$ = TOKEN\$(LvlPtr)
LOOP
CASE 2
' See if something of a higher precedence should be done first.
CALL sqjDesParse(3, x)
Op\$ = TOKEN\$(LvlPtr)
CALL sqjGetToken
CALL sqjDesParse(3, y)
CALL sqjApplyOp(Op\$, x, y)
Op\$ = TOKEN\$(LvlPtr)
LOOP
CASE 3
' See if something of a higher precedence should be done first.
CALL sqjDesParse(4, x)
Op\$ = TOKEN\$(LvlPtr)
DO WHILE fqjInList(MULTDIV, Op\$)
CALL sqjGetToken
CALL sqjDesParse(4, y)
CALL sqjApplyOp(Op\$, x, y)
Op\$ = TOKEN\$(LvlPtr)
LOOP
CASE 4
' See if something of a higher precedence should be done first.
CALL sqjDesParse(5, x)
Op\$ = TOKEN\$(LvlPtr)
IF fqjInList(POWER, Op\$) THEN
CALL sqjGetToken
CALL sqjDesParse(5, y)
CALL sqjApplyOp(Op\$, x, y)
END IF
CASE 5
Op\$ = ""
IF TypeToken(LvlPtr) = OperatorClass AND (fqjInList(ADDSUB,_
TOKEN\$(LvlPtr))) THEN
Op\$ = TOKEN\$(LvlPtr)
CALL sqjGetToken
END IF
CALL sqjDesParse(6, x)
' This handles negative prefixes
SELECT CASE Op\$
CASE "-"
x = -x
END SELECT
CASE 6
' This level handles parentheses
IF TOKEN\$(LvlPtr) = "(" AND TypeToken(LvlPtr) = OperatorClass_
THEN
CALL sqjGetToken
CALL sqjDesParse(1, x)
IF TOKEN\$(LvlPtr) <> ")" THEN
ErrorCode = eqjMismatchedParenthesis
END IF
CALL sqjGetToken
ELSE
SELECT CASE TypeToken(LvlPtr)
CASE DigitClass
x = VAL(TOKEN\$(LvlPtr))
CALL sqjGetToken
CASE FunctionClass
x = fqjSolveFormula(TOKEN\$(LvlPtr))
TypeToken(LvlPtr) = DigitClass
CALL sqjGetToken
END SELECT
END IF
END SELECT
END SUB
SUB sqjGetToken ()
TOKEN\$(LvlPtr) = ""
DO WHILE fqjInList(WHITESPACE, MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1))
PTR(LvlPtr) = PTR(LvlPtr) + 1
LOOP
Temp\$ = MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)
IF Temp\$ >= "0" AND Temp\$ <= "9" THEN
' Build up a number from its digits
DO WHILE INSTR(" ()" + OPERATOR\$, MID\$(EXPR\$(LvlPtr), PTR(LvlPtr),_
1)) = 0
TOKEN\$(LvlPtr) = TOKEN\$(LvlPtr) + MID\$(EXPR\$(LvlPtr),_
PTR(LvlPtr), 1)
PTR(LvlPtr) = PTR(LvlPtr) + 1
LOOP
TypeToken(LvlPtr) = DigitClass
EXIT SUB
END IF
IF INSTR("()" + OPERATOR\$, MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)) THEN
TypeToken(LvlPtr) = OperatorClass
TOKEN\$(LvlPtr) = MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)
PTR(LvlPtr) = PTR(LvlPtr) + 1
IF INSTR("()", TOKEN\$(LvlPtr)) THEN
EXIT SUB
ELSE
' see if it's a compound operator
IF INSTR(OPERATOR\$, MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)) THEN
Temp\$ = MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)
IF Temp\$ <> "-" THEN
TOKEN\$(LvlPtr) = TOKEN\$(LvlPtr) + Temp\$
PTR(LvlPtr) = PTR(LvlPtr) + 1
END IF
END IF
END IF
EXIT SUB
END IF
Temp\$ = MID\$(EXPR\$(LvlPtr), PTR(LvlPtr), 1)
IF Temp\$ >= "@" AND Temp\$ <= "Z" THEN
>>> Continued to next message
* OLX 2.1 TD * A program is just a big bug that happened to work....
```
Echo Basic Postings
Books at Amazon:
Back to BASIC: The History, Corruption, and Future of the Language
Hackers: Heroes of the Computer Revolution (including Tiny BASIC)
Go to: The Story of the Math Majors, Bridge Players, Engineers, Chess Wizards, Scientists and Iconoclasts who were the Hero Programmers of the Software Revolution
The Advent of the Algorithm: The Idea that Rules the World
Moths in the Machine: The Power and Perils of Programming
Mastering Visual Basic .NET | 1,287 | 3,755 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.36509 |
http://www.pokerlistings.com/strategy/beginner/how-to-determine-the-winning-hand?commentsstart6380=50 | 1,516,332,211,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887729.45/warc/CC-MAIN-20180119030106-20180119050106-00474.warc.gz | 530,310,215 | 15,895 | How to Determine the Winning Poker Hand
When playing poker with your family or friends, one of the things you're going to need to know is how to determine the winning hand in all scenarios.
Before we go any further, first you need to memorize or print out the order of poker hands.
Once you know that a flush beats a straight and three of a kind beats two pair, you're off to a good start.
The majority of poker hands are simple to determine a winner from.
If one player has a flush, and no one else has a flush or better, it doesn't take much thought to figure out who's the winner.
It's once things get a little bit more complicated that people start to get confused. First, you want to remember these rules of poker hands:
• You must make the best hand possible using exactly five cards
• All five cards are used in deciding the strength of the hand
• No cards outside of the best five have any bearing on the strength of the hand
If you're playing Texas Hold'em poker, players are allowed to use any combination of cards from their hand and/or the board cards.
This means if the absolute best five-card hand a player can make is by using the five cards on the board, then that is his or her final hand (this is known as playing the board).
Some Common Areas of Confusion
Here's a quick rundown of a couple common areas of confusion, and how to resolve the winner:
Two Players (or More) Have a Flush
If more than one player has a flush, you award the pot to the player with the highest flush. This includes all five cards, for example:
Board:
Player 1:
Player 2:
In this scenario, Player 1 wins the pot. The reason is that when you look at all five cards, Player 1 has the higher flush:
Player 1:
Player 2:
All the cards are the same, until the final fifth card. Since 7 is higher than 6, Player1 wins the entire pot.
If instead of the 2 on the board, that card was the T, both players would have the same flush (playing the board) and the pot would be split.
Two Players Have Two Pairs
When two players have two pairs, it can sometimes be confusing for people to know who won.
Take this example:
Board:
Player 1:
Player 2:
In this scenario, Player1 wins the entire pot. Two pair is always ranked by the value of the highest pair first, and only if that pair is the same for both players do you rank by the second pair.
If both of two pairs are identical, it will be the kicker that will decide the winner (the highest-value fifth card is the kicker).
In this scenario because the two paired on the river, Player 1 has two pair - A A 2 2 with the kicker K.
Player 2 has the lower two pair - K K Q Q with the kicker 3. Aces are higher than kings, so Player 1 wins the entire pot.
Who Wins?
Board:
Player 1:
Player 2:
Take a second to figure it out. This is a very bad beat, as once the river falls both players now have four of a kind with nines.
Only Player 1, who up until this point had nothing special, has the highest kicker with an ace.
Even though Player 2 flopped a full house - K K K 9 9 - once the fourth nine fell, he was now playing four-of-a-kind nines with a king kicker.
Player 1 wins the whole pot.
The Omaha Rule
The rules in determining the best hand in Omaha are exactly the same as in Texas Hold'em with one additional rule:
• Every player must make the best five-card hand using exactly two cards from his hand (you're dealt four cards in Omaha) and three cards from the board.
This means that if there are four hearts on the board and you only have one in your hand you do not have a flush.
You must always use exactly two cards from your hand.
More Beginner Strategy Articles:
19 December 2017
8 November 2017
When to Fire a Second Barrel on the Turn: A Simple Guide
27 September 2017
A Noob's Guide to 8-Game: Learn 8 Poker Games, Crush Them
5 September 2017
Error saving comment!
You need to wait 3 minutes before posting another comment.
Cody Gondek 2017-01-14 20:14:47
player two wins with A stra8 flush
Cody Gondek 2017-01-14 20:12:29
nooo player 2 wins
Cody Gondek 2017-01-14 20:11:38
player 2 wins
Cody Gondek 2017-01-14 20:11:02
that only applied on two pair
janice love 2017-01-08 18:22:38
Player one has 9,8. Player two has 7,7. The community cards are a A,7,4,6,7. Which player wins?
Devon Mitchell 2017-01-06 05:50:14
That's 5 9s
Devon Mitchell 2017-01-06 05:45:26
Player 2 has 2 pair jack higher than 8
Geneva Cardinas 2017-01-01 14:51:39
Player 1 has 8S 8C
Player 2 has JH 2D
Board 5H, 9H, 9D, JS, 10D
Who wins?
Robert M Howe 2016-12-16 03:41:25
P1: Q10
P2:Q9
Table: j,5,4,6,A
Who wins and why or is it a split pot??
Songyang Yuan 2016-12-07 17:28:51
P1 A8
P2 22
Board 28877
Who win?
Best Poker Sites - Editor`s Pick
Poker Site Ranking Bonus
1 Free Money Poker \$1,000 Review
• How Poker Got Its Bad Name: Sol Smith & the Steamboat Hustle
You know how poker keeps trying to get rid of its reputation of being a...
26 September 2017
• Beyond the Poker Table: Off the Beaten Path in Barcelona
For over a decade poker players have journeyed to Barcelona to play PokerStars events. It’s undoubtedly...
25 August 2017
• See More Blog Posts »
• Battle of Malta 2017: Nadav Lipszyc wins final table, takes home €200,000 prize
8 November 2017 39
• Poker Action and Side Events on Battle of Malta Day 3
6 November 2017 121
×
Sorry, this room is not available in your country. | 1,456 | 5,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-05 | latest | en | 0.968064 |
https://www.wired.com/2014/08/lotr-physics/ | 1,685,824,070,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00279.warc.gz | 1,139,988,488 | 168,806 | # Lord of the Rings Physics Homework
How long is the Balrog's whip? Pretty dang long.
I don't think I am going to give any Lord of the Rings spoilers here. The books are over 50 years old and even the first movie (the Peter Jackson version) was released in 2001. So let me just say that at one point, Gandalf fights a monster and essentially dies. That sucks for Gandalf (although I think he leveled-up in the process) but it is a great opportunity for some physics homework.
Just so that we all start with the same material, I am going to use the Gandalf vs. Balrog scene from the Peter Jackson version of Fellowship of the Ring and The Two Towers. The fight actually happens during Fellowship of the Ring but there are pieces of it shown in The Two Towers. If you do a quick search for "Gandalf vs Balrog", you should be able to find the video of it without too much difficulty.
No one is happy if I just give out homework questions. What about an example from the textbook? Well, there isn't a textbook. This is real world stuff. But ok, I'll start off with an example.
## How Long is the Balrog's Whip?
Here's the scene. Gandalf and the rest of the party are trying to escape from Moria. They cross a narrow bridge and then Gandalf makes a stand agains the Balrog. In order to prevent the Balrog from crossing, Gandalf breaks the bridge and the monster falls. But wait! At the last second, the Balrog's whip grabs a hold of Gandalf and pulls him into the chasm.
Let's get data from the video to estimate the length of the Balrog's whip. But first, a couple of assumptions.
• Even though this is Middle Earth, I will assume that the local gravitational field is just like on Earth. That means that a free falling object will have an acceleration of 9.8 m/s2.
• During this initial falling motion of the Balrog, I am going to assume that the air resistance is negligible so it accelerates with a constant acceleration.
Now for the only measurement from the video - time of fall. It's difficult to pinpoint exactly when the Balrog was in free fall as well as the time the whip hits Gandalf. I can use Tracker Video Analysis and mark a beginning and ending frame for the fall (with a conservative estimate of the start and stop times). This gives a falling time of about 13.8 seconds.
How far did the Balrog fall during this time interval? Well, if I stick with my original assumption that there isn't any air resistance I can use the following kinematic equation:
I can set the initial position at 0 meters and if the Balrog starts from rest, the initial velocity is 0 m/s. Putting in a time of 13.8 seconds and a value of g at 9.8 m/s2, I get a final position of -933 meters. That's just crazy.
The real crazy thing is my assumption that there is no air resistance. Without air resistance the Balrog would have a constant acceleration during this time. I can find the vertical velocity at the end of the 13.8 seconds with the definition of acceleration.
If I again assume the Balrog starts from rest, I get a final velocity of 135 m/s (over 300 mph). This is a problem. If the Balrog was moving this fast, then there should be a significant air resistance force. Suppose that I use the following model for the air resistance:
In this expression, ρ is the density of air. A is the cross sectional area of the object and C is a drag coefficient that depends on the shape of the object. Of course I don't really know C or even A for a Balrog. However, I do think that the Balrog has a terminal speed similar to that of Gandalf (who is human-like). At terminal speed, the air resistance force is equal to the weight of the object such that the net force is zero. When the net force is zero, the object doesn't change speed - thus the term "terminal velocity". If I guess a terminal velocity of 54 m/s (approximately true for humans) then I can solve for the things I don't know. Since I am assuming the Balrog falls the same way a human would, I am going to use a human mass of 68 kg (otherwise I would have to guess at the area and mass of a Balrog).
All the stuff on the left side of the equation is mostly unknown (I know the value of 1/2 and I can guess at the density of air). Since this stuff doesn't change, I am just going to call it K. Putting in my values for the terminal velocity and the mass, I get a value of K = 2.24 kg/m.
What now? Now, I can make a numerical calculation. If look at a falling Balrog in small time steps, I can approximate the forces as constant. With constant forces, I can calculate the change in velocity and change in position during this time interval. A new velocity means a new air resistance forces for the next time interval. So, I just keep doing this stuff over and over. Here is an older post with more details of a numerical calculation.
I'll skip all the details in the calculation - but, it's here if you want it. Here is a plot of the vertical position of a Balrog falling for 13.8 seconds.
You can see that the Balrog reaches a terminal velocity fairly quickly (after just about 3 seconds). Also, the falling distance with air resistance is 215 meters (700 feet).
So, how long is the Balrog's whip? Maybe my time is off for the start and finish of this falling motion. Maybe my air resistance model is wrong - or maybe the Balrog actually flies a little bit. I feel comfortable saying that the whip is at least 100 meters long and maybe as long as 200 meters.
Let me just look in the back of the book and check my answer for this homework problem. Oh snap. Only the even number problems have answers. I guess we will never know if I am correct. | 1,286 | 5,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.947517 |
https://www.physicsforums.com/threads/factoring-cubic-polynomial-help.431889/ | 1,521,312,162,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00231.warc.gz | 861,627,322 | 17,610 | # Factoring cubic polynomial help
1. Sep 24, 2010
### John H
1. The problem statement, all variables and given/known data
This is probably an easy question, but using the rational zero theorem I have not found any roots for this cubic polynomial.
Factor the Following
6x^3-37x^2-8x+12
2. Relevant equations
3. The attempt at a solution
I have used all my knowledge of factoring and have yet to factor this expression. I tried many different values a, such that f(a)=0 using remainder theorem, According to Descartes signs there should be one negative root.
2. Sep 24, 2010
### epenguin
Sure you copied it out right? It is fairly close to 6x^3-35x^2-8x+12 which works out quite nice.
Last edited: Sep 24, 2010
3. Sep 24, 2010
### John H
I just learned factoring yesterday, and we were told we could factor all (if i heard correctly) cubic functions, using the factor theorem, remainder theorem, long division, Descartes signs, rational roots theorem as our tools. Yet I am unable to factor this expression. I actually Just made one up. I wanted to know how you would even approach this.
4. Sep 24, 2010
### Office_Shredder
Staff Emeritus
There exists a formula which will tell you the roots of a cubic polynomial and allow you to factor it. However, it's extremely messy, you can see it here:
http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function
To give an example for this function, you can see the solutions on wolfram alpha:
http://www.wolframalpha.com/input/?i=6x^3-37x^2-8x%2B12+solve
They aren't particularly nice, and there are only two ways to find them really
1) Use the cubic root formula
2) Test various points to see where the polynomial is positive or negative, and narrow down the potential locations for zeros (hence approximating them in the process)
5. Sep 26, 2010
### epenguin
Basically you are asking for almost the main theme of algebra until sometime in the 19th Century.
How to approach?
If you are interested enough you can try to what I call hammering them, i.e. use the same approach you have probably learned to factoring quadratics. E.g. in your example you can suppose your cubic a product of three linear factors and the coefficients of x in the brackets are either (6, 1, 1) or (3, 2, 1) while the constant terms are either (12, 1, 1), (6, 2, 1), (3, 4, 1) or (3, 2, 2). I haven't tried all those for your equation but I think none of them work. Quite close is 6x3 - 35x2 - 8x + 12 which factorises as
((3x + 2)(2x - 1)(x - 6) I think.
Also if the roots are integral or rational fractions, they can be obtained by methods you seem to have done.
What most scientists would do is use a graphic scientific calculator or computer to plot the polynomial and see where it intersected the horizontal axis. Some have root finders or intersection finders and will give you a numerical value for a root. If this is a whole number you can check whether it is an exact solution. Sometimes because of the limitations of digital calculating it may give you instead of say 5.00000000, give you 5.00000001 or 4.999999999 etc. That is almost certainly 5 and you can check that it is.
But if you ask about how to solve them algebraically... well you first need to be fairly at ease handling imaginary and complex numbers. These are not that difficult and quite fun if anything in maths is.
Then you remember quadratics are solved by expressing them as the difference of two squares, so for this difference to =0 the squares are equal, and that gives you two solutions as a square has two square roots. In the case of the cubic one method is likewise to try and express it as the difference of two cubes. A cube has three cube roots (two of which may be complex numbers.) To actually get the numbers that make the cubes from the original coefficients you find you have to solve a quadratic equation. Then for the fourth degree you can likewise try to express it as again a difference of two squares but one of them is the square of a quadratic. Similarly to the previous case when you try to find from the original coefficients the numbers or expressions for the things squared, you find you have to solve a cubic equation.
The formulae are as Office Shredder says rather messy. For which reason, provided you understand complex numbers, it's worth trying to work themselves out yourself, otherwise the textbooks working (and there are lots of books) look like someone else's unmade bed.
To go further, to solve the fifth degree equation algebraically, it is one of the most celebrated theorems that this is impossible (algebraically being defined as by the arithmetic operations of additions etc. and root extractions). You can only do it algebraically for polynomial of degree higher than 4 if the polynomial is special.
But you can solve any polynomial numerically. That is, with numerical coefficients like your example there are procedures (also in the books called e.g. Horner's method) for getting a number however close to a root as you want. The electronic calculators are doing something like this, and I think (could be wrong) just home in from both sides of where the polynomial changes sign. Which is not very clever but they make up for that by doing it fast.
The difference between the algebraic and the numerical approaches is both smaller and bigger than it seems. It may seem the algebra gives you a neat exact formula as contrasted to a messy approximation of the numerical method. But we just said for the cubic and quartic they may be exact but not all that neat. More significantly they are no more exact than the numerical method. E.g. in quadratics you get a square root in the solution. √(b2 - 4ac) may sound like something exact but if that is √5 or √1.1 or √(most numbers) it just stands for an algorithm for an infinite calculation which you stop at some point. So the difference between algebraic and numerical is just the difference between one sort of potentially infinite approximation procedure and another not even very different sort! And BTW in scientific applications I think the algebraic solutions of the cubic and quartic are rarely used compared to the numerical, though I did once meet a chap who had used the quartic one for some aerodynamic problem.
On the other hand the theory I alluded to was revealing of mathematical structure. They spent centuries not understanding why they couldn't find an algebraic formula like those for degree 1 to 4 for degree 5 and above. The discovery of why opened new paths in maths. But you see at that point the perspective has shifted from your original one of finding a number satisfying an equation to understanding a mathematical structure.
I may have ventured some opinions with which not everyone will agree here. :uhh: | 1,560 | 6,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-13 | longest | en | 0.918042 |
https://www.emis.de/journals/JIS/VOL18/Vanderbei/vand3.html | 1,725,953,750,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00612.warc.gz | 719,035,277 | 1,979 | Journal of Integer Sequences, Vol. 18 (2015), Article 15.9.1
## Which Young Tableaux Can Represent an Outer Sum?
### Colin Mallows Flemington, NJ USA Robert J. Vanderbei Department of Operations Research and Financial Engineering Princeton University Princeton, NJ 08544 USA
Abstract:
Given two vectors, not necessarily of the same length, each having increasing elements, we form the matrix whose (i,j)-th element is the sum of the i-th element from the first vector and the j-th element from the second vector. Such a matrix is called an outer sum of the two vectors (a concept that is analogous to outer products). If we assume that all the entries of this matrix are distinct, then we can form another matrix of the same size but for which the (i,j)-th element is not the matrix element itself but rather the rank of this element in a sorted list of all the numbers in the first matrix. Such a matrix is called a Young tableau. We say that it "represents" the outer sum. In this paper, we address the question as to whether all Young tableaux can be generated this way. When one of the two dimensions is two, then the answer is yes. In all higher dimensional cases, the answer is no. We prove the positive result and give examples illustrating the negative result.
Full version: pdf, dvi, ps, latex
(Concerned with sequences A000108 A211400.)
Received April 1 2015; revised versions received July 23 2015; July 30 2015. Published in Journal of Integer Sequences, July 30 2015. | 346 | 1,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.894158 |
https://oeis.org/A202624 | 1,680,348,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949958.54/warc/CC-MAIN-20230401094611-20230401124611-00028.warc.gz | 502,683,354 | 5,683 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A202624 Array read by antidiagonals: T(n,k) = order of Fibonacci group F(n,k), writing 0 if the group is infinite, for n >= 2, k >= 1. 6
1, 2, 1, 3, 8, 8, 4, 3, 2, 5, 5, 24, 63, 0, 11, 6, 5, 0, 3, 22, 0, 7, 48, 5, 624, 0, 1512, 29, 8, 7, 342, 125, 4, 0, 0, 0, 9, 80, 0, 0, 7775, 0, 0, 0, 0, 10, 9, 8, 7 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 2,2 COMMENTS The Fibonacci group F(r,n) has presentation , where there are n relations, obtained from the first relation by applying the permutation (1,2,,n) to the subscripts and reducing subscripts mod n. Then T(n,k) = |F(n,k)|. T(7,5) was not known in 1998 (Chalk). REFERENCES Campbell, Colin M.; and Gill, David M. On the infiniteness of the Fibonacci group F(5,7). Algebra Colloq. 3 (1996), no. 3, 283-284. D. L. Johnson, Presentation of Groups, Cambridge, 1976, see table p. 182. Mednykh, Alexander; and Vesnin, Andrei; On the Fibonacci groups, the Turk's head links and hyperbolic 3-manifolds, in Groups-Korea '94 (Pusan), 231-239, de Gruyter, Berlin, 1995. Nikolova, Daniela B., The Fibonacci groups - four years later, in Semigroups (Kunming, 1995), 251-255, Springer, Singapore, 1998. Nikolova, D. B.; and Robertson, E. F., One more infinite Fibonacci group. C. R. Acad. Bulgare Sci. 46 (1993), no. 3, 13-15. Thomas, Richard M., The Fibonacci groups revisited, in Groups - St. Andrews 1989, Vol. 2, 445-454, London Math. Soc. Lecture Note Ser., 160, Cambridge Univ. Press, Cambridge, 1991. LINKS Brunner, A. M., The determination of Fibonacci groups, Bull. Austral. Math. Soc. 11 (1974), 11-14. A. M. Brunner, On groups of Fibonacci type, Proc. Edinburgh Math. Soc. (2) 20 (1976/77), no. 3, 211-213. C. M. Campbell and P. P. Campbell, Search techniques and epimorphisms between certain groups and Fibonacci groups, Irish Math. Soc. Bull. No. 56 (2005), 21-28. Chalk, Christopher P., Fibonacci groups with aspherical presentations, Comm. Algebra 26 (1998), no. 5, 1511-1546. C. P. Chalk and D. L. Johnson, The Fibonacci groups II, Proc. Roy. Soc. Edinburgh Sect. A 77 (1977), no. 12, 79-86. J. H. Conway et al., Advanced problem 5327, Amer. Math. Monthly, 72 (1965), 915; 74 (1967), 91-93. Helling, H.; Kim, A. C.; and Mennicke, J. L.; A geometric study of Fibonacci groups, J. Lie Theory 8 (1998), no. 1, 1-23. Derek F. Holt, An alternative proof that the Fibonacci group F(2,9) is infinite, Experiment. Math. 4 (1995), no. 2, 97-100. David J. Seal, The orders of the Fibonacci groups, Proc. Roy. Soc. Edinburgh, Sect. A 92 (1982), no. 3-4, 181-192. A. Szczepanski, The Euclidean representations of the Fibonacci groups, Quart. J. Math. 52 (2001), 385-389. EXAMPLE The array begins: k = 1 2 3 4 5 6 7 8 9 10 ... ---------------------------------------------------------- n=1: 0 0 0 0 0 0 0 0 0 0 ... n=2: 1 1 8 5 11 0 29 0 0 0 ... n=3: 2 8 2 0 22 1512 0 0 0 0 ... n=4: 3 3 63 3 0 0 0 0 ? 0 ... n=5: 4 24 0 624 4 0 0 0 0 0 ... n=6: 5 5 5 125 7775 5 0 0 0 0 ... n=7: 6 48 342 0 ? 7^6-1 6 0 0 0 ... n=8: 7 7 0 7 ? 0 8^7-1 7 0 0 ... n=9: 8 80 8 6560 0 0 0 9^8-1 8 0 ... n=10 9 9 999 4905 9 ? ? 0 10^9-1 9 ... ... For example, T(2,5) = 11, since the presentation defines the cyclic group of order 11. This example is due to John Conway. This table is based on those in Johnson (1976) and Thomas (1989), supplemented by values from Chalk (1998). We have ignored the n=1 row when reading the table by antidiagonals. CROSSREFS Cf. A037205 (a diagonal), A065530, A202625, A202626, A202627 (columns). Sequence in context: A205391 A352858 A078045 * A342224 A145490 A329180 Adjacent sequences: A202621 A202622 A202623 * A202625 A202626 A202627 KEYWORD nonn,tabl,more,nice AUTHOR N. J. A. Sloane, Dec 29 2011 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 April 1 06:57 EDT 2023. Contains 361673 sequences. (Running on oeis4.) | 1,565 | 4,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.686403 |
https://www.inchcalculator.com/height-converter/6ft-1in-to-inches/ | 1,723,585,667,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00725.warc.gz | 632,690,839 | 15,435 | What is 6' 1" in Inches?
Are you trying to find out how tall 6 feet 1 inch is in inches?
How tall is 6' 1" in inches:
73"
6' 1" is equal to 73 inches.
How to Convert 6' 1" to Inches
You can convert 6 ft 1 in to inches by following a few simple steps.
Step One: convert the number of feet to inches by multiplying by 12.
6 ft × 12 = 72 in
Step Two: add this value in inches to the remaining inches to get the total height in inches.
72 in + 1 in = 73 in
So, 6' 1" is 73 total inches. You can confirm this using our height converter.
Once you know the height in inches, you can multiply the result by 2.54 to figure out how tall 6' 1" is in centimeters.
6 ft 1 in to Inches Conversion Chart
Chart showing the height conversion for 6 ft 1 in and other close measurements to inches.
Feet & InchesTotal Inches
5' 8"68"
5' 9"69"
5' 10"70"
5' 11"71"
6' 0"72"
6' 1"73"
6' 2"74"
6' 3"75"
6' 4"76"
6' 5"77"
6' 6"78"
Our height comparison chart has a more comprehensive list of height conversions to convert heights to meters, centimeters, and inches. | 337 | 1,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.891328 |
https://math.stackexchange.com/questions/1908280/equation-of-plane-from-the-eigenvectors-of-covariance-matrix | 1,560,899,780,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998844.16/warc/CC-MAIN-20190618223541-20190619005541-00105.warc.gz | 508,184,775 | 34,916 | # Equation of plane from the EigenVectors of covariance matrix
I am working on some data which are 3D in dimensions. Using the PCA, I found the eigenvectors and eigenvalues which gives the direction of data spread. Sorry my maths is quite poor.
I want to plot the 3 planes (for visualization) that gives the 3 direction of the data spread for my report. I am using C++ thats why I cant use any plotting tools like matlab so I am writing my own.
How can I get the equation of the planes from the eigenvectors? The eigenvectors are direction so is it the normals of the plane? Equation of form Ax+by+cz+d = 0, what is will be the d value.
If what you are talking about are principal planes and you would like to find that out. Let $\pmb{R} \in \mathbb{R}^{3 \times 3}$ be your covariance matrix and call $\pmb{v}_1$, $\pmb{v}_2$, and $\pmb{v}_3$ the eigenvectors, then the plane normal to $\pmb{v}_k$ is $$\pmb{v}_k^{\text{T}} \begin{bmatrix} x - x_0\\ y - y_0\\ z - z_0 \end{bmatrix} = 0$$ where $(x_0,y_0,z_0)$ is a point on the plane. These values determine the $d$ you are talking about.
• [x,y,z] are the variables in your equation $Ax + by + cz + d = 0$ – Ahmad Bazzi Aug 30 '16 at 13:36 | 361 | 1,195 | {"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": 1, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-26 | latest | en | 0.865125 |
http://www.traditionaloven.com/tutorials/distance/convert-japan-jo-unit-to-taiwan-chi-unit.html | 1,529,454,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863259.12/warc/CC-MAIN-20180619232009-20180620012009-00196.warc.gz | 515,220,879 | 11,704 | Convert 丈 to chǐ | Japanese jō to Taiwanese chǐ
# length conversion
## Amount: 1 Japanese jō (丈) of length Equals: 10.00 Taiwanese chǐ (chǐ) in length
Converting Japanese jō to Taiwanese chǐ value in the length units scale.
TOGGLE : from Taiwanese chǐ into Japanese jō in the other way around.
## length from Japanese jō to Taiwanese chǐ Conversion Results:
### Enter a New Japanese jō Amount of length to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
Conversion calculator for webmasters.
## Length, Distance, Height & Depth units
Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
Convert length measuring units between Japanese jō (丈) and Taiwanese chǐ (chǐ) but in the other reverse direction from Taiwanese chǐ into Japanese jō.
conversion result for length: From Symbol Equals Result To Symbol 1 Japanese jō 丈 = 10.00 Taiwanese chǐ chǐ
# Converter type: length units
This online length from 丈 into chǐ converter is a handy tool not just for certified or experienced professionals.
First unit: Japanese jō (丈) is used for measuring length.
Second: Taiwanese chǐ (chǐ) is unit of length.
## 10.00 chǐ is converted to 1 of what?
The Taiwanese chǐ unit number 10.00 chǐ converts to 1 丈, one Japanese jō. It is the EQUAL length value of 1 Japanese jō but in the Taiwanese chǐ length unit alternative.
How to convert 2 Japanese jō (丈) into Taiwanese chǐ (chǐ)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
10.0010001 * 2 (or divide it by / 0.5)
QUESTION:
1 丈 = ? chǐ
1 丈 = 10.00 chǐ
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Japanese jō and Taiwanese chǐ ( 丈 vs. chǐ ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with length's values and properties.
International unit symbols for these two length measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Japanese jō is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Taiwanese chǐ is:
chǐ
### One Japanese jō of length converted to Taiwanese chǐ equals to 10.00 chǐ
How many Taiwanese chǐ of length are in 1 Japanese jō? The answer is: The change of 1 丈 ( Japanese jō ) unit of length measure equals = to 10.00 chǐ ( Taiwanese chǐ ) as the equivalent measure for the same length type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 丈 - Japanese jō for length amount, the rule is that the Japanese jō number gets converted into chǐ - Taiwanese chǐ or any other length unit absolutely exactly.
Conversion for how many Taiwanese chǐ ( chǐ ) of length are contained in a Japanese jō ( 1 丈 ). Or, how much in Taiwanese chǐ of length is in 1 Japanese jō? To link to this length Japanese jō to Taiwanese chǐ online converter simply cut and paste the following.
The link to this tool will appear as: length from Japanese jō (丈) to Taiwanese chǐ (chǐ) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 932 | 3,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-26 | latest | en | 0.853141 |
https://boards.straightdope.com/t/how-fast-does-ice-melt/337471 | 1,722,710,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00288.warc.gz | 109,483,256 | 5,088 | # How fast does ice melt?
I know that folks used to keep ice in ‘ice houses’ for months into warm weather, and I’ve seen snowmen and ploughed snow (both essentially ice) alongside roads last for a really long time. But is there some sort of easy rule of thumb for X volume of ice lasts Y amount of time at Z temperature?
Remember folks, I’m an English major, so all those formulas and talk of thermal conductivity induce out of body experiences for me. I know there are a pile of factors that influence this, but hypothetically if you could rule some out (being in a enviroment without direct sunlight that stayed at a constant temperature and had no precipatation) can you boil it down to some sort of constant rate?
Come to think of it, the hypothetical enviroment I just outlined sounds a whole lot like an ice house doesn’t it?
-rainy
I don’t know any of the formulas, so you are safe there, but I do (think I) know that it has everything to do with surface area. So irregular shapes will take varying amounts of time to melt.
To melt ice, energy must be transferred. The amount of energy will depend principally on the mass of the ice chunk. The rate at which the energy is transferred depends on a whole bunch of things such as temperature difference, air circulation, exposure to radiation, etc.
If you could hold enough things constant, you could certainly come up with a simplified rate of melting. “In this precisely controlled environment, a 1-kg cube of ice requires on average 1187 seconds to melt.” But I’m not sure how useful that would be.
In order to be useful in the real world, formulas that take real-world conditions into account have been developed. Such formulas were not principally designed to be sleep-inducing. But they can get a bit complex - the real world is often not a simple place. | 394 | 1,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-33 | latest | en | 0.960846 |
https://www.ics.uci.edu/~eppstein/cgt/color-game.html | 1,568,800,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573264.27/warc/CC-MAIN-20190918085827-20190918111827-00537.warc.gz | 898,295,991 | 1,787 | ```From: cet1@cus.cam.ac.uk (Chris Thompson)
Newsgroups: sci.math
Subject: Colouring planar maps as a 2-person game
Date: 2 Nov 1996 19:29:07 GMT
Organization: University of Cambridge, England
Keywords: graph planar colour game
```
```We seem to be having a lot of map-colouring related stuff on sci.math
recently, so here's something else on the same theme. Arising [oh no!
you've guessed!] from Martin Gardner's "Mathematical Games" column
in Scientific American. Relevant issues: Apr 1981, Jun 1981, Oct 1981.
The idea of the game is attributed to Steven Brams.
Given: a planar graph G, all faces initially uncoloured;
a palette of n colours;
two players, MIN and MAX.
The players take it it turns to colour a previously uncoloured face
of G, subject only to the constraint that adjacent faces must not
have the same colour. If either player cannot move while there are
still uncoloured faces, MAX wins. If all faces become coloured, MIN
wins. Usually, MIN plays first.
Question: is there some finite N such that if n >= N, then MIN can
win whatever G is? If so, what is the least such N?
The following example (due to Lloyd Shapley) shows that N >= 6.
Let G be (the skeleton of) a dodecahedron. MAX's strategy is to
always colour the face opposite MIN's last move, and in the same
colour. It's easy to see that this forces 6 different colours to
be used, so MAX wins if n <= 5.
After Gardner's column described this, Robert High discovered a
20-face map that MAX can win if n <= 6, so that N >= 7. There's
a picture of it in the October 1981 issue: I will post it if there
is enough interest. | 433 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-39 | latest | en | 0.920507 |
http://www.ssc.wisc.edu/~bhansen/forecast/July2016.htm | 1,508,430,057,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823350.23/warc/CC-MAIN-20171019160040-20171019180040-00376.warc.gz | 570,757,211 | 10,304 | # Wisconsin Unemployment Rate Forecast
## University of Wisconsin
22 July 2016
This memo reports a 12-month forecast for the seasonally adjusted Wisconsin unemployment rate. In addition to point forecasts (the expected future value of the unemployment rate), the memo also reports 50% and 80% forecast intervals (probable ranges for future values).
The unemployment rate in June 2016 was 4.2%, constant from May, after three monthly decreases, but roughly constant since March 2015.
The forecasts are summarized in Figure 1 and Table 1. The point forecast is for the unemployment rate to stay roughly constant for five months, and then slowly increase, to 4.9% by May 2017. The 80% forecast intervals show that there is considerable additional uncertainty. There is a possibility that the unemployment rate could increase as high as 6.2% by June 2017. It is also possible that the unemployment rate could decrease, to 3.7% by June 2017. The 50% forecast intervals refine this uncertainty, showing that it is unlikely the unemployment rate will decrease below 4.2%, or increase to over 4.7% by the end of 2016. Overall, the forecast is for the unemployment rate to increase over the next year.
A 50% forecast interval is designed to contain the future unemployment rate with 50% probability. It is just as likely for the rate to fall in this interval as out of it. This is the smallest possible interval which has even odds of containing the future rate. We can think of this interval as “likely” to contain the future rate.
An 80% forecast interval is designed to contain the future unemployment rate with 80% probability. We can think of this interval as “highly likely” to contain the future rate. The 80% interval is designed so that there is a 10% chance that the future value will be smaller than the forecast interval, and a 10% chance that the future value will be larger than the forecast interval.
To understand the economic reason behind these forecasts, the econometric model finds one salient features. The state unemployment rate is below its long-term average. Mean reversion predicts a slight increase, accounting for an increase of about 0.5 over the upcoming year. Building permits are also below their long-term average, accounting for an increase of about 0.4 over the upcoming year. The other variables contribute only small effects to the forecast.
Figure 1: Wisconsin Unemployment Rate Forecasts
TABLE 1: Wisconsin Unemployment Rate Forecasts
History Point Forecast 50% Interval Forecast 80% Interval Forecast 2016:1 4.6% 2016:2 4.6% 2016:3 4.5% 2016:4 4.4% 2016:5 4.2% 2016:6 4.2% 2016:7 4.2% (4.2%, 4.2%) (4.1%, 4.2%) 2016:8 4.2% (4.1%, 4.3%) (4.1%, 4.3%) 2016:9 4.3% (4.2%, 4.3%) (4.0%, 4.4%) 2016:10 4.3% (4.2%, 4.4%) (4.0%, 4.6%) 2016:11 4.4% (4.2%, 4.6%) (4.0%, 4.7%) 2016:12 4.5% (4.2%, 4.7%) (4.0%, 4.9%) 2017:1 4.5% (4.2%, 4.8%) (4.0%, 5.0%) 2017:2 4.6% (4.3%, 4.9%) (4.0%, 5.2%) 2017:3 4.7% (4.3%, 5.0%) (4.0%, 5.5%) 2017:4 4.8% (4.3%, 5.2%) (3.8%, 5.8%) 2017:5 4.9% (4.2%, 5.4%) (3.8%, 6.1%) 2017:6 4.9% (4.2%, 5.5%) (3.7%, 6.2%)
Previous Forecasts | 995 | 3,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-43 | latest | en | 0.95678 |
http://mathhelpforum.com/calculus/47927-countability-reals.html | 1,524,217,554,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937193.1/warc/CC-MAIN-20180420081400-20180420101400-00308.warc.gz | 211,084,685 | 10,369 | # Thread: countability of the reals
1. ## countability of the reals
We know that the real numbers $\displaystyle \bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
2. $\displaystyle [0,1]$ contains all the reals between $\displaystyle 0$ and $\displaystyle 1$ inclusive. Similarly, $\displaystyle [1,2]$ contains all the reals between $\displaystyle 1$ and $\displaystyle 2$ inclusive etc...
3. Originally Posted by particlejohn
Then the number of intervals are countable right?
Right. But each of these intervals is uncountable.
4. Hello,
Originally Posted by particlejohn
We know that the real numbers $\displaystyle \bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of intervals. So we would have for example: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
Hmm I'd say yes, because there's an interval corresponding to an integer. And $\displaystyle \mathbb{Z}$ is equipotent to $\displaystyle \mathbb{N}$, infinite and countable set.
Let $\displaystyle S=\{\dots \cup [-2,-1) \cup [-1,0) \cup [0,1) \cup \dots \}$
$\displaystyle S=\{A_i ~|~ \forall x \in A_i, ~i \le x < i+1 \}$ (actually i is the least integer contained in $\displaystyle A_i$)
Every element of S can be associated to an integer of $\displaystyle \mathbb{Z}$. Thus S is countable.
5. Originally Posted by particlejohn
We know that the real numbers $\displaystyle \bold{R}$ are uncountable. But what if we partitioned the real numbers into a infinite but countable number of closed intervals. So we would have for example: $\displaystyle [0,1] \cup [1,2] \cup [2,3], \dots$. Do the same thing for the negative reals. Then the number of intervals are countable right?
You cannot find a countable partition of $\displaystyle \mathbb{R}$ by countable sets.
If that is what you are asking. | 589 | 2,172 | {"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} | 3.984375 | 4 | CC-MAIN-2018-17 | latest | en | 0.827626 |
https://www.ncatlab.org/nlab/show/type+telescope | 1,660,230,410,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00214.warc.gz | 781,338,843 | 9,060 | # nLab type telescope
Contents
This entry is about a notion in formal logic (type theory). For the notion in homotopy theory see at mapping telescope.
# Contents
## Idea
In formal logic such as type theory a telescope $\Delta$ is a finite list of terms in context, whose elements/terms/inhabitants are substitutions.
## Definition
In dependent type theory, telescopes are defined inductively with the following rules
$\frac{}{\epsilon\; \mathrm{tel}} \qquad \frac{\Delta\; \mathrm{tel} \quad \Delta \vdash A \mathrm{type}}{(\Delta,x:A)\; \mathrm{tel}}$
The substitutions of telescopes are defined inductively with the following rules
$\frac{}{():\epsilon} \qquad \frac{\delta:\Delta \quad \Delta \vdash A \mathrm{type} \quad a:A[\delta]}{(\delta,a):(\Delta,x:A)}$
which are mutually defined with their action on terms and types.
$\frac{\Delta \vdash A\; \mathrm{type} \quad \delta:\Delta}{A[\delta]\; \mathrm{type}} \qquad \frac{\Delta \vdash a:A \quad \delta:\Delta}{a[\delta]:A[\delta]}$ | 278 | 1,001 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.630911 |
http://stackoverflow.com/questions/5282657/converting-sentences-into-first-order-logic?answertab=votes | 1,436,072,104,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097204.8/warc/CC-MAIN-20150627031817-00309-ip-10-179-60-89.ec2.internal.warc.gz | 249,426,738 | 18,475 | # Converting Sentences into first Order logic
in first order logic, i know the rules. However, whenever i convert some sentences into FOL, i get errors, I read many books and tutorials, do u have any tricks that can help me out,
some examples where i makes errors
Some children will eat any food
``````C(x) means “x is a child.”
F(x) means “x is food.”
Eat(x,y) x eats y
I would have written like this:
(∃x)(∀y) C(x) ∧ Eat(x,y)
edit: (∃x)(∀y) C(x) ∧ F(y) ∧ Eat(x,y)
But the book write it like this
(∃x)(C(x) ∧ (∀y)(F(y)→Eat(x,y)))
``````
Edit No2: 2nd Type of error i'm making: Turtles outlast Rabbits.
``````i'm writing it like this: ∀x,y Turtle(x) ∧ Rabbit(y) ∧ Outlast(x,y)
but according to the book ∀x,y Turtle(x) ∧ Rabbit(y) --> Outlast(x,y)
``````
Of course, I agree with the book, but is there any problem with my version !!
-
You should describe the errors... – gary Mar 12 '11 at 13:11
@gary comtois,Hi, I modified the question with one type of error i'm making – Noor Mar 12 '11 at 13:36
looks good, thanks – gary Mar 12 '11 at 14:34
From
xy [C(x) ∧ F(y) ∧ Eat(x, y)]
it follows that ∀y F(y), i.e. everything is food. ("There exists a child x such that for all y, y is food" and a bunch of other propositions hold.) It also follows that the child eats itself: if we denote the child by an arbitrary constant c and fill that in, we get
y [C(c) ∧ F(y) ∧ Eat(c, y)]
and since y is universally quantified, we can instantiate it by replacing it with c to get
C(c) ∧ F(c) ∧ Eat(c, c)
which is an undesirable state of affairs.
xy [Turtle(x) ∧ Rabbit(y) ∧ Outlasts(x, y)]
it follows that
x Turtle(x) ∧ ∀y Rabbit(y) ∧ ∀xy Outlasts(x, y)
I.e., everything is a turtle, everything is a rabbit, and everything outlasts everything, including itself.
The version in your book uses → to indicate that for every object y, if it is food, then it is eaten by x. You need a conditional to express sentences of the form "all Xs are Ys" or "every X does Y".
-
Can u point out the error in the 2nd example i placed in the edit Num2 – Noor Mar 13 '11 at 5:05
It's always the same error that you're making: there are no conditional statements in your examples. – CFP Mar 13 '11 at 9:14
@Noor: updated my answer. – larsmans Mar 13 '11 at 10:36
but if I say one Turtle outlast a Rabbit, then can i write it like thiy ∃x ∃y Turtle(x) ∧ Rabbit(y) ∧ Outlast(x,y) – Noor Mar 13 '11 at 11:11
@Noor. Yes. But universal quantifiers don't work that way. – larsmans Mar 13 '11 at 11:12
Whenever you have determiner every (or any or no) in an English sentence the corresponding FOL sentence should have both a universal quantifier and an implication in it. E.g. the translation template for the noun phrase every man would be:
``````∀ x (man(x) ⇒ ...)
``````
If your English sentence does not contain any determiners, then reformulate it so that every noun in it would come with a determiner. This way the mapping to FOL becomes clear. E.g. the ambiguous/vague sentence
``````Turtles outlast Rabbits.
``````
could be reformulated in several semantically different ways:
• Every turtle outlasts every rabbit.
• There are some turtles that outlast every rabbit.
• There are some turtles that outlast some rabbits.
• Most turtles outlast most rabbits.
• ...
Btw, there is an online tool APE that converts English sentences into FOL provided that you first reformulate your sentences so that they fall into the fragment of English that this tool supports. Note however that this tool returns a single FOL reading, i.e. it does not enumerate all the ambiguity the input might contain.
-
You didn't check whether `y` was food first. Considering your statement, let `a` be a children, ie. `C(a)` is true. Then `(∃x)(∀y) C(x) ∧ Eat(x,y)` implies `(∃x) C(x) ∧ Eat(x,a)`. In other words, you're stating that some children will eat anything, not only food.
-
Ok, now if include the F(y), is it good? – Noor Mar 12 '11 at 13:48
No, you're now saying that all y is food. – CFP Mar 12 '11 at 15:56
Can u point out the error in the 2nd example i placed in the edit Num2 – Noor Mar 13 '11 at 5:04
Yes, there's a problem with your second version. It reads: everything is turtle, and everything is rabbits, and everything outlasts everything. – CFP Mar 13 '11 at 9:13 | 1,251 | 4,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2015-27 | latest | en | 0.909399 |
https://www.joshuanava.biz/geometric/contents.html | 1,586,219,640,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00082.warc.gz | 994,202,547 | 12,286 | ## Construction Of Triangles Perimeter Altitude Vertical Angle Given
Preface to the second odition iii
Acknowledgment* iv
PART I GEOMETRIC DRAWING
Chepter
1 Scale» 3 The representative fraction; plain scales; diagonal scales; proportional scales
Exercises t 8
2 The construction of geometric figures from given data 10
Construction of the following tnanglea: Equilateral. given one of the sides Isosceles, given the perimeter and altitude Scalene, given base angles and altitude Scalene, given base, altitude end vertical angle Scalene, given penmeter and ratio of sides Scalene, given perimeter, altitude and vertical angle
Similar triangles with different penmeter Construction of the following quadrilaterals: Square, given one side Square, given the diagonal Rectangle, given the diagonal end one side Parallelogram, given two sides and an angle Rhombus, given the diagonal and length of sides Trapezium, given the parallel sides, the perpendicular distance between them and one angle
Construction of the following polygons: Regular hexagon, given the length of side Regular hexagon, given the diameter Regular octagon, given the diagonal (within a circle)
Regular octagon, given the diameter (within e square)
Reguter polygon, given tho length of side (three methods)
Reguler polygon, given the diagonal (within a circle)
Regular polygon, given the diameter
Exercises 2 19
page
3 Isometric projection 20 Conventionel isometric projection (isometric drawing); circles and curves in isometric;
true isometric projection, isometric scales. Exercises 3 26
4 The construction of circles to satisfy given conditions 29
To construct the circumference, given the diameter
To construct the diameter, given the circumference
To find the centre of a circle
To construct circles, given the following conditions:
To pass through three given points Inscribed circle of a regular polygon Circumscribed circle of a regular polygon Escribed circle to a regular polygon To pass through a fixed point and touch a line at a given point
To pass through two given points and touch a given line
To touch two given lines and pass through a given point
To touch another given circle and a given line To touch another circle and two tangents of that circle
To touch another circle and two given lines To pass through two given points and touch a given circle
Three circles which touch, given the position of their centres
Two circles within a third circle, ell three circles to louch
Any number of circles within a given orcle. all circles to touch
A number of equal circles within a given regular polygon
Equal circles around a given regular polygon Exercises 4 37
Tangency 39
Tangent from a point on the circumference w
Tangent from a point outside the circle Common tangent to two equal circle« (exterior)
Common tangent to two equal circle« (interior)
Common tangent to two unequal circle« (exterior)
Common tangent to two unequal circles (interior)
Exercises 5 42
6 Oblique projection 44 Oblique (aces at 30°. 45*. 60s. oblique scale
(1- J- I): Cavalier and Cabinet protections; circles and curves in oblique projection Exarcises 6 47
7 Enlarging and reducing plane figure*
and equivelent areas 50
Linear reductions m size; reduction of s<des by different proportions; reductions in area. Equivalent areas:
Rectangle equal in area to a given triangle Square equal in area to a given rectangle Square equal in area to a given triangle Triangle equal in area to a given polygon Circle which has the same area as two smaller circles added together To divide a triangle into three parts of equal area
To divide a polygon into a number of equal areas
Exercises 7 58
8 The blending of lines end curves 60 To blend a radius with two straight lines at nght angles
To bland a radius with two straight lines meeting at any angle To find the centre of an arc which passes through a given point and blends with a given line
To find the centre of an arc which blends with a line and a circle To find the centre of an arc which blends, externally and internally, with two circles To join two parallel lines with two equal radii, the sum of which equals the distance between the lines
To join two parallel lines with two equal radii
To join two parallel lines with two unequal radii
Exercises 8 66
9 loci 68 Loci of mechanisms, trammels, some other problems in loci
Extremes 9 76
page
10 Orthographic projection 78 Third Angle projection. First Angle projection: elevation and plan; front elevation, end elevation and plan, auxiliary elevations and auxiliary plans; prisms and pyramids;
cylinders and cones: sections.
Exercises 10 98
11 Conic sections—The Ellipse, the Parebola. the Hyperbola 102 The ellipse as a conic section; as a locus;
three constructions for the ellipse; the focus.
the normal and the tangent
The parabola as a conic section: as a locus;
within a rectangle; the focus and tangent
The hyperbola as a conic section; as a locus.
Exercises 11 110
12 Intersection of regular solids 112 Intersection of the following solids:
Two dissimilar square prisms at right angles Two dissimilar square prisms at an angle A hexagonal pnsm and a square prism at right angles
Two dissimilar hexagonal prisms at an angle A hexagonal prism and an octagonal pnsm at an angle
A square prism and a square pyramid at right angles
A square pyramid and a hexagonal prism at an angle
Dissimilar cylinders at right angles Dissimilar cylinders at an angle Dissimilar cylinders in different planes at an angle
Cylinder and square pyramid at right angles Cylinder and square pyramid at an angle Cylinder and hexagonal pyramid at an angle Cylinder and cone with cone enveloping the cylinder
Cylinder and cone, neither enveloping the other
Cylinder and cone with cylinder enveloping the cone
Cylinder and cone in different planes Cylinder and hemisphere Fillet curves
Exercises 12 129
13 Further orthographic projection Rabatment. traces. The straight line:
Projection of a line which is not parallel to any of the principal planes True length of a straight line and true angle between a straight line and vertical and horizontal planes
To find traces of a straight line given the plan and elevation
To draw the elevation and plan of a straight lint given the true length and the distances of the and* of the line from the principe! pianos
To construct the plan of a straight line given the distance of one end of the line from the XY line in the pian, the true length of the line and the elevation
To construct the elevation of a straight line given the distance of one end of the line from the XY line in the elevation, the true length of the line and the plan To construct the elevation of a straight line given the plan and the angle that the line makes with the horizontal plane To draw the plan of a line given the elevation and the angle that the line makes with the vertical plane The inclined plane:
Projection; traces; angles between planes; true shape of an inclined plane; true shapes of the sides of an oblique, truncated pyramid
The oblique plane: Projection ; traces.
To find the true angle between the oblique plane and the horizontal plana
To find the true angle between the oblique plane and the vertical plane
To find the true angle between the traces of an oblique plane
Exercises 13 141
14 Developments
Development of prisms: Square prism
Square prism with an oblique top Hexagonal prism with oblique ends Intersecting square and hexagonal prisms Intersecting hexagonal and octagonal prisms Development of cylinders: Cylinder
Cylinder with oblique top Cylinder cut obliquely at both ends Cylinder with circular cut-out An intersecting cylinder Both intersecting cylinders Development of pyramids Pyramid
Frustum of a square pyramid Hexagonal pyramid with top cut obliquely Hexagonal pyramid penetreted by a squere prism
Oblique hexagonal pyramid Development of cones: Cone
Frustum of e cone
Frustum of e cone cut obliquely
Cone with a cylindrical hole
Oblique cone
Exercises 14 156
16 Further problem« in loci 159
The cycloid; tangent and normal to the cycloid; epi- and hypo-cycloids; tangent and normal to the epi- and hypo-cycloids; inferior and superior trochoids; the involute; normal and tangent to the involute; the Archimedean spiral; the helix; round and square section coiled springs; square section threads.
Exercises 15 167
16 Freehand sketching 169 Pictorial sketching; sketching in Orthographic Projection; sketching as an aid to design
17 Some more problems solved by drawing 176 Areas of irregular shapes; the mid-ordinate rule Resolution of forces; a couple, calculation of some forces acting on a beam; forces acting at a point; equilibrant and resultant forces
Simple cam design; cam followers Exercises 17 190
PART II ENGINEERING DRAWING
18 Introduction 194 Type of projection
Sections Screw threads Nuts and bolts
Designation of ISO screw threads Types of bolts and screws
### Dimensioning
Conventional representees Machining symbols Surface roughness Abbreviations Framing and title blocks Assembly drewings Some engineering festerings: The stud and set bolt Locking devices Rivets and riveted joints Keys, keyweys end splines Conered loints Worked examples
Exercises 18 230
Appendix A 258
Sizes of ISO metric hexagon nuts, bolts and washers
Appendix B 260
Sizes of sloned and castle nuts with metnc threeds
Index 261
0 -1
## Pencil Drawing Beginners Guide
Easy Step-By-Step Lessons How Would You Like To Teach Yourself Some Of The Powerful Basic Techniques Of Pencil Drawing With Our Step-by-Step Tutorial. Learn the ABC of Pencil Drawing From the Experts.
Get My Free Ebook
### Responses
• Otto
How to draw the common tangent to two unequal circles?
9 years ago
• vanna
How to construct a similar triangle given perimeter?
8 years ago
• niklas
How to draw the isometric projection of a regular pentagon on a cylinder?
8 years ago
• Tero
How to construct an isosceles triangle given the perimeter and altitude?
4 years ago
• antony
How to construct a triangle given an altitude, vertical angle and the base?
4 years ago
• arsi
How to construct an isoceles triangle given its perimeter and altitude?
4 years ago
• aarne
How to construct an isosceles triangle with given base and altitude?
4 years ago
• Uwe
How to draw a cylinder meeting a cone neither enveloping the other step by step?
4 years ago
• annika
How to draw a triangle with a given altitude,base, vertical angle and perimeter?
4 years ago
• Vanessa
How to draw triangle with verticle angle?
4 years ago
• azelio
How to darw triangle to given perimeter altitude and vertical angle?
4 years ago
• kenneth
How to darw triangle to given base altitude and verticle angle?
4 years ago
• Aija
How to construct a triangle with given angles, perimeter and altitude?
4 years ago
• teegan
How to construct a triangle given the altitude, one side and the vertical angle?
4 years ago
• caoimhe
How to construct a triangle given perimeter height and vertical angle?
4 years ago
• camelia galbassi
How to construct a triangle given perimeter, altitude and vertical height?
4 years ago
• mikael
How to draw an isosceles triangle with perimeter and altitude?
4 years ago
• lisa jung
How to construct a triangle/angle at vertex/perimeter/altitude?
4 years ago
• Cody
HOW TO CONSTRUCT A TRIANGLE WHEN GIVEN THE BASE, ALTITUDE AND VERTICAL HEIGHT?
4 years ago
• teuvo
How to draw a triangle given the diagonal?
4 years ago
• bellina barese
How to draw an isosceles triangle given the perimeter and length?
4 years ago
• Erminio
How to construct a trinangle with a vertucal angle of 37.5?
4 years ago
• ronan
How to draw altitude for an isosceles trapezium using a setsquare and an scale?
3 years ago
• Ren
How to contruct a triangle when the altitude Nd vertical angle is given?
3 years ago
• Michelangelo Castiglione
How to contruct a triangle when the perimeter, altitude Nd vertical angle is given?
3 years ago
• heike
How to construct s triangle give the perimeter, altitude and vertical angle?
3 years ago
• Melanie
How to construct a triangle with given perimeter base angle and altitude?
3 years ago
• robin bennett
How to construct a triangle with vertical angle in engineering drawing?
3 years ago
• matthew
How to construct a triangle with perimeter, altitude and vertical angle in engineering drawing?
3 years ago
• rayyan
How to construct a triangle given the perimeters,altitude and the vertical angles?
3 years ago
• SWEN
How to construct vertical angles in triangles?
3 years ago
• Christina Bontrager
How to construct a triangle given base side, vertical angle and the altitude?
3 years ago
• Tanja Foerster
How to construct a triangle given a base, altitude, vertical angle, and drawing the same triangle?
3 years ago
• max
How to cibstruct a triangle with perimeter and angle ratuo?
3 years ago
• Alfrida
How to construct an equilateral triangle given the vertical height?
3 years ago
• AALIYAH
How to construct an isosceles triangle with perimeter ,altitude and vertical angle?
3 years ago
• ute
How to construct triangle given and perimeter, attitude and vertical angle?
3 years ago
• Thomas
How to do construction when perimeter and altitude is giving?
3 years ago
• asmara
How to contruct an isosceles triangle being given the perimeter and the altitude?
3 years ago
• Hugo Hay
How to construct a triangle given perimeter.altitude and verical angle?
3 years ago
• Hilda
How to form a triangle given perimeter vertical height and vertical angle?
3 years ago
• delma bergamaschi
How to construct a triangle when base and altutude are given?
3 years ago
• basso
How to construct an angle with a given perimeter and a vertical ratio?
3 years ago
• Aziza
How To Construct Triangles Given The Perimeter And Altitude?
3 years ago
• Ambrosia
How to constuct isosceles triangle given altidute and perimeter?
3 years ago
• ralph
How to construct a scalene triangle?
3 years ago
• kirsi v
How do u constuct a triangle with perimeter 115mm and a vertical angel of 45?
3 years ago
• ann
How to construct a triangle with a parimeter?
3 years ago
• leslie hammett
How to construct triangle with base 55mm, a latitude of 62mm with vertical angle of 37.5?
3 years ago
• robert
How to construct va triangle with altitude of 50mm and a vertical angel of 60°?
3 years ago
• marybeth
How do draw a triangle with perimeter 115mm height 40mm vertical angle 45?
2 years ago
• samwise
How to construct triangles given its vertical angle?
2 years ago
• jolanda siciliano
How to construct triangles given its perhmeter and altitude?
2 years ago
• jan stewart
How to constuct a triangle given altitude and perimeter?
2 years ago
• MYRTLE
How to construct an isoceles triangle of a given perimeter and an height?
2 years ago
How to draw a triangle one side vertical angle altitude given?
2 years ago
• Lete Mebrahtu
How to construct a triangle given perimeter, base and height?
2 years ago
• BELBA
How to construct a triangle in technical drawing?
2 years ago
• Semhar
How to constuct a triangl of perimeter 130mm altitude 40mm and a vertical angle of 90degre?
2 years ago
• renzo
How to construct a triangle given the perimeter the altitude and the vertival angle?
2 years ago
• phillipp
How to construct a triangle with base angles and an altitude with pictures?
2 years ago
• rowan
How to construct a triangle given the altitude,heights .perimeter,vertical angles and ratio pdf?
2 years ago
• Albert
How to construct a triangle show perimeter is=130mm attitude 400MM and vertical angle 60?
2 years ago
• anthony
How can i construct a truangle with base angles 60 and 45 degrers abd an altitude if 76mm?
2 years ago
• Tesmi
What is the procedure of constructing abtringle givenbverticle angle and altitude?
2 years ago
• Valdemar
How to construct similar triangle with given perimeter whenbase ,altitude and angle is given?
2 years ago
• nasih
How to construct a triangle given a perameter,verticale angle and an altitude?
2 years ago
• evelyn
How to construct a triangle with a perimeter of 115mm?
1 year ago
• thorsten
How to constract perimeter and attitude?
1 year ago
• Aliisa
How to construct a triangle of perimeter 115mm ,the altitude of 40mm and vertical angle of 45°?
6 months ago
• bernd
How to construct a triangle with a perimeter altitude and vertical angle is 45 degree?
6 months ago
• Anne
How to draw a tringle given the perimetre vertical angle and altitude?
6 months ago
• OUTI
How to construct an isosceles triangle given vertical height and perimeter?
6 months ago
• mira ruoho
How to construct a triangle given perimeter and alltitude?
5 months ago
• Melampus
How to construct a triangle given an angle and height in technical drawing?
2 months ago
• ralph
How to construct an isoscelence given the vertical height and vertical angle?
1 month ago | 3,929 | 16,832 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | latest | en | 0.766732 |
http://mathoverflow.net/questions/tagged/elliptic-curves?page=3&sort=newest&pagesize=15 | 1,469,529,325,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824757.8/warc/CC-MAIN-20160723071024-00062-ip-10-185-27-174.ec2.internal.warc.gz | 164,728,391 | 25,105 | # Tagged Questions
An elliptic curve is an algebraic curve of genus one with some additional properties. Questions with this tag will often have the top-level tags nt.number-theory or ag.algebraic-geometry in addition; note also the tag arithmetic-geometry as well as some related tags such as rational-points, abelian-...
1answer
225 views
### Conductor of a CM elliptic curve and its Grössencharacter
For a CM elliptic curve $E$ and its Grössencharakter, their conductors are both supported on bad primes of $E$. Moreover, by comparing their functional equation, there should be some obvious relations....
0answers
105 views
1answer
118 views
### The $p$-th power of the invariant derivative on an elliptic curve in characteristic $p$
I am not an expert in elliptic curves at all, so my question may naive and/or obvious. Let $E$ be an (affine) elliptic curve defined over a finite (or perfect) field of characteristic $p$. Since its ...
1answer
350 views
1answer
597 views
### Remark 4.23.4 in Hartshorne
Crosspost from math.stackexchange, since it's quite possible I might not get a response there. Remark 4.23.4 in Chapter IV of Hartshorne's Algebraic Geometry references a paper by Elkies that ...
1answer
184 views
### Equidistribution of representations by a binary cubic form
Let $f(x,y)\in\mathbb{Z}[x,y]$ be a binary cubic form with nonzero discriminant, and for a positive integer $m$ consider the integral representations $f(x,y)=m$. Assume that the number of ...
1answer
345 views
### Why there are only finitely many $\overline{\mathbb{Q}}$-isomorphism classes of elliptic curves with CM by $\mathcal{O}$?
For someone who does not have a very extensive knowledge of number theory, what is a good intuitive explanation as to why there are only finitely many $\overline{\mathbb{Q}}$ isomorphism classes of ...
0answers
297 views
### Result of Deuring, intuitive way to see it's true/quickest way to prove?
There is the following result of Deuring that goes as follows: Let $E/L$ be an elliptic curve defined over a number field $L$ with complex multiplication by an order $\mathcal{O}$ in an imaginary ...
1answer
173 views
### Question on paper of Stewart and Top about ranks of elliptic curves over Q(t)
I'm reading "On Ranks of Twists of Elliptic Curves and Power-Free Values of Binary Forms" by Stewart and Top, and struggling to understand the argument on pg 962 which shows that the rank of a ...
1answer
166 views
### isogeny clases of CM abelian varieties
Let $A$ be an abelian variety defined over $\overline{\mathbb{Q}}$ and with complex multiplication by a CM field $K$. Looking at the action of $K$ on $H^0(A, \Omega^1_A)$ one gets a CM type of $K$, ...
0answers
216 views | 682 | 2,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2016-30 | latest | en | 0.919181 |
https://www.traditionaloven.com/metal/precious-metals/palladium/convert-cubic-millimetre-mm3-palladium-to-troy-ounce-tr-oz-palladium.html | 1,656,972,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00461.warc.gz | 1,078,490,168 | 13,866 | ## Amount: cubic millimeter (mm3) of palladium volume Equals: 0.00039 troy ounces (oz t) in palladium mass
Calculate troy ounces of palladium per cubic millimeter unit. The palladium converter.
TOGGLE : from troy ounces into cubic millimeters in the other way around.
### Enter a New cubic millimeter Amount of palladium to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
## palladium from cubic millimeter to ounce (troy) Conversion Results :
Amount : cubic millimeter (mm3) of palladium
Equals: 0.00039 troy ounces (oz t) in palladium
Fractions: 39/100000 troy ounces (oz t) in palladium
CONVERT : between other palladium measuring units - complete list.
This calculator is based on a pure palladium with a density of ≅ 12 grams per one cubic centimeter, 12.023g/cm3 more precisely. Palladium, symbol Pd, is a rare metal and in the raw form found in the nature it always contains in the metal body a small amount of platinum. Palladium enjoys a widespread use; automobile industry, dentistry, jewellers make a swishy jewellery pieces by including this metal on their jewels, not to mention watchmakers', those who make quality luxury watches, and even professional musical instruments like transverse flutes for instance. I'd like to know whether the genuine TAG Heuer or Rolex Swiss watches have palladium in/on them. Palladium is listed in the group with the four expensive bullion metals (incl. platinum, gold and silver) where each of these four has own ISO currency code.
Convert palladium measuring units between cubic millimeter (mm3) and troy ounces (oz t) of palladium but in the other direction from troy ounces into cubic millimeters.
conversion result for palladium: From Symbol Equals Result To Symbol 1 cubic millimeter mm3 = 0.00039 troy ounces oz t
This online palladium from mm3 into oz t (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this palladium calculator are ...
With the above mentioned units calculating service it provides, this palladium converter proved to be useful also as a teaching tool:
1. in practicing cubic millimeters and troy ounces ( mm3 vs. oz t ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with palladium's density values including other physical properties this metal has.
International unit symbols for these two palladium measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for cubic millimeter is: mm3
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for ounce (troy) is: oz t
### One cubic millimeter of palladium converted to ounce (troy) equals to 0.00039 oz t
How many troy ounces of palladium are in 1 cubic millimeter? The answer is: The change of 1 mm3 ( cubic millimeter ) unit of a palladium amount equals = to 0.00039 oz t ( ounce (troy) ) as the equivalent measure for the same palladium type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of palladium can have a crucial/pivotal role in investments. If there is an exact known measure in mm3 - cubic millimeters for palladium amount, the rule is that the cubic millimeter number gets converted into oz t - troy ounces or any other unit of palladium absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many troy ounces ( oz t ) of palladium are contained in a cubic millimeter ( 1 mm3 ). Or, how much in troy ounces of palladium is in 1 cubic millimeter? To link to this palladium - cubic millimeter to troy ounces online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: palladium from cubic millimeter (mm3) to troy ounces (oz t) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,111 | 4,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.819342 |
https://www.tdisdi.com/how-much-weight-do-i-really-need/ | 1,571,204,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986664662.15/warc/CC-MAIN-20191016041344-20191016064844-00101.warc.gz | 1,108,035,790 | 39,455 | How Much Weight Do I Really Need?
by Ryan Conery:
Bringing the right amount of weight with you on a dive requires practice. However, it is important to get this dialed in. Having the correct weight makes it easier to get neutrally buoyant. It also helps with air consumption and it means you have to haul less weight to and from the dive site. Before we can determine our weighting requirements, we have to look at what factors are affecting us.
Neutral Buoyancy
In order to achieve neutral buoyancy, we must balance the upward force caused by water displacement with the downward force caused by our weight system. Archimedes’ Principle states that there is a buoyant force exerted on an object immersed in fluid. When an object gets placed in water, it will displace some of that liquid. Displacement means that the water is being pushed aside. That object is actually taking up space that was previously occupied by the water. Another way to picture this principle is by thinking about your bathtub. If you were to fill that tub up all the way to the top, what would happen if you tried to hop in? Of course, the water would overflow and be all over your floor, right? This is because you are displacing some of that water as you step into the tub. Now, the water that gets pushed aside wants to take its space back. This creates an upward buoyant force equal to the weight of the displaced water. For example, if we took a 1 cubic foot ball and submerged it in a bathtub, the ball would displace 1 cubic foot of water. Fresh water weighs 62.4lbs per cubic foot (salt water weighs 64lbs per cubic foot). This means that the ball is experiencing an upward buoyant force of 62.4lbs. If the ball weighs more than this, it will sink. If it weighs less, it will float.
In diving, we have the unique capability to utilize our buoyancy compensation device to ensure that we weigh the same amount as the water we are displacing. This allows us to achieve neutral buoyancy. Although we can get close to becoming perfectly neutral, we can never get rid of the slight rise and fall resulting from our breath cycle. As we inhale, we displace more water causing our surrounding environment to exert a larger upward buoyant force. So, we will ascend slightly as we breathe in. Vice versa, we displace less water as we exhale causing us to descend.
Archimedes’ Principle
Archimedes’ Principle controls how much weight we have to add to our system to reach this neutral state. Our bodies, wetsuit and BCD are naturally buoyant. We require ballast weight to overcome the initial positive force of buoyancy. There are many factors that affect how much weight we will need to carry, but, as a general rule of thumb, you may need anywhere from 5 to 10% of your body weight in lead. After weighing yourself, determine what your range of weights might be. If you weigh 200lbs for example, you may need to use between 10lbs and 20lbs of lead. This range of weight gives us a good starting point, but we will have to look at the factors that affect our buoyancy to really dial it in.
Exposure Protection
Exposure protection plays a large role in the weighting process as our suit is inherently buoyant. A thicker suit means that we are displacing more water and require more lead to enable us to sink. A 3mm wetsuit may only require 6 to 8% of your body weight in lead whereas a 7mm, wetsuit, or dry suit could require 10% of your body weight or more depending on your undergarments. One good experiment is to try on your suit and hop in a pool. Start by holding on to a 2lb block of lead. Keep adding weight in small increments until you begin to sink. This way, you can begin to estimate your weighting needs in a controlled environment.
Other Factors to consider for Neutral Buoyancy
Water type is another factor that affects the weight we will need to carry. Salt water is denser than fresh water which causes it to weigh more per cubic foot. This means that an object placed in salt water will experience a larger upward buoyant force than if the same object was placed in fresh water. Due to the increased density, you will need to carry additional weight when diving in salt water. Anywhere from 2 to 6lbs extra may be required depending on the salinity of the water.
The material your tank is composed of must also be considered. Aluminum tanks start out slightly negative at the beginning of a dive when they are full. However, as you consume air they will slowly become more buoyant. At the end of your dive, an aluminum tank will be anywhere from 2 to 4lbs positively buoyant. If you are diving aluminum tanks, you may want to add a couple pounds to your kit to compensate for the additional buoyancy. On the other hand, steel tanks start out around 6 to 8lbs negatively buoyant. As you consume air, they too will become more buoyant; however, they will stay on the negative side of the scale. You can use the additional weight a steel tank provides and remove some of the weight on your belt.
Experience improves Buoyancy
The last factor that can affect your buoyancy is experience. As you begin to dive more, you will notice that you start to need less weight. Often times, inexperienced divers have trouble getting all of the air out of their BC as they start their descent. Becoming more familiar with your gear and more comfortable underwater produces a more relaxed breathing cycle. Just by going out and diving, you will soon notice that you need less weight than you thought. Another test you may see divers do is called a float test. In full gear, empty the air from your BC. With your lungs half full, you should sit at eye level with the water. As you exhale, you should begin to sink. This is a great way to get a ballpark estimate of your weighting needs. However, remember that you might need to add a couple extra pounds to compensate for the increased buoyancy of a tank at the end of a dive. The best place to ensure your weighting is correct is at your safety stop. You should be able to maintain neutral buoyancy with little to no air in your BC.
Make sure, after every dive, you record the weight you used in your logbook. Also, try to log every factor that affected your buoyancy on that dive such as what suit you used, what the water type was and what kind of tank you had. Getting your weight dialed in is a learning process and it will require some adjustment. Keep at it, practice makes perfect.
/
/
/
/
/
/
/
/
0 replies | 1,415 | 6,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | longest | en | 0.959837 |
https://essaysprings.com/2022/05/13/please-follow-the-direction-and-have-answers-to-questions-and-a-diagram-if-needed-1-describe-your-sample-1-how-many-participants-are-male-and-female-2-what-percent-are-caucasian-african-america/ | 1,653,702,121,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00438.warc.gz | 292,207,849 | 27,082 | Chat with us, powered by LiveChat Please follow the direction and have answers to questions and a diagram if needed! 1. Describe your sample: 1. How many participants are male and female? 2. What percent are Caucasian, African America | The Best Academic Writing Website
Select Page
1. How many participants are male and female?
2. What percent are Caucasian, African American, Hispanic, Asian American, and Other?
3. What is the mean, median, and standard deviation for age?
4. What percent of participants are in each income level?
2. Are the following variables correlated?
Provide the correlation and state which variables are significantly correlated with each other (hint: look for the *).
1. Age
2. CrimeAttitudes
3. MHAttitudes
Part ll
Using the data set, test the following hypotheses You need to determine which test to use based on how many independent and dependent variables there are. (Hint: Think about how many groups you have to test):
Hypothesis 1: Males and females differ on their attitudes toward crime. (Use the variable labeled
CrimeAttitudes)
Hypothesis 2: Males have more negative attitudes towards mental health than females
(Use the variable labeled MHAttitudes).
Hypothesis 3: Attitudes towards mental health differs by ethnicity. (Use the variable labeled
MHAttitudes and Ethnicity)
Hypothesis 4: Attitudes towards crime varies by income level. (Use the variable labeled
CrimeAttitudes and Income)
For each of the 4 hypotheses above:
1. State the null and research hypothesis
2. State which test statistic should be used. Be specific.
3. Run the test on SPSS
4. Put all results in a table or list them in some way so that I can clearly see which numbers are your answers. Include the exact numbers you looked at to come up with your results. (do not copy and paste the printout. I want you to be able to pick out the correct numbers from the printouts, so only report the numbers you need. Points will be taken off for only copying the whole printout table without stating which specific numbers in the table you were looking at to get your answer)
5. State whether you reject or accept the null hypothesis
6. State your conclusion in a sentence
*Use p < .05 for all your tests to determine Significance
PART III
Write a summary discussing your project. Include a description of your sample (in words, not just numbers), a discussion of your results from your hypotheses including which tests you ran for each hypothesis, a discussion of your significant results, and a conclusion about what you feel the perceptions of the class are on there is a racial bias in the criminal
SPSS Project Variable Labels
Gender
1 = Male
2 = Female
Ethnicity
1 = Caucasian
2 = Hispanic
3 = Asian
4 = African American
5 = Other
Income
1 = Under \$15,000
2 = \$15,001 – \$30,000
3 = \$30,001 – \$50,000
4 = \$50.001 – \$75.000
5= Over \$75,000 | 661 | 2,898 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-21 | latest | en | 0.926849 |
https://www.theunitconverter.com/kilogram-square-centimeter-to-ounce-inch-square-second-conversion/ | 1,624,105,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00290.warc.gz | 939,465,555 | 14,455 | # kg∙cm² to oz∙in∙s² Conversion
## Kilogram Square Centimeter to Ounce Inch Square Second Conversion - Convert Kilogram Square Centimeter to Ounce Inch Square Second (kg∙cm² to oz∙in∙s²)
### You are currently converting Moment of Inertia units from Kilogram Square Centimeter to Ounce Inch Square Second
1 Kilogram Square Centimeter (kg∙cm²)
=
0.01416 Ounce Inch Square Second (oz∙in∙s²)
Kilogram Square Centimeter : The kilogram square centimeter is a derived unit of moment of inertia in the SI system. Its symbol is kg•cm². One kilogram square centimeter is equal to 10-4 kilogram square meter (kg•m²).
Ounce Inch Square Second : The pound inch square second is a unit of moment of inertia in the Imperial units and US Customary Units. Its official symbol is lbf•in•s². It is equal to 2.681170739793 pound foot². | 228 | 822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | longest | en | 0.704646 |
www.akbarian.org | 1,521,415,078,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646178.24/warc/CC-MAIN-20180318224057-20180319004057-00049.warc.gz | 350,158,296 | 10,805 | FD: Finite Difference Toolkit in Maple
# Tutorials
## 1 Heat Equation (fixed boundary, explicit FDA scheme)
We start with directly diving into a simple example of solving a PDE using FD. Consider the heat equation: $$\partial_t f(t,x) - \partial_x^2 f(t,x) = 0$$ and the following form for the finite differencing approximation to this equation: $$\frac{f^{n+1}_i-f^{n}_i}{h_t} - \frac{f^n_{i-1}-2f^n_{i}+f^n_{i-1}}{h_x^2} = 0$$ in which $$f^n_i$$ denotes discretized value of the function $$f(t^n,x_i)$$ at time step $$t^n = nh_t$$ and spatial position $$x_i = x_{\mathrm min} + i h_x$$. Note that this FDA is first order in time and second order in spatial domain. This equation requires two boundary conditions, which we denote by $$T_0$$ and $$T_1$$ (the temperatures at the two end of the domain). Suppose that we aim to solve this problem in the domain $$x\in(x_{\mathrm min},x_{\mathrm max})$$.
Here is the FD Maple code that generates the solver FORTRAN routine and the residual evaluators:
read "../../FD.mpl": Clean_FD(); Make_FD();
grid_functions := {f};
FD_table[t] := [[0],[0,1]];
HeatEq := diff(f(t,x),t) - diff(f(t,x),x,x);
init_f:= T0 + (T1-T0)*((x-xmin)/(xmax-xmin))^2;
Gen_Eval_Code(init_f,input="c",proc_name="init_f");
HeatDDS := [
{ i=[1,1,1] } = f(n+1,i) - T0 + myzero*x(i) ,
{ i=[2,Nx-1,1] } = Gen_Sten(HeatEq) ,
{ i=[Nx,Nx,1] } = f(n+1,i) - T1 +myzero*x(i)
];
A_Gen_Solve_Code(HeatDDS,{f(n+1,i)},input="d",proc_name="update_f");
Now we continue with describing the different elements of this code.
### 1.1 Set-up
This example illustrates some of the most important features of FD and its syntax. First, the commands:
Clean_FD(); Make_FD();
sets up some internal variables and creates the default finite differencing table. The second assignment:
grid_functions:={f};
defines f as a grid function, thus FD will attempt to discretize it accordingly. Anything not defined in the grid function set will be considered as a parameter or a known/given continuum function, i. e. $$g(x,y)$$ $$(g \notin {\mathrm GFS} )$$ will be discretized as $$g(x(i),y(j))$$ and FD will assume that the user will provide the function.
### 1.2 Changing the Finite Differencing Scheme
By default, FD uses centered second order finite difference scheme. Since here we are using a forward FDA for time derivative the second line is to update the FD_table content to forward in time:
FD_table[t] := [ [0] , [0,1] ];
The variable t is the built-in independent variable "time" which is used for time evolution problems. The first element [0] enforces FD to use the current time position for the discrete version of the function, i. e. $$f(t,x)$$ must be discretized as $$f(n,i)$$. The second element [0,1]] is to tell FD to use current and advance time level to discretize the first time derivative which is exactly what we need. Note that we have not provided elements for higher derivatives, thus FD would not be able to compute them. Here is another example to update the FD table to forward in time upto second derivative
FD_table[t] := [ [0] , [0,1,2] , [-1,0,1] ];
Here FD will use 3 point forward in time ( $$f^n_i , f^{n+1}_i , f^{n+2}_i$$ ) for first time derivative and centered scheme ( $$f^{n-1}_i , f^{n}_i , f^{n+1}_i$$ ) for the second time derivative of functions. A similar syntax can be used to specify how to discretize the derivatives in x, y and z direction.
Note that here we are changing the content of FD_table by specifying which points FD should use for which derivative. This is usually not the safe way to use FD, since the user should be careful about the accuracy of the FDA expressions and the fact that higher derivatives require more number of points. A better way, as it will be discussed later, is to specify the order of accuracy and the limitation on the points available (this matters at the points close to the boundary). In this method, FD will update the FD_table accordingly to achieve the desired accuracy.
### 1.3 Generating the Evaluator Code for the Initial Time Data
Next command is to create an evaluator FORTRAN routine that evaluates a function over a discretized domain. This kind of operation is used to initialize the grid functions.
Gen_Eval_Code(VDE/VCE,input="c/d",proc_name="somename");
The first argument is either a valid discrete expression (VDE) or a valid continuous expression (VCE). See the manual for the elaborated definition of this types. In a nutshell, a VCE is simply a function of fundamental independent variables $$(t,x,y,z)$$ in the continuous form such as f(t,x)+sin(x)+cos(xy^2)+... and VDE is a function appropriately indexed by numerical grid points such as: f(n,i)+sin(x(i))+g(i,j)+... VDEs are usually considered to be generated using FD package, and if human generated, FD parses the expression, to insure that it is indeed a valid discrete expression.
The second argument is to tell the routine if the expression that is being passed in is a discrete version (input="d") or is a continuous version (input="c"). Finally the third argument specifies the name of the FORTRAN routine that needs to be generated. (without the suffix .f)
This execution generates 3 files:
init_f.h init_f.f init_f_call
where init_f.h is the include C file to call the function from a C driver. init_f.f is the FORTRAN routine, and init_f_call is a typical call C expression that can be copy/pasted to the C driver routine. Note that all of the _call expressions generated by FD has res as their last argument. This needs to be modified manually by the user to the desired name(pointer). For example in the heat problem the initialization is supposed to initialize the current time level ( $$f^n_i$$ ). Therefore, res should be replaced with n_f which is the default name FD uses to denote the vector storing the current values of the function in the numerical domain.
### 1.4 Creating the Solver FORTRAN Code
The next step is to specify a FDA expression that needs to be solved in advanced time in the process of numerically solving the PDE. Furthermore, such a specification needs to be done differently for different points of the discrete domain. FD uses a specific syntax, similar to RNPL, to do so. Let's take a look at the discrete domain specifier (DDS) in this tutorial:
HeatDDS := [
{ i=[1,1,1] } = f(n+1,i) - T0 + myzero*x(i) ,
{ i=[2,Nx-1,1] } = Gen_Sten(HeatEq) ,
{ i=[Nx,Nx,1] } = f(n+1,i) - T1 +myzero*x(i)
];
The left hand side of the equations are sets of (upto 3 dimension) indexed equation such as:
{ i = [1,1,1] , j=[2,Ny,2] , k=... }
where it specifies which part of discrete domain the right hand side should be used for. One can think of each index equation such as i=[1,Nx-3,2] as a FORTRAN for loop. For instance:
$$i=[1,Nx-3,2] \longrightarrow \mathrm{do} \; i=1, \; Nx-3, \; 2$$
is a for loop with stepping 2 and in inclusive interval (1,Nx-3).
The right hand side can be a discrete or continuous valid expression which involves the variable for which FDA is going to be solved. Here the unknown variable is the advanced time f(n+1,i). Note that FD checks for consistency of LHS and RHS and all of the elements of a given DDS to ensure they have the same indexing and dimensionality. The existence of myzero*x(i) is necessary to ensure that the RHS contains at least one VDE after being converted to a solver expression. This is due to the fact that FD is using VDE/VCE as its data type, and a constant term such as T0 is not a valid discrete or continuous expression.
Finally, the solver code is generated via:
A_Gen_Solve_Code(HeatDDS,{f(n+1,i)},input="d",proc_name="update_f");
which takes a VDE/VCE, a set of unknown for which it will solve the VDE/VCE for, the type of input, and the desired FORTRAN function name. Similar to Gen_Res_Code, this call creates 3 files: update_f.f, update_f.h and update_f_call.
### 1.5 C Driver Code Structure:
Final step to solve this PDE is to write a driver routine, in C such a driver would have the following pseudo-code structure:
• Initialize memory for 2 time level for function f, namely: n_f, np1_f
• Call initializer function init_f to set up the initial time profile
• start taking a time step:
1. do 1 iteration of solver routine update_f
2. if using an explicit scheme one iteration of solver will find the solution at advanced time level, otherwise an iterative method is needed(see the wave example)
• output the new time level n+1 and continue to take the next time step
Note that FD uses n_f name for the current time level, and np1_f, np2_f ... for the advanced times, and nm1_f, nm2_f ... for the retarded time levels. Therefore it is easier to use similar names for the vectors in the C code. (See the code below)
The core part of the C driver looks like this:
/* Read parameters T0 T1 etc... */
/* Initialize numerical grid x */
/* Initialize memory for two time level of function f: n_f(1..Nx), np1_f(1..Nx) */
init_f_(x,&Nx,&T0,&T1,&xmax,&xmin,n_f);
for (i=0; i<time_steps; i++) {
norm_f = l2norm(Nx,n_f);
update_f_(n_f,np1_f,x,&Nx,&T0,&T1,&ht,&hx,&myzero,phys_bdy,np1_f);
residual_f_(n_f,np1_f,x,&Nx,&T0,&T1,&ht,&hx,&myzero,phys_bdy,&tres);
tol = tres/norm_f;
time = time + ht;
/* Output data as needed */
printf("step: %d time : %f res: %1.14e\n",i+1,time,tol);
/* Initialize n_f with the new np1_f for the next time step */
}
Here is the full c driver code that uses BBHUTIL libraries for I/O facilities.
The files of this tutorial are all located in FD/tutorials/heat. Here is a demonstration of the temperature profile being evolved using the code and XVS visualization tool.
Time evolution of the temperature, the solution obviously converges to the equilibrium temperature profile which is a linear function between the boundaries.
## 2 Wave Equation (periodic boundary, implicit FDA scheme)
Now let's focus on solving a 1-D wave equation using the implicit Crank-Nicolson scheme. This example introduces more features of FD, including manual FDA expression and imposing periodic boundary condition.
First, we convert the wave equation:
$$\partial^2_t f(t,x) = \partial^2_x f(t,x)$$
to first order in time by defining:
$$g(t,x) \doteq \partial_t f(t,x)$$
then the wave equation is:
$$\begin{cases} \partial_t f(t,x) = g(t,x)\\ \partial_t g(t,x) = \partial^2_x f(t,x)\\ \end{cases}$$
and we want to implement the following FDA equations:
$$\frac{f^{n+1}_i-f^{n}_i}{h_t} = \left( g^{n+1}_i + g^{n}_i \right) / 2$$
$$\frac{g^{n+1}_i-g^{n}_i}{h_t} = \left( \frac{f^{n+1}_{i-1} -2 f^{n+1}_{i} + f^{n+1}_{i+1} }{ {h_x}^2 } + \frac{f^{n}_{i-1} -2 f^{n}_{i} + f^{n}_{i+1} }{ {h_x}^2} \right) / 2$$
First, here is the FD code that solves the wave equation and also generates an independent residual evaluator (IRE) code for checking the solution (in the code f_t denotes the g defined in the equations):
read "/d/bh8/home/arman/FD/FD.mpl":Clean_FD();Make_FD();
grid_functions := {f,f_t};
eq1 := diff(f(t,x),t) = f_t(t,x);
eq2 := diff(f_t(t,x),t) = diff(f(t,x),x,x);
eq3 := diff(f(t,x),t,t) = diff(f(t,x),x,x);
Gen_Res_Code(lhs(eq3)-rhs(eq3),input="c",proc_name="ire_f");
FD_table[t] := [[0],[0,1]];
AVGT := a -> ( FD( a,[ [1],[0] ]) + FD( a,[ [0],[0] ]) )/2;
eq1_D:= Gen_Sten(lhs(eq1)) - AVGT(Gen_Sten(rhs(eq1)));
eq2_D:= Gen_Sten(lhs(eq2)) - AVGT(Gen_Sten(rhs(eq2)));
s_f:= [
{ i=[1,1,1] } = FD_Periodic(eq1_D,{i=1}) ,
{ i=[2,Nx-1,1] } = eq1_D,
{ i=[Nx,Nx,1] } = FD_Periodic(eq1_D,{i=Nx})
];
s_f_t:= [
{ i=[1,1,1] } = FD_Periodic(eq2_D,{i=1}),
{ i=[2,Nx-1,1] } = eq2_D,
{ i=[Nx,Nx,1] } = FD_Periodic(eq2_D,{i=Nx})
];
A_Gen_Solve_Code(s_f,{f(n+1,i)},input="d",proc_name="u_f",is_periodic=true);
A_Gen_Solve_Code(s_f_t,{f_t(n+1,i)},input="d",proc_name="u_f_t",is_periodic=true);
A_Gen_Res_Code(s_f,input="d",proc_name="res_f",is_periodic=true);
A_Gen_Res_Code(s_f_t,input="d",proc_name="res_f_t",is_periodic=true);
init_f:=A*exp(-(x-x0)^2/delx^2);
init_f_t:=idsignum*diff(init_f,x);
Gen_Eval_Code(init_f,input="c",proc_name="init_f");
Gen_Eval_Code(init_f_t,input="c",proc_name="init_f_t");
### 2.1 Independent Residual Evaluator (IRE)
First note that the equations eq1 and eq2 define the wave equation in a first order form, while eq3 defines the wave equation in a second order form. In the beginning of the code, we first create a residual evaluator which computes the residual of the wave equation:
eq3 := diff(f(t,x),t,t) = diff(f(t,x),x,x);
Gen_Res_Code(lhs(eq3)-rhs(eq3),input="c",proc_name="ire_f");
This is for testing purposes. Since FD uses second order FDA, this residual should converge by factor of 4, if the resolution is improved by a factor of 2. The driver C code in the following prints out this residual for monitoring. Note that since second time derivative requires 3 time level, the driver code should keep 3 time level during the evolution.
### 2.2 Writing a Manual FDA Operator
Next step is to create the time averaging operator we need in the Crank-Nicolson scheme. This is provided by defining the AVGT maple operator which in turn uses one of the fundamental operators in FD, named FD!
AVGT := a -> ( FD( a,[ [1],[0] ]) + FD( a,[ [0],[0] ]) )/2;
The syntax of FD is as following:
FD( VDE , [ [shift_t] , [shift_x,shift_y,shift_z] ] )
FD takes a VDE(Valid Discrete Expression) and returns the shifted version of it. As you can easily see, this operator can provide a manual way to generate any FDA expression. However, FD is not designed to be used extensively by the user, rather it is for imposing a non-trivial FDA expressions such as implicit schemes or the ones occur at the boundary points. See the manual of FD for detailed description of this function.
Next is to create the desired FDA expressions that defines the wave equation. This is done via: FD_table[t] := [[0],[0,1]];
eq1_D:= Gen_Sten(lhs(eq1)) - AVGT(Gen_Sten(rhs(eq1)));
eq2_D:= Gen_Sten(lhs(eq2)) - AVGT(Gen_Sten(rhs(eq2)));
Note that we change FD_table in the first line to enforce using forward in time derivatives for the first time derivatives occurring in the equation.
The user can look at the content of eq1_D and eq2_D which are the VDE version of the wave equation. Here is a snapshot of the maple session:
> eq1_D;
-f(n, i) + f(n + 1, i)
---------------------- - 1/2 f_t(n + 1, i) - 1/2 f_t(n, i)
ht
> eq2_D;
f_t(n, i) - f_t(n + 1, i) f(n + 1, i - 1) - 2 f(n + 1, i) + f(n + 1, i + 1)
- ------------------------- - 1/2 -------------------------------------------------
ht 2
hx
f(n, i - 1) - 2 f(n, i) + f(n, i + 1)
- 1/2 -------------------------------------
2
hx
>
### 2.3 Making an FDA Periodic
The next step is to write the Discrete Domain Specifiers (DSS) for the two equations:
s_f:= [
{ i=[1,1,1] } = FD_Periodic(eq1_D,{i=1}) ,
{ i=[2,Nx-1,1] } = eq1_D,
{ i=[Nx,Nx,1] } = FD_Periodic(eq1_D,{i=Nx})
];
s_f_t:= [
{ i=[1,1,1] } = FD_Periodic(eq2_D,{i=1}),
{ i=[2,Nx-1,1] } = eq2_D,
{ i=[Nx,Nx,1] } = FD_Periodic(eq2_D,{i=Nx})
];
Note that here at i=1 and i=Nx we convert the eq1/2_D to their periodic version. This is done via FD_Period function. The syntax of this function should be clear from the example. Any index i smaller than 1 at the point i=1 will be mapped to i+(Nx-1) and any index higher than Nx at {i=Nx} will be mapped to i-(Nx-1). Here is a screen-shot of the maple session.
> FD_Periodic(eq2_D,{i=1});
f_t(n, i) - f_t(n + 1, i) f(n + 1, i - 2 + Nx) - 2 f(n + 1, i) + f(n + 1, i + 1)
- ------------------------- - 1/2 ------------------------------------------------------
ht 2
hx
f(n, i - 2 + Nx) - 2 f(n, i) + f(n, i + 1)
- 1/2 ------------------------------------------
2
hx
>
Note how i-1 index is converted to i - 2 + Nx, which maps the point on the left of the left boundary to the point right to the right boundary. Of course point i=1 is mapped to i=Nx.
### 2.4 Typical Driver for an Implicit Scheme
Finally, the last part of the code generates the FORTRAN routines to do 1 step Newton-Gauss-Sidel relaxation operation to solve the wave equations:
A_Gen_Solve_Code(s_f,{f(n+1,i)},input="d",proc_name="u_f",is_periodic=true);
A_Gen_Solve_Code(s_f_t,{f_t(n+1,i)},input="d",proc_name="u_f_t",is_periodic=true);
A_Gen_Res_Code(s_f,input="d",proc_name="res_f",is_periodic=true);
A_Gen_Res_Code(s_f_t,input="d",proc_name="res_f_t",is_periodic=true);
Note that here we also need the residual evaluator routines to check see if the residual is below a certain threshold. A residual evaluator code is generated using A_Gen_Res_Code function. Also note that the is_periodic flag is turned on, since the VDE that is being passed into the routines is converted to a periodic form, otherwise the routines complain that the discrete expression is not valid. Finally, the initializer routines are required at the initial time:
init_f:=A*exp(-(x-x0)^2/delx^2);
init_f_t:=idsignum*diff(init_f,x);
Gen_Eval_Code(init_f,input="c",proc_name="init_f");
Gen_Eval_Code(init_f_t,input="c",proc_name="init_f_t");
Here is a pseudo-code for a C driver code to use these routines:
• Initialize memory for 2 time level for function f, namely: n_f, np1_f
• Call initializer function init_f to set up the initial time profile
• start taking a time step:
1. do 1 iteration of solver routine update_f and update_f_t
2. compute the residual using res_f and res_f_t to see if iteration is sufficient.
3. if residual is not below the desired threshold, goto 1.
• output the new time level n+1 and continue to take the next time step
the core part of the C code looks like this:
/* initialize the problem */
for (i=0; i<steps; i++) {
j=0;
tot_res = 1.0;
while (tot_res > 1.0e-9) {
u_f_(n_f,n_f_t,np1_f,np1_f_t,&Nx,&ht,phys_bdy,np1_f);
u_f_t_(n_f,n_f_t,np1_f,np1_f_t,&Nx,&ht,&hx,phys_bdy,np1_f_t);
res_f_(n_f,n_f_t,np1_f,np1_f_t,&Nx,&ht,phys_bdy,&res_f);
res_f_t_(n_f,n_f_t,np1_f,np1_f_t,&Nx,&ht,&hx,phys_bdy,&res_f_t);
tot_res=res_f/l2norm(Nx,np1_f) + res_f_t/l2norm(Nx,np1_f_t);
j++;
}
/* output data */
/* compute other diagnostic functions */
/* swap n and np1 time levels */
}
See the [C code ]. This tutorial can be found in FD/tutorials/wave1d_periodic folder. Here is a visualization of the solution solved using these routines:
Time evolution of the wave in a periodic boundary condition.
last update: Sun May 08 2016 | 5,275 | 18,344 | {"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": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-13 | latest | en | 0.745655 |
https://www.beyondthesize.com/from-giraffes-to-cargo-containers-understanding-100-feet/ | 1,716,221,150,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00661.warc.gz | 596,013,809 | 17,636 | # From Giraffes to Cargo Containers: Understanding 100 Feet
Comparing 100 Feet: Understanding Size with Familiar Objects
As we go about our daily lives, we come across numerous objects and creatures that vary in size. From towering buildings to tiny insects, we use size as a way to understand and interact with the world around us.
However, sometimes it can be challenging to comprehend the size of something, especially when it is measured in feet. In this article, we will explore eight comparisons of 100 feet to help us visualize and understand this length better.
Comparison 1: Boeing 737-500
A Boeing 737-500, a popular commercial jet airplane, has an average length of 100 feet. Imagine standing next to one of these aircraft, and you will notice just how massive it is.
This robust plane can hold up to over 140 passengers and can reach speeds of up to 586mph. It’s almost impossible to believe that something that large can fly, but with the aid of modern aviation technology, it can soar through the skies with ease.
Comparison 2: Half an Ice Hockey Rink
For sports enthusiasts, you’ll be familiar with the concept of an ice hockey rink. They’re commonly found in North America and are used for ice hockey games and other winter sports.
At 200 feet long, a standard rink is two times the length of 100 feet. To put this in perspective, imagine half of a standard ice hockey rink.
This comparison helps us visualize the relative size of 100 feet, and we can all agree; it’s much more significant than what we initially thought. Comparison 3: 3 School Buses
Anyone who has ever taken public transportation to school knows how large school buses can be.
A standard school bus is around 40 feet long, meaning for a comparison of 100 feet, we will need three of these buses lined up next to each other. This comparison is especially helpful for parents with young children or school administrators who can visualize the size of 100 feet within the confines of a school setting.
Comparison 4: Blue Whale
The blue whale is the largest animal on earth, and measuring up to 100 feet in length, they are a perfect comparison for this article. Blue whales are commonly found in the ocean, and their weight can reach up to 200 tons.
That’s equivalent to around 33 adult elephants! Although they are enormous, they feed on tiny krill and have a gentle disposition towards humans. Comparison 5: Semi-Truck Trailers or Cargo Containers
Semi-truck trailers or cargo containers are widely used for transportation, and a standard trailer measures around 53 feet long.
For a comparison of 100 feet, we will need two of these trailers connected end to end. Cargo containers are used to transport goods across long distances, and they are also stackable, allowing for easy storage.
With this comparison, we can visualize the space needed to transport goods efficiently over long distances. Comparison 6: 18 Adults
When we talk about 100 feet in terms of a human chain, it’s hard to picture how long that would be.
However, imagine 18 average-sized adults standing in a straight line, and you have a reference to compare the length of 100 feet. This comparison shows us how small we are in comparison to the world around us and reminds us that there’s still so much we have yet to explore and understand.
Comparison 7: Basketball Court or 10 Storey Building
For those that aren’t familiar with the metric system, 100 feet is roughly equivalent to 30 meters. This is roughly the size of a basketball court or a ten-story building.
The basketball court comparison is a useful one for sports enthusiasts who can visualize the size of a local recreational center. The ten-story building comparison is one that resonates with non-athletes, and it helps us understand how tall a building would be, which is the primary focus in the construction industry.
Comparison 8: Giraffes
Giraffes are fascinating creatures known for their long necks, but what most people don’t know is that they can grow up to 18 feet in height and 100 feet in length. Yes, that’s right! One hundred feet is the average length of nine adult giraffes, lined up next to each other.
This comparison serves as a reminder of the diversity of animals that exist on our planet, and how much more we need to learn about our animal kingdom.
## Conclusion
In conclusion, these eight comparisons help us understand the size of 100 feet in the context of familiar objects. From transportation and sports to animals, we can use these references to visualize and comprehend this vast length better.
Whether you’re an athlete or an animal enthusiast, understanding the size of objects and creatures is crucial for learning and exploration. Use these comparisons to develop a better appreciation for the world around us and all the fantastic things it has to offer.
## 3) Half an Ice Hockey Rink
Ice hockey is a popular team sport played on a large ice rink. The measurement and size of an official ice hockey rink are standard and governed by the National Hockey League (NHL) and International Ice Hockey Federation (IIHF).
An NHL rink, which is commonly found in North America, measures 200 feet long and 85 feet wide, while an IIHF rink, which is more common outside North America, measures 197 feet long and 98.4 feet wide. The size of an official ice hockey rink is enormous, and it can be difficult to visualize and understand the vast expanse of the space.
To put the size into perspective, imagine halving an NHL rink lengthways. This cut would result in the rink measuring 100 feet long and 85 feet wide, making it half the size of an official ice hockey rink.
If you’re wondering how long 100 feet is, this comparison provides a more tangible image in your mind. The half ice hockey rink comparison is a useful way of visualizing space and size.
It is an excellent reference for people who love sports, especially ice hockey, and those in the construction or architectural industry. You can imagine half of the rink being used for two smaller games to occur simultaneously, which provides the idea to the coaches as well.
## 4) 3 School Buses
School buses are ubiquitous to school districts across North America. A typical medium-sized school bus measures around 40 feet long and 8.5 feet wide.
The height of a school bus is usually around 12 feet. Transporting students to and from school safely is the primary function of school buses, and they have many safety features built-in.
With a measurement of 100 feet, it can be challenging to visualize the distance and grasp the size. To help us with visualization, imagine lining up three medium-sized school buses next to each other.
With this comparison, we can understand the length and width of 100 feet and the space an object of this length takes up. The importance of visualization is the ability to transfer a thought into a mental image that people can use to understand a subject better.
The school bus comparison works great for school administrators and parents with school-going children. By understanding the size of 100 feet, they can better plan and understand design layouts of parking, drop-off, and pick-up areas, bus loading zones, and other safety-related matters.
## Conclusion
Understanding the size and length of 100 feet is essential to better comprehend the objects and environment around us. The comparisons of half an ice hockey rink and three school buses provide visualization and understanding of how vast 100 feet can be.
These comparisons serve as an excellent reference point for people to understand the size of large objects or areas better. By being able to visualize and understand an object’s size, architects, builders, urban planners, and other professionals can better plan and design structures to fit the surrounding environment and function appropriately.
## 5) Blue Whale
The blue whale is the largest mammal on earth, and as such, it is an excellent comparison for objects that measure 100 feet in length. Blue whales can grow up to 100 feet long and can weigh up to 200 tons.
These gentle giants are found in all of the world’s oceans, often swimming, diving, and feeding in groups. To help visualize the size of a blue whale, you can compare one to a Boeing 737-500 airplane, which is also around 100 feet in length.
There are some remarkable facts about blue whales that you may not be aware of. For example, female blue whales are usually larger than males, growing up to around 90 feet long, while males are around 80 feet in length.
Additionally, baby blue whales, also known as calves, are born around 23 feet long and can weigh up to 3 tons. Blue whales are also among the loudest animals on the planet, with their vocalizations reaching up to 188 decibels, louder than the sound of a jet engine.
The blue whale comparison is an excellent visual reference for people who want to better understand the vastness of the ocean. It’s awe-inspiring to think about these gentle giants swimming deep in the vast ocean.
Additionally, it’s a reminder of the importance of marine conservation efforts to protect and preserve these magnificent creatures for future generations.
## 6) Semi-Truck Trailers and Cargo Containers
Semi-truck trailers and cargo containers are standard units used for transportation around the world. These containers come in various sizes and can transport a range of goods, from fresh produce and electronics to cars and heavy machinery.
A standard semi-truck trailer measures around 53 feet in length, with a cargo container of the same dimensions. The length of 100 feet can be visualized by connecting two of these trailers or containers end to end.
These units are built to withstand harsh weather and rough terrain, and are designed to protect their contents from damage. Cargo containers can also be shipped via cargo ships, trains, and planes, making them a versatile and popular mode of transportation.
The semi-truck trailer and cargo container comparison is significant for those in the transportation industry, logistics, and shipping. By understanding the size of 100 feet, these professionals can estimate the amount of cargo that can be transported at a given time and provide clients with accurate transport cost projections.
## Conclusion
The blue whale and semi-truck trailer/cargo container comparisons have provided us with an excellent visual reference to understand the vastness of the ocean and the transportation industry’s capacity. By picturing these massive objects, we can get a better idea of how enormous and intricate the world around us is.
The size of 100 feet is essential to various industries and can be used as an essential reference point in planning and decision-making.
The human body has long been used as a unit of measurement, with the average height of an adult human being being around 5.5 feet. With this information, we can visualize how long 100 feet is by lining up 18 adults in a straight line.
This visual representation provides an excellent reference point for the size of 100 feet, and we can see how many people it would take to span this length. This size comparison is significant for many industries, especially those in construction and urban planning.
By understanding the size of 100 feet, planners and developers can accurately estimate the amount of land required for a project, such as a park or shopping mall, and ensure that the space is optimized for the intended use.
## 8) Basketball Court and 10 Storey Building
A basketball court is a standard size level surface area measuring 94 feet in length and 50 feet in width. It’s used for basketball games, cardio workouts, and other various indoor activities.
With its rectangular shape, a basketball court is an excellent comparison for 100 feet in length, as it’s slightly shorter than the actual length at 94 feet. A 10 story building is also a useful comparison for understanding the length of 100 feet.
While there is no exact standard size for a 10 story building, the average height of such a building ranges between 100-120 feet tall. Therefore, a 10 story building would be an excellent comparison for understanding the height of 100 feet, while the width is arbitrary depending on the building’s size and shape.
These comparisons are significant for architects, construction workers, and anyone working in urban development. By understanding the size of a basketball court and a 10 story building, they can get a clear and tangible image in their mind of what 100 feet looks like.
It helps create accurate blueprints, plans, and designs for buildings and outdoor spaces, ensuring they are optimal for their intended use.
## Conclusion
In conclusion, human beings have used unit measurements for centuries, and this article’s comparisons for 100 feet show how we have standardized these units of measurements for easy reference. Whether it’s comparing it to the size of a basketball court, line of 18 adults, a 10 story building, or any other object, these comparisons provide a visual representation of size and space.
With a better understanding of size, people can create accurate, well-designed and functional spaces and improve upon design and construction projects.
## 9) Giraffes
Giraffes are some of the most unique and majestic animals in the world, and they make for an excellent comparison for objects that measure 100 feet long. Giraffes are the tallest mammals on earth, with adult males growing up to 18 feet tall and females up to 15 feet.
These gentle giants are commonly found in savannas, grasslands, and forests across Africa, and they feed on leaves from tall trees. Aside from their height, there are some fascinating facts about giraffes that you may not know.
For instance, they have one of the shortest sleep cycles of any mammal, with adult giraffes sleeping for only 20 minutes a day, while baby giraffes have sleep patterns similar to those of humans. Giraffes also have unique neck structures, with seven vertebrae like those in most mammals, but with each being much longer, around 10 inches in length.
To better understand the size of 100 feet, you can imagine lining up nine adult giraffes to represent the length. This comparison provides an excellent visual reference for people to understand the proportions of large objects or areas.
Additionally, it’s a reminder that our planet is rich in biodiversity, with many amazing creatures like giraffes that thrive in their ecosystems. The giraffe comparison is essential for those who love nature and wildlife.
It’s awe-inspiring to think about these gentle giants roaming on our planet, and it highlights the importance of conservation efforts to protect and preserve these magnificent creatures for future generations.
Conclusion:
Comparisons help us visualize and comprehend the world around us. The comparisons for 100 feet using blue whales, semi-truck trailers and cargo containers, basketball courts, school buses, and human chains help us understand the vastness of items that measure this length.
The comparisons with giraffes help us understand the size of such animals and appreciate our planet’s flora and fauna. These comparisons serve as an essential reference point for visualizing large objects or areas and how we relate and interact with them.
By understanding size, we can better plan and design structures, transport goods, build cities, and appreciate the world around us. | 3,139 | 15,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.939174 |
https://grandpaperwriters.com/question-answer-the-price-of-a-product-excluding-shipment-is-7-25-shipment-cost-from-omaha-to-b-is-2-75-and-omaha/ | 1,695,935,389,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00234.warc.gz | 308,175,598 | 11,353 | # Question & Answer: The price of a product excluding shipment is \$7.25 Shipment cost from Omaha to B is \$2 75 and Omaha…
The price of a product excluding shipment is \$7.25 Shipment cost from Omaha to B is \$2 75 and Omaha to A is \$5.15. Calculate the phantom cost far customers at B and freight absorption cost far seller to ship at A using the following figure. The price listed in the following figure is the actual selling price charged to customers. | 106 | 459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.951373 |
http://pithocrates.com/tag/3-phase-power/ | 1,563,870,127,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00230.warc.gz | 126,775,498 | 13,525 | # Electric Grid, Voltage, Current, Power, Phase Conductor, Neutral Conductor, 3-Phase Power, Transmission Towers and Corona Discharge
Posted by PITHOCRATES - August 15th, 2012
# Technology 101
## The Electric Grid is the Highways and Byways for Electric Power from the Power Plant to our Homes
Even our gasoline-powered cars operate on electricity. The very thing that ignites the air-fuel mixture is an electric spark. Pushed across an air-gap by a high voltage. Because that’s something that high voltages do. Push electrons with such great force that they can actually leave a conductor and travel through the air to another conductor. Something we don’t want to happen most of the time. Unless it’s in a spark plug in our gasoline engine. Or in some movie prop in a cheap science fiction movie.
No. When we use high voltage to push electrons through a conductor the last thing we want to happen is for the electrons to leave that conductor. Because we spend a pretty penny to push those electrons out of a power plant. And if we push the electrons out of the conductor they won’t do much work for us. Which is the whole point of putting electricity into the electric grid. To do work for us.
The electric grid. What exactly is it? The highways and byways for electric power. Power plants produce electric power. And send it to our homes. As well as our businesses. Power is the product of voltage and current. In our homes something we plug into a 120V outlet that draws 8 amps of current consumes 960 watts. Which is pretty big for a house. But negligible for a power plant generator producing current at 20,000 volts. For at 20,000 volts a generator only has to produce 0.48 amps (20,000 X 0.48 = 960). Or about 6% of the current at 120V.
## Between our Homes and the Power Plant we can Change that Current by Changing the Voltage
Current is money. Just as time is money. In fact current used over time helps to determine your electric bill. Where the utility charges you for kilowatt hours (voltage X current X time). (This would actually give you watt-hours. You need to divide by 1000 to get kilowatt hours.) The electric service to your house is a constant voltage. So it’s the amount of current you use that determines your electric bill. The more current you use the greater the power you use. Because in the power equation (voltage X current) voltage is constant while current increases.
Current travels in conductors. The size of the conductor determines a lot of costs. Think of automobile traffic. Areas that have high traffic volumes between them may have a very expensive 8-lane Interstate expressway interconnecting them. Whereas a lone farmer living in the ‘middle of nowhere’ may only have a much less expensive dirt road leading to his or her home. And so it is with the electric grid. Large consumers of electric power need an Interstate expressway. To move a lot of current. Which is what actually spins our electrical meters. Current. However, between our homes and the power plant we can change that current. By changing the voltage. Thereby reducing the cost of that electric power Interstate expressway.
The current flowing through our electric grid is an alternating current. It leaves the power plant. Travels in the conductors for about 1/120 of a second. Then reverses direction and heads back to the power plant. And reverses again in another 1/120 of a second. One complete cycle (travel in both directions) takes 1/60 of a second. And there are 60 of these complete cycles per second. Hence the alternating current. If you’re wondering how this back and forth motion in a wire can do any work just think of a steam locomotive. Or a gasoline engine. Where a reciprocating (back and forth) motion is converted into rotational motion that can drive a steam locomotive. Or an automobile.
## The Voltages of our Electric Grid balance the Cost Savings (Smaller Wires) with the Higher Costs (Larger Towers)
An electric circuit needs two conductors. When current is flowing away from the power plant in one it is flowing back to the power plant in the other. As the current changes direction is has to stop first. And when it stops flowing the current is zero. Using the power formula this means there are zero watts twice a cycle. Or 120 times a second. Which isn’t very efficient. However, if you bring two other sets of conductors to the work load and time the current in them properly you can remove these zero-power moments. You send the first current out in one set of conductors and wait 1/3 of a cycle. Then you send the second current out in the second set of conductors and wait another 1/3 cycle. Then you send the third current out in the third set of conductors. Which guarantees that when a current is slowing to stop to reverse direction there are other currents moving faster towards their peak currents in the other conductors. Making 3-phase power more efficient than single-phase power. And the choice for all large consumers of electric power.
Anyone who has ever done any electrical wiring in their home knows you can share neutral conductors. Meaning more than one circuit coming from your electrical panel can share the return path back to the panel. If you’ve ever been shocked while working on a circuit you switched off in your panel you have a shared neutral conductor. Even though you switched off the circuit you were working on another circuit sharing that neutral was still switched on and placing a current on that shared neutral. Which is what shocked you. So if we can share neutral conductors we don’t need a total of 6 conductors as noted above. We only need 4. Because each circuit leaving the power plant (i.e., phase conductor) can share a common neutral conductor on its way back to the power plant. But the interesting thing about 3-phase power is that you don’t even need this neutral conductor. Because in a balanced 3-phase circuit (equal current per phase) there is no current in this neutral conductor. So it’s not needed as all the back and forth current movement happens in the phase conductors.
Electric power travels in feeders that include three conductors per feeder. If you look at overhead power lines you will notice they all come in sets of threes when they get upstream of the final transformer that feeds your house. The lines running along your backyard will have three conductors across the top of the poles. As they move back to the power plant they pass through additional transformers that increase their voltage (and reduce their current). And the electric transmission towers get bigger. With some having two sets of 3-conductor feeders. The higher the voltage the higher off the ground they have to be. And the farther apart the phase conductors have to be so the high voltage doesn’t cause an arc to jump the ‘air gap’ between phase conductors. As you move further away from your home back towards the power plant the voltage will step up to values like 2.4kV (or 2,400 volts), 4.8kV and13.2kV that will typically take you back to a substation. And then from these substations the big power lines head back towards the power plant. On even bigger towers. At voltages of 115kV, 138kV, 230kV, 345kv, 500kV and as high as 765kV. When they approach the power plant they step down the voltage to match the voltage produced by its generators.
They select the voltages of our electric grid to balance the cost savings (smaller wires) with the higher costs (larger towers taking up more land). If they increase the voltage so high that they can use very thin and inexpensive conductors the towers required to transmit that voltage safely may be so costly that they exceed the cost savings of the thinner conductors. So there is an economic limit on voltage levels As well as other considerations of very high voltages (such as corona discharge where high voltages create such a power magnetic field around the conductors that it may ionize the air around it causing a sizzling sound and a fuzzy blue glow around the cable. Not to mention causing radio interference. As well as creating some smog-causing pollutants like ozone and nitrogen oxides.)
www.PITHOCRATES.com | 1,853 | 8,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.863319 |
http://demonstrations.wolfram.com/FamilyOfPlaneCurvesInTheExtendedGaussPlaneGeneratedByOneFunc/ | 1,526,982,589,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00417.warc.gz | 78,567,714 | 11,699 | # Family of Plane Curves in the Extended Gauss Plane Generated by One Function
This Demonstration shows four families of curves, each member of which satisfies an equation of the form , where, at any point on the parametrized curve, represents the arc length to that point and represents the spherical curvature at that point when the curve is projected onto the sphere via stereographic projection. If a curve on the sphere can be obtained from another via rotation of the sphere, then the two plane curves that are their stereographic projections are in the same family. The polar coordinates of an arbitrary point on the sphere define an axis of rotation; use that and an angle of rotation to get different curves in a family.
### DETAILS
The two-dimensional unit sphere , centered at the origin, is a model of the Gauss plane together with a point at infinity. This model is realized via a stereographic projection from the pole onto the plane through the equator. The group of rigid motions on coincides with the group of rotations that preserves . Any spherical curve on , parameterized by an arc length parameter , is defined up to a rigid motion on by a function called spherical curvature. The group of rigid motions on induces on the plane via the stereographic projection a subgroup of the Möbius group. The transformations of this subgroup are represented by the functions , , , where is the field of complex numbers. A relation between the spherical curvature of a curve on and the Euclidean curvature of its corresponding plane curve is an invariant under the group and determines any plane curve up to transformation from the group . For example, in snapshot 3, the function defines plane curves equivalent to the logarithmic spiral. Since the natural Cesàro equation for the curvature defines a Cornu's spiral in the Euclidean plane, the curve in snapshot 1 could be called a Cornu's spiral in the Möbius plane with a group of rigid motion .
### PERMANENT CITATION
Share: Embed Interactive Demonstration New! Just copy and paste this snippet of JavaScript code into your website or blog to put the live Demonstration on your site. More details » Download Demonstration as CDF » Download Author Code »(preview ») Files require Wolfram CDF Player or Mathematica. | 454 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2018-22 | latest | en | 0.905552 |
http://oeis.org/A179066 | 1,568,966,064,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573908.70/warc/CC-MAIN-20190920071824-20190920093824-00454.warc.gz | 143,240,963 | 5,127 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A179066 a(n) is the smallest positive number having the same digital root as n but no digit in common with n and not occurring earlier. 3
28, 11, 12, 13, 14, 15, 16, 17, 18, 37, 2, 3, 4, 5, 6, 7, 8, 9, 46, 38, 30, 31, 41, 33, 34, 35, 36, 1, 47, 21, 22, 50, 24, 25, 26, 27, 10, 20, 48, 58, 23, 51, 52, 53, 63, 19, 29, 39, 67, 32, 42, 43, 44, 72, 64, 74, 66, 40, 68, 78, 70, 71, 45, 55, 83, 57, 49, 59, 87, 61, 62 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The sequence certainly cannot be continued beyond N* = 1124578, because a(N*) could only be written using "0", "3", "6" and "9", and then its digital root A010888 is always 3, 6 or 9, while that of N* equals 1. It appears that a(n)=m <=> a(m)=n, i.e., a is its own inverse, a(a(n))=n, whenever a(m) is defined. a(261379)=4000555 is the first term for which a(m) is not defined. - Hans Havermann, Jan 24 2011 It seems that there is no invariant subset of the form {1,...,N} on which a is defined. Lars Blomberg has calculated all terms up to a(1124577)=3008889, the first 481185 terms being confirmed by Hans Havermann. - Eric Angelini, Mar 21 2011 LINKS M. F. Hasler, Table of n, a(n) for n = 1..13578 Lars Blomberg, Table of n, a(n) for n = 1..1124577 (complete sequence) E. Angelini, M. F. Hasler, B. Jubin, a(n) and n have the same digitsum but no digit in common, seqfan mailing list, Jan 04 2011 Hans Havermann, A179066 as a fractal EXAMPLE The digital root (A010888) of n=1 is 1 and the first not-yet-used number with the same digital root (1, 10, 19, 28, ...) not containing a "1" is 28. The digital root of n=2 is 2 and the first not-yet-used number with the same digital root (2, 11, 20, ...) not containing a "2" is 11. The digital root of n=12345 is 6 and the first not-yet-used number with the same digital root (6, 15, 24, ...) not containing a "1" or a "2" or a "3" or a "4" or a "5" is 60000. MATHEMATICA digitalRoot[n_] := Mod[n-1, 9] + 1; a[0] = 1; a[n_] := a[n] = For[d = IntegerDigits[n]; r = digitalRoot[n]; k = 1, True, k++, If[ FreeQ[ Array[a, n-1], k] && digitalRoot[k] == r && Intersection[d, IntegerDigits[k]] == {}, Return[k]]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Aug 13 2013 *) PROG (PARI) /* using the "proof by 9" idea from Benoît Jubin */ {A179066=[]; Nmax=9999 /* number of terms to compute */; S=sum(j=1, #A179066, 1<; NextA179066:=function(n, T); k:=DigitalRoot(n); while k in T or not IsEmpty(Set(Intseq(k)) meet Set(Intseq(n))) do k+:=9; end while; return k; end function; T:=[]; for n in [1..100] do a:=NextA179066(n, T); Append(~T, a); end for; T; // Klaus Brockhaus, Jan 26 2011 CROSSREFS Cf. A179105 for record values, A179110 for indices of record values. Sequence in context: A040761 A070659 A040760 * A033970 A033348 A040759 Adjacent sequences: A179063 A179064 A179065 * A179067 A179068 A179069 KEYWORD nonn,fini,full,base,nice AUTHOR Eric Angelini and M. F. Hasler, Jan 04 2011 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified September 20 03:54 EDT 2019. Contains 327210 sequences. (Running on oeis4.) | 1,223 | 3,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-39 | latest | en | 0.883816 |
https://www.interviewsansar.com/c-sharp-program-to-find-the-contiguous-sub-array-with-maximum-sum/ | 1,638,853,631,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00068.warc.gz | 883,843,220 | 13,156 | # C Sharp program to Find the contiguous sub array with Maximum sum
Given an array of N integers.Find the contiguous sub array with Maximum sum.
Contiguous sub array means sub-array with sequence index like (0,1,2..) or(1,2,3…)
INPUT
N:Size of the array
arr[N]: arr[0],arr[1]…. arr[N-1] number of integer elements
Output
Print the maximum sum of the contiguous sub-array in a separate line for each test case.
For Example:
Input array size is 5
array values { -1, 3, -4,5, 7 }
Output of Maximum sum of the contiguous sub-array is 12 of elements (5,7)
### Program to Find the contiguous sub array with Maximum sum
C# Code
```using System;
class SubarrayWithMaximumSum
{
static void Main(string[] args)
{
int maxvalue = Int32.MinValue;
//array with size N
int[] arr = new int[] { -1, 3, -4,5, 7 };
int sum = 0;
//logic to find the Subarray with Maximum sum
for (int i = 0; i < arr.Length; i++)
{
for (int j = i; j < arr.Length; j++)
{
sum = sum + arr[j];
if (maxvalue < sum)
{
//maxvalue stores the maximum sum of array wiht sequence index
maxvalue = sum;
}
else
{ //if sequence fails we will start with new index value as starting index for sub array
sum = 0;
break;
}
}
sum = 0;
}
Console.WriteLine("Sub array with Maximum sum= {0}",maxvalue);
}
}
```
Output
Sub array with Maximum sum= 12.
### Method-2
Algorithm
Initialize:
minvalue =0
maxvalue= minimum value of an integer
Loop for each element of the array
(a) sum = sum + arr[i];
(b) if (maxvalue < minvalue
maxvalue = maxvalue;
else
sum = 0;
```class SubarrayWithMaximumSum
{
static void Main(string[] args)
{
int maxvalue = Int32.MinValue;
//array with size N
int[] arr = new int[] { -1, 3, -4, 5, 7 };
int minvalue = 0;
//int start_index=0, end_index=0,start=0,end=0;
//logic to find the Subarray with Maximum sum
for (int i = 0; i < arr.Length; i++)
{
minvalue = minvalue + arr[i];
if (maxvalue < minvalue)
{
//maxvalue stores the maximum sum of array wiht sequence index
maxvalue = minvalue;
}
else
{ //if sequence fails we will start with new index value as starting index for sub array
minvalue = 0;
}
}
Console.WriteLine("Sub array with Maximum sum= {0}", maxvalue);
}
}
```
Output
Sub array with Maximum sum= 12. | 645 | 2,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.437275 |
https://cs.brown.edu/courses/cs1951x/docs/analysis/calculus/specific_functions.html | 1,695,571,284,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506658.2/warc/CC-MAIN-20230924155422-20230924185422-00018.warc.gz | 205,499,912 | 44,587 | # mathlibdocumentation
analysis.calculus.specific_functions
# Infinitely smooth bump function #
In this file we construct several infinitely smooth functions with properties that an analytic function cannot have:
• exp_neg_inv_glue is equal to zero for x ≤ 0 and is strictly positive otherwise; it is given by x ↦ exp (-1/x) for x > 0;
• real.smooth_transition is equal to zero for x ≤ 0 and is equal to one for x ≥ 1; it is given by exp_neg_inv_glue x / (exp_neg_inv_glue x + exp_neg_inv_glue (1 - x));
• f : cont_diff_bump_of_inner c, where c is a point in an inner product space, is a bundled smooth function such that
• f is equal to 1 in metric.closed_ball c f.r;
• support f = metric.ball c f.R;
• 0 ≤ f x ≤ 1 for all x.
The structure cont_diff_bump_of_inner contains the data required to construct the function: real numbers r, R, and proofs of 0 < r < R. The function itself is available through coe_fn.
• If f : cont_diff_bump_of_inner c and μ is a measure on the domain of f, then f.normed μ is a smooth bump function with integral 1 w.r.t. μ.
• f : cont_diff_bump c, where c is a point in a finite dimensional real vector space, is a bundled smooth function such that
• f is equal to 1 in euclidean.closed_ball c f.r;
• support f = euclidean.ball c f.R;
• 0 ≤ f x ≤ 1 for all x.
The structure cont_diff_bump contains the data required to construct the function: real numbers r, R, and proofs of 0 < r < R. The function itself is available through coe_fn.
noncomputable def exp_neg_inv_glue (x : ) :
exp_neg_inv_glue is the real function given by x ↦ exp (-1/x) for x > 0 and 0 for x ≤ 0. It is a basic building block to construct smooth partitions of unity. Its main property is that it vanishes for x ≤ 0, it is positive for x > 0, and the junction between the two behaviors is flat enough to retain smoothness. The fact that this function is C^∞ is proved in exp_neg_inv_glue.smooth.
Equations
noncomputable def exp_neg_inv_glue.P_aux :
Our goal is to prove that exp_neg_inv_glue is C^∞. For this, we compute its successive derivatives for x > 0. The n-th derivative is of the form P_aux n (x) exp(-1/x) / x^(2 n), where P_aux n is computed inductively.
Equations
noncomputable def exp_neg_inv_glue.f_aux (n : ) (x : ) :
Formula for the n-th derivative of exp_neg_inv_glue, as an auxiliary function f_aux.
Equations
The 0-th auxiliary function f_aux 0 coincides with exp_neg_inv_glue, by definition.
theorem exp_neg_inv_glue.f_aux_deriv (n : ) (x : ) (hx : x 0) :
has_deriv_at (λ (x : ), / x ^ (2 * n)) (exp_neg_inv_glue.P_aux (n + 1)) * real.exp (-x⁻¹) / x ^ (2 * (n + 1))) x
For positive values, the derivative of the n-th auxiliary function f_aux n (given in this statement in unfolded form) is the n+1-th auxiliary function, since the polynomial P_aux (n+1) was chosen precisely to ensure this.
theorem exp_neg_inv_glue.f_aux_deriv_pos (n : ) (x : ) (hx : 0 < x) :
(exp_neg_inv_glue.P_aux (n + 1)) * real.exp (-x⁻¹) / x ^ (2 * (n + 1))) x
For positive values, the derivative of the n-th auxiliary function f_aux n is the n+1-th auxiliary function.
theorem exp_neg_inv_glue.f_aux_limit (n : ) :
filter.tendsto (λ (x : ), / x ^ (2 * n)) (set.Ioi 0)) (nhds 0)
To get differentiability at 0 of the auxiliary functions, we need to know that their limit is 0, to be able to apply general differentiability extension theorems. This limit is checked in this lemma.
Deduce from the limiting behavior at 0 of its derivative and general differentiability extension theorems that the auxiliary function f_aux n is differentiable at 0, with derivative 0.
At every point, the auxiliary function f_aux n has a derivative which is equal to f_aux (n+1).
The successive derivatives of the auxiliary function f_aux 0 are the functions f_aux n, by induction.
@[protected]
The function exp_neg_inv_glue is smooth.
theorem exp_neg_inv_glue.zero_of_nonpos {x : } (hx : x 0) :
The function exp_neg_inv_glue vanishes on (-∞, 0].
theorem exp_neg_inv_glue.pos_of_pos {x : } (hx : 0 < x) :
The function exp_neg_inv_glue is positive on (0, +∞).
theorem exp_neg_inv_glue.nonneg (x : ) :
The function exp_neg_inv_glue is nonnegative.
noncomputable def real.smooth_transition (x : ) :
An infinitely smooth function f : ℝ → ℝ such that f x = 0 for x ≤ 0, f x = 1 for 1 ≤ x, and 0 < f x < 1 for 0 < x < 1.
Equations
theorem real.smooth_transition.one_of_one_le {x : } (h : 1 x) :
@[protected, simp]
@[protected, simp]
theorem real.smooth_transition.pos_of_pos {x : } (h : 0 < x) :
@[protected]
@[protected]
@[protected]
structure cont_diff_bump_of_inner {E : Type u_1} (c : E) :
Type
• r :
• R :
• r_pos : 0 < self.r
• r_lt_R : self.r < self.R
f : cont_diff_bump_of_inner c, where c is a point in an inner product space, is a bundled smooth function such that
• f is equal to 1 in metric.closed_ball c f.r;
• support f = metric.ball c f.R;
• 0 ≤ f x ≤ 1 for all x.
The structure cont_diff_bump_of_inner contains the data required to construct the function: real numbers r, R, and proofs of 0 < r < R. The function itself is available through coe_fn.
Instances for cont_diff_bump_of_inner
theorem cont_diff_bump_of_inner.R_pos {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
0 < f.R
@[protected, instance]
def cont_diff_bump_of_inner.inhabited {E : Type u_1} (c : E) :
Equations
noncomputable def cont_diff_bump_of_inner.to_fun {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
E →
The function defined by f : cont_diff_bump_of_inner c. Use automatic coercion to function instead.
Equations
@[protected, instance]
noncomputable def cont_diff_bump_of_inner.has_coe_to_fun {E : Type u_1} {c : E} :
(λ (_x : , E → )
Equations
@[protected]
theorem cont_diff_bump_of_inner.def {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) (x : E) :
f x = ((f.R - c) / (f.R - f.r)).smooth_transition
@[protected]
theorem cont_diff_bump_of_inner.sub {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) (x : E) :
f (c - x) = f (c + x)
@[protected]
theorem cont_diff_bump_of_inner.neg {E : Type u_1} (f : cont_diff_bump_of_inner 0) (x : E) :
f (-x) = f x
theorem cont_diff_bump_of_inner.one_of_mem_closed_ball {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} (hx : x ) :
f x = 1
theorem cont_diff_bump_of_inner.nonneg {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} :
0 f x
theorem cont_diff_bump_of_inner.nonneg' {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) (x : E) :
0 f x
A version of cont_diff_bump_of_inner.nonneg with x explicit
theorem cont_diff_bump_of_inner.le_one {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} :
f x 1
theorem cont_diff_bump_of_inner.pos_of_mem_ball {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} (hx : x f.R) :
0 < f x
theorem cont_diff_bump_of_inner.lt_one_of_lt_dist {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} (h : f.r < ) :
f x < 1
theorem cont_diff_bump_of_inner.zero_of_le_dist {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} (hx : f.R ) :
f x = 0
theorem cont_diff_bump_of_inner.support_eq {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
theorem cont_diff_bump_of_inner.tsupport_eq {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
@[protected]
theorem cont_diff_bump_of_inner.has_compact_support {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
theorem cont_diff_bump_of_inner.eventually_eq_one_of_mem_ball {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} (h : x f.r) :
@[protected]
theorem cont_diff_at.cont_diff_bump {E : Type u_1} {X : Type u_2} [ X] {n : ℕ∞} {c g : X → E} {f : Π (x : X), } {x : X} (hc : c x) (hr : (λ (x : X), (f x).r) x) (hR : (λ (x : X), (f x).R) x) (hg : g x) :
(λ (x : X), (f x) (g x)) x
cont_diff_bump is 𝒞ⁿ in all its arguments.
theorem cont_diff.cont_diff_bump {E : Type u_1} {X : Type u_2} [ X] {n : ℕ∞} {c g : X → E} {f : Π (x : X), } (hc : c) (hr : (λ (x : X), (f x).r)) (hR : (λ (x : X), (f x).R)) (hg : g) :
(λ (x : X), (f x) (g x))
@[protected]
theorem cont_diff_bump_of_inner.cont_diff {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {n : ℕ∞} :
f
@[protected]
theorem cont_diff_bump_of_inner.cont_diff_at {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} {n : ℕ∞} :
f x
@[protected]
theorem cont_diff_bump_of_inner.cont_diff_within_at {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {x : E} {n : ℕ∞} {s : set E} :
s x
@[protected]
theorem cont_diff_bump_of_inner.continuous {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) :
@[protected]
noncomputable def cont_diff_bump_of_inner.normed {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) (μ : measure_theory.measure E) :
E →
A bump function normed so that ∫ x, f.normed μ x ∂μ = 1.
Equations
theorem cont_diff_bump_of_inner.normed_def {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} (x : E) :
f.normed μ x = f x / (x : E), f x μ
theorem cont_diff_bump_of_inner.nonneg_normed {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} (x : E) :
0 f.normed μ x
theorem cont_diff_bump_of_inner.cont_diff_normed {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} {n : ℕ∞} :
(f.normed μ)
theorem cont_diff_bump_of_inner.normed_sub {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} (x : E) :
f.normed μ (c - x) = f.normed μ (c + x)
theorem cont_diff_bump_of_inner.normed_neg {E : Type u_1} {μ : measure_theory.measure E} (f : cont_diff_bump_of_inner 0) (x : E) :
f.normed μ (-x) = f.normed μ x
@[protected]
theorem cont_diff_bump_of_inner.integrable {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} [borel_space E] :
@[protected]
theorem cont_diff_bump_of_inner.integral_pos {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} [borel_space E] :
0 < (x : E), f x μ
theorem cont_diff_bump_of_inner.integral_normed {E : Type u_1} {c : E} (f : cont_diff_bump_of_inner c) {μ : measure_theory.measure E} [borel_space E] :
(x : E), f.normed μ x μ = 1
theorem cont_diff_bump_of_inner.integral_normed_smul {E : Type u_1} {X : Type u_2} [ X] {c : E} (f : cont_diff_bump_of_inner c) (μ : measure_theory.measure E) [borel_space E] (z : X) :
(x : E), f.normed μ x z μ = z
structure cont_diff_bump {E : Type u_1} [ E] (c : E) :
Type
• to_cont_diff_bump_of_inner :
f : cont_diff_bump c, where c is a point in a finite dimensional real vector space, is a bundled smooth function such that
• f is equal to 1 in euclidean.closed_ball c f.r;
• support f = euclidean.ball c f.R;
• 0 ≤ f x ≤ 1 for all x.
The structure cont_diff_bump contains the data required to construct the function: real numbers r, R, and proofs of 0 < r < R. The function itself is available through coe_fn.
Instances for cont_diff_bump
noncomputable def cont_diff_bump.to_fun {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
E →
The function defined by f : cont_diff_bump c. Use automatic coercion to function instead.
Equations
@[protected, instance]
noncomputable def cont_diff_bump.has_coe_to_fun {E : Type u_1} [ E] {c : E} :
(λ (_x : , E → )
Equations
@[protected, instance]
noncomputable def cont_diff_bump.inhabited {E : Type u_1} [ E] (c : E) :
Equations
theorem cont_diff_bump.R_pos {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
theorem cont_diff_bump.coe_eq_comp {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
theorem cont_diff_bump.one_of_mem_closed_ball {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) (hx : x ) :
f x = 1
theorem cont_diff_bump.nonneg {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) :
0 f x
theorem cont_diff_bump.le_one {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) :
f x 1
theorem cont_diff_bump.pos_of_mem_ball {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) (hx : x ) :
0 < f x
theorem cont_diff_bump.lt_one_of_lt_dist {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) (h : f.to_cont_diff_bump_of_inner.r < ) :
f x < 1
theorem cont_diff_bump.zero_of_le_dist {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) (hx : f.to_cont_diff_bump_of_inner.R ) :
f x = 0
theorem cont_diff_bump.support_eq {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
theorem cont_diff_bump.tsupport_eq {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
@[protected]
theorem cont_diff_bump.has_compact_support {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
theorem cont_diff_bump.eventually_eq_one_of_mem_ball {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) (h : x ) :
theorem cont_diff_bump.eventually_eq_one {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) :
@[protected]
theorem cont_diff_bump.cont_diff {E : Type u_1} [ E] {c : E} (f : cont_diff_bump c) {n : ℕ∞} :
f
@[protected]
theorem cont_diff_bump.cont_diff_at {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) {n : ℕ∞} :
f x
@[protected]
theorem cont_diff_bump.cont_diff_within_at {E : Type u_1} [ E] {c x : E} (f : cont_diff_bump c) {s : set E} {n : ℕ∞} :
s x
theorem cont_diff_bump.exists_tsupport_subset {E : Type u_1} [ E] {c : E} {s : set E} (hs : s nhds c) :
∃ (f : , s
theorem cont_diff_bump.exists_closure_subset {E : Type u_1} [ E] {c : E} {R : } (hR : 0 < R) {s : set E} (hs : is_closed s) (hsR : s ) :
∃ (f : ,
theorem exists_cont_diff_bump_function_of_mem_nhds {E : Type u_1} [ E] {x : E} {s : set E} (hs : s nhds x) :
∃ (f : E → ), f =ᶠ[nhds x] 1 (∀ (y : E), f y 1) f
If E is a finite dimensional normed space over ℝ, then for any point x : E and its neighborhood s there exists an infinitely smooth function with the following properties:
• f y = 1 in a neighborhood of x;
• f y = 0 outside of s;
• moreover, tsupport f ⊆ s and f has compact support;
• f y ∈ [0, 1] for all y.
This lemma is a simple wrapper around lemmas about bundled smooth bump functions, see cont_diff_bump`. | 4,778 | 13,833 | {"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} | 2.859375 | 3 | CC-MAIN-2023-40 | latest | en | 0.836632 |
https://www.diagrampedia.com/circuit-where-electricity-flows-freely-brainly/ | 1,713,027,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816820.63/warc/CC-MAIN-20240413144933-20240413174933-00009.warc.gz | 708,413,536 | 12,536 | # Circuit Where Electricity Flows Freely Brainly
By | January 7, 2023
Electricity is one of the most essential components of our lives, even though we may not always realize it. It powers the appliances, lights, and computers that we use every day. But understanding how electricity works can be difficult for many people. That’s why circuits provide an important way to visualize and understand electricity.
Circuits are pathways in which electricity can flow freely, and they are created by connecting electrical components with wires. The most basic circuit includes a power source, such as a battery or generator, connected to one or more components. When the power source is turned on, electricity flows through the circuit, providing energy or information to the component.
Circuits are used in a wide range of applications - from powering everyday items like kitchen appliances and cell phones to controlling large-scale projects like industrial robots and air traffic control systems. By providing a clear and organized way to visualize electricity, circuits make it easier to understand how electricity works and create powerful and reliable devices.
It's also possible to create a "circuit" where electricity flows freely without any components at all. This type of circuit is called a "closed loop," because the electricity is not connected to any device or component. This is useful for testing purposes and to study the flow of electricity in its purest form.
Circuits have revolutionized the way we use and understand electricity. By providing an organized way to visualize electricity, they make it easier to understand how electricity works and to create powerful and reliable devices. And by allowing electricity to flow freely, closed loop circuits also give us a deeper understanding of electricity itself.
Circuits And Electricity
Circuit Where Electricity Flows Freely Brainly Ph
Activity 3 Electric Circuits And Cur
Resistance Basic Concepts Of Electricity Electronics Textbook
Difference Between Electric Energy And Power With Calculator
Pulsed Electric Fields For Food Processing Technology Intechopen
Brainly Homework Math Solver Apps On Google Play
Why Can T Cur Flow Through An Open Circuit Quora
Physics Tutorial Requirements Of A Circuit
Alternating Cur Vs Direct Lesson For Kids Transcript Study Com
What Is Electricity Sparkfun Learn
Electric Circuits To Calculate The Size Of A Cur From Charge Flow And Time Taken Thursday August 06 Ppt
Wind Energy Systems And Applications Request Pdf
The Simple Circuit
Electric Circuits
2017stdiogamejam Words Txt At Master Stdiogamejam Github
What Is Electric Cur Brainly Ph
Circuits And Electricity
Electric Circuits
Help True Or False A Closed Switch Mans The Electricity Can Flow Freely Through Circuit Brainly Com | 540 | 2,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.930093 |
https://rashidfaridi.com/2008/06/27/topology-and-layers-in-gis/?replytocom=2109 | 1,582,901,080,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00517.warc.gz | 501,033,970 | 29,770 | ## Topology and Layers in GIS
TOPOLOGY & LAYERS (THEMES)
Topology refers to the spatial relationships between geographic features. It describes the relationships between connecting or
adjacent coverage features. Topological relationships are built from simple elements into complex elements: points
(simplest elements), arcs (sets of connected points), areas (sets of connected arcs), and routes (sets of sections, which are
arcs or portions of arcs).
Topology is useful in GIS because many spatial modeling operations don’t require coordinates, only topological information.
For example, to find an optimal path between two points requires a list of the arcs that connect to each other and the cost to
traverse each arc in each direction. Coordinates are only needed for drawing the path after it is calculated.
Components of Topology:
Topology has three basic components:
I. Connectivity (Arc – Node Topology):
o Points along an arc that define its shape are called Vertices.
o Endpoints of the arc are called Nodes.
o Arcs join only at the Nodes.
II. Area Definition / Containment (Polygon – Arc Topology):
o An enclosed polygon has a measurable area.
o Lists of arcs define boundaries and closed areas are maintained.
o Polygons are represented as a series of (x , y) coordinates that connect to define an area.
III. Contiguity:
o Every arc has a direction
o A GIS maintains a list of Polygons on the left and right side of each arc.
o The computer then uses this information to determine which features are next to one another.
Explanation of Topology:
TOPOLOGY & LAYERS (THEMES)
Topology refers to the spatial relationships between geographic features. It describes the relationships between connecting or
adjacent coverage features. Topological relationships are built from simple elements into complex elements: points
(simplest elements), arcs (sets of connected points), areas (sets of connected arcs), and routes (sets of sections, which are
arcs or portions of arcs).
Topology is useful in GIS because many spatial modeling operations don’t require coordinates, only topological information.
For example, to find an optimal path between two points requires a list of the arcs that connect to each other and the cost to
traverse each arc in each direction. Coordinates are only needed for drawing the path after it is calculated.
Components of Topology:
Topology has three basic components:
I. Connectivity (Arc – Node Topology):
o Points along an arc that define its shape are called Vertices.
o Endpoints of the arc are called Nodes.
o Arcs join only at the Nodes.
II. Area Definition / Containment (Polygon – Arc Topology):
o An enclosed polygon has a measurable area.
o Lists of arcs define boundaries and closed areas are maintained.
o Polygons are represented as a series of (x , y) coordinates that connect to define an area.
III. Contiguity:
o Every arc has a direction
o A GIS maintains a list of Polygons on the left and right side of each arc.
o The computer then uses this information to determine which features are next to one another.
Explanation of Topology:
yers (Themes):
Maps produced traditionally or by automation involved the separation of data into layers, that each contained different types of
features: rivers, roads, etc. These are then combined to form a map where layers have usually been printed in different colors.
To present all the data collected for a given area might require the production of several maps, as they could not all be printed together. Equally in GIS, data for an area are divided into layers or themes, divided by type but here for the dual purposes of display and analysis.
Themes
The database can be divided into as many layers as is necessary, where each layer contains one characteristic such as soils, land use, drainage, etc. The layers ‘overlay’ each other perfectly as a result of Georeferencing and enabling analysis between layers.
source
## About Rashid Faridi
I am Rashid Aziz Faridi ,Writer, Teacher and a Voracious Reader.
This entry was posted in GIS, Remote Sensing 101. Bookmark the permalink.
### 3 Responses to Topology and Layers in GIS
1. Tarun says:
Sir ,
for 3d modeling in GIS and what kind of topological structure we need and how to handle 3d model in GIS
Like
2. Hango says:
good explanations, but support with aid of accurate diagrams.
Like
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 941 | 4,463 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.920693 |
https://www.physicsforums.com/threads/strange-physics-question-help.732027/ | 1,516,104,008,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886416.17/warc/CC-MAIN-20180116105522-20180116125522-00035.warc.gz | 974,209,727 | 19,834 | # Strange physics question. Help
1. Jan 9, 2014
### vongola13th
1. The problem statement, all variables and given/known data
Two boxes each mass 10 kg are raised 2.0m to a shelf. The first one is lifted and second one
is pushed up a smooth ramp(frictionless). If the applied force on the second box is 50N, calculator the angle between the ramp
and the ground
2. Relevant equations
n/a
3. The attempt at a solution
Got work of the first box
(10)(9.8)(2)=W
=196J
Don't know what to do with it, tried 198=50(2)sintheta, but that won't work. Please show full solution, lost.
2. Jan 9, 2014
### rock.freak667
You'd need to account for the component of the 10 kg weight opposing the 50 N force.
Remember that sine = opposite/hypotenuse
so the distance the mass is sliding up the ramp is the hypotenuse.
3. Jan 9, 2014
### Staff: Mentor
[strike]You can determine the force to raise the first box via F=mg right?
The same amount of vertical force is needed to raise the second box on the ramp.
So try drawing a force diagram of a force pushing/pulling a block up a ramp.[/strike]
EDIT: disregard this post... see followon post on ramp mechanical advantage
Last edited: Jan 9, 2014
4. Jan 9, 2014
### vongola13th
so the force requried to life the first box is 98N,
so 98=sin(90)
=98
what about the other one, it wouldn't be 50=sintheta right?
5. Jan 9, 2014
### Staff: Mentor
Last edited: Jan 9, 2014
6. Jan 9, 2014
### vongola13th
Yea I looked at it and do I'm still lost, could you please tell me how to find it
7. Jan 9, 2014
### Staff: Mentor
We're here to help you when you get stuck not to do your homework. You need to show more work before we can help. I think what I gave would be sufficient to solve the problem.
8. Jan 9, 2014
### vongola13th
okay sorry, I figured out 196J=fxd
196=50*d
3.92=d
I now have the opposite and the hyp. I did sin^-1(2/3.92)=30degrees
I double checked it with 50*3.92*cos30degrees= 168J
This is not the 196 joules needed, where did I go wrong?
9. Jan 9, 2014
### Staff: Mentor
I think your check is wrong. How did you come up with it?
10. Jan 9, 2014
### vongola13th
Since I know the formula for work W=Fxdcostheta, I wanted to make sure the that about of work would be the same as for the case of the ramp and the case of the box being lifted up 90degrees. I figured that the side length must be 3.92 and used that as they hyp. I used sin to figure out the angle then checked it with that angle, but the work was 168 instead of the needed 196. I don't know what I did wrong. could you point it out?
11. Jan 10, 2014
### Staff: Mentor
But it isn't.
The "formula" is always W=Fd, where F is the force and d is the distance moved in the direction of that force. The formula never varies from this.
You have a force acting along the slope here. What is its magnitude, and through what distance does the point of application of that force move?
Your only "mistake" is to have become too complacent in your use of that "formula".
Stay alert, or you'll be tripped up. Again, and when you least expect it.
12. Jan 10, 2014
### lendav_rott
Sketch out the information you know with pen and paper. That way any assignment will become much less difficult to tackle.
Calculate the hypotenuse, which is the distance the 2nd box is being pushed up to reach the shelf.
On your sketch, mark every force you know of, that is acting on the 2nd box. There is gravity, the force it is being pushed by and the force the surface of the ramp is exerting on the box and luckily you won't have to account for friction.
If you think about it, the surface of the ramp does work as well all the way the box is being pushed. A = Fs or W=Fd if you prefer. What is the force the ramp is exerting on the box?
Also, do you understand why you are equating the potential energy of the 1st box already on the shelf to the work done by the force pushing the 2nd box to reach the shelf?
13. Jan 11, 2014
### Malverin
The work that you have to do is
$\textit{W}$= $\textit{m}$$\cdot$$\textit{g}$$\cdot$$\textit{h}$ = $\textit{50}$$\cdot\frac{\textit{h}}{\textit{sin Θ}}$ | 1,187 | 4,113 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 3.828125 | 4 | CC-MAIN-2018-05 | longest | en | 0.903745 |
https://nebusresearch.wordpress.com/2018/01/28/reading-the-comics-january-22-2018-breaking-workflow-edition/ | 1,686,260,071,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00291.warc.gz | 452,150,278 | 39,450 | # Reading the Comics, January 22, 2018: Breaking Workflow Edition
So I was travelling last week, and this threw nearly all my plans out of whack. We stayed at one of those hotels that’s good enough that its free Internet is garbage and they charge you by day for decent Internet. So naturally Comic Strip Master Command sent a flood of posts. I’m trying to keep up and we’ll see if I wrap up this past week in under three essays. And I am not helped, by the way, by GoComics.com rejiggering something on their server so that My Comics Page won’t load, and breaking their “Contact Us” page so that that won’t submit error reports. If someone around there can break in and turn one of their servers off and on again, I’d appreciate the help.
Hy Eisman’s Katzenjammer Kids for the 21st of January is a curiously-timed Tax Day joke. (Well, the Katzenjammer Kids lapsed into reruns a dozen years ago and there’s probably not much effort being put into selecting seasonally appropriate ones.) But it is about one of the oldest and still most important uses of mathematics, and one that never gets respect.
Morrie Turner’s Wee Pals rerun for the 21st gets Oliver the reputation for being a little computer because he’s good at arithmetic. There is something that amazes in a person who’s able to calculate like this without writing anything down or using a device to help.
Steve Kelley and Jeff Parker’s Dustin for the 22nd seems to be starting off with a story problem. It might be a logic problem rather than arithmetic. It’s hard to say from what’s given.
Mark Anderson’s Andertoons for the 22nd is the Mark Anderson’s Andertoons for the week. Well, for Monday, as I write this. It’s got your classic blackboard full of equations for the people in over their head. The equations look to me like gibberish. There’s a couple diagrams of aromatic organic compounds, which suggests some quantum-mechanics chemistry problem, if you want to suppose this could be narrowed down.
Greg Evans’s Luann Againn for the 22nd has Luann despair about ever understanding algebra without starting over from scratch and putting in excessively many hours of work. Sometimes it feels like that. My experience when lost in a subject has been that going back to the start often helps. It can be easier to see why a term or a concept or a process is introduced when you’ve seen it used some, and often getting one idea straight will cause others to fall into place. When that doesn’t work, trying a different book on the same topic — even one as well-worn as high school algebra — sometimes helps. Just a different writer, or a different perspective on what’s key, can be what’s needed. And sometimes it just does take time working at it all.
Richard Thompson’s Richard’s Poor Almanac rerun for the 22nd includes as part of a kit of William Shakespeare paper dolls the Typing Monkey. It’s that lovely, whimsical figure that might, in time, produce any written work you could imagine. I think I’d retired monkeys-at-typewriters as a thing to talk about, but I’m easily swayed by Thompson’s art and comic stylings so here it is.
Darrin Bell and Theron Heir’s Rudy Park for the 18th throws around a lot of percentages. It’s circling around the sabermetric-style idea that everything can be quantified, and measured, and that its changes can be tracked. In this case it’s comments on Star Trek: Discovery, but it could be anything. I’m inclined to believe that yeah, there’s an astounding variety of things that can be quantified and measured and tracked. But it’s also easy, especially when you haven’t got a good track record of knowing what is important to measure, to start tracking what amounts to random noise. (See any of my monthly statistics reviews, when I go looking into things like views-per-visitor-per-post-made or some other dubiously meaningful quantity.) So I’m inclined to side with Randy and his doubts that the Math Gods sanction this much data-mining.
## Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
## 8 thoughts on “Reading the Comics, January 22, 2018: Breaking Workflow Edition”
1. The teacher could completely confuse the kids beyond the concept of paper letters by making the question “Rochester sent letters to his friends Don, Mary,Mel and Frank”
Like
1. You are right, but it would also give Jim Scancarelli the chance to write this strip.
Like
2. What might the letter problem have been? I don’t know any old saws that start that way.
Like
1. Perhaps based on the postage needed to mail them all, (excepting of course the one he sent postage due to his friend Jack.)
Like
1. There’s certainly a viable problem in that, although given that the only bit of information is how long it takes one letter to be delivered I’m not sure it was going that way. But figuring a way to get all these letters delivered within a particular time for a particular budget is believable, if a little optimization-programming for elementary school students.
Like
2. I don’t know, but I’m going to assume it’s something like this person’s letter took a day longer than that person’s, and other person’s letter took twice as long as that person’s, and if you add up how many days all the letters took to get there it was this number. Something like that.
Like
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,211 | 5,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.952503 |
http://m.thermalfluidscentral.org/encyclopedia/index.php?title=Computational_methodologies_for_forced_convection&diff=cur&oldid=3575 | 1,596,552,058,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.94/warc/CC-MAIN-20200804131928-20200804161928-00265.warc.gz | 67,740,875 | 9,133 | # Computational methodologies for forced convection
(Difference between revisions)
Revision as of 04:56, 7 April 2010 (view source)← Older edit Current revision as of 01:16, 27 July 2010 (view source) (4 intermediate revisions not shown) Line 1: Line 1: - The convection problems that can be solved analytically – some of which were discussed in the preceding sections – are limited to the cases for which the boundary layer theory is valid. Even for cases where the boundary layer theory can be used to simplify the governing equations, analytical solutions can be obtained only if the similarity solution exists or if integral approximate solutions can be obtained. The analytical solution can be very complicated for cases with variable wall temperature or heat flux. When the boundary layer theory cannot be applied, or a simple analytical solution based on boundary layer theory cannot be obtained, a numerical solution becomes the desirable approach. Convection problems are complicated by the presence of advection terms in the governing equations. For incompressible flow, the problem is further complicated by the fact that there is not a governing equation for pressure. This section will cover (1) the numerical solution of the convection-diffusion equation with a known flow field, and (2) the algorithm to determine the flow field. + {{Comp Method for Forced Convection Category}} + The convection problems that can be solved analytically – some of which were discussed in the preceding sections – are limited to the cases for which the boundary layer theory is valid. Even for cases where the boundary layer theory can be used to simplify the governing equations, analytical solutions can be obtained only if the similarity solution exists or if integral approximate solutions can be obtained. The analytical solution can be very complicated for cases with variable wall temperature or heat flux. When the boundary layer theory cannot be applied, or a simple analytical solution based on boundary layer theory cannot be obtained, a numerical solution becomes the desirable approach. Convection problems are complicated by the presence of advection terms in the governing equations. For incompressible flow, the problem is further complicated by the fact that there is not a governing equation for pressure. This topic will cover (1) the numerical solution of the convection-diffusion equation with a known flow field, and (2) the algorithm to determine the flow field. The governing equation for a convection problem in the Cartesian coordinate system can be expressed into the following generalized form: The governing equation for a convection problem in the Cartesian coordinate system can be expressed into the following generalized form: Line 13: Line 14: |} |} - where $\varphi$ is a general variable that can represent the directional components of velocity (u, v, or w), temperature, or mass concentration. The general diffusivity $\Gamma$ can be viscosity, thermal conductivity, or mass diffusivity. S is the volumetric source term. Compared with the governing equation for heat conduction, the convection terms seem to be the only new terms introduced into eq. (4.200). As will become evident later, the contribution of the convection term on the overall heat transfer depends on the relative scale of convection over diffusion. Therefore, these two effects are always handled as one unit in the derivation of the discretization scheme (Patankar, 1980; Patankar, 1991). + where $\varphi$ is a general variable that can represent the directional components of velocity (''u'', ''v'', or ''w''), temperature, or mass concentration. The general diffusivity $\Gamma$ can be viscosity, thermal conductivity, or mass diffusivity. ''S'' is the volumetric source term. Compared with the governing equation for heat conduction, the convection terms seem to be the only new terms introduced into eq. (1). As will become evident later, the contribution of the convection term on the overall heat transfer depends on the relative scale of convection over diffusion. Therefore, these two effects are always handled as one unit in the derivation of the discretization scheme Patankar, S.V., 1980, Numerical Heat Transfer and Fluid Flow, Hemisphere, Washington, DC. Patankar, S.V., 1991, Computation of Conduction and Duct Flow Heat Transfer, Innovative Research.Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.. - While analytical solutions of convection problems presented in the preceding sections based on boundary layer theory are few, the numerical solutions of convection based on boundary layer theory have been abundant (Tao, 2001; Minkowycz et al., 2006). With significant advancement of computational capability, the numerical solution of convection problems can now be performed by solving the full Navier-Stokes equation. The algorithms that will be discussed in this section are therefore based on the solution of the full Navier-Stokes equation and the energy equation, not on the boundary layer equations. + + While analytical solutions of convection problems presented in the preceding sections based on boundary layer theory are few, the numerical solutions of convection based on boundary layer theory have been abundant Tao 2001, W.Q., Numerical Heat Transfer, 2nd Ed., Xi’an Jiaotong University Press, Xi’an, China (in Chinese). Minkowycz, M.J., Sparrow, E.M., and Murthy, J.Y., eds., 2006, Handbook of Numerical Heat Transfer, 2nd ed. John Wiley & Sons, Hoboken, NJ.. With significant advancement of computational capability, the numerical solution of convection problems can now be performed by solving the full Navier-Stokes equation. The algorithms that will be discussed in this section are therefore based on the solution of the full Navier-Stokes equation and the energy equation, not on the boundary layer equations. [[One-Dimensional Steady-State Convection and Diffusion]]
Line 21: Line 23: [[Numerical Simulation of Interfaces and Free Surfaces]]
[[Numerical Simulation of Interfaces and Free Surfaces]]
[[Application of Computational Methods]] [[Application of Computational Methods]] + + ==References== + {{Reflist}}
## Current revision as of 01:16, 27 July 2010
The convection problems that can be solved analytically – some of which were discussed in the preceding sections – are limited to the cases for which the boundary layer theory is valid. Even for cases where the boundary layer theory can be used to simplify the governing equations, analytical solutions can be obtained only if the similarity solution exists or if integral approximate solutions can be obtained. The analytical solution can be very complicated for cases with variable wall temperature or heat flux. When the boundary layer theory cannot be applied, or a simple analytical solution based on boundary layer theory cannot be obtained, a numerical solution becomes the desirable approach. Convection problems are complicated by the presence of advection terms in the governing equations. For incompressible flow, the problem is further complicated by the fact that there is not a governing equation for pressure. This topic will cover (1) the numerical solution of the convection-diffusion equation with a known flow field, and (2) the algorithm to determine the flow field.
The governing equation for a convection problem in the Cartesian coordinate system can be expressed into the following generalized form:
\begin{align} & \frac{\partial (\rho \varphi )}{\partial t}+\frac{\partial (\rho u\varphi )}{\partial x}+\frac{\partial (\rho v\varphi )}{\partial y}+\frac{\partial (\rho w\varphi )}{\partial z} \\ & =\frac{\partial }{\partial x}\left( \Gamma \frac{\partial \varphi }{\partial x} \right)+\frac{\partial }{\partial y}\left( \Gamma \frac{\partial \varphi }{\partial y} \right)+\frac{\partial }{\partial z}\left( \Gamma \frac{\partial \varphi }{\partial z} \right)+S \\ \end{align} (1)
where $\varphi$ is a general variable that can represent the directional components of velocity (u, v, or w), temperature, or mass concentration. The general diffusivity Γ can be viscosity, thermal conductivity, or mass diffusivity. S is the volumetric source term. Compared with the governing equation for heat conduction, the convection terms seem to be the only new terms introduced into eq. (1). As will become evident later, the contribution of the convection term on the overall heat transfer depends on the relative scale of convection over diffusion. Therefore, these two effects are always handled as one unit in the derivation of the discretization scheme [1][2][3].
While analytical solutions of convection problems presented in the preceding sections based on boundary layer theory are few, the numerical solutions of convection based on boundary layer theory have been abundant [4][5]. With significant advancement of computational capability, the numerical solution of convection problems can now be performed by solving the full Navier-Stokes equation. The algorithms that will be discussed in this section are therefore based on the solution of the full Navier-Stokes equation and the energy equation, not on the boundary layer equations.
## References
1. Patankar, S.V., 1980, Numerical Heat Transfer and Fluid Flow, Hemisphere, Washington, DC.
2. Patankar, S.V., 1991, Computation of Conduction and Duct Flow Heat Transfer, Innovative Research.
3. Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.
4. Tao 2001, W.Q., Numerical Heat Transfer, 2nd Ed., Xi’an Jiaotong University Press, Xi’an, China (in Chinese).
5. Minkowycz, M.J., Sparrow, E.M., and Murthy, J.Y., eds., 2006, Handbook of Numerical Heat Transfer, 2nd ed. John Wiley & Sons, Hoboken, NJ. | 2,114 | 9,831 | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.916261 |
https://www.sqlservercentral.com/forums/topic/urgent-how-to-extract-number-from-a-string/page/3/ | 1,618,980,614,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039508673.81/warc/CC-MAIN-20210421035139-20210421065139-00330.warc.gz | 1,091,025,637 | 23,154 | Urgent! - How to extract number from a string
• dva2007 - Tuesday, July 10, 2018 6:55 AM
After reading the article above I used dbo.DigitsOnlyEE which is very quick.
Exactly. That's why I wanted to see what you were using. Thank you for taking the time on the feedback here.
--Jeff Moden
RBAR is pronounced "ree-bar" and is a "Modenism" for Row-By-Agonizing-Row.
First step towards the paradigm shift of writing Set Based code:
________Stop thinking about what you want to do to a ROW... think, instead, of what you want to do to a COLUMN.
"Change is inevitable... change for the better is not".
"Dear Lord... I'm a DBA so please give me patience because, if you give me strength, I'm going to need bail money too!"
How to post code problems
How to Post Performance Problems
Create a Tally Function (fnTally)
• No problem. Thank you to all who took time out to respond.
DECLARE @NUMBERS AS VARCHAR(100)='',@ALPHA AS VARCHAR(100)='',@SPECIAL AS VARCHAR(100)=''
;WITH TREE AS
(
SELECT SUBSTRING(@EMAIL,1,1) AS CHR, CAST(1 AS INT) AS LVL
UNION ALL
SELECT SUBSTRING(@EMAIL,LVL+1,1) AS CHR, CAST(LVL+1 AS INT) AS LVL FROM TREE WHERE LEN(@EMAIL)>LVL
)
SELECT @NUMBERS=COALESCE(@NUMBERS + '', '')+ IIF(CHR LIKE '%[0-9]%', CHR, ''),@ALPHA=COALESCE(@ALPHA + '', '')+ IIF(CHR LIKE '%[a-zA-Z]%', CHR, ''),@SPECIAL=COALESCE(@SPECIAL + '', '')+ IIF(CHR NOT LIKE '%[a-zA-Z0-9]%', CHR, '') FROM TREE
SELECT @NUMBERS AS Num,@ALPHA AS Alph,@SPECIAL AS Spl
Viewing 3 posts - 31 through 33 (of 33 total) | 458 | 1,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-17 | longest | en | 0.778164 |
https://bbs.bccn.net/redirect.php?tid=506851&goto=lastpost | 1,632,333,546,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057371.69/warc/CC-MAIN-20210922163121-20210922193121-00305.warc.gz | 182,249,937 | 9,622 | | 网站首页 | 业界新闻 | 小组 | 交易 | 人才 | 下载频道 | 博客 | 代码贴 | 在线编程 | 编程论坛
```//假设数组a有1152个数，我想按照顺序每100个数算一个和，求一个平均值，该如何实现？
//假设数组a有1152个数，我想按照顺序每100个数算一个和，求一个平均值，该如何实现？
#include<stdio.h>
#include<time.h>
#include <stdlib.h>
int main()
{
int a[1152], i, k, c, s[13] = { 0 };
srand((unsigned int)time(NULL));
for (i = 0, k = 0, c = 0; i < 1152; i++) {
a[i] = rand() % 150 + 1;
s[k] += a[i];
c++;
if (c == 100) {
k++;
c = 0;
}
}
for (i = 0; i < k; i++){
printf(" %c %d",i==0?'(':'+', s[i]);
s[12] += s[i];
}
printf(" ) / %d = %d \n",k, s[12] / k);
return 0;
}```
//online parser: https://www.bccn.net/run/
```#include <stdio.h>
#include <string.h>
int main(int argc, char *argv[])
{
char c[12][100];
int i, j, sum;
memcpy(c, main, 1152);
for(i = 0; i < 12; i++) {
for(j = 0, sum = 0; j < sizeof(c[i]); j++) sum += c[i][j];
printf("%02d: sum = %d, avr = %d\n", i + 1, sum, sum / 100);
}
return 0;
}```
[此贴子已经被作者于2021-9-15 17:34编辑过]
```
#include <stdio.h>
#define N 1152
int main()
{
int sum = 0;
for (int i = 0, j; i < N; i++) {
sum += i;
j = (i + 1) % 100;
if (j == 0 || i == N - 1) {
if (j == 0) j = 100;
printf("%d / %d = %d\n\n", i, j, sum / j);
sum = 0;
}
else printf("%d + ", i);
}
return 0;
}
```
```#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
unsigned int num_all,num_one,num_current,group,i,j;
float *a,sum;
printf("请输入总个数及每组个数:");
scanf("%u %u",&num_all,&num_one);
a=malloc(num_all*sizeof(float));
srand((unsigned int)time(NULL)); //模拟数据
for(i=0; i<num_all; i++)
*(a+i)=rand();
group=num_all/num_one;
if(num_all%num_one!=0)
group+=1;
for(j=0; j<group; j++)
{
sum=0;
if(j+1==group)
{
if(num_all%num_one!=0)
num_current=num_all%num_one;
}
else
{
num_current=num_one;
}
for(i=0; i<num_current; i++)
{
sum+=*(a+j*num_one+i);
}
printf("第%u组和为%.2f,平均值为%.2f\n",j,sum,sum/num_current);
}
free(a);
return 0;
}
```
• 5
• 1/1页
• 1 | 810 | 1,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-39 | latest | en | 0.152412 |
http://mathhelpforum.com/algebra/219325-adding-subtracting-negative-numbers.html | 1,527,028,280,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864968.11/warc/CC-MAIN-20180522205620-20180522225620-00081.warc.gz | 188,933,496 | 9,686 | What I'm wondering is if when you make the numbers in the problem their absolute values, why does changing the sign work as well? Is it a property of absolute values that when both numbers become positive, you always change the sign? Or do you NOT always change the sign? Does making numbers in the problem their absolute value ALWAYS mean the sign in between them is - ? If so then this next problem doesn't make sense that you would ALWAYS use the sign of the bigger number as well, because the answer is negative. Does that mean that if the sign is already negative when you make them absolute values that you ALWAYS use the sign of the smaller number? I can do these problems in my head, but i want to be able to understand exactly why. Sorry I know i sound goofy asking this. Thank you guys SO much for helping with this.
ex. -4 - 7 =
is it |4| - |7| = -3 (i know this isn't right)
or |4| + |7| = 11
This isn't right either because its positive when it should be -11 . In these kinds of problems (when you change from subtracting to adding by changing the sign along with absolute values) do you always use the sign of the smaller number instead of the bigger one? So if the sign starts out as a + you use the sign of the larger number, but if the sign starts out as a - you use the sign of the smaller number? If so, does this rule ALWAYS apply?
2. ## Re: Adding/Subtracting Negative Numbers
your both the statements are correct.
in second case |4|-|7| = -3 because you are subtracting the absolute values.
In second case you are adding the absolute values. So there should be no confusion. -4-7=-11
Just remember that the absolute value or modulus returns positive value.
That is |-5| = 5; |-x| = x ; |a| = a etc.
Thus |-a| + |b| = a+b; |-a| - |b| = a-b; |-a| - |-b| = a - b etc | 444 | 1,787 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.929463 |
https://resources.quizalize.com/view/quiz/2nd-cycle-assessment1-mathematics-10-581fe749-7fc6-439b-a460-07419553e26c | 1,702,323,506,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00704.warc.gz | 546,403,053 | 17,718 | 2nd Cycle Assessment#1 Mathematics 10
Quiz by Jaylord Arguelles
Mathematics
Feel free to use or edit a copy
includes Teacher and Student dashboards
### Measure skillsfrom any curriculum
Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill.
• edit the questions
• save a copy for later
• start a class game
• automatically assign follow-up activities based on students’ scores
• assign as homework
• share a link with colleagues
• print as a bubble sheet
### Our brand new solo games combine with your quiz, on the same screen
Correct quiz answers unlock more play!
20 questions
• Q1
The term of an arithmetic sequence has a common ___________.
ratio
factor
term
difference
30s
• Q2
What is the formula to find the nth term of an arithmetic sequence?
30s
• Q3
It is a sequence in which each term is obtained by adding the preceding term by a common difference.
scrambled://ARITHMETIC
30s
• Q4
The sequence 4, 7, 10, 13, . . . is an arithmetic sequence
boolean://True
30s
• Q5
What is the common ratio?
2
1
34
30s
• Q6
Which of these sequences is an Arithmetic Sequence?
4, 10, 16, 22, …
10, 13, 16, 19, 22, ...
1, 1, 2, 3, 5,8, …
4, 8, 16, 32, …
11, 14, 17, 20,…
100, -50, 25,-12.5, …
30s
• Q7
It is a sequence in which each term is obtained by multiplying the preceding term by a non-zero fixed number called common ratio.
scrambled://GEOMETRIC
30s
• Q8
Geometric Sequence is a sequence in which each term is obtained by multiplying the preceding term by a non-zero fixed number called common ratio while Arithmetic Sequence is a sequence whose consecutive terms have a common difference.
boolean://true
30s
• Q9
Sort these sequences according to their type.
30s
• Q10
Match the words with its appropriate descriptions.
linking://Geometric Sequence|-1296, 216, -36, 6, …:Arithmetic Sequence|8.2, 8, 7.8, 7.6, …:Common Ratio|r:Common difference|d
30s
• Q11
It is a list of numbers or objects in a special order.
freetext://Sequence
30s
• Q12
It is the indicated sum of the terms of an arithmetic sequence,
Geometric Series
Arithmetics Sequence
Geometric Sequence
Arithmetics Series
30s
• Q13
It is the indicated sum of the terms of a geometric sequence.
Arithmetics Series
Geometric Sequence
Arithmetics Sequence
Geometric Series
30s
• Q14
30s
• Q15
30s
Teachers give this quiz to your class | 678 | 2,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.911252 |
https://socratic.org/questions/how-do-you-solve-7-sqrt-2x-1-10 | 1,575,681,808,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491871.35/warc/CC-MAIN-20191207005439-20191207033439-00137.warc.gz | 545,136,133 | 5,863 | # How do you solve 7 + sqrt[2x - 1] = 10?
Aug 5, 2015
$x = 5$
#### Explanation:
First, start by isolating the readical on one side of the equation. This can be done by adding $- 7$ to both sides
$\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{7}}} + \sqrt{2 x - 1} = 10 - 7$
$\sqrt{2 x - 1} = 3$
To get rid of the radical term, square both sides of the equation
${\left(\sqrt{2 x - 1}\right)}^{2} = {3}^{2}$
$2 x - 1 = 9$
$2 x = 10 \implies x = \frac{10}{2} = \textcolor{g r e e n}{5}$ | 220 | 547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 4.5 | 4 | CC-MAIN-2019-51 | latest | en | 0.76942 |
https://www.omnicalculator.com/fitness/hiking | 1,542,447,126,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00505.warc.gz | 941,845,372 | 17,559 | Distance
mi
Elevation gain
ft
%
Total walk if on flat
mi
Hiker's weight
lb
Backpack weight
lb
Burned while hiking up
kcal
Burned while hiking down
kcal
Total burned
kcal
Weight lost
lb
# Hiking Calculator
By
The hiking calculator allows you to calculate the amount of calories burned while hiking.
Years and seasons pass but hiking is still as popular as it's ever been. We tend to do a little hiking to stay fit, spend time with our loved ones, meet new folks and explore the natural beauty of our planet. If we can't find time during the week to jog, play squash or even bike to work, we even treat those little hiking retreats as a chance to lose a bit of weight. And it's definitely a good thing. Calculating calories burned during various kind of activities is a fairly simple matter as it revolves around the following equation:
`Calories Burned = MET * Weight (kg) * Time (hrs)`
MET, short for Metabolic Equivalent, is the amount of oxygen used for particular activities, such as running, walking up the stairs or cycling. Most common values you can find here. As an example, person weighing 70kg (154lbs) who jogs for an hour will burn approximately 70 * 7 * 1 = 490 calories. Although this result is only approximate, it can serve as a good starting point for our further calculations.
Back in 2002 a group of researchers published a paper in Journal of Applied Physiology entitled "Energy cost of walking and running at extreme uphill and downhill slopes". They put a group of volunteers on a treadmill and started testing how much oxygen they use at different gradients, starting from the flat walk or run and climbing higher and higher. After a thorough research they were able to determine how the slope of a hill or mountain affects the amount of oxygen used and, as a result, energy consumed. Having calculated that, they were able to come up with the equation which roughly allows to compare calories burned on a walk and a hike. From there, it was only a short way to use the aforementioned equation and calculate the energy levels for both climbing up and down.
We took the inspiration for this calculator from the very similar tool published several years ago on Hiking Science blog. And it goes like this - to get started enter the length of your trail as well as its elevation gain (the sum of every elevation hiked during the trail). This information should be available online as well as at the trail itself. Then add your body weight and your backpack's weight (skip if you're not taking one). Once you have those 4 numbers filled, all calculations will be performed and you'll see your approximate results.
Let's discuss a simple example to make things clear. Mark went on a hike last weekend which lasted for 12 miles (19,3km) and featured 2500 ft (762m) elevation gain. Mark is pretty slim himself, weighing 150 pounds (68kg) and carrying 18 pounds (8.15kg) of backpack on his back. As you can see, trail averaged around 7.89% grade. Below you can see what the way up and down would correspond to on flat terrain - to burn the same amount of calories Mark would have to walk 10.394 miles for the way up and 6 miles for the way down. In total it would result in 16.394 miles walked.
Now that we have those numbers, we can calculate the amount of calories burned on the way up and down. First, if Mark were to leave his backpack before the trail start, he would burn roughly 923 calories for the way up and 532 for the way down. In total - 1456 calories. Add the backpack and the numbers go up - 1005 for going up and 580 down. In total, Mark should expect to burn around 1585 calories during his hike.
Now, keep note that we based those formulas on a research done in a lab and not on a mountain slope. This means many outside factors were not taken into consideration such as Mark's stamina and individual predispositions, weather conditions, trail structure and speed of hike. Unfortunately, we simply don't have the sufficient amount of data so can only offer approximate values. Regardless, it should work as an interesting way to check the calories needed for each trail and could help hikers better prepare for their trips. Use the tool responsibly and don't forget to take some water and food with you, even if you're headed just for a short walk. Best of luck!
Piotr Małek
## Get the widget!
Hiking Calculator can be embedded on your website to enrich the content you wrote and make it easier for your visitors to understand your message.
It is free, awesome and will keep people coming back! | 1,001 | 4,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-47 | latest | en | 0.96451 |
http://www.slideshare.net/EC-Council/takedowncon-cryptanalysis | 1,429,413,402,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246637364.20/warc/CC-MAIN-20150417045717-00033-ip-10-235-10-82.ec2.internal.warc.gz | 793,002,618 | 35,045 | Upcoming SlideShare
×
Thanks for flagging this SlideShare!
Oops! An error has occurred.
×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply
# TakeDownCon Rocket City: Cryptanalysis by Chuck Easttom
181
views
Published on
Published in: Technology, Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total Views
181
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
17
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Transcript
• 1. Cryptanalysis
• 2. The Speaker Chuck Easttom ceasttom@cec-security.com www.CEC-Security.com
• 3. What cryptanalysis is NOT It’s not fast It’s not guaranteed It’s not easy It’s not what you see in movies the
• 4. Levels of Success Total break — the attacker deduces the secret key. Global deduction — the attacker discovers a functionally equivalent algorithm for encryption and decryption, but without learning the key. Instance (local) deduction — the attacker discovers additional plaintexts (or ciphertexts) not previously known. Information deduction — the attacker gains some Shannon information about plaintexts (or ciphertexts) not previously known. Distinguishing algorithm — the attacker can distinguish the cipher from a random permutation.
• 5. Resources Time — the number of "primitive operations" which must be performed. This is quite loose; primitive operations could be basic computer instructions, such as addition, XOR, shift, and so forth, or entire encryption methods. Memory — the amount of storage required to perform the attack. Data — the quantity of plaintexts and ciphertexts required.
• 6. Breaking Ciphers This means finding any method to decrypt the message that is more efficient than simple brute force attempts. Brute force is simply trying every possible key. If they algorithm uses a 128 bit key that means 2128 possible keys. In the decimal number system that is 3.402 * 1038 possible keys. If you are able to attempt 1 million keys every second it could still take as long as 10,790,283,070,806,014,188,970,529 years to break.
• 7. Breaking Ciphers Cryptanalysis is using other techniques (other than brute force) to attempt to derive the key. In some cases cryptographic techniques are used to test the efficacy of a cryptographic algorithm. Such techniques are frequently used to test hash algorithms for collisions. You must keep in mind that any attempt to crack any non-trivial cryptographic algorithm is simply an ‘attempt’. There is no guarantee of any method working. And whether it works or not it will probably be a long and tedious process. This should make sense to you. If cracking encryption where a trivial process, then encryption would be useless.
• 8. Frequency Analysis This is the basic tool for breaking most classical ciphers. In natural languages, certain letters of the alphabet appear more frequently than others. By examining those frequencies you can derive some information about the key that was used. This method is very effective against classic ciphers like Caesar, Vigenere, etc. It is far less effective against modern methods. In fact with modern methods, the most likely result is that you will simply get some basic information about the key, but you will not get the key. Remember in English the words’ the and and are the two most common three letter words. The most common single letter words are I and a. If you see two of the same letters together in a word, it is most likely ee or oo.
• 9. Known Plain Text/ Chosen Plain Text In this attack the attacker obtains the ciphertexts corresponding to a set of plaintexts of his own choosing. This can allow the attacker to attempt to derive the key used and thus decrypt other messages encrypted with that key. This can be difficult but is not impossible.
• 10. Cipher Text Only Ciphertext-only: The attacker only has access to a collection of cipher texts. This is much more likely than known plaintext, but also the most difficult. The attack is completely successful if the corresponding plaintexts can be deduced, or even better, the key. The ability to obtain any information at all about the underlying plaintext is still considered a success.
• 11. Related Key attack Related-key attack: Like a chosen-plaintext attack, except the attacker can obtain ciphertexts encrypted under two different keys. This is actually a very useful attack if you can obtain the plain text and matching cipher text.
• 12. Linear Cryptanalysis Linear cryptanalysis is based on finding affine approximations to the action of a cipher. It is commonly used on block ciphers. This technique was invented by Mitsarue Matsui. It is a known plaintext attack and uses a linear approximation to describe the behavior of the block cipher. Given enough pairs of plaintext and corresponding ciphertext, bits of information about the key can be obtained. Obviously the more pairs of plain text and cipher text one has, the greater the chance of success. Remember cryptanalysis is an attempt to crack cryptography. For example with the 56 bit DES key brute force could take up to 256 attempts. Linear cryptanalysis will take 243 known plaintexts. This is better than brute force, but still impractical for most situations.
• 13. Linear Cryptanalysis With this method, a linear equation expresses the equality of two expressions which consist of binary variables XOR’d. For example, the following equation, XORs sum of the first and third plaintext bits and the first ciphertext bit is equal to the second bit of the key: You can use this method to slowly recreate the key that was used.
• 14. Linear Cryptanalysis Now after doing this for each bit you will have an equation of the form we can then use Matsui's Algorithm 2, using known plaintextciphertext pairs, to guess at the values of the key bits involved in the approximation. For each set of values of the key bits on the right-hand side (referred to as a partial key), count how many times the approximation holds true over all the known plaintext-ciphertext pairs; call this count T. The partial key whose T has the greatest absolute difference from half the number of plaintext-ciphertext pairs is designated as the most likely set of values for those key bits
• 15. Differential Cryptanalysis Differential cryptanalysis is a form of cryptanalysis applicable to symmetric key algorithms. This was invented by Elii Biham and Adi Shamir. Essentially it is the examination of differences in an input and how that affects the resultant difference in the output. It originally worked only with chosen plaintext. Could also work with known plaintext and ciphertext only.
• 16. Differential Cryptanalysis By analyzing the changes in some chosen plaintexts, and the difference in the outputs resulting from encrypting each one, it is possible to recover some properties of the key.
• 17. Differential Cryptanalysis Differential Cryptanalysis is a Chosen Plaintext attack. By analyzing the Cipher, Differential Characteristics are discovered and used to discover information about the key. This technique doesn’t recover the key, but it attempts to reduce the number of possible keys so that it is possible to find the key in a reasonable amount of time.
• 18. Other methods Higher Order Differential Cryptanalysis Truncated Differential Cryptanalysis Impossible Differential Cryptanalysis Boomerang Attack Mod-n cryptanalysis Boomerang Attack
• 19. Other Techniques • • • • Seeking clues Using other passwords Learning about the subject Tricking the person into giving you the password
• 20. Questions ceasttom@cec-security.com www.CEC-Security.com | 1,643 | 7,786 | {"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} | 2.765625 | 3 | CC-MAIN-2015-18 | longest | en | 0.848948 |
http://van.physics.illinois.edu/qa/listing.php?id=802 | 1,503,536,853,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126017.3/warc/CC-MAIN-20170824004740-20170824024740-00271.warc.gz | 448,892,045 | 6,161 | # Q & A: Calculating Efficiency
Q:
If a machining operation has a run-time standard of say, one hour per hundred parts produced and a setup time of also one hour, in order to calculate the efficiency for say, a work order of one thousand parts, should both the setup time and the run time be divided by one hundred (to get it down to one part) and then multiplied by one thousand to determine what the standard, expressed as a percent, should be for that particular work order? Or should the setup time be kept apart from the run-time (since the setup time will be the same no matter what the size of the work order)and remain at one hour and then added to the result of the calculation for the run-time standard? Thank you. Lee
- Lee
Steel Products, Marion, Iowa, U.S.
A:
Lee -
It depends on how you're looking at it. If you're trying to figure out the production rate once you've started running, it would simply be 100 parts per hour (or 1/100 = 0.010 hours per part). But if you want to know the production rate for a particular order, you need to know the total number of parts being produced (1000, in your example) and the total amount of time it would take to make them (1 hour setup + 10 hours run-time = 11 hours total). So the production rate for that particular order would be 1000 parts / 11 hours = 90.9 parts per hour. (Or 1/90.9 = 0.011 hours per part).
To figure this out in percent, you first have to know what the maximum possible is. In this case, we can say that the highest concievable rate of production would be 100 parts per hour. So we just divide the particular rate of production by this (and multiply by 100%) to get the percent efficiency for that particular order:
(90.9 parts/hr) / (100 parts/hr) * 100% = 90.9% efficiency
OR...
(parts/hr produced) / (max. parts/hr possible) * 100%
One thing that I think is interesting to notice is that because of your one hour set-up time, your percent efficiency will get higher and higher as the size of your order increases. Take, for example, orders of 100 parts, 1000 parts, and 10,000 parts:
For 100 parts, it will take 1 hour to setup and 1 hour to run, so the rate of production is 100 parts / 2 hours, or 50 parts per hour. So the efficiency is:
(50 parts/hr) / (100 parts/hr) * 100% = 50% efficiency
We already figured out that for 1000 parts, it's 90.9% efficiency.
And for 10,000 parts, it takes 1 hour to setup and 100 hours to run, so the rate of production is 10,000 parts / 101 hours, or 99 parts per hour. So we get:
(99 parts/hr) / (100 parts/hr) * 100% = 99% efficiency
And all together:
100 parts: 50% efficiency
1000 parts: 91% efficiency
10,000 parts: 99% efficiency
-Tamara
(published on 10/22/2007) | 707 | 2,706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-34 | latest | en | 0.940314 |
https://ask-public.com/5519/ | 1,652,772,273,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00184.warc.gz | 170,168,488 | 11,534 | The resistance of armature winding depends on?
212 views
The resistance of armature winding depends on length of conductor, cross-sectional area of the conductor, number of conductors.
Related Questions
A 200 V dc shunt motor is running at 1000 rpm and drawing a current of 10 A. Its armature winding resistance is 2 Ohm. It. is braked by plugging. The resistance to be connected in series with armature to restrict armature current to 10 A, is
Last Answer : A 200 V dc shunt motor is running at 1000 rpm and drawing a current of 10 A. Its armature winding resistance is 2 Ohm. It. is braked by plugging. The resistance to be connected in series with armature to restrict armature current to 10 A, is 36 Ohm
331 views 1 answer
Last Answer : It is the wire wound around a aluminium metal it is found commonly in transformers or motors
106 views 1 answer
Aluminium is not used as winding wire for the armature of a dc machine because (1) aluminium has low resistivity (2) a large winding space is taken by aluminium conductors and creates ... ) the thermal conductivity of aluminium is low (4) aluminium has low conductivity as compared to copper
Last Answer : Aluminium is not used as winding wire for the armature of a dc machine because (2) a large winding space is taken by aluminium conductors and creates jointing problems (3) the thermal conductivity of aluminium is low (4) aluminium has low conductivity as compared to copper
621 views 1 answer
A 200 V DC machine supplies 20 A at 200 V as a generator. The armature resistance is 0.2 Ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10%, the ratio of motor speed to generator speed is
Last Answer : A 200 V DC machine supplies 20 A at 200 V as a generator. The armature resistance is 0.2 Ohm. If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10%, the ratio of motor speed to generator speed is 0.87
127 views 1 answer
The normal value of the armature resistance of a d.c. motor is?
Last Answer : The normal value of the armature resistance of a d.c. motor is 0.5.
2.9k views 1 answer
Last Answer : Armature reaction in a generator results in demagnetisation of leading pole tip and magnetisation of trailing pole tip.
216 views 1 answer
Last Answer : Armature coil is short circuited by brushes when it lies along neutral axis.
187 views 1 answer
Last Answer : Open circuited armature coil of a D.C. machine is identified by the scarring of the commutator segment to which open circuited coil is connected and indicated by a spark completely around the commutator.
159 views 1 answer
Last Answer : The armature core of a D.C. generator is usually made of silicon steel.
153 views 1 answer
Last Answer : The demagnetising component of armature reaction in a D.C. generator reduces generator e.m.f.
123 views 1 answer
Last Answer : Armature reaction of an unsaturated D.C. machine is crossmagnetising.
353 views 1 answer
Last Answer : The commutator segments are connected to the armature conductors by means of copper lugs.
202 views 1 answer
205 views 0 answers
Last Answer : Wave winding is composed of that even number which is exact multiple of poles + 2.
90 views 1 answer
Last Answer : Lap winding is composed of any even number of conductors.
129 views 1 answer
Last Answer : In D.C. generators, lap winding is used for low voltage, high current.
94 views 1 answer
Last Answer : In the case of lap winding resultant pitch is difference of front and back pitches.
122 views 1 answer
Last Answer : In case of D.C. machine winding, number of commutator segments is equal to number of armature coils.
665 views 1 answer
Last Answer : In lap winding, the number of brushes is always same as the number of poles.
323 views 1 answer
Define armature reaction in an alternator. Explain the effect of armature reaction at various P.F. of loads of alternator.
Last Answer : Armature Reaction: The effect of armature flux on main flux is called as armature reaction. Armature Reaction at Various Power Factors: When armature is loaded, the armature flux modifies ... lagging p.f. For capacitive loads the effect of flux is partly distorting and partly strengthening.
421 views 1 answer
In a synchronous machine, if the field flux axis ahead of the armature field axis in the direction of rotation, the machine is operating as (1) synchronous motor (2) synchronous generator (3) asynchronous motor (4) asynchronous generator
Last Answer : In a synchronous machine, if the field flux axis ahead of the armature field axis in the direction of rotation, the machine is operating as synchronous generator
601 views 1 answer
In dc machines, the armature windings are placed on the rotor because of the necessity for (1) electromechanical energy conversion (2) generation of voltage (3) commutation (4) development of torque
Last Answer : 3 Commutation
291 views 2 answers
In a 3 -phase alternator, if there is only magnetizing armature reaction, the load is : (A) capacitive (B) inductive (C) resistive (D) inductive and resistive
Last Answer : In a 3 -phase alternator, if there is only magnetizing armature reaction, the load is : capacitive
119 views 1 answer
The armature current on symmetrical 3 phase short circuit of a synchronous machine (salient pole) (A) has q -axis current only (B) has d -axis current only (C) has both d and q axis currents (D) cannot be divided between q and d axis currents
Last Answer : The armature current on symmetrical 3 phase short circuit of a synchronous machine (salient pole) has d -axis current only
193 views 1 answer
The induced emf in the armature of a lap -wound four -pole dc machine having 100 armature conductors rotating at 600 rpm and with 1 Wb flux per pole is (1) 1000V (2) 100V (3) 600V (4) 10,000V
Last Answer : The induced emf in the armature of a lap -wound four -pole dc machine having 100 armature conductors rotating at 600 rpm and with 1 Wb flux per pole is 1000V
370 views 1 answer
What is the effect of armature reaction? A) Demagnetization of the main field B) Demagnetization of the brushes C) Cross-magnetization of the brushes D) None of these
Last Answer : What is the effect of armature reaction? A) Demagnetization of the main field B) Demagnetization of the brushes C) Cross-magnetization of the brushes D) None of these
372 views 1 answer
The armature reaction in a synchronous generator supplying leading power factor load is (A) magnetizing (B) demagnetizing (C) demagnetizing and cross-magnetizing (D) magnetizing and cross-magnetizing
Last Answer : The armature reaction in a synchronous generator supplying leading power factor load is magnetizing and cross-magnetizing
96 views 1 answer
The induced e.m.f. in the armature conductors of a d.c. motor is?
Last Answer : The induced e.m.f. in the armature conductors of a d.c. motor is sinusoidal.
825 views 1 answer
Last Answer : When the armature of a D.C. motor rotates, e.m.f. induced is back e.m.f.
203 views 1 answer
Last Answer : The armature voltage control of D.C. motor provides constant torque drive.
92 views 1 answer
Last Answer : The armature torque of the D.C. shunt motor is proportional to armature current.
210 views 1 answer
Last Answer : In a D.C. shunt motor, under the conditions of maximum power, the current in the armature will be more than full-load current.
367 views 1 answer
Last Answer : In a A.C. series motor armature coils are usually connected to commutator through resistance.
161 views 1 answer
Last Answer : In a synchronous motor, the armature current has large values for high and low excitation.
138 views 1 answer
Last Answer : The armature current of the synchronous motor has large values for low and high excitation.
107 views 1 answer
Why is the armature of DC generator laminated?
Last Answer : The armature of DC generator is laminated to reduce eddy current losses.
176 views 1 answer
Armature control of DC motor
Last Answer : Armature control of DC motor
162 views 2 answers
Last Answer : The critical resistance of the D.C. generator is the resistance of field.
555 views 1 answer
Last Answer : In a D.C. generator, the critical resistance refers to the resistance of field.
136 views 1 answer
State advantages of short pitch winding over full pitch winding in alternators.
Last Answer : Advantages of Short Pitch Winding over Full Pitch Winding in Alternators: 1) Short pitching reduces the amount of copper needed for end connection when compared with full pitched coil. 2) They ... reduced, thereby increasing the efficiency. 4) The power quality of generated emf is improved.
478 views 1 answer
Damper winding in a synchronous motor
Last Answer : Damper winding in a synchronous motor Serves to start the motor.
87 views 1 answer
A certain transformer has 200 turns in the primary winding, 50 turns in the secondary winding and 120 volts applied to the primary. What is the voltage across the secondary? (A) 30 Volts (B) 480 Volts (C) 120 Volts (D) 60 Volts
Last Answer : A certain transformer has 200 turns in the primary winding, 50 turns in the secondary winding and 120 volts applied to the primary. What is the voltage across the secondary? (A) 30 Volts (B) 480 Volts (C) 120 Volts (D) 60 Volts
223 views 1 answer
A _________ is used to measure the stator % winding temperature of the generator. (A) Thermocouple (B) Pyrometer (C) Resistance thermometer (D) Thermometer
Last Answer : A Resistance thermometer is used to measure the stator % winding temperature of the generator.
3.4k views 1 answer
A 36 slot, 4 -pole, dc machine has a simplex lap winding with two conductors per slot. The back pitch and front pitch adopted could be respectively
Last Answer : A 36 slot, 4 -pole, dc machine has a simplex lap winding with two conductors per slot. The back pitch and front pitch adopted could be respectively 19,17
588 views 1 answer
Damper winding is provided in 3 phase synchronous motors to :
Last Answer : Damper winding is provided in 3 phase synchronous motors to : provide starting torque and prevent hunting
93 views 1 answer
Last Answer : In D.C. machines fractional pitch winding is used to reduce the sparking.
355 views 1 answer
Last Answer : The starting winding of a single-phase motor is placed in stator.
115 views 1 answer
Last Answer : If starting winding of a single-phase induction motor is left in the circuit, it will draw excessive current and overheat.
234 views 1 answer
Last Answer : Starting winding of a single phase motor of a refrigerator is disconnected from the circuit by means of a magnetic relay.
215 views 1 answer
Last Answer : Centrifugal switch disconnects the auxiliary winding of the motor at about 70 to 80 percent of synchronous speed. | 2,645 | 10,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.932244 |
https://www.lotterypost.com/thread/304124/2 | 1,498,594,723,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00218.warc.gz | 890,017,482 | 20,928 | Welcome Guest
You last visited June 27, 2017, 3:06 pm
All times shown are
Eastern Time (GMT-5:00)
27 minus the sum PLUS
Topic closed. 102 replies. Last post 12 months ago by Soledad.
Page 2 of 7
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 17, 2016, 11:49 pm - IP Logged
KS draw date 6/17
27 minus the sum 978 24=sum 03 58 3-4 rundown 978 becomes 977 4 9 3 7 4 7 3 9 4 7 3 7
927 18=sum 09 54 4 3 1 2 2 3 4 2 1 1
184 13=sum 14 69 7 4 3 4 7 6 3 2
2 7 7 4 9 5
9 5 4 5
vtrack indicated: 5 is digits 4,9
1 is digits 0,5 Why? There is no 0 in rundown. 9 can be it.
pairs: 03 13 39 49 55 68 89 pairs: 92 52 44 02
04 14 34 48 56 69 97 57 47 07
05 19 38 45 58 95 54 42 05
06 18 59 94 55 04
08 15 35 46 99 09
09 16 36 00
39
Once again run the filter. There is no 5 in the pyramid today so 5 pairs go away. Digit 5 could still be in the draw but will match with another pair to get it. I'm not expecting double, so those pairs can be eliminated. Other pairs are eliminated as a result of the filter.
So we have 12 pairs from 27 minus, and 6 pairs from 3-4 rundown. We have no matches.
Turn tail and run?
No, it just means we have to make a decision on which pairs and rundown to finish with. I pick 3-4 rundown. Why? Probably familiarity and the fact that in May this workout had 24 days of at least one hit in the two draws. Some had both.
Now we eliminated the O pairs due to filter. This leads me to certainly believe it is vtrack 5. All the pairs left have a 4 or 9 in them.
Match to digits in 3-4 rundown. You get:
245, 247, 249 297. 295, 792, 795, 794, 945, 472, 475. Certainly a reasonable number of combo's.
Draws for 6/17 Mid: 457 Quit now? Maybe, maybe not. Why? In May when the second draw for the day was hit it was the other vtrack. So now we are looking at just the digit 5. 5 combo's.........sure why not?
EVE:954
That's why this rundown is amazing. And even better when matched with the 27 minus. Why? In May when the pairs matched.....always a hit.
Did you do your tracking today ??
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 18, 2016, 12:06 am - IP Logged
I have a real hard time understanding black apple...i understand a little of th 34 rundown..but what does he mean by number line???? i will appreciate if you can clarify that for me. Hes a good man but he dosent speak english..ive tried asking him before but his answers are too confusing...thanks...
Took me a long time to get the hang of it. But it is worth it. Practice, practice, practice. Spend the time to do a back test for your state for an entire 30 days. Get really familiar with it.
Read the post about the number line carefully. Copy and paste it in a document and eliminate the colors and the different fonts.
Read it again. It will make sense. If in doubt, remember 3 digits. 4,1 1,4 4,7 7,4 8.5 5.8 Just an example, there are others. In fact....9,2 2,9. Why? 9,0,1,2....
Copy, laminate it, and use it.
Balancing the number is crucial. When I first started using the 3-4 rundown, for what ever reason, my search didn't take me to the beginning. I found it about a year after starting to use it. I only did the rundown using the draw. It worked...... a lot. But when I made the change, it made a huge difference.
It will give you the vtracks to use. It will often give the root to use. It will even give the pairs. I've wheeled the digits. I've matched them in pairs and then combo's. Both will work.
The workout is best for two draws, and my experience shows evening draw is the one to work from. IF it doesn't hit a draw for the day, most times, it's just a day early. It'll hit the next one.
OH, yeah. Repeat pairs. It'll make sense.
Did you do your tracking today ??
New Mexico
United States
Member #86099
January 29, 2010
11477 Posts
Offline
Posted: June 18, 2016, 12:14 am - IP Logged
KS draw date 6/17
27 minus the sum 978 24=sum 03 58 3-4 rundown 978 becomes 977 4 9 3 7 4 7 3 9 4 7 3 7
927 18=sum 09 54 4 3 1 2 2 3 4 2 1 1
184 13=sum 14 69 7 4 3 4 7 6 3 2
2 7 7 4 9 5
9 5 4 5
vtrack indicated: 5 is digits 4,9
1 is digits 0,5 Why? There is no 0 in rundown. 9 can be it.
pairs: 03 13 39 49 55 68 89 pairs: 92 52 44 02
04 14 34 48 56 69 97 57 47 07
05 19 38 45 58 95 54 42 05
06 18 59 94 55 04
08 15 35 46 99 09
09 16 36 00
39
Once again run the filter. There is no 5 in the pyramid today so 5 pairs go away. Digit 5 could still be in the draw but will match with another pair to get it. I'm not expecting double, so those pairs can be eliminated. Other pairs are eliminated as a result of the filter.
So we have 12 pairs from 27 minus, and 6 pairs from 3-4 rundown. We have no matches.
Turn tail and run?
No, it just means we have to make a decision on which pairs and rundown to finish with. I pick 3-4 rundown. Why? Probably familiarity and the fact that in May this workout had 24 days of at least one hit in the two draws. Some had both.
Now we eliminated the O pairs due to filter. This leads me to certainly believe it is vtrack 5. All the pairs left have a 4 or 9 in them.
Match to digits in 3-4 rundown. You get:
245, 247, 249 297. 295, 792, 795, 794, 945, 472, 475. Certainly a reasonable number of combo's.
Draws for 6/17 Mid: 457 Quit now? Maybe, maybe not. Why? In May when the second draw for the day was hit it was the other vtrack. So now we are looking at just the digit 5. 5 combo's.........sure why not?
EVE:954
That's why this rundown is amazing. And even better when matched with the 27 minus. Why? In May when the pairs matched.....always a hit.
Theos working on a program.
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 18, 2016, 12:19 am - IP Logged
Awesome. It would make it faster. Not sure how to automate the process of balancing the number. Matching digits for pairs and combos doesn't seem as if it would be too bad.
BTW, all that make sense? I worry about explaining clearly enough.
Did you do your tracking today ??
New Mexico
United States
Member #86099
January 29, 2010
11477 Posts
Offline
Posted: June 18, 2016, 12:48 am - IP Logged
Awesome. It would make it faster. Not sure how to automate the process of balancing the number. Matching digits for pairs and combos doesn't seem as if it would be too bad.
BTW, all that make sense? I worry about explaining clearly enough.
I wasted too much time responding to blogs. Never again !
You are doing your best but I'm not up to speed on what you are doing.
Maybe a step by step with the process, that may work.
United States
Member #116477
September 11, 2011
406 Posts
Offline
Posted: June 18, 2016, 6:50 am - IP Logged
Me neither i need to know what "balancing numbers" means..ive got a lot figured out but im trying to fgure out "number lines" blackapple makes no mention of v tracs on his post i do not understand them i dont know why you put them in your post. Also i notice when i take yesterdays number and do that 34 rundown no doubling ect i get a lot of pairs and 3 numbers very often....yes can you please show the balance and number line ideas? think thats the key that blackappl shows but again he speaks a foreign language. What is Balance the numbers???????????????
New Mexico
United States
Member #86099
January 29, 2010
11477 Posts
Offline
Posted: June 18, 2016, 9:46 am - IP Logged
KS draw date 6/17
27 minus the sum 978 24=sum 03 58 3-4 rundown 978 becomes 977 4 9 3 7 4 7 3 9 4 7 3 7
927 18=sum 09 54 4 3 1 2 2 3 4 2 1 1
184 13=sum 14 69 7 4 3 4 7 6 3 2
2 7 7 4 9 5
9 5 4 5
vtrack indicated: 5 is digits 4,9
1 is digits 0,5 Why? There is no 0 in rundown. 9 can be it.
pairs: 03 13 39 49 55 68 89 pairs: 92 52 44 02
04 14 34 48 56 69 97 57 47 07
05 19 38 45 58 95 54 42 05
06 18 59 94 55 04
08 15 35 46 99 09
09 16 36 00
39
Once again run the filter. There is no 5 in the pyramid today so 5 pairs go away. Digit 5 could still be in the draw but will match with another pair to get it. I'm not expecting double, so those pairs can be eliminated. Other pairs are eliminated as a result of the filter.
So we have 12 pairs from 27 minus, and 6 pairs from 3-4 rundown. We have no matches.
Turn tail and run?
No, it just means we have to make a decision on which pairs and rundown to finish with. I pick 3-4 rundown. Why? Probably familiarity and the fact that in May this workout had 24 days of at least one hit in the two draws. Some had both.
Now we eliminated the O pairs due to filter. This leads me to certainly believe it is vtrack 5. All the pairs left have a 4 or 9 in them.
Match to digits in 3-4 rundown. You get:
245, 247, 249 297. 295, 792, 795, 794, 945, 472, 475. Certainly a reasonable number of combo's.
Draws for 6/17 Mid: 457 Quit now? Maybe, maybe not. Why? In May when the second draw for the day was hit it was the other vtrack. So now we are looking at just the digit 5. 5 combo's.........sure why not?
EVE:954
That's why this rundown is amazing. And even better when matched with the 27 minus. Why? In May when the pairs matched.....always a hit.
Are you putting the numbers like this:
238 sum 13
27-= 14
mirror 69.
Now are you putting the 69 in the 3,4 blackapple program? If so can you show this and how it I'd associated with vtracs.
United States
Member #116477
September 11, 2011
406 Posts
Offline
Posted: June 18, 2016, 5:35 pm - IP Logged
Read my leeps too!y!ou tell us how important it is to balance numbers..can ANYBODY out there tell us how we do this?? Ive asked several times...only silence
is there anyone out there who can explain what Blackapple means by NUMBER LINES?? and BALANCING THE NUMBERS?somebody must know but nobodies talkin...!!!!
cali
Colombia
Member #170335
November 26, 2015
39 Posts
Offline
Posted: June 18, 2016, 7:38 pm - IP Logged
KS draw date 6/17
27 minus the sum 978 24=sum 03 58 3-4 rundown 978 becomes 977 4 9 3 7 4 7 3 9 4 7 3 7
927 18=sum 09 54 4 3 1 2 2 3 4 2 1 1
184 13=sum 14 69 7 4 3 4 7 6 3 2
2 7 7 4 9 5
9 5 4 5
vtrack indicated: 5 is digits 4,9
1 is digits 0,5 Why? There is no 0 in rundown. 9 can be it.
pairs: 03 13 39 49 55 68 89 pairs: 92 52 44 02
04 14 34 48 56 69 97 57 47 07
05 19 38 45 58 95 54 42 05
06 18 59 94 55 04
08 15 35 46 99 09
09 16 36 00
39
Once again run the filter. There is no 5 in the pyramid today so 5 pairs go away. Digit 5 could still be in the draw but will match with another pair to get it. I'm not expecting double, so those pairs can be eliminated. Other pairs are eliminated as a result of the filter.
So we have 12 pairs from 27 minus, and 6 pairs from 3-4 rundown. We have no matches.
Turn tail and run?
No, it just means we have to make a decision on which pairs and rundown to finish with. I pick 3-4 rundown. Why? Probably familiarity and the fact that in May this workout had 24 days of at least one hit in the two draws. Some had both.
Now we eliminated the O pairs due to filter. This leads me to certainly believe it is vtrack 5. All the pairs left have a 4 or 9 in them.
Match to digits in 3-4 rundown. You get:
245, 247, 249 297. 295, 792, 795, 794, 945, 472, 475. Certainly a reasonable number of combo's.
Draws for 6/17 Mid: 457 Quit now? Maybe, maybe not. Why? In May when the second draw for the day was hit it was the other vtrack. So now we are looking at just the digit 5. 5 combo's.........sure why not?
EVE:954
That's why this rundown is amazing. And even better when matched with the 27 minus. Why? In May when the pairs matched.....always a hit.
someone managed to put this in an Excel file?
Excel teachers, with all due respect it would be good to put him in an Excel ...
Bakersfield, Ca
United States
Member #89877
April 17, 2010
207 Posts
Offline
Posted: June 20, 2016, 11:42 am - IP Logged
Theos working on a program.
This is a hard one for me. Correct me if I'm wrong, I am assuming that in the 3-4 rundown you are using the last digit as the vtrac number, also the pairs for the 3-4 rundown are coming from the bottom 2 rows of the pyramid.
Next,( the number drawn), if the digits are less than 3 from each other we use those digits according to BA's number line.
To some of you readers, these are all questions not the way it works
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 20, 2016, 12:13 pm - IP Logged
This is a hard one for me. Correct me if I'm wrong, I am assuming that in the 3-4 rundown you are using the last digit as the vtrac number, also the pairs for the 3-4 rundown are coming from the bottom 2 rows of the pyramid.
Next,( the number drawn), if the digits are less than 3 from each other we use those digits according to BA's number line.
To some of you readers, these are all questions not the way it works
Yes, the bottom two rows hold all the good stuff.
I will put the workout for 6/17 here again just for demo purposes.
978 becomes 977. Theo, you are correct on the number balancing. There is no difference of 3 so you use the 978, drop last digit and double the second.......977.
4 9 3 7 4 7 3 9 4 7 3 7
4 3 1 2 2 3 4 2 1 1
7 4 3 4 7 6 3 2
2 7 7 4 9 5
95 4 5
Now the bottom row of digits indicates the vtrac. We have a 4,9.............indicates vtrac 5. We also have a 5. There is no 0 digit in the 3-4 rundown. 9 is sometimes it. So this would indicate vtrac 1......0 or 5.
Yes, pairs come from the bottom digits matched to all digits in bottom two rows.
Match the digits to all others in bottom row. Remember 9 can be 0, so include it. If it's not in the actual combo, the filter process will eliminate it out (the Pair).
Next thing I've found is part of the filter process can be......the digits in the pair must be adjacent in the rundown. Ex.: 95, 97,94, You may recall in the previous post we had a pair 47. We kept it....why? In the left set of numbers, 7 is in the second row, and just above it is the 4. I have had a pair come from that situation.
Blackapple has mentioned many times the bottom row can be the root sum digit also. I have found it can OFTEN be one of the others also. This is another point that can be used in reduction. It's not perfect, but MOST times works.
While this thread is about the 27 minus the sum and the 3-4 rundown, don't forget why. When a pair matches both workouts, there is a hit. The 3-4 rundown produces plenty of hits on its own, but with 27 minus match....it is a virtual lock.
Did you do your tracking today ??
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 20, 2016, 12:16 pm - IP Logged
Another note on this, trying not to confuse.....
Sometimes the combo is "RIGHT" there. Example.....the 457 and 954 are aligned in the workout. This happens. If you see it.........double up, triple up time???
Did you do your tracking today ??
New Mexico
United States
Member #86099
January 29, 2010
11477 Posts
Offline
Posted: June 20, 2016, 12:22 pm - IP Logged
Another note on this, trying not to confuse.....
Sometimes the combo is "RIGHT" there. Example.....the 457 and 954 are aligned in the workout. This happens. If you see it.........double up, triple up time???
Thus the need for a program. To save time and backtest this needs to either have someone show more examples or have a program. Vtrac are fine but I don't know.
United States
Member #42083
June 27, 2006
771 Posts
Offline
Posted: June 20, 2016, 12:30 pm - IP Logged
Read my leeps too!y!ou tell us how important it is to balance numbers..can ANYBODY out there tell us how we do this?? Ive asked several times...only silence
is there anyone out there who can explain what Blackapple means by NUMBER LINES?? and BALANCING THE NUMBERS?somebody must know but nobodies talkin...!!!!
When doing the 3-4 rundown, you are using previous draw combo. I prefer evening because the rundown will usually give at least one of the combo's for the next two draws, if not both of them. I don't see that as much with using a mid draw.
If there is not a difference of 3 between 2 digits in the combo, you simply drop the last digit and double the second. Example,: 978
No difference so becomes 977.
However, if the is a difference of 3 digits between 2 digits in the combo, you balance. Examples, 41-14
47-74 85-58 03-30 The two digits that are 3 apart are compared on the number line. the digit that is stand alone comes to the front.
Number line is thus:1-8-5-2-9-6-3-0-7-4-1
So in the example Blackapple uses....368. The 3 and 6 are three digits apart. 8 comes to the front, 6 comes before the 3. Double the middle digit.....866.
Why is it important to do this. The digit alignment in the workout. There are a few times it seems it doesn't matter, but it really does. digits align differently if you don't. We know the pairs coming from the rundown are adjacent. This is usually different if you compare doing the balancing and not doing the balancing. Sometimes the digits are just plain different.
When I first started doing the rundown, I did not do this. Since doing it my hits have dramatically improved.
Did you do your tracking today ??
New Mexico
United States
Member #86099
January 29, 2010
11477 Posts
Offline
Posted: June 20, 2016, 12:36 pm - IP Logged
Can you work with theo and get a program that works with the entire progrm including the 3,4 vtrac and the final result?, picks.
Page 2 of 7 | 5,808 | 20,022 | {"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} | 3.15625 | 3 | CC-MAIN-2017-26 | latest | en | 0.224864 |
https://circuit-diagramz.com/digital-alarm-clock-using-4026-logic-gates-schematic-circuit-diagram/ | 1,680,106,976,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00327.warc.gz | 211,448,588 | 33,497 | # Digital alarm clock using 4026 logic gates Schematic Circuit Diagram
## Description and concept behind digital clock
This digital alarm clock project uses 4026 IC which is a decade counter as well as the seven-segment driver. Seven segment display is used for displaying numbers from 0 to 9 and it will display numbers when the enable pin of 4026 is high on the rising edge of the clock ie the circuit starts counting and displaying results when the enable pin is made high. Each seven-segment display needs very little current to glow the LED inside it so we have to use a 500-ohm resistor on each segment so that we would not be blown off the LED.
7806 is a voltage regulator IC this part of the circuit is used to maintain the power supply to the circuit but if you are using a simulation tool like Proteus you will not need this circuit.
Another important thing to be noted is counting less than 9 can be achieved by connecting that pin to the reset. For example, to count from 0 to 5 6 should be connected to the reset pin so that if the counter reaches to threshold ie 5 then it goes to reset in the next counting step.
And the last thing to remember is this circuit uses an external clock source of 1 Hz frequency. If you want to make your own clock please refer to this diagram of 555 timer ic. This will produce an exact 1 Hz clock. if this is not a 1 Hz frequency please maintain it by changing the variable resistor or capacitor.
Schematic Diagram:
CD4060B consists of an oscillator section and 14 ripple-carry binary counter stages. The oscillator configuration allows the design of either RC or crystal oscillator circuits. RESET input is provided which resets the counter to the all-O’s state and disables the oscillator. A high level on the RESET line accomplishes the reset function. All counter stages are master-slave flip-flops. The state of the counter is advanced one step in binary order on the negative transition of I (and O). All inputs and outputs are fully buffered. Schmitt trigger action on the input-pulse line permits unlimited input-pulse rise and fall times.
Check Also
Close
Close
Close | 454 | 2,132 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | longest | en | 0.874665 |
http://irvinemclean.com/maths/appa16.htm | 1,506,339,828,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818691476.47/warc/CC-MAIN-20170925111643-20170925131643-00664.warc.gz | 182,691,497 | 1,704 | 16th Powers
Base Euler length Euler representation 4 8246 (88,x3,7734x2,424x1) 5 3605 (14x4,2146x3,1222x2,223x1) 6 851 (5,5,479x4,20x3,308x2,39x1) 7 443 (10x6,32x5,29x4,336x3,19x2,17x1) 8 528 (6x7,27x6,32x5,238x4,56x3,7x2,162x1) 9 341 (3x8,23x7,86x6,10x5,21x4,147x3,38x2,13x1) 10 324 (30x8,46x7,9x6,9x5,63x4,47x3,94x2,26x1) 11 239 (10,10,10x9,20x8,52x7,20x6,36x5,16x3,4x2,79x1) 12 222 (11,13x10,9,22x8,26x7,7x6,33x5,51x4,22x3,2,45x1) 13 137 (12,12,11,11,13x10,34x9,37x8,9x7,11x6,8x5,5x4,8x3,4x2,4x1) 14 171 (3x13,11,10x10,18x9,5x8,29x7,10x6,17x4,70x3,2,7x1) 15 154 (14,13,13,15x12,11,11,17x10,10x9,21x8,6x6,5,5,11x4,21x3,18x2,28x1) 16 137 (15,15,5x13,7x12,12x11,12x10,5x9,28x8,3x7,4x6,5,5,13x4,28x3,9x2,7x1) 17 136 (16,12x14,3x13,6x12,16x11,19x10,21x9,32x8,10x7,6,6x5,8x3,1) 18 104 (17,16,7x15,3x14,13,3x12,13x11,4x8,10x7,16x5,23x4,7x3,6x2,9x1) 19 103 (18,18,16,16,15,13,13,12,12,8x11,3x10,9,10x7,23x6,33x5,6x4,3,9x1) 20 96 (10x17,7x16,3x15,5x14,13x13,8x11,4x9,8,4x7,14x6,8x5,4,4,14x3,2,2,2) 21 95 (20,18,18,9x17,10x15,10x14,10x13,12,6x10,8x9,3x8,17x7,5,5,7x4,8x3,1) 22 89 (21,3x19,4x18,3x17,3x16,4x15,3x13,11,6x10,4x9,10x8,17x7,6,6,18x5,7x3,3x1) 23 93 (3x21,3x19,5x18,7x17,16,15,14,12,12,11,11,9x10,5x9,7,7,5,5,5x4,31x3,5x2,9x1) 24 92 (22,4x21,2x20,6x19,6x17,2x16,3x14,6x13,11x12,3x11,10,5x8,7,7,6,5,5,5,4,25x3,2,9x1) | 998 | 1,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-39 | latest | en | 0.392764 |
http://mathhelpforum.com/algebra/105846-realitvely-easy-question.html | 1,526,832,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863626.14/warc/CC-MAIN-20180520151124-20180520171124-00483.warc.gz | 176,305,875 | 9,507 | 1. ## Realitvely Easy Question
$\displaystyle y = 2x$
$\displaystyle 5 = x^2 + y^2$
Alright guys, I missed a note, and im tryign to figure out what to do...
I beleive I use substitution, but im not quite sure how exactly. Can someone quicly run through it for me? Thanks guys.
2. $\displaystyle 5=x^2+(2x)^2=x^2+4x^2=5x^2\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$
3. You are my hero. | 135 | 391 | {"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} | 3.3125 | 3 | CC-MAIN-2018-22 | latest | en | 0.757982 |
https://community.fabric.microsoft.com/t5/Desktop/DAX-formula-to-return-data-for-last-6-months/m-p/80101 | 1,695,993,727,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00653.warc.gz | 198,721,205 | 86,038 | cancel
Showing results for
Did you mean:
Microsoft
## DAX formula to return data for last 6 months
I have a series of line and clustered column charts that I currently have showing only the last 6 months of data (column for each month). At this point I have to change the filters on these charts every month in order to update them to only show the last 6 months of data.
Is there any way to use a DAX formula to create a calculated column that I could use as a filter to only return last 6 months?
I used a DAX formula to create a column that I use as a filter for last 30 days (see below), so I was hoping to create something similar for returning the last 6 months. Using the last 168 days obviously doesn't work because it doesn't account for every day of the first month.
Last30days = IF(AND('Date'[Date]>=[Today]-30,'Date'[Date]<=[Today]),1,0)
1 ACCEPTED SOLUTION
Memorable Member
@TannerBuck7 is a bit late so it must be a more efficient way but for now
Specially to avoid when year changes you will need a monthindex
Create a YearMonth Column ( if you don't have )
```YearMonth =
('Calendar'[YearKey] * 100 ) + MONTH('Calendar'[DateKey])```
Then create the month index column
```MonthIndex =
VAR MonthRow = 'Calendar'[YearMonth]
RETURN
CALCULATE (
DISTINCTCOUNT ( 'Calendar'[YearMonth] );
FILTER ( 'Calendar'; 'Calendar'[YearMonth] <= MonthRow )
)```
and finally the Last6Months
```Last6Months =
VAR TodayMonthIndex =
CALCULATE (
MAX ( 'Calendar'[MonthIndex] );
FILTER ( 'Calendar'; TODAY () = 'Calendar'[DateKey] )
)
VAR monthtocheck = Calendar[MonthIndex]
RETURN
IF (
AND ( monthtocheck >= TodayMonthIndex - 5; monthtocheck <= TodayMonthIndex );
1;
0
)```
More steps but when the year change you will still see six months before
Hope it helps
Konstantinos Ioannou
18 REPLIES 18
Anonymous
Not applicable
I am using teh same method but i am gettign the error . please check here
i have created month idnex column and year month column i already have in my table
@NHar @Anonymous @TannerBuck7 @konstantinos please tell me the solution for this
New Member
Here is a much simpler solution than the accepted one:
Last6Months = (YEAR(TODAY()) * 12 + MONTH(TODAY())) - (YEAR(Agenda[Date])*12+MONTH(Agenda[Date])) < 6
New Member
How would I go about not including the current month you're in?
Eg: It's June now, using this (WIth May selected) it still includes June's results. This is the Measure's DAX:
``Last6Months = (YEAR(TODAY()) * 12 + MONTH(DATEVALUE("01 " & IssueData[Month Selected] & " 2022 " ))) - (YEAR(IssueData[Closed Date])*12+(MONTH(DATEVALUE("01 " & IssueData[Month Selected] & " 2022 " ))) < 6) ``
The bit below is a measure that stores the month selected:
``````Month Selected =
IF (
HASONEVALUE ( IssueData[Closed Date].[Month] ),
VALUES(IssueData[Closed Date].[Month])
)``````
Frequent Visitor
I was able to solve this a slightly different way. In my own example I needed Last six complete months. To create the date ranges I was looking for, I created a seperate table and related that back to my calendar. This also gave me the ability to utilize both a drop down, and a Date Slicer that was only active when "Custom" was choosen in the drop down.
``````Date Periods =
Union(
DATESMTD('Calendar'[Dates]), "Type", "MTD", "Order", 1
),
DATESQTD('Calendar'[Dates]), "Type", "QTD", "Order", 2
),
DATESYTD('Calendar'[Dates]), "Type", "YTD", "Order", 3
),
PREVIOUSMONTH(DATESMTD('Calendar'[Dates])), "Type", "Last Month", "Order", 4
),
Previousquarter(DATESQTD('Calendar'[Dates])), "Type", "Last QTR", "Order", 5
),
PREVIOUSYEAR(DATESYTD('Calendar'[Dates])), "Type", "Last Year", "Order", 6
),
DATESINPERIOD('Calendar'[Dates], TODAY() - 30, 30, Day), "Type", "Last 30 Days", "Order", 7
),
DATESINPERIOD('Calendar'[Dates], TODAY() - 90, 90, Day), "Type", "Last 90 Days", "Order", 8
),
DATESINPERIOD('Calendar'[Dates], EOMONTH(TODAY(), -1), -6, MONTH), "Type", "Last 6 Months", "Order", 9
),
CALENDAR(MIN('Calendar'[Dates]), MAX('Calendar'[Dates])), "Type", "Custom", "Order", 10
)
)``````
New Member
Hi, when i run this code, i get the following error
"The expression refers to multiple columns. Multiple columns cannot be converted to a scalar value."
Can you please give an example - with the table. I am not able to understand how addcolumns is working here.
Anonymous
Not applicable
Elegant solution! Exactly the sort of thing I was working on. Thanks for sharing.
New Member
This 1 line solution did the job for me:
`Last 6 Months = IF(LedgerMovementFin[Posting Period] > EDATE(TODAY(), -6), LedgerMovementFin[Amount], 0)`
Anonymous
Not applicable
This worked for me as well, but I would like to know why. Could you explain Edate fuction? I couldn't actually find documentation on it.
Memorable Member
@TannerBuck7 You can try add a calculated column with something similar adjusted to your needs, also use it as slicer or filter in formulas
When refresh it auto adjust the periods based on TODAY()
```Date Periods =
VAR Datediff =
1
* ( 'Calendar'[Date] - TODAY () )
RETURN
SWITCH (
TRUE;
AND ( Datediff <= 0; Datediff >= -180 ); "Last 6 Months";
Datediff < 180; "Older than 6 Months"
)```
Konstantinos Ioannou
Microsoft
If I am understanding this correctly the formula will just return data for the last 180 days right? The problem is that I need to be able to see data for every day of all 6 months (which will usually be more than 180 days) - with this formula it will not include some days of the first month depending on the current date.
Let me know If I misunderstood the formula or If I need to further clarify the issue at hand.
Microsoft
I want to create a calculated column that returns true or false depending on whether it is within the 6 month range or not.
I tried this: Last6Months = IF(MONTH('Date') >= (MONTH('Date') -6), "True", "False")
but it returned this error: "The Error refers to multiple columns. Multiple columns cannot be converted to a scalar value."
Memorable Member
@TannerBuck7 Not sure what you are trying. Can you define the period of 6 months?
In general when we say last six months we mean -180 days so we have standard period.
For example today is July 21st do you want to show February to today & when change to 1st of August show March to August?
Regarding your formula change it to ( which is 6 calendar months before & the future also since is bigger than
` `
`Last6months =IF(MONTH('Calendar'[DateKey]) >=MONTH(TODAY()) -6;TRUE();FALSE())`
If you try to describe with more details what are you trying to achive & what tables you have would help
Konstantinos Ioannou
Microsoft
Yes so since we are in July I would want to show data from February 1st to today, and as soon as we get to August 1st I would want the data to automatically change to show March 1st to the current date in August.
Memorable Member
@TannerBuck7 is a bit late so it must be a more efficient way but for now
Specially to avoid when year changes you will need a monthindex
Create a YearMonth Column ( if you don't have )
```YearMonth =
('Calendar'[YearKey] * 100 ) + MONTH('Calendar'[DateKey])```
Then create the month index column
```MonthIndex =
VAR MonthRow = 'Calendar'[YearMonth]
RETURN
CALCULATE (
DISTINCTCOUNT ( 'Calendar'[YearMonth] );
FILTER ( 'Calendar'; 'Calendar'[YearMonth] <= MonthRow )
)```
and finally the Last6Months
```Last6Months =
VAR TodayMonthIndex =
CALCULATE (
MAX ( 'Calendar'[MonthIndex] );
FILTER ( 'Calendar'; TODAY () = 'Calendar'[DateKey] )
)
VAR monthtocheck = Calendar[MonthIndex]
RETURN
IF (
AND ( monthtocheck >= TodayMonthIndex - 5; monthtocheck <= TodayMonthIndex );
1;
0
)```
More steps but when the year change you will still see six months before
Hope it helps
Konstantinos Ioannou
Anonymous
Not applicable
Hello,
How would I change this to show prior 6 months but starting from end of last month. For example 12/1/2019 - 05/31/2020.
Frequent Visitor
You can change this
AND ( monthtocheck >= TodayMonthIndex - 5, monthtocheck <= TodayMonthIndex )
to
AND ( monthtocheck >= TodayMonthIndex - 6, monthtocheck < TodayMonthIndex )
Frequent Visitor
Thanks for the comments, it is very valueable.
When i use this solution on my table, it sums all the values.
In my situation i want to have the last value available for the 6 different months
I used in a calculated table the following calculated column:
MaxDate = MAXX(RELATEDTABLE(TestBalances);TestBalances[Date])
The only problem here is that it takes the very last value, of a range of dates.
I want to have the last value of a month
date table sample:
date yearmonthnumber
01-jan-2016 2016_01
02-jan-2016 2016_01
..
31-jan-2016 2016_01
01-feb-2016 2016_02
...
balance table sample:
date balance type
01-jan-2016 2000 Green
15-jan-2016 2005 Green
26-jan-2016 2009 Green
01-feb-2016 2100 Green
15-feb-2016 2105 Green
24-feb-2016 2109 Green
01-mch-2016 2200 Green
15-mch-2016 2205 Green
17-mch-2016 2209 Green
The result that I want to have is:
26-jan-2016 2009 Green
24-feb-2016 2109 Green
17-mch-2016 2209 Green
Important note: when there is no value of the balance for a month, it should take the last available balance.
Any suggestions?
John
Microsoft
Thank you for your prompt response.
If I understand this correctly won't this mean that the formula will just return data for the last 180 days? The problem with this strategy is that I need month totals for every month in the last 6 month time range. If I just return the last 180 days it will only give me data for part of the first month in the time range, depending on what day of the month the 180 days start at.
Let me know If I misunderstood your solution or if I need to clarify what I am trying to do.
Announcements
#### Power BI September 2023 Update
Take a look at the September 2023 Power BI update to learn more.
#### Learn Live: Event Series
Join Microsoft Reactor and learn from developers.
#### Exclusive opportunity for Women!
Join us for a free, hands-on Microsoft workshop led by women trainers for women where you will learn how to build a Dashboard in a Day!
#### Power Platform Conference-Power BI and Fabric Sessions
Join us Oct 1 - 6 in Las Vegas for the Microsoft Power Platform Conference.
Top Solution Authors
Top Kudoed Authors | 2,799 | 10,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-40 | latest | en | 0.810934 |
https://cs.stackexchange.com/questions/153445/algorithm-for-maximum-non-crossing-edge-set-in-bipartite-graph-with-a-fixed-perm | 1,660,425,220,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571987.60/warc/CC-MAIN-20220813202507-20220813232507-00591.warc.gz | 198,072,362 | 65,117 | # Algorithm for maximum non-crossing edge set in bipartite graph with a fixed permutation
I'm trying to identify an algorithm to solve this computational problem
Input:
• Bipartite graph (V, W, E), with E ⊆ V×W
• A fixed order for both V and W: V = (v1, ..., vn) and W = (w1, ..., wm)
Output:
• The subset of E of maximum cardinality in which there are no crossing edges.
The mental image is that the two vertex sets are arranged in two parallel arrays with edges between them, which may or may not cross. More formally, edges (vk, wl) and (vp, wq) cross if either 1) k < l and p > q, or 2) k > l and p < q.
Does anyone know about any published literature on this problem?
It seems that there is a much larger body of literature on the problem of choosing permutations of V and W to minimize edge crossing, which is NP-Hard. However, I'm working with a fixed permutation for both sets.
I've also identified a few papers that count the number of edge crossings under a fixed permutation:
However, none of these papers discuss the problem of minimizing edge crossings by removing edges.
There are a few more papers in which the algorithm selects a maximum non-crossing matching under a fixed permutation.
This is closer to the mark since it's maximizing the cardinality of an edge set, but I don't want to restrict the edge set to being a matching. Any leads?
Without loss of generality, assume every vertex has degree 1. Create $$\deg(u)$$ copies of a vertex $$u$$, then sort the created vertices in the order of the opposite vertex of the corresponding edge.
Then, the bipartite graph can be represented by a permutation as in permutation graphs. Now, a set of edges in the bipartite graph is non-crossing if and only if the corresponding sub-sequence of the permutation is increasing. This longest increasing subsequence problem is solvable in $$O(n \log n)$$ time, therefore the original problem is solvable in $$O((|V| + |W| + |E|) \log (|V| + |W| + |E|))$$ time.
• That's perfect. Thanks! Aug 7 at 17:31 | 493 | 2,022 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-33 | longest | en | 0.941908 |
https://www.proprofs.com/quiz-school/story.php?title=mental-rotation-task_8jo | 1,713,673,273,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00206.warc.gz | 869,520,378 | 99,673 | # Mental Rotation Trivia Questions
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Romained
R
Romained
Community Contributor
Quizzes Created: 1 | Total Attempts: 520
Questions: 10 | Attempts: 523
Settings
Choose the figure which is identical to the original figure, aside from its orientation. Take your time, this is not timed. I'm looking at accuracy depending upon gender.
• 1.
### A B C DWhich image is the same as the original image, aside from its orientation?
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
B. Figure B
Explanation
The question asks for the image that is the same as the original image, aside from its orientation. This means that the image should have the same content and shape as the original image, but may be rotated or flipped. Looking at the options, Figure B is the only image that has the same content and shape as the original image, but is rotated 90 degrees counterclockwise. Therefore, Figure B is the correct answer.
Rate this question:
• 2.
### A B C DWhich image is the same as the original image, aside from its orientation?
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
A. Figure A
C. Figure C
Explanation
The question asks which image is the same as the original image, aside from its orientation. This means we need to find the image that has the same content as the original image, but may be rotated or flipped. The correct answer states that Figure A and Figure C are the same as the original image.
Rate this question:
• 3.
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
A. Figure A
• 4.
### A B C DWhich image is the same as the original image, aside from its orientation?
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
D. Figure D
Explanation
The question asks which image is the same as the original image, aside from its orientation. Since the original image is not specified, we can assume that it is not shown among the options. Therefore, the correct answer is Figure D, as it is the only option that is not the original image but has a different orientation.
Rate this question:
• 5.
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
C. Figure C
• 6.
### A B C DWhich image is the same as the original image, aside from its orientation?
• A.
Figure B
• B.
Figure C
• C.
Figure D
• D.
Figure A
A. Figure B
Explanation
The question asks which image is the same as the original image, aside from its orientation. This means that we need to find an image that is identical to the original image, but has been rotated or flipped. Looking at the options, Figure B is the only image that matches this description. It is the same as the original image, but it has been rotated 90 degrees clockwise. Therefore, Figure B is the correct answer.
Rate this question:
• 7.
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
A. Figure A
C. Figure C
• 8.
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
B. Figure B
• 9.
### A B C DWhich image is the same as the original image, aside from its orientation?
• A.
Figure A
• B.
Figure B
• C.
Figure C
• D.
Figure D
A. Figure A
Explanation
Figure A is the same as the original image, aside from its orientation.
Rate this question:
• 10.
• A.
Male
• B.
Female
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Mar 22, 2023
Quiz Edited by
ProProfs Editorial Team
• Oct 21, 2016
Quiz Created by
Romained
Related Topics | 1,043 | 4,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.904707 |
http://www.ehow.co.uk/how_7575961_calculate-absolute-neutrophil-count.html | 1,490,914,918,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218203536.73/warc/CC-MAIN-20170322213003-00098-ip-10-233-31-227.ec2.internal.warc.gz | 501,932,381 | 17,438 | # How to calculate absolute neutrophil count
Written by richard barker, dvm
• Share
• Tweet
• Share
• Pin
• Email
White blood cells, or WBCs, fight infection within the body. The blood WBC count increases or decreases in response to infection and conditions such as cancer or the use of certain drugs. The relative and absolute numbers of different types of blood WBCs often reflect the type, duration or severity of disease. A number of factors -- including bacterial infections, cancer, stress and chemotherapy -- cause an increase or decrease in numbers of blood neutrophils, a type of WBC. Laboratories report blood neutrophil levels as a percentage of the total WBC count and as a neutrophil concentration or absolute neutrophil count. Information needed to calculate the absolute neutrophil count includes the WBC count and the percentage of WBCs that are neutrophils.
Skill level:
Easy
### Things you need
• WBC count
• Neutrophil count
• Band count
• Calculator
Show MoreHide
## Instructions
1. 1
Obtain an absolute WBC count and the percentage of the WBC count attributed to each different type of WBC from your doctor. Laboratories perform either a machine or manual analysis of a blood sample to attain these numbers, often reported as a Complete Blood Count with Differential, or CBC with Differential.
2. 2
Add together the percentages of segmented neutrophil cells, often listed as neutrophils or polys, and band neutrophils, usually listed as bands. Segmented neutrophils are mature neutrophils; bands are immature neutrophils. The sum represents the total percentage of blood WBC's that are neutrophils.
3. 3
Multiply the sum by the WBC count, then divide by 100. The result will give the absolute neutrophil count. Report the result as the number of neutrophils per microliter of blood.
For example, if the WBC is 10,000 cells per microliter, the percentage of neutrophils is 10 and of bands is five, then the absolute neutrophil count is [(10 + 5) x 10,000]/100, or 1500 neutrophils per microliter of blood.
#### Tips and warnings
• Use the correct number for WBC; WBC is often reported as a number times ten to the third power. In the example above, the WBC would be reported as 10.
• Most laboratories report the absolute neutrophil count in the CBC with Differential results. Calculation is not needed in these cases.
### Don't Miss
#### Resources
• All types
• Articles
• Slideshows
• Videos
##### Sort:
• Most relevant
• Most popular
• Most recent
By using the eHow.co.uk site, you consent to the use of cookies. For more information, please see our Cookie policy. | 614 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | latest | en | 0.906448 |
http://diyourself.ru/technology/how-to-supercharge-a-cheap-radio-control-2-do-it.html | 1,521,438,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646375.29/warc/CC-MAIN-20180319042634-20180319062634-00118.warc.gz | 83,049,720 | 3,860 | # Do-It-Yourself How to supercharge a cheap radio control 2 - do it yourself
This is a more descriptive instructible than the last one about making a faster radio control. This instructible is only for RCs that are 2 wheel drive or only have 2 motors. (If you are not entertained by it as a slow rc and don't mind the possibility that if you mess up too much then your rc could be broken then continue. If you reduce your RC's value I am not responsible.) First measure the gear box and get an approximate length of the area you have for the motor. If you can not see if you DO have about an inch and a half from the motor edge of the tire then go to this instructible, for instructions on how to mount a bigger motor; if you want a harder job that will make a faster RC if you use the right motor.(http://www.instructables.com/id/how-to-supercharge-a-cheap-radio-control/)
## Step 1: MORE POWER
Then, find out the voltage the car uses also find the Mah rating if it is rechargeable. If the voltage is 6 volts and under then double the voltage in battery rating is required. If it is 6 volts and over then give it 5-6 more volts of battery rating. If the RC is alkaline battery powered you will need to make it rechargeable battery powered. If your batteries are not rated at at least 1000 Mah then double the Mah is needed of the batteries you will use in the car.
## Step 2: Batteries in series
To get the voltage you figured you need you must put the batteries in series (do not do this with lithium batteries). Like the pucture conect the negative to the positive. Then connect an additional wire to the negative. And one to the positive if there is not already one there. The wires labeled b are the ones already attached to the circuit board.
## Step 3: Soldering
Here is the relay I use. ( http://www.instructables.com/id/relay-that-clears-pulses/) Cut the wires connected to the motor and connect them to the relay's motor part. Connect the wires labeled r on the first picture, to the N and P on the relay second picture.(n to negative wire p to positive wire) The black lines on the second picture are the wires connecting parts of the same relay as in the instructible above. Then connect the wires labeled M+ and M- to the motor's negative and positive wires you cut. | 527 | 2,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-13 | latest | en | 0.91186 |
https://math.answers.com/math-and-arithmetic/Is_87_a_multiple_of_four | 1,723,239,325,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00099.warc.gz | 311,249,364 | 48,010 | 0
# Is 87 a multiple of four?
Updated: 7/25/2024
Wiki User
13y ago
NO!!!
By the very fact that it is an ODD NUmber, then it is not a multiple of '4'.
lenpollock
Lvl 16
2w ago
Wiki User
13y ago
4 is not a multiple of 87, though it is a multiple of 88.
Earn +20 pts
Q: Is 87 a multiple of four?
Submit
Still have questions?
Related questions
### Is 87 a multiple of 6?
No, 87 is not a multiple of 6.
### What is the least common multiple of 87 and 29?
Since 87 is a multiple of 29, it is automatically the LCM.
### What is the LCM of 87 and 3?
Since 87 is a multiple of 3, it is automatically the LCM.
### What is the least common multiple of 5 and 87?
The LCM of 5 and 87 is 435.
### What is the LCM of 116 and 87?
Least Common Multiple (LCM) for 116 87 is 348.
### What is 87 percent of four hundred thousand?
87% of 400,000 is 348,000
### Is 87 in the table of four?
No, because 4 does not divide equally into 87.
2
### What number is divisible by 87?
Take any multiple of 87, and you get a number that is divisible by 87 - for example:87 x 0 87 x 1 87 x 2 87 x 3 ... 87 x (-1) 87 x (-2) ...
### Is the least common multiple of 21 70 and 87 5990?
No. The LCM of 21 70 and 87 is 6,090.
Four: 1 3 29 87.
87 | 418 | 1,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.904722 |
https://www.mycollegehive.com/q.php?q=QUjUEZXSJLh735XDeCZK | 1,610,730,598,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00229.warc.gz | 995,989,071 | 16,232 | mycollegehive
Given that y varies inversely as the square of x. If x =3 wh...
waec-2018 ssce-exam-2018 math waec-mathematics-2018
0
Given that y varies inversely as the square of x. If x =3 when y =100. Find the equation connecting x and y
A: yx2 = 300
B: yx2 = 900
C: y=100x/9
D: y=900x2
98 views
Share Follow
University of Benin Nigeria
31 August 2020
University of Benin Nigeria
06 November 2020
0
y ∝ 1/x2
y = k/x2 ------------- eqn(1)
where k is the proportionality constant
from eqn(1), k = yx2 ------------ eqn(2)
if x = 3 and y = 200,
k = 100(3)2
= 900
Therefore, the equation connecting x and y will be:
from eqn(1),
y = 900/x2
yx2 = 900
Share
### Groups
How to insert math formulas/equations
### Related Tags
waec-2018
1 followers
29 questions
ssce-exam-2018
1 followers
29 questions
math
13 followers
1260 questions
waec-mathematics-2018
1 followers
28 questions | 312 | 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} | 3.390625 | 3 | CC-MAIN-2021-04 | latest | en | 0.731346 |
https://www.cake.co/conversations/3pxPLxl/deck-sizes-in-card-games/4NgPb5l | 1,566,436,172,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316555.4/warc/CC-MAIN-20190822000659-20190822022659-00555.warc.gz | 755,764,644 | 28,986 | Cake
• Oh yea I forgot about that. In comparison, Hearthstone will let you pick up to 2 of each card except for legendaries of which you can only have 1.
• One thing I need to add is that these game (Eternal, Magic as well I think) allow you to have up to four copies of the same card in your deck.
I know nothing about Magic or similar card games, but I saw that mathematics was one of the topics assigned and I was intrigued by the question
When does it make sense to have more than 75 cards?
It sometimes helps to mess around with a smaller or simpler problem and then apply the rule(s) discovered to the actual problem.
Letβs say you have three cards labeled π¦, π, π
I could label them 1,2,3 or a,b,c but I like using emojis.
If the synergy comes from dealing the π¦ card first followed by the π card, then there is 1 out of 6 possibilities that you will get dealt that sequence:
π¦π
π¦π
ππ¦
ππ
ππ¦
ππ
So 1:6
Another way to look at it is that you had 3 possibilities for the first card you drew and two possibilities for the second card you drew.
And 3 times 2 equals 6 combinations. If you had 4 cards, youβd have 4 possibilities for the first card times 3 possibilities for the second card, which would equal 12 combinations.
So 76 cards would equal 76*75=5,700 possible combinations.
Back to the 3 cards. What if the sequence dealt didnβt matter for synergy, that is
π¦π has the same value as
ππ¦
Then in a 3 card deck there are 2 successful draws out of 6 possible combinations. So 2:6 or 1:3.
Your chances are doubled if the order doesnβt matter.
What if your 3 card deck had two of the same card? That is your 3 cards are π¦ππ.
π¦π
π¦π
ππ¦
ππ
ππ¦
ππ
If order doesnβt matter, you have 4 successful chances out of 6 or 2:3.
If there are π cards that can be combined with a number of different cards
π¦π
ππ¦
ππ
to create synergies, then the value of adding the π card goes up even more.
Hope that was both understandable and somewhat helpful.
• @apm I think yours is a good example to show the general problem we're talking about, if expanded slightly.
In that example
> our minimum deck size is three
> we're drawing a hand of two cards
> drawing {duck, pizza} is a good event, drawing {X, cake} in our initial hand is bad, but we still want to draw cake at some point.
If our deck consists of just {duck, pizza, cake}, then as you point out we have a probability of 1/3 to get the good event. We have a 2/3 probability of immediately drawing {X, cake} - but we only want cake later.
Replacing cake with another pizza (deck is {duck, pizza, pizza}), we increase the "good draw" probability to 2/3, but at the same time we decrease our probability of drawing cake to 0 - and we want to eventually draw cake.
So, if instead of replacing cake with pizza, we add another pizza on top of that - making our deck {duck, pizza, pizza, cake} - we will still have a 1/3 probability of drawing {duck, pizza}, while our probability of drawing cake early is reduced to 1/2 while still being available at some point.
Adding another duck for {duck, duck, pizza, pizza, cake} increases our "good draw" probability to 2/5, while decreasing the probability for an early cake draw to also 2/5. On the other hand, the fact that we've added another card also means that there's a 1/5 probability of cake being the fifth card in our deck. If we're only ever drawing four cards, this can mean that we're not drawing cake at all
So, in this toy example, depending on what value we want to give to the events "initial good draw", "eventual cake draw" and "no cake draw at all", it might be sensible (or not) to play with a deck consisting of four or five cards instead of just three.
At the same time though, we're making pretty big assumptions again and stay very vague in some regards, so I'm not convinced that this outcome can just be generalized to any deck size and any value for synergistic effects between cards.
• @Eric Looking at this and squinting hard, it looks as if the mathematically proper way to deal with this would be to do a discrete sum over all possible deck permutations and positions ("position" meaning number of cards drawn over time) using the value you suggest, then dividing by some factor that depends on deck size, to get a closed formula for something like an Average Deck Value (ADV).
Then, manipulating the inequality ADV(deck_small) < ADV(deck_large) (basically stating that we're looking for a situation where the larger deck is better than the smaller deck) to remove constant factors etc. might lead to some more clues under which circumstances this can be true, if at all.
• @Factotum I was thinking about trying to have a small card set example. Suppose there is no synergy and you have 6 cards with their card value equal to their number, 1, 2, ... 6. You could create a deck with 5 cards D1, and you could create a deck with 6 cards D2. So the logical way to build D1 is to put cards 2 through 6 in the deck, and leave out card 1 which has the lowest value.
In this hypothetical situation we start with 0 cards in our hand. Then the draw potential for each deck would be:
```D1 draw potential = (2 + 3 + 4 + 5 + 6) / 5 = 4
D2 draw potential = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5```
If you could generalize this to where you are comparing a deck of size n vs n+1, where the deck of n+1 has all of the exact same cards, plus one more which is guaranteed to be smaller, we want to prove something like:
sum(n): the sum of value in deck 1 (avg value * n)
c: an n + 1 card that is lower value than the lowest value in the deck
of n cards.
prove: draw potential for the larger deck is worse than the smaller deck.
```1: sum(n) / n > (sum(n) + c) / (n + 1)
2: (n + 1) * sum(n) > (sum(n) + c) * n
3: n * sum (n) + sum(n) > n * sum(n) + c * n
4: sum(n) > c * n
5: sum(n) / n > c```
Seen in this form, we can see that as long as the card c has a value less than the average value of the smaller deck, then it reduces the average value of the larger deck. Meaning lower draw potential.
I think this can be generalized to include synergy as well, it just makes things more complicated.
Edit: Math
• @Eric Thanks. I think this is similar to the original reasoning of just cutting the "worst" card from your oversized deck. This is definitely correct for card games without synergy.
In the meantime, I've thought about your last post and my response to it. One thing led to another, and now I have this crazy formula for the "Average Deck Value"... I think I've got some explaining to do! :D | 1,818 | 6,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2019-35 | longest | en | 0.943947 |
https://aimsciences.org/article/doi/10.3934/ipi.2012.6.133 | 1,566,608,022,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027319155.91/warc/CC-MAIN-20190823235136-20190824021136-00206.warc.gz | 352,191,174 | 14,742 | # American Institute of Mathematical Sciences
February 2012, 6(1): 133-146. doi: 10.3934/ipi.2012.6.133
## The order of convergence for Landweber Scheme with $\alpha,\beta$-rule
1 Department of Mathematics, Shanghai Maritime University, Shanghai 200135, China 2 LMAM, School of Mathematical Sciences, Peking University, Beijing 100871, China
Received February 2011 Revised September 2011 Published February 2012
The Landweber scheme is widely used in various image reconstruction problems. In previous works, $\alpha,\beta$-rule is suggested to stop the Landweber iteration so as to get proper iteration results. The order of convergence of discrepancy principal (DP rule), which is a special case of $\alpha,\beta$-rule, with constant relaxation coefficient $\lambda$ satisfying $0<\lambda\sigma_1^2<1,~(\|A\|_{V,W}=\sigma_1>0)$ has been studied. A sufficient condition for convergence of Landweber scheme is that the value $\lambda_m\sigma_1^2$ should be lied in a closed interval, i.e. $0<\varepsilon\leq\lambda_m\sigma_1^2\leq2-\varepsilon$, $(0<\varepsilon<1)$. In this paper, we mainly investigate the order of convergence of the $\alpha,\beta$-rule with variable relaxation coefficient $\lambda_m$ satisfying $0 < \varepsilon\leq\lambda_m \sigma_1^2 \leq 2-\varepsilon$. According to the order of convergence, we can conclude that $\alpha,\beta$-rule is the optimal rule for the Landweber scheme.
Citation: Caifang Wang, Tie Zhou. The order of convergence for Landweber Scheme with $\alpha,\beta$-rule. Inverse Problems & Imaging, 2012, 6 (1) : 133-146. doi: 10.3934/ipi.2012.6.133
##### References:
show all references
##### References:
[1] Matania Ben–Artzi, Joseph Falcovitz, Jiequan Li. The convergence of the GRP scheme. Discrete & Continuous Dynamical Systems - A, 2009, 23 (1&2) : 1-27. doi: 10.3934/dcds.2009.23.1 [2] Maurizio Grasselli, Morgan Pierre. Convergence to equilibrium of solutions of the backward Euler scheme for asymptotically autonomous second-order gradient-like systems. Communications on Pure & Applied Analysis, 2012, 11 (6) : 2393-2416. doi: 10.3934/cpaa.2012.11.2393 [3] Josef Diblík, Klara Janglajew, Mária Kúdelčíková. An explicit coefficient criterion for the existence of positive solutions to the linear advanced equation. Discrete & Continuous Dynamical Systems - B, 2014, 19 (8) : 2461-2467. doi: 10.3934/dcdsb.2014.19.2461 [4] Stefan Kindermann, Andreas Neubauer. On the convergence of the quasioptimality criterion for (iterated) Tikhonov regularization. Inverse Problems & Imaging, 2008, 2 (2) : 291-299. doi: 10.3934/ipi.2008.2.291 [5] Jiann-Sheng Jiang, Kung-Hwang Kuo, Chi-Kun Lin. Homogenization of second order equation with spatial dependent coefficient. Discrete & Continuous Dynamical Systems - A, 2005, 12 (2) : 303-313. doi: 10.3934/dcds.2005.12.303 [6] Benoît Merlet, Morgan Pierre. Convergence to equilibrium for the backward Euler scheme and applications. Communications on Pure & Applied Analysis, 2010, 9 (3) : 685-702. doi: 10.3934/cpaa.2010.9.685 [7] François Dubois. Third order equivalent equation of lattice Boltzmann scheme. Discrete & Continuous Dynamical Systems - A, 2009, 23 (1&2) : 221-248. doi: 10.3934/dcds.2009.23.221 [8] Bertram Düring, Daniel Matthes, Josipa Pina Milišić. A gradient flow scheme for nonlinear fourth order equations. Discrete & Continuous Dynamical Systems - B, 2010, 14 (3) : 935-959. doi: 10.3934/dcdsb.2010.14.935 [9] Desmond J. Higham, Xuerong Mao, Lukasz Szpruch. Convergence, non-negativity and stability of a new Milstein scheme with applications to finance. Discrete & Continuous Dynamical Systems - B, 2013, 18 (8) : 2083-2100. doi: 10.3934/dcdsb.2013.18.2083 [10] Bahareh Akhtari, Esmail Babolian, Andreas Neuenkirch. An Euler scheme for stochastic delay differential equations on unbounded domains: Pathwise convergence. Discrete & Continuous Dynamical Systems - B, 2015, 20 (1) : 23-38. doi: 10.3934/dcdsb.2015.20.23 [11] Xinfu Chen, Bei Hu, Jin Liang, Yajing Zhang. Convergence rate of free boundary of numerical scheme for American option. Discrete & Continuous Dynamical Systems - B, 2016, 21 (5) : 1435-1444. doi: 10.3934/dcdsb.2016004 [12] Hedy Attouch, Alexandre Cabot, Zaki Chbani, Hassan Riahi. Rate of convergence of inertial gradient dynamics with time-dependent viscous damping coefficient. Evolution Equations & Control Theory, 2018, 7 (3) : 353-371. doi: 10.3934/eect.2018018 [13] Harbir Antil, Mahamadi Warma. Optimal control of the coefficient for the regional fractional $p$-Laplace equation: Approximation and convergence. Mathematical Control & Related Fields, 2019, 9 (1) : 1-38. doi: 10.3934/mcrf.2019001 [14] Wenqing Bao, Xianyi Wu, Xian Zhou. Optimal stopping problems with restricted stopping times. Journal of Industrial & Management Optimization, 2017, 13 (1) : 399-411. doi: 10.3934/jimo.2016023 [15] Huijiang Zhao, Yinchuan Zhao. Convergence to strong nonlinear rarefaction waves for global smooth solutions of $p-$system with relaxation. Discrete & Continuous Dynamical Systems - A, 2003, 9 (5) : 1243-1262. doi: 10.3934/dcds.2003.9.1243 [16] Anna Maria Cherubini, Giorgio Metafune, Francesco Paparella. On the stopping time of a bouncing ball. Discrete & Continuous Dynamical Systems - B, 2008, 10 (1) : 43-72. doi: 10.3934/dcdsb.2008.10.43 [17] Maike Schulte, Anton Arnold. Discrete transparent boundary conditions for the Schrodinger equation -- a compact higher order scheme. Kinetic & Related Models, 2008, 1 (1) : 101-125. doi: 10.3934/krm.2008.1.101 [18] Yongchao Liu, Hailin Sun, Huifu Xu. An approximation scheme for stochastic programs with second order dominance constraints. Numerical Algebra, Control & Optimization, 2016, 6 (4) : 473-490. doi: 10.3934/naco.2016021 [19] Tadahisa Funaki, Yueyuan Gao, Danielle Hilhorst. Convergence of a finite volume scheme for a stochastic conservation law involving a $Q$-brownian motion. Discrete & Continuous Dynamical Systems - B, 2018, 23 (4) : 1459-1502. doi: 10.3934/dcdsb.2018159 [20] Arnaud Debussche, Jacques Printems. Convergence of a semi-discrete scheme for the stochastic Korteweg-de Vries equation. Discrete & Continuous Dynamical Systems - B, 2006, 6 (4) : 761-781. doi: 10.3934/dcdsb.2006.6.761
2018 Impact Factor: 1.469 | 1,941 | 6,233 | {"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} | 2.59375 | 3 | CC-MAIN-2019-35 | latest | en | 0.773519 |
https://en.wikipedia.org/wiki/History_of_geometry | 1,713,692,431,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.87/warc/CC-MAIN-20240421071342-20240421101342-00399.warc.gz | 210,774,431 | 58,366 | # History of geometry
Geometry (from the Ancient Greek: γεωμετρία; geo- "earth", -metron "measurement") arose as the field of knowledge dealing with spatial relationships. Geometry was one of the two fields of pre-modern mathematics, the other being the study of numbers (arithmetic).
Classic geometry was focused in compass and straightedge constructions. Geometry was revolutionized by Euclid, who introduced mathematical rigor and the axiomatic method still in use today. His book, The Elements is widely considered the most influential textbook of all time, and was known to all educated people in the West until the middle of the 20th century.[1]
In modern times, geometric concepts have been generalized to a high level of abstraction and complexity, and have been subjected to the methods of calculus and abstract algebra, so that many modern branches of the field are barely recognizable as the descendants of early geometry. (See Areas of mathematics and Algebraic geometry.)
## Early geometry
The earliest recorded beginnings of geometry can be traced to early peoples, such as the ancient Indus Valley (see Harappan mathematics) and ancient Babylonia (see Babylonian mathematics) from around 3000 BC. Early geometry was a collection of empirically discovered principles concerning lengths, angles, areas, and volumes, which were developed to meet some practical need in surveying, construction, astronomy, and various crafts. Among these were some surprisingly sophisticated principles, and a modern mathematician might be hard put to derive some of them without the use of calculus and algebra. For example, both the Egyptians and the Babylonians were aware of versions of the Pythagorean theorem about 1500 years before Pythagoras and the Indian Sulba Sutras around 800 BC contained the first statements of the theorem; the Egyptians had a correct formula for the volume of a frustum of a square pyramid.
### Egyptian geometry
The ancient Egyptians knew that they could approximate the area of a circle as follows:[2]
Area of Circle ≈ [ (Diameter) x 8/9 ]2.
Problem 50 of the Ahmes papyrus uses these methods to calculate the area of a circle, according to a rule that the area is equal to the square of 8/9 of the circle's diameter. This assumes that π is 4×(8/9)2 (or 3.160493...), with an error of slightly over 0.63 percent. This value was slightly less accurate than the calculations of the Babylonians (25/8 = 3.125, within 0.53 percent), but was not otherwise surpassed until Archimedes' approximation of 211875/67441 = 3.14163, which had an error of just over 1 in 10,000.
Ahmes knew of the modern 22/7 as an approximation for π, and used it to split a hekat, hekat x 22/x x 7/22 = hekat;[citation needed] however, Ahmes continued to use the traditional 256/81 value for π for computing his hekat volume found in a cylinder.
Problem 48 involved using a square with side 9 units. This square was cut into a 3x3 grid. The diagonal of the corner squares were used to make an irregular octagon with an area of 63 units. This gave a second value for π of 3.111...
The two problems together indicate a range of values for π between 3.11 and 3.16.
Problem 14 in the Moscow Mathematical Papyrus gives the only ancient example finding the volume of a frustum of a pyramid, describing the correct formula:
${\displaystyle V={\frac {1}{3}}h(a^{2}+ab+b^{2})}$
where a and b are the base and top side lengths of the truncated pyramid and h is the height.
### Babylonian geometry
The Babylonians may have known the general rules for measuring areas and volumes. They measured the circumference of a circle as three times the diameter and the area as one-twelfth the square of the circumference, which would be correct if π is estimated as 3. The volume of a cylinder was taken as the product of the base and the height, however, the volume of the frustum of a cone or a square pyramid was incorrectly taken as the product of the height and half the sum of the bases. The Pythagorean theorem was also known to the Babylonians. Also, there was a recent discovery in which a tablet used π as 3 and 1/8. The Babylonians are also known for the Babylonian mile, which was a measure of distance equal to about seven miles today. This measurement for distances eventually was converted to a time-mile used for measuring the travel of the Sun, therefore, representing time.[3] There have been recent discoveries showing that ancient Babylonians may have discovered astronomical geometry nearly 1400 years before Europeans did.[4]
### Vedic India geometry
The Indian Vedic period had a tradition of geometry, mostly expressed in the construction of elaborate altars. Early Indian texts (1st millennium BC) on this topic include the Satapatha Brahmana and the Śulba Sūtras.[5][6][7]
According to (Hayashi 2005, p. 363), the Śulba Sūtras contain "the earliest extant verbal expression of the Pythagorean Theorem in the world, although it had already been known to the Old Babylonians."
The diagonal rope (akṣṇayā-rajju) of an oblong (rectangle) produces both which the flank (pārśvamāni) and the horizontal (tiryaṇmānī) <ropes> produce separately."[8]
They contain lists of Pythagorean triples,[9] which are particular cases of Diophantine equations.[10] They also contain statements (that with hindsight we know to be approximate) about squaring the circle and "circling the square."[11]
The Baudhayana Sulba Sutra, the best-known and oldest of the Sulba Sutras (dated to the 8th or 7th century BC) contains examples of simple Pythagorean triples, such as: ${\displaystyle (3,4,5)}$, ${\displaystyle (5,12,13)}$, ${\displaystyle (8,15,17)}$, ${\displaystyle (7,24,25)}$, and ${\displaystyle (12,35,37)}$[12] as well as a statement of the Pythagorean theorem for the sides of a square: "The rope which is stretched across the diagonal of a square produces an area double the size of the original square."[12] It also contains the general statement of the Pythagorean theorem (for the sides of a rectangle): "The rope stretched along the length of the diagonal of a rectangle makes an area which the vertical and horizontal sides make together."[12]
According to mathematician S. G. Dani, the Babylonian cuneiform tablet Plimpton 322 written c. 1850 BC[13] "contains fifteen Pythagorean triples with quite large entries, including (13500, 12709, 18541) which is a primitive triple,[14] indicating, in particular, that there was sophisticated understanding on the topic" in Mesopotamia in 1850 BC. "Since these tablets predate the Sulbasutras period by several centuries, taking into account the contextual appearance of some of the triples, it is reasonable to expect that similar understanding would have been there in India."[15] Dani goes on to say:
"As the main objective of the Sulvasutras was to describe the constructions of altars and the geometric principles involved in them, the subject of Pythagorean triples, even if it had been well understood may still not have featured in the Sulvasutras. The occurrence of the triples in the Sulvasutras is comparable to mathematics that one may encounter in an introductory book on architecture or another similar applied area, and would not correspond directly to the overall knowledge on the topic at that time. Since, unfortunately, no other contemporaneous sources have been found it may never be possible to settle this issue satisfactorily."[15]
In all, three Sulba Sutras were composed. The remaining two, the Manava Sulba Sutra composed by Manava (fl. 750-650 BC) and the Apastamba Sulba Sutra, composed by Apastamba (c. 600 BC), contained results similar to the Baudhayana Sulba Sutra.
## Greek geometry
### Classical Greek geometry
For the ancient Greek mathematicians, geometry was the crown jewel of their sciences, reaching a completeness and perfection of methodology that no other branch of their knowledge had attained. They expanded the range of geometry to many new kinds of figures, curves, surfaces, and solids; they changed its methodology from trial-and-error to logical deduction; they recognized that geometry studies "eternal forms", or abstractions, of which physical objects are only approximations; and they developed the idea of the "axiomatic method", still in use today.
#### Thales and Pythagoras
Thales (635-543 BC) of Miletus (now in southwestern Turkey), was the first to whom deduction in mathematics is attributed. There are five geometric propositions for which he wrote deductive proofs, though his proofs have not survived. Pythagoras (582-496 BC) of Ionia, and later, Italy, then colonized by Greeks, may have been a student of Thales, and traveled to Babylon and Egypt. The theorem that bears his name may not have been his discovery, but he was probably one of the first to give a deductive proof of it. He gathered a group of students around him to study mathematics, music, and philosophy, and together they discovered most of what high school students learn today in their geometry courses. In addition, they made the profound discovery of incommensurable lengths and irrational numbers.
#### Plato
Plato (427-347 BC) was a philosopher, highly esteemed by the Greeks. There is a story that he had inscribed above the entrance to his famous school, "Let none ignorant of geometry enter here." However, the story is considered to be untrue.[16] Though he was not a mathematician himself, his views on mathematics had great influence. Mathematicians thus accepted his belief that geometry should use no tools but compass and straightedge – never measuring instruments such as a marked ruler or a protractor, because these were a workman's tools, not worthy of a scholar. This dictum led to a deep study of possible compass and straightedge constructions, and three classic construction problems: how to use these tools to trisect an angle, to construct a cube twice the volume of a given cube, and to construct a square equal in area to a given circle. The proofs of the impossibility of these constructions, finally achieved in the 19th century, led to important principles regarding the deep structure of the real number system. Aristotle (384-322 BC), Plato's greatest pupil, wrote a treatise on methods of reasoning used in deductive proofs (see Logic) which was not substantially improved upon until the 19th century.
### Hellenistic geometry
#### Euclid
Euclid (c. 325-265 BC), of Alexandria, probably a student at the Academy founded by Plato, wrote a treatise in 13 books (chapters), titled The Elements of Geometry, in which he presented geometry in an ideal axiomatic form, which came to be known as Euclidean geometry. The treatise is not a compendium of all that the Hellenistic mathematicians knew at the time about geometry; Euclid himself wrote eight more advanced books on geometry. We know from other references that Euclid's was not the first elementary geometry textbook, but it was so much superior that the others fell into disuse and were lost. He was brought to the university at Alexandria by Ptolemy I, King of Egypt.
The Elements began with definitions of terms, fundamental geometric principles (called axioms or postulates), and general quantitative principles (called common notions) from which all the rest of geometry could be logically deduced. Following are his five axioms, somewhat paraphrased to make the English easier to read.
1. Any two points can be joined by a straight line.
2. Any finite straight line can be extended in a straight line.
3. A circle can be drawn with any center and any radius.
4. All right angles are equal to each other.
5. If two straight lines in a plane are crossed by another straight line (called the transversal), and the interior angles between the two lines and the transversal lying on one side of the transversal add up to less than two right angles, then on that side of the transversal, the two lines extended will intersect (also called the parallel postulate).
Concepts, that are now understood as algebra, were expressed geometrically by Euclid, a method referred to as Greek geometric algebra.
#### Archimedes
Archimedes (287-212 BC), of Syracuse, Sicily, when it was a Greek city-state, was one of the most famous mathematicians of the Hellenistic period. He is known for his formulation of a hydrostatic principle (known as Archimedes’ principle) and for his works on geometry, including Measurement of the Circle and On Conoids and Spheroids. His work On Floating Bodies is the first known work on hydrostatics, of which Archimedes is recognized as the founder. Renaissance translations of his works, including the ancient commentaries, were enormously influential in the work of some of the best mathematicians of the 17th century, notably René Descartes and Pierre de Fermat.[17]
#### After Archimedes
After Archimedes, Hellenistic mathematics began to decline. There were a few minor stars yet to come, but the golden age of geometry was over. Proclus (410-485), author of Commentary on the First Book of Euclid, was one of the last important players in Hellenistic geometry. He was a competent geometer, but more importantly, he was a superb commentator on the works that preceded him. Much of that work did not survive to modern times, and is known to us only through his commentary. The Roman Republic and Empire that succeeded and absorbed the Greek city-states produced excellent engineers, but no mathematicians of note.
The great Library of Alexandria was later burned. There is a growing consensus among historians that the Library of Alexandria likely suffered from several destructive events, but that the destruction of Alexandria's pagan temples in the late 4th century was probably the most severe and final one. The evidence for that destruction is the most definitive and secure. Caesar's invasion may well have led to the loss of some 40,000-70,000 scrolls in a warehouse adjacent to the port (as Luciano Canfora argues, they were likely copies produced by the Library intended for export), but it is unlikely to have affected the Library or Museum, given that there is ample evidence that both existed later.[18]
Civil wars, decreasing investments in maintenance and acquisition of new scrolls and generally declining interest in non-religious pursuits likely contributed to a reduction in the body of material available in the Library, especially in the 4th century. The Serapeum was certainly destroyed by Theophilus in 391, and the Museum and Library may have fallen victim to the same campaign.
## Classical Indian geometry
In the Bakhshali manuscript, there is a handful of geometric problems (including problems about volumes of irregular solids). The Bakhshali manuscript also "employs a decimal place value system with a dot for zero."[19] Aryabhata's Aryabhatiya (499) includes the computation of areas and volumes.
Brahmagupta wrote his astronomical work Brāhma Sphuṭa Siddhānta in 628. Chapter 12, containing 66 Sanskrit verses, was divided into two sections: "basic operations" (including cube roots, fractions, ratio and proportion, and barter) and "practical mathematics" (including mixture, mathematical series, plane figures, stacking bricks, sawing of timber, and piling of grain).[20] In the latter section, he stated his famous theorem on the diagonals of a cyclic quadrilateral:[20]
Brahmagupta's theorem: If a cyclic quadrilateral has diagonals that are perpendicular to each other, then the perpendicular line drawn from the point of intersection of the diagonals to any side of the quadrilateral always bisects the opposite side.
Chapter 12 also included a formula for the area of a cyclic quadrilateral (a generalization of Heron's formula), as well as a complete description of rational triangles (i.e. triangles with rational sides and rational areas).
Brahmagupta's formula: The area, A, of a cyclic quadrilateral with sides of lengths a, b, c, d, respectively, is given by
${\displaystyle A={\sqrt {(s-a)(s-b)(s-c)(s-d)}}}$
where s, the semiperimeter, given by: ${\displaystyle s={\frac {a+b+c+d}{2}}.}$
Brahmagupta's Theorem on rational triangles: A triangle with rational sides ${\displaystyle a,b,c}$ and rational area is of the form:
${\displaystyle a={\frac {u^{2}}{v}}+v,\ \ b={\frac {u^{2}}{w}}+w,\ \ c={\frac {u^{2}}{v}}+{\frac {u^{2}}{w}}-(v+w)}$
for some rational numbers ${\displaystyle u,v,}$ and ${\displaystyle w}$.[21]
Parameshvara Nambudiri was the first mathematician to give a formula for the radius of the circle circumscribing a cyclic quadrilateral.[22] The expression is sometimes attributed to Lhuilier [1782], 350 years later. With the sides of the cyclic quadrilateral being a, b, c, and d, the radius R of the circumscribed circle is:
${\displaystyle R={\sqrt {\frac {(ab+cd)(ac+bd)(ad+bc)}{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}}.}$
## Chinese geometry
The first definitive work (or at least oldest existent) on geometry in China was the Mo Jing, the Mohist canon of the early philosopher Mozi (470-390 BC). It was compiled years after his death by his followers around the year 330 BC.[23] Although the Mo Jing is the oldest existent book on geometry in China, there is the possibility that even older written material existed. However, due to the infamous Burning of the Books in a political maneuver by the Qin Dynasty ruler Qin Shihuang (r. 221-210 BC), multitudes of written literature created before his time were purged. In addition, the Mo Jing presents geometrical concepts in mathematics that are perhaps too advanced not to have had a previous geometrical base or mathematic background to work upon.
The Mo Jing described various aspects of many fields associated with physical science, and provided a small wealth of information on mathematics as well. It provided an 'atomic' definition of the geometric point, stating that a line is separated into parts, and the part which has no remaining parts (i.e. cannot be divided into smaller parts) and thus forms the extreme end of a line is a point.[23] Much like Euclid's first and third definitions and Plato's 'beginning of a line', the Mo Jing stated that "a point may stand at the end (of a line) or at its beginning like a head-presentation in childbirth. (As to its invisibility) there is nothing similar to it."[24] Similar to the atomists of Democritus, the Mo Jing stated that a point is the smallest unit, and cannot be cut in half, since 'nothing' cannot be halved.[24] It stated that two lines of equal length will always finish at the same place,[24] while providing definitions for the comparison of lengths and for parallels,[25] along with principles of space and bounded space.[26] It also described the fact that planes without the quality of thickness cannot be piled up since they cannot mutually touch.[27] The book provided definitions for circumference, diameter, and radius, along with the definition of volume.[28]
The Han Dynasty (202 BC-220 AD) period of China witnessed a new flourishing of mathematics. One of the oldest Chinese mathematical texts to present geometric progressions was the Suàn shù shū of 186 BC, during the Western Han era. The mathematician, inventor, and astronomer Zhang Heng (78-139 AD) used geometrical formulas to solve mathematical problems. Although rough estimates for pi (π) were given in the Zhou Li (compiled in the 2nd century BC),[29] it was Zhang Heng who was the first to make a concerted effort at creating a more accurate formula for pi. Zhang Heng approximated pi as 730/232 (or approx 3.1466), although he used another formula of pi in finding a spherical volume, using the square root of 10 (or approx 3.162) instead. Zu Chongzhi (429-500 AD) improved the accuracy of the approximation of pi to between 3.1415926 and 3.1415927, with 355113 (密率, Milü, detailed approximation) and 227 (约率, Yuelü, rough approximation) being the other notable approximation.[30] In comparison to later works, the formula for pi given by the French mathematician Franciscus Vieta (1540-1603) fell halfway between Zu's approximations.
### The Nine Chapters on the Mathematical Art
The Nine Chapters on the Mathematical Art, the title of which first appeared by 179 AD on a bronze inscription, was edited and commented on by the 3rd century mathematician Liu Hui from the Kingdom of Cao Wei. This book included many problems where geometry was applied, such as finding surface areas for squares and circles, the volumes of solids in various three-dimensional shapes, and included the use of the Pythagorean theorem. The book provided illustrated proof for the Pythagorean theorem,[31] contained a written dialogue between of the earlier Duke of Zhou and Shang Gao on the properties of the right angle triangle and the Pythagorean theorem, while also referring to the astronomical gnomon, the circle and square, as well as measurements of heights and distances.[32] The editor Liu Hui listed pi as 3.141014 by using a 192 sided polygon, and then calculated pi as 3.14159 using a 3072 sided polygon. This was more accurate than Liu Hui's contemporary Wang Fan, a mathematician and astronomer from Eastern Wu, would render pi as 3.1555 by using 14245.[33] Liu Hui also wrote of mathematical surveying to calculate distance measurements of depth, height, width, and surface area. In terms of solid geometry, he figured out that a wedge with rectangular base and both sides sloping could be broken down into a pyramid and a tetrahedral wedge.[34] He also figured out that a wedge with trapezoid base and both sides sloping could be made to give two tetrahedral wedges separated by a pyramid.[34] Furthermore, Liu Hui described Cavalieri's principle on volume, as well as Gaussian elimination. From the Nine Chapters, it listed the following geometrical formulas that were known by the time of the Former Han Dynasty (202 BCE–9 CE).
Areas for the[35]
Volumes for the[34]
Continuing the geometrical legacy of ancient China, there were many later figures to come, including the famed astronomer and mathematician Shen Kuo (1031-1095 CE), Yang Hui (1238-1298) who discovered Pascal's Triangle, Xu Guangqi (1562-1633), and many others.
## Islamic Golden Age
By the beginning of the 9th century, the "Islamic Golden Age" flourished, the establishment of the House of Wisdom in Baghdad marking a separate tradition of science in the medieval Islamic world, building not only Hellenistic but also on Indian sources.
Although the Islamic mathematicians are most famed for their work on algebra, number theory and number systems, they also made considerable contributions to geometry, trigonometry and mathematical astronomy, and were responsible for the development of algebraic geometry.[citation needed]
Al-Karaji (born 953) completely freed algebra from geometrical operations and replaced them with the arithmetical type of operations which are at the core of algebra today.[citation needed]
In some respects, Thābit ibn Qurra is critical of the ideas of Plato and Aristotle, particularly regarding motion. It would seem that here his ideas are based on an acceptance of using arguments concerning motion in his geometrical arguments. Another contribution Thabit made to geometry was his generalization of the Pythagorean theorem, which he extended from special right triangles to all triangles in general, along with a general proof.[36]
Ibrahim ibn Sinan ibn Thabit (born 908), who introduced a method of integration more general than that of Archimedes, and al-Quhi (born 940) were leading figures in a revival and continuation of Greek higher geometry in the Islamic world. These mathematicians, and in particular Ibn al-Haytham, studied optics and investigated the optical properties of mirrors made from conic sections.[citation needed]
Astronomy, time-keeping and geography provided other motivations for geometrical and trigonometrical research. For example, Ibrahim ibn Sinan and his grandfather Thabit ibn Qurra both studied curves required in the construction of sundials. Abu'l-Wafa and Abu Nasr Mansur both applied spherical geometry to astronomy.[citation needed]
A 2007 paper in the journal Science suggested that girih tiles possessed properties consistent with self-similar fractal quasicrystalline tilings such as the Penrose tilings.[37][38]
## Renaissance
The transmission of the Greek Classics to medieval Europe via the Arabic literature of the 9th to 10th century "Islamic Golden Age" began in the 10th century and culminated in the Latin translations of the 12th century. A copy of Ptolemy's Almagest was brought back to Sicily by Henry Aristippus (d. 1162), as a gift from the Emperor to King William I (r. 1154–1166). An anonymous student at Salerno travelled to Sicily and translated the Almagest as well as several works by Euclid from Greek to Latin.[39] Although the Sicilians generally translated directly from the Greek, when Greek texts were not available, they would translate from Arabic. Eugenius of Palermo (d. 1202) translated Ptolemy's Optics into Latin, drawing on his knowledge of all three languages in the task.[40] The rigorous deductive methods of geometry found in Euclid's Elements of Geometry were relearned, and further development of geometry in the styles of both Euclid (Euclidean geometry) and Khayyam (algebraic geometry) continued, resulting in an abundance of new theorems and concepts, many of them very profound and elegant.
Advances in the treatment of perspective were made in Renaissance art of the 14th to 15th century which went beyond what had been achieved in antiquity. In Renaissance architecture of the Quattrocento, concepts of architectural order were explored and rules were formulated. A prime example of is the Basilica di San Lorenzo in Florence by Filippo Brunelleschi (1377–1446).[41]
In c. 1413 Filippo Brunelleschi demonstrated the geometrical method of perspective, used today by artists, by painting the outlines of various Florentine buildings onto a mirror. Soon after, nearly every artist in Florence and in Italy used geometrical perspective in their paintings,[42] notably Masolino da Panicale and Donatello. Melozzo da Forlì first used the technique of upward foreshortening (in Rome, Loreto, Forlì and others), and was celebrated for that. Not only was perspective a way of showing depth, it was also a new method of composing a painting. Paintings began to show a single, unified scene, rather than a combination of several.
As shown by the quick proliferation of accurate perspective paintings in Florence, Brunelleschi likely understood (with help from his friend the mathematician Toscanelli),[43] but did not publish, the mathematics behind perspective. Decades later, his friend Leon Battista Alberti wrote De pictura (1435/1436), a treatise on proper methods of showing distance in painting based on Euclidean geometry. Alberti was also trained in the science of optics through the school of Padua and under the influence of Biagio Pelacani da Parma who studied Alhazen's Optics.
Piero della Francesca elaborated on Della Pittura in his De Prospectiva Pingendi in the 1470s. Alberti had limited himself to figures on the ground plane and giving an overall basis for perspective. Della Francesca fleshed it out, explicitly covering solids in any area of the picture plane. Della Francesca also started the now common practice of using illustrated figures to explain the mathematical concepts, making his treatise easier to understand than Alberti's. Della Francesca was also the first to accurately draw the Platonic solids as they would appear in perspective.
Perspective remained, for a while, the domain of Florence. Jan van Eyck, among others, was unable to create a consistent structure for the converging lines in paintings, as in London's The Arnolfini Portrait, because he was unaware of the theoretical breakthrough just then occurring in Italy. However he achieved very subtle effects by manipulations of scale in his interiors. Gradually, and partly through the movement of academies of the arts, the Italian techniques became part of the training of artists across Europe, and later other parts of the world. The culmination of these Renaissance traditions finds its ultimate synthesis in the research of the architect, geometer, and optician Girard Desargues on perspective, optics and projective geometry.
The Vitruvian Man by Leonardo da Vinci(c. 1490)[44] depicts a man in two superimposed positions with his arms and legs apart and inscribed in a circle and square. The drawing is based on the correlations of ideal human proportions with geometry described by the ancient Roman architect Vitruvius in Book III of his treatise De Architectura.
## Modern geometry
### The 17th century
In the early 17th century, there were two important developments in geometry. The first and most important was the creation of analytic geometry, or geometry with coordinates and equations, by René Descartes (1596–1650) and Pierre de Fermat (1601–1665). This was a necessary precursor to the development of calculus and a precise quantitative science of physics. The second geometric development of this period was the systematic study of projective geometry by Girard Desargues (1591–1661). Projective geometry is the study of geometry without measurement, just the study of how points align with each other. There had been some early work in this area by Hellenistic geometers, notably Pappus (c. 340). The greatest flowering of the field occurred with Jean-Victor Poncelet (1788–1867).
In the late 17th century, calculus was developed independently and almost simultaneously by Isaac Newton (1642–1727) and Gottfried Wilhelm Leibniz (1646–1716). This was the beginning of a new field of mathematics now called analysis. Though not itself a branch of geometry, it is applicable to geometry, and it solved two families of problems that had long been almost intractable: finding tangent lines to odd curves, and finding areas enclosed by those curves. The methods of calculus reduced these problems mostly to straightforward matters of computation.
### The 18th and 19th centuries
#### Non-Euclidean geometry
The very old problem of proving Euclid's Fifth Postulate, the "Parallel Postulate", from his first four postulates had never been forgotten. Beginning not long after Euclid, many attempted demonstrations were given, but all were later found to be faulty, through allowing into the reasoning some principle which itself had not been proved from the first four postulates. Though Omar Khayyám was also unsuccessful in proving the parallel postulate, his criticisms of Euclid's theories of parallels and his proof of properties of figures in non-Euclidean geometries contributed to the eventual development of non-Euclidean geometry. By 1700 a great deal had been discovered about what can be proved from the first four, and what the pitfalls were in attempting to prove the fifth. Saccheri, Lambert, and Legendre each did excellent work on the problem in the 18th century, but still fell short of success. In the early 19th century, Gauss, Johann Bolyai, and Lobachevsky, each independently, took a different approach. Beginning to suspect that it was impossible to prove the Parallel Postulate, they set out to develop a self-consistent geometry in which that postulate was false. In this they were successful, thus creating the first non-Euclidean geometry. By 1854, Bernhard Riemann, a student of Gauss, had applied methods of calculus in a ground-breaking study of the intrinsic (self-contained) geometry of all smooth surfaces, and thereby found a different non-Euclidean geometry. This work of Riemann later became fundamental for Einstein's theory of relativity.
It remained to be proved mathematically that the non-Euclidean geometry was just as self-consistent as Euclidean geometry, and this was first accomplished by Beltrami in 1868. With this, non-Euclidean geometry was established on an equal mathematical footing with Euclidean geometry.
While it was now known that different geometric theories were mathematically possible, the question remained, "Which one of these theories is correct for our physical space?" The mathematical work revealed that this question must be answered by physical experimentation, not mathematical reasoning, and uncovered the reason why the experimentation must involve immense (interstellar, not earth-bound) distances. With the development of relativity theory in physics, this question became vastly more complicated.
#### Introduction of mathematical rigor
All the work related to the Parallel Postulate revealed that it was quite difficult for a geometer to separate his logical reasoning from his intuitive understanding of physical space, and, moreover, revealed the critical importance of doing so. Careful examination had uncovered some logical inadequacies in Euclid's reasoning, and some unstated geometric principles to which Euclid sometimes appealed. This critique paralleled the crisis occurring in calculus and analysis regarding the meaning of infinite processes such as convergence and continuity. In geometry, there was a clear need for a new set of axioms, which would be complete, and which in no way relied on pictures we draw or on our intuition of space. Such axioms, now known as Hilbert's axioms, were given by David Hilbert in 1894 in his dissertation Grundlagen der Geometrie (Foundations of Geometry).
#### Analysis situs, or topology
In the mid-18th century, it became apparent that certain progressions of mathematical reasoning recurred when similar ideas were studied on the number line, in two dimensions, and in three dimensions. Thus the general concept of a metric space was created so that the reasoning could be done in more generality, and then applied to special cases. This method of studying calculus- and analysis-related concepts came to be known as analysis situs, and later as topology. The important topics in this field were properties of more general figures, such as connectedness and boundaries, rather than properties like straightness, and precise equality of length and angle measurements, which had been the focus of Euclidean and non-Euclidean geometry. Topology soon became a separate field of major importance, rather than a sub-field of geometry or analysis.
#### Geometry of more than 3 dimensions
The 19th century saw the development of the general concept of Euclidean space by Ludwig Schläfli, who extended Euclidean geometry beyond three dimensions. He discovered all the higher-dimensional analogues of the Platonic solids, finding that there are exactly six such regular convex polytopes in dimension four, and three in all higher dimensions.
In 1878 William Kingdon Clifford introduced what is now termed geometric algebra, unifying William Rowan Hamilton's quaternions with Hermann Grassmann's algebra and revealing the geometric nature of these systems, especially in four dimensions. The operations of geometric algebra have the effect of mirroring, rotating, translating, and mapping the geometric objects that are being modeled to new positions.
### The 20th century
Developments in algebraic geometry included the study of curves and surfaces over finite fields as demonstrated by the works of among others André Weil, Alexander Grothendieck, and Jean-Pierre Serre as well as over the real or complex numbers. Finite geometry itself, the study of spaces with only finitely many points, found applications in coding theory and cryptography. With the advent of the computer, new disciplines such as computational geometry or digital geometry deal with geometric algorithms, discrete representations of geometric data, and so forth.
## Notes
1. ^ Howard Eves, An Introduction to the History of Mathematics, Saunders: 1990 (ISBN 0-03-029558-0), p. 141: "No work, except The Bible, has been more widely used...."
2. ^ Ray C. Jurgensen, Alfred J. Donnelly, and Mary P. Dolciani. Editorial Advisors Andrew M. Gleason, Albert E. Meder, Jr. Modern School Mathematics: Geometry (Student's Edition). Houghton Mifflin Company, Boston, 1972, p. 52. ISBN 0-395-13102-2. Teachers Edition ISBN 0-395-13103-0.
3. ^ Eves, Chapter 2.
4. ^
5. ^ A. Seidenberg, 1978. The origin of mathematics. Archive for the history of Exact Sciences, vol 18.
6. ^
7. ^ Most mathematical problems considered in the Śulba Sūtras spring from "a single theological requirement," that of constructing fire altars which have different shapes but occupy the same area. The altars were required to be constructed of five layers of burnt brick, with the further condition that each layer consist of 200 bricks and that no two adjacent layers have congruent arrangements of bricks. (Hayashi 2003, p. 118)
8. ^ (Hayashi 2005, p. 363)
9. ^ Pythagorean triples are triples of integers ${\displaystyle (a,b,c)}$ with the property: ${\displaystyle a^{2}+b^{2}=c^{2}}$. Thus, ${\displaystyle 3^{2}+4^{2}=5^{2}}$, ${\displaystyle 8^{2}+15^{2}=17^{2}}$, ${\displaystyle 12^{2}+35^{2}=37^{2}}$ etc.
10. ^ (Cooke 2005, p. 198): "The arithmetic content of the Śulva Sūtras consists of rules for finding Pythagorean triples such as (3, 4, 5), (5, 12, 13), (8, 15, 17), and (12, 35, 37). It is not certain what practical use these arithmetic rules had. The best conjecture is that they were part of religious ritual. A Hindu home was required to have three fires burning at three different altars. The three altars were to be of different shapes, but all three were to have the same area. These conditions led to certain "Diophantine" problems, a particular case of which is the generation of Pythagorean triples, so as to make one square integer equal to the sum of two others."
11. ^ (Cooke 2005, pp. 199–200): "The requirement of three altars of equal areas but different shapes would explain the interest in transformation of areas. Among other transformation of area problems the Hindus considered in particular the problem of squaring the circle. The Bodhayana Sutra states the converse problem of constructing a circle equal to a given square. The following approximate construction is given as the solution.... this result is only approximate. The authors, however, made no distinction between the two results. In terms that we can appreciate, this construction gives a value for π of 18 (3 − 22), which is about 3.088."
12. ^ a b c (Joseph 2000, p. 229)
13. ^ Mathematics Department, University of British Columbia, The Babylonian tabled Plimpton 322.
14. ^ Three positive integers ${\displaystyle (a,b,c)}$ form a primitive Pythagorean triple if ${\displaystyle c^{2}=a^{2}+b^{2}}$ and if the highest common factor of ${\displaystyle a,b,c}$ is 1. In the particular Plimpton322 example, this means that ${\displaystyle 13500^{2}+12709^{2}=18541^{2}}$ and that the three numbers do not have any common factors. However some scholars have disputed the Pythagorean interpretation of this tablet; see Plimpton 322 for details.
15. ^ a b
16. ^ Cherowitzo, Bill. "What precisely was written over the door of Plato's Academy?" (PDF). www.math.ucdenver.edu/. Archived (PDF) from the original on 2013-06-25. Retrieved 8 April 2015.
17. ^ "Archimedes". Encyclopedia Britannica.
18. ^ Luciano Canfora; The Vanished Library; University of California Press, 1990. - google books
19. ^ (Hayashi 2005, p. 371)
20. ^ a b (Hayashi 2003, pp. 121–122)
21. ^ (Stillwell 2004, p. 77)
22. ^ Radha Charan Gupta [1977] "Parameshvara's rule for the circumradius of a cyclic quadrilateral", Historia Mathematica 4: 67–74
23. ^ a b Needham, Volume 3, 91.
24. ^ a b c Needham, Volume 3, 92.
25. ^ Needham, Volume 3, 92-93.
26. ^ Needham, Volume 3, 93.
27. ^ Needham, Volume 3, 93-94.
28. ^ Needham, Volume 3, 94.
29. ^ Needham, Volume 3, 99.
30. ^ Needham, Volume 3, 101.
31. ^ Needham, Volume 3, 22.
32. ^ Needham, Volume 3, 21.
33. ^ Needham, Volume 3, 100.
34. ^ a b c Needham, Volume 3, 98–99.
35. ^ Needham, Volume 3, 98.
36. ^ Sayili, Aydin (1960). "Thabit ibn Qurra's Generalization of the Pythagorean Theorem". Isis. 51 (1): 35–37. doi:10.1086/348837. S2CID 119868978.
37. ^ Peter J. Lu and Paul J. Steinhardt (2007), "Decagonal and Quasi-crystalline Tilings in Medieval Islamic Architecture" (PDF), Science, 315 (5815): 1106–1110, Bibcode:2007Sci...315.1106L, doi:10.1126/science.1135491, PMID 17322056, S2CID 10374218, archived from the original (PDF) on 2009-10-07.
38. ^ Supplemental figures Archived 2009-03-26 at the Wayback Machine
39. ^ d'Alverny, Marie-Thérèse. "Translations and Translators", in Robert L. Benson and Giles Constable, eds., Renaissance and Renewal in the Twelfth Century, 421–462. Cambridge: Harvard Univ. Pr., 1982, pp. 433–4.
40. ^ M.-T. d'Alverny, "Translations and Translators," p. 435
41. ^ Howard Saalman. Filippo Brunelleschi: The Buildings. (London: Zwemmer, 1993).
42. ^ "...and these works (of perspective by Brunelleschi) were the means of arousing the minds of the other craftsmen, who afterwards devoted themselves to this with great zeal."
Vasari's Lives of the Artists Chapter on Brunelleschi
43. ^ "Messer Paolo dal Pozzo Toscanelli, having returned from his studies, invited Filippo with other friends to supper in a garden, and the discourse falling on mathematical subjects, Filippo formed a friendship with him and learned geometry from him."
Vasarai's Lives of the Artists, Chapter on Brunelleschi
44. ^ The Secret Language of the Renaissance - Richard Stemp | 9,543 | 41,400 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 22, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.959946 |
https://matplotlib.org/3.0.3/gallery/lines_bars_and_markers/multicolored_line.html | 1,723,231,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640768597.52/warc/CC-MAIN-20240809173246-20240809203246-00881.warc.gz | 301,631,975 | 4,866 | You are reading an old version of the documentation (v3.0.3). For the latest version see https://matplotlib.org/stable/gallery/lines_bars_and_markers/multicolored_line.html
# Multicolored linesΒΆ
This example shows how to make a multi-colored line. In this example, the line is colored based on its derivative.
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.collections import LineCollection
from matplotlib.colors import ListedColormap, BoundaryNorm
x = np.linspace(0, 3 * np.pi, 500)
y = np.sin(x)
dydx = np.cos(0.5 * (x[:-1] + x[1:])) # first derivative
# Create a set of line segments so that we can color them individually
# This creates the points as a N x 1 x 2 array so that we can stack points
# together easily to get the segments. The segments array for line collection
# needs to be (numlines) x (points per line) x 2 (for x and y)
points = np.array([x, y]).T.reshape(-1, 1, 2)
segments = np.concatenate([points[:-1], points[1:]], axis=1)
fig, axs = plt.subplots(2, 1, sharex=True, sharey=True)
# Create a continuous norm to map from data points to colors
norm = plt.Normalize(dydx.min(), dydx.max())
lc = LineCollection(segments, cmap='viridis', norm=norm)
# Set the values used for colormapping
lc.set_array(dydx)
lc.set_linewidth(2)
line = axs[0].add_collection(lc)
fig.colorbar(line, ax=axs[0])
# Use a boundary norm instead
cmap = ListedColormap(['r', 'g', 'b'])
norm = BoundaryNorm([-1, -0.5, 0.5, 1], cmap.N)
lc = LineCollection(segments, cmap=cmap, norm=norm)
lc.set_array(dydx)
lc.set_linewidth(2)
line = axs[1].add_collection(lc)
fig.colorbar(line, ax=axs[1])
axs[0].set_xlim(x.min(), x.max())
axs[0].set_ylim(-1.1, 1.1)
plt.show()
Keywords: matplotlib code example, codex, python plot, pyplot Gallery generated by Sphinx-Gallery | 516 | 1,783 | {"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} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.576715 |
https://az.beautycolorcode.com/convert/length/foot-to-kilometer | 1,701,210,778,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00365.warc.gz | 151,785,050 | 11,537 | # The feet to kilometers converter
Enter the length in Foot below to get the value converted to Kilometer.
## How to convert feet to kilometers
• 1 Foot is equal to 0.0003048 kilometers
• 1 Kilometer is equal to 3280.83989501136 feet
• The distance d in kilometers (km) is equal to the distance d in feet(ft) multiplied by 0.0003048, use the formula below:
dkm = dft * 0.0003048
• Example: convert 20 feet to kilometers
dkm = 20ft * 0.0003048 = 0.006096km
## The feet to kilometers conversion table
feetkilometers
0.01 ft0.000003048 km
0.1 ft0.00003048 km
1 ft0.0003048 km
2 ft0.0006096 km
3 ft0.0009144 km
4 ft0.0012192 km
5 ft0.001524 km
6 ft0.0018288 km
7 ft0.0021336 km
8 ft0.0024384 km
9 ft0.0027432 km
10 ft0.003048 km
20 ft0.006096 km
30 ft0.009144 km
40 ft0.012192 km
50 ft0.01524 km
60 ft0.018288 km
70 ft0.021336 km
80 ft0.024384 km
90 ft0.027432 km
100 ft0.03048 km | 338 | 892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.64827 |
https://www.codingninjas.com/codestudio/library/cherry-pickup | 1,653,804,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00338.warc.gz | 780,931,594 | 27,453 | # Cherry Pickup
Firdausia Fatima
Last Updated: May 13, 2022
## Introduction
One of the most exciting aspects of computer problems is how many various ways they can be solved. Based on several factors, one is superior to the other.
And sorting through all of them to find the finest one is a journey (though not one that will turn you into an alchemist), but one that will teach you a lot.
This is a problem that will completely transform the way you think and widen your range of solutions.
Now that you know how cool this problem is, let’s jump right into it.
## Problem Statement
You are given an N X N matrix where each cell has either of three states.
0 - This block is free
1 - This block contains a cherry
-1 - This block contains a thorn
Maximize the number of cherries on the round trip from [0, 0] to [N - 1, N - 1] and then back to [0, 0].
### Constraints:
• You can only take right and down steps while moving from [0, 0] to [N - 1, N - 1].
• And while coming back from [N - 1, N - 1] to [0,0], you can only take left and up steps.
• You can pass through the free block.
• While passing through the cherry block, you can collect the cherry, but after this, it’s a free block.
• You cannot pass through the block with thorns.
If you cannot complete the round trip, return -1.
### Example:
Let’s consider this 3 X 3 matrix.
As shown in the above image, we can pick a maximum of 3 cherries. Hence, the answer for the above matrix is 3.
Now that we’ve defined the problem. Let’s dive into the solutions.
One possible solution is to find the path with the most cherries from top-left to bottom-right and then find the path with the most cherries from bottom-right to top-left.
But, unfortunately for us, this one is incorrect, as you can see in the following example.
If we go from top-left to bottom-right and then back to top-left, taking the maximum cherry count path then,
Path from [0, 0] to [3, 3] = [(0, 0), (0, 1), (1, 1), (2, 1), (3, 1), (3, 2), (3, 3)]
Path from [3, 3] to [0, 0] = [(3, 3), (2, 3), (1, 3), (0, 3), (0, 2), (0, 1), (0, 0)]
Cherries Collected = 6
Now, let’s take the optimal cherry path.
Path from [0, 0] to [3, 3] = [(0, 0), (0, 1), (1, 1), (1, 2), (1, 3), (2, 3), (3, 3)]
Path from [3, 3] to [0, 0] = [(3, 3), (3, 2), (3, 1), (2, 1), (2, 0), (1, 0), (0, 0)]
Cherries Collected = 7
Hence, choosing the maximum Cherry path will not always give maximum cherries in the round trip.
## Approach 1
We can solve this problem using Backtracking. Traverse every path from top-left to bottom-right, and for each path, find all paths back to top-right and find the maximum cherry paths. It’s like the cartesian product of all paths down to all paths up.
So we will have two functions( ‘traverseDown’) that will traverse from top-right to bottom-left and one ( ‘traverseUp’ ) that will traverse back from bottom-right to top-left.
### Program
``````#include <iostream>
#include <climits>
#include <vector>
using namespace std;
int maxCherryCount = INT_MIN;
// Function that will try all paths from bottom-right to top-left.
void traverseUp(int r, int c, int n, vector<vector<int>> &arr, int cherryCollected)
{
// Check if the block is valid.
if (r < 0 || c < 0 || r >= n || c >= n || arr[r][c] == -1)
{
return;
}
// If we are back to top-left, that means we have completed the traversal, so we update maxCherryCount.
if (r == 0 && c == 0)
{
maxCherryCount = max(maxCherryCount, cherryCollected);
}
// Store cherries in the block.
int cherries = arr[r][c];
// Now collect the cherry
arr[r][c] = 0;
// Traverse left and up.
traverseUp(r - 1, c, n, arr, cherryCollected + cherries);
traverseUp(r, c - 1, n, arr, cherryCollected + cherries);
// Backtrack.
arr[r][c] = cherries;
}
// Function to traverse all paths from top-left to bottom-right.
void traverseDown(int r, int c, int n, vector<vector<int>> &arr, int cherryCollected)
{
// Check if the block is valid.
if (r < 0 || c < 0 || r >= n || c >= n || arr[r][c] == -1)
{
return;
}
// Once we have reached the bottom-right now, traverse all paths back to the top-left.
if (r == n - 1 && c == n - 1)
{
traverseUp(r, c, n, arr, cherryCollected);
}
// Store cherries in the block.
int cherries = arr[r][c];
// Collect cherries.
arr[r][c] = 0;
// Traverse right and down.
traverseDown(r + 1, c, n, arr, cherryCollected + cherries);
traverseDown(r, c + 1, n, arr, cherryCollected + cherries);
// Backtrack.
arr[r][c] = cherries;
}
int main()
{
int n;
cout << "Enter the dimension of the matrix (N X N): ";
cin >> n;
vector<vector<int>> arr(n, vector<int>(n));
cout << "Enter the cherry matrix:\n";
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}
traverseDown(0, 0, n, arr, 0);
cout << "Maximum cherries collected: " << maxCherryCount << endl;
return 0;
}``````
### Input
``````Enter the dimension of the matrix (N X N): 4
Enter the cherry matrix:
0 1 0 0
0 1 0 1
1 1 -1 0
-1 1 1 1``````
### Output
``Maximum cherries collected: 8``
### Time Complexity
O( 2^(4N-4) ), where ‘N’ is the dimension of the matrix.
Total steps in a round trip = 4N-4, For moving from [0, 0] to [N - 1, N - 1] we take 2N - 2 steps ( Down steps = N - 1, Right steps = N - 1), so to complete a round trip double the steps of the one-way path.
For each step, we have two directions to choose from, and hence time complexity is 2 ^ (4N - 4).
### Space Complexity
O(4- 4), where ‘N’ is the dimension of the matrix.
Because of the stack space used in recursion, the maximum number of function calls in the stack is 4- 4.
Let's see if we can apply Dynamic Programming to the above solution. One requirement for DP is that problems be solved using the solution to smaller subproblems (previously calculated), which is not the case here because we cannot divide the matrix into solved and unsolved sections where the solution to the unsolved part is somehow dependent on previously solved sections. As a result, we'll have to think up a new approach.
## Approach 2
This time, we'll travel both downward and upward at the same time. So we'll have two robots that go from [0, 0] to [N - 1, N - 1], and when they both reach the bottom right, the MAX_CHERRY_COUNTvariable will be updated. Because one robot travelling in a round trip equals two robots travelling in a downward direction since the second robot’s path can be considered as the backward path for the first robot, let’s look at an example.
The number of possible directions both robots can go is - Down Down, Right Right, Right Down, Down Right.
And once we get which direction will give us the maximum number of cherries, we can add them to the current count of cherries and return them.
Let’s try to code it now.
### Program
``````#include <iostream>
#include <climits>
#include <vector>
using namespace std;
// Function for two robots traversing simultaneously.
int solver(int r1, int c1, int r2, int c2, int n, vector<vector<int>> &arr)
{
// Check for boundaries. Since we are never decrementing the variables, we omitted the check for r1 < 0.
if (r1 >= n || c1 >= n || c2 >= n || r2 >= n || arr[r1][c1] == -1 || arr[r2][c2] == -1)
{
return INT_MIN;
}
// If we've reached the bottom-right block, return its possible value.
if (r1 == n - 1 && c1 == n - 1)
{
return arr[r1][c1];
}
int cherries = 0;
// If robots are on the same block, we've to collect the cherry only once.
if (r1 == r2 && c1 == c2)
{
cherries += arr[r1][c1];
}
else
{
cherries += arr[r1][c1] + arr[r2][c2];
}
// There are four possibility for two robots to move. Either both will move right, or both will move down, or one will move right, and second will move down, or one will move right and second will move down.
int rr = solver(r1, c1 + 1, r2, c2 + 1, n, arr);
int rd = solver(r1, c1 + 1, r2 + 1, c2, n, arr);
int dd = solver(r1 + 1, c1, r2 + 1, c2, n, arr);
int dr = solver(r1 + 1, c1, r2, c2 + 1, n, arr);
// Add maximum possible case to the cherries in this block and return.
cherries += max(max(rr, dd), max(rd, dr));
return cherries;
}
int main()
{
int n;
cout << "Enter the dimension of the matrix (N X N): ";
cin >> n;
vector<vector<int>> arr(n, vector<int>(n));
cout << "Enter the cherry matrix:";
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}
int maxxCherryCount = solver(0, 0, 0, 0, n, arr);
cout << "Maximum Cherries collected : " << maxxCherryCount << endl;
return 0;
}``````
### Input
``````Enter the dimension of the matrix (N X N): 4
Enter the cherry matrix:
0 0 1 0
1 0 1 0
-1 -1 0 1
1 -1 0 1``````
### Output
``Maximum Cherries collected: 5``
### Time Complexity
O(2 ^ (4 * N - 4)), where ‘N’ is the dimension of the matrix.
Total steps in each traversal from top-left to bottom-right = 2*(2N - 2) => both robots.
And we have four possibilities (rr, dd, rd, dr).
### Space Complexity
O(4N-4), where ‘N’ is the dimension of the matrix.
Because of stack space used in recursion. The maximum of function calls in the stack is 4- 4.
## Approach 2 + DP
Let’s see how and why we can add DP to the above approach.
When we are at the start, robot 1 is at [r1,c1] and robot 2 is at [r2,c2]. We call four functions one for each case ( Right Right, Down Down, Right Down, Down Right ).
And we choose the case whose cherry count is maximum. So if we are in the same situation again, we can use the previously stored result to calculate the answer.
Since there are four variables, we’ll have to make a four-dimensional cache.
All we have to change in the above solution is to cache the result before returning and see if the result is calculated before, then return the result.
Another small improvement is that since both robots are starting from [0,0] and moving one step either right or down, r1+c1 = r2+c2 on every step.
So we can remove c2 from the function call, and it’s not variable now, so the cache dimension is reduced to three.
### Program
``````#include <iostream>
#include <climits>
#include <vector>
#include <map>
using namespace std;
// Function for two robots traversing simultaneously.
int solver(int r1, int c1, int r2, int n, vector<vector<int>> &arr, map<string, int> &dp)
{
// Calculate c2.
int c2 = r1 + c1 - r2;
// Check for boundaries. Since we are never decrementing the variables, we omitted the check for r1<0.
if (r1 >= n || c1 >= n || c2 >= n || r2 >= n || arr[r1][c1] == -1 || arr[r2][c2] == -1)
{
return INT_MIN;
}
// If we've reached the bottom-right block, return its value.
if (r1 == n - 1 && c1 == n - 1)
{
return arr[r1][c1];
}
string key = to_string(r1) + " " + to_string(c1) + " " + to_string(r2);
if (dp.find(key) != dp.end())
{
return dp[key];
}
int cherries = 0;
// If robots are on the same block, we've to collect the cherry only once.
if (r1 == r2 && c1 == c2)
{
cherries += arr[r1][c1];
}
else
{
cherries += arr[r1][c1] + arr[r2][c2];
}
// There are four possiblity for 2 robots to move. Either both will move right, or both will move down, or one will move right and second will move down, or one will move right and second will move down.
int rr = solver(r1, c1 + 1, r2, n, arr, dp);
int rd = solver(r1, c1 + 1, r2 + 1, n, arr, dp);
int dd = solver(r1 + 1, c1, r2 + 1, n, arr, dp);
int dr = solver(r1 + 1, c1, r2, n, arr, dp);
// Add Maxx possible case to the cherries in this block.
cherries += max(max(rr, dd), max(rd, dr));
// Cache the result and return.
return dp[key] = cherries;
}
int main()
{
int n;
cout << "Enter the dimension of the matrix (N X N): ";
cin >> n;
vector<vector<int>> arr(n, vector<int>(n));
cout << "Enter the cherry matrix:\n";
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
cin >> arr[i][j];
}
}
unordered_map<string, int> dp;
int maxxCherryCount = solver(0, 0, 0, n, arr, dp);
cout << "Maximum cherries collected: " << maxxCherryCount << endl;
return 0;
}``````
### Input
``````Enter the dimension of the matrix (N X N): 4
Enter the Cherry matrix:
0 0 1 0
1 0 1 0
-1 -1 0 1
1 -1 0 0``````
### Output
``Maximum Cherries collected: 4``
### Time Complexity
O( N^3 ), where ‘N’ is the dimension of the matrix.
Because we have N ^ 3 dynamic programming states.
### Space Complexity
O( N^3 ), where is the dimension of the matrix.
Because we have N ^ 3 dynamic programming states, the size required for caching will be O(N ^ 3)
## Key Takeaways
Solving questions like these teach you not only computer science concepts but also how to approach complex problems. We’ve got tons of such problems and blogs on the Coding Ninjas Platform. Also, recently Coding Ninjas has released a specially designed test series for acing Interviews- CodeStudio Test Series.
Thanks for reading. I hope you’ve gained a lot from this blog.
By Firdausia Fatima | 3,830 | 12,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.923031 |
https://github.com/csdms-contrib/slepian_echo/blob/master/cube2moll.m | 1,542,788,620,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039747369.90/warc/CC-MAIN-20181121072501-20181121094501-00216.warc.gz | 622,911,441 | 15,044 | # csdms-contrib/slepian_echo
Switch branches/tags
Nothing to show
Fetching contributors…
Cannot retrieve contributors at this time
36 lines (32 sloc) 1.05 KB
function [lonm,latm,vd]=cube2moll(v) % [lonm,latm,vd]=cube2moll(v) % % Transforms a cubed-sphere model to Mollweide projection by % fine-scale interpolation to a regular grid % % INPUT: % % v NxNx6 matrix with model values % % OUTPUT: % % lonm,latm Mollweide longitudes and latitudes % vd Corresponding data values % % Last modified by fjsimons-at-alum.mit.edu, 06/28/2010 % Find the coordinates of the cubed sphere that are being used [x,y,z]=cube2sphere(nextpow2(size(v,1))); % Turn them into regular spherical coordinates [lon,lat]=cart2sph(x,y,z); lon(lon<0)=lon(lon<0)+2*pi; % Make Mollweide coordinates fit for the purpose lons=linspace(0,2*pi,4*size(v,2)); lats=linspace(pi/2,-pi/2,4*size(v,1)); [lons,lats]=meshgrid(lons,lats); [lonm,latm]=mollweide(lons,lats,pi); % Remember that this was flipped, see PLOTONCUBE for i=1:6; v(:,:,i)=flipud(v(:,:,i)); end warning off MATLAB:griddata:DuplicateDataPoints % Perform the interpolation ON THE SPHERE! vd=griddata(lon(:),lat(:),v(:),lons,lats); warning on MATLAB:griddata:DuplicateDataPoints | 372 | 1,204 | {"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} | 2.78125 | 3 | CC-MAIN-2018-47 | latest | en | 0.445387 |
http://www.goddardconsulting.ca/matlab-monte-carlo-spread.html | 1,723,191,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640762343.50/warc/CC-MAIN-20240809075530-20240809105530-00516.warc.gz | 36,091,934 | 2,375 | ## Pricing a Spread Option in MATLAB
A Spread option is an example of an option that has a payoff that is both path dependent and is dependent on multiple assets. This makes it ideally suited for pricing using the Monte-Carlo approach as discussed in the Monte-Carlo Methods tutorial.
The Simulating Multiple Asset Paths in MATLAB tutorial gives an example of a MATLAB function for generating the types of correlated multiple asset paths required for option pricing using Monte-Carlo methods. That tutorial is expanded here where MATLAB code for pricing a Spread option is presented.
### MATLAB Script: Spread
The following is code for generating a user specified number of correlated asset paths for two assets and then using those paths to price a given Spread option. The payoff of the option is given by
Equation 1: Payoff for a given Spread Option
where Δ(t)max is the maximum spread between the two assets over the lifetime of the option and Δallowable is a constant pre-specified maximum allowable spread.
```% Script to price a spread option using a monte-carlo approach
% Some parts of this could be vectorized, but has not been done here
% so that it's easier to understand what's going on.
% Define required parameters
S0 = [50 52];
r = 0.02;
mu = [0.03 0.04];
sig = [0.1 0.15];
corr = [1 0;0 1];
dt = 1/365;
etime = 50; % days to expiry
T = dt*etime;
nruns = 100000;
% generate the paths
S = AssetPathsCorrelated(S0,mu,sig,corr,dt,etime,nruns);
% The payoff is path dependent
payoff = nan*ones(nruns,1);
for idx = 1:nruns
maxDifference = diff(squeeze(S(:,idx,:)),1,2);
payoff(idx) = 0;
else
end
end
% Determine the option price
oPrice = mean(payoff)*exp(-r*T)
```
### Example Usage
The following shows the results of executing the Spread script.
```oPrice =
1.3074
```
Other MATLAB based Monte-Carlo tutorials are linked off the Software Tutorials page. | 471 | 1,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.838161 |
https://issuhub.com/view/index?id=207&pageIndex=185 | 1,719,150,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00889.warc.gz | 273,853,995 | 4,696 | # Pattern Recognition and Machine Learning
(Jeff_L) #1
``3.5. The Evidence Approximation 165``
``````a Bayesian approach, like any approach to pattern recognition, needs to make as-
sumptions about the form of the model, and if these are invalid then the results can
be misleading. In particular, we see from Figure 3.12 that the model evidence can
be sensitive to many aspects of the prior, such as the behaviour in the tails. Indeed,
the evidence is not defined if the prior is improper, as can be seen by noting that
an improper prior has an arbitrary scaling factor (in other words, the normalization
coefficient is not defined because the distribution cannot be normalized). If we con-
sider a proper prior and then take a suitable limit in order to obtain an improper prior
(for example, a Gaussian prior in which we take the limit of infinite variance) then
the evidence will go to zero, as can be seen from (3.70) and Figure 3.12. It may,
however, be possible to consider the evidence ratio between two models first and
then take a limit to obtain a meaningful answer.
In a practical application, therefore, it will be wise to keep aside an independent
test set of data on which to evaluate the overall performance of the final system.``````
### 3.5 The Evidence Approximation
``````In a fully Bayesian treatment of the linear basis function model, we would intro-
duce prior distributions over the hyperparametersαandβand make predictions by
marginalizing with respect to these hyperparameters as well as with respect to the
parametersw. However, although we can integrate analytically over eitherwor
over the hyperparameters, the complete marginalization over all of these variables
is analytically intractable. Here we discuss an approximation in which we set the
hyperparameters to specific values determined by maximizing themarginal likeli-
hood functionobtained by first integrating over the parametersw. This framework
is known in the statistics literature asempirical Bayes(Bernardo and Smith, 1994;
Gelmanet al., 2004), ortype 2 maximum likelihood(Berger, 1985), orgeneralized
maximum likelihood(Wahba, 1975), and in the machine learning literature is also
called theevidence approximation(Gull, 1989; MacKay, 1992a).
If we introduce hyperpriors overαandβ, the predictive distribution is obtained
by marginalizing overw,αandβso that``````
``p(t|t)=``
``````∫∫∫
p(t|w,β)p(w|t,α,β)p(α, β|t)dwdαdβ (3.74)``````
``````wherep(t|w,β)is given by (3.8) andp(w|t,α,β)is given by (3.49) withmNand
SNdefined by (3.53) and (3.54) respectively. Here we have omitted the dependence
on the input variablexto keep the notation uncluttered. If the posterior distribution
p(α, β|t)is sharply peaked around valueŝαand̂β, then the predictive distribution is
obtained simply by marginalizing overwin whichαandβare fixed to the valueŝα
and̂β, so that``````
``p(t|t)p(t|t,α,̂̂β)=``
``````∫
p(t|w,β̂)p(w|t,α,̂̂β)dw. (3.75)`````` | 791 | 2,934 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.914252 |
https://www.jagranjosh.com/articles/wbjee-important-questions-and-preparation-tips-integrals-1514273257-1 | 1,679,820,998,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00769.warc.gz | 968,815,726 | 33,667 | # WBJEE: Important Questions and Preparation Tips – Integrals
In this article, WBJEE aspirants will find chapter notes of chapter Integrals including all important topics like indefinite integral, definite integral, integration using partial fractions, integration by parts, definite integral, properties of definite integral etc for WBJEE entrance examination 2018. About 2-4 questions are being asked from this topic in the examination.
WBJEE 2018: Integrals
Engineering aspirants always search chapter notes related to every topic of any subject i.e., Physics, Chemistry and Mathematics when exams are around the corner. Chapter notes help aspirants to revise all important concepts and formula without wasting their time.
Find chapter notes of chapter Integrals based on the latest syllabus for coming WBJEE entrance examination 2018. Aspirants always get 2-4 questions from this topic in the examination.
1. These notes contain all important topics related to Integrals like indefinite integral, definite integral, integration using partial fractions, integration by parts, definite integral, properties of definite integral etc.
2. Each concept in these notes is explained in a very detailed manner by Subject Experts of Mathematics.
3. With the help of these notes WBJEE aspirants can easily understand any concept related to Integrals very easily.
4. These notes contain important concepts, formulae and some previous year solved questions.
5. With the help of previous year questions, engineering aspirants can easily understand the difficulty level of the examination and can prepare accordingly.
West Bengal Joint Entrance Examination (WBJEE) is a state level common entrance test organized by West Bengal Joint Entrance Examinations Board for admission to the Undergraduate Level Engineering and Medical Courses through a common entrance test in the State of West Bengal.
Important Concepts:
WBJEE: Important Questions and Preparation Tips – Binomial Theorem
WBJEE: Important Questions and Preparation Tips – Application of Derivatives
Some previous years solved questions are given below:
Question 1:
Solution 1:
Hence, the correct option is (C).
WBJEE: Important Questions and Preparation Tips – Circles
Question 2:
Solution 2:
Hence, the correct option is (a).
WBJEE: Important Questions and Preparation Tips – Differential Equations
Question 3:
Solution 3:
Hence, the correct option is (C).
WBJEE: Important Questions and Preparation Tips – Area under Curve
Question 4:
Solution 4:
Hence, the correct option is (A).
WBJEE: Important Questions and Preparation Tips – Parabola
Question 5:
Solution 5:
Hence, the correct option is (C).
WBJEE: Important Questions and Preparation Tips – Limits
WBJEE 2018: Notification, Application, Dates, Eligibility, Exam Pattern, Syllabus
## Related Categories
खेलें हर किस्म के रोमांच से भरपूर गेम्स सिर्फ़ जागरण प्ले पर | 619 | 2,906 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.883896 |
https://mathlake.com/Supplementary-And-Complementary-Angles | 1,701,611,528,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.23/warc/CC-MAIN-20231203125921-20231203155921-00036.warc.gz | 434,235,795 | 5,834 | # Supplementary And Complementary Angles
Supplementary angles and complementary angles are defined with respect to the addition of two angles. If the sum of two angles is 180 degrees then they are said to be supplementary angles, which forms a linear angle together. Whereas if the sum of two angles is 90 degrees, then they are said to be complementary angles, and they form a right angle together.
When two line segments or lines meet at a common point (called vertex), at the point of intersection an angle is formed. When a ray is rotated about its endpoint, then the measure of its rotation in an anti-clockwise direction is the angle formed between its initial and final position.
In fig. 1 if the ray $$\small \overrightarrow{OP}$$ is rotated in the direction of the ray $$\small \overrightarrow{OQ}$$, then the measure of its rotation represents the angle formed by it. In this case, the measure of rotation that is the angle formed between the initial side and the terminal side is represented by Ɵ.
## Complementary Angles
When the sum of two angles is 90°, then the angles are known as complementary angles. In other words, if two angles add up to form a right angle, then these angles are referred to as complementary angles. Here we say that the two angles complement each other.
Suppose if one angle is x then the other angle will be 90o – x. Hence, we use these complementary angles for trigonometry ratios, where on ratio complement another ratio by 90 degrees such as;
• sin (90°- A) = cos A and cos (90°- A) = sin A
• tan (90°- A) = cot A and cot (90°- A) = tan A
• sec (90°- A) = cosec A and cosec (90°- A) = sec A
Hence, you can see here the trigonometric ratio of the angles gets changed if they complement each other.
Complementary Angles
In the above figure, the measure of angle BOD is 60o and angle AOD measures 30o. On adding both of these angles we get a right angle, therefore ∠BOD and ∠AOD are complementary angles.
The following angles in Fig. 3 given below are complementary to each other as the measure of the sum of both the angles is 90o. ∠POQ and ∠ABC are complementary and are called complements of each other.
Complementary Angles Example
For example: To find the complement of 2x + 52°, subtract the given angle from 90 degrees.
90o – (2x + 52o) = 90o – 2x – 52o = -2x + 38o
The complement of 2x + 52o is 38o – 2x.
Facts of complementary angles: Two right angles cannot complement each other Two obtuse angles cannot complement each other Two complementary angles are acute but vice versa is not possible
## Supplementary Angles
When the sum of two angles is 180°, then the angles are known as supplementary angles. In other words, if two angles add up, to form a straight angle, then those angles are referred to as supplementary angles.
The two angles form a linear angle, such that, if one angle is x, then the other the angle is 180 – x. The linearity here proves that the properties of the angles remain the same. Take the examples of trigonometric ratios such as;
• Sin (180 – A) = Sin A
• Cos (180 – A) = – Cos A (quadrant is changed)
• Tan (180 – A) = – Tan A
Supplementary Angles
In Fig. 4 given above, the measure of ∠AOC is 60o and ∠AOB measures 120o. On adding both of these angles we get a straight angle. Therefore, ∠AOC and ∠AOB are supplementary angles, and both of these angles are known as a supplement of each other.
Also, learn:
### Difference between Complementary and Supplementary Angles
Complementary Angles Supplementary Angles Sum is equal to 90 degrees Sum is equal to 180 degrees Two angles complement each other Two angles supplement each other It is not defined for linear pair of angles It is defined for linear pair of angles Meant only for right angles Meant only for straight angles
How to remember easily the difference between Complementary angle and supplementary angles?
• C letter of Complementary stands for “Corner” (A right angle, 90o)
• S letter of Supplementary stands for “Straight” ( a straight line, 180o)
## Solved Examples
The example problems on supplementary and complementary angles are given below:
Example 1:
Find the complement of 40 degrees.
Solution:
As the given angle is 40 degrees, then,
Complement is 50 degrees.
We know that Sum of Complementary angles = 90 degrees
So, 40° + 50° = 90°
Example 2:
Find the Supplement of the angle 1/3 of 210°.
Solution:
Step 1: Convert 1/3 of 210°
That is, 1/3 x 210° = 70°
Step 2: Supplement of 70° = 180° – 70° = 110°
Therefore, Supplement of the angle 1/3 of 210° is 110°
Example 3:
The measures of two angles are (x + 25)° and (3x + 15)°. Find the value of x if angles are supplementary angles.
Solution:
We know that, Sum of Supplementary angles = 180 degrees
So,
(x + 25)° + (3x + 15)° = 180°
4x + 40° = 180°
4x = 140°
x = 35°
The value of x is 35 degrees.
Example 4:
The difference between two complementary angles is 52°. Find both the angles.
Solution:
Let, First angle = m degrees, then,
Second angle = (90 – m)degrees {as per the definition of complementary angles}
Difference between angles = 52°
Now,
(90° – m) – m = 52°
90° – 2m = 52°
– 2m = 52° – 90°
-2m = -38°
m = 38°/2°
m = 19°
Again, Second angle = 90° – 19° = 71°
Therefore, the required angles are 19°, 71°.
### Practice Questions
1. Check if 65° and 35° are complementary angles.
2. Is 80° and 100° supplementary?
3. Find the complement of 54°.
4. Find the supplement of 99°.
5. If one angle measures 50° and is supplementary to another angle. Then find the value of another angle. Also, state what type of angle it is?
## Frequently Asked Questions – FAQs
### What are complementary angles? Give example.
When the sum of two angles is equal to 90 degrees, they are called complementary angles. For example, 30 degrees and 60 degrees are complementary angles.
### What are supplementary angles? Give examples.
When the sum of the measure of two angles is equal to 180 degrees, they are called supplementary angles. For example, 70 degrees and 110 degrees are supplementary.
### How to find complementary angles?
Since, the sum of complementary angles equals 90 degrees, therefore if we know the measure of one angle, then we can find the unknown angle easily.
For example, if one of the two angles is 45 degrees, then;
x + 45 = 90
x = 90 – 45 = 45°
### What is the complementary angle of 40 degrees?
The complementary angle of 40 degrees is:
90 – 40 = 50 degrees
### How to find supplementary angles?
To find the angle which is supplementary to another angle, subtract the given angle from 180 degrees.
For example, if one angle is 60 degrees, then another angle is 180 – 60 = 120 degrees. | 1,767 | 6,688 | {"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} | 4.65625 | 5 | CC-MAIN-2023-50 | longest | en | 0.933169 |
http://mathhelpforum.com/calculus/50905-displacement-vector-problem.html | 1,496,092,206,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612553.95/warc/CC-MAIN-20170529203855-20170529223855-00200.warc.gz | 272,840,101 | 10,550 | 1. ## displacement vector problem
2. Since the airplane is travelling at a constant height, then your j component stays the same.
Now, you know that the airplane is travelling at a constant velocity. You are given that in 30 seconds, it travels $9.24 \cdot 10^3$ metres. So, $v = \frac{d}{t} = \frac{9.24 \cdot 10^3 \ \text{metres}}{30.0 \ \text{s}} = 308 \ ms^{-1}$
So how far will it have travelled in 70 seconds? This will be your i-component.
Now use that position vector and find its magnitude and angle with respect to the x-axis.
3. Originally Posted by skeeter
$9.24 \times 10^3$ meters in 30 sec is $3.08 \times 10^2$ meters per second.
can you now determine the horizontal displacement at t = 70 seconds?
the vertical component does not change.
finish the problem ... find the magnitude and direction of the position vector at t = 70 seconds.
i still have it wrong
21560 m
it tells me Your answer is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation.
4. Show us what you have done. It should work out.
5. Originally Posted by o_O
Show us what you have done. It should work out.
i got it now.
it is the resultant, not the horizontal thing i got before. | 330 | 1,237 | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-22 | longest | en | 0.913551 |
https://www.zhygear.com/description-of-involute-profile-of-the-most-complete-gear-transmission/ | 1,723,716,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00124.warc.gz | 830,064,520 | 28,615 | # Description of involute profile of the most complete gear transmission
#### The basic requirements of gear transmission for tooth profile curve
First, the transmission should be stable; second, the bearing capacity should be strong
#### The formation and property of involute
1.The formation of involute
When a moving line (occurrence line) rolls purely along a fixed circle (base circle), the trace of any point K on the moving line is called the involute of the circle.
2.Properties of involute
It can be seen from the formation of involute:
(1) The segment KB on which the occurrence line rolls over the base circle is equal to the arc length AB on the base circle.
(2) The normal of any point K on the involute must be tangent to the base circle.
The radius of curvature of each point on the involute is not equal.
The farther away the point is from the base circle, the larger the radius of curvature and the straighter the involute. vice versa.
(4) The shape of the involute determines the size of the base circle.
The base circle is the same, and the shape of the involute is the same.
When the radius of the base circle is infinite, the involute will become a straight line, and the gear will become a rack.
(5) There is no involute in the base circle.
#### Basic law of involute profile engagement
In order to keep the instantaneous transmission ratio constant, no matter where the tooth profile of the two wheels contacts, the common normal line passing through the contact point must intersect the connecting center line of the two wheels at a fixed point.
#### Meshing characteristics of involute profile
1.Constant transmission ratio
2.The transmission ratio of the two gears is inversely proportional to the radius of the two pitch circles and the radius of the two base circles. Because the pitch radius and base radius of two meshing gears are fixed values, the transmission ratio can be kept constant
3.Separability of transmission
When the center distance of two wheels changes slightly, the instantaneous transmission ratio will remain unchanged, which is called the separability of involute gear transmission.
4.Due to the manufacturing and installation errors of gears, there are some errors between the actual center distance and the design center distance of involute gears. However, due to the separability, the transmission ratio can still remain unchanged. Engagement angle is fixed
Cos α′ = RB1 / R1 ′ = Rb2 / r2 ′ = constant
It shows that the engagement angle α ‘of the involute profile is a fixed value.
Since the meshing angle is constant, the pressure direction between the tooth profiles will not change, which is very beneficial to the stability of gear transmission.
Scroll to Top | 564 | 2,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-33 | latest | en | 0.88569 |
https://git.libre-soc.org/?p=soc.git;a=blob;f=src/soc/fu/mul/formal/proof_main_stage.py;h=0cf767f341668663a4f4c736af5aea5625c1c0d2;hb=HEAD | 1,718,721,547,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00592.warc.gz | 243,457,255 | 9,535 | 1 # Proof of correctness for multiplier
2 # Copyright (C) 2020 Michael Nolan <mtnolan2640@gmail.com>
3 # Copyright (C) 2020 Luke Kenneth Casson Leighton <lkcl@lkcl.net>
4 # Copyright (C) 2020 Samuel A. Falvo II <kc5tja@arrl.net>
6 """Formal Correctness Proof for POWER9 multiplier
8 notes for ov/32. similar logic applies for 64-bit quantities (m63)
10 m31 = exp_prod[31:64]
11 comb += expected_ov.eq(m31.bool() & ~m31.all())
13 If the instruction enables the OV and OV32 flags to be
14 set, then we must set them both to 1 if and only if
15 the resulting product *cannot* be contained within a
16 32-bit quantity.
18 This is detected by checking to see if the resulting
19 upper bits are either all 1s or all 0s. If even *one*
20 bit in this set differs from its peers, then we know
21 the signed value cannot be contained in the destination's
22 field width.
24 m31.bool() is true if *any* high bit is set.
25 m31.all() is true if *all* high bits are set.
27 m31.bool() m31.all() Meaning
28 0 x All upper bits are 0, so product
29 is positive. Thus, it fits.
30 1 0 At least *one* high bit is clear.
31 Implying, not all high bits are
32 clones of the output sign bit.
33 Thus, product extends beyond
34 destination register size.
35 1 1 All high bits are set *and* they
36 match the sign bit. The number
37 is properly negative, and fits
38 in the destination register width.
40 Note that OV/OV32 are set to the *inverse* of m31.all(),
41 hence the expression m31.bool() & ~m31.all().
42 """
45 from nmigen import (Module, Signal, Elaboratable, Mux, Cat, Repl,
46 signed)
47 from nmigen.asserts import Assert, AnyConst, Assume, Cover
48 from nmutil.formaltest import FHDLTestCase
49 from nmutil.stageapi import StageChain
50 from nmigen.cli import rtlil
52 from soc.decoder.power_fields import DecodeFields
53 from soc.decoder.power_fieldsn import SignalBitRange
55 from soc.fu.mul.pipe_data import CompMULOpSubset, MulPipeSpec
56 from soc.fu.mul.pre_stage import MulMainStage1
57 from soc.fu.mul.main_stage import MulMainStage2
58 from soc.fu.mul.post_stage import MulMainStage3
60 from soc.decoder.power_enums import MicrOp
61 import unittest
64 # This defines a module to drive the device under test and assert
65 # properties about its outputs
66 class Driver(Elaboratable):
67 def __init__(self):
68 # inputs and outputs
69 pass
71 def elaborate(self, platform):
72 m = Module()
73 comb = m.d.comb
75 rec = CompMULOpSubset()
77 # Setup random inputs for dut.op
78 comb += rec.insn_type.eq(AnyConst(rec.insn_type.width))
79 comb += rec.fn_unit.eq(AnyConst(rec.fn_unit.width))
80 comb += rec.is_signed.eq(AnyConst(rec.is_signed.width))
81 comb += rec.is_32bit.eq(AnyConst(rec.is_32bit.width))
82 comb += rec.imm_data.imm.eq(AnyConst(64))
83 comb += rec.imm_data.imm_ok.eq(AnyConst(1))
85 # set up the mul stages. do not add them to m.submodules, this
86 # is handled by StageChain.setup().
87 pspec = MulPipeSpec(id_wid=2)
88 pipe1 = MulMainStage1(pspec)
89 pipe2 = MulMainStage2(pspec)
90 pipe3 = MulMainStage3(pspec)
92 class Dummy: pass
93 dut = Dummy() # make a class into which dut.i and dut.o can be dropped
94 dut.i = pipe1.ispec()
95 chain = [pipe1, pipe2, pipe3] # chain of 3 mul stages
97 StageChain(chain).setup(m, dut.i) # input linked here, through chain
98 dut.o = chain[-1].o # output is the last thing in the chain...
100 # convenience variables
101 a = dut.i.ra
102 b = dut.i.rb
103 o = dut.o.o.data
104 xer_ov_o = dut.o.xer_ov.data
105 xer_ov_ok = dut.o.xer_ov.ok
107 # For 32- and 64-bit parameters, work out the absolute values of the
108 # input parameters for signed multiplies. Needed for signed
109 # multiplication.
111 abs32_a = Signal(32)
112 abs32_b = Signal(32)
113 abs64_a = Signal(64)
114 abs64_b = Signal(64)
115 a32_s = Signal(1)
116 b32_s = Signal(1)
117 a64_s = Signal(1)
118 b64_s = Signal(1)
120 comb += a32_s.eq(a[31])
121 comb += b32_s.eq(b[31])
122 comb += a64_s.eq(a[63])
123 comb += b64_s.eq(b[63])
125 comb += abs32_a.eq(Mux(a32_s, -a[0:32], a[0:32]))
126 comb += abs32_b.eq(Mux(b32_s, -b[0:32], b[0:32]))
127 comb += abs64_a.eq(Mux(a64_s, -a[0:64], a[0:64]))
128 comb += abs64_b.eq(Mux(b64_s, -b[0:64], b[0:64]))
130 # For 32- and 64-bit quantities, break out whether signs differ.
131 # (the _sne suffix is read as "signs not equal").
132 #
133 # This is required because of the rules of signed multiplication:
134 #
135 # a*b = +(abs(a)*abs(b)) for two positive numbers a and b.
136 # a*b = -(abs(a)*abs(b)) for any one positive number and negative
137 # number.
138 # a*b = +(abs(a)*abs(b)) for two negative numbers a and b.
140 ab32_sne = Signal()
141 ab64_sne = Signal()
142 comb += ab32_sne.eq(a32_s ^ b32_s)
143 comb += ab64_sne.eq(a64_s ^ b64_s)
145 # setup random inputs
146 comb += [a.eq(AnyConst(64)),
147 b.eq(AnyConst(64)),
148 ]
150 comb += dut.i.ctx.op.eq(rec)
152 # check overflow and result flags
153 result_ok = Signal()
154 enable_overflow = Signal()
156 # default to 1, disabled if default case is hit
157 comb += result_ok.eq(1)
159 # Assert that op gets copied from the input to output
160 comb += Assert(dut.o.ctx.op == dut.i.ctx.op)
161 comb += Assert(dut.o.ctx.muxid == dut.i.ctx.muxid)
163 # Assert that XER_SO propagates through as well.
164 comb += Assert(dut.o.xer_so == dut.i.xer_so)
166 # main assertion of arithmetic operations
167 with m.Switch(rec.insn_type):
169 ###### HI-32 #####
171 with m.Case(MicrOp.OP_MUL_H32):
172 comb += Assume(rec.is_32bit) # OP_MUL_H32 is a 32-bit op
174 exp_prod = Signal(64)
175 expected_o = Signal.like(exp_prod)
177 # unsigned hi32 - mulhwu
178 with m.If(~rec.is_signed):
179 comb += exp_prod.eq(a[0:32] * b[0:32])
180 comb += expected_o.eq(Repl(exp_prod[32:64], 2))
181 comb += Assert(o[0:64] == expected_o)
183 # signed hi32 - mulhw
184 with m.Else():
185 # Per rules of signed multiplication, if input signs
186 # differ, we negate the product. This implies that
187 # the product is calculated from the absolute values
188 # of the inputs.
189 prod = Signal.like(exp_prod) # intermediate product
190 comb += prod.eq(abs32_a * abs32_b)
191 comb += exp_prod.eq(Mux(ab32_sne, -prod, prod))
192 comb += expected_o.eq(Repl(exp_prod[32:64], 2))
193 comb += Assert(o[0:64] == expected_o)
195 ###### HI-64 #####
197 with m.Case(MicrOp.OP_MUL_H64):
198 comb += Assume(~rec.is_32bit)
200 exp_prod = Signal(128)
202 # unsigned hi64 - mulhdu
203 with m.If(~rec.is_signed):
204 comb += exp_prod.eq(a[0:64] * b[0:64])
205 comb += Assert(o[0:64] == exp_prod[64:128])
207 # signed hi64 - mulhd
208 with m.Else():
209 # Per rules of signed multiplication, if input signs
210 # differ, we negate the product. This implies that
211 # the product is calculated from the absolute values
212 # of the inputs.
213 prod = Signal.like(exp_prod) # intermediate product
214 comb += prod.eq(abs64_a * abs64_b)
215 comb += exp_prod.eq(Mux(ab64_sne, -prod, prod))
216 comb += Assert(o[0:64] == exp_prod[64:128])
218 ###### LO-64 #####
219 # mulli, mullw(o)(u), mulld(o)
221 with m.Case(MicrOp.OP_MUL_L64):
223 with m.If(rec.is_32bit): # 32-bit mode
224 expected_ov = Signal()
225 prod = Signal(64)
226 exp_prod = Signal.like(prod)
228 # unsigned lo32 - mullwu
229 with m.If(~rec.is_signed):
230 comb += exp_prod.eq(a[0:32] * b[0:32])
231 comb += Assert(o[0:64] == exp_prod[0:64])
233 # signed lo32 - mullw
234 with m.Else():
235 # Per rules of signed multiplication, if input signs
236 # differ, we negate the product. This implies that
237 # the product is calculated from the absolute values
238 # of the inputs.
239 comb += prod.eq(abs32_a[0:64] * abs32_b[0:64])
240 comb += exp_prod.eq(Mux(ab32_sne, -prod, prod))
241 comb += Assert(o[0:64] == exp_prod[0:64])
243 # see notes on overflow detection, above
244 m31 = exp_prod[31:64]
245 comb += expected_ov.eq(m31.bool() & ~m31.all())
246 comb += enable_overflow.eq(1)
247 comb += Assert(xer_ov_o == Repl(expected_ov, 2))
249 with m.Else(): # is 64-bit; mulld
250 expected_ov = Signal()
251 prod = Signal(128)
252 exp_prod = Signal.like(prod)
254 # From my reading of the v3.0B ISA spec,
255 # only signed instructions exist.
256 #
257 # Per rules of signed multiplication, if input signs
258 # differ, we negate the product. This implies that
259 # the product is calculated from the absolute values
260 # of the inputs.
261 comb += Assume(rec.is_signed)
262 comb += prod.eq(abs64_a[0:64] * abs64_b[0:64])
263 comb += exp_prod.eq(Mux(ab64_sne, -prod, prod))
264 comb += Assert(o[0:64] == exp_prod[0:64])
266 # see notes on overflow detection, above
267 m63 = exp_prod[63:128]
268 comb += expected_ov.eq(m63.bool() & ~m63.all())
269 comb += enable_overflow.eq(1)
270 comb += Assert(xer_ov_o == Repl(expected_ov, 2))
272 # not any of the cases above, disable result checking
273 with m.Default():
274 comb += result_ok.eq(0)
276 # check result "write" is correctly requested
277 comb += Assert(dut.o.o.ok == result_ok)
278 comb += Assert(xer_ov_ok == enable_overflow)
280 return m
283 class MulTestCase(FHDLTestCase):
284 def test_formal(self):
285 module = Driver()
286 self.assertFormal(module, mode="bmc", depth=2)
287 self.assertFormal(module, mode="cover", depth=2)
288 def test_ilang(self):
289 dut = Driver()
290 vl = rtlil.convert(dut, ports=[])
291 with open("main_stage.il", "w") as f:
292 f.write(vl)
295 if __name__ == '__main__':
296 unittest.main() | 2,802 | 9,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.682613 |
statsassignmenthelp49383.acidblog.net | 1,544,454,780,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00062.warc.gz | 268,258,331 | 5,988 | # Stata Project Help - An Overview
Evaluation of Variance can be a take a look at used to carry out Evaluation on numerous types of components that have an impact on a data established. It may be used to ascertain if a romantic relationship exist in between two or even more groups. It is mostly utilised when a statistical computer software will not be readily available and computing needs to be accomplished by hand.
I need help in my STATA project. Additional particulars will probably be shared with suitable bidders and we are able to examine cost. I need this accomplished by 23rd November 2017.
An algorithm with the distribution from the Roy's premier root underneath the null hypothesis was derived in [8] while the distribution below the alternative is examined in.[9]
The pool of statistical details accessible offers college students an notion concerning the matter on which They may be to study, and what perform has been a person just before that investigation.
)At school, all we did was excel this, excel that. And around the check he would as a selected calculation about a thing that is further than the textbook and his powerpoints. 0 individuals identified this helpful 0 men and women didn't come across this useful report this rating Load Additional No ratings found – look at all ratings for this professor.
We had been amazed and delighted with the caliber of the tutoring our daughter gained at MyTutor. She employed two tutors for a handful of classes near to her GCSEs As well as in both equally occasions the tutors were being well-well prepared and able to offer very helpful previous-minute recommendations and tips.
Study summary figures and frequencies routinely while you perform details preparing, specially when you create new variables or change the framework within your information. See if Whatever you get is plausible. If the results adjust, ensure you can make clear why.
It As a result offers some safety from model mis-specification, in that so assuming that among the list of two types is correctly specified, our estimates are consistent. For our simple case in point, this can be carried out utilizing:
When you're typing within the command window a command may be assuming that desired. Inside a do-file you will likely want to break lengthy commands into strains to improve readability.
The STATA can be accessed by help of desktop or perhaps laptops in a computer lab. Given that That is an integrated program, hence it is on the market in labs mostly.
Stata is available for Home windows, Unix, and Mac desktops. This tutorial focuses on the Windows Edition, but the vast majority of contents applies to the other platforms also. The regular Model is called Stata/IC (or Intercooled Stata) and might deal with up to two,047 variables. There's a Distinctive edition known as Stata/SE that will deal with nearly 32,766 variables (and in addition permits extended string variables and larger matrices), and also a Variation for multicore/multiprocessor desktops referred to as Stata/MP, which will allow bigger datasets which is substantially faster.
Our function is rather in depth and easy to grasp. The coed can use it in visit our website upcoming for reference or making ready for their tests
A fantastic assistance, Tremendous user friendly. We have been thrilled to have found these tutoring expert services; the tutor we have been working with is actually exceptional.
We employed many tutors leading around our son's GCSE examinations and in just about every circumstance they helped be certain why not find out more he realized a good go quality. With significant calibre tutor's, the service is cost efficient, practical, and interactive. | 733 | 3,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-51 | latest | en | 0.966431 |
https://web2.0calc.com/questions/the-sequence-beginning-with-3-7-follows-this-rule | 1,600,773,529,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400205950.35/warc/CC-MAIN-20200922094539-20200922124539-00144.warc.gz | 706,011,723 | 7,102 | +0
# The sequence beginning with 3, 7... follows this rule:
-3
354
3
+956
The sequence beginning with 3, 7... follows this rule:
Apr 14, 2019
#1
+2856
+1
So since 7 is the second term, we multiply 2 by 4 and add it to 7. We get 15
Our sequence so far: 3,7,15,
Then we multiply 4 and 4 to get 16 as 15 is the FOURTH term then add 16 to 15. We get 31.
Do you notice a pattern? 3,7,15,31.
Yes 2n+1 can represent each change for the terms.
So we keep calculating...
Apr 14, 2019
edited by Guest Apr 14, 2019
edited by CalculatorUser Apr 14, 2019
#2
+111455
+2
A little confusing, but....
Term number 1 2 3 4 5 6 7
3 7 15 27 43 63 87
Note that the 3rd term = previous term + 4 times the previous term number = 7 + 4(2) = 15
4th term = 15 + 4(3) = 27
5th term = 27 + 4(4) = 43
6th term = 43 + 4(5) = 63
7th term = 63 + 4(6) = 87
Notice something else.....the difference between terms is 4, 8, 12, 16, 20, 24
Just for the heck of it, what would be the 8th term ????
Apr 14, 2019
#3
+956
-4
If we continue this same rule for the next term in the sequence then...
The 8th term would be: 8th term = 87 + 4(7) = 115
--7H3_5H4D0W
GAMEMASTERX40 Apr 22, 2019 | 501 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-40 | latest | en | 0.820925 |
https://codegolf.stackexchange.com/questions/47701/fractal-points-on-a-line | 1,716,232,552,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00252.warc.gz | 148,259,624 | 47,486 | Fractal Points on a Line
Sometimes, when I'm really bored (really bored), I like to draw a line segment and draw points on it.
First, I draw a line segment of a certain size, which is 2^N for some value of N. The line will be represented by a series of . characters.
................
Then, I plot a point at the left end. Points will be represented by X characters.
X...............
Then, I follow a pattern. Starting at the most recently plotted point (which I'll call A), I advance to the next plotted point (B) on the line (wrapping around as necessary). Then, I advance to the next plotted point on the line (C). Then, I plot a new point half-way in between this third point (C) and the next already plotted point (D).
Whenever you wrap around the line, the "middle" is determined in wrapping manner. The newly plotted point is always to the right of C.
Let's say that the following line was my current line. Here is how I would plot the next two points. For this example, I'll label each important point with a letter.
X...A...X.X...X.
^
X...A...B.X...X.
^
X...A...B.C...X.
^
X...A...B.C...D.
^
X...X...X.X.A.X.
^
X...X...X.X.A.B.
^
C...X...X.X.A.B.
^
C...D...X.X.A.B.
^
X.A.X...X.X.X.X.
^
Returning to the previous example, the next point will be plotted in the middle of the line.
X.......X.......
This is perhaps a little bit of a special case: advancing to the next point simply leaves you where you started. The only useful halfway point is the "cyclic" halfway point (the halfway point on the line), as opposed plotting a point on top of itself.
Below is the series of points that I would plot on the line from here to the end.
X.......X.......
X.......X...X...
X.......X.X.X...
X...X...X.X.X...
X...X...X.XXX...
X.X.X...X.XXX...
X.X.X...XXXXX...
There is no longer any room to plot the next point, as it would have to be wedged between two adjacent points, so I have reached the maximum depth for the given value of N = 4. The last line in the above list is "complete."
The Challenge
The goal is to write the shortest program/named function that will print/return the completed line for a given value of N. The above shows N = 4.
Input
Input will be a single non-negative integer N. The length of the generated line will then be 2^N.
Output
Output will be the completed line of length 2^N, formed by . and X characters. A trailing newline doesn't matter.
Example I/O
0
X
1
XX
2
X.XX
3
X.X.XXX.
4
X.X.X...XXXXX...
5
X.X.X...X...X...X.XXX.XXX.......
Python 2, 137
n=input()
l=[0]*2**n;i=0
while~i%2:i=i/2%2**n;l[i]=1;i=sum([k for k,v in enumerate(l*4)if(k>i)*v][1:3])
print''.join(['.X'[x]for x in l])
Quite straightforward.
Pyth, 49
More or less a translation. Main difference is that I don't use a list representing the line, I use a string.
J*\.^2QW!%Z2K%/Z2lJ=JXJK\X=Zs<tf&q\X@JT>TKU*4J2)J
Try it online.
Q=input(); Z=0 #implicit
J*\.^2Q J = "."*(2^Q)
W!%Z2 while !(Z%2):
K%/Z2lJ K=(Z/2)%(len(J))
=JXJK\X change J, the point at index K is changed to a "X"
f U*4J filter all elements T in [0, 1, 2, ..., 4*len(J)-1]:
&q\X@JT>TK where J[T]=='X' and T>K
<t 2 only us the second and third element
=Zs and store the sum to Z
)J end while and print J
Clip, 95
[z?zF#2(z*2*:(#2(z'.'X]'X]]n[Fa[b[q[j[r?=rirFrsrb'X]b]][t[u+*<ut*Blb/+tu2]]g+2jqg+3jq]#qa]%b'X}
~2\?,[]0{.@|$4*.@?2+1$>2<.+~<!4$,*++.2/\1&!}do;{&!'X.'1/=}+% The number of loop iterations required seems to be A061419, but the do-loop is shorter than calculating that. A divmod would save a char inside the do loop. The part which feels most wasteful is the output conversion, but I don't see how to improve it. CJam, 55 53 51 50 bytes 2ri#:N'.*0aN{):U+__Nf++$_U#))>2<1bNe|2/N%|}*{'Xt}/
Try it online here
Java, 209207195 191 bytes
I'm surprised I was able to get it this short. There is probably still room for improvement. As usual, suggestions will be appreciated :)
This returns a char[]. Call using a(n).
char[]a;int b,c,d,e=2;char[]a(int f){java.util.Arrays.fill(a=new char[b=1<<f],'.');for(a[0]=88;d+1<e;c=(d+e)/2,a[c%b]=88)e=b(d=b(b(c)));return a;}int b(int f){for(;;)if(a[++f%b]>87)return f;}
Indented:
char[] a;
int b, c, d, e = 2;
char[] a(int f){
java.util.Arrays.fill(
a = new char[
b = 1 << f
]
, '.'
);
for(
a[0] = 88;
d + 1 < e;
c = (d + e) / 2,
a[c % b] = 88
)
e = b(
d = b(
b(c)
)
);
return a;
}
int b(int f){
for (;;)
if (a[++f % b] > 87)
return f;
}
12 bytes thanks to Peter :)
4 bytes thanks to TNT ;)
• (c%b+b)%b? Are you expecting c to be negative? Mar 12, 2015 at 21:18
• c=0 and d=0 can be shortened to just c and d. int types defined at the class level are automatically initialized to 0.
– TNT
Mar 13, 2015 at 13:35
(a:b)%(c:d)|a<c=a:b%(c:d)|1<2=c:(a:b)%d
f i=putStr\$map(!(0:g[0,n..]))[0..n-1]where n=2^i;e!l|eeleml='X'|1<2='.';g(_:_:c:d:r)|m==c=[]|1<2=mod m n:g((d:r)%[m,m+n..])where m=div(c+d)2
Usage: f 5. Output: X.X.X...X...X...X.XXX.XXX........
Unfortunately Haskell doesn’t have a merge function in the standard libraries, so I have to provide my own (-> %). Fortunately I have to merge only infinite lists, so I don’t have to cover the base cases, i.e. empty lists. It still costs 40 bytes.
How it works: instead of setting the Xs directly in an array, I keep a list of positions where they are. Furthermore I do not wrap around at 2^N but keep on increasing the positions towards infinity (e.g. for N=2 with an X at the front, the position list looks like [0,4,8,12,16,20,…]). I take the 3rd and 4th element (c and d), calculate the new position (c+d)/2, keep it for the output list, merge the old position list from position 4 (the d) on with a new one starting with (c+d)/2 and recur. I stop when (c+d)/2 equals c. Finally I add a 0 to the output list and print Xs at the given positions and . elsewhere.
step by step example, N=2
step position list (c+d)/2 output lists to merge (pos. list for next round)
list old list from d on / new list from (c+d)/2
#1 [0,4,8,12,16,…] 10 [10] [12,16,20,24,…] / [10,14,18,22,…]
#2 [10,12,14,16,18,…] 15 [10,15] [16,18,20,22,…] / [15,19,23,27,…]
#3 [15,16,18,19,20,…] 18
stop here, because c equals (c+d)/2
add 0 to the output list: [0,10,15]
take all elements modulo 2^N: [0,2,3]
print X at position 0, 2 and 3
Mathematica, 110102112 108
a=Array["."&,n=2^Input[]];a[[Mod[Round@{n/2,n}//.{x_,y_,z___}/;y-x>1:>{z,x+n,(x+y)/2+n,y+n},n]+1]]="X";""<>a
` | 2,226 | 6,610 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-22 | latest | en | 0.916106 |
https://fr.slideserve.com/dacia/cs3410-hw1-review | 1,632,326,479,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00387.warc.gz | 306,459,252 | 20,747 | CS3410 HW1 Review
CS3410 HW1 Review
Télécharger la présentation
CS3410 HW1 Review
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
1. CS3410 HW1 Review 2014, 2, 21
2. Agenda • We will go through the HW1 questions together • TAs will then walk around to help
3. Question 1: Karnaugh Map • Sum of products: • Karnaugh map minimization: • Cover all 1’s • Group adjacent blocks of 2n 1’s that yield a regular shape • Encode common features ab c 00 01 11 10 0 0 0 0 1 1 1 1 0 1
4. Rules for Karnaugh Map Minimization • Minterms can overlap • Minterms can span 1, 2, 4, 8 … cells • The map can wrap around ab ab c c 00 01 11 10 10 00 01 11 0 0 1 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0
5. Question 2: Numbers & Arithmetic • Binary translation: • Base conversion via repetitive division • From binary to Hex and Oct • Negating a number (2’s complement) • Overflow • Overflow happened iff carry into msb != carry out of msb
6. Question 4: FSM
7. Question 4: FSM (cont.) Spam filter x 2 0 1 2 0 1 0 1 ….. x2 x0 x1 Output Okay
8. Question 4: FSM (cont.) Spam filter x 2 0 1 2 0 1 0 1 ….. x2 x0 x1 Output Okay SPAM
9. Question 4: FSM (cont.) State (x1=0, x2=0) and (x1=0, x2=1) have exactly the same transitions AND output. So they are NOT distinct states.
10. Question 7: Performance • Instruction mix for some program P, assume: • 25% load/store ( 3 cycles / instruction) • 60% arithmetic ( 2 cycles / instruction) • 15% branches ( 1 cycle / instruction) • CPI: • 3 * .25 + 2 * .60 + 1 * .15 = 2.1 • CPU Time = # Instructions x CPI x Clock Cycle Time • Assuming 400k instructions, 30 MHz : • 400k * 2.1 / 30 = 28000 µs (1 µs = 1 microsecond = 1/1M S)
11. Question 8 Registers and Control are in parallel
12. Question 8 (cont.) • Refer to section 1.6 in the text book • 4.3.1: The clock cycle time is determined by the critical path (the load instruction) • 4.3.2: • Speedup = • Execution time = cycle time * num of instructions • Speedup < 1 means we are actually slowing down Execution time (old) Execution time (new)
13. Question 8 (cont.) • 4.3.3: • Cost-performance ratio (CPR) = • The higher, the better • This question asks for a “comparison” of CPR CPR ratio= = * Performance Cost CPR (old) Cost (new) Perf (old) CPR (new) Cost (old) Perf (new) 1 speedup
14. Question 9/10/11: Assembler Code • Writing the instructions in a human-readable format • http://www.cs.cornell.edu/courses/CS3410/2014sp/MIPS_Vol2.pdf • Core instruction set • http://www.cs.cornell.edu/courses/CS3410/2014sp/project/pa1/pa1.html
15. Assembler Code • When writing the assembler code: • Decide which register stores which variable • Typically you should use \$t0~\$t9 and \$s0~\$s7 (You don’t need to understand their difference now) • Decide which instruction you want to use • Get familiar with the core instruction set • Get familiar with some basic patterns
16. Basic Assembler Coding Patterns • Arithmetic • C code: • Assembler: a = b + c; #a: \$s0, b: \$s1, c:\$s2 ADD \$s0, \$s1, \$s2
17. Basic Assembler Coding Patterns • Brunch • C code: • Assembler: if(a < b) //DO A... else //DO B... #a: \$s0, b: \$s1 SLT \$t0, \$s0, \$s1 BEQ \$t0, \$zero, POINTB #DO A... POINTB: #DO B...
18. Basic Assembler Coding Patterns • While loop • C code: • Assembler: while(a < b) //Do something... #a: \$s0, b: \$s1 LOOP: SLT \$t0, \$s0, \$s1 BEQ \$t0, \$zero, EXIT #Do something... J LOOP EXIT: #Out of the loop...
19. Basic Assembler Coding Patterns • Array access • C code: • Assembler: intmyArray[10]; a = myArray[2]; #a: \$s0, myArray: \$s1 LW \$s0, 8(\$s1)
20. C Programming • Have you tried the hello-world? • Use csuglab machines. It is easier. • How to read input from the terminal? • scanf: • You need a buffer for it int scanf ( const char * format, ... );
21. Good Luck! Questions? | 1,297 | 3,862 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.697939 |
http://metamath.tirix.org/mpests/h1de2bi | 1,718,641,096,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861733.59/warc/CC-MAIN-20240617153913-20240617183913-00885.warc.gz | 22,766,613 | 4,387 | # Metamath Proof Explorer
## Theorem h1de2bi
Description: Membership in 1-dimensional subspace. All members are collinear with the generating vector. (Contributed by NM, 19-Jul-2001) (Revised by Mario Carneiro, 15-May-2014) (New usage is discouraged.)
Ref Expression
Hypotheses h1de2.1 ${⊢}{A}\in ℋ$
h1de2.2 ${⊢}{B}\in ℋ$
Assertion h1de2bi ${⊢}{B}\ne {0}_{ℎ}\to \left({A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)↔{A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\right)$
### Proof
Step Hyp Ref Expression
1 h1de2.1 ${⊢}{A}\in ℋ$
2 h1de2.2 ${⊢}{B}\in ℋ$
3 his6 ${⊢}{B}\in ℋ\to \left({B}{\cdot }_{\mathrm{ih}}{B}=0↔{B}={0}_{ℎ}\right)$
4 2 3 ax-mp ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}=0↔{B}={0}_{ℎ}$
5 4 necon3bii ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0↔{B}\ne {0}_{ℎ}$
6 1 2 h1de2i ${⊢}{A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\to \left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}$
7 6 adantl ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}$
8 7 oveq2d ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)$
9 2 2 hicli ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\in ℂ$
10 9 recclzi ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ$
11 ax-hvmulass ${⊢}\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\wedge {B}{\cdot }_{\mathrm{ih}}{B}\in ℂ\wedge {A}\in ℋ\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)$
12 9 1 11 mp3an23 ${⊢}\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)$
13 10 12 syl ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)$
14 ax-1cn ${⊢}1\in ℂ$
15 14 9 divcan1zi ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({B}{\cdot }_{\mathrm{ih}}{B}\right)=1$
16 15 oveq1d ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}=1{\cdot }_{ℎ}{A}$
17 13 16 eqtr3d ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)=1{\cdot }_{ℎ}{A}$
18 ax-hvmulid ${⊢}{A}\in ℋ\to 1{\cdot }_{ℎ}{A}={A}$
19 1 18 ax-mp ${⊢}1{\cdot }_{ℎ}{A}={A}$
20 17 19 syl6eq ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)={A}$
21 20 adantr ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({B}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{A}\right)={A}$
22 8 21 eqtr3d ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)={A}$
23 1 2 hicli ${⊢}{A}{\cdot }_{\mathrm{ih}}{B}\in ℂ$
24 ax-hvmulass ${⊢}\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\wedge {A}{\cdot }_{\mathrm{ih}}{B}\in ℂ\wedge {B}\in ℋ\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)$
25 23 2 24 mp3an23 ${⊢}\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)$
26 10 25 syl ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}=\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)$
27 mulcom ${⊢}\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\wedge {A}{\cdot }_{\mathrm{ih}}{B}\in ℂ\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right)=\left({A}{\cdot }_{\mathrm{ih}}{B}\right)\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)$
28 10 23 27 sylancl ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right)=\left({A}{\cdot }_{\mathrm{ih}}{B}\right)\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)$
29 23 9 divreczi ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}=\left({A}{\cdot }_{\mathrm{ih}}{B}\right)\left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)$
30 28 29 eqtr4d ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right)=\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}$
31 30 oveq1d ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right)\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}$
32 26 31 eqtr3d ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}$
33 32 adantr ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left(\frac{1}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}\left(\left({A}{\cdot }_{\mathrm{ih}}{B}\right){\cdot }_{ℎ}{B}\right)=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}$
34 22 33 eqtr3d ${⊢}\left({B}{\cdot }_{\mathrm{ih}}{B}\ne 0\wedge {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to {A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}$
35 34 ex ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left({A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\to {A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\right)$
36 23 9 divclzi ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ$
37 2 elexi ${⊢}{B}\in \mathrm{V}$
38 37 snss ${⊢}{B}\in ℋ↔\left\{{B}\right\}\subseteq ℋ$
39 2 38 mpbi ${⊢}\left\{{B}\right\}\subseteq ℋ$
40 occl ${⊢}\left\{{B}\right\}\subseteq ℋ\to \perp \left(\left\{{B}\right\}\right)\in {\mathbf{C}}_{ℋ}$
41 39 40 ax-mp ${⊢}\perp \left(\left\{{B}\right\}\right)\in {\mathbf{C}}_{ℋ}$
42 41 choccli ${⊢}\perp \left(\perp \left(\left\{{B}\right\}\right)\right)\in {\mathbf{C}}_{ℋ}$
43 42 chshii ${⊢}\perp \left(\perp \left(\left\{{B}\right\}\right)\right)\in {\mathbf{S}}_{ℋ}$
44 h1did ${⊢}{B}\in ℋ\to {B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)$
45 2 44 ax-mp ${⊢}{B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)$
46 shmulcl ${⊢}\left(\perp \left(\perp \left(\left\{{B}\right\}\right)\right)\in {\mathbf{S}}_{ℋ}\wedge \frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\wedge {B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)\to \left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)$
47 43 45 46 mp3an13 ${⊢}\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\in ℂ\to \left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)$
48 36 47 syl ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)$
49 eleq1 ${⊢}{A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\to \left({A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)↔\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)$
50 48 49 syl5ibrcom ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left({A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\to {A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)\right)$
51 35 50 impbid ${⊢}{B}{\cdot }_{\mathrm{ih}}{B}\ne 0\to \left({A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)↔{A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\right)$
52 5 51 sylbir ${⊢}{B}\ne {0}_{ℎ}\to \left({A}\in \perp \left(\perp \left(\left\{{B}\right\}\right)\right)↔{A}=\left(\frac{{A}{\cdot }_{\mathrm{ih}}{B}}{{B}{\cdot }_{\mathrm{ih}}{B}}\right){\cdot }_{ℎ}{B}\right)$ | 5,012 | 10,331 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 55, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.426711 |
https://answerbun.com/mathematics/determinant-of-a-linear-transform-between-two-different-vector-spaces-with-the-same-dimension/ | 1,670,375,758,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711121.31/warc/CC-MAIN-20221206225143-20221207015143-00372.warc.gz | 134,379,649 | 16,768 | # Determinant of a linear transform between two different vector spaces with the same dimension
Mathematics Asked by lyrin on November 28, 2020
Let $T:V to W$ be a linear transform where $V$ and $W$ have the same dimension. If $V = W$, the determinant of $T$ is independent of the choice of basis. It measures the $T$’s dilation ratio of the volume spaned by the basis vectors and the ratio is indepedent of the basis because the two space have the same basis. However, if we use different bases for $V$ and $W$ (although the two spaces are isomorphic anyway), the determinant may depend on the choice of the bases. Is that right? In this case, what kind of meaning does the determinant have?
(Rewritten - the original post went off into unnecessary territory, but I had leave and didn't have time to prune it appropriately then.)
Determinants are only defined on transformations of a finite-dimensional vector space into itself. They are not defined for transformations into another vector space, not even if that vector space is isomorphic to the first.
Most commonly, determinants are first defined on square matrices. To define them on transformations from an $n$-dimensional vector space $V$ over the field $Bbb F$ (if you don't understand that, just consider $Bbb F$ to be either $Bbb R$ or $Bbb C$, depending on whether your vector space is real or complex), you first find a basis for $V$. By that basis, you can identify every vector $v in V$ with an ordered $n$-tuple $(v_1, v_2, ..., v_n) in Bbb F^n$. This identification defines an invertible linear transformation $R : V to Bbb F^n$. If $T : V to V$ is a linear transformation, then $RTR^{-1}=Rcirc T circ R^{-1}$ defines a linear transformation from $Bbb F^n to Bbb F^n$. I.e., $RTR^{-1}$ is an $n times n$ matrix. As such its determinant is already defined. So we define $det T := det(RTR^{-1})$.
Now extrinsically, that definition depends on $R$ - which itself depends on the basis originally chosen. But it can be shown that the exact same value for $det T$ is obtained for any basis, so in fact $det T$ does not depend on the basis.
But one must be careful here: we make use of $R$ twice. Once to convert from $Bbb F^n$ to $V$ and then convert back from $V$ to $Bbb F^n$. Suppose we decided to choose different bases for those two conversions. We could try $RTS^{-1}$ where $S$ is defined by some other basis. But while it can be shown that $det STS^{-1} = det RTR^{-1}$, in general $det RTS^{-1} ne det RTR^{-1}$. For example, if the vectors for the basis defining $R$ happen to be $2$ times the vectors in the basis defining $S$, then $det RTS^{-1} = 2^ndet RTR^{-1}$. So the magic that allows you to get the same value of $det T$ regardless of the basis chosen requires the bases used for both transformations to be the same.
However, if $T : V to W$ with $V ne W$, then even when $V$ and $W$ are of the same dimension and therefore both isomorphic to $Bbb F^n$, you cannot choose the same bases for transforming $Bbb F^n to V$ and for transforming $W to Bbb F^n$. If $V$ and $W$ had the same basis, they would necessarily be the same vector space. Because we have no way of relating the two transformations $S : V to Bbb F^n$ and $R : W to Bbb F^n$, we cannot define a single $det T$ in this case. If $S'$ and $R'$ are tranformations defined by different bases, then in general, $det RTS^{-1} ne det R'TS'^{-1}$
So you can only talk about the determinant of a transformation $T : V to V$. The determinant of a transformation $T : V to W$ is not well-defined, even when $V$ and $W$ have the same dimension.
Correct answer by Paul Sinclair on November 28, 2020
You can in fact define the determinant for maps between different (non-canonically isomorphic) vector spaces. Let $$V,W$$ be vector space of dimension $$n$$. The determinant of a linear map $$T:Vto W$$ is the induced map on top wedge $$mathrm{det}(T):bigwedge^n(V)tobigwedge^n(W);v_1wedgecdotswedge v_n mapsto Tv_1wedgecdotswedge Tv_n$$. This function takes values in the vector space $$mathrm{Hom}(bigwedge^n(V),bigwedge^n(W))$$ which has dimension 1. When we have chosen an isomorphism of $$Vcong W$$ we get an isomorphism $$mathrm{Hom}(bigwedge^n(V),bigwedge^n(W)) cong mathrm{End}(bigwedge^n(V)) =mathbb{F}$$ which then gives the determinant as you may recognise it. Without this choice of isomorphism we cannot find a canonical isomorphism with $$mathbb{F}$$ but we can still say intelligent things about our determinant function. For example it is zero if the rank of $$T$$ is less than $$n$$.
Answered by Callum on November 28, 2020
## Related Questions
### How to prove that $x(t) = cos{(frac{pi}{8}cdot t^2)}$ aperiodic?
2 Asked on February 16, 2021 by yk1
### Most General Unifier computation
1 Asked on February 16, 2021 by milano
### How to check if a number can be represented as the sum of two consecutive perfect cubes.
2 Asked on February 16, 2021 by sqrt
### How to compute $mathbb{P}(B(tau)leq a)$, $B(t)$ is standard Brownian motion.
1 Asked on February 15, 2021 by wheea
### Convergence radius for the Taylor series of a function of two variables.
0 Asked on February 15, 2021 by elster
### Probability distribution of time at which 2 random walks meet
0 Asked on February 15, 2021 by deltachief
### Looking for closed-forms of $int_0^{pi/4}ln^2(sin x),dx$ and $int_0^{pi/4}ln^2(cos x),dx$
7 Asked on February 15, 2021 by anastasiya-romanova
### Proving $sum_{cyc}frac{(a-1)(c+1)}{1+bc+c}geq 0$ for positive $a$, $b$, $c$ with $abc=1$.
2 Asked on February 14, 2021 by darius-chitu
### Proof that $frac{2^{-x}-1}{x} = sum_{n=0}^{infty} frac{ (-1)^{n+1}x^n(ln2)^{n+1}}{(n+1)!}$
3 Asked on February 14, 2021 by sugaku
### $24ml+1=k^2$ has no solution for all $l=1 dots m$
1 Asked on February 14, 2021 by gevorg-hmayakyan
### I can’t find the analytical expression of the sum from this probability problem
1 Asked on February 14, 2021
### Slightly confused about the commutator subgroup
1 Asked on February 14, 2021
### Isomorphism from ring to ring. Exercise from General Topology (J. Kelley)
1 Asked on February 14, 2021 by flowian
### Polynomial with root $α = sqrt{2}+sqrt{5}$ and using it to simplify $α^6$
4 Asked on February 14, 2021
### $sigma(n)$ is injective?
1 Asked on February 14, 2021 by ferphi
### Polynomial multiplication is associative
0 Asked on February 14, 2021
### Find area of equilateral $Delta ABC$
2 Asked on February 14, 2021 by ellen-ellen
### Convergence in distribution of $N(0, 1/n)$
1 Asked on February 13, 2021
### Diagonal of parallelogram and parallelepiped
0 Asked on February 13, 2021 by return | 1,929 | 6,648 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.89983 |
https://365datascience.com/question/probability-numerical-outcomes/ | 1,713,134,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00300.warc.gz | 69,053,261 | 41,427 | 25 Oct 2021
Posted on:
24 Oct 2021
1
# Probability: Numerical Outcomes
how are these probablities calculated..??
A = 10 P(A) = 0.5
B = 20 P(B) = 0.4
C = 100 P(C) = 0.1
2 answers ( 0 marked as helpful)
Instructor
Posted on:
25 Oct 2021
1
Hey Bharath,
Thank you for your question!
These probabilities are arbitrary and just given as an example, you are not expected to derive them.
Kind regards,
365 Hristina
Posted on:
25 Oct 2021
0
ok | 150 | 454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.874466 |
https://bakerst221b.com/docs/ttp/malware/analyse/basics/assebly/assembly-operations/ | 1,709,537,687,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00106.warc.gz | 113,570,692 | 12,489 | # Operations
Created: 27.09.2020
In this article I’m describing all assembly operations that I’ve encountered myseld and also wasn’t lazy anough to put down an explanation about here. However, I won’t be paying much attention to some operation that I consider straightforward, like `ADD`. I’m going to put a flag for each operation indicating corresponding arch: `arm` or `x86` (just learning ARM myself for iOS analysis).
Most of instruction have the following anatomy: `instruction <destination operand>, <source operand>`. Some operations look like this: `instruction <source operand>` when `<destination operand>` is always the same register (default). An example: `MUL`. When `MUL`ing, you always multiply `eax` on some value.
## Moving Data
### MOV
`x86`
The god of assembly operations. Its main purpose is pretty obvious: to move stuff from one place to another (like the Tower of Hanoi). Let’s suppose that initially we have EAX = 0x89, EBX = 0x11 and ESI = 0x4037C0.
Instruction Description
`mov eax, ebx` Copy data from EBX into EAX. Now EAX = EBX = 0x89.
`mov ebx, 0x4037C4` Copy 0x4037C4 and put it into EBX. Now EAX = 0x89 and EBX = 0x4037C4
`mov eax, [ebx]` Copy data from address stored in EBX into EAX. If EBX = 0x4037C4, CPU goes to 0x4037C4 address, looks out for the data at this address, say, 0x77 and put it into EAX. Now EBX = 0x4037C4 and EAX is now 0x77.
`mov eax, [0x4037C4]` Since EBX = 0x4037C4, this operation is equivalent to the previous one.
`mov ebx, [esi+eax*4]` First, CPU calculates the address: ESI + EAX*4 = 0x4037C4 + 0x77*4 = 0x4037C4 + 1DC = 4039A0. Then, CPU goes to 0x4039A0 and copies the value at this address into EBX. Say, we have 0x33 there. So, EBX is not 0x33.
So, to conclude, you can `MOV` a value directly, a value at address, a value in another register, an address itself (which is technically also a value) or a value at address using expression (the last example). Whenever there is an address “in assembly’s mind”, you’ll see square brackets []. Whenever the value - no brackets. It’s something that in the higher levels of abstraction is usually called a reference type ([address]) when a reference is copies and a value type (value) when the value is copied. In case of reference types, whenever you change it, it changes elsewhere. For example, consider `mov eax, 0x4037C4` and `mov ebx, 0x4037C4`. If we `mov [eax], 0x42`, the `ebx` is also `0x42` now since it point to the same memory address.
### LEA
`x86`
For smarties, it’s called “load effective address”. Usually used for arrays and complex address calculations. Let’s assume that initially we have EAX = 0x89, EBX = 0x11 and ESI = 0x4037C0.
Instruction Description
`lea ebx, [eax*5 + 5]` `eax*5 + 5` is equivalent to `5*(eax+1)` (ordinary mathematical manipulation). `(0x89 + 1 )*5 = 0x99 * 5 = 2FD`. In not `lea` it would require 4 operations instead: `inc eax` (0x89 + 1); `mov ecx, 5` `mul ecx` (0x99 * 5); `mov ebx, eax`. For `mul` operation see below in Arithmetic section.
### MOV vs LEA
Let’s compare these two:
MOV LEA
`mov eax,[ebx+8]` `lea eax,[ebx+8]`
The first instruction (the one with `mov`) does the following: “Add 8 to the value at `ebx`, go to this address and store the value found in `eax`”. So, it calculates the address and gets the value at this address to store in `eax`.
The second instruction (the one with `lea`) does the following: “Add 8 to the value at `ebx` and store the result in `eax`”. So, it calculates the address and puts the address into `eax`.
To conclude, `lea` stores address (reference types) and `mov` usually stores values (value types). However, note that `mov` can move addresses as well, since address is also just an integer, i.e. value.
### MOVSXD
`x86`
Example: `movsxd rsi, [rbp+8h]`
Copies the contents of the `<src_operand>` to the `<dst_operand>` and sign extends the value to 16 or 32 bits. The size of the converted value depends on the operand-size attribute. In 64-bit mode, the instructionโs default operation size is 32 bits.
## Arithmetic
`x86`
`add eax, 5` - adds a value to a value in register, address or to another value.
### SUB
`x86`
Affected flags: `CF`, `ZF`
`CF` =1 if `<destination operand>` is less than the `<source operand>`, i.e. after substraction there is a negative number.`ZF` = `1` if `<destination operand>` = `<source operand>` and the result is zero.
Let’s assume that `EAX` = `0x99` and `EBX` = `0x2`.
instruction description
`sub eax, 0x99` Now `eax` is `0`, therefore `ZF` = `1`.
`sub ebx, 0x10` This results in negative number, therefore, `CF` = `1`.
### SBB
`x86`
Affected by flags: `CF`
Almoust the same but a little tricky. It’s affected by `CF` flag. If `CF` = 0, then `sbb eax, 0x10` is equivalent to `sub eax, 0x10` (which is `eax - 0x10`). If `CF` = 1, then it means: `eax = eax - 0x10 - 1`.
### MUL and DIV
`x86`
Affected flags:
`CF` = `OF` = `0` if the high-order bits of the product are 0.
Both of these instructions operate on a predefined register. For example, `mul ecx` is actually `mul eax, ecx` i.e. `eax` * `ecx`. The result is stored in register `AX`, `DX:AX`, or `EDX:EAX` (depending on the operand size). The high-order bits of the product are in `AH`, `DX`, or `EDX`, respectively.
### IMUL and IDIV
`x86`
Affected flags: `CF`, `OF`
Same as `MUL` and `DIV` but operate on signed values.`CF` = `OF` = `1` if significant bits are carried into the upper half of the result. `cdq` instruction is usully used before `IDIV`. It converts a double to quad, quote:
The CDQ instruction copies the sign (bit 31) of the value in the EAX register into every bit position in the EDX register.
Forms:
1. Like `MUL` and `DIV` when the `<src operand>` is used inly.
2. `imul edi, esi` when both `<dst operand>` and `<src operand>` are used
3. `imul edi, esi, edx`, when beside the `<dst operand>` there are two `<src operand>`. The operations are as follows: `esi`*`edx` = `edi`.
`<dst operand>` is always a register or memory address. `<src operand>` can be a register, an address or a value. When a value is used, it is sign-extended to the length of the destination operand format.
NB โ The length of the product is calculated to twice the length of the operands. With the one-operand form, the product is stored exactly in the destination. With the two- and three- operand forms, however, result is truncated to the length of the destination before it is stored in the destination register. This is why the `CF` or `OF` flag should be tested to ensure that no significant bits are lost.
If this instruction is used for unsigned operations (since the lower half of the product is the same regardless if the operands are signed or unsigned. ), the `CF` and `OF` flags cannot be used to determine if the upper half of the result is non-zero.
## Shifts
### ROR and ROL
`x86`
Affected flags: `CF` , `OF`
Rotate the integer n-time to the left or right. When you see such an instruction, very often it is an indication of encryption. To better understand both I need an example. Let’s take an 8-bit binary value. The initial state is:
`0 1 0 0 1 1 1 0 `
See the `1` at the beginning of this number, second bit from the left (let’s call him Matt). We will then locate him after `ROR` and `ROL`.
Let’s now `ROR` (rotate bits right) by 1. Every bit is moved to the right by one position. Our first state (for future reference):
`0 0 1 0 0 1 1 1`
Where the hell is Matt now? Now, this bit is the third bit from the left.
Let’s now `ROL` (rotate bits left) by 1. Every bit is moved to the left by one position. Our second state:
`1 0 0 1 1 1 0 0`
Where the hell is Matt now? This bit is the first bit from the left, so, he’s become the most significant bit in this number ๐ .
Let’s now `ROL` the last number by 1 again. Every bit is again moved to the left. But Matt has nowhere to move! He’s falling nowhere…
Where the hell is Matt now? He seemed to have got drowned, but he managed it through the swamp and emerged… But now…Matt used to be the most significant bit ๐when in the second state , but now he’s just a ๐ฉ, the least significant bit. As you can see, he’s the first from the end.This is the third state:
`0 0 1 1 1 0 0 1`
Let’s make him worthy again and give him his newly acquired and recently lost regalia. Let’s `ROR` him by 1 again and get back to the second state (unforunately he’ll have to dive into the swamp again):
`1 0 0 1 1 1 0 0`
When moving from the second to the third state Matt has been in a swamp, or in a wormhole ๐ if you prefer a space metaphor. Let me introduce our wormhole - `CF` flag. The spirit of Matt was printed on this flag. In other, less eloquent words, when falling from the edge into the swamp, his value (`1`) was copied into `CF`. So as any other bit that would “fall”. For example, if we get back to the third (and the most unfortunate for Matt) state (`0 0 1 1 1 0 0 1`). Matt’s spirit is still there, therefore `CF` is still `1`. Let’s `ROL` this number by 1 once again, Martha (who’s now the most significant, i.e. the first bit of the number) falls into the swamp, gets copied into `CF` and emerges at the end as the least significant bit ๐ฉ, making Matt now the second least significant bit, i.e. the second bit from the end (which is not that bad now)โ. Now we have the forth state:
`0 1 1 1 0 0 1 0`
and the `CF` = 0 now bearing Martha’s spirit.
The processor restricts the count to a number between 0 and 31 by masking all the bits in the count operand except the 5 least-significant bits.
### RCR and RCL
`x86`
Affected flags: `CF` , `OF`
It’s pretty much the same, with just one small difference. `CF` flag is now taken into account, it’s not just a wormhole ๐ anymore. Let’s consider the third state from the previous examples:
`0 0 1 1 1 0 0 1`
Let `CF` be 1 now (may be it was set by some preceding operation like `ROL`).
If we now `RCL`, the fourth state will be as follows:
`CF` = Martha = 0.
`0 1 1 1 0 0 1 1`
The value that was in `CF` is now at the end of our number (`1`), and it’s the most significant bit is now in `CF`. Everyone else has just shifted to the left by 1 bit. It’s as if we were operating not on a 8-bit value, but on a 9 bit value:
MAIN value CF
`0 1 1 1 0 0 1 1` `0`
which results in something like that: `0 1 1 1 0 0 1 1` `0`.
Let’s now `RCR` back to the third state. `CF` = `0`, now it is moved to it’s place (most significant bit) `0 0 1 1 1 0 0 1` and since it was Matt (`1`) who’s falling from the cliff, `CF` = `1`. Everyone else has just shifted to the right.
Another flag, which behaviour is quite peculiar, is `OF`. It only changes when we shift by 1. When we whift by 2 or more - nothing’s happening to it. After CPU’s performed the rotates, it calculated `OF` like this. For left rotates (`RCL` and `ROL`), the `OF` = `CF XOR the most-significant bit`. For right rotates, the `OF` = `most-significant-bit-1 XOR most-significant-bit-2`. For the example above with ` RCL`, when we enetered the fourth state:
`CF` = `0` and the number itself is `0 1 1 1 0 0 1 1`.
`OF` = `0 XOR 0` = `0`
For RCR operation leading us back to the third state: `0 0 1 1 1 0 0 1`. Never mind `CF` since it’s not included in the calculations. The two most significant bits after rotation are `0` and `0` (the first two digits). `OF` = `0 XOR 0` = again `0`.
### SHL and SHR
`x86`
Affected flags: `CF`
Shifts bits by the value specified in second operand to the left or to the right. The last bit dropped off is written to `CF` “before death โ ๏ธ “. Example:
`1 0 1 0 1 1 0 1`
Let’s `SHR` the above number: `0 1 0 1 0 1 1 0 `.
Let’s now `SHR` once again: `0 0 1 0 1 0 1 1 `.
The main rule here: for each `SHR` add a `0` at the beginning and remove one digit from the end. The same is for `SHL`: for each `SHL` add `0` to the end and remove one digit from the beginning.
Above number is 8 bit long. So we can pop 8 bits by shifting in one direction (`SHR` for example only). When the last digit is poped off its value is written to `CF`, in the example above it was `1`, hence now `CF` = `1`.
#### Useful tip
`SHL` can be used as an optimized multiplication and division by `2^n`”. Here is an example:
SHL equivalent
`shl eax, 1` eax * 2
`shl eax, 2` eax * 4
`shl eax, 3` eax * 8
### ROL/ROR vs SHL/SHR
What’s the difference between the two (well, even four)? When I inspected my old notes, I’ve got a little confused because I’d totally forgotten that. Tha’ts why I’ve included this section for future, should my memory fail me once again.
The difference between the two is pretty much the same as the difference between “rolling” and “shifting”. Say, we have a password padlock for a suitcase and set our passcode to `1234`.
Then we shuffle it and have `5432`. How to open it then? We rotate each dial until we get to our passcode digits: the dial with `5` is rotated 4 times to get `1`, the dial with `4` is rotated 6 times and etc. No one would expect that when we rotate a dial on the lock, it disapears after reaching the end. But that would be the case if the operation in the padlock’s intestines was shifting. And that what’s happening to the shifted bits when shifting:
So, in `ROR/ROL` instruction no bits are lost, all of the bits of the original number are preserved. They are just rolling like those numbers in the lock ๐. But with `SHL/SHR` instruction the numbers are dimped into a ๐ wormhole and never seen again. If we shift long enogh, we turn any number to a bunch of zeroes until the only footprint left would be a `CF` flag which will hold the last shifted and dropped off bit. But even this would be overwritten with `0` shoud you shift one last time…
## Comparisons
### TEST
`x86`
Affected flags: `ZF`
A beautiful instruction in that it’s so simply and lightweight. It does the same as `AND` but operands are not changed. The result is in `ZF` (either `0` or `1`).
AND test
`and eax, eax` `test eax`
Interesting! ๐ฎ โThe above operations are identical, but the second takes less CPU cycles. It’s usually used to test, whether the value is `0`.
### CMP
`x86`
Affected flags: `CF`, `ZF`
This one is like `SUB`. It’s almoust the same as `SUB eax, edx`, for example. This instruction, just like the previous one, doesn’t change operands, however:
ZF CF
dst = src `1` `0`
dst < src `0` `1`
dst > src `0` `0`
When dst = src, dst - src = 0 therefore we set `ZF` (zero flag) to `1`. When dst < src, dst - src = negative number therefor `CF` (carry flag) = `1`. When everything is primitive (dst > src), dst - src = positive number, hence no flags are changed.
โ When some of the flags were changed suring some previously performed operation, are they reverted to the states above according to the values in dst and src? Example, if `ZF` = 1 before our `CMP dst, src` where dst < src, will `ZF` be set to `0` after this instruction is executed?
## Buffers
### REP
`x86`
Affected flags: `ZF`
This class of instruction is comprosed of different kinds of loops. It uses RSI or ESI as the source (ESI means “source index”) and EDI or RDI as the destination (EDI means “destination index”). ECX (counter) is used as a … surprise-surprise… a counter. There are several types of `REP` instruction:
instruction description
`rep`
`repe` or `repz`
`repne` or `repnz`
`REP` family is never seen alone. It’s always followed by some operation. Why? Because basically it’s a repetition. You can’t repeat nothing. There repeatiotions are performed on buffers (strings, for example). There are 4 possible operations seen with `rep`:
instruction description C++ analogr
`repe cmpsb` Compare two buffers `memcmp`
`rep stosb` Set all bytes to some value in `AL` `memset`
`rep movsb` Copy a buffer `memcpy`
`repne scasb` Search for a byte
`repe cmpsb`. To better illustrate, I’ve written the below pseudocode:
``````function bool compare(){
edi = ['d','s', 't', '1'];
esi = ['s', 'r', 'c', '1'];
for(ecx = len(edi); ecx >= 0; ecx--){
if (edi[ecx] != esi[ecx]) return false;
}
return true;
}
``````
`ECX` is set to the buffer’s length, `ESI` - is a pointer to the first buffer, `EDI` - the pointer to the second. The loop runs until `ECX` = 0 or the bytes compared are different. The above loop will run 2 times and return false when `ecx` = `2` since `edi[2]='t'` and `esi[2] = 'c'` which means that the buffers are different and there is no need to run the loop further.
`rep stosb`. Destination buffer - `EDI`, source - `AH`. `ECX` is a counter.
``````function buffer[] init(){
ah = 'a';
edi = [];
for (ecx = len(edi); ecx >= 0; ecx--){
edi[ecx] = ah;
}
return edi;
}
``````
The above loop will run 3 times. Upon function return `edi` = `['a', 'a', 'a']`. Very often is seen after `xor eax, eax`, since `xor` something on itself returns that something being filled with `0`, i.e. it means zeroing out a value. And we need to make sure there is no garbage lurking in `EAX` before setting it the desired value (in our example, `'a'`) to be later used to set `edi` to `a`. Just to remind, `al` is the lowerst byte of `EAX` register.
`rep movsb`. `ESI` - source buffer, `EDI` - destination buffer, `ECX` - counter.
``````function void copy(){
esi = ['s', 'r', 'c'];
edi = ['d', 's', 't'];
for (ecx = len(edi); ecx >= 0; ecx--) {
edi[ecx] = esi[ecx];
}
return;
}
``````
The above loop will run 3 times. At the end, `edi=esi = ['s', 'r', 'c']`.
`repne scasb`. `EDI` - buffer address, `AL` - byte to search. `ECX` - counter.
``````function bool search(){
edi = ['d', 's', 't'];
al = 'd';
// len(edi) = 3
for (ecx = len(edi); ecx >= 0; ecx--) {
//this will return true on the 3rd iteration, when ecx = 1
if edi[ecx] == al return true;
}
return false;
}
``````
The above loop will run 3 times, and on the 3rd time being run it’ll return `true`, because `edi[1]` = `'t'`.
## Jumps
### JMP and friends
`x86`
Affected by flags: `ZF`, `OF`
In general, these instruction have this skeleton: `jmp location`.
instruction description note
`jmp` unconditional jump, meaning “Jump Forest, jump!” no matter what
`jz` Jump if `ZF` = 1 (the result of previous instruction was `0`)
`jnz` opposite to `jz`. Jump if `ZF` = 0 (the result of previous instruction was not `0`)
`je` if the result of preceding `cmp op1, op2` was 0 (the operands were equal)
`jne` opposite to `je`. Jump if the result of preceding `cmp op1, op2` was not 0 (the operands were not equal)
`jg`, `ja` Jump if the result of preceding `cmp op1, op2` was a positive integer (op1 > op2, is greater). `ja` for unsigned comarison.
`jge`, `jae` like `jg` or `ja` combined with `je`. Jump if the result of preceding `cmp op1, op2` was a positive integer or 0 (op1 >= op2, is greater or equal). `jae` for unsigned comarison.
`jl`, `jb` opposite to `jg`, `ja`. Jump if the result of preceding `cmp op1, op2` was a negative integer (op1 < op2, is less). `jb` for unsigned comarison.
`jle`, `jbe` like `jl` or `jb` combined with `je`. Jump if the result of preceding `cmp op1, op2` was a positive integer or 0 (op1 <= op2, is less or equal). `jbe` for unsigned comarison.
`jo` jump if the result of the previous instruction set `OF` to `1`
`js` jump if the result of the previous instruction set `SF` to `1`
`jecxz` jump if `ecx` = 0
Getting familiar with jumps. Below is the table of examples. Try to quickly determine the location of the jump. Answers are listed right below the table, so spoiler alert! โ
## Logical
### AND
`x86`
Interesting! :open_mouth :Can be used to clear some bits with a mask. For example, if you have `1100 1011` and you need to zero all bits out. All you need to do, is to `and` with `0000 0000`. To determine whether an integer is even or not, mask it with `0000 0001`. Even numbers have `0` at the end, and uneven - `1`. `and`ing an even number with `0000 0001` will result in `0` and `and`ing an uneven - with `1`. Also, you can make a number less by 2, 4, 8 etc by applying a corresponding mask:
`1101` substract 2
`0111` Subtract 8
`1011` substract 4
### OR
Set to `1` if either of the bits is `1`. Repeat for each bit of the first operand and the second operand. Writes to the destination operand.
Interesting! ๐ฎ Can be used to set all bits to `1` with `1111`. โ For example, we have `1110 or 1111` = `1111`. Basically, any value `or`ed by `1111` is `1111`.
### XOR
Exclusive OR. 1 if the first operand’s bit is not equal to the second’s.
Interesting! ๐ฎ A quick way to set `eax` to `0`. Operation’s `xor eax, eax` opcode is `33 C0` (2 bytes) while `mov eax, 0` - opcode `b8 00 00 00 00 ` which is 5 bytes (costy ๐ด ).
Also, an interesting observation to investigate further: If I mask any value with `1111` , I get an operation equal to substraction (unsigned):
`1010 xor 1111` is `0101` (5 in decimal)
`1011 xor 1111` is `0100` (4d)
`1100 xor 1111` is `0011` (3d) | 6,218 | 20,817 | {"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} | 2.640625 | 3 | CC-MAIN-2024-10 | latest | en | 0.856074 |
http://focm2014.dm.uba.ar/viewAbstract.php?code=760 | 1,556,190,233,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578716619.97/warc/CC-MAIN-20190425094105-20190425120105-00439.warc.gz | 64,852,792 | 3,146 | FoCM 2014 conference
Workshop A6 - Real Number Complexity
December 12, 15:00 ~ 15:30 - Room C11
## Can everything be computed? - On the Solvability Complexity Index and Towers of Algorithms
### University of Cambridge, United Kingdom - ach70@cam.ac.uk
This talk addresses some of the fundamental barriers in the theory of computations. Many computational problems can be solved as follows: a sequence of approximations is created by an algorithm, and the solution to the problem is the limit of this sequence (think about computing eigenvalues of a matrix for example). However, as we demonstrate, for several basic problems in computations (computing spectra of infinite dimensional operators, solutions to linear equations or roots of polynomials using rational maps) such a procedure based on one limit is impossible. Yet, one can compute solutions to these problems, but only by using several limits. This may come as a surprise, however, this touches onto the boundaries of computational mathematics. To analyze this phenomenon we use the Solvability Complexity Index (SCI). The SCI is the smallest number of limits needed in order to compute a desired quantity. In several cases (spectral problems, inverse problems) we provide sharp results on the SCI, thus we establish the absolute barriers for what can be achieved computationally. For example, we show that the SCI of spectra of self-adjoint infinite matrices is equal to two, thus providing the first algorithms to compute such spectra in two limits. Moreover, we establish barriers on error control. We prove that no algorithm can provide error control on the computational spectral problem or solutions to infinite-dimensional linear systems. In particular, one can get arbitrarily close to the solution, but never knowing when one is "epsilon" away. In addition, we provide bounds for the SCI of spectra of classes of Schrodinger operators, thus we affirmatively answer the long standing question on whether or not these spectra can actually be computed. Finally, we show how the SCI provides a natural framework for understanding barriers in computations. In particular, we demonstrate how the impossibility result of McMullen on polynomial root finding with rational maps in one limit, and the framework of Doyle and McMullen on solving the quintic in several limits, can be put in the SCI framework.
Joint work with Jonathan Ben-Artzi (University of Cambridge), Olavi Nevanlinna (Aalto University), Markus Seidel (TU Chemnitz). | 500 | 2,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.902865 |
http://www.caraudio.com/forums/archive/index.php/t-132442.html | 1,438,691,040,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042990611.52/warc/CC-MAIN-20150728002310-00334-ip-10-236-191-2.ec2.internal.warc.gz | 349,426,531 | 3,867 | PDA
View Full Version : Ported box plans for 2 12" L7s
KK07
12-23-2005, 01:44 AM
Hey there, I have a guy here in town ready to build a ported box for me for my pair of 12" L7s. I just need the most ideal box plans for my set up, I'd like the total volume to be around 5 cu feet (2.5 cubes each) and have it tuned to around 36 hz. The max dimensions for my trunk are 31" wide x 10" tall x 34" deep. Polecat suggested the 2.5 cubes each and this is listed as a "street bass" enclosure in the Kicker manual. I have about 1000 watts rms total to play with so 500 watts each.
Now my question is, when you multiply 31x10x34 and divide by 1728 you get ~6.0 cubes net, but how does port area factor into this? In the Kicker manual they have this:
Street Bass 2.5 cu feet + port; 2.5 x 13.25" port, 16.5" long
When port area + sub displacement are added into the enclosure volume, should I end up with about 5 cubes net? Sorry about all of the questions, I'm just looking for the most ideal efficient enclosure for my car (97 Grand Am). I've already messaged JMac and saywhat? so hopefully one of them can help me but if someone else has time to design one, please let me know, I have paypal ready, thanks a lot!
Jordantyler
12-23-2005, 01:51 AM
i would send an email to saywaht? he doesnt like PM's
KK07
12-23-2005, 01:56 AM
i would send an email to saywaht? he doesnt like PM's
Yeah, sorry, I should have said that, I sent him an e-mail.
Alright, I'm just going to attempt the port thing here:
Okay 31x10x34 = 10540 / 1728 = ~6.1 cubes net
So now I'd take those specs kicker gives for port right? Multiply those out?
2.5x13.25x16.5 = 546.56 x number of ports? (2) = 1093.125 / 1728 = 0.63 cubes
So would I go 6.1 cubes - 0.63 cubes - sub displacement and that would equal the net volume?
Please tell me if this is completely wrong haha, I'm learning as I go here so hopefully I'm close to being on the right track, thanks for your help!
squeak12
12-23-2005, 02:31 AM
Those are outside dimensions. You need to account for wood thickness which is
.75 so it goes like this
29.5*8.5*32.5= 8149
8194 / 1728 = 4.72
4.72 - (.63*2) = 3.46
You will have just over 3.25 cubic feet after all displacement. You want 5 cu. ft. correct?
KK07
12-23-2005, 02:50 AM
Those are outside dimensions. You need to account for wood thickness which is
.75 so it goes like this
29.5*8.5*32.5= 8149
8194 / 1728 = 4.72
4.72 - (.63*2) = 3.46
You will have just over 3.25 cubic feet after all displacement. You want 5 cu. ft. correct?
Yeah I wanted 5 cu feet :crap:, so 3.25 cubes is the most I can get with those dimensions? Those dimensions = no trunk at all for me so I can't add anymore :(. Dammit...
squeak12
12-23-2005, 03:24 AM
What do you have a Mustang or something?
saywhat?
12-23-2005, 11:39 AM
hey man. i just emailed you back about some possibilities before reading this thread. I think your best bet is to sell off one of your L7's and fit the correct box to one of them with the power you have. Im almost certain you would be as loud or louder than two squeezed in that tiny box.
saywhat?
12-23-2005, 11:40 AM
by the way, thanks for the props ms. jordan, always great to hear from ya.
KK07
12-23-2005, 02:09 PM
I have a 97 Grand Am, I always though I had a decent sized trunk, bah. I guess I could try selling one of them but I really wanted a 2 sub set up.
KK07
12-23-2005, 11:45 PM
Is there any realistic way of fitting the box into the trunk through the trunk opening? I was planning on removing the back seat to fit it in there but there are these little sections that support the back seat that will get in the way. I actually have like 36 x 15 x 40 but I guess my installer was factoring in a method of actually getting it in the trunk. I'm pretty sure I could bring the height number up a couple of inches...I don't know though, anyone have any opinions on this?
saywhat?
12-23-2005, 11:52 PM
hey man, get on aim if u have it and we can try to figure something out.
KK07
12-24-2005, 12:43 AM
hey man, get on aim if u have it and we can try to figure something out.
Hey, I'm on, I tried messaging you but it says you're offline haha.
Tayta_Mayne
12-24-2005, 01:00 AM
I might buy one of your L7's if you wanna sell one. pm me with a price if your gonna sell maybe we can work a trade out, or just cash if you want | 1,316 | 4,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2015-32 | longest | en | 0.940964 |
https://www.r-bloggers.com/na-na-na-na-hey-hey-hey-goodbye/ | 1,586,213,859,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00298.warc.gz | 1,084,164,479 | 60,788 | # NA NA NA NA, Hey Hey Hey, Goodbye
September 5, 2015
By
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
## The Problem
Suppose you are doing a method comparison for which some results are above or below the linear range of your assay(s). Generally, these will appear in your spreadsheet (gasp!) program as (< x) or (> y) or, in the case of our mass spectrometer, “No Peak”. When you read these data into R using `read.csv()`, R will turn then into factors, which I personally find super–annoying and which inspired this conference badge (see bottom right) as I learned from University of British Columbia prof Jenny Bryan.
For this reason, when we read the data in, it is convenient to choose the option `stringsAsFactors = FALSE`. In doing so, the data will be treated as strings and be in the character class. But for regression comparison purposes, we need to make the data numeric and all of the (< x) and (> y) results will be converted to NA. In this post, we want to address a few questions that follow:
• How do we find all the NA results?
• How can we replace them with a numeric (like 0)?
• How can we rid ourselves of rows containing NA?
## Finding NA's
Let's read in the data which comes from a method comparison of serum aldosterone between our laboratory and Russ Grant's laboratory (LabCorp) published here. I'll read in the data with stringsAsFactors = FALSE. These are aldosterone results in pmol/L. To convert to ng/dL, divide by 27.7.
```myData<-read.csv("Comparison.csv", sep=",", stringsAsFactors = FALSE)
str(myData)```
```## 'data.frame': 96 obs. of 3 variables:
## \$ Sample.Num: int 1 2 3 4 5 6 7 8 9 10 ...
## \$ Aldo.Us : chr "462.3" "433.2" "37.7" "137.7" ...
## \$ Aldo.Them : num 457.2 418.1 42.1 133.9 27.4 ...```
`head(myData)`
```## Sample.Num Aldo.Us Aldo.Them
## 1 1 462.3 457.2
## 2 2 433.2 418.1
## 3 3 37.7 42.1
## 4 4 137.7 133.9
## 5 5 29.4 27.4
## 6 6 552.1 639.7```
You can see the problem immediately, our data (“Aldo.Us”) is a character vector. This is not good for regression. Why did this happen? We can find out:
`myData\$Aldo.Us`
```## [1] "462.3" "433.2" "37.7" "137.7" "29.4" "552.1" "41.6"
## [8] "158.7" "1198" "478.4" "160.7" "167.9" "211.6" "493.3"
## [15] "195.6" "649.8" "644" "534.1" "212.7" "413.3" "150.7"
## [22] "451.2" "25.8" "118.8" "496.1" "486.1" "846.8" "139.9"
## [29] "No Peak" "98.3" "113.8" "230.7" "530.2" "26.6" "390.3"
## [36] "782.8" "886.7" "83.4" "44" "71.2" "657" "321.6"
## [43] "188.6" "451.2" "485.3" "No Peak" "144.9" "249.6" "682"
## [50] "601.9" "330.5" "216.6" "500.3" "20.5" "271.5" "196.7"
## [57] "309.4" "235.7" "171.7" "124.9" "293.6" "345.4" "243.5"
## [64] "75.1" "508.3" "442.4" "531.3" "317.4" "647.9" "562"
## [71] "366.5" "37.1" "231.6" "73.7" "526.3" "No Peak" "165.6"
## [78] "105.8" "77.8" "211.6" "125.8" "76.5" "58.2" "111.9"
## [85] "238.5" "31.6" "156.8" "191.7" "402.5" "108.9" "183.7"
## [92] "314.4" "90" "98.9" "144.9" "971.4"```
Ahhh…it's the dreaded “No Peak”. This is what the mass spectrometer has put in its data file. So, let's force everything to numeric:
`myData\$Aldo.Us <- as.numeric(myData\$Aldo.Us)`
`## Warning: NAs introduced by coercion`
We see the warnings about the introduction of NAs. And we get:
`myData\$Aldo.Us`
```## [1] 462.3 433.2 37.7 137.7 29.4 552.1 41.6 158.7 1198.0 478.4
## [11] 160.7 167.9 211.6 493.3 195.6 649.8 644.0 534.1 212.7 413.3
## [21] 150.7 451.2 25.8 118.8 496.1 486.1 846.8 139.9 NA 98.3
## [31] 113.8 230.7 530.2 26.6 390.3 782.8 886.7 83.4 44.0 71.2
## [41] 657.0 321.6 188.6 451.2 485.3 NA 144.9 249.6 682.0 601.9
## [51] 330.5 216.6 500.3 20.5 271.5 196.7 309.4 235.7 171.7 124.9
## [61] 293.6 345.4 243.5 75.1 508.3 442.4 531.3 317.4 647.9 562.0
## [71] 366.5 37.1 231.6 73.7 526.3 NA 165.6 105.8 77.8 211.6
## [81] 125.8 76.5 58.2 111.9 238.5 31.6 156.8 191.7 402.5 108.9
## [91] 183.7 314.4 90.0 98.9 144.9 971.4```
`summary(myData\$Aldo.Us)`
```## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## 20.5 118.8 230.7 305.5 478.4 1198.0 3```
Now we have 3 NAs. We want to find them and get rid of them. From the screen we could figure out where the NAs were and manually replace them. This is OK on such a small data set but when you start dealing with data sets having thousands or millions of rows, approaches like this are impractical. So, let's do it right.
If we naively try to use an equality we find out nothing.
`which(myData\$Aldo.Us==NA)`
`## integer(0)`
Hunh? Whasgoinon?
This occurs because NA means “unknown”. Think about it this way. If one patient's result is NA and another patient's result is NA, then are the results equal? No, they are not (necessarily) equal, they are both unknown and so the comparison should be unknown also. This is why we do not get a result of TRUE when we ask the following question:
`NA==NA`
`## [1] NA`
So, when we ask R if unknown #1 is equal to unknown #2, it responds with “I dunno.”, or “NA”. So if we want to find the NAs, we should inquire as follows:
`is.na(myData\$Aldo.Us)`
```## [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [12] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [23] FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
## [34] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [45] FALSE TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [56] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [67] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE FALSE
## [78] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
## [89] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE```
or, for less verbose output:
`which(is.na(myData\$Aldo.Us))`
`## [1] 29 46 76`
## Hey Hey! Ho Ho! Those NAs have got to go!
Now we know where they are, in rows 29, 46, and 76. We can replace them with 0, which is OK but may pose problems if we use weighted regression (i.e. if we have a 0 in the x-data and we weight data by 1/x). Alternatively, we can delete the rows entirely.
To replace them with 0, we can write:
`myData\$Aldo.Us[which(is.na(myData\$Aldo.Us))] <- 0`
and this is equivalent:
`myData\$Aldo.Us[is.na(myData\$Aldo.Us)] <- 0`
To remove the whole corresponding row, we can write:
`myDataBeGoneNA <- myData[-which(is.na(myData\$Aldo.Us)),]`
or:
`myDataBeGoneNA <- myData[!is.na(myData\$Aldo.Us),]`
### Complete Cases
What if there were NA's hiding all over the place in multiple columns and we wanted to banish any row containing one or more NA? In this case, the `complete.cases()` function is one way to go:
`complete.cases(myData)`
```## [1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [23] TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE TRUE TRUE TRUE
## [34] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [45] TRUE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [56] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [67] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE TRUE
## [78] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
## [89] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE```
This function shows us which rows have no NAs (the ones with TRUE as the result) and which rows have NAs (the three with FALSE). We can banish all rows containing any NAs generally as follows:
`myDataBeGoneNA <- myData[complete.cases(myData),]`
This data set now has 93 rows:
`nrow(myDataBeGoneNA)`
`## [1] 93`
You could peruse the excluded data like this:
`myData[!complete.cases(myData),]`
```## Sample.Num Aldo.Us Aldo.Them
## 29 29 NA 6.6
## 46 46 NA 7.0
## 76 76 NA 5.7```
### na.omit()
Another way to remove incomplete cases is the `na.omit()` function (as Dr. Shannon Haymond pointed out to me). So this works too:
`myDataBeGoneNA <- na.omit(myData)`
## Row Numbers are Actually Names
In all of these approaches, you will notice something peculiar. Even though we have excluded the three rows, the row numbering still appears to imply that there are 96 rows:
`tail(myDataBeGoneNA)`
```## Sample.Num Aldo.Us Aldo.Them
## 91 91 183.7 170.4
## 92 92 314.4 307.6
## 93 93 90.0 214.0
## 94 94 98.9 75.1
## 95 95 144.9 129.3
## 96 96 971.4 807.7```
but if you check the dimensions, there are 93 rows:
`nrow(myDataBeGoneNA)`
`## [1] 93`
Why? This is because the row numbers are not row numbers; they are numerical row names. When you exclude a row, none of the other row names change. This was bewildering to me in the beginning. I thought my exclusions had failed somehow.
## Now we can move on
Once this is done, you can go on and do your regression, which, in this case, looks like this.
Finally, if you are ever wondering what fraction of your data is comprised of NA, rather than the absolute number, you can do this as follows:
`mean(is.na(myData\$Aldo.Us))`
`## [1] 0.03125`
If you applied this to the whole dataframe, you get the fraction of NA's in the whole dataframe (again–thank you Shannon):
`mean(is.na(myData))`
`## [1] 0.01041667`
## Final Thought:
`is.na(newunderthesun)`
`## [1] TRUE`
-Dan
R-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 3,625 | 10,148 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-16 | longest | en | 0.838671 |
https://civil4m.com/threads/cash-flow-statement-project-value-net-present-value-etc-in-a-contracting-system.1503/ | 1,600,933,752,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400214347.17/warc/CC-MAIN-20200924065412-20200924095412-00138.warc.gz | 292,701,518 | 19,847 | # Cash flow statement,project value, net present value etc in a contracting system
#### Srinivasan
Prime CE
Cash Flow
Cash flow refers to a contractors income and out go of cash. The net cash flow is the difference between disbursement and the income over a period of a time. A positive cash flow indicates that cash income is exceeding disbursement and a negative cash flow signifies just the opposite. And the contractor must maintain a cash balance sufficient to meet pay rolls, pay for materials, make equipment payments, and satisfy other financial obligations as they become due.
PROJECT CASH FLOWS
Cash Flow Stream –
Basic Components Initial Investment
Operating Cash Flows Terminal Cash Flow
Cash forecast:
A Cash forecast is a schedule that summarizes the estimated cash receipts, estimated disbursement, and available cash balances for some period into the future. The preparation of the cash forecast begins with the collection of detailed information regarding future cash income and expenditures.
APPRAISAL (DECISION MAKING) CRITERIA
Evaluating Project‘s Financial Feasibility Two broad categories
1. Non discounting criteria
Pay Back Period Average Rate of Return
2. Discounting criteria
Net Present Value - NPV
Internal Rate of Return - IRR
PRESENT VALUE [PV]
Present Value of a Single amount: PV = Fn[ 1/(1+r)]n
Where,
P = Present Value
F = Future Value
r= Discount rate per period
n = Number of periods
Present Value of a Cash Flow Stream n
PVn = Ct_t=0 ( 1+r)t
PVn = Present Value of a Cash Flow stream ,
Ct= Cash Flow occurring at time t
r= Discount rate,n= Duration of Cash Flow stream
NET PRESENT VALUE - NPV
Sum of the Present Value of all expected Cash flows [inflows as well as outflows] associated with an investment / project, taken one year at a time and discounted by a factor which represents the opportunity cost of capital.
NPV = Ct, t=0 (1+r) t
Where,
Ct = Cash Flow occurring at the end of year t, t = 0....n)
[A cash inflow will have a positive sign, whereas a cash outflow has a negative sign]
n = Life of the project
r = Rate of Discount
INTERNAL RATE OF RETURN [IRR]
The discount rate which results in NPV = O It is the discount rate in the equation
n
Ct__ = 0 (NPV)
t=1 ( 1+r)t
Where
Ct = Cash flow at the end of year t
r = Internal Rate of Return [Discount rate]
n = Life of the project
ACCEPT, IF NPV IS POSITIVE
REJECT, IF NPV IS NEGATIVE
INDIFFEENT , IF NPV IS ZERO
Computer Usage:
The computer is invaluable and indispensable element in the conduct of a successful construction business.
Areas of computer usage:
1. Financial accounting: The computer performs basic accounting functions such as accounts receivable, accounts payable, general ledger, inventory, cash forecast, financial statements, subcontractor control.
2. Equipment accounting: Computer maintains records of equipment depreciation, ownership and operating costs, hours of operation, maintenance, spare parts, production rates and units costs.
3. Payroll: Using time card input, the computer prepares payroll checks, periodic and special payroll and tax reports, the payroll register, and updates the employee master files.
4. Purchasing: The computer participates in the tabulation and handing of bits, purchase order preparation, expediting and shop drawings.
sivaprasad7 and MAR
Excellent
Replies
0
Views
271
Replies
0
Views
656
Replies
2
Views
616
Replies
0
Views
575
Replies
4
Views
741 | 797 | 3,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-40 | latest | en | 0.886169 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.