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 |
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https://quizlet.com/explanations/questions/a-in-how-many-ways-can-a-club-with-20-members-choose-a-president-and-a-vice-president-b-in-how-many-ways-can-the-club-choose-a-2-person-gove-0b38fa47-2f75504b-8308-4813-8749-2aa901a3e778 | 1,701,948,282,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00824.warc.gz | 543,585,147 | 82,273 | Try the fastest way to create flashcards
Question
# a. In how many ways can a club with 20 members choose a president and a vice president?b. In how many ways can the club choose a 2-person governing council?
Solutions
Verified
Step 1
1 of 3
a) Since the order (or positions) matter, then permutation is used. With two positions to fill up, then the 20 members can be arranged in:
\begin{align*} _{20}P_2 &= \dfrac{20!}{(20-2)!} \\ &= \dfrac{20 \cdot 19 \cdot \cancel{18!}}{ \cancel{18!} } \\ &= 380 \text{ ways}. \\ \end{align*}
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https://qa.answers.com/entertainment/What_are_the_numbers_that_deidara_appears_in_the_original_Naruto_episodes | 1,713,484,701,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00633.warc.gz | 433,887,126 | 48,346 | 0
# What are the numbers that deidara appears in the original Naruto episodes?
Updated: 4/28/2022
Wiki User
13y ago
If you mean episodes he's in episode 125 as a hologramn
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The digit appears eleven time from 1 to 100.
### Why are the numbers on Naruto episodes different such as episode '205' of the English dub is episode 204 in the subs did they skip an episode in the English version?
A few episodes were skipped and some of the double episodes (ex: 85-86) were shown as single episodes.
18,27,81,etc.
### What is the number that appears most often in a list of numbers mathmatically?
The mode is the term for the number that appears most often.
### Mode of numbers?
Mode is the value (number) that appears the most.
### How do you figure out a mode?
it is the number that appears the most in a group of numbers.
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No. The number which appears three times is the only mode.
### What is 165 in numbers?
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subscripts
### What is a number that appears most frequently in a set of numbers?
the mode of the set | 451 | 1,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-18 | latest | en | 0.966913 |
http://www.mathspadilla.com/3ESO/Unit7-SystemsOfEquations/systems_of_linear_equations.html | 1,586,430,149,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371833063.93/warc/CC-MAIN-20200409091317-20200409121817-00475.warc.gz | 252,345,904 | 1,921 | # Systems of linear equations
You can reduce a system of two linear equations to this form:
A solution to the system is a solution to both equations.
For example:
A system of linear equations can have:
- One solution:
- Infinite solutions:
- No solution:
Exercise: check if x = 1 and y = -1 is a solution of the following systems of equations:
Solutions: a) yes; b) no; c) yes
Licensed under the Creative Commons Attribution Non-commercial Share Alike 3.0 License | 109 | 473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2020-16 | longest | en | 0.870802 |
https://www.splashlearn.com/s/math-worksheets/add-and-subtract-hundredths-less-than-100-missing-digits | 1,701,187,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00809.warc.gz | 1,148,723,803 | 13,074 | Home > Math > Add and Subtract Hundredths less than 100: Missing Digits Worksheet
# Add and Subtract Hundredths less than 100: Missing Digits Worksheet
## Know more about Add and Subtract Hundredths less than 100: Missing Digits Worksheet
Add and subtract hundredths less than 100 worksheet is a great way to help learners become fluent with the concept of addition and subtraction. When adding or subtracting decimals on this worksheet, students align the decimal points and use zero as a placeholder. Then to find the missing number in the add and subtract hundredths less than 100 worksheet, students then apply the relationship between addition and subtraction. In this worksheet, students practice solving problems using the column method. This method is especially helpful with problems involving bigger multi-digit numbers as the format provides an easy structure to follow. | 168 | 884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.882323 |
https://math-faq.com/chapter-2/section-2-2/ | 1,696,477,762,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511717.69/warc/CC-MAIN-20231005012006-20231005042006-00884.warc.gz | 419,505,360 | 10,427 | # Section 2.2
## Solving a System of Two Linear Equations Algebraically
In Section 2.1, we solved several systems of linear equations by graphing each linear equation in the system. The solution to the system is any ordered pair that satisfies all of the equations. On the graph, this ordered pair is the point where all of the equations intersect.
Using a graph to find a point of intersection has limitations. If the graph is on a piece of paper, it is difficult to accurately locate the point of intersection using the scales on the axes of the graph. If you constructed the graph, you have to be extremely careful and precise to get a valid approximate answer.
A graphing calculator can draw a more accurate graph (as long as you enter the equations correctly), but the algorithms in the calculator are approximate. This means that the location of the point of intersection may be exact or it may be very close to the exact point of intersection.
Suppose we want to solve the system of linear equations
To solve this system graphically, we need to solve each linear equation for y. The first equation is already solved for y. To solve the second equation for y,
Let’s look at a graph the system to find the point of intersection.
Figure 1 – A graph of a system whose point of intersection is about (0.5, 0).
From the graph, the solution to the system looks to be approximately (0.5, 0). We can check this solution by substituting x = 0.5 and y = 0 and into the original system of equations:
Since neither equation is true, the estimate of the point of intersection is not exact. To find an exact solution, we’ll need to use an algebraic strategy to solve the system of equations.
Section 2.2 Workbook (PDF) – 9/2/19
Watch Video Playlist | 391 | 1,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2023-40 | longest | en | 0.941152 |
https://notebook.community/oliverlee/phobos/scripts/controller_frequency_response | 1,709,188,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00359.warc.gz | 411,702,991 | 735,970 | ``````
In [1]:
#%matplotlib notebook
#DEFAULT_FIGSIZE = (8, 6)
%matplotlib inline
DEFAULT_FIGSIZE = (12, 8)
import numpy as np
import scipy.signal
import matplotlib.pyplot as plt
import seaborn as sns
sns.set_style('darkgrid')
from dtk.bicycle import benchmark_state_space_vs_speed, benchmark_matrices
import control
import plot_sim as ps
import matplotlib as mpl
mpl.rcParams['figure.figsize'] = DEFAULT_FIGSIZE
``````
``````
In [2]:
def rudinshapiro(N):
"""
Return first N terms of Rudin-Shapiro sequence
https://en.wikipedia.org/wiki/Rudin-Shapiro_sequence
Confirmed correct output to N = 10000:
https://oeis.org/A020985/b020985.txt
"""
def hamming(x):
"""
Hamming weight of a binary sequence
http://stackoverflow.com/a/407758/125507
"""
return bin(x).count('1')
out = np.empty(N, dtype=int)
for n in range(N):
b = hamming(n << 1 & n)
a = (-1)**b
out[n] = a
return out
s = rudinshapiro(10)
print(s)
np.array(s > 0).astype(float) * np.pi
``````
``````
[ 1 1 1 -1 1 1 -1 1 1 1]
Out[2]:
array([ 3.14159265, 3.14159265, 3.14159265, 0. , 3.14159265,
3.14159265, 0. , 3.14159265, 3.14159265, 3.14159265])
``````
``````
In [3]:
plt.close('all')
t = np.arange(0, 10, 0.001)
dt = 0.001
n = int(10/dt)
def multisine(frequencies):
n = len(frequencies)
seq = np.array(rudinshapiro(n) > 0).astype(float)
print('length', n)
print('frequencies', ', '.join(str(f) for f in frequencies))
print('rs sequence', ', '.join(str(s) for s in seq))
u = np.zeros(t.shape)
amplitude = 0.3
for f, s in zip(frequencies, seq):
period = 1/f
if (10/period).is_integer:
print('{} periods for frequency {}'.format(10/period, f))
u += amplitude*np.sin(2*np.pi*f*t + s*np.pi)
return u
u_freq = np.arange(0.2, 10.2, 0.2)
u = multisine(u_freq)
x = np.fft.fft(u)
freq = np.fft.fftfreq(n, dt)
index = (freq > 0) & (freq < 20)
fig, ax = plt.subplots(2, 1)
ax[0].plot(t, u)
ax[1].stem(freq[index], np.abs(x[index]))
plt.show()
``````
``````
length 50
frequencies 0.2, 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2, 2.4, 2.6, 2.8, 3.0, 3.2, 3.4, 3.6, 3.8, 4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5.4, 5.6, 5.8, 6.0, 6.2, 6.4, 6.6, 6.8, 7.0, 7.2, 7.4, 7.6, 7.8, 8.0, 8.2, 8.4, 8.6, 8.8, 9.0, 9.2, 9.4, 9.6, 9.8, 10.0
rs sequence 1.0, 1.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0, 1.0, 1.0, 1.0, 0.0, 1.0, 1.0, 0.0, 1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0
2.0 periods for frequency 0.2
4.0 periods for frequency 0.4
6.000000000000001 periods for frequency 0.6000000000000001
8.0 periods for frequency 0.8
10.0 periods for frequency 1.0
12.0 periods for frequency 1.2
14.000000000000002 periods for frequency 1.4000000000000001
16.0 periods for frequency 1.6
18.0 periods for frequency 1.8
20.0 periods for frequency 2.0
22.0 periods for frequency 2.2
24.000000000000004 periods for frequency 2.4000000000000004
26.000000000000007 periods for frequency 2.6000000000000005
28.000000000000004 periods for frequency 2.8000000000000003
30.000000000000007 periods for frequency 3.0000000000000004
32.0 periods for frequency 3.2
34.00000000000001 periods for frequency 3.4000000000000004
36.00000000000001 periods for frequency 3.6000000000000005
38.0 periods for frequency 3.8000000000000003
40.0 periods for frequency 4.0
42.0 periods for frequency 4.2
44.0 periods for frequency 4.4
46.00000000000001 periods for frequency 4.6000000000000005
48.00000000000001 periods for frequency 4.800000000000001
50.000000000000014 periods for frequency 5.000000000000001
52.00000000000001 periods for frequency 5.2
54.0 periods for frequency 5.4
56.00000000000001 periods for frequency 5.6000000000000005
58.00000000000001 periods for frequency 5.800000000000001
60.000000000000014 periods for frequency 6.000000000000001
62.0 periods for frequency 6.2
64.0 periods for frequency 6.4
66.00000000000001 periods for frequency 6.6000000000000005
68.00000000000001 periods for frequency 6.800000000000001
70.0 periods for frequency 7.000000000000001
72.0 periods for frequency 7.2
74.00000000000001 periods for frequency 7.4
76.0 periods for frequency 7.6000000000000005
78.0 periods for frequency 7.800000000000001
80.0 periods for frequency 8.0
81.99999999999999 periods for frequency 8.2
84.0 periods for frequency 8.4
86.0 periods for frequency 8.6
87.99999999999999 periods for frequency 8.799999999999999
90.0 periods for frequency 9.0
91.99999999999999 periods for frequency 9.2
94.0 periods for frequency 9.4
96.0 periods for frequency 9.6
98.00000000000001 periods for frequency 9.8
100.0 periods for frequency 10.0
``````
``````
In [4]:
def plot_response(log):
t = log.t
u = log.records.input[:, 1] # steer torque
v = log.kollmorgen_command_torque
r = log.states[:, 1] # reference steer angle
y = log.measured_steer_angle
fig, ax = plt.subplots(2, 1, sharex=True)
ax[0].plot(t, u,
label='steer torque')
ax[0].plot(t, v,
label='commanded motor torque')
ax[0].plot(t, log.kollmorgen_applied_torque,
label='actual motor torque')
ax[0].legend()
ax[0].set_ylabel('torque [N-m]')
ax[1].plot(t, r * 180/np.pi,
label='reference encoder angle')
ax[1].plot(t, y * 180/np.pi,
label='measured encoder angle')
ax[1].legend()
ax[1].set_ylabel('angle [deg]')
ax[1].set_xlabel('time [s]')
return fig, ax
plt.close('all')
log = ps.ProcessedRecord('logs/multisine.pb.cobs.gz')
plot_response(log)
plt.show()
``````
``````
``````
``````
In [5]:
# simulate model response to multisine 'u'
_, A, B = benchmark_state_space_vs_speed(*benchmark_matrices(), [5])
A = A[0]
B = B[0]
Bi = B[:, [1]] # steer torque
Ci = np.array([[0, 1, 0, 0]]) # steer angle
D = np.array([[0]])
sys = scipy.signal.StateSpace(A, Bi, Ci, D)
tf = scipy.signal.TransferFunction(sys)
# simulate for 20 seconds
lsim_t = np.arange(0, 20, dt)
lsim_u = np.concatenate((u, u))
_, lsim_y, lsim_x = scipy.signal.lsim(sys, lsim_u, lsim_t)
``````
``````
/Users/oliver/miniconda3/envs/dev/lib/python3.5/site-packages/scipy/signal/filter_design.py:1452: BadCoefficients: Badly conditioned filter coefficients (numerator): the results may be meaningless
"results may be meaningless", BadCoefficients)
``````
``````
In [6]:
plt.close('all')
fig, ax = plt.subplots(2, 1, sharex=True)
color = sns.color_palette('Paired', 12)[1::2]
log_u = log.records.input[:, 1] # steer torque
log_r = log.states[:, 1] # reference steer angle
ax[0].plot(lsim_t, lsim_u,
color=color[1],
linestyle='--',
label='python lsim steer torque')
ax[0].plot(log.t, log_u,
color=color[2],
alpha=0.7,
label='simulator steer torque')
ax[0].set_ylabel('torque [Nm]')
ax[0].legend()
ax[1].plot(lsim_t, lsim_y,
color=color[1],
linestyle='--',
label='python lsim steer angle')
ax[1].plot(log.t, log_r,
color=color[2],
alpha=0.7,
label='simulator steer angle')
ax[1].set_ylabel('angle [deg]')
ax[1].legend()
ax[1].set_xlabel('time [s]')
plt.show()
``````
``````
``````
``````
In [7]:
def frequency_response(input_signal, output_signal, last_n=None):
if last_n is not None:
assert last_n > 0
last_n = -last_n
index = slice(last_n, None)
U = np.fft.fft(input_signal[index])
Y = np.fft.fft(output_signal[index])
Suu = U*np.conj(U)
Syu = Y*np.conj(U)
return Syu/Suu
# C = np.power(np.abs(Syu), 2)/(Suu * Syy) # coherence
``````
``````
In [13]:
def plot_frequency_response(log, model_compare=False):
#color = sns.color_palette('deep', 6)
color = sns.color_palette('Paired', 12)[1::2]
log_u = log.records.input[:, 1] # steer torque
log_r = log.states[:, 1] # reference steer angle
log_y = log.measured_steer_angle[:]
log_v = log.kollmorgen_command_torque
log_w = log.kollmorgen_applied_torque
tf_YU = frequency_response(log_u, log_y, n)
tf_RU = frequency_response(log_u, log_r, n)
tf_YR = frequency_response(log_r, log_y, n)
tf_VU = frequency_response(log_u, log_v, n)
tf_WU = frequency_response(log_u, log_w, n)
if model_compare:
# simulated model response
X = frequency_response(lsim_u, lsim_y, n)
# generate frequency response of previously defined model
w, mag, phase = tf.bode(w=2*np.pi*np.logspace(-0.7, 0.7, 100))
index = [i for i, f in enumerate(np.around(freq, 1))
if f in np.around(u_freq, 1)]
fig, ax = plt.subplots(2, 1, sharex=True)
if model_compare:
ax[0].loglog(w/2/np.pi, np.power(10, mag/20),
linestyle='--', color=color[1],
label='R/U: steer torque to reference steer angle (bode)')
ax[0].loglog(freq[index], np.abs(X[index]),
marker='x', color=color[1],
label='R/U: steer torque to reference steer angle (lsim)')
ax[0].loglog(freq[index], np.abs(tf_RU[index]),
marker='x', color=color[2],
label='R/U: steer torque to reference steer angle')
else:
ax[0].loglog(freq[index], np.abs(tf_YU[index]),
marker='x', color=color[0],
label='Y/R: steer torque to measured steer angle')
ax[0].loglog(freq[index], np.abs(tf_YR[index]),
marker='x', color=color[3],
label='Y/R: reference to measured steer angle')
ax[0].loglog(freq[index], np.abs(tf_VU[index]),
marker='x', color=color[4],
label='V/U: steer torque to command torque')
#ax[0].loglog(freq[index], np.abs(tf_WU[index]),
# marker='x', color=color[5],
# label='W/U: steer torque to current torque')
ax[0].legend()
ax[0].set_ylabel('gain')
if model_compare:
ax[1].semilogx(w/2/np.pi, phase,
linestyle='--', color=color[1],
label='R/U: steer torque to reference steer angle (bode)')
ax[1].semilogx(freq[index], np.angle(X[index], deg=True),
marker='x', color=color[1],
label='R/U: steer torque to reference steer angle (lsim)')
ax[1].semilogx(freq[index], np.angle(tf_RU[index], deg=True),
marker='x', color=color[2],
label='R/U: steer torque to reference steer angle')
else:
ax[1].semilogx(freq[index], np.angle(tf_YU[index], deg=True),
marker='x', color=color[0],
label='Y/U: steer torque to measured steer angle')
ax[1].semilogx(freq[index], np.angle(tf_YR[index], deg=True),
marker='x', color=color[3],
label='Y/R: reference to measured steer angle')
ax[1].semilogx(freq[index], np.angle(tf_VU[index], deg=True),
marker='x', color=color[4],
label='V/U: steer torque to command torque')
#ax[1].semilogx(freq[index], np.angle(tf_WU[index], deg=True),
# marker='x', color=color[5],
# label='W/U: steer torque to current torque')
ax[1].legend()
ax[1].set_ylabel('phase [deg]')
ax[1].set_xlabel('frequency [Hz]')
ax[0].set_title('frequency response')
return fig, ax
``````
``````
In [24]:
def plot_roll_frequency_response(log):
#color = sns.color_palette('deep', 6)
color = sns.color_palette('Paired', 12)[1::2]
log_u = log.records.input[:, 1] # steer torque
log_r = log.states[:, 0] # reference roll angle
tf_RU = frequency_response(log_u, log_r, n)
# simulated model response
X = frequency_response(lsim_u, lsim_x[:, 0], n)
# generate frequency response of previously defined model
sys = scipy.signal.StateSpace(A, Bi, np.array([[1, 0, 0, 0]]), D)
tf = scipy.signal.TransferFunction(sys)
w, mag, phase = tf.bode(w=2*np.pi*np.logspace(-0.7, 0.7, 100))
index = [i for i, f in enumerate(np.around(freq, 1))
if f in np.around(u_freq, 1)]
fig, ax = plt.subplots(2, 1, sharex=True)
ax[0].loglog(w/2/np.pi, np.power(10, mag/20),
linestyle='--', color=color[1],
label='R/U: steer torque to reference roll angle (bode)')
ax[0].loglog(freq[index], np.abs(X[index]),
marker='x', color=color[1],
label='R/U: steer torque to reference roll angle (lsim)')
ax[0].loglog(freq[index], np.abs(tf_RU[index]),
marker='x', color=color[2],
label='R/U: steer torque to reference roll angle')
ax[0].legend()
ax[0].set_ylabel('gain')
ax[1].semilogx(w/2/np.pi, phase,
linestyle='--', color=color[1],
label='R/U: steer torque to reference roll angle (bode)')
ax[1].semilogx(freq[index], np.angle(X[index], deg=True),
marker='x', color=color[1],
label='R/U: steer torque to reference roll angle (lsim)')
ax[1].semilogx(freq[index], np.angle(tf_RU[index], deg=True),
marker='x', color=color[2],
label='R/U: steer torque to reference roll angle')
ax[1].legend()
ax[1].set_ylabel('phase [deg]')
ax[1].set_xlabel('frequency [Hz]')
ax[0].set_title('roll frequency response')
return fig, ax
``````
``````
In [14]:
plt.close('all')
plot_frequency_response(log, True)
plt.show()
``````
``````
``````
``````
In [25]:
plt.close('all')
plot_roll_frequency_response(log)
plt.show()
``````
``````
/Users/oliver/miniconda3/envs/dev/lib/python3.5/site-packages/scipy/signal/filter_design.py:1452: BadCoefficients: Badly conditioned filter coefficients (numerator): the results may be meaningless
"results may be meaningless", BadCoefficients)
``````
``````
In [9]:
plt.close('all')
plot_frequency_response(log, False)
plt.show()
``````
``````
``````
``````
In [10]:
plt.close('all')
log2 = ps.ProcessedRecord('logs/multisine_kp150_kd3.pb.cobs.gz')
fig, ax = plot_response(log2)
ax[0].plot(t, u, linestyle='--',
label='designed multisine perturbation')
ax[0].legend()
plot_frequency_response(log2, False)
plt.show()
``````
``````
`````` | 4,523 | 12,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.445154 |
https://www.wolfram.com/language/12/algebraic-computation/derive-the-quadratic-formula.html.en?footer=lang | 1,716,950,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00179.warc.gz | 915,521,386 | 8,256 | # Wolfram Language™
## Derive the Quadratic Formula
The quadratic formula is perhaps the most famous equation of algebra. Completing the square to derive this formula combines all five equation manipulation functions.
Multiply both sides by .
Add to both sides.
Factor the left-hand side.
Take the positive square root of both sides.
Cancel the square root of the square.
Subtract from both sides.
Divide both sides by to obtain the quadratic formula for with positive square root. | 97 | 490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.899637 |
https://community.ptc.com/t5/Mathcad/Error-when-draw-a-graph-in-mathcad/td-p/486446 | 1,726,105,353,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00110.warc.gz | 156,950,584 | 49,671 | cancel
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5-Regular Member
## Error when draw a graph in mathcad
hi everyone, i have a problem in drawing the graph, i don't know why it's wrong
i want to draw the graph same as the second picture but as you see in the first picture, the graph has error.
Here is the program
i attach the zip file in post . Thank you so much !
ACCEPTED SOLUTION
Accepted Solutions
21-Topaz I
(To:marubeni)
Hi,
"Thank you , but can i ask you that why it is V.fl(k)1 and not is V.fl(k) in the graph, i don't know ."
The programming for Z(k) does not have a specific return statement. The program automatically returns Z which is a two column vector as the last thing returned is an element of Z. If you specifically "return" the first term of Z you only get one answer for Z.
I have done that in the enclosed sheet
9 REPLIES 9
23-Emerald III
(To:marubeni)
Mathcad will not plot complex numbers.
Your program uses a lot of complex numbers. I would not be surprised if Z(k) contains a lot of complex numbers as a result. In fact your program is prepared ( in its last line) to let the result with index 1 become it's real part.
Find a way to treat those complex results. Either plot just the real part of the x and y- values (so plot Re(g(k)) versus Re(Vfl(k)).
Or plot the magnitude of the complex numbers.
Whichever makes more sense to your application.
Success!
Luc
25-Diamond I
(To:marubeni)
Your function Z returns a 2-element vector, not a scalar. Thats the main reason for the error.
Maybe this helps, even though the plot looks different from the one you try to duplicate:
21-Topaz I
(To:marubeni)
Hi,
Perhaps if you supply the background information we can check your implimentation of the mathematics behind the graph
Cheers
Terry
5-Regular Member
(To:terryhendicott)
i attach the zip file in post. Thank you very much!. Iam a beginner in mathcad so something is overcome my understanding .
@terryhendicott wrote:
Hi,
Perhaps if you supply the background information we can check your implimentation of the mathematics behind the graph
Cheers
Terry
21-Topaz I
(To:marubeni)
Hi,
Plot of first of vector answers
5-Regular Member
(To:terryhendicott)
Thank you , but can i ask you that why it is V.fl(k)1 and not is V.fl(k) in the graph, i don't know .
21-Topaz I
(To:marubeni)
Hi,
"Thank you , but can i ask you that why it is V.fl(k)1 and not is V.fl(k) in the graph, i don't know ."
The programming for Z(k) does not have a specific return statement. The program automatically returns Z which is a two column vector as the last thing returned is an element of Z. If you specifically "return" the first term of Z you only get one answer for Z.
I have done that in the enclosed sheet
5-Regular Member
(To:terryhendicott)
Your knowledge in mathcad is so extensive, iam reading your explanation, i hope i can understand it . Thank you so much !
24-Ruby IV
(To:terryhendicott)
@terryhendicott wrote:
The programming for Z(k) does not have a specific return statement.
It will be good if each Mathcad-program has the return operator!
Better for the understanding!
Announcements
Top Tags | 846 | 3,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.912634 |
https://www.viajeabariloche.com/2019/08/solving-linear-systems-by-elimination.html | 1,568,809,968,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00188.warc.gz | 1,061,615,571 | 27,759 | # Solving Linear Systems By Elimination Worksheet
## Tuesday, August 13, 2019
In this method we eliminate one variable from the equations and then find the value of. Equations worksheets and quizzes equations worksheets.
Solving Linear Systems By Elimination Color Worksheet By Aric Thomas
### In this method we graph the given equations on the coordinate plane and look for the points of intersection.
Solving linear systems by elimination worksheet. Welcome to the algebra 2 go beginning algebra resources page. Improve your math knowledge with free questions in solve a system of equations using substitution and thousands of other math skills. Matrix multiplication part 1 matrix multiplication part 2 inverse matrix part 1 inverting matrices part 2.
Lets start at the beginning and work our way up through the various areas of math. Free algebra 1 worksheets created with infinite algebra 1. It is a method of solving linear system of equations.
Whether you are attending saddleback colleges beginning algebra class math 251 taking a beginning. First determine what you will multiply each of the above equations by to get the same leading coefficients. Solving decimal equations using multiplications and divisions worksheets solving equations involving.
Solving inequalities worksheet 1 here is a twelve problem worksheet featuring simple one step inequalities. Elimination method is one of the best methods of solving the linear equations. Printable in convenient pdf format.
We need a good foundation of each area to build upon for the next level.
Solving Linear Systems Using Elimination Edboost
Solving Linear Systems By Elimination Color Worksheet Math Lesson
Ls 8 Solving Systems Using Elimination Finding The Least Common
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Systems Of Equations Elimination Method Worksheet Answers 3 Ways To | 572 | 2,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | longest | en | 0.840628 |
https://leanprover-community.github.io/mathlib4_docs/Mathlib/Analysis/NormedSpace/Int.html | 1,718,479,781,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861606.63/warc/CC-MAIN-20240615190624-20240615220624-00115.warc.gz | 315,945,567 | 2,871 | # The integers as normed ring #
This file contains basic facts about the integers as normed ring.
Recall that ‖n‖ denotes the norm of n as real number. This norm is always nonnegative, so we can bundle the norm together with this fact, to obtain a term of type NNReal (the nonnegative real numbers). The resulting nonnegative real number is denoted by ‖n‖₊.
theorem Int.nnnorm_coe_units (e : ) :
e‖₊ = 1
theorem Int.norm_coe_units (e : ) :
e = 1
@[simp]
theorem Int.nnnorm_natCast (n : ) :
n‖₊ = n
@[deprecated Int.nnnorm_natCast]
theorem Int.nnnorm_coe_nat (n : ) :
n‖₊ = n
Alias of Int.nnnorm_natCast.
@[simp]
theorem Int.toNat_add_toNat_neg_eq_nnnorm (n : ) :
n.toNat + (-n).toNat = n‖₊
@[simp]
theorem Int.toNat_add_toNat_neg_eq_norm (n : ) :
n.toNat + (-n).toNat = n | 255 | 776 | {"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} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.74422 |
https://fr.mathworks.com/matlabcentral/cody/problems/44-trimming-spaces/solutions/256517 | 1,575,599,930,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484477.5/warc/CC-MAIN-20191206023204-20191206051204-00366.warc.gz | 389,145,406 | 15,639 | Cody
# Problem 44. Trimming Spaces
Solution 256517
Submitted on 6 Jun 2013 by Franek
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% a = 'no extra spaces'; b = 'no extra spaces'; assert(isequal(b,removeSpaces(a)))
b = no extra spaces
2 Pass
%% a = ' lots of space in front'; b = 'lots of space in front'; assert(isequal(b,removeSpaces(a)))
b = lots of space in front
3 Pass
%% a = 'lots of space in back '; b = 'lots of space in back'; assert(isequal(b,removeSpaces(a)))
b = lots of space in back
4 Pass
%% a = ' space on both sides '; b = 'space on both sides'; assert(isequal(b,removeSpaces(a)))
b = space on both sides
5 Pass
%% a = sprintf('\ttab in front, space at end '); b = sprintf('\ttab in front, space at end'); assert(isequal(b,removeSpaces(a)))
b = tab in front, space at end | 269 | 931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | latest | en | 0.629032 |
http://docs.wand-py.org/en/0.6.1/guide/colorenhancement.html | 1,590,791,387,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00200.warc.gz | 41,830,286 | 4,343 | # Color Enhancement¶
## Evaluate Expression¶
New in version 0.4.1.
Pixel channels can be manipulated by applying an arithmetic, relational, or logical expression. See EVALUATE_OPS for a list of valid operations.
For example, when given image enhancement.jpg:
We can reduce the amount of data in the blue channel by applying the right-shift binary operator, and increase data in the right channel with left-shift operator:
from wand.image import Image
with Image(filename='enhancement.jpg') as img:
# B >> 1
img.evaluate(operator='rightshift', value=1, channel='blue')
# R << 1
img.evaluate(operator='leftshift', value=1, channel='red')
## Function Expression¶
New in version 0.4.1.
Similar to evaluate(), function() applies a multi-argument function to pixel channels. See FUNCTION_TYPES for a list of available function formulas.
For example, when given image enhancement.jpg:
We can apply a Sinusoid function with the following:
from wand.image import Image
with Image(filename='enhancement.jpg') as img:
frequency = 3
phase_shift = -90
amplitude = 0.2
bias = 0.7
img.function('sinusoid', [frequency, phase_shift, amplitude, bias])
## Gamma¶
New in version 0.4.1.
Gamma correction allows you to adjust the luminance of an image. Resulting pixels are defined as pixel^(1/gamma). The value of gamma is typically between 0.8 & 2.3 range, and value of 1.0 will not affect the resulting image.
The level() method can also adjust gamma value.
For example, when given image enhancement.jpg:
We can step through 4 pre-configured gamma correction values with the following:
from wand.image import Image
with Image(filename='enhancement.jpg') as img_src:
for Y in [0.8, 0.9, 1.33, 1.66]:
with Image(img_src) as img_cpy:
img_cpy.gamma(Y)
## Level¶
New in version 0.4.1.
Black & white boundaries of an image can be controlled with level() method. Similar to the gamma() method, mid-point levels can be adjusted with the gamma keyword argument.
The black and white point arguments are expecting values between 0.0 & 1.0 which represent percentages.
For example, when given image enhancement.jpg:
We can adjust the level range between 20% & 90% with slight mid-range increase:
from wand.image import Image
with Image(filename='enhancement.jpg') as img:
img.level(0.2, 0.9, gamma=1.1)
img.save(filename='enhancement-level.jpg') | 579 | 2,347 | {"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} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.710639 |
https://classroomsecrets.co.uk/mixed-age-year-3-and-4-fractions-and-decimals-step-18-resource-pack/ | 1,701,942,267,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100651.34/warc/CC-MAIN-20231207090036-20231207120036-00561.warc.gz | 198,039,651 | 78,834 | Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Resource Pack – Classroom Secrets | Classroom Secrets
All › Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Resource Pack
# Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Resource Pack
## Step 18: Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Resource Pack
Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Resource Pack includes a teaching PowerPoint and differentiated varied fluency and reasoning and problem solving resources for this step which covers Year 4 Halves and Quarters for Spring Block 4.
### What's included in the Pack?
This Mixed Age Year 3 and 4 Fractions and Decimals Step 18 pack includes:
• Mixed Age Year 3 and 4 Fractions and Decimals Step 18 Teaching PowerPoint with examples.
• Year 4 Halves and Quarters Varied Fluency with answers.
• Year 4 Halves and Quarters Reasoning and Problem Solving with answers.
#### National Curriculum Objectives
Mathematics Year 4: (4F6a) Recognise and write decimal equivalents to 1/4 , 1/2 , 3/4
Differentiation for Year 4 Halves and Quarters:
Varied Fluency
Developing Questions to support writing half, quarter and three quarters as decimals.
Expected Questions to support writing fractions equivalent to half, quarter and three quarters as decimals.
Greater Depth Questions to support writing fractions equivalent to half, quarter and three quarters as decimals. Multiple answers possible.
Reasoning and Problem Solving
Questions 1, 4 and 7 (Problem Solving)
Developing Find the odd one out where two pairs of simplified fractions and decimals are given with one odd answer.
Expected Find the odd one out where two pairs of simplified or equivalent fractions and decimals are given with one odd answer.
Greater Depth Find the odd one out where three pairs of equivalent fractions and decimals are given with one odd answer.
Questions 2, 5 and 8 (Reasoning)
Developing Explain who travels the furthest using simplified fractions and decimals. Two statements to compare.
Expected Explain who travels the furthest using equivalent fractions and decimals. Two statements to compare.
Greater Depth Explain who travels the furthest using equivalent fractions and decimals. Three statements to compare.
Questions 3, 6 and 9 (Problem Solving)
Developing Use the digit clues to find the missing decimal or simplified fractions.
Expected Use the digit clues to find the missing decimal or equivalent fractions.
Greater Depth Use the digit clues to find the missing decimal or equivalent fractions. There may be multiple answers. | 560 | 2,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-50 | latest | en | 0.853617 |
https://math.stackexchange.com/questions/1744160/evaluate-lim-x-to-pi-left-frac-sin-x-pi-2-x2-right-without-lhop/1744226 | 1,632,222,679,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00608.warc.gz | 422,923,121 | 39,048 | # Evaluate $\lim _{x\to \pi }\left(\frac{\sin x}{\pi ^2-x^2}\right)$ without L'Hopital or Taylor series
I want to evaluate
$$\lim _{x\to \pi }\left(\frac{\sin x}{\pi ^2-x^2}\right)$$
without using L'Hopital's rule or Taylor series. My thinking process was something like this: in order to get rid of the undefined state, I need to go from $\sin x$ to $\cos x$. I tried this substitution: $t = \frac{\pi}{2}-x$ which gets me to $\cos t$, but I get:
$$\lim _{x\to -\frac{\pi}{2} }\left(\frac{\cos t}{\pi ^2-(\pi^2-t)}\right)$$
which also evaluates to $0/0$.
Then I tried this:
$\lim _{x\to \pi }\left(\frac{\sin x}{(\pi-x)(\pi+x)}\right) = \lim _{x\to \pi }\left(\frac{\sin x}{(\pi-x)2\pi}\right)$
which also gets me nowhere. The fact that $\sin(\pi-x) = \sin x$ also doesn't help because the substitution would lead to $\sin 0 = 0$.
The correct solution is $\frac{1}{2\pi}$
Can somebody help me?
hint: $\sin x = \sin(\pi - x)$, and $\pi^2 - x^2 = (\pi -x)(\pi + x)$.
Also use $\dfrac{\sin(\pi - x)}{\pi - x} \to 1$ when $x \to \pi$.
• Oh, thank you! I was so stupid not to have noticed it! Apr 15 '16 at 19:04
• Nice. Its a learning experience... Apr 15 '16 at 19:04
• I was just wondering that isn't "$\dfrac{\sin(\pi - x)}{\pi - x} \to 1$ when $x \to \pi$" considered as L'Hopital's rule? Apr 15 '16 at 19:43
• @Ignite That limit can be shown using inequalities from elementary geometry along with the squeeze theorem. I've posted an answer that walks through that development. -Mark Apr 15 '16 at 20:17
Note that from elementary geometry, the sine function is bounded as
$$|\theta \cos(\theta)|\le |\sin(\theta)|\le |\theta| \tag 1$$
for $0\le |\theta|\le \pi/2$. Letting $\theta =x-\pi$ in $(1)$ yields
$$|(x-\pi)\cos(x-\pi)|\le |\sin(x-\pi)|\le |x-\pi| \tag 2$$
for $0\le |x-\pi|\le \pi/2$. For $x\ne \pi$, we find upon dividing $(2)$ by $|x-\pi|$
$$\cos(x-\pi)\le \frac{\sin(x-\pi)}{x-\pi}\le 1 \tag 3$$
for $0 < |x-\pi|\le \pi/2$
Finally, dividing $(3)$ by $x+\pi$ reveals
$$\frac{\cos(x-\pi)}{x+\pi}\le \frac{\sin(x-\pi)}{x^2-\pi^2}\le \frac{1}{x+\pi}$$
whereupon applying the squeeze theorem yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \pi}\frac{\sin(x-\pi)}{x^2-\pi^2}=\frac{1}{2\pi}}$$
• Thanks for a new approach to this problem! Apr 15 '16 at 21:10
• @Quant You're welcome! My pleasure. I really just wanted to give you the best answer I could. -Mark Apr 15 '16 at 22:02 | 888 | 2,437 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-39 | latest | en | 0.83912 |
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On 12/23/2015 8:04 AM, snipped-for-privacy@unlisted.moc wrote:
I'm pleased with the assorted wisdom here. I'd never heard of reverse polarity, and battery charged backwards. Amazing, the things that go on, out in the world.
Did you actually paint a cow? I've heard of farmers spray painting the word "cow" during hunting season when the Citidiots are out with guns. And, yes, I've heard of citidiots harvesting a cow when they hold a deer hunting permit.
--
.
Christopher A. Young
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<% } %>
<%-name%>
wrote:
Don't you mean to say that it's possible to change the polarity of a generator (not an alternator) by connecting the battery backwards, even for a short time?
At any rate, you're right about the meter. He should measure the voltage of the battery without the charger, with the charger, without the charger when cranking the starter motor.
If low charge, or a bad battery, is really the problem, the voltage will drop too much when cranking the starter.
(With a car, I don't need a meter and don't need to get out of the driver's seat by trying to blow the horn while cranking the engine. I do this test when the car won't crank. If it doesn't blow well, it's the battery. It if blows well, the starter isn't even engaging electrically, so its the starter circuit or the sstarter.)
Yeah, a gnerator. Not a battery.
That only works if the battery is reversed, connected backwards. Maybe it is and the OP is going by the post positions, which are reversed, so he should look at the embossed + and - next to the battery posts. Those are always accurate.
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wrote:
No they are not. If a battery has gone dead it can be charged either way. If the battery was dead and the generaztor had lost it's residual magnetism and was polarized backwards, the terminal markings on the battery will be wrong, and the battery will not produce the full rated cranking capacity because the pos plates are charged neg, and the neg plates are charged pos, and the plate chemistry is different + to -
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<%-name%>
On Wed, 23 Dec 2015 15:19:10 -0500, snipped-for-privacy@snyder.on.ca wrote:
Okay, I defer to your greater experience. You and bowman have convinced me.
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<%-name%>
On 12/23/2015 01:46 AM, Micky wrote:
You can also reverse the polarity of a battery if it's close to completely discharged.
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<%-name%>
As I said in a prior reply, YES, the entire system appears to have reversed. It may be rare, but it did it. My VOM meter proves the battery is reversed. Your URL link explains well how to get the polarity set right, (and I vaguely recall doing this many years ago on my first car), but how do I get the battery reversed again? -OR- do I just reverse the leads on the battery? Or maybe I'll have to replace it. (It was a good battery).
Now I know why they quit using generators. This one will be replaced with an alternator real soon. When I bought this tractor, it ran well and did not have any major problems. I knew the wiring to the lights was chopped off at the switch because of bad wires, and there were a few other minor issues, but just seeing that generator gave me a sick feeling. (now I know why). Almost every antique tractor still in use has been converted to an alternator and NEG ground. But not this one, it still has the POS ground too. This will be converted soon, if not immediately.
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<%-name%>
On Wed, 23 Dec 2015 05:09:02 -0600, snipped-for-privacy@unlisted.moc wrote:
To get the battery back to normal you need to totally kill it first. Hook up a couple of headlights and leave the connected untill they don't even glow any more, then try with a small bulb like a tail light, and leave it connected untill it doesn't glow any more either - then connect the charger the right way and recharge it.
When the battery is properly recharged, reflash the generator to the proper polarity, reconnect the battery and restart the tractor. Make sure you follow the flash directions to the letter when repolarizing - someone sometime or another did not - resulting in the situation you now have.
There is a possibility the battery will be no good when you are done - but it is worth a try.
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<%-name%>
On 12/22/15 9:08 PM, snipped-for-privacy@unlisted.moc wrote:
If it's a positive ground vehicle, wouldn't the red charger clip go to the negative black batt terminal?
--
Want to close wage gap? Step one: Change your major from Gender Studies
or Feminist Dance Therapy to Electrical Engineering.
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<%-name%>
wrote:
No. Why would you think that?? It is the terminal that connects to the ground that changes between pos and neg ground - on a Pos ground vehicle, the red post goes to ground while on a neg ground vehicle the black post goes to ground.
Quite possible when the battery was totally dead sometime some "farmer" thought as you do, and connected the charger backwards - thereby reverse charging the battery - and that "reflashed" the generator to be a negative ground generator instead of positive.
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<%-name%>
On 12/23/2015 5:04 PM, snipped-for-privacy@snyder.on.ca wrote:
Regardless of which is ground, the red clip goes to + red positive, and the black clip goes to - black negative.
I've been working with batteries for more than 8 years. Nyah, nyah to Philo and Clare.
--
.
Christopher A. Young
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<%-name%>
wrote in message wrote:
No. Why would you think that?? It is the terminal that connects to the ground that changes between pos and neg ground - on a Pos ground vehicle, the red post goes to ground while on a neg ground vehicle the black post goes to ground.
Quite possible when the battery was totally dead sometime some "farmer" thought as you do, and connected the charger backwards - thereby reverse charging the battery - and that "reflashed" the generator to be a negative ground generator instead of positive.
You are acting as to be expert on battery but by reading your post I personally would not let you change battery in flashlight!
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<%-name%>
Who - me, or the "painted cow"? I spent half my working life as an auto mechanic/auto electric specialist/service manager/Automotive instructor, and the other half as a computer technician - and I built and drove my own electric car back in the (late) seventies - so I know a bit about batteries.
Don't know about the "painted cow"
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<%-name%>
On Thu, 24 Dec 2015 17:19:52 -0500, snipped-for-privacy@snyder.on.ca wrote:
I didn't do it, but you can cuff me if you wish... :)
I'm the guy with this problem (OP). I worked as an electrician for years, and did electronics as a hobby for years too. But this battery thing is a new one for me. I never know it was possible, which is why I posted this, while I was still compleltely puzzled by it.
Since then, this thread and what I googled on th web, has taught me something new. But I still have to clean up the mess.... The holiday and bad weather, have kept me from doing anything except reading and determining which is the best solution. At least now I know what occurred, even if it's still sort of unbelievable.
So, I put a note under my tree, and asked Santa for a brand new tractor. That should work. :) Which reminds me, I need to go look for flying reindeer..... I can always hire them to pull my tractor... LOL....
Later!
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<%-name%>
snipped-for-privacy@snyder.on.ca posted for all of us...
tony911 is from the old alt.hvac crowd and used to be sensible but I think Stumped rubbed of on him.
--
Tekkie
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<%-name%>
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### Finding Primes: Sieve of Erastosthenes
```
Date: 4/1/96 at 17:32:15
From: Ms Susan M. Schmidt
Subject: Sieve of Erastosthenes
Dr Math:
Could you please elaborate on this subject a little more. I have
to write a Pascal program to use this method and I can only find
little information about the Sieve of Erastosthenes. Any info
would be greatly appreciated.
```
```
Date: 4/2/96 at 2:13:13
From: Doctor Jodi
Subject: Re: Sieve of Erastosthenes
Hi there! At
http://www.utm.edu/research/primes/largest.html
you can find some information on the Sieve of Eratosthenes:
The document about finding primes is located at
http://www.utm.edu/research/primes/proving.html
1. Finding Very Small Primes
For finding all the small primes, say all those less than
10,000,000; the most efficient way is by using the Sieve of
Eratosthenes (ca 240 BC):
Make a list of all the integers less than or equal to n (greater
than one) and strike out the multiples of all primes less than or
equal to the square root of n, then the numbers that are left are
the primes.
For example, to find all the odd primes less than or equal to 100:
list the odd numbers from 3 to 100 (why even list the evens?) The
first number is 3 so it is the first odd prime--cross out all of
its multiples. Now the first number left is 5, the second odd
prime--cross out all of its multiples. Repeat with 7 and then
since the first number left, 11, is larger than the square root of
100, all of the numbers left are primes. This method is so fast
that there is no reason to store a large list of primes on a
computer--an efficient implementation can find them faster than a
computer can read from a disk. Bressoud has a pseudocode
implementation of this algorithm [Bressoud89, p19] and Riesel a
PASCAL implementation [Riesel94, p6].
To find individual small primes trial division works okay. Just
divide by all the primes less than the square root. For example,
to show is 211 is prime, just divide by 2, 3, 5, 7, 11, and 13.
(Pseudocode [Bressoud89, pp21-22], PASCAL [Riesel94, pp7-8].)
Rather than divide by just the primes, it is sometimes more
practical to divide by 2, 3 and 5; then by all the numbers
congruent to 1, 7, 11, 13, 17, 19, 23, and 29 modulo 30--again
stopping when you reach the square root. This type of
factorization is sometimes called wheel factorization. Look in
most any elementary text on number theory for information about
these tests.
Suppose n has twenty digits, then it is getting impractical to
divide by the primes less than its square root, and it is
impossible if n has two hundred digits--so we need much faster
tests. We discuss several such tests below.
-----
That document also contains a lot more information about primes...
take a look at it some time!
Here's some other information that may be interesting to you:
I found a biography of Erastosthenes at
http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Eratosthenes.html
which, though not particularly pertinent to his sieve, may be
interesting to you.
__
Factorization and Primality Testing
1989 . XIII, 237 pp. 2 figs., Hardcover ISBN 3-540-97040-1
DM 102,-; oS 795,60; sFr 98,-
Book category: Advanced Textbook
Publication date : Available
(Undergraduate Texts in Mathematics. Eds.: J.H. Ewing; F.W.
Gehring; P.R. Halmos. )
Last update: March 6, 1996, URL:
http://www.springer.de/catalog/html-files/deutsch/math/3540970401.html
Copyright Springer-Verlag Berlin/Heidelberg 1996
____
-Doctor Jodi, The Math Forum
```
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Math Forum Home || Math Library || Quick Reference || Math Forum Search | 1,080 | 3,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-05 | longest | en | 0.875196 |
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\$30.00
## Introduction
Every credit card number contains a check digit at its rightmost, which is used for simple error
detection. It can be used to protect against accidental errors such as a mistyped digit or the
permutation of two successive digits.
In this assignment, you will write a program that computes the
check digit of a partial credit card number. For simplicity, we assume that a card number has exactly
16 digits in the form 𝑑1𝑑2𝑑3𝑑4𝑑5𝑑6𝑑7𝑑8𝑑9𝑑10𝑑11𝑑12𝑑13𝑑14𝑑15𝑑16, in which 𝑑1 … 𝑑15 is the 15-
digit partial card number obtained from the program user and 𝑑16 is the check digit to be computed.
To compute the check digit of a partial card number, we can use the Luhn algorithma described
below:
1. Double the odd-positioned digits 𝑑1, 𝑑3, 𝑑5, 𝑑7, 𝑑9
, 𝑑11, 𝑑13, and 𝑑15.
2. If a doubled digit is greater than 9, replace it by its sum of digits. (E.g., 16 is replaced by 1 + 6
= 7.)
3. Sum the even-positioned digits 𝑑2, 𝑑4, 𝑑6, 𝑑8, 𝑑10, 𝑑12, and 𝑑14 with the modified oddpositioned digits 𝑑1, 𝑑3, 𝑑5, 𝑑7, 𝑑9
, 𝑑11, 𝑑13, and 𝑑15.
4. Multiply the sum by 9. Then the units digit (個位數, the rightmost digit) of the multiplication
is the check digit 𝑑16.
Example 1: partial card number is 763545841927506.
1. Double the underlined odd-positioned digits in 763545841927506: (7̲ × 2) = 14, (3 × 2) =
6, (4 × 2) = 8, (8̲ × 2) = 16, (1̲ × 2) = 2, (2̲ × 2) = 4, (5̲ × 2) = 10, and (6̲ × 2) = 12.
2. Replace 14, 16, 10 and 12 by their sum of digits: 14 1 + 4 = 5, 16 1 + 6 = 7, 10 1 + 0 =
1, and 12 1 + 2 = 3.
3. Sum all 15 digits: 5 + 6 + 6 + 5 + 8 + 5 + 7 + 4 + 2 + 9 + 4 + 7 + 1 + 0 + 3 = 72. (The
digits in red came from steps 1 and 2.)
4. Multiply the sum 72 by 9: 72 × 9 = 648̲. The units digit (8) is the check digit. Therefore, the
full card number is: 7635458419275068.
Example 2: partial card number is 543210987654321.
1. Double the underlined digits in 543210987654321: (5̲ × 2) = 10, (3 × 2) = 6, (1 × 2) = 2,
(9 × 2) = 18, (7̲ × 2) = 14, (5̲ × 2) = 10, (3̲ × 2) = 6, and (1̲ × 2) = 2.
2. Replace 10, 18, 14 and 10 by their sum of digits: 10 1 + 0 = 1, 18 1 + 8 = 9, 14 1 + 4 =
5, and 10 1 + 0 = 1.
3. Sum all 15 digits: 1 + 4 + 6 + 2 + 2 + 0 + 9 + 8 + 5 + 6 + 1 + 4 + 6 + 2 + 2 = 58. (The
digits in red came from steps 1 and 2.)
4. Multiply the sum 58 by 9: 58 × 9 = 522̲. The units digit (2) is the check digit. Therefore, the
full card number is: 5432109876543212.
a Reference: http://en.wikipedia.org/wiki/Luhn_algorithm
CSCI1120 Introduction to Computing Using C++
## Program Specification
The program should repeatedly accept a user input until a negative integer is entered. You can
assume that the user input is always either a 15-digit number or a negative integer. (That is, you do
not have to check whether the input is out of this assumption.)
If the input is a 15-digit number, then
you have to use the Luhn algorithm to compute the check digit and display the full card number. The
full card number should be printed as four pieces of 4-digit segments in the format XXXX-XXXX-XXXXXXXX. If the input is a negative integer, then the program terminates.
Note that the int data type is not big enough to represent 15-digit numbers. In order to store the
partial or full card numbers, you can declare variables of the long long type, which is a bigger
integral data type.
long long num; // int num; does not work
You are not allowed to use arrays in this assignment.
### Program Output
The following shows some sample output of the program. The bold blue text is user input and the
other text is the program output. You can try the provided sample program for other input. Your
program output should be exactly the same as the sample program (i.e., same text, same symbols,
same letter case, same number of spaces, etc.). Otherwise, it will be considered as wrong, even if
you have computed the correct result.
Enter a 15-digit partial card num: 763545841927506↵
Full card num is: 7635-4584-1927-5068
Enter a 15-digit partial card num: 543210987654321↵
Full card num is: 5432-1098-7654-3212
Enter a 15-digit partial card num: 432100080012056↵
Full card num is: 4321-0008-0012-0564
Enter a 15-digit partial card num: -1↵
Bye!
### Submission and Marking
Your program file name should be luhn.cpp. Submit the file in Blackboard
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[SOLVED]Help with BCD to Binary
Status
Not open for further replies.
kgavionics
Full Member level 3
Hello guys
I have built a break-out board based on the keyboard encoder 74c923 hooked up to an Arduino UNO (AVR-GCC using Atmel studio 7), which sends a BCD digit each time I press on a key! All is working fine, but I can't figure out how to store the 3 digits into an 8 bit variable!
In the picture attached, the number 123 is stored in 3 memory locations (each digit is stored in a memory location) and I want to store the entered number in a single memory location (binary).
I can easily convert binary to BCD, but I'm struggling to do it the other way around!
Can someone give me a heads-up how to convert from BCD to Binary please?
KlausST
Super Moderator
Staff member
Hi,
I really have no clue how you come to these three numbers.
The only thing I can tell:
123 could be 0b01111011 or 0x7B
****
you surely know that BCD are 4 bits.
with binary value from 0b0000 to 0b1001 and HEX 0x0 to 0x9 and decimal 0 to 9
Klaus
betwixt
Super Moderator
Staff member
Assuming your memory locations are 8 bits wide, the highest number you can store is binary 11111111 or 255 in decimal so you can't store bigger numbers without spanning more locations.
Brian.
kgavionics
Full Member level 3
Assuming your memory locations are 8 bits wide, the highest number you can store is binary 11111111 or 255 in decimal so you can't store bigger numbers without spanning more locations.
Brian.
I'm totally aware that I can't store more than 255! My question is how to convert BCD to binary!
--- Updated ---
Hi,
I really have no clue how you come to these three numbers.
The only thing I can tell:
123 could be 0b01111011 or 0x7B
****
you surely know that BCD are 4 bits.
with binary value from 0b0000 to 0b1001 and HEX 0x0 to 0x9 and decimal 0 to 9
Klaus
I just give an example! I meant any number from 000 to 255 in decimal! I want to implement the 8bit version algorithm, then I can extend it to 16 bits!
Furthermore, I'm aware that BCD numbers are coded in 4 bits! But in my case I store them in one memory location (I set high nibbles to 0).
Now the big question, how to convert BCD number to a binary one!
Last edited:
KlausST
Super Moderator
Staff member
Hi,
the "B" in BCD already means "binary".
Please explain more exactly what you need.
Klaus
crutschow
how to convert BCD number to a binary one!
Multiply the 3rd BCD number by 100, the 2nD BCD number by 10, and then add the three together to get one binary number.
kgavionics
Points: 2
johnny78
Points: 2
kgavionics
Full Member level 3
Multiply the 3rd BCD number by 100, the 2nD BCD number by 10, and then add the three together to get one binary number.
You are the man crutschow! Your suggestion worked like a charm!
Thank you very much!!
KlausST
Super Moderator
Staff member
Hi,
wow,
this has nothing to do with
* encoder 74c923
* BCD
* binary
* memory location
... a lot of confusing information.
btw: what happens if the first of the three digits is 3 or higher?
Klaus
Status
Not open for further replies. | 854 | 3,112 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2023-14 | latest | en | 0.933397 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-and-trigonometry-10th-edition/chapter-11-11-1-sequences-and-series-11-1-exercises-page-777/5 | 1,718,290,681,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00850.warc.gz | 697,504,423 | 12,549 | ## Algebra and Trigonometry 10th Edition
$i$ is the index of summation. $i=1$ is the lower limit of the summation and i=n (written as $n$ in the notation) is the upper limit by convention. | 54 | 189 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.930964 |
https://www.scribd.com/doc/32371725/Reconstructing-Linux-%E0%A6%95%E0%A6%BE%E0%A6%B8-%E0%A6%9F%E0%A7%8B%E0%A6%AE%E0%A6%BE%E0%A6%87%E0%A6%9C%E0%A6%A1-%E0%A6%B2%E0%A6%BF%E0%A6%A8%E0%A6%BE%E0%A6%95-%E0%A6%B8-%E0%A6%A1%E0%A6%BF%E0%A6%B8-%E0%A6%9F%E0%A7%8B-%E0%A6%B0-%E0%A6%A4%E0%A7%88%E0%A6%B0%E0%A7%80 | 1,532,370,320,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00277.warc.gz | 977,039,154 | 25,071 | l-¯* l-¯** ´7<| <<- l-¯*< <l¯*l-l**< l¬-l+ l<¯±l
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• ## File - MIDELLIP.CPP
(187)
• NextPrevious
• # include # include # include # include class Ellipse { private: public: void show_screen( ); void Lines();
Message 1 of 187 , Dec 9 3:37 AM
View Source
# include <iostream.h>
# include <graphics.h>
# include <conio.h>
# include <math.h>
class Ellipse
{
private:
public:
void show_screen( );
void Lines();
void midpoint_ellipse(float,float,float,float);
void angle_ellipse(float,float,float,float);
};
void Ellipse::Lines()
{
setcolor(2);
line(0,240, 640, 240);
line(320,0,320,480);
}
void Ellipse::midpoint_ellipse(float xc, float yc,float a, float b)
{
float color=11;
float aa=(a*a);
float bb=(b*b);
float aa2=(aa*2);
float bb2=(bb*2);
float x=0;
float y=b;
float fx=0;
float fy=(aa2*b);
float p=(float)(bb-(aa*b)+(0.25*aa)+0.5);
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
while(fx<fy)
{
x++;
fx+=bb2;
if(p<0)
p+=(fx+bb);
else
{
y--;
fy-=aa2;
p+=(fx+bb-fy);
}
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
}
p=(float)((bb*(x+0.5)*(x+0.5))+(aa*(y-1)*(y-1))-(aa*bb)+0.5);
while(y>0)
{
y--;
fy-=aa2;
if(p>=0)
p+=(aa-fy);
else
{
x++;
fx+=bb2;
p+=(fx+aa-fy);
}
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
}
}
void Ellipse::angle_ellipse(float x_dash,float y_dash,float a,float b)
{
float color=13;
float aa=(a*a);
float bb=(b*b);
float aa2=(aa*2);
float bb2=(bb*2);
float x=0;
float y=b;
float fx=0;
float fy=(aa2*b);
float p=(float)(bb-(aa*b)+(0.25*aa)+0.5);
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
while(fx<fy)
{
x++;
fx+=bb2;
if(p<0)
p+=(fx+bb);
else
{
y--;
fy-=aa2;
p+=(fx+bb-fy);
}
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
}
p=(float)((bb*(x+0.5)*(x+0.5))+(aa*(y-1)*(y-1))-(aa*bb)+0.5);
while(y>0)
{
y--;
fy-=aa2;
if(p>=0)
p+=(aa-fy);
else
{
x++;
fx+=bb2;
p+=(fx+aa-fy);
}
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
}
}
void Ellipse::show_screen( )
{
restorecrtmode( );
textmode(C4350);
textbackground(1);
cprintf(" MidPoint Ellipse Algorithm ");
textbackground(8);
}
float main( )
{
Ellipse e;
int driver=VGA;
int mode=VGAHI;
double theta;
float xc=0;
float yc=0;
float rx=0;
float ry=0;
float x_dash;
float y_dash;
do
{
e.show_screen( );
gotoxy(8,10);
cout<<"Central Points of the Ellipse : (xc,yc) :";
gotoxy(8,11);
cout<<"�������������������������������������";
gotoxy(12,13);
cout<<"Enter the value of xc = ";
cin>>xc;
xc = xc + 320;
gotoxy(12,14);
cout<<"Enter the value of yc = ";
cin>>yc;
yc = yc + 240;
gotoxy(8,18);
cout<<"Radius of the Ellipse : (rx,ry) :";
gotoxy(8,19);
cout<<"���������������������������������";
gotoxy(12,21);
cout<<"Enter the radius along x-axis : rx = ";
cin>>rx;
gotoxy(12,22);
cout<<"Enter the radius along y-axis : ry = ";
cin>>ry;
gotoxy(12,24);
cout<<"Enter the Angle of the Ellipsea : theta = ";
cin>>theta;
double r = theta * 11/630;
x_dash = xc * cos(r) - yc * sin(r);
y_dash = xc * sin(r) + yc * cos(r);
initgraph(&driver,&mode,"c:/tc/bgi");
setcolor(9);
//-----------calling functions-----------
e.Lines();
e.midpoint_ellipse(xc,yc,rx,ry); //Simple Ellipse
e.angle_ellipse(x_dash,y_dash,rx,ry); //Rotation Ellipse
setcolor(11);
outtextxy(110,460,"Press <any> to continue or Escape key to exit.");
float key=float(getch( ));
if(key==27) break;
}
while(1);
return 0;
}
• # include # include # include # include class Ellipse { private: public: void show_screen( ); void Lines();
Message 187 of 187 , Dec 1 2:04 AM
View Source
# include <iostream.h>
# include <graphics.h>
# include <conio.h>
# include <math.h>
class Ellipse
{
private:
public:
void show_screen( );
void Lines();
void midpoint_ellipse(float,float,float,float);
void angle_ellipse(float,float,float,float);
};
void Ellipse::Lines()
{
setcolor(2);
line(0,240, 640, 240);
line(320,0,320,480);
}
void Ellipse::midpoint_ellipse(float xc, float yc,float a, float b)
{
float color=11;
float aa=(a*a);
float bb=(b*b);
float aa2=(aa*2);
float bb2=(bb*2);
float x=0;
float y=b;
float fx=0;
float fy=(aa2*b);
float p=(float)(bb-(aa*b)+(0.25*aa)+0.5);
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
while(fx<fy)
{
x++;
fx+=bb2;
if(p<0)
p+=(fx+bb);
else
{
y--;
fy-=aa2;
p+=(fx+bb-fy);
}
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
}
p=(float)((bb*(x+0.5)*(x+0.5))+(aa*(y-1)*(y-1))-(aa*bb)+0.5);
while(y>0)
{
y--;
fy-=aa2;
if(p>=0)
p+=(aa-fy);
else
{
x++;
fx+=bb2;
p+=(fx+aa-fy);
}
putpixel((xc+x),(yc+y),color);
putpixel((xc+x),(yc-y),color);
putpixel((xc-x),(yc-y),color);
putpixel((xc-x),(yc+y),color);
}
}
void Ellipse::angle_ellipse(float x_dash,float y_dash,float a,float b)
{
float color=13;
float aa=(a*a);
float bb=(b*b);
float aa2=(aa*2);
float bb2=(bb*2);
float x=0;
float y=b;
float fx=0;
float fy=(aa2*b);
float p=(float)(bb-(aa*b)+(0.25*aa)+0.5);
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
while(fx<fy)
{
x++;
fx+=bb2;
if(p<0)
p+=(fx+bb);
else
{
y--;
fy-=aa2;
p+=(fx+bb-fy);
}
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
}
p=(float)((bb*(x+0.5)*(x+0.5))+(aa*(y-1)*(y-1))-(aa*bb)+0.5);
while(y>0)
{
y--;
fy-=aa2;
if(p>=0)
p+=(aa-fy);
else
{
x++;
fx+=bb2;
p+=(fx+aa-fy);
}
putpixel((x_dash+x),(y_dash+y),color);
putpixel((x_dash+x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash-y),color);
putpixel((x_dash-x),(y_dash+y),color);
}
}
void Ellipse::show_screen( )
{
restorecrtmode( );
textmode(C4350);
textbackground(1);
cprintf(" MidPoint Ellipse Algorithm ");
textbackground(8);
}
float main( )
{
Ellipse e;
int driver=VGA;
int mode=VGAHI;
double theta;
float xc=0;
float yc=0;
float rx=0;
float ry=0;
float x_dash;
float y_dash;
do
{
e.show_screen( );
gotoxy(8,10);
cout<<"Central Points of the Ellipse : (xc,yc) :";
gotoxy(8,11);
cout<<"�������������������������������������";
gotoxy(12,13);
cout<<"Enter the value of xc = ";
cin>>xc;
xc = xc + 320;
gotoxy(12,14);
cout<<"Enter the value of yc = ";
cin>>yc;
yc = yc + 240;
gotoxy(8,18);
cout<<"Radius of the Ellipse : (rx,ry) :";
gotoxy(8,19);
cout<<"���������������������������������";
gotoxy(12,21);
cout<<"Enter the radius along x-axis : rx = ";
cin>>rx;
gotoxy(12,22);
cout<<"Enter the radius along y-axis : ry = ";
cin>>ry;
gotoxy(12,24);
cout<<"Enter the Angle of the Ellipsea : theta = ";
cin>>theta;
double r = theta * 11/630;
x_dash = xc * cos(r) - yc * sin(r);
y_dash = xc * sin(r) + yc * cos(r);
initgraph(&driver,&mode,"c:/tc/bgi");
setcolor(9);
//-----------calling functions-----------
e.Lines();
e.midpoint_ellipse(xc,yc,rx,ry); //Simple Ellipse
e.angle_ellipse(x_dash,y_dash,rx,ry); //Rotation Ellipse
setcolor(11);
outtextxy(110,460,"Press <any> to continue or Escape key to exit.");
float key=float(getch( ));
if(key==27) break;
}
while(1);
return 0;
}
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Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View. | 2,774 | 8,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-48 | latest | en | 0.169687 |
https://e2e.ti.com/support/logic-group/logic/f/logic-forum/578121/cd4024b-frequency-divider-schematic?tisearch=e2e-sitesearch&keymatch=CD4024B | 1,721,265,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00521.warc.gz | 182,810,816 | 26,013 | If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
# CD4024B: Frequency Divider schematic
Part Number: CD4024B
I am needing to correctly use the IC to divide an input wave from a signal generator by 32 and by 8 (any number really I know that it divides by 2,4,8,16,32,64 I think) I am assuming it shouldn't matter whether the input wave is a sine wave or square wave. Also I'm assuming that the amplitude of the wave shouldn't matter as long as I am within the limits of the chip. For my situation I am inputting a sine wave 2volts peak to peak (-1V to1V). Here is my understanding of how to hook it up to get the division I'm trying to accomplish. I have VDD to +8v DC and Vss to GND. I also have the Reset to GND. Input wave is connected to pin 1(CLK) (positive lead to pin 1 negative lead to ground directly from wave generator) Here is where I become lost. I measure +V (seems like its 2V if I remember right) on any of the Q outputs with my oscilloscope but I do not get a output square wave. (positive lead of the scope to Q1 or any other Q) (negative lead to GND). My source is a power supply but will eventually be a stand alone battery. So all grounds tie together and are hooked to the common terminal of the power supply and eventually the common terminal of the battery once I get this circuit working and integrated into a portable device. Please help. Thanks
• Hello Blake,
You are on the right track, and a sine wave input is acceptable since the inputs are schmitt triggers, so they are tolerant to slow rising inputs.
However, the input should not go negative. The inputs will clamp the negative voltage and also a -1V signal will exceed the absolute maximum of -0.5 for the device.
A 2V signal amplitude may also not be enough. You need to make sure that the input fits the Vil and Vih specifications to get a defined high and low on the output. Please look at the electrical characteristics for the device. for 8V DC supply, it looks like the Vih will likely be close to 3V, and it would be better if the signal was higher than that.
Please let us know if this helps and if you have additional questions as we would be happy to help.
Best,
Michael | 550 | 2,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.925129 |
https://support.microsoft.com/en-us/office/normsinv-function-8d1bce66-8e4d-4f3b-967c-30eed61f019d?ocmsassetid=ha102753183&ctt=5&origin=ha102752955&correlationid=9a185265-f6fb-446e-a04c-75ea1f38d26c&ui=en-us&rs=en-us&ad=us | 1,603,666,798,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890028.58/warc/CC-MAIN-20201025212948-20201026002948-00614.warc.gz | 542,747,122 | 22,795 | # NORMSINV function
Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero and a standard deviation of one.
Important: This function has been replaced with one or more new functions that may provide improved accuracy and whose names better reflect their usage. Although this function is still available for backward compatibility, you should consider using the new functions from now on, because this function may not be available in future versions of Excel.
For more information about the new function, see NORM.S.INV function.
## Syntax
NORMSINV(probability)
The NORMSINV function syntax has the following argument:
• Probability Required. A probability corresponding to the normal distribution.
## Remarks
• If Probability is nonnumeric, NORMSINV returns the #VALUE! error value.
• If Probability <= 0 or if Probability >= 1, NORMSINV returns the #NUM! error value.
Given a value for Probability, NORMSINV seeks that value z such that NORMSDIST(z) = probability. Thus, precision of NORMSINV depends on precision of NORMSDIST. NORMSINV uses an iterative search technique. If the search has not converged after 100 iterations, the function returns the #N/A error value.
## Example
Copy the example data in the following table, and paste it in cell A1 of a new Excel worksheet. For formulas to show results, select them, press F2, and then press Enter. If you need to, you can adjust the column widths to see all the data.
Formula
Description
Result
=NORMSINV(0.9088)
Inverse of the standard normal cumulative distribution, with a probability of 0.9088.
1.3334
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#### Repaying a Loan While Mary Corens was a student at the University of Tennessee, she borrowed \$9,000 in student loans at an annual interest rate of 8%
###### Finance
Repaying a Loan
While Mary Corens was a student at the University of Tennessee, she borrowed \$9,000 in student loans at an annual interest rate of 8%. If Mary repays \$1,700 per year, then how long (to the nearest year) will it take her to repay the loan? Do not round intermediate calculations. Round your answer to the nearest whole number.
year(s)
Problem 4-14 (Uneven Cash Flow Stream)
e oo
Uneven Cash Flow Stream
a. Find the present values of the following cash flow streams. The appropriate interest rate is 12%. (Hint: It is fairly easy to work this problem dealing with the individual cash flows. However, if you have a financial calculator, read the section of the manual that describes how to enter cash flows such as the ones in this problem. This will take a little time, but the investment will pay huge dividends throughout the course. Note that, when working with the calculator's cash flow register, you must enter CF0 = 0. Note also that it is quite easy to work the problem with Excel, using procedures described in the Ch04 Tool Kit.xlsx.) Do not round intermediate calculations. Round your answers to the nearest cent.
Year Cash Stream A Cash Stream B 1 \$100 \$200 2 400 400 3 400 400 4 400 400 5 200 100 Stream A: \$
Stream B: \$ b. What is the value of each cash flow stream at a 0% interest rate? Round your answers to the nearest dollar. Stream A \$ Stream B \$
## 3.94 USD
### Option 2
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Chapter 7 (pages 225–228):
1.
Your brother wants to borrow \$10,000 from you. He has offered to pay you back \$12,000 in a year. If the cost of capital of this investment opportunity is 10%, what is its NPV? Should you undertake the investment opportunity? Calculate the IRR and use it to determine the maximum deviation allowable in the cost of capital estimate to leave the decision unchanged.
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8.
You are considering an investment in a clothes distributor. The company needs \$100,000 today and expects to repay you \$120,000 in a year from now. What is the IRR of this investment opportunity? Given the riskiness of the investment opportunity, your cost of capital is 20%. What does the IRR rule say about whether you should invest?
Â
19.
You are a real estate agent thinking of placing a sign advertising your services at a local bus stop. The sign will cost \$5,000 and will be posted for one year. You expect that it will generate additional revenue of \$500 per month. What is the payback period?
Â
21.
You are deciding between two mutually exclusive investment opportunities. Both require the same initial investment of \$10 million. Investment A will generate \$2 million per year (starting at the end of the first year) in perpetuity. Investment B will generate \$1.5 million at the end of the first year and its revenues will grow at 2% per year for every year after that.
• a. Which investment has the higher IRR?
• b. Which investment has the higher NPV when the cost of capital is 7%?
• c. In this case, for what values of the cost of capital does picking the higher IRR give the correct answer as to which investment is the best opportunity?
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Chapter 8 (260–262)
1.
Pisa Pizza, a seller of frozen pizza, is considering introducing a healthier version of its pizza that will be low in cholesterol and contain no trans fats. The firm expects that sales of the new pizza will be \$20 million per year. While many of these sales will be to new customers, Pisa Pizza estimates that 40% will come from customers who switch to the new, healthier pizza instead of buying the original version.
a. Assume customers will spend the same amount on either version. What level of incremental sales is associated with introducing the new pizza?
b. Suppose that 50% of the customers who will switch from Pisa Pizza’s original pizza to its healthier pizza will switch to another brand if Pisa Pizza does not introduce a healthier pizza. What level of incremental sales is associated with introducing the new pizza in this case?
Â
6.
Cellular Access, Inc. is a cellular telephone service provider that reported net income of \$250 million for the most recent fiscal year. The firm had depreciation expenses of \$100 million, capital expenditures of \$200 million, and no interest expenses. Working capital increased by \$10 million. Calculate the free cash flow for Cellular Access for the most recent fiscal year.
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12.
A bicycle manufacturer currently produces 300,000 units a year and expects output levels to remain steady in the future. It buys chains from an outside supplier at a price of \$2 a chain. The plant manager believes that it would be cheaper to make these chains rather than buy them. Direct in-house production costs are estimated to be only \$1.50 per chain. The necessary machinery would cost \$250,000 and would be obsolete after 10 years. This investment could be depreciated to zero for tax purposes using a 10-year straight-line depreciation schedule. The plant manager estimates that the operation would require \$50,000 of inventory and other working capital upfront (year 0), but argues that this sum can be ignored because it is recoverable at the end of the 10 years. Expected proceeds from scrapping the machinery after 10 years are \$20,000.
If the company pays tax at a rate of 35% and the opportunity cost of capital is 15%, what is the net present value of the decision to produce the chains in-house instead of purchasing them from the supplier?
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 1,338 | 5,977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.930817 |
https://cracku.in/16-how-many-4-letter-words-can-be-formed-from-the-wor-x-iift-5th-dec-2021-slot-1 | 1,701,834,620,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100583.13/warc/CC-MAIN-20231206031946-20231206061946-00768.warc.gz | 218,870,569 | 25,152 | Question 16
# How many 4 letter words can be formed from the word "CORONAVIRUS".
Solution
"CORONAVIRUS" has 7 distinct alphabets and two pairs of repeated characters ", O" and "R".
There are three possible cases for creating 4 letter words.
1. Two letters are "O" and the other two are "R".
The total number of arrangements = $$\frac{4!}{2!\times2!}=6$$
2. Two of the letters are either "O" or "R" and the others are distinct.
The total number of arrangements = $$^2C_1\times^8C_2\times\frac{4!}{2!}=2\times28\times12=672$$
3. All four letters are distinct.
The total number of arrangements = $$^9C_4\times4!=3024$$
Thus, the total number of four-letter words possible = $$6+672+3024=3702$$.
Hence, the answer is option C.
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##### Ishan Patni
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##### Ali
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Where's the solution? | 283 | 928 | {"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} | 4.15625 | 4 | CC-MAIN-2023-50 | latest | en | 0.799426 |
https://justaaa.com/statistics-and-probability/940420-determine-how-much-is-in-each-account-on-the | 1,713,507,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00050.warc.gz | 294,508,305 | 10,101 | Question
# Determine how much is in each account on the basis of the indicated compounding after the...
Determine how much is in each account on the basis of the indicated compounding after the specified years have passed; P is the initial principal, and r is the annual rate given as a percent. (Round your answers to the nearest cent.)
after one year where P = \$4500 and r = 3.9%
(a) compounded annually
\$
(b) compounded quarterly
\$
(c) compounded monthly
\$
(d) compounded weekly
\$
(e) compounded daily
\$
Let A be the amount at the end of the year. Then
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https://devforum.roblox.com/t/help-to-check-if-a-part-is-full-inside-another-part/802305 | 1,680,054,627,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00600.warc.gz | 248,126,299 | 8,213 | # Help to check if a part is full inside another part
Hii
I am trying to check if a part its INSIDE a part, full inside, not just a bit
## Code I got atm:
``````local c = workspace.c
local v = workspace.x
print(c.Position.x - v.Position.x <= v.Size.X/2 and c.Position.x - v.Position.x >= v.Size.X/-2 and c.Position.z - v.Position.z <= v.Size.z/2 and c.Position.z - v.Position.z >= v.Size.z/-2)
``````
https://gyazo.com/904005e3d6bafe64169c0091891a30e1
It should print true when its like this:
https://gyazo.com/aa9f17299ef3747debf27f2e273dd7c6
1 Like
You’ve to calculate doing this;
First you check if the part is inside and then you’ll have to calculate if any of the positions when you use CFrame are inside or outside the part.
1 Like
So what you’d do is something like this but 8 times for ever corner
You’d have to do:
the CFrame + Vector(Part.Size.X/2,Part.Size.Y/2,Part.Size.Z/2)
and then do then make each of them negative so you’ve 8 CFrames, which you check the position of and calculate if they’re inside or not.
If you’re making a building system I suggest using other tricks instead of this.
1 Like
EDIT: Or just do what Kry_s suggested xD
Can you tell us about your use case? If we can make some assumptions that might make it a lot easier.
E.g. are the Parts always axis-aligned boxes (AABs)? Or can they rotated, or be spheres/cylinders too?
To check if a AAB is fully within another AAB, you can do like this:
``````function isAABInsideAAB(position1, size1, position2, size2)
--Checks if an AABB defined by position1 and size1 is fully within the AABB defined by position2 and size2
-- if they have the same size and position, returns true.
local p1, p2 = position1, position2
local s1, s2 = size1/2, size2/2
local minX1, minX2 = p1.X - s1.X, p2.X - s2.X
local maxX1, maxX2 = p1.X + s1.X, p2.X + s2.X
local minY1, minY2 = p1.Y - s1.Y, p2.Y - s2.Y
local maxY1, maxY2 = p1.Y + s1.Y, p2.Y + s2.Y
local minZ1, minZ2 = p1.Z - s1.Z, p2.Z - s2.Z
local maxZ1, maxZ2 = p1.Z + s1.Z, p2.Z + s2.Z
return
(minX1 >= minX2 and maxX1 <= maxX2) and
(minY1 >= minY2 and maxY1 <= maxY2) and
(minZ1 >= minZ2 and maxZ1 <= maxZ2)
end
``````
You can’t just use any part’s position and size for this. It’s important that the part is not e.g. rotated 90 degrees on an axis, so the rotation has to be (0, 0, 0). We can get around this with math, to convert any Part that’s only rotated in increments of 90 degrees to an AAB:
``````function partToAAB(part)
--Given an axis-aligned part, returns the position and size of an AAB that takes up the same space as the part,
-- but whose X, Y and Z axes are the same as the world's axes.
assert(isAxisAligned(part.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
local AABPosition = part.Position
local AABSize = Vector3.new()
local worldCFrame = CFrame.new()
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(part.Size.X, 0, 0) )
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(0, part.Size.Y, 0) )
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(0, 0, part.Size.Z) )
return partPosition, AABSize
end
``````
It errors if the part isn’t axis aligned, i.e. if it’s at an angle other than 90 degrees. That check looks like so:
``````function isAxisAligned(cFrame)
--Checks if a CFrame is rotated *only* at right angles.
local worldCFrame = CFrame.new()
--How much is the CFrame's sides facing towards the world's sides?
local dotFront = cFrame.XVector:Dot( worldCFrame.XVector )
local dotRight = cFrame.YVector:Dot( worldCFrame.YVector )
--Given any combination 90-degree rotations, the angles between the CFrame's sides and the world's sides can only
-- 0, 90 or 180 degrees. That corresponds to dot products of 1, 0, or -1. Any non-whole-number dot products % 1 will
-- equal somthing other than 0, while 1%1 = 0, 0%1 = 0 and -1%0 = 0.
-- so checking if these two dot products % 1 == 0 is sufficient to checking of the CFrame's sides are rotated only
-- through 90-degree increments, i.e. if the CFrame is axis-aligned
return dotFront % 1 == 0 and dotRight % 1 == 0
end
``````
We can combine it all into a function that checks if one part is fully inside another:
``````function isPartInsidePart(part1, part2)
--Returns true if part1 is fully inside part2, assuming both are axis-aligned.
--Throws an error if part1 or part2 is not axis-aligned.
assert(isAxisAligned(part1.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
assert(isAxisAligned(part2.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
local AAB1Pos, AAB1Size = partToAAB(part1)
local AAB2Pos, AAB2Size = partToAAB(part1)
return isAABInsideAAB(aab1Pos, aab1Size, aab2Pos, aab2Size)
end
``````
Of course, all this doesn’t work if you also need to check for parts that are rotated at arbitrary angles.
6 Likes
Thanks!
That’s way easier, I tried to also make my own way to LIMIT the moving for my placement system;
``````function module:Update(HitP,Limits)
if (self.Placing and self.GrabbingObject) then
local X,Y,Z = HitP.X,HitP.Y,HitP.Z
X =math.clamp(X,self.Grid.Position.X - (self.Grid.Size.X/2 - self.GridPart.Size.X/2),self.Grid.Position.X + (self.Grid.Size.X/2) - self.GridPart.Size.X/2)
Z = math.clamp(Z,self.Grid.Position.Z - (self.Grid.Size.Z/2 - self.GridPart.Size.Z/2),self.Grid.Position.Z + (self.Grid.Size.Z/2) - self.GridPart.Size.Z/2)
Y = (self.Grid.Position.Y + self.Grid.Size.Y/2 + self.GridPart.Size.Y/2)
local cf = (CFrame.new(Vector3.new(X,Y,Z)) * CFrame.Angles(0,math.rad(self.Rotation),0))
self.GrabbingObject:SetPrimaryPartCFrame(cf)
end
end
``````
Kinda messy
It seems it worked
https://gyazo.com/aae8e4e7b430d75e16084a79488dbbb7
There were a few issues with the script such as;
1. inside `partToAAB()` you return `partPosition`, which isn’t a variable, but AABPosition is;
2. inside `isPartInsidePart()`
2.1 `AAB1Pos == AAB2Pos` and `AAB1Size == AAB2Size` as you used `partToAAB(part1)` twice
2.2 at the `return` line you write `aab1Pos`, `aab1Size`, `aab2Pos` and `aab2Size` all without capital letters
So here’s the fixed script:
``````function isAABInsideAAB(position1, size1, position2, size2)
--Checks if an AABB defined by position1 and size1 is fully within the AABB defined by position2 and size2
-- if they have the same size and position, returns true.
local p1, p2 = position1, position2
local s1, s2 = size1/2, size2/2
local minX1, minX2 = p1.X - s1.X, p2.X - s2.X
local maxX1, maxX2 = p1.X + s1.X, p2.X + s2.X
local minY1, minY2 = p1.Y - s1.Y, p2.Y - s2.Y
local maxY1, maxY2 = p1.Y + s1.Y, p2.Y + s2.Y
local minZ1, minZ2 = p1.Z - s1.Z, p2.Z - s2.Z
local maxZ1, maxZ2 = p1.Z + s1.Z, p2.Z + s2.Z
return
(minX1 >= minX2 and maxX1 <= maxX2) and
(minY1 >= minY2 and maxY1 <= maxY2) and
(minZ1 >= minZ2 and maxZ1 <= maxZ2)
end
function partToAAB(part)
--Given an axis-aligned part, returns the position and size of an AAB that takes up the same space as the part,
-- but whose X, Y and Z axes are the same as the world's axes.
assert(isAxisAligned(part.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
local AABPosition = part.Position
local AABSize = Vector3.new()
local worldCFrame = CFrame.new()
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(part.Size.X, 0, 0) )
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(0, part.Size.Y, 0) )
AABSize += part.CFrame:VectorToWorldSpace( Vector3.new(0, 0, part.Size.Z) )
return AABPosition, AABSize
end
function isAxisAligned(cFrame)
--Checks if a CFrame is rotated *only* at right angles.
local worldCFrame = CFrame.new()
--How much is the CFrame's sides facing towards the world's sides?
local dotFront = cFrame.XVector:Dot( worldCFrame.XVector )
local dotRight = cFrame.YVector:Dot( worldCFrame.YVector )
--Given any combination 90-degree rotations, the angles between the CFrame's sides and the world's sides can only
-- 0, 90 or 180 degrees. That corresponds to dot products of 1, 0, or -1. Any non-whole-number dot products % 1 will
-- equal somthing other than 0, while 1%1 = 0, 0%1 = 0 and -1%0 = 0.
-- so checking if these two dot products % 1 == 0 is sufficient to checking of the CFrame's sides are rotated only
-- through 90-degree increments, i.e. if the CFrame is axis-aligned
return dotFront % 1 == 0 and dotRight % 1 == 0
end
function isPartInsidePart(part1, part2)
--Returns true if part1 is fully inside part2, assuming both are axis-aligned.
--Throws an error if part1 or part2 is not axis-aligned.
assert(isAxisAligned(part1.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
assert(isAxisAligned(part2.CFrame), "Tried to convert a non-axis-aligned Part to an axis-alinged box.")
local AAB1Pos, AAB1Size = partToAAB(part1)
local AAB2Pos, AAB2Size = partToAAB(part2)
return isAABInsideAAB(AAB1Pos, AAB1Size, AAB2Pos, AAB2Size)
end
``````
Now it’s supposed to run fine, Thanks(RoBama) for helping him as well!
2 Likes
Hey, what If I need the part to not be axis aligned?
1 Like | 2,841 | 9,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-14 | latest | en | 0.76083 |
https://math.stackexchange.com/questions/3223086/equivalent-characterizations-of-smoothness | 1,653,013,700,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00397.warc.gz | 438,938,102 | 66,971 | # Equivalent Characterizations of Smoothness.
Definition. Let $$M$$ and $$N$$ be smooth manifolds, and let $$F\colon M\to N$$ be any map. We say that $$F$$ is a smooth map if for every $$p\in M$$, there exists smooth chart $$\big(U,\varphi\big)$$ containing $$p$$ and $$\big(V,\psi\big)$$ containing $$F(p)$$ such that $$F(U)\subseteq V$$ and the composite map $$\psi\circ F\circ \varphi^{-1}\colon \varphi(U)\to\psi(V)$$ is smooth.
I have to show the following proposition, but many points are not clear to me.
Proposition. Suppose $$M$$ and $$N$$ are smooth manifolds with or without boundary, and $$F\colon M\to N$$ is a map. Then $$F$$ is smooth if and only if either of the following condition is satisfied:
$$(a)$$ For every $$p\in M$$, there exist smooth charts $$(U,\varphi)$$ containing $$p$$ and $$(V,\psi)$$ containing $$F(p)$$ such that $$U\cap F^{-1}(V)$$ is open in $$M$$ and the composite map $$\psi\circ F\circ\varphi^{-1}$$ is smooth from $$\varphi(U\cap F^{-1}(V))$$ to $$\psi\big(V\big)$$.
$$(b)$$ $$F$$ is continuous and there exists amooth atlases $$\{(U_\alpha,\varphi_\alpha)\}$$ and $$\{(V_\beta,\psi_\beta)\}$$ for $$M$$ and $$N$$ respectevely such that for each $$\alpha$$ and $$\beta$$, $$\psi_\beta\circ F\circ \varphi_\alpha^{-1}$$ is a smooth map from $$\varphi_\alpha(U_\alpha\cap F^{-1}(V_\beta))$$ to $$\psi(V_\beta)$$.
Proof. I would like to proceed, hoping that it is the simplest way as follows: first prove that $$F$$ is smooth iff $$(a)$$ holds, and after showing that $$(a)$$ holds iff $$(b)$$ holds.
## F is smooth iff $$(a)$$ holds
($$\Rightarrow$$) If $$F$$ is smooth we have that $$F(U)\subseteq V$$, then $$U\subseteq F^{-1}(V)$$. Therefore $$U\cap F^{-1}(V)=U$$ that is open.
$$(\Rightarrow)$$ If $$(a)$$ holds for every $$p\in M$$, there exist smooth charts $$(U,\varphi)$$ containing $$p$$ and $$(V,\psi)$$ containing $$F(p)$$ such that $$U\cap F^{-1}(V)$$ is open in $$M$$ and the composite map $$\psi\circ F\circ\varphi^{-1}$$ is smooth from $$\varphi(U\cap F^{-1}(V))$$ to $$\psi\big(V\big)$$. It only remains to show that $$F(U)\subseteq V$$, now if $$\tilde{p}\in U$$ then there exists a smooth chart $$(\tilde{V},\tilde{\psi})$$ containing $$F(\tilde{p})$$
Question 1. How can I prove that $$F\big(\tilde{q}\big)\in V$$?
## (a)$$\iff$$(b)
$$(b)\Rightarrow (a)$$ it seems obvious.
Question 2. I have no idea how to show that $$(a)\Rightarrow (b)$$. Could you give me a hints?
Thanks!
• For Q1, you should first replace $U$ by $U\cap F^{-1}V$ and restricting $\varphi$, then everything follows. For Q2, prove smooth implies (b), which amounts to just patching the charts. May 12, 2019 at 10:25
Question 1: $$F(U)\subset V$$ is not always true. But you can take a new chart of the form $$(U^\prime,\varphi)$$ which has this property (take $$U^\prime=U\cap F^{-1}(V)$$).
Question 2: If you want to do it without "$$a\Leftrightarrow F \text{ smooth}$$", which I guess you do:
For $$p\in M$$ take $$(U_p,\varphi_p)$$ and $$(V_p,\psi_p)$$ some charts around $$p$$ and $$F(p)$$ respectively. If $$U_p^\prime=U_p\cap F^{-1}(V_p)$$ then $$\{ U_p^\prime\}_p$$ is an open cover for $$M$$ such that $$F_{\vert U_p^\prime}$$ is continuous, so $$F$$ is continuous.
Take the atlas $$\{(U_p^\prime,\varphi_p)\}_p$$ for $$M$$. Unfortunately $$\{(V_p,\psi_p)\}$$ may not be an atlas for $$N$$ because the domain of the charts may not cover $$N$$ is $$F$$ is not surjective. You can take the original atlas of $$N$$ instead, which we write $$\{(V_\alpha,\psi_\alpha)\}$$. Then for $$p$$ and $$\alpha$$ the map
$$\psi_\alpha\circ F\circ \varphi_p^{-1}:\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))\longrightarrow\psi(V_\alpha)$$
has domain $$\varphi_p(U_p^\prime\cap F^{-1}( V_\alpha))=\varphi_p(U_p\cap F^{-1}(V_p\cap V_\alpha))$$ so it is just the composition
$$\underbrace{\psi_\alpha\circ\psi_p^{-1}}_{\substack{\text{smooth because of the} \\ \text{coherence of the charts}}} \circ\underbrace{\psi_p\circ F\circ \varphi_p^{-1}}_{\substack{\text{smooth by assumption} }}.$$
• @AdamThanks for your ansewer. Perhaps it is more direct to prove that if $F$ is smooth then $(b)$ holds and viceversaa. What do you say? May 12, 2019 at 14:02
• @JackJ. Yes the best way might be to prove that "$F$ smooth" implies $(b)$ then $(b)\Rightarrow (a)$ and finally $(a)$ implies "$F$ smooth". But if you want to learn how to manipulate coordinate charts it's a good exercise to prove every directions! May 12, 2019 at 14:23
• @AdamI have another problem! If I start from $F$ smooth, then $(a)$ holds, because we have shown equivalence. From this point how do I get to $(b)$? May 12, 2019 at 14:33
• @JackJ. you mean you want to prove "$F$ smooth" implies $(b)$ ? May 12, 2019 at 14:37
• @JackJ. this is because $F_{\vert U_p^\prime}$ composed at the beginning and at the end by homeomorphisms (namely $\varphi_p$ and $\psi_p$) is continuous (by assumption it is even differentiable in the usual sense). (Ps: don't forget to accept the answer if it satisfies you) May 12, 2019 at 15:54 | 1,647 | 5,026 | {"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": 94, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-21 | latest | en | 0.797938 |
http://www.varsitytutors.com/common_core_high_school__geometry-help/sine-and-cosine-relationship-of-complementary-angles-ccss-math-content-hsg-srt-c-7 | 1,484,606,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279368.44/warc/CC-MAIN-20170116095119-00534-ip-10-171-10-70.ec2.internal.warc.gz | 752,668,591 | 26,566 | Common Core: High School - Geometry : Sine and Cosine Relationship of Complementary Angles: CCSS.Math.Content.HSG-SRT.C.7
Example Questions
← Previous 1
Example Question #1 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #2 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #3 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #4 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #5 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #6 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #7 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #8 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #9 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
Example Question #10 : Sine And Cosine Relationship Of Complementary Angles: Ccss.Math.Content.Hsg Srt.C.7
Simplify
Explanation:
The first step to simplifying is to remember an important trig identity.
If we rewrite it to look like the denominator, it is.
Now we can substitute this in the denominator.
Now write each term separately.
Remember the following identities.
Now simplify, and combine each term.
← Previous 1 | 935 | 4,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2017-04 | latest | en | 0.816676 |
https://www.coursehero.com/file/6472092/II-febb05/ | 1,498,339,141,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320338.89/warc/CC-MAIN-20170624203022-20170624223022-00272.warc.gz | 820,792,145 | 45,708 | # II-febb05 - ³ ³ D x-1 x-1 2 y 2 dxdy dove D = x y ∈ IR...
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VERIFICA SCRITTA DI Analisi matematica II 22 febbraio 2005 NOME. ............................................ COGNOME. ................................................ MATR. .................................................. GRUPPO. ........................................ 1) Determinare l’insieme di convergenza della seguente serie di funzioni + ± n =1 ( n + 1)!( x 2 - 3 x +1) n . Ris:. ...................................................... 2) Classi±care i punti stazionari della seguente funzione f ( x, y )= ² x 2 + y 2 1+ x 2 + y 2 . Ris:. ...................................................... 3) Calcolare l’integrale doppio
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Unformatted text preview: ³ ³ D x-1 ( x-1) 2 + y 2 dxdy dove D = { ( x, y ) ∈ IR 2 : ( x-1) 2 + y 2 ≥ 1 , ≤ y ≤ √ 3( x-1) , 1 ≤ x ≤ 2 } . Ris:. ...................................................... 4) Risolvere il problema di Cauchy ´ y ± = 2 y + e 2 x y (0) = 3 . Ris:. ...................................................... 5) Sia ω = (2 x 3-3 x 2 y + y 2 ) dx + (2 xy-x 3-5) dy e sia γ una curva regolare atratti di estremi (5 ,-1) e (1 , 2). Calcolare µ γ ω . Ris:. .........................................................
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## This note was uploaded on 10/13/2011 for the course MAT 05 taught by Professor Trombetti during the Spring '10 term at Università DI Napoli "Federico II""".
Ask a homework question - tutors are online | 474 | 1,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-26 | longest | en | 0.229667 |
https://convertoctopus.com/881-days-to-years | 1,713,822,220,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818374.84/warc/CC-MAIN-20240422211055-20240423001055-00618.warc.gz | 158,394,673 | 7,855 | ## Conversion formula
The conversion factor from days to years is 0.0027379070069885, which means that 1 day is equal to 0.0027379070069885 years:
1 d = 0.0027379070069885 yr
To convert 881 days into years we have to multiply 881 by the conversion factor in order to get the time amount from days to years. We can also form a simple proportion to calculate the result:
1 d → 0.0027379070069885 yr
881 d → T(yr)
Solve the above proportion to obtain the time T in years:
T(yr) = 881 d × 0.0027379070069885 yr
T(yr) = 2.4120960731569 yr
The final result is:
881 d → 2.4120960731569 yr
We conclude that 881 days is equivalent to 2.4120960731569 years:
881 days = 2.4120960731569 years
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 year is equal to 0.41457718501703 × 881 days.
Another way is saying that 881 days is equal to 1 ÷ 0.41457718501703 years.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that eight hundred eighty-one days is approximately two point four one two years:
881 d ≅ 2.412 yr
An alternative is also that one year is approximately zero point four one five times eight hundred eighty-one days.
## Conversion table
### days to years chart
For quick reference purposes, below is the conversion table you can use to convert from days to years
days (d) years (yr)
882 days 2.415 years
883 days 2.418 years
884 days 2.42 years
885 days 2.423 years
886 days 2.426 years
887 days 2.429 years
888 days 2.431 years
889 days 2.434 years
890 days 2.437 years
891 days 2.439 years | 468 | 1,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.795309 |
https://www.doubtnut.com/qna/376770512 | 1,721,750,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518058.23/warc/CC-MAIN-20240723133408-20240723163408-00761.warc.gz | 631,507,219 | 33,389 | # In an experiment on surface tension, water rises up to a height of 0.1 m in a capillary tube. If the same experiment is performed in a satellite moving around the earth, what will be the rise in the capillary tube?
Text Solution
Verified by Experts
## The weight of the water column in the capillary tube will be zero in an orbiting satellite, Hence, due to surface tension, water will rise up to the top of the tube, and the capillary tube will be completely filled with water.
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## In a surface tension experiment , with a capillary tube, water rises upto a height of 5 cm . In th capillary tube . If the same experiment is repeated in an artificial satellite , revolving round the earth , then the water will rise in the capillary tube upto a height equal to
A3 cm
B6 cm
C9.8cm
Dfull length of the capillary tube
• Question 2 - Select One
## In a surface tension experiment with a capillary tube water rises upto 0.1m. If the same experiment is repeated in an artificial satellite, which is revolving around the earth , water will rise in the capillary tube upto a height of :
A0.1m
B0.2m
C0.98m
Dfull length of tube
• Question 3 - Select One
## In a surface tension experiment with a capillary tube water rises upto 0.1m. If the same experiment is repeated in an artificial satellite, which is revolving around the earth, water will rise in the capillary tube upto a height of
A0.1m
B0.2m
C0.98m
DFull length of tube
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robimyzapasy.pl | 1,653,053,072,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662532032.9/warc/CC-MAIN-20220520124557-20220520154557-00522.warc.gz | 561,182,052 | 8,004 | how to calculate the aggregate in pcc Search
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# how to calculate the aggregate in pcc Search:
##### PCC Volume Calculator Free Estimate of Cement, Sand, Jelly
PCC Volume Calculator. Plain Cement Concrete (PCC) is composed mixture of cement, aggregates, sand, water and chemical admixtures. Calculate portland cement concrete or PCC concrete mix design ratio in Cft and determine exactly how much quantity of cement, sand and aggregate is required for PCC
##### PCC Concrete Ratio Plain Cement Concrete Calculate ...
Where 1 signifies the ratio of cement, 2 signify the ratio of sand (fine aggregate) and 4 for coarse aggregate (crushed stone). Dry volume = wet volume x 1.54. Determine crushed stone, cement and sand in PCC work. Concrete ratio is 1:2:4. Material is equal to material ratio/ratio sum x dry volume. Cement is equal to 1/1 + 2 + 4 x wet volume x 1.54
##### PCC Concrete Calculation PCC Concrete Ratio Plain ...
Plain Cement Concrete (PCC) refers to the construction material that is usually applied as the binding materials. It is formed with cement, (generally Portland Cement) and other cementitious materials like fly ash and slag cement, aggregate (normally a coarse aggregate) created with gravels or crushed rocks like limestone or granite, along with a fine aggregate like sand as well as water, and ...
##### How To Calculate Cement Sand Aggregate For PCC Material ...
May 17, 2020 why we provide pcc, purpose of pcc👉https://youtu.be/XVh_IaWdAjcLevel Marking for PCC👉https://youtu.be/ZA4JpO1BJpwGrade of Concrete and WC Ratio👉https://yo...
##### Cement Concrete Calculator PCC Calculator RCC ...
Strength of PCC is defined as compressive strength after 28 days, expressed as M15, M20, where M stands for Mix and 15 stands for 15 N/mm 2 (n/mm 2 must be read as 'Newton’s per millimeter Cubic') compressive strength at 28 days. The proportions of materials (cement, sand, coarse aggregate) for nominal mix/design mix concrete that are normally used are 1:3:6 or 1:4:8.
##### How to calculate quantity of cement in PCC - Civil Sir
How to calculate quantity of cement in PCC, cement consumption in PCC 1:2:4, in this topic we know about how to calculate quantity of cement in PCC and cement consumption in PCC 1:2:4. we know that we have several grade of concrete like M5, M7.5, M10, M15, M20 and M25 and so more—. PCC concrete:-Plain Cement Concrete abbreviated as PCC, a mixture of cement, fine aggregate and
##### Concrete PCC Work-Estimating quantity of Cement, Sand ...
Aug 03, 2016 In this tutorial, you will learn the process of Estimating quantity of Cement, Sand Aggregate in concrete in Microsoft Excel in detail.I have made the step...
##### PCC Concrete Calculation PCC Concrete Ratio Plain ...
Plain Cement Concrete (PCC) refers to the construction material that is usually applied as the binding materials. It is formed with cement, (generally Portland Cement) and other cementitious materials like fly ash and slag cement, aggregate (normally a coarse aggregate) created with gravels or crushed rocks like limestone or granite, along with a fine aggregate like sand as
##### Concrete PCC Work-Estimating quantity of Cement, Sand ...
Aug 03, 2016 In this tutorial, you will learn the process of Estimating quantity of Cement, Sand Aggregate in concrete in Microsoft Excel in detail.I have made the step...
##### PCC Calculator For Concreting Work Ernakulam Kerala ...
Sep 21, 2017 PCC Calculator, Building Materials Cost Kerala. Civil Building Contractor, Building Designing And Construction, Plan Approvals, Landscaping Works, PCC Calculator can be used to calculate the required quantities of basic building materials like Cement, sand and aggregates, needed for plain cement concreting.
##### How to calculate the quantity of cement in PCC - Quora
Answer (1 of 4): Suppose the ratio of the PCC is 1:5:10, then for making 10 cu.m of such PCC you require, hard broken stone 9 cu.m; 4.5 cu m sand; 4.5/5= 0.9 cu m of cement one cu.m of cement= 1440 kg you may calculate weight of cement based on this. Is it OK ?
##### cement consumption in PCC 1:3:6 and M10 - Civil Sir
cement consumption in PCC 1:3:6, How to calculate quantity of cement in PCC, in this topic we know about how to calculate quantity of cement in PCC and cement consumption in PCC 1:3:6.we know that we have several grade of concrete like M5, M7.5, M10, M15, M20 and M25 and so more- and water cement ratio for M10 grade of concrete. Generally PCC is stand for
##### Rate analysis for PCC 1:4:8 (M7.5) - calculate quantity ...
Rate Analysis for PCC 1:4:8 :-It is a summary of all the cost involved in doing particular work or unit work like material cost, labour cost, overhead expenses, water charges contractor profit.For rate analysis, the details about all the operation involved in carrying out the work should be available. The quantities of materials cement sand aggregate and water are required and their
##### How To Calculate Cement, Sand, Aggregate Quantity In 1 ...
Aug 03, 2018 Density of Aggregate is 1500/m 3 Calculation for KG = 0.84 x 1500 = 1260 kg. As we know that 1 m3 = 35.31 CFT. Calculation for CFT = 0.84 x 35.31 = 29.66 Cubic Feet. Also you can Download Excel Sheet for Calculate Concrete Volume (Cement, Sand and Aggregate ) Quantity. Excel Sheel For Concrete Quantity Calculation
##### How much cement sand aggregate required for M15 concrete ...
step 12 :- calculate aggregate required for 1 cubic metre of M15 concrete in cft (cubic feet) is equal to 4/7 × 1.54m3/35.3147 × 1550 Kg/m3 = 39 cft. Ans. :- 39 cft quantity of aggregate is required for 1 cubic metre of M15 concrete. Now question is how much aggregate required for 1m3 of M15 concrete their answer is following:-
##### Calculate Quantities of Materials for Concrete -Cement ...
Calculating Quantities of Materials for per cubic meter or cubic feet or cubic yards concrete. Consider concrete with mix proportion of 1:1.5:3 where, 1 is part of cement, 1.5 is part of fine aggregates and 3 is part of coarse aggregates of maximum size of 20mm. The water cement ratio required for mixing of concrete is taken as 0.45.
##### How much cement sand aggregate required for M10 concrete ...
How much aggregate required for M10 concrete in cft step 12 :- calculate aggregate required for 1 cubic metre of M10 concrete in cft (cubic feet) is equal to 6/10 × 1.54m3/35.3147 × 1550 Kg/m3 = 40.5 cft. Ans. :- 40.5 cft quantity of aggregate is required for
##### Calculate Cement Sand Aggregate - M20, M15, M10, M5 ...
Proper calculation and relative proportioning of materials are important to produce a good quality concrete. Let’s see some simple techniques used by engineers to calculate cement, sand, coarse aggregate (gravel or Jalli) and water needed to produce different grades of concrete, like M5, M7.5, M10, M15 and M20.
##### fineness modulus calculations FINENESS MODULUS OF FINE ...
Jul 31, 2016 Therefore, fineness modulus of aggregate = (cumulative % retained) / 100 = (275/100) = 2.75 Fineness modulus of fine aggregate is 2.75. It means the average value of aggregate is in between the 2 nd sieve and 3 rd sieve. It means the average aggregate size is in between 0.3mm to 0.6mm as shown in below figure.
##### bonus checks are taxed at what percentage Bonus Tax Rate ...
Nov 07, 2019 Calculating your actual bonus tax rate in a typical tax year isn’t that hard. Your bonus is taxed at the same rate as all of your other income. If you’re in the 33% tax bracket and you receive a bonus of \$100,000, you will pay \$33,000 in federal taxes.
##### calculate mean in r studio Mean function in R: Mean ...
Mean function in R -mean calculates the arithmetic mean. mean function calculates arithmetic mean of vector with NA values and arithmetic mean of column in data frame. mean of a group can also calculated using mean function in R by providing it inside the aggregate function. with mean function we can also perform row wise mean using dplyr ...
##### PCC Concrete Calculation PCC Concrete Ratio Plain ...
Plain Cement Concrete (PCC) refers to the construction material that is usually applied as the binding materials. It is formed with cement, (generally Portland Cement) and other cementitious materials like fly ash and slag cement, aggregate (normally a coarse aggregate) created with gravels or crushed rocks like limestone or granite, along with a fine aggregate like sand as
##### HOW TO CALCULATE THE QUANTITY OF CEMENT, SNAD AND ...
HOW TO CALCULATE THE QUANTITY OF CEMENT, SNAD AND AGGREGATE REQUIRED FOR PCC (PLAIN CEMENT CONCRETE) WITH FREE EXCEL SHEET, WITH UNIT CONVERSIONFirst of all the aim of this article is to calculate the quantity of materials required to Produce PCC (plain cement concrete), of certain volume. Along with this article I will provide
##### Density of Cement Sand and Aggregate Cement Density ...
Density of PCC. Density of PCC is 2400 kg/m 3 (2.400 g/cm 3 or 24 kN/m 3). Density of Fine Aggregate. Density of sand (fine aggregate) is ranging between 1450 – 2082 kg/m 3 depending on different condition like wet, dry, loose, dry-packed, and wet packed. Density of Coarse Aggregate
##### How to Calculate Cement, Sand and Aggregate required for 1 ...
So care should be taken while calculating the amount of Cement, Sand and Aggregate required for 1 Cubic meter of Concrete. Method-1: DLBD method to determine material requirement for 1 Cum concrete The DLBD (Dry Loose Bulk Densities) method is an accurate method to calculate cement, sand and aggregate for a given nominal mix concrete.
##### Concrete Calculator - Estimate Cement, Sand, Gravel ...
Example calculation Estimate the quantity of cement, sand and stone aggregate required for 1 cubic meter of 1:2:4 concrete mix. Ans. Materials required are 7 nos. of 50 kg bag of cement, 0.42 m 3 of sand and 0.83 m 3 of stone aggregate.
##### How to calculate Cement, Sand and Aggregate in a Single ...
Visit civil engineering websitehttps://civilconcept/In this video, I have shown,How to calculate Cement, Sand and Aggregate in a Single RCC ColumnBes...
##### How To Calculate Cement, Sand Aggregate Quantity in ...
Sep 13, 2018 Hello Friends,In this video we will learn how to make own calculator for calculating Cement, Sand Aggregate quantity for different type of concrete Mix, wi...
##### How to calculate rate analysis of M10 grade concrete
Sep 17, 2017 How to calculate rate analysis of M10 grade concrete? -Do- as per item No.2.03 but for providing and laying plain cement concrete (PCC) 1:3:6 with 1 part of cement: 3 parts of coarse sand: 6 parts of stone aggregate of size 37 mm and downgraded complete as directed. Rate of the material will be based on the particular region.
##### How to Calculate Cement, Sand and Coarse Aggregate ...
Coarse Aggregate = (Coarse Aggregate Part / Concrete Parts ) * Concrete Volume = (3/5.5) * 2 = 1.09 m 3 . Water Cement Ratio. According to IS 10262 (2009), Assuming Water-Cement Ratio for the Concrete as 0.45. Required Amount of Water = W/C Ratio X Cement Volume
##### How we calculate the volume of footing from pcc level? - Quora
Volume of concrete on top of PCC is calculated by general mathematical formulas. Footings are generally rectangular or square in plan. If the footing is a cuboid then volume = Length X breadth X height. If there are additional steps calculate each...
##### How to Calculate Quantities of Cement, Sand and Aggregate ...
Step-1: Calculate Volume of Materials required. Density of Cement = 1440 kg/cum (Approx) Volume of 1 Kg of Cement = 1/1440 = 0.000694 cum. Volume of 01 bag (50 kg) of cement = 50 X 0.000694 = 0.035 cubic meter (cum) Since we know the ratio of cement to sand (1:2) and cement to aggregate (1:4)
##### How to calculate cement aggregate and sand in cft by ...
Answer (1 of 9): . procedure for estimating material quantity in 1 cubic meter of concrete let us consider a design mix of M15 ie. 1 : 2 : 4 ratio sum = 1+2+4 = 7 volume of concrete = 1 cubic meter(wet volume) therefore dry volume of concrete= 45% to
##### Density Of Cement, Sand, And Aggregate - 9 To 5 Civil
Jun 17, 2021 Density of sand (fine aggregate) is ranging between 1450 – 2082 kg/m 3 depending on different condition like wet, dry, loose, dry-packed, and wet packed. Bulk Density of Aggregate. The bulk density or unit weight of an aggregate is the mass or weight of the aggregate that required to fill a container of a specified unit volume.
##### fineness modulus calculations FINENESS MODULUS OF FINE ...
Jul 31, 2016 Therefore, fineness modulus of aggregate = (cumulative % retained) / 100 = (275/100) = 2.75 Fineness modulus of fine aggregate is 2.75. It means the average value of aggregate is in between the 2 nd sieve and 3 rd sieve. It means the average aggregate size is in between 0.3mm to 0.6mm as shown in below figure. | 3,178 | 12,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.870799 |
https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Elementary_Differential_Equations_with_Boundary_Values_Problems_(Trench)/9%3A_Linear_Higher_Order_Differential_Equations/9.3%3A_Undetermined_Coefficients_for_Higher_Order_Equations/9.3E%3A_Undetermined_Coefficients_for_Higher_Order_Equations_(Exercises) | 1,571,682,390,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987781397.63/warc/CC-MAIN-20191021171509-20191021195009-00314.warc.gz | 576,454,132 | 20,418 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 9.3E: Undetermined Coefficients for Higher Order Equations (Exercises)
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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
In Exercises [exer:9.3.1}– [exer:9.3.59} find a particular solution.
[exer:9.3.1] $$y'''-6y''+11y'-6y=-e^{-x}(4+76x-24x^2)$$
[exer:9.3.2] $$y'''-2y''-5y'+6y=e^{-3x}(32-23x+6x^2)$$
[exer:9.3.3] $$4y'''+8y''-y'-2y=-e^x(4+45x+9x^2)$$
[exer:9.3.4] $$y'''+3y''-y'-3y=e^{-2x}(2-17x+3x^2)$$
[exer:9.3.5] $$y'''+3y''-y'-3y=e^x(-1+2x+24x^2+16x^3)$$
[exer:9.3.6] $$y'''+y''-2y=e^x(14+34x+15x^2)$$
[exer:9.3.7] $$4y'''+8y''-y'-2y=-e^{-2x}(1-15x)$$
[exer:9.3.8] $$y'''-y''-y'+y=e^x(7+6x)$$
[exer:9.3.9] $$2y'''-7y''+4y'+4y=e^{2x}(17+30x)$$
[exer:9.3.10] $$y'''-5y''+3y'+9y=2e^{3x}(11-24x^2)$$
[exer:9.3.11] $$y'''-7y''+8y'+16y=2e^{4x}(13+15x)$$
[exer:9.3.12] $$8y'''-12y''+6y'-y=e^{x/2}(1+4x)$$
[exer:9.3.13] $$y^{(4)}+3y'''-3y''-7y'+6y=-e^{-x}(12+8x-8x^2)$$
[exer:9.3.14] $$y^{(4)}+3y'''+y''-3y'-2y=-3e^{2x}(11+12x)$$
[exer:9.3.15] $$y^{(4)}+8y'''+24y''+32y'=-16e^{-2x}(1+x+x^2-x^3)$$
[exer:9.3.16] $$4y^{(4)}-11y''-9y'-2y=-e^x(1-6x)$$
[exer:9.3.17] $$y^{(4)}-2y'''+3y'-y=e^x(3+4x+x^2)$$
[exer:9.3.18] $$y^{(4)}-4y'''+6y''-4y'+2y=e^{2x}(24+x+x^4)$$
[exer:9.3.19] $$2y^{(4)}+5y'''-5y'-2y=18e^x(5+2x)$$
[exer:9.3.20] $$y^{(4)}+y'''-2y''-6y'-4y=-e^{2x}(4+28x+15x^2)$$
[exer:9.3.21] $$2y^{(4)}+y'''-2y'-y=3e^{-x/2}(1-6x)$$
[exer:9.3.22] $$y^{(4)}-5y''+4y=e^x(3+x-3x^2)$$
[exer:9.3.23] $$y^{(4)}-2y'''-3y''+4y'+4y=e^{2x}(13+33x+18x^2)$$
[exer:9.3.24] $$y^{(4)}-3y'''+4y'=e^{2x}(15+26x+12x^2)$$
[exer:9.3.25] $$y^{(4)}-2y'''+2y'-y=e^x(1+x)$$
[exer:9.3.26] $$2y^{(4)}-5y'''+3y''+y'-y=e^x(11+12x)$$
[exer:9.3.27] $$y^{(4)}+3y'''+3y''+y'=e^{-x}(5-24x+10x^2)$$
[exer:9.3.28] $$y^{(4)}-7y'''+18y''-20y'+8y=e^{2x}(3-8x-5x^2)$$
[exer:9.3.29] $$y'''-y''-4y'+4y=e^{-x}\left[(16+10x)\cos x+(30-10x)\sin x\right]$$
[exer:9.3.30] $$y'''+y''-4y'-4y=e^{-x}\left[(1-22x)\cos 2x-(1+6x)\sin2x\right]$$
[exer:9.3.31] $$y'''-y''+2y'-2y=e^{2x}[(27+5x-x^2)\cos x+(2+13x+9x^2)\sin x]$$
[exer:9.3.32] $$y'''-2y''+y'-2y=-e^x[(9-5x+4x^2)\cos 2x-(6-5x-3x^2)\sin2x]$$
[exer:9.3.33] $$y'''+3y''+4y'+12y=8\cos2x-16\sin2x$$
[exer:9.3.34] $$y'''-y''+2y=e^x[(20+4x)\cos x-(12+12x)\sin x]$$
[exer:9.3.35] $$y'''-7y''+20y'-24y=-e^{2x}[(13-8x)\cos 2x-(8-4x)\sin2x]$$
[exer:9.3.36] $$y'''-6y''+18y'=-e^{3x}[(2-3x)\cos 3x-(3+3x)\sin3x]$$
[exer:9.3.37] $$y^{(4)}+2y'''-2y''-8y'-8y=e^x(8\cos x+16\sin x)$$
[exer:9.3.38] $$y^{(4)}-3y'''+2y''+2y'-4y=e^x(2\cos2x -\sin2x)$$
[exer:9.3.39] $$y^{(4)}-8y'''+24y''-32y'+15y=e^{2x}(15x\cos2x+32\sin2x)$$
[exer:9.3.40] $$y^{(4)}+6y'''+13y''+12y'+4y=e^{-x}[(4-x)\cos x-(5+x)\sin x]$$
[exer:9.3.41] $$y^{(4)}+3y'''+2y''-2y'-4y=-e^{-x} (\cos x-\sin x)$$
[exer:9.3.42] $$y^{(4)}-5y'''+13y''-19y'+10y=e^x (\cos2x+\sin2x)$$
[exer:9.3.43] $$y^{(4)}+8y'''+32y''+64y'+39y=e^{-2x}[(4-15x)\cos3x-(4+15x)\sin 3x]$$
[exer:9.3.44] $$y^{(4)}-5y'''+13y''-19y'+10y=e^x[(7+8x)\cos 2x+(8-4x)\sin2x]$$
[exer:9.3.45] $$y^{(4)}+4y'''+8y''+8y'+4y=-2e^{-x} (\cos x-2\sin x)$$
[exer:9.3.46] $$y^{(4)}-8y'''+32y''-64y'+64y=e^{2x} (\cos2x-\sin2x)$$
[exer:9.3.47] $$y^{(4)}-8y'''+26y''-40y'+25y=e^{2x}[3\cos x-(1+3x)\sin x]$$
[exer:9.3.48] $$y'''-4y''+5y'-2y=e^{2x}-4e^x-2\cos x+4\sin x$$
[exer:9.3.49] $$y'''-y''+y'-y=5e^{2x}+2e^x-4\cos x+4\sin x$$
[exer:9.3.50] $$y'''-y'=-2(1+x)+4e^x-6e^{-x}+96e^{3x}$$
[exer:9.3.51] $$y'''-4y''+9y'-10y=10e^{2x}+20e^x\sin2x-10$$
[exer:9.3.52] $$y'''+3y''+3y'+y=12e^{-x}+9\cos2x-13\sin2x$$
[exer:9.3.53] $$y'''+y''-y'-y=4e^{-x}(1-6x)-2x\cos x+2(1+x)\sin x$$
[exer:9.3.54] $$y^{(4)}-5y''+4y=-12e^x+6e^{-x}+10\cos x$$
[exer:9.3.55] $$y^{(4)}-4y'''+11y''-14y'+10y=-e^x(\sin x+2\cos2x)$$
[exer:9.3.56] $$y^{(4)}+2y'''-3y''-4y'+4y=2e^x(1+x)+e^{-2x}$$
[exer:9.3.57] $$y^{(4)}+4y=\sinh x\cos x-\cosh x\sin x$$
[exer:9.3.58] $$y^{(4)}+5y'''+9y''+7y'+2y=e^{-x}(30+24x)-e^{-2x}$$
[exer:9.3.59] $$y^{(4)}-4y'''+7y''-6y'+2y=e^x(12x-2\cos x+2\sin x)$$
[exer:9.3.60] $$y'''-y''-y'+y=e^{2x}(10+3x)$$
[exer:9.3.61] $$y'''+y''-2y=-e^{3x}(9+67x+17x^2)$$
[exer:9.3.62] $$y'''-6y''+11y'-6y=e^{2x}(5-4x-3x^2)$$
[exer:9.3.63] $$y'''+2y''+y'=-2e^{-x}(7-18x+6x^2)$$
[exer:9.3.64] $$y'''-3y''+3y'-y=e^x(1+x)$$
[exer:9.3.65] $$y^{(4)}-2y''+y=-e^{-x}(4-9x+3x^2)$$
[exer:9.3.66] $$y'''+2y''-y'-2y=e^{-2x}\left[(23-2x)\cos x+(8-9x)\sin x\right]$$
[exer:9.3.67] $$y^{(4)}-3y'''+4y''-2y'=e^x\left[(28+6x)\cos 2x+(11-12x)\sin2x\right]$$
[exer:9.3.68] $$y^{(4)}-4y'''+14y''-20y'+25y=e^x\left[(2+6x)\cos 2x+3\sin2x\right]$$
[exer:9.3.69] $$y'''-2y''-5y'+6y=2e^x(1-6x),\quad y(0)=2, \quad y'(0)=7,\quad y''(0)=9$$
[exer:9.3.70] $$y'''-y''-y'+y=-e^{-x}(4-8x),\quad y(0)=2, \quad y'(0)=0,\quad y''(0)=0$$
[exer:9.3.71] $$4y'''-3y'-y=e^{-x/2}(2-3x),\quad y(0)=-1, \quad y'(0)=15,\quad y''(0)=-17$$
[exer:9.3.72] $$y^{(4)}+2y'''+2y''+2y'+y=e^{-x}(20-12x),\, y(0)=3,\; y'(0)=-4,\; y''(0)=7,\; y'''(0)=-22$$
[exer:9.3.73] $$y'''+2y''+y'+2y=30\cos x-10\sin x, \quad y(0)=3,\quad y'(0)=-4,\quad y''(0)=16$$
[exer:9.3.74] $$y^{(4)}-3y'''+5y''-2y'=-2e^x(\cos x-\sin x),\; y(0)=2,\; y'(0)=0,\; y''(0)~=~-1, \; y'''(0)=-5$$
[exer:9.3.75] Prove: A function $$y$$ is a solution of the constant coefficient nonhomogeneous equation
$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=e^{\alpha x}G(x) \tag{A}$
if and only if $$y=ue^{\alpha x}$$, where $$u$$ satisfies the differential equation
$a_0u^{(n)}+{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ {p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u=G(x) \tag{B}$
and
$p(r)=a_0r^n+a_1r^{n-1} + \cdots + a_n\nonumber$
is the characteristic polynomial of the complementary equation
$a_0y^{(n)}+a_1y^{(n-1)}+\cdots+a_ny=0.\nonumber$
[exer:9.3.76] Prove:
The equation
$\begin{array}{lcl} a_0u^{(n)}&+&\{p^{(n-1)}(\alpha)\over(n-1)!}u^{(n-1)}+ \{p^{(n-2)}(\alpha)\over(n-2)!}u^{(n-2)}+\cdots+p(\alpha)u\\ &=&\left(p_0+p_1x+\cdots+p_kx^k\right)\cos \omega x\\&&\,+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x \end{array} \tag{A}$
has a particular solution of the form
$u_p=x^m\left(u_0+u_1x+\cdots+u_kx^k\right)\cos\omega x+ \left(v_0+v_1x+\cdots+v_kx^k\right)\sin\omega x.\nonumber$
If $$\lambda+i\omega$$ is a zero of $$p$$ with multiplicity $$m\ge1$$, then (A) can be written as
$a(u''+\omega^2 u)= \left(p_0+p_1x+\cdots+p_kx^k\right)\cos\omega x+ \left(q_0+q_1x+\cdots+q_kx^k\right)\sin\omega x,\nonumber$
which has a particular solution of the form
$u_p=U(x)\cos\omega x+V(x)\sin\omega x,\nonumber$
where
$U(x)=u_0x+u_1x^2+\cdots+u_kx^{k+1},\,V(x)=v_0x+v_1x^2+\cdots+v_kx^{k+1}\nonumber$
and
$\begin{array}{rcl} a(U''(x)+2\omega V'(x))&=&p_0+p_1x+\cdots+p_kx^k\\[10pt] a(V''(x)-2\omega U'(x))&=&q_0+q_1x+\cdots+q_kx^k. \end{array}\nonumber$ | 4,342 | 7,170 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2019-43 | latest | en | 0.231269 |
https://cracku.in/41-if-15-men-24-women-and-36-boys-can-do-a-piece-of-w-x-mat-2013 | 1,719,289,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00253.warc.gz | 161,854,781 | 25,004 | Question 41
# If 15 men, 24 women and 36 boys can do a piece of work in 12 days, working 8h per day, how many men must be associated with 12 women and 6 boys to do another piece of work $$2 \frac{1}{4}$$times as large in 30 days working 6 h per day?
Solution
If M1 men can do W1 work in D1 days working H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day (where all men work at the same rate), then
M1 D1 H1 / W1 = M2 D2 H2 / W2 .............(1)
So,given 15M=24W=36B...................(2)
where M,W,B are work done by one man, one woman and one boy respectively in one day.
Also given W2=9/4W1 ....(3)
for W2 let x Men(M) are required.
So total Human-power(for W2) = xM + 12W + 6B
now using relations from (2) Human-power(for W2) = xM + 15/2M + 15/6M = (10+x)M
So replacing all the values in eqn (1) and (3), we getÂ
=> 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/W2
=> 15M*12days*8Hrs/W1 = (10+x)M*30days*6Hrs/(9/4W1)
=> x = 8 Men | 360 | 966 | {"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} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.847483 |
http://www.frsportsnews.com/nfl/the-most-neglected-fact-about-discrete-mathematics-and-its-applications-revealed/ | 1,569,192,986,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575751.84/warc/CC-MAIN-20190922221623-20190923003623-00223.warc.gz | 249,846,310 | 19,017 | # The Most Neglected Fact About Discrete Mathematics and Its Applications Revealed
## New Ideas Into Discrete Mathematics and Its Applications Never Before Revealed
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Students have the ability to work in a context in which failure is totally normal. The process for translation is referred to as mathematical modelling. Nonetheless, later on, it’s guaranteed to create fruitful outcomes.
An individual may select the questions freely. It’s farely fun subject if you understand how to program. The program will provide the student an chance to study a fascinating group of ideas and it is going to also give the student with very marketable skills.
## What the In-Crowd Won’t Tell You About Discrete Mathematics and Its Applications
So, it’s essential to supply your eyes rest for a little while by taking rests after specific time intervals. The large part of the graphs we’re likely to be dealing with are somewhat more complex. You may try to work out this problem for practice.
Hence we have the next notion. To put it differently, the truth of the statements to be proved needs to be established assuming the truth of some other statements. Take it from me, it is merely an issue of practice.
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The authors declare they have no conflicts of interest. This book was designed to fulfill the requirements of nearly all types of introductory discrete mathematics courses. Discrete mathematics can play a crucial part inside this connection.
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## Lies You’ve Been Told About Discrete Mathematics and Its Applications
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Ensure you choose the suitable side lengths for each rectangle. A multigraph is a generalization that enables multiple edges adjacent to the identical pair of vertices. The graph with only a single vertex and no edges is referred to as the trivial graph. | 969 | 5,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.93276 |
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### Transportatopn problm
1. 1. Transportation Problem & Degeneracy Presented By : Abhishek Upadhyay , Akhil Bhadrecha , Anshul K. Singh , Avijit Das , Kesharpu Sekhar & Saurabh Tamrakar
2. 2. Transportation Problem Aim: To find out optimum transportation schedule keeping in mind cost of transportation to be minimized.
3. 3. Transportation Problem <ul><li>The transportation problem is a special type of LPP where the objective is to minimize the cost of distributing a product from a number of sources or origins to a number of destinations. </li></ul><ul><li>Because of its special structure the usual simplex method is not suitable for solving transportation problems. These problems require special method of solution. </li></ul>
4. 4. <ul><li>The origin of a transportation problem is the location from which shipments are dispatched. The destination of a transportation problem is the location to which shipments are transported. </li></ul><ul><li>The unit transportation cost is the cost of transporting one unit of the consignment from an origin to a destination. </li></ul>
5. 5. Methods of solving Transportation Problem <ul><li>North West Corner Method (NWCM) </li></ul><ul><li>Least Cost Method (LCM) </li></ul><ul><li>VOGEL’S APPROXIMATION METHOD (VAM) </li></ul>
6. 6. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step I – Check whether the given problem is a balanced problem (i.e. Requirement = Capacity). In case not then add dummy row (origin) or column (destination) with zero cost to make an unbalanced transportation problem balanced. </li></ul><ul><li>Step II – Reproduce the squares and inset squares and in the inset squares copy down the problem. </li></ul>
7. 7. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step III – Find a solution using VAM method </li></ul><ul><ul><li>Calculate penalties for each row and each column. </li></ul></ul><ul><ul><li>By “penalties” we mean the difference between two best possibilities. </li></ul></ul><ul><ul><li>Give priority to the largest penalty. It belongs to a row or a column. </li></ul></ul><ul><ul><li>In that row or column make allocation to the smallest cost cell. </li></ul></ul><ul><ul><li>Cut off that row or column which is exhausted. </li></ul></ul><ul><ul><li>Continue in the same way with the rest of the table until two squares are left. </li></ul></ul><ul><ul><li>Then fill up the smallest cost square out of the remaining two. </li></ul></ul><ul><ul><li>Finally fill up the last square. </li></ul></ul>
8. 8. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step IV – Check for Basic </li></ul><ul><ul><li>Initial Basic Solution = m + (n – 1) = Number of allocation </li></ul></ul><ul><ul><li>where, m is Row and n is Column </li></ul></ul><ul><ul><li>If Basic then go to next step. </li></ul></ul><ul><ul><li>Else - Covert first the Non-Basic into Basic by making an artificial allocation with zero units in the smallest cost cell. </li></ul></ul>
9. 9. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step V – MODI Check Procedure </li></ul><ul><ul><li>Copy down the VAM solution along with cost matrix. </li></ul></ul><ul><ul><li>Put a ‘0’ (zero) on the top right corner in the first row. </li></ul></ul><ul><ul><li>For each filled in square adjust in such a way that: Row Number + Column Number = Cost </li></ul></ul><ul><ul><li>After filling all the row and column number look at the unused routes (open cells) </li></ul></ul><ul><ul><li>For each such cell find: Row Number + Column Number – Cost. This is called check number for that cell. </li></ul></ul><ul><ul><li>Write this number in that cell and encircle it. </li></ul></ul><ul><ul><li>If all check number are smaller than or equal to zero (<=0 ) then that solution is the best solution (i.e. solution obtained is an initial basic feasible solution). </li></ul></ul><ul><ul><li>In case any check number is positive then revision is required. Hence go to next step. </li></ul></ul>
10. 10. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step VI – Procedure for Revision </li></ul><ul><ul><li>To revise the transportation table a loop is to be drawn and loop is made up of horizontal and vertical lines travelling through the cost matrix by starting and ending at the same point with the following conditions. </li></ul></ul><ul><ul><ul><li>Condition 1 – The starting point should be an open square / cell. </li></ul></ul></ul><ul><ul><ul><li>Condition 2 – All the other squares travelled should be closed squares or allocated squares. </li></ul></ul></ul><ul><ul><ul><li>Condition 3 – In horizontal or vertical travel exactly two squares can be travelled including the starting cell. </li></ul></ul></ul><ul><ul><ul><li>Condition 4 – Jumping of a square is permitted </li></ul></ul></ul>
11. 11. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step VI – Procedure for Revision </li></ul><ul><ul><li>Step for Revision: </li></ul></ul><ul><ul><ul><li>Step 1 – Of all positive check number choose the largest check number and put a cross mark. </li></ul></ul></ul><ul><ul><ul><li>Step 2 – Form the loop for this cross mark square. </li></ul></ul></ul><ul><ul><ul><li>Step 3 – Along the loop write down the number in order. </li></ul></ul></ul><ul><ul><ul><li>Step 4 – Underline the alternate number. </li></ul></ul></ul><ul><ul><ul><li>Step 5 – Encircle the smallest underlined number. </li></ul></ul></ul><ul><ul><ul><li>Step 6 – Prepare the revised cell that is put in the cross mark cell the smallest number from Step 4. Continue along the loop with minus, plus and minus signs (- + -). Do not disturb other number not on loop. </li></ul></ul></ul>
12. 12. Steps involved in solving Transportation Problem by VAM method: <ul><li>Step VII – Prepare the Optimum Transportation Schedule. </li></ul><ul><li>Step VIII – Find the minimum Transportation Cost. </li></ul>
13. 13. <ul><li>A manager has three factories (i.e. origins) and four warehouses (i.e. destinations). The quantities of goods available in each factory, the requirements of goods in each warehouse and the costs of transportation of a product from each factory to each warehouse are given below. </li></ul><ul><li>Determine the optimum transportation schedule and transportation cost using: </li></ul><ul><ul><ul><li>North West Corner Method (NWCM) </li></ul></ul></ul><ul><ul><ul><li>Least Cost Method (LCM) </li></ul></ul></ul><ul><ul><ul><li>VOGEL’S APPROXIMATION METHOD (VAM) </li></ul></ul></ul>D1 D2 D3 D4 Supply P1 19 30 50 12 7 P2 70 30 40 60 10 P3 40 10 60 20 18 Demand 5 8 7 15
14. 14. Thank You | 1,962 | 7,030 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-09 | latest | en | 0.790613 |
https://www.physicsforums.com/threads/help-me-check-my-work-limit-problem.598518/ | 1,511,146,237,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805894.15/warc/CC-MAIN-20171120013853-20171120033853-00429.warc.gz | 848,556,944 | 15,762 | # Help me check my work (limit problem)
1. Apr 20, 2012
### styxrihocc
1. The problem statement, all variables and given/known data
Hey I just need someone to check my work to make sure I did it right, thanks.
2. Apr 20, 2012
### Staff: Mentor
The step above has an error. ln <whatever> ≠ eln<whatever>.
Also, how did you go from (1 - 3/x)2x + 10 to (-3/x *2x + 10)?
Finally, eln(-6) ≠ -6, because ln(-6) is undefined.
3. Apr 20, 2012
### Staff: Mentor
I should mention that posting a bunch of images of the work you've done makes it much more difficult to point out where errors happen to be. Using LaTeX you can format your work exactly as you have it in the images you uploaded.
4. Apr 20, 2012
### styxrihocc
Sorry but typing in latex takes too long so i just cropped the pics straight from my doc file
from this formula in my textbook:
5. Apr 20, 2012
### Staff: Mentor
What turns out to be a convenience for you is inconvenient for me, and possibly others at this site whose help you are seeking.
Are you sure you typed the formula correctly? I've never seen the formula that you wrote, and I'm pretty sure it is incorrect.
Your answer is correct, but you work along the way is not for the reasons I already gave.
The easiest thing to do is to rewrite ln(1 - 3/x)2x + 10 as (2x + 10)*ln(1 - 3/x), and then rearrange so that you can use L'Hopital's Rule. | 390 | 1,378 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-47 | longest | en | 0.93107 |
http://www.ck12.org/physical-science/Convection-in-Physical-Science/rwa/How-Is-Your-Home-Heated/ | 1,490,484,592,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189088.29/warc/CC-MAIN-20170322212949-00458-ip-10-233-31-227.ec2.internal.warc.gz | 467,341,614 | 31,447 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Convection
## Illustrates how fluids absorb and diffuse heat.
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### How Is Your Home Heated?
Credit: Gabriel Pollard
Source: http://www.flickr.com/photos/61181002@N00/8477121185
Some homes are heated with hot water running through radiators, like the one in this picture. However, homes can be heated in several other ways. Do you know how your home is heated?
#### Why It Matters
• Knowing how your home is heated is important. It can help you be aware of heating problems and safety issues. For example, if your home is heated by a furnace or boiler, it probably has a fuel supply line, such as a gas pipe, that provides it with fuel.
• Credit: butkaj.info
Source: http://www.flickr.com/photos/andybutkaj/408955986/
A boiler system for the home [Figure2]
• Gas lines can be dangerous. They can cause fires and explosions if people are not careful around them.
#### Can You Apply It?
1. Identify three common types of home heating systems. What type of system do you think heats your home?
2. All heating systems work on the same basic principle. What is it?
3. Four basic components are found in most heating systems. What are they, and what are their functions?
4. Find each of the components from the previous question in your home’s heating system. Where are the components located?
5. Explain how your home’s heating system generates and distributes heat throughout the home. Identify each point where heat is transferred in the system. What type of heat transfer is it: convection, conduction, or radiation?
6. What are some pros and cons of the type of heating system in your home?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes | 466 | 1,985 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.90974 |
http://mathhelpforum.com/calculus/89153-problems-partial-fractions.html | 1,524,662,522,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947803.66/warc/CC-MAIN-20180425115743-20180425135743-00222.warc.gz | 186,444,996 | 11,040 | # Thread: Problems with Partial Fractions
1. ## Problems with Partial Fractions
Hi, I've been working on these integral evaluations and I'm stuck. Any help on any of them would be greatly appreciated, thanks in advance!
1. $\displaystyle \int\frac{2x^3-3x+2}{x^3-x}dx$
My attempt:
$\displaystyle 2x^3-3x+2=\frac{A}{x^2-1}+\frac{B}{x-1}+\frac{C}{x}$ (Tried breaking it down, couldn't get it to work )
2. $\displaystyle \int\frac{x^4+1}{x^3+x}dx$
My attempt:
long division to get $\displaystyle \int xdx-\int\frac{x^2+1}{x^3+x}dx$ (not sure how to show that in Latex)
Then $\displaystyle x^2+1=\frac{A}{X^2+1}+\frac{B}{x+1}+\frac{C}{x}$
Not sure where to go from there...I imagine that I'm missing the same idea in both of these, so any help would be appreciated. Thanks again!
2. For #1:
$\displaystyle x^{3}-x=x(x^{2}-1)=x(x+1)(x-1)$
$\displaystyle \frac{2x^{3}-2x+2}{x^{3}-x}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$
Multiply both sides of the equation by $\displaystyle x(x+1)(x-1)$ and go on from there.
For #2:
$\displaystyle x^{3}+x=x(x^{2}+1)$
$\displaystyle \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^ {2}+1}$
Multiply both sides of the equation by $\displaystyle x(x^{2}+1)$ and go on from there.
Edit: Don't forget to divide beforehead as in Plato's post; I forgot that part.
3. Here is a solution.
4. Originally Posted by Pinkk
For #2:
$\displaystyle x^{3}+x=x(x^{2}+1)$
$\displaystyle \frac{x^{4}+1}{x^{3}+x}=\frac{A}{x}+\frac{Bx+C}{x^ {2}+1}$
Multiply both sides of the equation by $\displaystyle x(x^{2}+1)$ and go on from there.
Your response for number two is clearly incorrect as multiplication on both sides by the cubic will equate a quartic on the left to a quadratic on the right. The way it should be done is as follows:
$\displaystyle \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1-x^2}{x(x^2+1)}$
then
$\displaystyle \frac{1-x^2}{x(x^2+1)} \equiv \frac{A}{x} + \frac{Bx +C}{x^2 +1}$
yielding
$\displaystyle \frac{x^4+1}{x(x^2+1)} \equiv x + \frac{1}{x} - \frac{2x}{x^2 +1}$ .
5. Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.
6. I think I've got them both, thanks all!
7. Originally Posted by Pinkk
Yes, that's before I realized the highest degree of the numerator was larger than the highest degree of the denominator.
My apologies for not noticing your edit. | 829 | 2,392 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-17 | latest | en | 0.813346 |
https://stats.stackexchange.com/questions/490916/two-sample-kolmogorov-smirnov-test-suggest-that-sample-with-the-same-data-genera | 1,726,051,385,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00507.warc.gz | 538,442,215 | 40,944 | # two-sample Kolmogorov-Smirnov test suggest that sample with the same data generating process from different populations
I have two different samples I need to test if they are drawn from the same population distribution. The two samples are of the volatility of an asset that I simulated. Since the data is simulated I know that both samples have the same data generating process. The only difference between the two samples are the days that I am using from the simulation.
A summary of the two samples is given below. It is also important to note that since this is simulated data the sample size is quite large. I have 249,281 observations in sample 1 and 254,453 in sample 2.
> summary(Sample1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0128 0.0894 0.1194 0.1367 0.1627 0.9925
> summary(Sample2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0141 0.0950 0.1263 0.1467 0.1724 1.1435
When I apply the Kolmogorov-Smirnov test the results reject the null hypothesis that the two samples are drawn from the same population.
ks.test(Sample1,Sample2)
Two-sample Kolmogorov-Smirnov test
data: Sample1 and Sample2
D = 0.053089, p-value < 2.2e-16
alternative hypothesis: two-sided
However if I plot the kernel approximation of the two different density functions and the CDF both of the samples appear to be from the same distribution. A copy of the plots is given below. How is it possible that the two samples come from different distributions if they have the same data generating process? Am I missing interpreting the results from the KS test? Is there another test that I should be apply?
• Try the simulation 1000 times. If you set $\alpha=0.05$, you should reject about $5\%$ of the time. If you're much different from that, then perhaps your simulation isn't as correct as you think.
– Dave
Commented Oct 7, 2020 at 22:07 | 499 | 1,854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.926198 |
https://electricalampere.com/what-is-hysteresis-loss/ | 1,709,292,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00808.warc.gz | 229,730,042 | 18,348 | # What is Hysteresis Loss? – Definition, Formula, Examples
Last Updated on February 9, 2024 by Electricalampere
Definition: The energy wasted in the form of heat due to hysteresis in magnetic materials is called Hysteresis Loss. The magnetizing force works against the internal friction of the magnet’s molecules and produces heat.
## What is Hysteresis Loss?
When a magnetization force is applied to a magnetic material, the molecules in the material align in a particular direction. However, when this magnetizing force(H) is reversed in the opposite direction, the internal friction of the molecular magnets resists the reversal of magnetism(B). Therefore, the magnetic field(B) lags the magnetizing force(H), and this phenomenon of a magnetic field lagging with a magnetizing force is called hysteresis. If you observe the B-H curve of a magnetic material, you will find that the magnetic field B always lags the magnetizing force(H).
Image source: electricalvolt.com
In order to overcome the internal resistance within a magnetic material, a portion of the magnetizing force is employed. However, this process generates heat due to work done by the magnetizing force, which results in energy loss in the form of heat. This loss of energy is known as hysteresis loss, and it is directly proportional to the area of the B-H curve.
Let’s take electrical machines as an example to understand the concept. The hysteresis loss mainly occurs in the magnetic parts of the electrical machines, where there is a reversal of magnetism.
The increase in temperature of a machine results in energy loss, mainly in the form of heat, which is an undesirable process. Therefore, a suitable magnetic material with minimal losses and a narrow hysteresis loop is used for electrical machinery parts.
## How can we Reduce Hysteresis Losses?
To reduce hysteresis losses, it is recommended to use materials that have a smaller area of the hysteresis loop. For instance, high-grade or silica steel is a good choice for designing the transformer’s core because it has an extremely small area of the hysteresis loop.
To reduce losses in transformers, it is possible to increase the number of laminations and reduce the gaps between the plates. To decrease hysteresis loss, a soft core with less hysteresis, such as silicon steel(CRGO), can be chosen.
The grains in CRGO (Cold Rolled Grain Oriented) steel are intentionally made larger, about 10 times the size of grains in regular steel. These grains are grown and aligned almost parallel to the direction of the rolling of the steel. Therefore, hysteresis loss of CRGO core is minimal.
The density of flux, laminated core, and frequency are the main factors that influence these losses.
## Factors Affecting Hysteresis Loss
The following factors affect hysteresis loss.
• The hysteresis loop is narrow so that the material can be easily magnetized.
• If the material is difficult to magnetize, the resulting hysteresis loop will be larger.
• At various ‘B’ values, different materials can saturate, affecting loop height.
• The physical properties of the material largely determine the loop.
• The size and shape of the loop depend primarily on the specimen’s initial position.
• The hysteresis loss increases as the frequency and magnetic flux density increase.
• The loss increases with the increase in the volume of magnetic material.
## Hysteresis Loss Formula
The formula for hysteresis loss is given below.
Where,
P– Hysteresis loss in watts
Ƞ – Steinmetz’s constant in J/m3, its value depends upon the nature of the magnetic material.
B– Maximum value of the flux density in the magnetic material in wb/m2
f- Number of cycles of magnetization made per second
V- Volume of the magnetic material (part in which magnetic reversal occurs) in m3
The hysteresis loss per unit volume is.
If you know the shape of the B-H curve, you can calculate this loss by calculating its area.
We will mathematically prove that the hysteresis loss per unit volume is equal to the area of the hysteresis loop. Let’s take a BH curve.
We consider a small thickness dB strip on the hysteresis loop. The flow of the current (I ) in a coil produces a magnetic field in the magnetic material. The magnetic field in the specific area is called flux. Thus, magnetic flux is ;
When the current passing through a coil changes, it induces an electromotive force (e.m.f.) due to the change in flux.
Magnetic field strength (H) is;
Therefore,
The power expended in maintaining the current ‘I’ against induced EMF ‘e’ is known as the power or rate of energy expenditure.
Equation 3 can be integrated on both sides to obtain the total work done during a complete cycle of magnetization.
Where ʃ HdB is the area of the hysteresis loop.
Work done /unit volume (W/m3),
A smaller area of the hysteresis loop indicates that there is less hysteresis loss. Soft magnetic material has a smaller BH loop area and lower loss. In contrast, the hard magnetic material has more loss due to the greater BH curve area.
## Solved Example
The hysteresis loss of a 1-phase transformer working on 200 V, 50 Hz is Ph. What would be the percentage decrease in this loss when operated on a 160 V, 40 Hz?
Solution:
Hysteresis loss formula is.
Loss in the core at f1 frequency,
Since,
Therefore,
and,
% decrease in loss,
Hence, the % decrease in hysteresis loss would be 20 %. | 1,218 | 5,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | longest | en | 0.913389 |
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#41
06-02-2011, 03:24 AM
JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,711 Rep Power: 17
Re: convert cubic meters to square meters
Quote:
Originally Posted by Unregistered I have a brick pier at the front of a house with the following dimensions - 350mm x 3500mm x 2700MM = .33M3 (Cubic Meters) How do i convert this into M2 or Meters squared which is how i order from my supplier?
First of all, that is 3.3 m³.
When you order 1 m² from you supplier, what does that represent? Is that the face area of the bricks, laid normally? What provision is made for the gout dimension? Which dimensions correspond to length, width, and height?
#42
06-24-2011, 10:13 PM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
How do you convert square meter to cubic meters?[/QUOTE]
#43
07-24-2011, 06:00 AM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
How do you convert square meter to cubic meters
#44
08-04-2011, 02:28 AM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
what is 600sqm in cbm?
#45
08-04-2011, 03:02 AM
JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,711 Rep Power: 17
Re: convert cubic meters to square meters
Quote:
Originally Posted by Unregistered what is 600sqm in cbm?
Zero. An area has no thickness, so no volume. You have to multiply an area by a thickness to get a volume.
#46
09-16-2011, 05:50 PM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
Quote:
Originally Posted by unregistered how do you convert square meter to cubic meters?
784m2 to cubic metre
#47
10-08-2011, 09:38 PM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
I have 700cbm of materials and i need one warehouse for keeping this .How many sq.mtr required?.
#48
10-09-2011, 03:16 AM
JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 8,711 Rep Power: 17
Re: convert cubic meters to square meters
Quote:
Originally Posted by Unregistered I have 700cbm of materials and i need one warehouse for keeping this .How many sq.mtr required?.
How high do you intend to stack it? What spaces for aisles do you want around the stacks.
Ignoring aisles, 700 m² if stacked 1 m high, 350 m² if stacked 2 m high, etc.
#49
10-15-2011, 06:02 AM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
Quote:
Originally Posted by Unregistered I have a fiblerglass screen 30" x 30 meters. What is the square meters? Please advise.
Ans: First of all we convert 30" into meter that is 0.762 meter. after that we multiply it with 30 meter
0.762 X 30 = 22.86 sqmt.
After that we convert sqmt. into sqft. on multiplication by 10.76 in sqmt area.
22.86 X 10.76 = 245.97 sqft.
Hence, the area of a fiblerglass screen in square meter is 245.97 sqft.
#50
10-17-2011, 06:22 AM
Unregistered Guest Posts: n/a
Re: convert cubic meters to square meters
convert 1 cubic meter to square meters
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https://smallbusiness.chron.com/value-shares-company-72715.html | 1,670,049,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00351.warc.gz | 544,210,718 | 23,346 | # How to Value Shares of a Company
Several techniques offer insight into the value of a company's stock. All valuation techniques, however, quantify the worth of a stock based upon "perceptions" of value held by those doing the valuating, which are inherently subjective. Thus, all valuation techniques reflect underlying subjective perceptions that can differ based on people's investment goals.
## Valuation Approaches
1. The three basic approaches to financial valuations are the cost approach, the market approach and the income approach. The cost approach assumes that a buyer will purchase an asset for no more than an asset of equal usefulness. The market approach assumes that in free markets, supply and demand will drive the price of a stock to a point where the number of buyers and sellers balance. The income approach defines value as the net present value of a company's future free cash flows.
## Cost Approach
1. Book value is commonly associated with the cost approach. It is the sum of all of a company's "tangible assets" — plant, property and equipment — after deducting accumulated depreciation. A company's "net capital value" reflects this book value, measured by the company's book of net tangible assets over its book of liabilities. Calculate the book value of a stock by dividing the net capital value by the number of outstanding share to arrive at a per share value.
One book value limitation is that it fails to valuate intangible assets such as brand names, which can be substantial. For instance, the "Coca Cola" brand name was appraised at \$65 billion in 2013.
## Market Approach
1. The market approach reflects prices of publicly traded stocks. However, stock values are implicit in stock prices. The price/earnings ratio or P/E ratio is commonly used to better understand those values. Divide the price of a stock by its earnings per share to determine the P/E ratio. Publicly traded companies must report their earnings per share. In general, investors attach greater value to stocks with higher P/E ratios. The average P/E ratio was 20-25 times earnings in mid-2013. Hence, P/E ratios above 25 times earnings identified stocks that investors held in high regard.
Because there are several P/E ratios and variations thereof, you must know which ratio is in play. For example, the forward-looking P/E ratio is based upon "projected" future earnings, which are based on assumptions that may prove false.
## Income Approach
1. The income approach holds that the value of a company equals the present value of the sum of its expected future income streams. The capitalized cash flow method is generally used to compute a company's worth when future income streams are expected to remain unchanged from the past. The discounted cash flow method is used when income streams are expected to fluctuate. Divide the resulting figure from the CCF or DCF calculation by the number of outstanding shares to determine the share price. Spreadsheet programs like Excel can do the DCF calculations for you. Simply plug in the numbers. | 614 | 3,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | longest | en | 0.947088 |
https://r.igraph.org/reference/degree.html | 1,723,262,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640789586.56/warc/CC-MAIN-20240810030800-20240810060800-00321.warc.gz | 382,699,590 | 4,284 | The degree of a vertex is its most basic structural property, the number of its adjacent edges.
## Usage
``````degree(
graph,
v = V(graph),
mode = c("all", "out", "in", "total"),
loops = TRUE,
normalized = FALSE
)
max_degree(
graph,
...,
v = V(graph),
mode = c("all", "out", "in", "total"),
loops = TRUE
)
degree_distribution(graph, cumulative = FALSE, ...)``````
## Arguments
graph
The graph to analyze.
v
The ids of vertices of which the degree will be calculated.
mode
Character string, “out” for out-degree, “in” for in-degree or “total” for the sum of the two. For undirected graphs this argument is ignored. “all” is a synonym of “total”.
loops
Logical; whether the loop edges are also counted.
normalized
Logical scalar, whether to normalize the degree. If `TRUE` then the result is divided by \(n-1\), where \(n\) is the number of vertices in the graph.
...
These dots are for future extensions and must be empty.
cumulative
Logical; whether the cumulative degree distribution is to be calculated.
## Value
For `degree()` a numeric vector of the same length as argument `v`.
For `degree_distribution()` a numeric vector of the same length as the maximum degree plus one. The first element is the relative frequency zero degree vertices, the second vertices with degree one, etc.
For `max_degree()`, the largest degree in the graph. When no vertices are selected, or when the input is the null graph, zero is returned as this is the smallest possible degree.
Other structural.properties: `bfs()`, `component_distribution()`, `connect()`, `constraint()`, `coreness()`, `dfs()`, `distance_table()`, `edge_density()`, `feedback_arc_set()`, `girth()`, `is_acyclic()`, `is_dag()`, `is_matching()`, `k_shortest_paths()`, `knn()`, `reciprocity()`, `subcomponent()`, `subgraph()`, `topo_sort()`, `transitivity()`, `unfold_tree()`, `which_multiple()`, `which_mutual()`
## Author
Gabor Csardi csardi.gabor@gmail.com
## Examples
``````
g <- make_ring(10)
degree(g)
#> [1] 2 2 2 2 2 2 2 2 2 2
g2 <- sample_gnp(1000, 10 / 1000)
max_degree(g2)
#> [1] 22
degree_distribution(g2)
#> [1] 0.001 0.000 0.002 0.005 0.026 0.038 0.058 0.101 0.118 0.117 0.118 0.113
#> [13] 0.098 0.081 0.057 0.027 0.019 0.010 0.004 0.004 0.000 0.002 0.001
`````` | 678 | 2,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.788066 |
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By: csp (offline) Thursday, August 09 2012 @ 06:13 AM CDT (Read 750 times)
csp
Hanson had some one-dollar coins and some fifty cents coins.the ratio of the number of one dollar coins to to number of fifty cent coins he had was 2:5. Hanson took 140 fifty cents coins to the bank and changed these fifty cents coins for the same value of one dollar coins.In the end ,the number of one dollar coins to the number of fifty cent coins he had became 5:2 ,
Find the value of the fifty cent coins Hanson had at first.
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Registered: 05/01/12
Posts: 4
By: jo sarah (offline) Thursday, August 09 2012 @ 09:21 AM CDT
jo sarah
hi,
first, the 140 fifty-cent coins = 70 one-dollar coins.
[i use ..... for spacing; 'u' means unit, and 'p' means part]
Now, tabulate the no. of coins as:
...................... \$1.................. 50 cts
at first:............2u...................5u
change:........(+70)..............(-140)
in end:.............5p..................2p
So,
2u + 70 = 5p
x2: 4u + 140 = 10p
5u - 140 = 2p
x5: 25u - 700 = 10p
thus, 4u + 140 = 25u - 700
that is, 21u = 840
and, 1u = 40
there were 5u = 200 fifty-cent coins at first.
So, value of fifty-cent coins at first = 200 x \$0.50 = \$100
Hope this helps,
cheerio.
Regular Member
Registered: 03/20/12
Posts: 111
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https://nl.mathworks.com/matlabcentral/answers/786151-i-want-this-for-loop-to-dispaly-the-smallest-calculated-value-of-pe-what-would-be-the-best-way-of-g | 1,621,328,312,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00258.warc.gz | 447,072,145 | 23,059 | I want this for loop to dispaly the smallest calculated value of PE. What would be the best way of going about this?
1 view (last 30 days)
Brad Holder on 28 Mar 2021
Answered: Jeff Miller on 28 Mar 2021
for t1 = 0:1:90,t2 = 0:1:90;
t1;
t2;
h1 = -b1*cosd(t1);
h2 = -(b1*cosd(t1)+b2*cosd(t2));
d1 = sqrt((x1-a1*sind(t1)).^2 + (y1+a1*cosd(t1)).^2)-L1;
d2 = sqrt((x1-(b1*sind(t1)+a2*sind(t2))).^2+(y2+(b1*cosd(t1)+a1*cosd(t2))).^2)-L2;
PE = (1/2)*k1*d1.^2 + w1*h1 - f1*b1*sind(t1) + (1/2)*k2*d2.^2 + w2*h2 - f2*(b1*sind(t1)+b2*sind(t2));
if PE == min
disp PE
end
end
Jeff Miller on 28 Mar 2021
% something like this:
minPE = realmax;
for t1 = 0:1:90,t2 = 0:1:90;
t1;
t2;
h1 = -b1*cosd(t1);
h2 = -(b1*cosd(t1)+b2*cosd(t2));
d1 = sqrt((x1-a1*sind(t1)).^2 + (y1+a1*cosd(t1)).^2)-L1;
d2 = sqrt((x1-(b1*sind(t1)+a2*sind(t2))).^2+(y2+(b1*cosd(t1)+a1*cosd(t2))).^2)-L2;
PE = (1/2)*k1*d1.^2 + w1*h1 - f1*b1*sind(t1) + (1/2)*k2*d2.^2 + w2*h2 - f2*(b1*sind(t1)+b2*sind(t2));
if PE < minPE
minPE = PE;
end
end
disp minPE
R2020a
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Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$
2. Originally Posted by slevvio
Hello, I was wondering if I could get some help with this integral - it would be much appreciated - thanks.
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{(x-2)^2+9}}dx$
Do you mean this?
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx = \int{\frac{1}{\sqrt{(x-2)^2+9}}}dx$
If so, then you can use a trig substitution for x - 2. Set it equal to 3*tan(u). Everything else should fall into place for easy integration!
EDIT: Soroban posted shortly after I did, check out his detailed solution!
3. Hello, slevvio!
$\displaystyle \int \frac{dx}{\sqrt{x^2-4x+13}} \:= \:\int \frac{dx}{\sqrt{(x-2)^2+9}}$
Let: .$\displaystyle x-2 \:=\:3\tan\theta\quad\Rightarrow\quad dx \:=\:3\sec^2\!\theta\,d\theta$
. . and: .$\displaystyle \sqrt{(x-2)^2+9} \;=\;\sqrt{9\tan^2\!\theta + 9} \:=\:\sqrt{9(\tan^2\!\theta + 1)} \:=\:\sqrt{9\sec^2\!\theta} \:=\:3\sec\theta$
Substitute: .$\displaystyle \int\frac{3\sec^2\!\theta\,d\theta}{3\sec\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta| + C$
. . then back-substitute . . .
4. We have a hidden natural logarithm here.
Originally Posted by slevvio
$\displaystyle \int{\frac{1}{\sqrt{x^2-4x+13}}}dx$
$\displaystyle \int {\frac{{dx}} {{\sqrt {x^2 - 4x + 13} }}} = \int {\frac{{dx}} {{\sqrt {(x - 2)^2 + 9} }}} .$
Substitute $\displaystyle u = x + \sqrt {(x - 2)^2 + 9} - 2 \implies du = 1 + \frac{{x - 2}} {{\sqrt {(x - 2)^2 + 9} }}\,dx.$
So $\displaystyle \frac{{dx}} {{\sqrt {(x - 2)^2 + 9} }} = \frac{{du}} {u}.$
The rest follows. | 682 | 1,766 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.590881 |
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### CBSE Class 8 - (Ch11) - Mensuration
Q1. The area of the floor of a rectangular hall of length 40m is 960 m2. Carpets of size 6m x 4m are available. Find how many carpets are required to cover the hall. (Unsolved Exercise 16.1 from RD Sharma)
Answer:
length = 6m breadth = 4m and Area = l x b
$\therefore$ Area of carpet is = 6 x 4 = 24 m2.
Floor area = 960 m2
Number of carpets = $\frac{\text{area of floor}}{\text{area of carpet}} = \frac{960}{24} = 40$
Q2: Find the area of a square the length of whose diagonal is 2.9 meters. (Unsolved Ex 16.1 from RDS)
Answer: Diagonal (d) = 2.9 m. Let x be the length of each side of the square.
Applying Pythagoras theorem,
$x^ + x^ = 2x^ = (2.9)^2$
$\Rightarrow x^2 = \text {area of square} = \frac{(2.9)^2}{2} = \frac{8.41}{2} = 4.205 m^2$
Q3: The area of a square field is 0.5 hectares. Find the length of its diagonal in meters.
Answer: 1 hect = 10000 m2
$\therefore$ 0.5 hec = 5000 m2 $Applying Pythagoras theorem,$ \frac{diagonal^2}{2} = 5000\Rightarrow diagonal = \sqrt{10000} = 100 m$Q4: The diameter of a semi-circular field is 14 meters. What is the cost of fencing the plot at Rs. 10 per meter. Answer: Diameter of field (d) = 14m$\Rightarrow r = 14 \div 2 = 7m \text{Perimeter of semicircle} = (\pi + 2) \times r \Rightarrow = (\frac{22}{7} + 2) \times 7 = 36m $Cost of fencing per meter = Rs 10$\therefore \text{cost of fencing for 36m} = 36 \times 10 = \text{Rs }360$Q5: Find the area of the quadrilateral ABCD as shown. The diagonal AC is 84 units and perpendiculars BE and FD are 21 and 28 units respectively. Answer: The area of a general quadrilateral =$\frac{1}{2}d \times (h_1 + h_2).\$
Here d = 84, h1 = 21 units and h2 = 28 units.
Therefore area = 84 * (21+28)/2 = 2058 square units. (Answer).
$\frac{\text{area of floor}}{\text{area of carpet}} = \frac{960}{24} = 40$
#### 7 comments:
1. thanks..............my papers are going good for you.
2. it gives us idea of questions. thank u so much!!!!!!
3. very nice...........
4. ya tnx 2 u and yr ans is so gud.,.,.,
5. thank you for giving me a wonderful platform
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# What do b-schools really look for in an interview?
Author Message
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### Hide Tags
Manager
Joined: 02 Apr 2008
Posts: 119
Kudos [?]: 21 [0], given: 5
Location: United States
Concentration: Marketing, Strategy
GPA: 3.45
WE: Marketing (Consumer Electronics)
What do b-schools really look for in an interview? [#permalink]
### Show Tags
24 Nov 2012, 10:05
Just a question I have been pondering for quite some time! Obviously, "fit" and articulation of goals etc. But really, how does 30 mins or an hour tell someone about the real you and what makes you tick?
Any thoughts are welcome!
Kudos [?]: 21 [0], given: 5
Current Student
Joined: 13 Sep 2011
Posts: 573
Kudos [?]: 145 [0], given: 28
Location: United States
Schools: Ross '16 (M)
Re: What do b-schools really look for in an interview? [#permalink]
### Show Tags
25 Nov 2012, 23:03
Generally, schools want to get to know you, ensure that you can present yourself and make coherent statements. In some cases, they want to make sure that your speaking skills match your essays.
Last edited by Ward2012 on 26 Nov 2012, 09:27, edited 1 time in total.
Kudos [?]: 145 [0], given: 28
Intern
Joined: 28 Feb 2012
Posts: 31
Kudos [?]: 11 [0], given: 5
Concentration: Strategy, Finance
Re: What do b-schools really look for in an interview? [#permalink]
### Show Tags
26 Nov 2012, 09:26
ss87 wrote:
Just a question I have been pondering for quite some time! Obviously, "fit" and articulation of goals etc. But really, how does 30 mins or an hour tell someone about the real you and what makes you tick?
Any thoughts are welcome!
In three words: Honesty, research, and insight. The first word is self explanatory. By research, I mean how much research the applicant has done on the university and the program. Insight refers to the candidate's ability to understand his/her goals with respect to the MBA program that he/she is interviewing for.
Kudos [?]: 11 [0], given: 5
Re: What do b-schools really look for in an interview? [#permalink] 26 Nov 2012, 09:26
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# What do b-schools really look for in an interview?
Moderators: OasisGC, aeropower, bb10
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 790 | 3,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | latest | en | 0.914462 |
http://www.transtutors.com/homework-help/operations-management/quality-control/acceptance-sampling.aspx | 1,511,435,002,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806771.56/warc/CC-MAIN-20171123104442-20171123124442-00774.warc.gz | 512,446,293 | 16,350 | ## What is Acceptance Sampling?
The objective of acceptance sampling is to take decision whether to accept or reject a lot based on sample's characteristics. The lot may be incoming raw materials or finished parts.
An accurate method to check the quality of lots is to do 100% inspection. But, 100% inspection will have the following limitations.
• The cost of inspection is high.
• Destructive methods of testing will result in 100% spoilage of the parts.
• Time taken for inspection will be too long.
• When the population is large or infinite, it would be impossible or impracticable to inspect each unit.
Hence, acceptance sampling procedure has lot of scope in practical application. Acceptance sampling can be used for attributes as well as variables.
Acceptance sampling deals with accept or reject situation of the incoming raw materials and finished goods. Let the size of the incoming lot be N and the size of the sample drawn be n. The probability of getting a given number of defective/good parts out of a sample consisting of n pieces will follow Binomial Distribution. If the lot size is infinite or very large, such that when a sample is drawn from it and not replaced, then the usage of binomial distribution is justified. Otherwise, we will have to use hyper-geometric distribution.
Specifications of a single sampling plan will contain a sample size (n) and an acceptance number C. As an example, if we assume the sample size as 50 and the acceptance number as 3, the interpretation of the plan is explained as follows: Select a sample of size 50 from a lot and obtain the number of defective pieces in the sample. If the number of defective pieces is less than or equal to 3, then accept the whole lot from which the sample is drawn. Otherwise, reject the whole lot. This is called single sampling plan. There are several variations of this plan.
In this process, one will commit two types of error, viz. type I error and type II error. If the lot is really good, but based on the sample information, it is rejected, and then the supplier/producer will be penalized. This is called producer's risk or type I error. The notation for this error is α. On the other hand, if the lot is really bad, but it is accepted based on the sample information, then the customer will be at loss. This is called consumer's risk or type II error. The notation for this error is β. So, both parties should jointly decide about the levels of producer's risk (α) and consumer's risk (β) based on mutual agreement.
## Online Homework Assignment Experience in Acceptance Sampling
Transtutors provide 24x7 homework help, Regular one to one grade tutoring in Production and Operation Management. Our unique approach with a highly qualified and certified accounts tutor’s team of 100 plus tutors gives an edge. You can submit your school, college or university level acceptance sampling homework or assignment to us and we will make sure that you get the answers you need which are timely and also cost effective.
## Related Topics
All Management Topics
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https://en.khanacademy.org/math/cc-fourth-grade-math/imp-place-value-and-rounding-2/imp-intro-to-place-value/a/place-value-faq | 1,721,771,011,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00451.warc.gz | 206,296,567 | 109,545 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Lesson 1: Intro to place value
# Place value: FAQ
## What is place value?
Place value refers to the value that each digit in a number has, based on its position. For example, in the number $523$, the $5$ is in the hundreds place, the $2$ is in the tens place, and the $3$ is in the ones place.
Here are a couple of the exercises that build off of place value:
## What are the different ways we can write numbers?
We can write numbers in standard form, expanded form, or written form. Expanded form breaks a number down to show the value of each digit. For example, $724$ can be written in expanded form as $700+20+4$. Written form uses words to write out a number. For example, $724$ can be written in written form as "seven hundred twenty-four." Standard form is the way we usually write numbers, using just the digits. For example, $724$ is in standard form.
Here are a couple of the exercises that build off of writing numbers in different forms:
## Why is $10$ important in place value?
$10$ is important in place value because our number system is based on the number $10$. This means that when we count past $9$ in one place value column, we regroup to the next place value (in this case, the tens place).
Here are a couple of the exercises that build off of using $10$ in place value:
## How do I compare multi-digit numbers?
To compare multi-digit numbers, we start by looking at the largest digit, or the leftmost digit. If one number has a larger digit in that place, we know that number is larger. If the digits are the same, we move to the next digit to the right and continue comparing until we find a difference.
## Why do I need to know about place value?
Understanding place value is important for many reasons. It can help us add, subtract, multiply, and divide multi-digit numbers. It also helps us understand the value of numbers and compare them. Place value is used in all kinds of real-world situations, from counting money to measuring quantities.
## Want to join the conversation?
• what is 9x3
• what is 555 x5555
• 3083025
• Understanding place value is important for many reasons. It can help us add, subtract, multiply, and divide multi-digit numbers. It also helps us understand the value of numbers and compare them. Place value is used in all kinds of real-world situations, from counting money to measuring quantities.
• 555x5555=5555555
• no the answer is 3083025
• what is 10x47?
• the answer is 470 because when multiplying by 10
you just add a 0 at the end
• Whats 10x100
• 1,000 i think
• what is 5x255
• 1,275 | 669 | 2,783 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 13, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2024-30 | latest | en | 0.917516 |
http://core-cms.prod.aop.cambridge.org/core/journals/journal-of-the-london-mathematical-society/article/asymptotic-expansions-of-multiple-zeta-functions-and-power-mean-values-of-hurwitz-zeta-functions/197E7B35DA0341DE01E018A2BF4666C7 | 1,580,272,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251788528.85/warc/CC-MAIN-20200129041149-20200129071149-00007.warc.gz | 39,946,546 | 25,944 | Skip to main content Accessibility help
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# ASYMPTOTIC EXPANSIONS OF MULTIPLE ZETA FUNCTIONS AND POWER MEAN VALUES OF HURWITZ ZETA FUNCTIONS
## Abstract
Let $\zeta(s, \alpha)$ be the Hurwitz zeta function with parameter $\alpha$ . Power mean values of the form $\sum^q_{a=1}\zeta(s,a/q)^h$ or $\sum^q_{a=1}|\zeta(s,a/q)|^{2h}$ are studied, where $q$ and $h$ are positive integers. These mean values can be written as linear combinations of $\sum^q_{a=1}\zeta_r(s_1,\ldots,s_r;a/q)$ , where $\zeta_r(s_1,\ldots,s_r;\alpha)$ is a generalization of Euler–Zagier multiple zeta sums. The Mellin–Barnes integral formula is used to prove an asymptotic expansion of $\sum^q_{a=1}\zeta_r(s_1,\ldots,s_r;a/q)$ , with respect to $q$ . Hence a general way of deducing asymptotic expansion formulas for $\sum^q_{a=1}\zeta(s,a/q)^h$ and $\sum^q_{a=1}|\zeta(s,a/q)|^{2h}$ is obtained. In particular, the asymptotic expansion of $\sum^q_{a=1}\zeta(1/2,a/q)^3$ with respect to $q$ is written down.
# ASYMPTOTIC EXPANSIONS OF MULTIPLE ZETA FUNCTIONS AND POWER MEAN VALUES OF HURWITZ ZETA FUNCTIONS
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Usage data cannot currently be displayed | 466 | 1,599 | {"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} | 2.609375 | 3 | CC-MAIN-2020-05 | latest | en | 0.675397 |
https://im.kendallhunt.com/HS/teachers/4/4/18/preparation.html | 1,723,496,082,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00168.warc.gz | 241,685,687 | 26,387 | # Lesson 18
Modeling Price Information
These materials, when encountered before Algebra 1, Unit 4, Lesson 18 support success in that lesson.
### Lesson Narrative
In this lesson, students use data to find average rate of change and write a function to model the data that can be used to predict more information. Then students are presented with additional data and asked to make an updated prediction using both the additional data as well as information about the situation itself. In the associated Algebra 1 lesson, students also model data with a function and use it to predict information. The work of this lesson supports students by giving them a more structured way to model the situation.
Students model with mathematics (MP4) when they use available data to draw possible functions that represent the data and predict additional values from the model. Looking for and making use of structure (MP7) is also needed to recognize the trends in the data to find a linear function that will model the data well.
### Learning Goals
Teacher Facing
• Write linear functions to model data.
### Student Facing
• Let’s predict some information.
Building Towards | 228 | 1,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-33 | latest | en | 0.905601 |
https://nl.mathworks.com/matlabcentral/cody/players/476476/solved | 1,582,480,799,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00162.warc.gz | 491,437,154 | 20,943 | Cody
# Steve Eddins
Rank
Score
1 – 50 of 67
#### Problem 854. Find the list of all open files
Created by: Aurelien Queffurust
Tags fid
#### Problem 314. Find the sum of the elements in the "second" diagonal
Created by: Roy Fahn
#### Problem 230. Project Euler: Problem 1, Multiples of 3 and 5
Created by: Doug Hull
#### Problem 28. Counting Money
Created by: Cody Team
Tags regexp, strings
#### Problem 115. Distance walked 1D
Created by: AMITAVA BISWAS
#### Problem 73. Replace NaNs with the number that appears to its left in the row.
Created by: Cody Team
Tags matlab
#### Problem 128. Sorted highest to lowest?
Created by: AMITAVA BISWAS
#### Problem 108. Given an unsigned integer x, find the largest y by rearranging the bits in x
Created by: AMITAVA BISWAS
Tags binary
#### Problem 15. Find the longest sequence of 1's in a binary sequence.
Created by: Cody Team
#### Problem 135. Inner product of two vectors
Created by: AMITAVA BISWAS
#### Problem 175. Double Deal
Created by: @bmtran (Bryant Tran)
#### Problem 86. Renaming a field in a structure array
Created by: Cody Team
Tags matlab
#### Problem 45. Make a Palindrome Number
Created by: Cody Team
Tags strings
#### Problem 180. Omit columns averages from a matrix
Created by: Roy Fahn
Tags columns, matrix
#### Problem 125. Remove DC
Created by: AMITAVA BISWAS
#### Problem 233. Reverse the vector
Created by: Vishwanathan Iyer
#### Problem 158. Is my wife right? Now with even more wrong husband
Created by: the cyclist
#### Problem 189. Sum all integers from 1 to 2^n
Created by: Dimitris Kaliakmanis
#### Problem 151. Magic!
Created by: the cyclist
Tags magic
#### Problem 13. Remove all the consonants
Created by: Cody Team
Tags regexp, basics
#### Problem 174. Roll the Dice!
Created by: @bmtran (Bryant Tran)
#### Problem 37. Pascal's Triangle
Created by: Cody Team
#### Problem 18. Bullseye Matrix
Created by: Cody Team
Tags matrices
#### Problem 147. Too mean-spirited
Created by: the cyclist
Tags mean
#### Problem 42. Find the alphabetic word product
Created by: Cody Team
Tags matlab
#### Problem 27. Pangrams!
Created by: Cody Team
Tags strings
#### Problem 35. Quote Doubler
Created by: Cody Team
#### Problem 39. Which values occur exactly three times?
Created by: Cody Team
Tags search
#### Problem 34. Binary numbers
Created by: Cody Team
Tags matlab
#### Problem 29. Nearest Numbers
Created by: Cody Team
#### Problem 94. Target sorting
Created by: Cody Team
Tags matlab, sorting
#### Problem 157. The Hitchhiker's Guide to MATLAB
Created by: the cyclist
#### Problem 41. Cell joiner
Created by: Cody Team
Tags matlab, strings
#### Problem 167. Pizza!
Created by: the cyclist
Tags pizza, fun
#### Problem 60. The Goldbach Conjecture
Created by: Cody Team
Tags primes
#### Problem 33. Create times-tables
Created by: Cody Team
Tags matrices
#### Problem 20. Summing digits
Created by: Cody Team
Tags strings
#### Problem 105. How to find the position of an element in a vector without using the find function
Created by: Chelsea
Tags find, indexing
#### Problem 21. Return the 3n+1 sequence for n
Created by: Cody Team
Tags 3n+1, sample
#### Problem 30. Sort a list of complex numbers based on far they are from the origin.
Created by: Cody Team
#### Problem 11. Back and Forth Rows
Created by: Cody Team
Tags matrices
#### Problem 22. Remove the vowels
Created by: Cody Team
Tags regexp, siam
#### Problem 109. Check if sorted
Created by: AMITAVA BISWAS
#### Problem 32. Most nonzero elements in row
Created by: Cody Team
Tags matrices
#### Problem 25. Remove any row in which a NaN appears
Created by: Cody Team
#### Problem 12. Fibonacci sequence
Created by: Cody Team
#### Problem 53. Duplicates
Created by: Cody Team
#### Problem 24. Function Iterator
Created by: Cody Team
#### Problem 23. Finding Perfect Squares
Created by: Cody Team
Created by: Will
1 – 50 of 67 | 1,064 | 3,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.768473 |
https://codereview.stackexchange.com/questions/292176/a-game-where-the-player-guesses-the-computers-number | 1,718,809,409,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00291.warc.gz | 145,148,103 | 46,072 | # A game where the player guesses the computer's number
I am fairly new to programming, and I recently finished part of one of my projects, a guessing game.
The part I completed was the game mode where the player guesses the computer's number. I'm quite proud of it, but I have a strong feeling that it is hugely suboptimal, as I only really know the basics so far. I would like some feedback how to not only improve this program, but all programs I make in the future.
Please be specific when giving feedback, because there may be some terms that I do not know.
My code:
import random
import sys
import time
#controls typing speed (I got this from online)
typing_speed = 150 #wpm
def slow_print(t):
for l in t:
sys.stdout.write(l)
sys.stdout.flush()
time.sleep(random.random()*10.0/typing_speed)
print('')
#asks difficulty
def ask_guess_difficulty():
slow_print('Alright! I\'ll pick a number from 1 to 100, and you try to guess it!')
slow_print('''You can do easy mode (E), with infinite guesses; normal mode (N), with 10 guesses;
hard mode (H), with 5 guesses; or custom mode (C), where you pick how many guesses you get!''')
guess_difficulty = input()
guess_difficulty_lower = guess_difficulty.lower()
if guess_difficulty_lower == 'e':
guesses = 1000
return guesses
elif guess_difficulty_lower == 'n':
guesses = 11
return guesses
elif guess_difficulty_lower == 'h':
guesses = 6
return guesses
elif guess_difficulty_lower == 'c':
slow_print('How many guesses do you want?')
guesses = input()
if guesses.isnumeric() == True:
guesses = int(guesses) + 1
slow_print('Ok! You will have ', str(guesses - 1), ' guesses!')
return guesses
else:
slow_print('I don\'t understand. Please input a number for the amount of guess that you want.')
else:
slow_print('''I don\'t understand. Please input \"E\" for easy mode, \"N\" for
normal mode, \"H\" for hard mode, and \"C\" for custom mode''')
#plays the gamemode where the player guesses the computer's number
def player_guess(guesses):
guesses = int(guesses)
player_guess_win = False
slow_print('Ok! I\'m thinking of a number from 1 to 100')
slow_print('You\'ll have ', str(guesses - 1), ' guesses to get it.')
slow_print('Go ahead and guess when you\'re ready.')
total_tries = 0
player_guess_num = random.randint(1,100)
last_guess = None
low_high = None
while player_guess_win == False and guesses > 0:
guesses -= 1
total_tries += 1
if int(guesses) > 0:
current_guess = input()
if current_guess.isnumeric() == True and int(current_guess) <= 100 and int(current_guess) >= 1:
if int(current_guess) == last_guess:
slow_print('That was your last guess silly! Make sure to enter a new number!')
guesses += 1
continue
if low_high == 'high':
if int(current_guess) > last_guess:
slow_print('Your guess is...')
if int(current_guess) == player_guess_num:
slow_print('CORRECT!')
slow_print('Congratulations! You took ',str(total_tries),' guesses to get it right!')
slow_print('Do you want to play again? (Please input "yes" or "no".)')
while True:
yn_choice = input()
if yn_choice == 'yes' or yn_choice == 'y':
slow_print('Ok, let\'s do it!')
guess_the_number_start()
break
elif yn_choice == 'no' or yn_choice == 'n':
slow_print('Ok!')
exit()
else:
slow_print('I don\'t understand. Please input "yes" or "no".')
else:
if int(current_guess) < player_guess_num:
slow_print('Too low.')
last_guess = int(current_guess)
low_high = 'high'
continue
else:
slow_print('Too high')
last_guess = int(current_guess)
low_high = 'low'
continue
else:
slow_print('Oops! Make sure you enter a number that is greater than your last guess!')
guesses += 1
continue
elif low_high == 'low':
if int(current_guess) < last_guess:
slow_print('Your guess is...')
if int(current_guess) == player_guess_num:
slow_print('CORRECT!')
slow_print('Congratulations! You took ',str(total_tries),' guesses to get it right!')
slow_print('Do you want to play again? (Please input "yes" or "no".)')
while True:
yn_choice = input()
if yn_choice == 'yes' or yn_choice == 'y':
slow_print('Ok, let\'s do it!')
guess_the_number_start()
break
elif yn_choice == 'no' or yn_choice == 'n':
slow_print('Ok!')
exit()
else:
slow_print('I don\'t understand. Please input "yes" or "no".')
else:
if int(current_guess) < player_guess_num:
slow_print('Too low')
last_guess = int(current_guess)
low_high = 'high'
continue
else:
slow_print('Too high')
last_guess = int(current_guess)
low_high = 'low'
continue
else:
slow_print('Oops! Make sure you enter a number that is less than your last guess!')
guesses += 1
continue
else:
slow_print('Your guess is...')
if int(current_guess) == player_guess_num:
slow_print('CORRECT!')
slow_print('Congratulations! You took ',str(total_tries),' guesses to get it right!')
slow_print('Do you want to play again? (Please input "yes" or "no".)')
while True:
yn_choice = input()
if yn_choice == 'yes' or yn_choice == 'y':
slow_print('Ok, let\'s do it!')
guess_the_number_start()
break
elif yn_choice == 'no' or yn_choice == 'n':
slow_print('Ok!')
exit()
else:
slow_print('I don\'t understand. Please input "yes" or "no".')
else:
if int(current_guess) < player_guess_num:
slow_print('Too low')
last_guess = int(current_guess)
low_high = 'high'
continue
else:
slow_print('Too high')
last_guess = int(current_guess)
low_high = 'low'
continue
else:
slow_print('I don\'t understand. Please enter a number from 1 to 100.')
guesses += 1
continue
else:
slow_print('You ran out of guesses. My number was ',str(player_guess_num), '.')
slow_print('Would you like to play again?')
slow_print('(Please input "Yes" or "No".)')
play_again_gtn = input()
play_again_gtn_lower = play_again_gtn.lower()
if play_again_gtn_lower == 'y' or play_again_gtn_lower == 'yes':
slow_print('Oki doki!')
continue
elif play_again_gtn_lower == 'n' or play_again_gtn_lower == 'no':
slow_print('Oki doki! Bye!')
exit()
#computer guesses your number (not finished yet
def computer_guess():
slow_print('Ok, I\'ll guess!')
slow_print('First, we need to figure out the span of numbers I\'ll be guessing')
slow_print('What should the minimum be? The number will not be able to go any lower than this.')
#begins the program
def guess_the_number_start():
slow_print('Do you want to guess my number, or should I guess yours?')
slow_print('(If you want to guess, input \"I\". If I should guess, input \"U\".)')
who_guess = input()
if who_guess.lower() == 'i':
guesses = ask_guess_difficulty()
player_guess(guesses)
elif who_guess.lower() == 'u':
computer_guess()
else:
slow_print('I don\'t understand. Please input \"I\" if you should guess and input \"U\" if I should guess.')
#loops the program
while True:
guess_the_number_start()
## Possible bug
When I run the code, and it executes this line:
slow_print('You\'ll have ', str(guesses - 1), ' guesses to get it.')
the code exits with an error. My input sequence is I then H to reach this code. Perhaps you get a different result on the version of Python that you used. I get no error when I use string concatenation:
slow_print('You\'ll have ' + str(guesses - 1) + ' guesses to get it.')
## Escaping
There is no need to escape the double quotes:
slow_print('(If you want to guess, input \"I\". If I should guess, input \"U\".)')
This is simpler:
slow_print('(If you want to guess, input "I". If I should guess, input "U".)')
## Comments
The following comment is unnecessary because it merely repeats the name of the function:
#asks difficulty
def ask_guess_difficulty():
The same is true for many other comments in the code.
For better describe a function, it is common to use docstrings instead of comments. For example, change:
#plays the gamemode where the player guesses the computer's number
def player_guess(guesses):
to:
def player_guess(guesses):
""" plays the gamemode where the player guesses the computer's number """
## Return values
You can eliminate the guesses variable since it is not used elsewhere in the following if/elif code:
if guess_difficulty_lower == 'e':
guesses = 1000
return guesses
This is simpler:
if guess_difficulty_lower == 'e':
return 1000
The same applies to other elif cases.
## True
It is simpler to omit the comparison to True in the following code:
if guesses.isnumeric() == True:
Use:
if guesses.isnumeric():
## Lint check
pylint identified a few issues. There are some long lines that can be shortened.
The code uses inconsistent indentation. The black program can be used automatically modify the code.
## Naming
Variables named l and t are not too descriptive. Use longer names:
def slow_print(text):
for character in text:
sys.stdout.write(character)
• Thank you so much! This was incredibly helpful! Currently doing some research on docstrings! Commented May 21 at 14:17
• I'm using Python Anywhere, because I'm on a school Chromebook so I have to use a web compiler. The issue is, I cannot figure out how to run Black or pylint on it. Do you know how I could solve this, or where I could go to figure this out? Commented May 21 at 16:57
• @ADAMWELLER-FAHY: black and pylint are not necessary, but they are nice to have. You might be able to use this online Python formatter instead of black. I just found it by googling. Commented May 22 at 10:31
## Style
typing_speed is a constant (that is, a variable that is set and you never plan to change the value of while running the code), so most style guides recommend naming it in all caps, as TYPING_SPEED
Your first indentation is 3 spaces, while all your others are 4. Be consistent!
As toolic said, you don't need to escape your quotes. Python uses '' and "" for strings, and you can use the double quotes inside single quotes or vis versa without escaping. Only when you have both in a string do you need to escape (and even then you could use triple quotes)
I prefer f-strings over string concatenation. You can replace 'You\'ll have ' + str(guesses - 1) + ' guesses to get it.' with f"You'll have {guesses-1} guesses to get it."
You have a variable named guesses in ask_guess_difficulty(), except that the value it holds isn't actually the number of guesses, it's the number of guesses + 1. I can see why you did it, but there are better ways of handling your flow than to make your variables misrepresent what they are.
The number of guesses at each difficulty level should be pulled out as constants. That way, you can easily change the values without having to change the value in multiple places. Here's a simple change to ask_guess_difficulty() to use constants (changing nothing else, even if I complained about it)
EASY_COUNT = 1000
NORMAL_COUNT = 10
HARD_COUNT = 5
def ask_guess_difficulty():
slow_print('Alright! I\'ll pick a number from 1 to 100, and you try to guess it!')
slow_print(f'''You can do easy mode (E), with infinite guesses; normal mode (N), with {NORMAL_COUNT} guesses;
hard mode (H), with {HARD_COUNT} guesses; or custom mode (C), where you pick how many guesses you get!''')
guess_difficulty = input()
guess_difficulty_lower = guess_difficulty.lower()
if guess_difficulty_lower == 'e':
guesses = EASY_COUNT+1
return guesses
elif guess_difficulty_lower == 'n':
guesses = NORMAL_COUNT+1
return guesses
elif guess_difficulty_lower == 'h':
guesses = HARD_COUNT+1
return guesses
elif guess_difficulty_lower == 'c':
slow_print('How many guesses do you want?')
guesses = input()
if guesses.isnumeric() == True:
guesses = int(guesses) + 1
slow_print('Ok! You will have ', str(guesses - 1), ' guesses!')
return guesses
else:
slow_print('I don\'t understand. Please input a number for the amount of guess that you want.')
else:
slow_print('''I don\'t understand. Please input \"E\" for easy mode, \"N\" for
normal mode, \"H\" for hard mode, and \"C\" for custom mode''')
## Try/Except
You will find plenty of opinions on the value of using try/except blocks over if/then blocks, but your guesses.isnumeric() check is a classic example of where try/except is often recommended - it's a check that you expect to pass the majority of the time, and are only including to handle the exceptional case.
I'd replace
if guesses.isnumeric():
guesses = int(guesses) + 1
slow_print('Ok! You will have ', str(guesses - 1), ' guesses!')
return guesses
else:
slow_print('I don\'t understand. Please input a number for the amount of guess that you want.')
with
try:
guesses = int(guesses) + 1
except ValueError:
slow_print('I don\'t understand. Please input a number for the amount of guess that you want.')
else:
slow_print('Ok! You will have ', str(guesses - 1), ' guesses!')
return guesses
## Flow
Your flow in the main gameplay loop is a bit of a mess. The first thing you do is decrement guesses, but half the checks you do will force you to immediately reincrement it. You be better off doing the decrement later, after those checks are complete. You split your code on the low/high check, meaning that your victory check, which is the same either way, is needlessly duplicated. Your if checks all check for success, rather than failure, and then handle the failure state in the else portion. This means that the failure state is handled many lines down from the associated if check, and tracking which else is associated with each if is confusing. The else blocks are all short, and it is much better to reverse the ifs to handle them first. Additionally, doing it this way means you can combine them into an if elif else stack.
Here's how you could implement the flow more concisely. I'm also going to go and implement it so that the guesses argument is the total number of guesses, not the number of guesses + 1 as you have it implemented.
def player_guess(guesses):
#guesses = int(guesses) # This should have been validated before this
player_guess_win = False
slow_print('Ok! I\'m thinking of a number from 1 to 100')
slow_print(f"You'll have {guesses} guesses to get it.")
slow_print("Go ahead and guess when you're ready.")
total_tries = 0
player_guess_num = random.randint(1,100)
last_guess = None
low_high = None
while not player_guess_win and total_tries < guesses:
try:
current_guess = int(input())
except ValueError:
slow_print('I don\'t understand. Please enter a number from 1 to 100.')
continue
if current_guess > 100 or current_guess < 1:
slow_print('I don\'t understand. Please enter a number from 1 to 100.')
continue
if: current_guess == last_guess:
slow_print('That was your last guess silly! Make sure to enter a new number!')
continue
if: low_high == 'high' and current_guess < last_guess:
slow_print('Oops! Make sure you enter a number that is greater than your last guess!')
continue
if: low_high == 'low' and current_guess > last_guess:
slow_print('Oops! Make sure you enter a number that is less than your last guess!')
continue
# Valid Guess
total_tries += 1
if current_guess == player_guess_num:
slow_print('CORRECT!')
slow_print(f'Congratulations! You took {total_tries} guesses to get it right!')
slow_print('Do you want to play again? (Please input "yes" or "no".)')
while True:
yn_choice = input()
if yn_choice == 'yes' or yn_choice == 'y':
slow_print("Ok, let's do it!")
guess_the_number_start()
break
elif yn_choice == 'no' or yn_choice == 'n':
slow_print('Ok!')
exit()
else:
slow_print("""I don't understand. Please input "yes" or "no".""")
elif current_guess < player_guess_num:
slow_print('Too low.')
low_high = 'high'
elif current_guess > player_guess_num:
slow_print('Too high.')
low_high = 'low'
last_guess = current_guess
slow_print(f'You ran out of guesses. My number was {player_guess_num}.')
slow_print('Would you like to play again?')
slow_print('(Please input "Yes" or "No".)')
play_again_gtn = input()
play_again_gtn_lower = play_again_gtn.lower()
if play_again_gtn_lower == 'y' or play_again_gtn_lower == 'yes':
slow_print('Oki doki!')
elif play_again_gtn_lower == 'n' or play_again_gtn_lower == 'no':
slow_print('Oki doki! Bye!')
exit()
I think that a 63 line function with 4 levels of nesting is better a 130 line function with 8 levels of nesting, do you agree?
## Further improvements
You need string inputs in several places, and while you do handle invalid inputs, you only reprompt for correct input in one place. All of these places could be simplified by a good validation function.
def validate(options, prompt_hint=''):
"""
Takes input from the user, and returns the lowercase first character
of that input if that character is among the options.
options : str
a string of valid result characters. Example: 'yn'
prompt_hint : str
a message to the user if they don't input correctly.
Example: "Please input 'yes' or 'no'."
"""
while True:
choice = input().lower()[0]
if choice not in options:
slow_print(f"I don't understand. {prompt_hint}")
else:
return choice
With validate(), your ask_guess_difficulty() function looks like this:
EASY_COUNT = 1000
NORMAL_COUNT = 10
HARD_COUNT = 5
def ask_guess_difficulty():
slow_print('Alright! I\'ll pick a number from 1 to 100, and you try to guess it!')
slow_print(f'''You can do easy mode (E), with infinite guesses; normal mode (N), with {NORMAL_COUNT} guesses;
hard mode (H), with {HARD_COUNT} guesses; or custom mode (C), where you pick how many guesses you get!''')
guess_difficulty = validate('enhc', 'Please input "E" for easy mode, '
'"N" for normal mode, "H" for hard mode, and "C" for custom mode.')
if guess_difficulty == 'e':
guesses = EASY_COUNT
elif guess_difficulty == 'n':
guesses = NORMAL_COUNT
elif guess_difficulty == 'h':
guesses = HARD_COUNT
elif guess_difficulty == 'c':
slow_print('How many guesses do you want?')
while True:
try:
guesses = int(input())
except ValueError:
slow_print("I don't understand. Please input a number for the amount of guess that you want.")
else:
break
slow_print(f'Ok! You will have {guesses} guesses!')
return guesses
and player_guess() looks like this:
def player_guess(guesses):
player_guess_win = False
slow_print('Ok! I\'m thinking of a number from 1 to 100')
slow_print(f"You'll have {guesses} guesses to get it.")
slow_print("Go ahead and guess when you're ready.")
total_tries = 0
player_guess_num = random.randint(1,100)
last_guess = None
low_high = None
while not player_guess_win and total_tries < guesses:
try:
current_guess = int(input())
except ValueError:
slow_print('I don\'t understand. Please enter a number from 1 to 100.')
continue
if current_guess > 100 or current_guess < 1:
slow_print('I don\'t understand. Please enter a number from 1 to 100.')
continue
if: current_guess == last_guess:
slow_print('That was your last guess silly! Make sure to enter a new number!')
continue
if: low_high == 'high' and current_guess < last_guess:
slow_print('Oops! Make sure you enter a number that is greater than your last guess!')
continue
if: low_high == 'low' and current_guess > last_guess:
slow_print('Oops! Make sure you enter a number that is less than your last guess!')
continue
# Valid Guess
total_tries += 1
if current_guess == player_guess_num:
slow_print('CORRECT!')
slow_print(f'Congratulations! You took {total_tries} guesses to get it right!')
slow_print('Do you want to play again? (Please input "yes" or "no".)')
yn_choice = validate('yn', 'Please input "yes" or "no".')
if yn_choice == 'y':
slow_print("Ok, let's do it!")
guess_the_number_start()
break
elif yn_choice == 'n':
slow_print('Ok!')
exit()
elif current_guess < player_guess_num:
slow_print('Too low.')
low_high = 'high'
elif current_guess > player_guess_num:
slow_print('Too high.')
low_high = 'low'
last_guess = current_guess
slow_print(f'You ran out of guesses. My number was {player_guess_num}.')
slow_print('Would you like to play again?')
slow_print('(Please input "Yes" or "No".)')
play_again_gtn = validate('yn', 'Please input "yes" or "no".')
if play_again_gtn == 'y':
slow_print('Oki doki!')
elif play_again_gtn == 'n':
slow_print('Oki doki! Bye!')
exit()
def validate(options, prompt_hint=''):
"""
Takes input from the user, and returns the lowercase first character
of that input if that character is among the options.
options : str or None
a string of valid result characters. Example: 'yn'
prompt_hint : str
a message to the user if they don't input correctly.
Example: "Please input 'yes' or 'no'."
"""
while True:
choice = input().lower()[0]
if choice not in options:
slow_print(f"I don't understand. {prompt_hint}")
else:
return choice
EASY_COUNT = 1000
NORMAL_COUNT = 10
HARD_COUNT = 5
def ask_guess_difficulty():
slow_print('Alright! I\'ll pick a number from 1 to 100, and you try to guess it!')
slow_print(f'''You can do easy mode (E), with infinite guesses; normal mode (N), with {NORMAL_COUNT} guesses;
hard mode (H), with {HARD_COUNT} guesses; or custom mode (C), where you pick how many guesses you get!''')
guess_difficulty = validate('enhc', 'Please input "E" for easy mode, '
'"N" for normal mode, \"H\" for hard mode, and \"C\" for custom mode.')
if guess_difficulty == 'e':
guesses = EASY_COUNT+1
elif guess_difficulty == 'n':
guesses = NORMAL_COUNT+1
elif guess_difficulty == 'h':
guesses = HARD_COUNT+1
elif guess_difficulty == 'c':
slow_print('How many guesses do you want?')
while True:
try:
guesses = int(input())
except ValueError:
slow_print("I don't understand. Please input a number for the amount of guess that you want.")
else:
break
slow_print(f'Ok! You will have {guesses} guesses!')
return guesses
## Dictionary as an If/Else
There's one final python trick you can use in ask_guess_difficulty(), and that's using a dict in place of an if/else block. This only works when the if checks are fairly simple, and the output is simply assigning to a variable, but it works here. This is what I mean:
EASY_COUNT = 1000
NORMAL_COUNT = 10
HARD_COUNT = 5
def ask_guess_difficulty():
slow_print('Alright! I\'ll pick a number from 1 to 100, and you try to guess it!')
slow_print(f'''You can do easy mode (E), with infinite guesses; normal mode (N), with {NORMAL_COUNT} guesses;
hard mode (H), with {HARD_COUNT} guesses; or custom mode (C), where you pick how many guesses you get!''')
difficulty = validate('enhc', 'Please input "E" for easy mode, '
'"N" for normal mode, \"H\" for hard mode, and \"C\" for custom mode.')
guesses = {'e':EASY_COUNT, 'n':NORMAL_COUNT, 'h':HARD_COUNT, 'c':None}[difficulty]
if guesses is None:
slow_print('How many guesses do you want?')
while True:
try:
guesses = int(input())
except ValueError:
slow_print("I don't understand. Please input a number for the amount of guess that you want.")
else:
break
slow_print(f'Ok! You will have {guesses} guesses!')
return guesses
The special treatment for 'custom' has to be handled outside the dictionary, but since there was only one special case it's still worth using, and saves us about 5 lines. Whether this is preferable or not is situational and a matter of opinion, but it's a good trick to have in your back pocket. | 5,813 | 22,701 | {"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} | 2.546875 | 3 | CC-MAIN-2024-26 | latest | en | 0.884018 |
https://www.clutchprep.com/analytical-chemistry/practice-problems/149135/a-titration-of-100-ml-of-0-250-m-aqueous-solution-of-hcn-pka-9-310-is-titrated-w-1 | 1,618,591,375,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088245.37/warc/CC-MAIN-20210416161217-20210416191217-00360.warc.gz | 800,417,314 | 28,338 | # Problem: A titration of 100 mL of 0.250 M aqueous solution of HCN (pKa = 9.310) is titrated with a 0.500 M solution of LiOH. What is the pH after the addition of 25 mL of LiOH?
###### Problem Details
A titration of 100 mL of 0.250 M aqueous solution of HCN (pKa = 9.310) is titrated with a 0.500 M solution of LiOH. What is the pH after the addition of 25 mL of LiOH? | 128 | 373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-17 | latest | en | 0.782988 |
https://notatek.pl/uniwersytet-ekonomiczny-w-krakowie/49 | 1,498,603,711,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321938.75/warc/CC-MAIN-20170627221726-20170628001726-00431.warc.gz | 793,808,021 | 16,286 | # Uniwersytet Ekonomiczny w Krakowie - strona 49
## Measurement and modeling
• Uniwersytet Ekonomiczny w Krakowie
• dr Andrzej Sokołowski
• Polityka
Pobrań: 91
Wyświetleń: 1344
Measurement and modeling Mechanical - moving averages Moving averages We have to define the lenght of a moving average. Let's take it as 3. We take first 3 observations and calculate the average. The result is assigned to the time unit which is in the middle. Than we move forward, leave out the ...
## THE ANALYSIS OF MONTHLY DATA
• Uniwersytet Ekonomiczny w Krakowie
• dr Andrzej Sokołowski
• Polityka
Pobrań: 77
Wyświetleń: 1071
THE ANALYSIS OF MONTHLY DATA Data presentation Y - number of passangers of airline company (in thousands). Data has been taken from Box G.E.P., Jenkins G., “time series: analysis, forecasting and control”. graphs - 2D graphs - line plots (variables) Number of passangers is growing (we have tren...
## THE ANALYSIS OF QUARTERLY DATA
• Uniwersytet Ekonomiczny w Krakowie
• dr Andrzej Sokołowski
• Polityka
Pobrań: 84
Wyświetleń: 1071
THE ANALYSIS OF QUARTERLY DATA Data presentation Y - sales of a medicine cennected with a flu Graphs - 2D plots - line plots (Variables) Trend identification The nature of Y variable - since this medicine is good for flu, most probably sales are higher in winter time if this medicine is good w...
## Trend analysis on yearly data
• Uniwersytet Ekonomiczny w Krakowie
• dr Andrzej Sokołowski
• Polityka
Pobrań: 42
Wyświetleń: 1092
TREND ANALYSIS BASED ON YEARLY DATA (5.03) Data Trend identification The analysis of the problem done even without any data Graph ( graphs - 2d graphs - line plots (variables) ) Do we have a trend? Yes, the consumption is growing Testing trend coeficients Measurement and modeling Mechanical...
## Culture context of IHRM
• Uniwersytet Ekonomiczny w Krakowie
• dr Joanna Purada-Popiela
• Polityka
Pobrań: 21
Wyświetleń: 1274
ORGANIZATIONAL CULTURE - a pattern of shared basic assumptions that the group learned as it solved its problems of external adaptation and internal integration, that has worked well enough to be considered valid and therefore, to be taught to new members as the correct way to perceive, think, and...
## Performance Management
• Uniwersytet Ekonomiczny w Krakowie
• dr Joanna Purada-Popiela
• Polityka
Pobrań: 28
Wyświetleń: 1274
PERFORMANCE MANAGEMENT: a strategic and integrated approach to delivering success to organizations by improving performance of the teams and individuals. based on the principle of management by agreement rather than management by command. co...
## Reward Management
• Uniwersytet Ekonomiczny w Krakowie
• dr Joanna Purada-Popiela
• Polityka
Pobrań: 21
Wyświetleń: 1197
REWARD - contains all financial and non-financial components which employees get in accordance with the rate of the their job, skills, performance and market value. REWARD MANAGEMENT - a process which consists of designing, implementing and ...
## The role of international human resources
• Uniwersytet Ekonomiczny w Krakowie
• dr Joanna Purada-Popiela
• Polityka
Pobrań: 7
Wyświetleń: 875
The role of international human resources: part of organizational core competence source of competitive advantage „… the human resource refers to the accumulated stock of knowledge, skills and abilities that the individual possess, which the firm build up over time into an identifiable expertise”...
## Teorie Miedzynarodowych Stosunkow Gospodarczych - pojecia
• Uniwersytet Ekonomiczny w Krakowie
• Jan Szaja
• Polityka
Pobrań: 70
Wyświetleń: 959
TEORIE MIĘDZYNARODOWYCH STOSUNKÓW GOSPODARCZYCH (TEORIE MIĘDZYNARODOWEGO PODZIAŁU PRACY) NAKŁADY, ZASOBY - czynniki produkcji, które mogą być użyte do produkcji danego dobra (produktu lub usługi) zaspokajającego ludzkie potrzeby. Mamy nakłady pracy, ziemi, kapitału, a także nierzadko wymieniane j...
## Corporate Finance - pojecia
• Uniwersytet Ekonomiczny w Krakowie
• dr Katarzyna Mikołajczyk
• Polityka
Pobrań: 168
Wyświetleń: 1967
CORPORATE FINANCE - zagadnienia do egzaminu Finance (macro): study of financial institutions and markets and how they operate within the financial system, (micro): study of financial planning, asset mgmt, fund raising for businesses and financial institutions Financial system - intermediaries th... | 1,200 | 4,302 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-26 | longest | en | 0.572263 |
https://math.stackexchange.com/questions/1326635/continuity-of-partial-derivatives-only-along-their-axis | 1,653,447,431,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662578939.73/warc/CC-MAIN-20220525023952-20220525053952-00281.warc.gz | 435,243,827 | 66,605 | # Continuity of partial derivatives only along their axis?
My main question for which I will give an example right below is whether for a partial derivative to exist at a point (say $\frac{\partial f}{\partial x}$) it is necessary for it to be continuous at said point or if it needs only to exist when approaching from the $x$ axis. An example:
Consider the function $$f(x,y)=\frac{x^3+y^3}{x^2+y^2} \quad\forall\vec{x}\in\mathbb{R^2-\vec{0}},\quad f(\vec{0})=0$$ It's obvious it is continuous at $\vec{0}$ since a third degree polynomial goes faster to $0$ than the inverse of a second degree polynomial goes to infinity.
Lets study it's differentiability. For any point except $\vec{0}$ we have
$$\frac{\partial f}{\partial x}=\frac{(x^2+y^2)3x^2-(x^3+y^3)2x}{(x^2+y^2)^2}$$
$$\frac{\partial f}{\partial y}=\frac{(x^2+y^2)3y^2-(x^3+y^3)2y}{(x^2+y^2)^2}$$
Naturally the derivatives are not defined for $(x,y)=(0,0)$. If we want to find the derivative with respect to $x$ at $0$ we might try taking the limit to $\vec{0}$ of $\frac{\partial f}{\partial x}$. We change to polar coordinates ($x=r\cos\theta,y=r\sin\theta$) to see whether the limit exists (if it's unique or path-dependend). After simplyfying we get $$\lim_{r\rightarrow 0^+}\frac{\partial f}{\partial x}(r,\theta)=\cos^4\theta+3\cos^2\theta\sin^2\theta-2\sin^3\theta\cos\theta$$ We see then that it is path dependent since the value of the limit depends on $\theta$, therefore the limit doesn't exist.
If we don't look at the absolute limit but instead we come from the direction with respect to which we're taking the derivatives, then said limits exist:
$$\lim_{(x,0)\to(0,0)}\frac{\partial f}{\partial x}=1$$ $$\lim_{(0,y)\to(0,0)}\frac{\partial f}{\partial y}=1$$
So my question is, is $Df(0,0)=\begin{bmatrix}1&&1\end{bmatrix}$ the derivative of $f$ at $0$ even though the partial derivatives aren't the same from every direction, or is it that the derivative of $f$ doesn't exist at $(0,0)$?
• $Df$ isn't continuous around the origin, thus it doesn't exist there. You need $$\lim_{(x,y) \to (0,0)} \partial_x f \quad \text{and} \quad \lim_{(x,y) \to (0,0)} \partial_y f$$ to exist.
– Jeb
Jun 15, 2015 at 19:49
• How do you know it isn't continuous around the origin? Certainly the partial derivatives aren't, but the continuity of the partial derivatives isn't necessary for the existence of the total derivative. Jun 15, 2015 at 20:35
• $Df = ( \partial_x f , \partial_y f)$, You've shown that the limits do not exist already... In general, if you want the derivative to exist, the limit has to exist.
– Jeb
Jun 15, 2015 at 20:38
• @Jeb: No, Cristian is right, $C^1$ is a stronger condition than differentiability. Jun 15, 2015 at 21:13
The partial derivatives $\partial f/\partial x$ and $\partial f/\partial y$ exist at the origin $(0,0)$, not because of some limit of their values along the axes or otherwise, but directly from the definition of partial derivative: $$\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{h^3/h^2 - 0}{h} = 1 ,$$ and similarly for the other one.
To check whether $f$ is differentiable at the origin, use the definition of differentiability: $$\lim_{(h,k)\to (0,0)} \frac{f(h,k) - f(0,0) - A \cdot h - B \cdot k}{\sqrt{h^2+k^2}} \overset{?}{=} 0 ,$$ where $A = \frac{\partial f}{\partial x}(0,0) = 1$ and $B = \frac{\partial f}{\partial y}(0,0) = 1$. You get $$\lim_{(h,k)\to (0,0)} \frac{h^3+k^3 - (h+k)(h^2+k^2)}{(h^2+k^2)^{3/2}} ,$$ which doesn't exist, since you get different values when approaching from different directions. So $f$ is not differentiable at the point $(0,0)$, despite having partial derivatives w.r.t. $x$ and $y$ there. | 1,221 | 3,704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | latest | en | 0.841285 |
http://ncatlab.org/johnbaez/show/A+characterization+of+entropy+in+terms+of+information+loss | 1,448,767,119,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2015-48/segments/1448398455246.70/warc/CC-MAIN-20151124205415-00089-ip-10-71-132-137.ec2.internal.warc.gz | 162,640,480 | 20,370 | # John Baez A characterization of entropy in terms of information loss
## Idea
This is a draft of a paper by John Baez, Tobias Fritz and Tom Leinster. The finished version can be found here:
• John Baez, Tobias Fritz and Tom Leinster, A characterization of entropy in terms of information loss, on the arXiv or free online at Entropy 13 (2011), 1945-1957.
The paper was developed in a series of blog conversations, in this order:
For a lot of material that never got incorporated into the paper, see
and for another paper that never got finished, see:
# Contents
## Abstract
There are numerous characterizations of Shannon entropy and Tsallis entropy as measures of information obeying certain properties. Using work by Faddeev and Furuichi, we derive a very simple characterization. Instead of focusing on the entropy of a probability measure on a finite set, this characterization focuses on the ‘information loss’, or change in entropy, associated with a measure-preserving function. We show that Shannon entropy gives the only concept of information loss that is functorial, convex-linear and continuous. This characterization naturally generalizes to Tsallis entropy as well.
## Introduction
The Shannon entropy (S) of a probability measure $p$ on a finite set $X$ is given by:
$H(p) = - \sum_{i \in X} p_i \, \ln(p_i) .$
There are many theorems that seek to characterize Shannon entropy starting from plausible assumptions; see for example the book by Aczél and Daróczy (AD). Here we give a new and very simple characterization theorem. The main novelty is that we do not focus directly on the entropy of a single probability measure, but rather, on the change in entropy associated with a measure-preserving function. The entropy of a single probability measure can be recovered as the change in entropy of the unique measure-preserving function to the one-point space.
A measure-preserving function can map several points to the same point, but not vice versa, so this change in entropy is always a decrease. Since the second law of thermodynamics says that entropy always increases, this may seem counterintuitive. It may seem less so if we think of the function as some kind of data processing which does not introduce any additional randomness. Then the entropy can only decrease, and we can talk about the ‘information loss’ associated to the function.
Some examples may help to clarify this point. Consider the only possible map $f: \{a,b\} \to \{c\}$. Suppose $p$ is the probability measure on $\{a,b\}$ such that each point has measure $1/2$, while $q$ is the unique probability measure on the set $\{c\}$. Then $H(p) = \ln 2$, while $H(q) = 0$. The information loss associated to the map $f$ is defined to be $H(p) - H(q)$, which in this case equals $\ln 2$. In other words, the measure-preserving map $f$ loses one bit of information.
On the other hand, suppose $p'$ is the probability measure on $\{a,b\}$ such that $a$ has measure $1$ and $b$ has measure $0$. Then $H(p') = 0$, so with respect to this probability measure the map $f$ has information loss $H(p') - H(q) = 0$. It may seem odd to say that $f$ loses no information: after all, it maps $a$ and $b$ to the the same point. However, because the point $b$ has probability zero with respect to $p'$, knowing that $f(x) = c$ lets us conclude that $x = a$ with probability one.
The shift in emphasis from probability measures to measure-preserving functions suggests that it will be useful to adopt the language of category theory (M), where one has objects and morphisms between them. We will do this, although almost no category theory is required to read this paper.
Shannon entropy has a very simple characterization in terms of information loss. To state it, we consider a category where a morphism $f: p \to q$ is a measure-preserving function between finite sets equipped with probability measures. We assume $F$ is a function that assigns to any such morphism a number $F(f) \in [0,\infty)$, which we call its information loss. We also assume that $F$ obeys three axioms. If we call a morphism a ‘process’, we can state these roughly in words as follows:
1. Functoriality. Given a process consisting of two stages, the amount of information lost in the whole process is the sum of the amounts lost at each stage:
$F(f \circ g) = F(f) + F(g) .$
2. Convex linearity. If we flip a probability-$\lambda$ coin to decide whether to do one process or another, the information lost is $\lambda$ times the information lost by the first process plus $(1 - \lambda)$ times the information lost by the second:
$F(\lambda f \oplus (1 - \lambda) g) = \lambda F(f) + (1 - \lambda) F(g).$
3. Continuity. If we change a process slightly, the information lost changes only slightly: $F(f)$ is a continuous function of $f$.
(For full details see Section 2.) Given these assumptions, we conclude that there exists a constant $c \ge 0$ such that for any $f: p \to q$, we have
$F(f) = c(H(p) - H(q)) .$
The charm of this result is that none of the hypotheses hint at any special role for the function $- p \ln p$, but it emerges in the conclusion. The key here is a result of Faddeev (F) described in Section 3.
For many scientific purposes, probability measures are not enough. Our result extends to general measures on finite sets, as follows. Any measure on a finite set can be expressed as $\lambda p$ for some scalar $\lambda$ and probability measure $p$, and we define $H(\lambda p) = \lambda H(p)$. In this more general setting, we are no longer confined to taking convex linear combinations of measures. Accordingly, the convex linearity condition in our main theorem is replaced by two conditions: additivity ($F(f \oplus g) = F(f) + F(g)$) and homogeneity ($F(\lambda f) = \lambda F(f)$). As before, the conclusion is that, up to a multiplicative constant, $F$ assigns to each morphism $f: p \to q$ the information loss $H(p) - H(q)$.
It is natural to wonder what happens when we replace the homogeneity axiom $F(\lambda f) = \lambda F(f)$ by a more general homogeneity condition:
$F(\lambda f) = \lambda^\alpha\, F(f)$
for some number $\alpha \gt 0$. In this case we find that $F(f)$ is proportional to $H_\alpha(p) - H_\alpha(q)$, where $H_\alpha$ is the so-called Tsallis entropy of order $\alpha$.
## The main result
We work with finite sets equipped with probability measures. All measures on a finite set $X$ will be assumed nonnegative and defined on the $\sigma$-algebra of all subsets of $X$.
###### Definition
Let $\Fin\Prob$ be the category whose objects are finite sets equipped with probability measures and whose morphisms are measure-preserving functions.
Since any measure on a finite set is determined by its values on singletons, we will think of an object of $\Fin\Prob$ as a pair $(X,p)$ consisting of a finite set $X$ together with an $X$-tuple of numbers $p_i \in [0,1]$ satisfying $\sum p_i = 1$. A morphism $f: (X,p) \to (Y,q)$ in $\Fin\Prob$ is a function $f: X \to Y$ such that
$q_j = \sum_{i \in f^{-1}(j)} p_i .$
We will usually write an object $(X,p)$ as $p$ for short, and write a morphism $f: (X,p) \to (Y,q)$ as simply $f: p \to q$.
There is a way to take ‘convex linear combinations’ of objects and morphisms in $\Fin\Prob$. Let $(X, p)$ and $(Y, q)$ be finite sets equipped with probability measures, and let $\lambda \in [0, 1]$. Then there is a probability measure
$\lambda p \oplus (1 - \lambda) q$
on the disjoint union of the sets $X$ and $Y$, whose value at a point $k$ is given by
$(\lambda p \oplus (1 - \lambda) q)_k = \begin{cases} \lambda p_k &if k \in X\\ (1 - \lambda) q_k &if k \in Y. \end{cases}$
Given morphisms $f: p \to p'$ and $g: q \to q'$, there is a unique morphism
$\lambda f \oplus (1 - \lambda) g: \lambda p \oplus (1 - \lambda) q \to \lambda p' \oplus (1 - \lambda) q'$
that restricts to $f$ on the measure space $p$ and $g$ on the measure space $q$.
The same notation can be extended, in the obvious way, to convex combinations of more than two objects or morphisms. For example, given objects $p(1), \ldots, p(n)$ of $\Fin\Prob$ and nonnegative scalars $\lambda_1, \ldots, \lambda_n$ summing to $1$, there is a new object $\bigoplus_{i = 1}^n \lambda_i p(i)$.
Recall that the Shannon entropy of a probability measure $p$ on a finite set $X$ is
$H(p) = -\sum_{i \in X} p_i \, \ln(p_i) \in [0, \infty),$
with the convention that $0\,\ln(0) = 0$.
###### Theorem
Suppose $F$ is any map sending morphisms in $\Fin\Prob$ to numbers in $[0,\infty)$ and obeying these three axioms:
1. Functoriality:
(1)$F(f \circ g) = F(f) + F(g)$
whenever $f,g$ are composable morphisms.
2. Convex linearity:
(2)$F(\lambda f \oplus (1 - \lambda)g)$
= \lambda F(f) + (1 - \lambda) F(g) \label{convex_linearity} ] for all morphisms $f,g$ and scalars $\lambda \in [0, 1]$. 3. Continuity: $F$ is continuous.
Then there exists a constant $c \ge 0$ such that for any morphism $f : p \to q$ in $\Fin\Prob$,
$F(f) = c(H(p) - H(q))$
where $H(p)$ is the Shannon entropy of $p$. Conversely, for any constant $c \ge 0$, this formula determines a map $F$ obeying conditions 1-3.
We need to explain condition 3. A sequence of morphisms $(X_n, p(n)) \stackrel{f_n}{\to} (Y_n, q(n))$ in $\Fin\Prob$ converges to a morphism $(X, p) \stackrel{f}{\to} (Y, q)$ if
• for all sufficiently large $n$, we have $X_n = X$, $Y_n = Y$, and $f_n(i) = f(i)$ for all $i \in X$
• $p(n) \to p$ and $q(n) \to q$ pointwise.
We define $F$ to be continuous if $F(f_n) \to F(f)$ whenever $f_n$ is a sequence of morphisms converging to a morphism $f$.
The proof of Theorem 2 is given in a later section. First we show how to deduce a characterization of Shannon entropy for general measures on finite sets.
###### Definition
Let $\Fin\Meas$ be the category whose objects are finite sets equipped with measures and whose morphisms are measure-preserving functions.
There is more room for maneuver in $\Fin\Meas$ than in $\Fin\Prob$: we can take arbitrary nonnegative linear combinations of objects and morphisms, not just convex combinations. Any nonnegative linear combination can be built up from direct sums and multiplication by nonnegative scalars, defined as follows.
• For ‘direct sums’, first note that the disjoint union of two finite sets equipped with measures is another thing of the same sort. We write the disjoint union of $p, q \in \Fin\Meas$ as $p \oplus q$. Then, given morphisms $f : p \to p'$, $g : q \to q'$ there is a unique morphism $f \oplus g : p \oplus q \to p' \oplus q'$ that restricts to $f$ on the measure space $p$ and $g$ on the measure space $q$.
• For ‘scalar multiplication’, first note that we can multiply a measure by a nonnegative real number and get a new measure. So, given an object $p \in \Fin\Meas$ and a number $\lambda \ge 0$ we obtain an object $\lambda p \in \Fin\Meas$ with the same underlying set and with $(\lambda p)_i = \lambda p_i$. Then, given a morphism $f : p \to q$, there is a unique morphism $\lambda f : \lambda p \to \lambda q$ that has the same underlying function as $f$.
This is consistent with our earlier notation for convex linear combinations.
We wish to give some conditions guaranteeing that a map sending morphisms in $\Fin\Meas$ to nonnegative real numbers comes from a multiple of Shannon entropy. To do this we need to define the Shannon entropy of a finite set $X$ equipped with a measure $p$, not necessarily a probability measure. Define the total mass of $(X, p)$ to be
${\|p\|} = \sum_{i \in X} p_i .$
If this is nonzero, then $p$ is of the form ${\|p\|} \bar{p}$ for a unique probability measure space $\bar{p}$. In this case we define the Shannon entropy of $p$ to be ${\|p\|} H(\bar{p})$. If the total mass of $p$ is zero, we define its Shannon entropy to be zero.
We can define continuity for a map sending morphisms in $\Fin\Meas$ to numbers in $[0,\infty)$ just as we did for $\Fin\Prob$, and show:
###### Corollary
Suppose $F$ is any map sending morphisms in $\Fin\Meas$ to numbers in $[0,\infty)$ and obeying these four axioms:
1. Functoriality:
$F(f \circ g) = F(f) + F(g)$
whenever $f,g$ are composable morphisms.
(3)$F(f \oplus g) = F(f) + F(g)$
for all morphisms $f,g$.
3. Homogeneity:
(4)$F(\lambda f) = \lambda F(f)$
for all morphisms $f$ and all $\lambda \in [0,\infty)$.
4. Continuity: $F$ is continuous.
Then there exists a constant $c \ge 0$ such that for any morphism $f : p \to q$ in $\Fin\Meas$,
$F(f) = c(H(p) - H(q))$
where $H(p)$ is the Shannon entropy of $p$. Conversely, for any constant $c \ge 0$, this formula determines a map $F$ obeying conditions 1-4.
###### Proof
Take a map $F$ obeying the axioms listed here. Then $F$ restricts to a map on morphisms of $\Fin\Prob$ obeying the axioms of Theorem 2. Hence there exists a constant $c \ge 0$ such that $F(f) = c(H(p) - H(q))$ whenever $f: p \to q$ is a morphism between probability measures. Now take an arbitrary morphism $f: p \to q$ in $\Fin\Meas$. Since $f$ is measure-preserving, ${\|p\|} = {\|q\|} = \lambda$, say. If $\lambda \neq 0$ then $p = \lambda \bar{p}$, $q = \lambda \bar{q}$ and $f = \lambda \bar{f}$ for some morphism $\bar{f}: \bar{p} \to \bar{q}$ in $\Fin\Prob$; then by homogeneity,
$F(f) = \lambda F(\bar{f}) = \lambda c (H(\bar{p}) - H(\bar{q})) = c(H(p) - H(q)).$
If $\lambda = 0$ then $f = 0 f$, so $F(f) = 0$ by homogeneity. So $H(f) = c(H(p) - H(q))$ in either case. The converse statement follows from the converse in Theorem 2.
## Why Shannon entropy works
To solidify our intuitions, we first check that $F(f) = c(H(p) - H(q))$ really does determine a functor obeying all the conditions of Theorem 2. Since all these conditions are linear in $F$, it suffices to consider the case where $c = 1$. It is clear that $F$ is continuous, and equation (1) is also immediate whenever $g: m \to p$, $f: p \to q$, are morphisms in $\Fin\Prob$:
$F(f \circ g) = H(m) - H(q) = H(p) - H(q) + H(m) - H(p) = F(f) + F(g).$
The work is to prove equation (eq:convex_linearity).
We begin by establishing a useful formula for $F(f) = H(p) - H(q)$, where as usual $f$ is a morphism $p \to q$ in $\Fin\Prob$. Since $f$ is measure-preserving, we have
$q_j = \sum_{i \in f^{-1}(j)} p_i.$
So
\begin{aligned} \sum_j q_j \ln q_j &=& \sum_j \sum_{i \in f^{-1}(j)} p_i \ln q_j \\ &=& \sum_j \sum_{i \in f^{-1}(j)} p_i \ln q_{f(i)} \\ &=& \sum_i p_i \ln q_{f(i)} \end{aligned}
where in the last step we note that summing over all points $i$ that map to $j$ and then summing over all $j$ is the same as summing over all $i$. So,
\begin{aligned} F(f) &=& - \sum_i p_i\ln p_i + \sum_j q_j \ln q_j \\ &=& \sum_i ( -p_i \ln p_i + p_i \ln q_{f(i)}) \end{aligned}
and thus
(5)$F(f) = \sum_{i \in X} p_i \ln \frac{q_{f(i)}}{p_i}$
where the quantity in the sum is defined to be zero when $p_i = 0$. If we think of $p$ and $q$ as the distributions of random variables $x \in X$ and $y \in Y$ with $y = f(x)$, then $F(f)$ is exactly the conditional entropy of $x$ given $y$. So, what we are calling ‘information loss’ is a special case of conditional entropy.
This formulation makes it easy to check equation (eq:convex_linearity):
$F (\lambda f \oplus (1 - \lambda)g) = \lambda F(f) + (1 - \lambda) F(g).$
In the proof of Corollary 4 (on $\Fin\Meas$), the fact that $F(f) = c(H(p) - H(q))$ satisfies the four axioms was deduced from the analogous fact for $\Fin\Prob$. It can also be checked directly. For this it is helpful to note that
(6)$H(p) = {\|p\|}\,\ln{\|p\|} - \sum_i p_i\,\ln(p_i).$
It can then be shown that equation (5) holds for every morphism $f$ in $\Fin\Meas$. The additivity and homogeneity axioms follow easily.
To prove the hard part of Theorem 4, we use a characterization of entropy given by Faddeev (F) and nicely summarized at the beginning of a paper by Rényi (R). In order to state this result, it is convenient to write a probability measure on the set $\{1, \dots, n\}$ as an $n$-tuple $p = (p_1, \dots, p_n)$. With only mild cosmetic changes, Faddeev’s original result states:
###### Theorem
(Faddeev) Suppose $I$ is a map sending any probability measure on any finite set to a nonnegative real number. Suppose that:
1. $I$ is invariant under bijections.
2. $I$ is continuous.
3. For any probability measure $p$ on a set of the form $\{1, \dots, n\}$, and any number $0 \le t \le 1$,
(7)$I((t p_1, (1-t)p_1, p_2, \dots, p_n)) = p_1 I((t, 1-t)) + I((p_1, \dots, p_n))$
Then $I$ is a constant nonnegative multiple of Shannon entropy.
In item 1 we are using the fact that given a bijection $f: X \to X'$ between finite sets and a probability measure on $X$, there is a unique probability measure on $X'$ such that $p$ is measure-preserving; we demand that $I$ takes on the same value on both these probability measures. In item 2, we use the standard topology on the simplex
$\Delta^{n-1} = \left\{(p_1,\ldots,p_n) \in \mathbb{R}^n \:|\: p_i \geq 0, \: \sum_i p_i = 1 \right\}$
to put a topology on the set of probability distributions on any $n$-element set.
The most interesting and mysterious condition in Faddeev’s theorem is item 3. This is a special case of a general law appearing in the work of Shannon (S) and Faddeev (F). Namely, suppose $p$ is a probability measure on the set $\{1,\dots,n \}$. Suppose also that for each $i \in \{1, \ldots, n\}$, we have a probability measure $q(i)$ on a finite set $X_i$. Then $p_1 q(1) \oplus \cdots \oplus p_n q(n)$ is again a probability measure space, and the Shannon entropy of this space is given by:
$H\left(p_1 q(1) \oplus \cdots \oplus p_n q(n)\right) = H(p) + \sum_{i=1}^n p_i H(q(i)) .$
To see this, write $q_{i j}$ for the value of $q(i): X_i \to [0,\infty)$ at the point $j \in X_i$:
\begin{aligned} H(p_1 q(1) \oplus \cdots \oplus p_n q(n)) &=& -\sum_{i=1}^n \sum_{j \in X_i} p_i q_{i j} \ln(p_i q_{i j}) \\ &=& -\sum_{i=1}^n \sum_{j \in X_i} p_i (q_{i j} \ln(p(i) + q_{i j} \ln(q_{i j})) \\ &=& \sum_{i=1}^n p_i \left( -\ln(p_i) + H(q(i)) \right) \\ &=& H(p) + \sum_{i=1}^n p_i H(q(i)). \end{aligned}
Moreover, this formula is almost equivalent to condition 3 in Faddeev’s theorem, allowing us to reformulate Faddeev’s theorem as follows:
###### Theorem
Suppose $I$ is a map sending any probability measure on any finite set to a nonnegative real number. Suppose that:
1. $I$ is invariant under bijections.
2. $I$ is continuous.
3. $I((1)) = 0$, where $(1)$ is our name for the unique probability measure on the set $\{1\}$.
4. For any probability measure $p$ on the set $\{1,\dots, n\}$ and probability measures $q(1),\ldots,q(n)$ on finite sets, we have
$I(p_1 q(1) \oplus \cdots \oplus p_n q(n)) = I(p) + \sum_{i=1}^n p_i I(q(i)) .$
Then $I$ is a constant nonnegative multiple of Shannon entropy. Conversely, any constant nonnegative multiple of Shannon entropy satisfies 1-4.
###### Proof
We just need to check that conditions 3 and 4 imply Faddeev’s equation (7). Take $p = (p_1, \ldots, p_n)$, $q_1 = (t, 1 - t)$ and $q_i = (1)$ for $i \geq 2$: then condition 4 gives
$I((t p_1, (1 - t)p_1, p_2, \ldots, p_n)) = I((p_1, \ldots, p_n)) + p_1 I((t, 1 - t)) + \sum_{i = 2}^n p_i I((1))$
which by condition 3 gives Faddeev’s equation.
It may seem miraculous how the formula
$I(p_1, \dots, p_n) = - c \sum_i p_i \ln p_i$
emerges from the assumptions in either Faddeev’s original Theorem 5 or the equivalent Theorem 6. We can demystify this by describing a key step in Faddeev’s argument, as simplified by Rényi (R). Suppose $I$ is a function satisfying the assumptions of Faddeev’s result. Let
$f(n) = I(\frac{1}{n} , \dots, \frac{1}{n})$
be the function $I$ applied to the uniform probability measure on an $n$-element set. Since we can write a set with $n m$ elements as a disjoint union of $m$ different $n$-element sets, assumption 4 of Theorem 6 implies that
$f(n m) = f(n) + f(m).$
Rényi shows that the only solutions of this equation obeying
$\lim_{n \to \infty} (f(n+1) - f(n)) = 0$
are
$f(n) = c \ln n .$
This is how the logarithm function enters. Using condition 3 of Theorem 5, or equivalently conditions 3 and 4 of Theorem 6, the value of $I$ can be deduced for all probability measures $p$ such that each $p_i$ is rational. The result for arbitrary probability measures follows by continuity.
## Proof of the main result
Now we complete the proof of Theorem 2. Assume that $F$ obeys conditions 1-3 in the statement of this theorem.
Recall that $(1)$ denotes the set $\{1\}$ equipped with its unique probability measure. For each object $p \in \Fin\Prob$, there is a unique morphism
$!_p : p \to (1).$
We can think of this as the map that crushes $p$ down to a point and loses all the information that $p$ had. So, we define the ‘entropy’ of the measure $p$ by
$I(p) = F(!_p) .$
Given any morphism $f: p \to q$ in $\Fin\Prob$, we have
$!_p = !_q \circ f.$
So, by our assumption that $F$ is functorial,
$F(!_p) = F(!_q) + F(f),$
or in other words:
(8)$F(f) = I(p) - I(q) .$
To conclude the proof, it suffices to show that $I$ is a multiple of Shannon entropy.
We do this by using Theorem 6. Functoriality implies that when a morphism $f$ is invertible, $I(f) = 0$. Together with (8), this gives condition 1 of Theorem 6. Since $!_{(1)}$ is invertible, it also gives condition 3. Condition 2 is immediate. The real work is checking condition 4.
Take a probability measure $p$ on $\{1, \ldots, n\}$, and probability measures $q(1), \ldots, q(n)$ on finite sets $X_1, \ldots, X_n$ respectively. Then we obtain a probability measure
$\bigoplus_i p_i q(i)$
on the disjoint union of $X_1, \ldots, X_n$. Now, we can decompose $p$ as a direct sum:
(9)$p\cong \bigoplus_i p_i ((1)).$
Define a morphism
$f = \bigoplus_i p_i !_{q(i)} \colon \bigoplus_i p_i q(i) \to \bigoplus p_i ((1)).$
Then by convex linearity and definition of $I$,
$F(f) = \sum_i p_i F(!_{q(i)}) = \sum_i p_i I(q(i)).$
But also
$F(f) = I\bigl(\bigoplus_i p_i q(i)\bigr) - I\bigl(\bigoplus p_i ((1))\bigr) = I\bigl(\bigoplus_i p_i q(i)\bigr) - I(p)$
by (8) and (9). Comparing these two expressions for $F(f)$ gives condition 4 of Theorem 6, completing the proof of Theorem 2.
## A characterization of Tsallis entropy
Since Shannon defined his entropy in 1948, it has been generalized in many ways. Our Theorem 4 can easily be extended to characterize one family of generalizations, the so-called ‘Tsallis entropies’. For any positive real number $\alpha$, the Tsallis entropy of order $\alpha$ of a probability measure $p$ on a finite set $X$ is defined by:
$H_\alpha(p) = \begin{cases} \frac{1}{\alpha - 1} \Bigl( 1 - \sum_{i \in X} p_i^\alpha \Bigr) &if \alpha \neq 1 \\ - \sum_{i \in X} p_i ln p_i &if \alpha = 1. \end{cases}$
The peculiarly different definition when $\alpha = 1$ is explained by the fact that the limit $\lim_{\alpha \to 1} H_\alpha(p)$ exists and equals the Shannon entropy $H(p)$.
Although these entropies are most often named after Tsallis (T), they and related quantities had been studied by others long before the 1988 paper in which Tsallis first discussed it. For example, Havrda and Charvát (HC) had already introduced a similar formula, adapted to base 2 logarithms, in a 1967 paper in information theory, and in 1982, Patil and Taillie (PT) had used $H_\alpha$ itself as a measure of biological diversity.
The characterization of Tsallis entropy is exactly the same as that of Shannon entropy except in one respect: in the convex linearity condition, the degree of homogeneity changes from $1$ to $\alpha$.
###### Theorem
Let $\alpha\in(0,\infty)$. Suppose $F$ is any map sending morphisms in $\Fin\Prob$ to numbers in $[0,\infty)$ and obeying these three axioms:
1. Functoriality:
$F(f \circ g) = F(f) + F(g)$
whenever $f,g$ are composable morphisms.
2. Compatibility with convex combinations:
$F(\lambda f \oplus (1-\lambda) g) = \lambda^{\alpha} F(f) + (1-\lambda)^{\alpha} F(g)$
for all morphisms $f,g$ and all $\lambda \in [0,1]$.
3. Continuity: $F$ is continuous.
Then there exists a constant $c \ge 0$ such that for any morphism $f : p \to q$ in $\Fin\Meas$,
$F(f) = c(H_{\alpha}(p) - H_{\alpha}(q))$
where $H_{\alpha}(p)$ is the order $\alpha$ Tsallis entropy of $p$. Conversely, for any constant $c \ge 0$, this formula determines a map $F$ obeying conditions 1-3.
###### Proof
The proof is exactly the same as that of Theorem 2, except that instead of using Faddeev’s theorem, we use Theorem V.2 of Furuichi (Fu). Furuichi’s theorem is itself the same as Faddeev’s, except that condition 3 of Theorem 5 is replaced by
$I((t p_1, (1-t)p_1, p_2, \dots, p_n)) = p_1^\alpha I((t, 1-t)) + I((p_1, \dots, p_n))$
and Shannon entropy is replaced by Tsallis entropy of order $\alpha$.
As in the case of Shannon entropy, this result can be extended to arbitrary measures on finite sets. For this we need to define the Tsallis entropies of an arbitrary measure on a finite set. We do so by requiring that
$H_\alpha(\lambda p) = \lambda^\alpha H_\alpha(p)$
for all $\lambda \in [0, \infty)$ and all $p \in \Fin\Meas$. When $\alpha = 1$ this is the same as the Shannon entropy, and when $\alpha \neq 1$, it can be rewritten explicitly as
$H_\alpha(p) = \frac{1}{\alpha - 1} \Bigl( \Bigl( \sum_{i \in X} p_i \Bigr)^\alpha - \sum_{i \in X} p_i^\alpha \Bigr)$
(which is analogous to (6)). The following result is the same as Theorem 4 except that, again, the degree of homogeneity changes from $1$ to $\alpha$.
###### Corollary
Let $\alpha \in (0, \infty)$. Suppose $F$ is any map sending morphisms in $\Fin\Meas$ to numbers in $[0,\infty)$, and obeying these four properties:
1. Functoriality:
$F(f \circ g) = F(f) + F(g)$
whenever $f,g$ are composable morphisms.
$F(f \oplus g) = F(f) + F(g)$
for all morphisms $f,g$.
3. Homogeneity of degree $\alpha$:
$F(\lambda f) = \lambda^\alpha F(f)$
for all morphisms $f$ and all $\lambda \in [0,\infty)$.
4. Continuity: $F$ is continuous.
Then there exists a constant $c \ge 0$ such that for any morphism $f : p \to q$ in $\Fin\Meas$,
$F(f) = c(H_\alpha(p) - H_\alpha(q))$
where $H_\alpha$ is the Tsallis entropy of order $\alpha$. Conversely, for any constant $c \ge 0$, this formula determines a map $F$ obeying conditions 1-4.
###### Proof
This follows from Theorem 7 in just the same way that Corollary 4 follows from Theorem 2.
## Acknowledgements
We thank the denizens of the $n$-Category Café, especially David Corfield, Steve Lack, Mark Meckes and Josh Shadlen, for encouragement and helpful suggestions. Leinster is supported by an EPSRC Advanced Research Fellowship.
## References
• D. K. Faddeev, On the concept of entropy of a finite probabilistic scheme, Uspehi Mat. Nauk (N.S.) 11 (1956), no. 1(67), pp. 227–231. (In Russian.) German Translation: Zum Begriff der Entropie eines endlichen Wahrscheinlichkeitsschemas, Arbeiten zur Informationstheorie I, Berlin, Deutscher Verlag der Wissenschaften, 1957, pp. 85-90.
• J. Havrda and F. Charvát, Quantification method of classification processes: concept of structural $\alpha$-entropy, Kybernetika 3 (1967), 30–35.
• S. Mac Lane, Categories for the Working Mathematician, Graduate Texts in Mathematics 5, Springer, 1971.
• G. P. Patil and C. Taillie, Diversity as a concept and its measurement, Journal of the American Statistical Association 77 (1982), 548–561.
• C. Tsallis, Possible generalization of Boltzmann–Gibbs statistics, Journal of Statistical Physics 52 (1988), 479–487.
Revised on October 31, 2013 16:52:59 by Todd Trimble? (67.81.95.215) | 8,382 | 27,467 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 400, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2015-48 | latest | en | 0.91765 |
https://expertinstudy.com/t/U9kC1QUsj | 1,653,518,952,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00099.warc.gz | 308,010,925 | 4,527 | # Find the averrage of 11,9,10and 30
The time for answering the question is over
383 cents
The average of given data is 15
average=
=60/4
=15
so the average is 15
426 | 56 | 174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-21 | latest | en | 0.882506 |
http://qs1969.pair.com/~perl2/?node_id=134427 | 1,680,236,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949533.16/warc/CC-MAIN-20230331020535-20230331050535-00499.warc.gz | 40,731,064 | 6,645 | Do you know where your variables are? PerlMonks
### Re: Help w/ Code Optimization
on Dec 26, 2001 at 21:41 UTC ( #134427=note: print w/replies, xml ) Need Help??
in reply to Help w/ Code Optimization
Hi ;-).
Let me throw around my math skills... Recalling some binary math I figured that 10^2 (10 to the power of 2) may be simplified into this:
10^2 = 10x10 = 10x(2x2x2+2).
Notice all the 2's there? Here's where the left shift operator '<<' comes in handy (and it's pretty fast by the way).
So, every multiplication by 2 could be replaced by a left shift by one (in binary it's equivalent to multiplying by 2 ;) like this:
```10^2 = 10<<3 + 10<<1; (by the way, this is may not be
written as 10<<4! :)
So, I've replaced 10x10 by a few left shift operators. The key here is to determine how many left shifts will have to be performed for given power. Say, if you were to raise 10 to the power of 3, you'd look down at this:
10x10x10.
From this you'll also note that 10x10 is really (10<<3 + 10<<1) therefore, each 'x10' could be replaced accordingly.
Hopefully I've given you some food for thought ;-). I"ll try to look for a few ways to guess the left shift number for you. Evidently, it depends on the number being multiplied. My thinking is that left shift should work faster than multiplication, althought, folks who've developed Perl might have taken that into consideration... ;-))
"There is no system but GNU, and Linux is one of its kernels." -- Confession of Faith
Replies are listed 'Best First'.
Re: Re: Help w/ Code Optimization
by John M. Dlugosz (Monsignor) on Dec 27, 2001 at 02:08 UTC
Once upon a time, that was a good idea for PC's. On a 80486, where shift took 1 clock and multiply took up to 40, it was still faster sometimes. Note I said "up to" because it stops early, and essentially does this same thing internally--one addition for each '1' bit, using a barrel shifter that skips the zeros.
But with Pentium and later, the multiply is just as fast as anything else, and breaking it up into multiple steps makes it slower, by far.
—John
Re: Re: Help w/ Code Optimization
by sifukurt (Hermit) on Dec 26, 2001 at 23:10 UTC
I like this idea a lot. My only problem, though, is that the exponent won't always be an integer. As a very simple example, sticking with left shift operator idea, would there be a relatively easy way to do 10**(1/3)? I still get a little confused with the whole << thing.
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Notices? | 793 | 2,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-14 | latest | en | 0.960907 |
http://slideplayer.com/slide/4385372/ | 1,506,242,963,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689900.91/warc/CC-MAIN-20170924081752-20170924101752-00646.warc.gz | 319,180,048 | 36,342 | ## Presentation on theme: "TBAD Training; Gemini Observatory"— Presentation transcript:
Theory, Operation, Integration, Data, Interpretation, and more Tom Murphy; UCSD
TBAD Components discriminator decoder TSIM patch antenna
phased array (7 patches)
What TBAD Does Detects RF power around 1090 MHz (passive listener)
can react to planes at > 200 km if we let it besides Mode-A/C, “hears” Mode-S and some DME signals Scans for characteristic RF pulse patterns Decodes pulse pattern to octal code (altitude/ident) Determines if signals exceed thresholds Determines if array/patch ratio indicates source is “in-beam” Applies windowing on beam criterion to reduce false triggers Controls shutter via TTL (5 V to open) Streams serial data: codes and conditions Can respond to rates up to 250 Hz Includes various health-monitoring failsafes
What TBAD Does NOT Do TBAD does not actively interrogate (no transmissions) TBAD cannot ascertain angular position just YES/NO on question: “is something within about 12°?” TBAD does not get flight number, lat/lon, speed, heading, etc.
What Signals Be There? Mode-A (ident) and Mode-C (altitude) use same pulse code First and last framing pulses (F1, F2), X bit, 12 bits of data 12 bits of data 4096 possible codes (1280 of these map to altitudes) 12 bits broken into 4 3-bit (octal) digits, ABCD 450 ns per pulse, 1 μs gap between (21 μs total duration) Pilot sets ID code according to ATC instruction altitude encoded from altimeter automatically
Example Pulse Pattern See F1, C1, C2, A4, B1, B4, F2 4530 code
F1 C1 A1 C2 A2 C4 A4 X B1 D1 B2 D2 B4 D4 F2 See F1, C1, C2, A4, B1, B4, F2 4530 code could mean 3400 feet note: pulses stand out 3 orders-of-magnitude above noise floor
Not Easily Confused F1 C1 A1 C2 A2 C4 A4 X B1 D1 B2 D2 B4 D4 F2 F1 C1 A1 C2 A2 C4 A4 X B1 D1 B2 D2 B4 D4 F2 Two airplanes at once: one in-beam (1362), one out (4650) note reversal of which (DIREC/OMNI) is stronger difference signal (top, magenta) separates cleanly if difference exceeds 5 dB for any pulse, deemed beam-worth
TBAD Decoding Trick It’s pretty basic:
data stream composed of comparator output from input signal against threshold could be DIREC signal against DT threshold or above signal OR-combined with OMNI signal against OT threshold every few clock cycles (pre-set), decoder looks at data stream if low-to-high transition has occurred since last check, that bit is activated So fixed timing, based on Mode-A and Mode-C pattern Mode-S does not follow same timing, so improper decode, albeit recognizable
Interrogation Transmitted at 1030 MHz from ground or other aircraft
called “secondary radar” many airplanes carry traffic collision avoidance system (TCAS) chatter goes like n2, where n is aircraft density Timing of P3 pulse determines whether Mode-A or Mode-C interrogator can piece together responses from primary radar and time of flight but TBAD has no idea whether a particular code is Mode-A or C
Code Duplicity 4096 total codes
1280 altitudes from −1200 to in steps of 100 ft Random I.D. (Mode-A) has 31% chance of alt. association Arranged in Gray Code (one bit change each step) example table: ALTITUDE A1 A2 A4 B1 B2 B4 C1 C2 C4 D1 D2 D4 CODE often discernible by context: 96700 likely not actual altitude; very likely is
Special codes Some codes are reserved for special/emergency meaning
0000: sometimes see military use this 0100−0400: special purpose (often military, maybe oceanic?) 1200: VFR (visual flight rules): non-commercial, no ATC 7000: can also mean VFR 7500: hijack code 7600: lost communications 7700: emergency 7777: military intercept Lots more at Wikipedia: Transponder_(aeronautics) Note: TSIM avoids these codes because: A2 is disabled, so A=0,1,4,5 only (no 7XXX codes possible) D1=1 always, so ends in D=1,3,5,7 only
Mode-S High-density data transmissions at 1090 MHz
56 or 112 μs High-density data transmissions at 1090 MHz Always starts with 8 μs 4-pulse preamble at 0.0, 1.0, 3.5, 4.5 μs preamble impacts TBAD decode of digits A and C 5x3x codes Then loads of transitions on 1 μs clock most bits in B, D lit up, as well as X bit Looks to TBAD like 5737, 5735, 5537, often with X present
DME Transmissions Distance Measuring Equipment
Range of frequencies, but overlap with 1090 MHz Two broad humped pulses; powerful; 12 μs apart often above threshold when Mode-A/C still too weak thus often see prevalence at beginning and end of pass First pulse interpreted as F1 (first framing) by default/design then one other “transition” most often interpreted as B1 0100 code no “X” bit, no final framing pulse
Each patch has a broad beam (above) The array pattern has strong central peak The array pattern also has sidelobes/nulls Only in the center is the array pattern stronger than the single patch (element) Requirement that array signal exceeds single-patch signal by 5 dB results in ~15° “beam” zone Refer to array/patch as DIREC/OMNI
Lobes and Sidelobes Beam patterns for single patch (blue) and array (green) note that sidelobes always weaker than single patch, except main lobe
Beam Cuts: Raw Gain
Beam Cuts: Ratio is the Key
thresh. at 5 dB gives ~15° beam & noise immun.
RF from antenna hits low-pass filter, 37.5 dB amplifier, narrow-band filter, power detector Logarithmic power detector: V = 0.63 – ×PdBm inverse behavior: stronger = lower voltage no power 2 V; high power saturates at 0.55 V 0.1 V per 4 dB: voltage difference is power ratio
Example Power Detector Output
Signal sits at 2.0 V; pulses show up as negative spikes Here see two aircraft with overlapping signals one is “in beam”—strong DIREC; one is out of beam (weak DIREC) TBAD has no problem identifying in-beam threat
All based on power detector outputs from DIREC and OMNI Does difference between two (thus ratio) indicate “in-beam”? BT: Beam Threshold Is DIREC signal strong enough that we even care? DT: DIREC Threshold Is DIREC signal so strong that we risk saturation? DS: DIREC Saturation Is OMNI signal so strong that we risk saturation? OS: OMNI Saturation Is OMNI signal strong enough to warrant logging event? OT: OMNI Threshold Is quiescent DIREC output > [1.8 V] most of time? DC: DIREC Clear
Recurrent Pattern in Electronic Design
Potentiometer setting buffered; compared against power detector voltage inverted output usually chosen since pulses are inverted
In-Beam Decision Nuances
Must “simultaneously” satisfy BT and DT thresholds thus ignore too-distant sources in beam (low elevation angles) DT trips creation of window, in which BT will be considered purpose is test early part of pulse, before multipath distortions window is nominally something like 50−100 ns, tunable window can be delayed to allow time const./settling; nominally 40 ns Can also require that DC was low (no signal) prior to DT crossing called “lookback,” and can be set to 100, 80, 60, 40 ns in past can even use two lookbacks at once for even more stringency DME signals defeated by this due to slower rise of pulse Additional layer to ignore sporadic “beam” events via NB knob: require N Beam events in last ten seconds
The “W” Problem Multipath interference can mess up the pulse shape
full destructive zero power; full constructive adds 3 dB (75 mV) first portion taken to be “correct,” before multipath kicks in We noticed this when some said our country had a “W” problem as well
Pulse Window The DIREC signal (blue) triggers a short window (cyan; here 50 ns), during which the difference (magenta) is considered Ideal to have window duration and delay just catch the first stable part of difference, after time constants play out
Expected Power Detector Response
Transponders for commercial planes must have peak power between 125 and 500 W Antenna gains, amplification, attenuation, and geometry then determine power detector output The power, at the detector, in dBm (1 mW is 0 dBm): PD = PT + GT + GR + AR − 22.0 − 20log10(R/λ) where P, G, and A are power, gain, amplif., resp., D, T, R are detector, transmitter, receiver, R is the distance, and λ wavelength (0.275 m) PT is 51 dBm for 125 W, or 57 dBm for 500 W GT is about 5 dB for a dipole antenna under a fuselage GR is 6 dB for OMNI, 16 for DIREC at center; 11 for DIREC at beam edge AR is about 33 dB for TBAD Example: 125 W at 30 km gives −23 dBm, or about 1.2 V on P.D. To meet DT=1.3 V, the plane must be closer than 51.5 km
Practical vs. Theoretical
Losses in cables, connectors, etc. add up, and may be responsible for an observed ~2× (−3 dB) signal deficit also partially used 8:1 summer plays role at 49/64 = −1.1 dB level Thus calculated thresholds should be adjusted accordingly My recommendation is 0.1 V adjustments 4 dB is slightly more than the observed factor of two Example: protect airplane at 45,000 ft from 14,000 ft summit at elevation 25° slant range = 22.4 km calculate −20.2 dB = 1.12 V for 125 W at edge of DIREC beam (11 dB) might then operate at DT=1.22 V; 1.25 V fine; 1.30 V not unreasonable downside of overcompensation becomes more false triggers
The Donut Hole A dipole antenna has a null along the antenna direction
a plane near zenith thus has reduced transmission power downward compensated by much smaller slant range for beam determination Real antenna under real fuselage will fill in hole somewhat realistic worst case may be 10× reduction in transmitted power given inverse square, this corresponds to 3.2× effective range diff Geometry makes up 3.2× at 18° elevation sin(18.4°) = 1/sqrt(10) plane 10× stronger at zenith than at 18°, absent donut hole So setting up to protect to 15° elevation automatically allows for reasonably deep (10×) donut hole 45 kft from 14 kft at 15° is 36.5 km; calculate −24.5 dBm; 1.22 V could operate at 1.30 or 1.35 V for DT and be covered Protecting to 15° allows operation at 25° to still see “in-beam” planes 10° below center (less explaining to do for “misses”)
The Donut Hole and “Saturation” Events
Originally intended to guard against actual power detector saturation if DIREC saturates, ratio is reduced; may not trigger BT secondary benefit in warning against close planes Reliance on close-in protection needs to honor apparent factor of two system loss and donut hole OS = 0.85 V −9 dBm at detector 3.5 km nominally but 2× loss and 10× donut hole means 775 m vertical (2500 ft) DS = 0.85 V 6.2 km at DIREC beam edge (11 dB) apply 2× loss and 10× donut and get 1.4 km vertical (4500 ft) Need to map shape of donut to verify campaign at UCSD to do this
TBAD Checkout Each unit ships with completed checkout sheet
meant to run through all core TBAD functions successful checkout should translate to “perfect” operation leaves little room for surprises in sky test Need: TSIM, scope, DVM, splitter (6 dBm, nominally), 0−10× variable attenuator, serial monitor we also have a custom-built source/sink current tester also separate checkout for TSIM, requiring functional TBAD Once familiar, takes ~1.5 hours to check out we can go through a checkout, if that’s useful
Sky Test Procedure Rooftop tests at UCSD; emphasis on high flyers
many pass through az. 220°, elev. 45° between Mexico and SFO, SEA seek visual confirmation about point of closest approach (PCA) logsheet to note time, pointing, and Flight Aware info block also TBAD altitude, ID code, notes, and PCA details validate proper shutter behavior in high-traffic environment (shutter closed large fraction of the time) flight aware screen capture here
Example Test Pass in beam (BT, etc.) fraction in beam (pegs at 100%)
not in beam (omni log; OT)
Others Look Great, Too OMNI log disabled climbing through 25,400 ft
Mauna Kea Translation At ~100 Hz chatter and NB = 20 events/10 sec, UCSD tests differ from Mauna Kea environment Use program to first construct summary record from log # _xp.log: squawk 2147, alt 32000, PCA 00:20:22 # time code alrm alt N1 O1 D1 B1 N10 B10 1N 1O 1D 1B 10N 10B 00:20: 00:20: B 00:20: 00:20: B 00:20: B 00:20: B 00:20: B Then assess what event would look like (forwards and backwards) at Mauna Kea event rates and NB settings by default, use rate of 2.0 events/second, NB = 3 scale events to this lower rate
Example Scaled Rate Output
With rate = 2.0 Hz and NB = 3: # _xp.log: squawk 2147, alt 32000, PCA 00:20:22 For PCA at 00:20:22, translated to Mauna Kea with NB=3 and rate = 2.0 Hz Running forward in time... BT shutter close at 00:20:04.571, 17.4 sec before PCA Shutter release at 00:20:56.253, 34.3 sec after PCA Running backward in time... BT shutter close at 00:20:36.256, 14.3 sec 'before' PCA Shutter release at 00:19:44.577, 37.4 sec 'after' PCA In either direction, protected > 14 seconds in advance Shutter held closed for ~35 seconds after PCA both directions about 15 sec still in beam; 10 seconds before beam events clear out of running counter; 10 second hold-off of shutter once condition clear
Data Stream & Monitoring
TBAD sends out serial packet (115,200 baud, 8N1) 14 characters plus termination characters shutter; code; alarms; knob/power; format bits; checksum meant to be human/machine readable
Example Lines Approaching flight over UCSD, triggered DS, in-beam
:27: s6320.DBEF.FF :27: s EF..A DME :27: s EF..A DME :27: s6327.DBEF.FF :27: s5737.DBEF.FFC ModeS :27: s6320.DBEF.FF :27: s EF.FD ModeS :27: s5735.DBEFXF ModeS :27: s6327.DBEF.FCE :27: s6320.DBEF.FF Only green text straight from TBAD time stamp and some interpretation layered on by monitor software shutter closed; plane at 9000 ft, code 6327 DME just two pulses; real codes have ‘F.F’; ‘X’ in Mode-S
Monitor Program Python program listens to serial stream and logs, with minor interpretation listens via terminal server/IP or serial/USB if on laptop Can also grab telescope telemetry for handy insertion of az/el implementations at APO, Palomar, Keck Non-essential for TBAD protection, but vital for validation and health check Can provide template 257 line tbad-ts.py geared for terminal server or serial-over-USB version of the same can also share observatory-specific telemetry-gathering code though likely not useful/exportable
More Examples An example from APO: note azim/elev/dome-status in log
:13: o PF.FD O :13: o PF.FD O :13: o PF.FD O :13: o PF..AC DME O :13: o PFXF ModeS O :13: o PFXF ModeS O :13: o5777..BPFX ModeS O :13: o5777..BPFX ModeS O :13: o0573..BPF.FE O :13: o1624..BPF.FE O :13: o0573..BPF.FE O :13: o1624..BPF.FE O :13: o1624..BPF.FE O :13: s1624..BPF.FE O :13: s0573..BPF.FEA O ID=0573 at 32,000 ft enters beam, with some Mode-S, DME shutter closes near end; ~half-second sequence
Special Codes Mode-A and Mode-C are octal
only 0 to 7 possible Embellish with 8’s and 9’s for special occasions shutter open gets o8888 shutter closed due to fault gets s9999 shutter closed due to high background gets s9998 One-minute keep-alive verifies operational state code is i0000 (informational) Any single spike (within ~20 μs decode period) is ignored would look like 0000 with first framing and no other pulses distinguished from military zeros in that latter have final framing pulse called a glitch; reported, but not acted upon electrical discharges (including lightning) tend to follow this pattern
APO Validation Roughly 3-month campaign at APO; Flight Explorer validation 74 nights of operation 108 alarms for aircraft (most in-beam, 5 from OMNI saturation) Only one F.E. case close to beam went unreported but this case fraught with suspicious F.E. anomalies, so deem unreliable Several robust detections not reported in F.E. (military, private?) Flight Explorer gives lat, lon, alt, speed, flight number 03/21/2014, 08:54:35, 1010, Aircraft entered area, JBU278, APO, SFO, FLL, A320, 07:02, 11:37, 349, 563, , , 100 03/21/2014, 08:55:26, 1010, Aircraft entered area, JBU278, APO, SFO, FLL, A320, 07:02, 11:37, 349, 563, , , 100 flt#,zone,orig,dest,type,depart,arrive,flight-level,knots,lat,lon,heading Jet Blue 278 Airbus 320 SFO to FLL; ft; 563 knots, heading 100° Unreliable time steps: ~5 min offset; jumpy
Data Comparison TBAD data Flight Explorer data
TBAD time stamps are as good/reliable as log computer airplane ID code and altitude are known by virtue of TBAD’s decoding telescope pointing is known data sampling is dense: many dozens of records per second, typically airplane position or angle is not known from TBAD data flight number, heading, speed, etc. are not known from TBAD data aside from the “B” designation indicating a source within 12° of boresight, TBAD presents no information on where an airplane is on the sky Flight Explorer data the flight is easily and uniquely identified (unless blocked, as some are) excellent information on position, speed, heading, and altitude time stamps are unreliable and offset by an unknown amount data are sparse: roughly one sample per minute the squawk code is not known for the flight
Flight Explorer Time Steps
Following Jet Blue 278 flight speed calc dt calc hdg calc JBU278: JBU278: JBU278: JBU278: JBU278: F.E. reported speed and time steps are speed and dt columns But using lat/lon positions and F.E. time stamps, get calculated speeds lurching all over (passenger whiplash?) If using the lat/lon and reported speed, calculate actual time steps much closer to 60 seconds suspect this is “right” and adjust timestamps toward mean in analysis Heading based on F.E. and calculated from lat/lon agree well Conclusion: not a single time stamp is reliable
One of the more difficult pattern-matching tasks I have done especially hard in busy environment; considerably easier at Mauna Kea Altitude is the only shared information and not a unique identifier; confused with Mode-A or other planes Variable time offset a real bane Python code: inbeam-apo.py (1600 lines) disentangles Got 17 TBAD groups 1. 00:20:52: Possible matches from 1 temporal candidates N707LM at level with code 0566, 6.0 deg away (alt) B 2. 01:10:39: Possible matches from 1 temporal candidates AAL1093 at level with code 2330, 26.9 deg away (alt) 3. 01:40:53: Possible matches from 2 temporal candidates UAL632 at level with code 2454, 14.6 deg away (alt) 4. 01:48:39: Possible matches from 3 temporal candidates SWA427 at level with code 4041, 33.8 deg away (alt) 5. 02:05:39: Possible matches from 3 temporal candidates Clean separation: beg: 0.0/18.3/0.1; end: 0.0/52.8/0.0 ASA763 at level with code 0721, 41.0 deg away (alt) AAL949 at level with code 0557, 60.3 deg away (alt) etc.
Summary Output At the end is a summary: And a shutter report
17 groups; 18 identified; 10 excluded; 1 no-match; 0 missed; 4 in-beam And a shutter report DID NOT detect shutter closure for AAL2497 (15.7 deg) at 06:02:06 Shutter closed for SWA4354 ( 1.6 deg) at 04:06:31 due to ..B; dt = sec Shutter closed for DAL1080 ( 6.5 deg) at 03:03:06 due to ..B; dt = sec Shutter closed for N707LM ( 6.0 deg) at 00:22:13 due to ..B; dt = sec DID NOT detect shutter closure for UAL632 (14.5 deg) at 01:42:18 Shutter closed for BLOCKED_A ( 9.3 deg) at 02:54:23 due to ..B; dt = sec the NOT shutter events were outside of protected zone And any flights in Flight Explorer not in TBAD MISSED AAL1507 at 04:14:59 level at 31924, PCA 48.4 deg (PA 180.5) which is due to congestion/overlap in group 12: 12. 04:12:52: Possible matches from 2 temporal candidates On this night, four in-beam events Can plot TBAD’s sky view (moves with telescope)
What Does TBAD See? Red: TBAD triggered B; Blue: TBAD logged/saw; Black: analysis software did not disentangle
Typical & Super-Busy nights
By comparison, Keck had to wait 6 months to see first non-horizon beam crosser (medical emergency diversion LAX-SYD HNL!)
within 10°, perfect record of in-beam detection; soft-ish edge
Inferred Beam Size within 10°, perfect record of in-beam detection; soft-ish edge
Squawk ID Codes for Unidentified Traffic
Who do They Belong To?
TSIM The Transponder SIMulator creates Mode-A/C style pulse trains at variable strength avoids emergency codes, VFR, throws in ‘X’, omits F2 framing pulse 32 signal levels, from ~0 dBm to −64 dBm in steps of 2 dB, or 0.05 V on power detector Launched free-space via patch antenna make sure polarization matches: feedpoint low/high, not left/right ABCD code communicates mode (AB), signal strength (CD) Various mode/parameter settings explore range of TBAD responses Medley mode created to usefully explore full range of TBAD performance verify consistent operation across time; shutter response correct threshold behavior always at same power codes
TSIM Modes Modes based on 10 or 12 second “frames” 0: Medley
12 if expect shutter closure early in frame: time to clear 0: Medley sampling to be covered later 1: Full Ramp visits every power in sweep from low to high and back; repeats 12 s useful for exploring thresholds, repeatability, calibration 2: Weak Ramp partial ramp to value set by parameter knob, up to half range repeats on 12 second interval used in medley as pause to clear shutter without triggering anew
TSIM Modes, Continued 3: N Moderate 4: High-Rate Moderate
send out a number of moderate signals, meant to be > DT number depends on parameter knob; matches TBAD NB knob useful to see if TBAD triggers on N beam events 4: High-Rate Moderate 50 Hz output of moderate (meant to be > DT) signals useful to verify no bottlenecks in TBAD data stream/logging 5: High-Rate Background Combines previous two modes: N moderate pulses based on parameter knob, amid high rate of weak (sub DT) pulses useful to check that in-beam response unperturbed by high rate background events only fulfils purpose if OT threshold brought up to catch weak signals
TSIM Modes, Continued 6: Variable Rate 7: Empty 8: Strong Packet
moderate-level (> DT) signals emerge at rate depending on parameter knob rate is 10 Hz plus 10 times the parameter knob reading useful to check response to high rates, or steady stream of signals 7: Empty everybody needs a rest, and this allows TSIM to be ON, without making a peep 8: Strong Packet a single strong signal packet is emitted, with a 12 second rest following strength adjustable (2 dB steps) by parameter knob in upper half of range useful to check that saturation events trigger on a single instance
TSIM Modes, Continued Modes 9-15: X per Frame 0: Medley Mode
X = 2, 5, 10, 20, 50, 100, 500 in 10 second frame strength adjustable over full range, in 4 dB steps useful for steady stream of signals with adjustable strength and rate primary test modes for bench checkouts 0: Medley Mode cycles through most useful modes, checking: thresholds/sensitivity via full ramp shutter performance via variable number of moderate signals saturation performance at two levels for OS and DS check high rate performance high rate background (background usually invisible if OT “normal”) weak ramps interspersed to allow shutter to re-open parameter knob meant to match TBAD NB knob static installation can run test by applying AC power remotely | 6,154 | 22,848 | {"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} | 2.625 | 3 | CC-MAIN-2017-39 | longest | en | 0.749331 |
https://electronics.stackexchange.com/questions/410457/theres-conflicts-in-the-definitions-of-xnor?noredirect=1 | 1,585,556,464,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00218.warc.gz | 459,403,140 | 28,449 | # There's Conflicts in the Definitions of XNOR? [duplicate]
XNOR has two definitions:
$$\1^{st}:XNOR=NOT(XOR)\$$
$$\2^{nd}: XNOR(A,B)=\overline{A}\cdot\overline{B}+A\cdot B\$$
The problem is that these definitions are not equal in Odd inputs, the second definition acts as xor at odd inputs.
is the second really a definition or is it just the SOP of the first definition at 2 inputs?
• If you want to ask about odd numbers of inputs, you should make your definitions of the function for an odd number of inputs. – The Photon Dec 4 '18 at 21:23
• These are equal. The number of inputs here are not Odd. XOR with number of inputs greater than 2 has two conflicting definitions indeed. electronics.stackexchange.com/questions/93713/… – Eugene Sh. Dec 4 '18 at 21:23
XNOR is a negation of the Exclusuve OR on any number of inputs. Thus if you have n inputs, you perform | 243 | 873 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-16 | latest | en | 0.86018 |
https://www.lookoutlanding.com/2010/2/20/1318516/sabremetrics-101-a-win-is-a-win-is | 1,679,384,200,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943637.3/warc/CC-MAIN-20230321064400-20230321094400-00302.warc.gz | 999,650,363 | 42,347 | Filed under:
# Sabermetrics 101: A Win is A Win is 10 Runs (Or Thereabouts)
The runs to win conversion is the cornerstone of understanding how to measure value with modern metrics. It's also not quite as clear as it might appear.
Prerequisites for understanding: Expected win/loss, value, environment.
Prerequisites from derivation: Environment, data, expected win/loss.
How Do We Get There?
Ten. Ten runs per win. I suppose that's all you really need to know, but it's always good knowing how we come up with things otherwise they seem totally arbitrary. First of all, ten runs for every win isn't strictly true. It's close enough to suit us, for the most part, but knowing some of the little details will come in handy later. Let's start what we know about the run/win relationship already: Pythagorean win expectancy.
With Pythagorean expectancy as part of our toolbox, we can derive some interesting results. First of all, it's obvious that the league as a whole will have a .500 winning percentage, will score a league average number of runs a game, and concede a league average number of runs per game. What happens when you bump up runs per game and keep runs conceded static? The result is an increase in winning percentage (and therefore wins by the same amount, if we're just looking at one game). If a certain change in runs is equivalent to a change in wins, we have our runs per win. This value changes slightly as we go up and down the win percentage scale, but it sits around ten for the overwhelming majority of the time, so that's what we'll use from now on. Except when...
A Special Case
... a player has a major impact on the local run environment. When a game features a #5 pitcher, there will probably be more total runs score in said game. The opposite is true if an ace is on the mound for one team. Since pitchers can have such a large impact on the environment, and our method of deriving the runs per win relationship depends on the environment, the numbers will ideally be re-run for pitchers. The effect serves to amplify the distance a pitcher is from the mean (in wins). Note also that a change in league run environment will alter (perhaps significantly) the runs to wins conversion.
But Why?
If we are using a statistic based on runs, why do we need to convert to wins? Apart from the special case for pitchers outlined above, it seems a bit strange to put so much weight on a translation that essentially involves dividing by ten. The reason this is done is to keep our focus on what teams actually value: winning games. There's no logical reason for runs to have a value outside of their power to win or lose games, so in order to actually assign a value to a player, the win conversion is necessary, either implicitly or explicitly. Keeping the unique pitchers on the same scale as the hitters and average pitchers is another consideration, although I believe that it's less fundamentally important.
To Sum Up
• Using ten runs per win suits us perfectly for the most part (thanks to the typical run environment we operate in).
• Extremely good or bad pitchers have an effect on the runs per win conversion, lowering the amount of runs per win for good pitchers and raising them for bad ones.
• We convert to wins because there's no logical reason that runs might reflect player value, while wins clearly benefit teams.
What Follows
WAR. | 712 | 3,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-14 | longest | en | 0.961503 |
https://www.jiskha.com/display.cgi?id=1332471965 | 1,503,101,086,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105195.16/warc/CC-MAIN-20170818233221-20170819013221-00386.warc.gz | 913,262,405 | 4,325 | # MaTh AlgeBRA
posted by .
Type or write the essay question into a Microsoft Word document or on a separate sheet of paper and submit your completed essay to your teacher.
Let f(x) = 2x^2 + x - 3 and g(x) = x - 1. Perform the function operation and then find the domain.
f(x)/g(x)
Note: x^2 is x squared
• MaTh AlgeBRA -
Do synthetic division or long division
Domain is all reals except x=1 because it gives you a 0 denominator.
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More Similar Questions | 685 | 2,774 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-34 | latest | en | 0.946333 |
https://rdrr.io/github/hojsgaard/doBy/man/esticon.html | 1,571,385,040,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677964.40/warc/CC-MAIN-20191018055014-20191018082514-00135.warc.gz | 668,393,417 | 16,145 | # esticon: Contrasts for lm, glm, lme, and geeglm objects In hojsgaard/doBy: Groupwise Statistics, LSmeans, Linear Contrasts, Utilities
## Description
Computes linear functions (i.e. weighted sums) of the estimated regression parameters. Can also test the hypothesis, that such a function is equal to a specific value.
## Usage
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'gls' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'geeglm' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'lm' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'glm' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'mer' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'coxph' esticon(obj, L, beta0, conf.int = TRUE, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'lme' esticon(obj, L, beta0, conf.int = NULL, level = 0.95, joint.test = FALSE, ...) ## S3 method for class 'esticon_class' coef(object, ...) ## S3 method for class 'esticon_class' summary(object, ...) ## S3 method for class 'esticon_class' tidy(x, conf.int = FALSE, conf.level = 0.95, ...) ## S3 method for class 'esticon_class' confint(object, parm, level = 0.95, ...)
## Arguments
obj Regression object (of type lm, glm, lme, geeglm). L Matrix (or vector) specifying linear functions of the regresson parameters (one linear function per row). The number of columns must match the number of fitted regression parameters in the model. See 'details' below. beta0 A vector of numbers conf.int TRUE level The confidence level joint.test Logical value. If TRUE a 'joint' Wald test for the hypothesis L beta = beta0 is made. Default is that the 'row-wise' tests are made, i.e. (L beta)i=beta0i. If joint.test is TRUE, then no confidence inteval etc. is calculated. ... Additional arguments; currently not used. object An esticon_class object. x A linear contrast object (as returned by esticon(). conf.level The desired confidence level. parm a specification of which parameters are to be given confidence intervals, either a vector of numbers or a vector of names. If missing, all parameters are considered.
## Details
Let the estimated parameters of the model be
β_1, β_2, …, β_p
A linear function of the estimates is of the form
l=λ_1 β_1+λ_2 β_2+ …+λ_p β_p
where λ_1, λ_2, …,λ_p is specified by the user.
The esticon function calculates l, its standard error and by default also a 95 pct confidence interval. It is sometimes of interest to test the hypothesis H_0: l=β_0 for some value β_0 given by the user. A test is provided for the hypothesis H_0: l=0 but other values of β_0 can be specified.
In general, one can specify r such linear functions at one time by speficying L to be an r\times p matrix where each row consists of p numbers λ_1,λ_2,…, λ_p. Default is then that β_0 is a p vector of 0s but other values can be given.
It is possible to test simulatneously that all speficied linear functions are equal to the corresponding values in β_0.
For computing contrasts among levels of a single factor, 'contrast.lm' may be more convenient.
## Value
Returns a matrix with one row per linear function. Columns contain estimated coefficients, standard errors, t values, degrees of freedom, two-sided p-values, and the lower and upper endpoints of the 1-alpha confidence intervals.
## Author(s)
Søren Højsgaard, [email protected]
## Examples
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 data(iris) lm1 <- lm(Sepal.Length ~ Sepal.Width + Species + Sepal.Width : Species, data=iris) ## Note that the setosa parameters are set to zero coef(lm1) ## Estimate the intercept for versicolor lambda1 <- c(1, 0, 1, 0, 0, 0) esticon(lm1, L=lambda1) ## Estimate the difference between versicolor and virgica intercept ## and test if the difference is 1 lambda2 <- c(0, 1, -1, 0, 0, 0) esticon(lm1, L=lambda2, beta0=1) ## Do both estimates at one time esticon(lm1, L=rbind(lambda1, lambda2), beta0=c(0, 1)) ## Make a combined test for that the difference between versicolor and virgica intercept ## and difference between versicolor and virginica slope is zero: lambda3 <- c(0, 0, 0, 0, 1, -1) esticon(lm1, L=rbind(lambda2, lambda3), joint.test=TRUE) # Example using esticon on coxph objects (thanks to Alessandro A. Leidi). # Using dataset 'veteran' in the survival package # from the Veterans' Administration Lung Cancer study if (require(survival)){ data(veteran) sapply(veteran, class) levels(veteran\$celltype) attach(veteran) veteran.s <- Surv(time, status) coxmod <- coxph(veteran.s ~ age + celltype + trt, method='breslow') summary(coxmod) # compare a subject 50 years old with celltype 1 # to a subject 70 years old with celltype 2 # both subjects on the same treatment AvB <- c(-20, -1, 0, 0, 0) # compare a subject 40 years old with celltype 2 on treat=0 # to a subject 35 years old with celltype 3 on treat=1 CvB <- c(5, 1, -1, 0, -1) est <- esticon(coxmod, L=rbind(AvB, CvB)) est ##exp(est[, c(2, 7, 8)]) }
hojsgaard/doBy documentation built on Oct. 11, 2019, 9:55 a.m. | 1,700 | 5,495 | {"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} | 2.796875 | 3 | CC-MAIN-2019-43 | latest | en | 0.557446 |
https://paylawexam.com/how-are-taxes-on-income-from-non-profit-organizations-calculated | 1,723,150,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640740684.46/warc/CC-MAIN-20240808190926-20240808220926-00088.warc.gz | 363,085,493 | 29,024 | # How are taxes on income from non-profit organizations calculated?
How are taxes on income from non-profit organizations go to this website The New York City Tax Authority (NYCTA) has been studying the situation closely for the last four years as part of its annual audit of the city’s annual property tax bill. Instead of doing their research, NYCTA, along with its outside consultants and its other advisers, is comparing the number of property owners at any one time to the number of percentage of property owners who don’t manage over three-quarters of the budget to figure out what percentage of them are at lower than required minimum tax rates (MSL, or one quarter of a property owner) and one quarter of the city’s sales taxes. It does this by comparing the percentage owners have had some action they’ve taken to change the MSL, which means taking more money from the property owner and paying off more property taxes. The NYCTA can calculate the MSL for either the remaining three quarters or the remaining five quarters. Their estimate for the three quarters is 3.6%. More than 500 people died before the five-quarter (3.2%) period was counted. Further, if any of those people’s businesses were at a state income tax rate of 8 percent, or more, they could have easily replaced the state tax rate of 20 percent at the visit the website DNR level, so these numbers will be more accurate. Moreover, even if some of those businesses’ income was used to compute the MSL, they’re still going to have increased MSL and decrease them as a percentage of property owners. Instead, we calculated the MSL for two pieces of the city’s sales tax. The first was the City’s percent of property owners who manage over three-quarters of the city’s business over three years. The second was read this percentage of them who manage over five-quarter of the city’s business. And in these figures, assuming that the percentage used in equations 1How are taxes on income from non-profit organizations calculated? by John Paulson. December 2, 2012 – 11:35 am http://www.newgov.gov/taj-questions/113976-finance-of-income-taxaments-substitute-an-ipso facto-income-unit/ John Paulson, M.D., is a professor of economics in Georgia Southern University’s School of Public Health’s Department of Public Health. His recent work as a policy analyst and analyst in a Fortune 500 organization has helped determine how it can improve infrastructure and reduce the nation’s growing health care risks.
## Is Doing Homework For Money Illegal?
(M.D. Paulson is a professor of economics, associate professor of economics, and a senior leader in government at Georgia Southern University’s School of Public Health.) In his recent book, “How We Live!”, Ron Paul suggested that taxes on cost-saving measures should be reduced simply because of current revenue. Paul’s new book has shown that the cost of living and the number of uninsured are both rising, and that both increases and decreases, when those costs are tied to the tax system and society’s broader economic activities. We believe that the question of the size of the costs of government spending on health care, and taxation of that, should not be decided by theory alone. And in addition Mr. Paulson argues that through today’s tax system there is a greater danger of “spillovers.” These include those who are contributing to the financial burden to other households carrying the same amount of income that is being taxed. By doing so, we’ve already been shown the possible dangers of tax traps, and we’re also showing that they exist. We are therefore now looking towards improving some of the ways we can encourage those who are not providing for their lifestyles to reach these potential goal-setting goals. Does reducing the size of that number of households actually help the nation? Mr. Paulson also suggests that, as part of a U.S. economic strategy, we should step up our efforts and invest in existing programs and programs to encourage greater the capacity of the average American to make sufficient Visit Website The next step is to move into using strategies to encourage greater the density of the average American and a more flexible top end of the political equation. Perhaps the next great step to the future in economic policy will be the right fiscal policy. However, one of the great challenges with the political policy arena is how to make those policies fit into the present era. In order for Americans to be able to elect leaders and support health care, not some more modest Democratic politician, but an elected, more ambitious, more socially conservative, and more politically liberal. When the next man stands down is that of the next president, which suggests one is not quite sure how to manage change while making him a good president.
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https://algebraworksheets.co/easy-pre-algebra-worksheets/ | 1,660,150,659,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00610.warc.gz | 123,907,179 | 13,984 | # Easy Pre Algebra Worksheets
Easy Pre Algebra Worksheets – Algebra Worksheets are designed to make math easy for students. This study area focuses on the study of mathematical symbols and the rules that govern their manipulation. It is the thread that runs through the heart that runs through all math which includes physics and geometry. It is an integral part of your education. This section will teach you how to use algebra, focusing on the most crucial subjects. If you’re in school and want to enhance your abilities, you’ll find that these worksheets are a great starting point.
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# 10 Physics questions: Acceleration, speed, air resistance, density, temperature, work
Not what you're looking for?
1. You push with a 50-N horizontal force on a 10 kg mass resting on a horizontal surface against a horizontal friction force of 20N. What is the acceleration?
2. A railroad diesel engine weighs six times as much as a freight car. If the diesel engine coasts a 7km per hour into a freight car that is intially at rest, how fast do the tow coast after they couple together?
3. A car moving at 20km.h skids 10m with locked brakes. How far will the car skid with locked brakes at 120km/h?
4. A cannonball shot with an intial velocity of 150m/s at an angle of 45 degrees follows a parabolic path and hits a balloon at the top of its trajectory. Ignoring air resistance, how fast is the cannonball going when it hits the balloon?
5. A vacationer floats lazily in the ocean with 93% of his body below the surface. The density of ocean water is 1,025 kg/m^3. What is the vacationer's average density?
6. What will be the final temperature of 100g of 25 degree C water when 100 g of 45 degree C iron nails are submerge in it. (The specific heat of iron is 0.12 cal/g. The specific heat of water is 1.00 cal/g.)
7. An electric iron connected to a 100-V source draws 10A of current. How much heat (in joules) does it generate in a minute?
8. A weight suspended from a spring is seen to bob up and down over a distance of 40 centimeters twice each second. What is frequency? Its period? Its amplitude? Include units.
9. Balance the chemical equation and be sure to show your work!
___ CH4 + ____O2 → ____CO2 + _____H2O
10. How many grams of ozone, O3 (48 amu), can be produced from 96 grams of oxygen, O2(32amu), in the following reaction - show all work!
3 O2 → 2 O3
##### Solution Summary
With excellent explanation and calculations, the problems are solved.
##### Solution Preview
1. You push with a 50-N horizontal force on a 10 kg mass resting on a horizontal surface against a horizontal friction force of 20N. What is the acceleration?
In order to find the acceleration, first we need to calculate the net force on the object.
F1= 50N F2= -20N
FT= F1+F2
FT= 50-20
FT= 30 N (1N= 1kg.m/s2)
Now we can calculate the object's acceleration "a" by dividing the net force to its mass.
m= 10 kg
a= FT/ m
= 30 / 10
a = 3 m/s2
2. A railroad diesel engine weighs six times as much as a freight car. If the diesel engine coasts a 7km per hour into a freight car that is intially at rest, how fast do the tow coast after they couple together?
We don't know the masses of diesel engine and the freight car. However we can show them with variables in terms of each other.
Mass of railroad diesel engine m1= 6m
Mass of freight= m2= m
mT = total mass when they couple = m+6m= 7m
V1= Initial speed of diesel engine = 7km/h
V2= Initial speed of freight car = 0 (at rest)
VT = final speed
We are going to use the momentum formula:
m1*V1 + m2*V2 = mT*VT
6m * 7 + m*0 = 7m* VT
VT *7m = 42m
VT= 42m/7m
VT= 6 km/h is the final speed of the total mass.
3. A car moving at 20km.h skids 10m with locked brakes. How far will the car skid with locked ...
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##### The Moon
Test your knowledge of moon phases and movement.
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##### Intro to the Physics Waves
Some short-answer questions involving the basic vocabulary of string, sound, and water waves.
##### Introduction to Nanotechnology/Nanomaterials
This quiz is for any area of science. Test yourself to see what knowledge of nanotechnology you have. This content will also make you familiar with basic concepts of nanotechnology.
##### Variables in Science Experiments
How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz. | 1,102 | 4,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.909478 |
https://www3.nd.edu/~pbui/teaching/cse.30872.su24/challenge07.html | 1,722,952,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00593.warc.gz | 846,751,423 | 4,259 | Many summers ago, Caleb went to a day camp where he learned how to play chess. He quickly learned the basic rules of chess and grew to love playing the game (almost as much as playing Minecraft). On some evenings, he will challenge the instructor to a game after dinner (and sometimes even wins!).
In chess each piece can move in a different manner. For instance, the rook, Caleb's favorite piece, can move any number of spaces vertically or horizontally. A knight, however, can only move in a `L`-shape motion: two steps horizontally followed by one vertically, or one step horizontally then two vertically:
Unfortunately, Caleb really dislikes the knight piece because its movement is so confusing. To help Caleb practice visualize the movement of a knight, the instructor proposes the following problem to his son:
Imagine you place a knight chess piece on a phone dialpad. Suppose you dial keys on the keypad using only hops a knight can make. Every time the knight lands on a key, we dial that key and make another hop. The starting position counts as being dialed.
What are the distinct numbers you can dial in `N` hops from a particular starting position?
Because solving this problem efficiently requires the use of recursion, Caleb asks that you help him out by programming a solution that solves this challenge.
## Input¶
You will be given a series of starting positions and hops in the following format:
``````START HOPS
...
START HOPS
``````
### Example Input¶
``````6 3
0 4
``````
## Output¶
For each pair of starting positions and hops, output all possible numbers starting with the specified position and of the specified hops formed on the dialpad by using only the chess knight's motion.
### Example Output¶
``````604
606
616
618
672
676
0404
0406
0434
0438
0492
0494
0604
0606
0616
0618
0672
0676
``````
Note: Output the numbers in ascending order and ensure there is a single blank line between the outputs of the pairs of starting positions and hops.
#### Programming Challenges¶
This is inspired by this blog post: Google Interview Questions Deconstructed: The Knightâs Dialer and this 935. Knight Dialer on LeetCode.
#### Algorithmic Complexity¶
For each input test case, your solution should have the following targets:
Time Complexity `O(N!)`, where `N` is the number of hops. Space Complexity `O(D)`, where `D` is the number of distinct numbers from starting point.
Your solution may be below the targets, but it should not exceed them.
## Submission¶
``````\$ cd path/to/cse-30872-su24-assignments # Go to assignments repository
\$ git checkout master # Make sure we are on master
\$ git pull --rebase # Pull any changes from GitHub
\$ git checkout -b challenge07 # Create and checkout challenge07 branch
\$ \$EDITOR challenge07/program.cpp # Edit your code
\$ git commit -m "challenge07: done" # Commit your changes
\$ git push -u origin challenge07 # Send changes to GitHub
``````
To check your code, you can use the `.scripts/check.py` script or curl:
``````\$ .scripts/check.py
Checking challenge07 program.cpp ...
Result Success
Time 0.02
Score 6.00 / 6.00
\$ curl -F source=@challenge07/program.cpp https://dredd.h4x0r.space/code/cse-30872-su24/challenge07
{"result": "Success", "score": 6, "time": 0.016405344009399414, "value": 6, "status": 0}
``````
#### Pull Request¶
Once you have committed your work and pushed it to GitHub, remember to create a pull request and assign it to the instructor. | 838 | 3,558 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.908216 |
https://community.datawatch.com/thread/1454 | 1,503,489,494,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120194.50/warc/CC-MAIN-20170823113414-20170823133414-00091.warc.gz | 751,161,997 | 111,997 | 6 Replies Latest reply: May 15, 2014 9:53 AM by Steve Caiels
# Julian Date
I have a database with real julian dates, ex:2450816= jan 2 1998, 2450817= jan 3 1998, how can I convert this format to our calendar dates?
Any help would be greatly appreciated.
Maria
• ###### Julian Date
Monarch V7 or V8 includes a Julian date conversion function. See: JulianToDate function in Monarch V7 or V8 help.
The use of Julian dates are usually rare but in the (U.S.) military, the use of Julian dates are surprisingly common. For Monarch V6 and prior, a (VERY) long formula may do the trick, but I understand some exceptions might exist such as the presence of a leap year. Monarch V8 has many more benefits that justify upgrading beyond Julian date conversions smile.gif[/img]
• ###### Julian Date
Mike:
I have Monarch v8. JulianToDate assumes the date format "YYYYDDD" instead of "julian". My database uses the real julian calendar, which is not the format supported by monarch.
• ###### Julian Date
Todd:
Monarch cannot read it, it gives me a "null" result.
• ###### Julian Date
Maria
Try this formula - note you need to replace the field with the field name you are using.
Note that the input field needs to be a numeric data type and this field needs to be a date calculated field.
Ctod(Str(Int((Int(80(Int((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)-Int(1461(Int(4000((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)1)/1461001))/4)31)) /2447))+2-12(Int((Int(80(Int((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)-Int(1461(Int(4000((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)+1)/1461001))/4) 31))/2447))/11))),2,0)"/"Str((Int((Int()68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)-Int(1461(Int(4000((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)1)/146100 1))/4)31))-Int(2447(Int(80(Int((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)-Int(1461(Int(4000((Int()+68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)1)/1461001)) /4)31))/2447))/80),2,0)"/"Str(Int(100((Int((4(Int()68569))/146097))-49)(Int(4000((Int()68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)1)/1461001))(Int((Int(80*(Int((Int()685 69)-Int((146097(Int((4(Int()68569))/146097))3)/4)-Int(1461(Int(4000((Int()68569)-Int((146097(Int((4(Int()68569))/146097))3)/4)1)/1461001))/4)+31))/2447))/11))),4,0))
Please note that for some reason the forum software inserts a space in one of the instances of 68569, so make sure you delete that after you have pasted it.
Gareth
Originally posted by Maria1121:
I have a database with real julian dates, ex:2450816= jan 2 1998, 2450817= jan 3 1998, how can I convert this format to our calendar dates?
Any help would be greatly appreciated.
Maria /b[/quote]
• ###### Julian Date
Gareth
Nice easy formula. NOT!
One limitation is that Monarch has to be in m-d-y format for this to work, but otherwize this is very cool.
Regards,
smile.gif[/img]
• ###### Julian Date
Hi,
I have an expression that works for the few dates you have used.But PLEASE look into this more before using it live as I’m not an expert on Julian dates.
{01/01/1998}+(jd-2450815)
I don’t THINK there will be any issues regarding leap years, but you may wish to check that carefully.
Regards
Steve | 1,034 | 3,289 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-34 | longest | en | 0.529214 |
https://dalimyscha.web.app/151.html | 1,721,465,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00003.warc.gz | 158,526,756 | 3,127 | # Nbragg's law derivation pdf
Derivation of bragg s law bb d c a, tthhee distancedistance sincesince bc cd, cd, wwee havehave. Thenthen after after substitutionsubstitution givesgives n. Keplers laws of planetary motion and his theory of gravitation. Xray diffraction, braggs law and laue equation last updated. In physics, braggs law, or wulffbraggs condition, a special case of laue diffraction, gives the. Bragg s law statement bragg s law is a special case of laue diffraction which determines the angles of coherent and incoherent scattering from a crystal lattice. There are at least three ways to derive the bragg scattering condition, applicable to electromagnetic waves, usually xrays, scattering from a crystalline solid. B to expect maximum diffracted intensity for a particular family of crystal. The concept of braggs law assumes that the scattering is.
Proceeding like newton with a discrete time approach of motion and a geometrical representation of velocity and acceleration, we obtain keplers laws. Newton used classical geometry and the emerging techniques of differential and in tegral calculus to give mathematical derivations from general. Derivation ofbragg s law, also known as the bragg scattering condition. Poiseuilles law derivation peters education website.
Bragg s law means that the diffraction can occur only when the following equation is satis. If d sin 2 d n braggs law the beams coming from both objects are in phase, they. When xrays are incident on a particular atom, they make an electronic cloud move just like an electromagnetic wave. Deriving braggs law braggs law can easily be derived by considering the conditions necessary to make the phases of the beams coincide when the incident angle equals and reflecting angle. A simple derivation of keplers laws without solving. And, when the path difference, \d\ is equal to a whole number, \n\, of wavelength, a constructive interference will occur. Derivation of bragg s law sin sin t t hkl hkl x d d x path difference. This gives the criterion for constructive interference. A rigorous derivation from the more general laue equations is available see page. Bragg condition a laue diffraction peak corresponding to a change in the wave vector given by the reciprocal lattice vector corresponds to a bragg reflection from the family of direct lattice planes perpendicular to. Consider a solid cylinder of fluid, of radius r inside a hollow. An introduction to electricity and strength of materials with peter eyland. These considerations lead directly to lambertbeers law.
The law states that when the xray is incident onto a crystal surface, its angle of incidence, \\theta\, will reflect back with a same angle of scattering, \\theta\. Xray diff ractio n, braggs law a nd a pplicat ion s. Chapter 1 bragg s law first of all, let us study the bragg s law. Direct derivation of braggs law reflection from the first plane the scattered waves will be in phase whatever the distribution of the point scatterers in the first plane if the angle of the reflected wave vector, k h, is also equal to.
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https://homework.cpm.org/category/CC/textbook/cc1/chapter/5/lesson/5.3.1/problem/5-71 | 1,606,917,751,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141708017.73/warc/CC-MAIN-20201202113815-20201202143815-00269.warc.gz | 292,686,296 | 15,057 | ### Home > CC1 > Chapter 5 > Lesson 5.3.1 > Problem5-71
5-71.
Bianca is trying to find the area of this rectangle. She already measured one side as $10$ cm. Which other length(s) could she measure to use in her area calculation? Explain your reasoning.
How do you find area of a rectangle?
If $10$ cm is the base of the rectangle, which lengths could be the height?
You might want to rotate your paper so that the placement of the rectangle is more familiar.
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# Mean Standard Error Interpretation
## Contents
When the standard error is large relative to the statistic, the statistic will typically be non-significant. The answer to the question about the importance of the result is found by using the standard error to calculate the confidence interval about the statistic. I could not use this graph. Hutchinson, Essentials of statistical methods in 41 pages ^ Gurland, J; Tripathi RC (1971). "A simple approximation for unbiased estimation of the standard deviation". check over here
The true standard error of the mean, using σ = 9.27, is σ x ¯ = σ n = 9.27 16 = 2.32 {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt Please help. You nearly always want some measure of uncertainty - though it can sometimes be tough to figure out the right one. Accessed September 10, 2007. 4.
## What Is A Good Standard Error
QUESTION 3: Since the SEM is not calculated directly but estimated from the SD of a sample, what effect does departure from a normal distribution of the sample have on calculation Authors Carly Barry Patrick Runkel Kevin Rudy Jim Frost Greg Fox Eric Heckman Dawn Keller Eston Martz Bruno Scibilia Eduardo Santiago Cody Steele menuMinitab® 17 SupportWhat is the standard error of This can artificially inflate the R-squared value.
The second sample has three observations that were less than 5, so the sample mean is too low. Here's how I try to explain it (using education research as an example). Thanks for the question! Standard Error Regression How to cite this article: Siddharth Kalla (Sep 21, 2009).
McHugh. How To Interpret Standard Error In Regression Or decreasing standard error by a factor of ten requires a hundred times as many observations. The exceptions to this generally do not arise in practice. http://blog.minitab.com/blog/adventures-in-statistics/regression-analysis-how-to-interpret-s-the-standard-error-of-the-regression Of the 100 sample means, 70 are between 4.37 and 5.63 (the parametric mean ±one standard error).
You can probably do what you want with this content; see the permissions page for details. The Standard Error Of The Estimate Measures The Variability Of The The standard deviation of the age was 9.27 years. In a regression, the effect size statistic is the Pearson Product Moment Correlation Coefficient (which is the full and correct name for the Pearson r correlation, often noted simply as, R). But there is still variability.
## How To Interpret Standard Error In Regression
In cases where the standard error is large, the data may have some notable irregularities.Standard Deviation and Standard ErrorThe standard deviation is a representation of the spread of each of the Means ±1 standard error of 100 random samples (n=3) from a population with a parametric mean of 5 (horizontal line). What Is A Good Standard Error Available at: http://www.scc.upenn.edu/čAllison4.html. Standard Error Example National Center for Health Statistics (24).
Two data sets will be helpful to illustrate the concept of a sampling distribution and its use to calculate the standard error. check my blog Thanks for the beautiful and enlightening blog posts. Large S.E. Maybe the estimated coefficient is only 1 standard error from 0, so it's not "statistically significant." But what does that mean, if you have the whole population? The Standard Error Of The Estimate Is A Measure Of Quizlet
These authors apparently have a very similar textbook specifically for regression that sounds like it has content that is identical to the above book but only the content related to regression This is a meaningful population in itself. Therefore, it is essential for them to be able to determine the probability that their sample measures are a reliable representation of the full population, so that they can make predictions http://facetimeforandroidd.com/standard-error/mean-standard-deviation-and-standard-error-calculator.php Using a sample to estimate the standard error In the examples so far, the population standard deviation σ was assumed to be known.
If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative How To Interpret Standard Deviation However, you can’t use R-squared to assess the precision, which ultimately leaves it unhelpful. You'll Never Miss a Post!
## The standard error can include the variation between the calculated mean of the population and once which is considered known, or accepted as accurate.
I actually haven't read a textbook for awhile. The ages in that sample were 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. You use standard deviation and coefficient of variation to show how much variation there is among individual observations, while you use standard error or confidence intervals to show how good your Standard Error Vs Standard Deviation Coefficient of determination The great value of the coefficient of determination is that through use of the Pearson R statistic and the standard error of the estimate, the researcher can
S represents the average distance that the observed values fall from the regression line. Suppose the sample size is 1,500 and the significance of the regression is 0.001. By taking the mean of these values, we can get the average speed of sound in this medium.However, there are so many external factors that can influence the speed of sound, have a peek at these guys Sometimes we can all agree that if you have a whole population, your standard error is zero.
This figure is the same as the one above, only this time I've added error bars indicating ±1 standard error. The effect of the FPC is that the error becomes zero when the sample size n is equal to the population size N. For example, you have all the inpatient or emergency room visits for a state over some period of time. At a glance, we can see that our model needs to be more precise.
Most of these things can't be measured, and even if they could be, most won't be included in your analysis model. We wanted inferences for these 435 under hypothetical alternative conditions, not inference for the entire population or for another sample of 435. (We did make population inferences, but that was to Journal of the Royal Statistical Society. Why I Like the Standard Error of the Regression (S) In many cases, I prefer the standard error of the regression over R-squared.
If one survey has a standard error of $10,000 and the other has a standard error of$5,000, then the relative standard errors are 20% and 10% respectively. The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error. Further, as I detailed here, R-squared is relevant mainly when you need precise predictions. BREAKING DOWN 'Standard Error' The term "standard error" is used to refer to the standard deviation of various sample statistics such as the mean or median.
This statistic is used with the correlation measure, the Pearson R. JSTOR2682923. ^ Sokal and Rohlf (1981) Biometry: Principles and Practice of Statistics in Biological Research , 2nd ed. However, the sample standard deviation, s, is an estimate of σ. Radford Neal says: October 25, 2011 at 2:20 pm Can you suggest resources that might convincingly explain why hypothesis tests are inappropriate for population data?
That is, for a sample with mean 5.00 and SEM 0.50, is it correct to conclude the true population mean lies between 4.50 and 5.50 with probability 68.3%? Sieve of Eratosthenes, Step by Step Why does the find command blow up in /run/? It's sort of like the WWJD principle in causal inference: if you think seriously about your replications (for the goal of getting the right standard error), you might well get a | 1,709 | 7,884 | {"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} | 3.890625 | 4 | CC-MAIN-2019-04 | latest | en | 0.849532 |
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by skyturnred
Tags: basis
P: 116 1. The problem statement, all variables and given/known data This is the question on my assignment: In each case below, given a vector space V , find a basis B for V containing the linearly independent set S ⊂ B. It has a bunch of different cases but I think that if you help me with the following two, I will learn enough to do the others. The first case is the following: (a) V = R$^{4}$, S = {(1,0,0,1),(0,1,1,0),(2,1,1,1)}. and the next case is: (b) V = M2×2 = the vector space of all 2 × 2 matrices, and S= [1, 1; 1, 0] [0, 1; 1, 1] [1, 0; 1, 1] 2. Relevant equations 3. The attempt at a solution My problem with BOTH cases is this: I only know how to find a basis given a bunch of vectors that form a span (in other words, I know how to find the linearly dependent ones and kick them out of the equation). But I DONT understand how to find the missing parts of the basis given what the basis is SUPPOSED to span. Can someone please walk me through this? And my SECOND problem is with case (b): I cannot visualize how matrices can span something. I understand vectors, but not matrices. And since I can't understand it, I can't approach it to find the basis. Thanks SO much in advance for your help! | 363 | 1,282 | {"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} | 3.21875 | 3 | CC-MAIN-2014-41 | latest | en | 0.941994 |
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### Finding the order of the automorphism group of the abelian group of order 8.
So I am given an abelian group of order $8$ such that for all non-identity elements $x^2 = e$ (all elements have order two). So I know the answer is gonna be $168$, but I gotta prove this. So far I ...
154 views
### Linear algebra of finite abelian groups
Let $\phi:G \to H$ be a surjective homomorphism of finite abelian groups, and let $g_1, \ldots, g_n$ be an irredundant set of generators (from now on, a basis) for $G$. be a basis for $G$, meaning a ...
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### Dimension of subspace fixed by subgroup representation.
If $G$ is an abelian group with cyclic subgroup $H$ and $(\rho,V)$ is a (permutation) representation of $G$. Then I can form a representation of $H$ by considering the composition ...
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### number of differents vector space structures over the same field $\mathbb{F}$ on an abelian group
My question here raised another one. How many differents vector space structures over a field $\mathbb{F}$ we may have on an abelian group? I know that there are abelian groups that we can not endow ...
### Is $\mathbb{Z}(p^{\infty})$ a vector space over some field $\mathbb{F}$?
I don't know how to write in good English, so I will follow Hungerford's word from his book Algebra. The following relation on the additive group $\mathbb{Q}$ of rational numbers is a congruence ...
Ciao! Let $A$ be a finite abelian group, and let $\psi : A \times A \to \mathbb{Q}/\mathbb{Z}$ be an alternating, non-degenerate bilinear form on $A$. Maybe I should say what I mean by these ... | 435 | 1,626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2014-41 | latest | en | 0.863935 |
https://stclairdrake.net/5p-14-8p-4/ | 1,653,286,622,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662555558.23/warc/CC-MAIN-20220523041156-20220523071156-00469.warc.gz | 617,182,251 | 4,155 | ### Rearrange:
Rearrange the equation by subtracting what is come the ideal of the equal sign from both sides of the equation : 5*p-14-(8*p+4)=0
## Step 1 :
Pulling out choose terms :
1.1 pull out like factors:-3p - 18=-3•(p + 6)
Equation at the finish of action 1 :
## Step 2 :
Equations i m sorry are never true:2.1Solve:-3=0This equation has actually no solution. A a non-zero consistent never equals zero.
Solving a single Variable Equation:2.2Solve:p+6 = 0Subtract 6 from both sides of the equation:p = -6
## One equipment was found :
p = -6
You are watching: 5p-14=8p+4
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See more: Convert 1 Qt Equals How Many Oz In A Quart, Your Questions Answered
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Helper III
## Formula to pull in previous day balance
Hello,
I am looking to pull the previous run date (second column) USD balance (7th column) if the "if value date/Run date" column is equal to "Y".
For example, the new column should contain a value of -800,000 (previous day USD balance) in the first row and in the second row a value of 0 since the if statement is equal to "N". Please note, this should be applied by flex_acct codes as there are many different codes (ex:75104, 75015) on the report (first column)
Also, this report runs 5 days a week and therefore does not include weekend dates so it would need to find the previous run date and not previous date.
1 ACCEPTED SOLUTION
Anonymous
Not applicable
Hi @gmasta1129 ,
``````Measure =
VAR cur_rk = [RK]
VAR cur_flexacct =
SELECTEDVALUE ( 'Table'[flex_acct] )
VAR cur_cd =
SELECTEDVALUE ( 'Table'[if ValueDaate/Run Date] )
VAR tmp =
FILTER (
ALL ( 'Table' ),
'Table'[flex_acct] = cur_flexacct
&& [RK] = cur_rk - 1
)
VAR _val =
CALCULATE ( MAX ( 'Table'[usd_balance] ), tmp )
VAR cur_rd =
SELECTEDVALUE ( 'Table'[Run Date] )
VAR _pre_date =
CALCULATE ( MAX ( 'Table'[Run Date] ), tmp )
VAR cd =
DATEDIFF ( cur_rd, _pre_date, DAY )
RETURN
IF ( DAY ( cur_rd ) = 1 && cd <> 1, 0, IF ( cur_cd = "N", 0, _val ) )
``````
Please refer the attached .pbix file.
Best regards,
Community Support Team_ Binbin Yu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
8 REPLIES 8
Helper III
@Anonymous The formula above is pulling all zeros even when there is a balance in the previous day.
I think the issue is the return formula.
RETURN IF ( DAY ( cur_rd ) = 1 && cd <> 1, 0, IF ( cur_cd = "N", 0, _val ) )
` `
It is correct in this scenario since the last negative balance was on 8/2 and there was no balance on 10/31
But this scenario has a balance as of 10/31 (previous day) and it is pulling 0 when I should be seeing -3.8 million
Helper III
Hello @Anonymous ,
This worked exactly as expected except for one scenario. If there is a zero balance for a certain run date in a specific flex acct then the flex acct does not pull onto the report.
Therefore the formula is finding the latest balance from the latest run date and pulling that value in. Can the formula be tweaked where it finds the previous "Run Date" balance and if there is no balance, then pull in 0?
For example, in the screenshot below, flex_acct 71513, the USD balance on run date 11/1/2022 is -8,785.07 but there is no run date for 10/31/2022. The last run date is 10/24/2022 which means the balance as of 10/31/2022 is 0. For our reporting purposes any flex acct with a zero balance does not pull into the report.
Your formula which is under column "USD Balance (Previous Day)" is pulling in the latest run date balance which is 10/24/2022 of -999,519,51. This is incorrect since the previous day from 11/1/2022 is 10/31/2022. because the balance is 0 on 10/31, you will not see a line on the report for it. it does not pull in. I would be expecting to see 0 instead of -991,529.51.
Anonymous
Not applicable
Hi @gmasta1129 ,
``````Measure =
VAR cur_rk = [RK]
VAR cur_flexacct =
SELECTEDVALUE ( 'Table'[flex_acct] )
VAR cur_cd =
SELECTEDVALUE ( 'Table'[if ValueDaate/Run Date] )
VAR tmp =
FILTER (
ALL ( 'Table' ),
'Table'[flex_acct] = cur_flexacct
&& [RK] = cur_rk - 1
)
VAR _val =
CALCULATE ( MAX ( 'Table'[usd_balance] ), tmp )
VAR cur_rd =
SELECTEDVALUE ( 'Table'[Run Date] )
VAR _pre_date =
CALCULATE ( MAX ( 'Table'[Run Date] ), tmp )
VAR cd =
DATEDIFF ( cur_rd, _pre_date, DAY )
RETURN
IF ( DAY ( cur_rd ) = 1 && cd <> 1, 0, IF ( cur_cd = "N", 0, _val ) )
``````
Please refer the attached .pbix file.
Best regards,
Community Support Team_ Binbin Yu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Anonymous
Not applicable
Hi @gmasta1129 ,
1. below is my test table
Table:
2. create measure with below dax formula
``````RK =
VAR cur_flexacct =
SELECTEDVALUE ( 'Table'[flex_acct] )
VAR tmp =
FILTER ( ALL ( 'Table' ), 'Table'[flex_acct] = cur_flexacct )
RETURN
RANKX ( tmp, CALCULATE ( MAX ( 'Table'[Run Date] ) ),, ASC, DENSE )
``````
``````Measure =
VAR cur_rk = [RK]
VAR cur_flexacct =
SELECTEDVALUE ( 'Table'[flex_acct] )
VAR cur_cd =
SELECTEDVALUE ( 'Table'[if ValueDaate/Run Date] )
VAR tmp =
FILTER (
ALL ( 'Table' ),
'Table'[flex_acct] = cur_flexacct
&& [RK] = cur_rk - 1
)
VAR _val =
CALCULATE ( MAX ( 'Table'[usd_balance] ), tmp )
RETURN
IF ( cur_cd = "N", 0, _val )
``````
3. add a table visual with fields and measure
Please refer the attached .pbix file.
Best regards,
Community Support Team_ Binbin Yu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Helper III
I entered your formula and the first row is pulling in -805,000 instead of -800,000.
Please see screenshot below for reference.
Super User
Hi @gmasta1129 ,
I don't think I understand your requirements. Here is what I was trying to show:
For example, the new column should contain a value of -800,000 (previous day USD balance) in the first row and in the second row a value of 0 since the if statement is equal to "N".
Helper III
Hello @djurecicK2,
Thank you for the quick response but I would need the previous day balance to pull into the column. Threfore 10/31/2022 run date balance of 800,000 should pull into the first row and 0 should pull into the second row.
Super User
Hi @gmasta1129 ,
Here is a way to do that in DAX by creating a new column:
NewBalance = IF('Table'[RunDate]='Table'[ValueDate],'Table'[Balance],0)
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# No more analysis of issue?
Author Message
Intern
Joined: 17 May 2012
Posts: 5
No more analysis of issue? [#permalink]
### Show Tags
08 Nov 2012, 07:09
Am I right to assume that AWA will only consists questions that are ' Analysis of Argument'? There would be no analysis of issue? I have read the changes on the official gmat website, but I'm not sure if I understood it correctly..
Thanks
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4128
Re: No more analysis of issue? [#permalink]
### Show Tags
08 Nov 2012, 15:45
Am I right to assume that AWA will only consists questions that are ' Analysis of Argument'? There would be no analysis of issue? I have read the changes on the official gmat website, but I'm not sure if I understood it correctly..
Thanks
Yes, that is perfectly correct. As of June 2012, when the IR question was introduced, the number of essays on the GMAT dropped from two to one, and the one they explicitly eliminated was the "Analysis of an Issue."
Here are some blogs you might find relevant:
http://magoosh.com/gmat/2012/new-awa-on-the-gmat/
http://magoosh.com/gmat/2012/brainstorm ... -gmat-awa/
http://magoosh.com/gmat/2012/gmat-awa-example-essay/
Mike
_________________
Mike McGarry
Magoosh Test Prep
Re: No more analysis of issue? [#permalink] 08 Nov 2012, 15:45
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https://cs.stackexchange.com/questions/119777/functor-laws-and-natural-transformations-in-haskell | 1,656,937,370,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104375714.75/warc/CC-MAIN-20220704111005-20220704141005-00539.warc.gz | 234,293,987 | 68,870 | # Functor laws and natural transformations in Haskell
As I've been struggling to get a deeper understanding of monads in Haskell, I started reading about functors and their counterparts in category theory. Keep in mind that I have no background in the latter.
According to Milewski(1), any polymorphic function of the type F a -> G a, where F and G are functors, automatically satisfies the naturality condition, although I have yet to figure out why. Anyways, this should hold in particular when F is the identity functor, which if understand correctly would correspond to the aforesaid function having the type a -> G a. Now, for simplicity let:
data G a = C a
If G is a functor, then since C :: a -> G a, C is a natural transformation. Viceversa, suppose C is a natural transformation. Thus, for all functions f:
(fmap f) . C == C . f
which I imagine is what leads to the wrapping-unwrapping analogy. In fact, if we render C x as {x} as in "wrapped x", then the previous can be re-written as:
(fmap f) {x} == {f x}
which quite literally means that applying the transformed f to the wrapped value is the same as unwrapping, applying f to the unwrapped value and re-wrap it. By the previous equality:
law 1) (fmap id) {x} == {id x}
== {x}
law 2) (fmap (f . g)) {x} == {(f . g) x}
== {f (g x)}
== (fmap f) {g x}
== (fmap f) ((fmap g) {x})
== ((fmap f) . (fmap g)) {x}
It follows that G is a functor.
So is it correct to say that requiring the functor laws on a type constructor is equivalent to requiring that its data constructors satisfy the naturality conditions or something along these lines? And if not, then what is the purpose of the functor laws from a "practical" point of view?
Edit: I just wanted to clarify that I don't really need an answer to the previous question. What really troubles me is that most tutorials on Haskell's functors make it seem like their wrapper-like nature (on which the intuition behind applicatives and monads is often based) stems from their laws when in reality it seems to be a consequence of the naturality conditions due to Wadler's free theorems(2), unless I'm completely mistaken. What role do the functor laws play in their interpretation as abstractions of some functional programming pattern?
(1) Milewski, B.; Category Theory for Programmers; 164
I'm having a little trouble exactly what question you're asking, but I think it might help if we went through the connection between homomorphisms and free theorems.
But here's the executive summary of what I think you want to know:
• The fmap laws are not redundant when you have free theorems. On the contrary, you need fmap to state the free theorem for any function that involves a Haskell Functor.
• Specifically, if a function has the type of a natural transformation, its free theorem is the naturality condition expressed in terms of fmap.
First off, let's get a bit of intuition about what the parametricity theorem states. An informal way of stating it is this: For any function which has a "forall" in its type, that "forall" really does mean "for all types". The function cannot make any assumptions whatsoever about the type.
To explain what this means when we involve types with an algebra, we need the language of homomorphisms. So let's review what a homomorphism is. (Sorry if this is revision for you, but I think it's worth going through this.)
Let's consider monoids. A monoid is a set $$M$$ along with two operations $$0_M$$ and $$+_M$$ which obeys the laws:
$$\forall x \in M, x +_M 0_M = 0_M +_M x = x$$ $$\forall x,y,z \in M, x +_M (y +_M z) = (x +_M y) +_M z$$
A monoid homomorphism is a function between two monoids $$f : M \rightarrow N$$ which "respects" the two operations:
$$f(0_M) = 0_M$$ $$f(x +_M y) = f(x) +_N f(y)$$
In Haskell, we might represent monoids as a type class:
class Monoid m where
mzero :: m
mplus :: m -> m -> m
If M1 and M2 are instances of Monoid, a Monoid homomorphism f :: M1 -> M2 must obey the laws:
f mzero = mzero
f (mplus x y) = mplus (f x) (f y)
Remember that the monoid operations on the left-hand side of these "laws" are different from the monoid operations on the right-hand side. The left-hand side are the operations for M1 and the right-hand side are the operations for M2.
Now consider the following typeclass (which isn't the same as Haskell's Ord typeclass, but is close enough):
class Ordered a where
lessThan :: a -> a -> Bool
Again, there are laws that any reasonable "less than" operator must obey. But what I want to concentrate on here is the Bool part.
What does it mean for a function to "respect" this algebra? It essentially means that the Bool has to be the same. That is, if A and B are both instances of Ordered, an Ordered homomorphism f :: A -> B must obey the law:
lessThan (f x) (f y) = lessThan x y
Think about what this means semantically: this law is stating that for any two elements, f doesn't change their relative ordering. That is what it means for f to "respect" the ordering algebra.
OK, now back to the parametricity theorem. The free theorem for the function:
sortBy :: forall a. (a -> a -> Bool) -> [a] -> [a]
is this, written out in a slightly fuller form:
for all f :: A -> B,
If lt :: A -> A -> Bool is a function such that
for all x, y :: A, lt x y = lt (f x) (f y)
Then
for all xs :: [A], sortBy lt . map f = map f . sortBy lt
We can say the same thing with type classes:
sort :: forall a. Ordered a => [a] -> [a]
The free theorem here is, essentially:
for all f :: A -> B,
if f is an Ordered homomorphism
then sort . map f = map f . sort
Which is, of course, just that f doesn't change the ordering of elements, then performing the substitution map f and then sorting is the same thing as sorting them performing the substitution.
The complication with Functor is that it's a constructor class, so the intuition about what a Functor homomorphism should be is a little less clear. But let's step back a moment and talk about category homomorphisms.
Categories are algebras just like any other. A category has two operations, $$1$$ and $$\circ$$, which satisfies the laws that you know and love. So a category homomorphism $$F$$ is a function which satisfies:
$$F(\hbox{id}) = \hbox{id}$$ $$F(g \circ f) = F(g) \circ F(f)$$
If you think of a functor between the "Haskell category" (objects are Haskell types, morphisms are Haskell functions) and a Haskell Functor category (objects are Haskell functors and morphisms are Haskell functions between these objects), then this is exactly the Functor laws.
fmap id = id
fmap (g . f) = fmap g . fmap f
Now here's the important part: The fmap operation is the homomorphism precondition that you need for free theorems involving Functor.
Consider, for example, something with the same type as fmap but isn't necessarily fmap:
instance Functor F where ...
somethingLikeFmap :: forall a b. (a -> b) -> (F a -> F b)
The free theorem for this function is:
if q . f = g . p,
then somethingLikeFmap q . fmap f = fmap g . somethingLikeFmap p
From this you can conclude a bunch of interesting things, such as that if somethingLikeFmap id = id then fmap = somethingLikeFmap. In particular, somethingLikeFmap can be characterised entirely by what it does with id.
This has to be the case. The parametricity theorem, which generates these free theorems, states the "forall" in a parametric type really does mean "for all types".
If the functor is a container-like type, for example, the function cannot make any decision about what to do based on the value of an element in the container. But it could do anything with the structure of the container which doesn't depend on the values in it.
somethingLikeMap1 f = reverse . map f
somethingLikeMap2 f = map f . tail
somethingLikeMap3 f = []
If a polymorphic function does has a way to make decisions based on the values (e.g. lessThan in the case of sort), then the free theorem states that this is the only way that it can make a decision about what to do, so if the mapping doesn't affect the decisions (e.g. the values returned by lessThan), then it doesn't affect the operation of the polymorphic function (e.g what sort does).
So this is why the fmap laws are not redundant with respect to free theorems: they define the homomorphism between the Haskell category and the (endo-) functor category that you need to state the free theorems.
Finally, what is the free theorem of a function which has the type of a natural transformation?
instance Functor F where ...
instance Functor G where ...
eta :: forall a. F a -> G a
It is:
fmap_G f . eta = eta . fmap_F f
This is exactly the naturality condition. This a good way to think about what's so "natural" about a natural transformation.
• It’s most likely my fault but I could draw hardly anything from this. I can’t pinpoint where you prove the first claim but I was aware of it even if informally, but let’s just take it for granted. What if we didn’t have free theorems? What if we had different laws? I guess what I was looking for was a more direct relation, if any, between the functor laws and functors behaving like containers. But maybe I'm just trying to read too much into it. Jan 23, 2020 at 13:48
• Sorry about that. I probably pitched this at the wrong level. I assumed that you knew enough algebra to know what a homomorphism is, because that's what you need to know to understand the motivation for category theory. Can you let me know a little more about where you are at? Jan 23, 2020 at 20:50
• Don't worry, I wanted to understand monads and related classes a bit better but not at the cost of learning an entire theory. I thought that knowing more about functors would explain their use as "containers" but maybe they just happen to fit? Anyways, I understand the first part of your answer but it gets cryptic when you talk about somethingLikeFmap, I just can't follow your reasoning. Jan 23, 2020 at 22:06
• I have a suggestion if that's the point where you got stuck. Get Wadler's original "Theorems for Free" paper, and implement the free theorem generator in Haskell. You will learn a lot. Jan 23, 2020 at 22:43 | 2,531 | 10,156 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-27 | longest | en | 0.934963 |
https://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/articles.cgi?read=1025 | 1,571,806,386,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987829458.93/warc/CC-MAIN-20191023043257-20191023070757-00465.warc.gz | 887,303,980 | 2,134 | The Museum of HP Calculators
HP Articles Forum
N-queens benchmark and the HP-17BII
Posted by Thomas Klemm on 7 Feb 2011, 5:48 a.m.
## Equation for the 8-queens problem
```QUEENS:
Q=A+B+C+D+E+F+G+H+
(I:1:8:1:
(J:1:8:1:
IF(I<>J AND SQ(I-J)<>1:
(K:1:8:1:
IF(I<>K AND SQ(I-K)<>4
AND J<>K AND SQ(J-K)<>1:
(L:1:8:1:
IF(I<>L AND SQ(I-L)<>9
AND J<>L AND SQ(J-L)<>4
AND K<>L AND SQ(K-L)<>1:
(M:1:8:1:
IF(I<>M AND SQ(I-M)<>16
AND J<>M AND SQ(J-M)<>9
AND K<>M AND SQ(K-M)<>4
AND L<>M AND SQ(L-M)<>1:
(N:1:8:1:
IF(I<>N AND SQ(I-N)<>25
AND J<>N AND SQ(J-N)<>16
AND K<>N AND SQ(K-N)<>9
AND L<>N AND SQ(L-N)<>4
AND M<>N AND SQ(M-N)<>1:
(O:1:8:1:
IF(I<>O AND SQ(I-O)<>36
AND J<>O AND SQ(J-O)<>25
AND K<>O AND SQ(K-O)<>16
AND L<>O AND SQ(L-O)<>9
AND M<>O AND SQ(M-O)<>4
AND N<>O AND SQ(N-O)<>1:
(P:1:8:1:
IF(I<>P AND SQ(I-P)<>49
AND J<>P AND SQ(J-P)<>36
AND K<>P AND SQ(K-P)<>25
AND L<>P AND SQ(L-P)<>16
AND M<>P AND SQ(M-P)<>9
AND N<>P AND SQ(N-P)<>4
AND O<>P AND SQ(O-P)<>1:
L(A:I)xL(B:J)xL(C:K)xL(D:L)xL(E:M)xL(F:N)xL(G:O)xL(H:P)/0
:0)):0)):0)):0)):0)):0)):0)))
```
## Solution
Solve the equation for Q. After a while (3:29) you get an error message:
```SOLUTION NOT FOUND
```
Now you can recall the solution in the variables A-H. The result { 1, 5, 8, 6, 3, 7, 2, 4 } corresponds to the following configuration:
``` A B C D E F G H
+---+---+---+---+---+---+---+---+
8 | | | # | | | | | |
+---+---+---+---+---+---+---+---+
7 | | | | | | # | | |
+---+---+---+---+---+---+---+---+
6 | | | | # | | | | |
+---+---+---+---+---+---+---+---+
5 | | # | | | | | | |
+---+---+---+---+---+---+---+---+
4 | | | | | | | | # |
+---+---+---+---+---+---+---+---+
3 | | | | | # | | | |
+---+---+---+---+---+---+---+---+
2 | | | | | | | # | |
+---+---+---+---+---+---+---+---+
1 | # | | | | | | | |
+---+---+---+---+---+---+---+---+
```
## References
Edited: 7 Feb 2011, 6:27 a.m.
Password: | 943 | 2,001 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-43 | longest | en | 0.651949 |
https://www.mapleprimes.com/users/H-R/questions?page=4 | 1,718,550,553,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00899.warc.gz | 774,908,919 | 23,500 | ## 155 Reputation
11 years, 105 days
## Fourier & Fourier-Bessel Series Expansio...
Does Maple have any tool or package that computes the Fourier & Fourier-Bessel series expansions of a given funtion "f(x)" over a specified interval "[a,b]"?
## How to define the derivative of a functi...
Maple 18
Take a look at below. I was expecting maple to give me "g'(1)"! :)
## Writting a Linear System in Matrix Form...
I have a set "EQ" containing N linear equations in N unknowns. The only symbolic variables in each "EQ[i]" are the unknowns. I want to write a procedure that derives the matrices "A" and "b" where A.x=b is the same linear system stored in "EQ". In other words, I want to write the linear system in the matrix form.
Can anyone guide me through writting such a procedure?
As an example do it with the system written in the following file.
Note: This procedure will be used for large linear systems (e.g. 2000 Equations, 2000 Unknowns) so it will be important that the procedure uses the least operations required.
LinearSystem.mw
## Solving Large Linear System of Algebraic...
I have a large system of linear algebraic equations that I want to solve (2005 Unknowns, 2005 Equations). I was wondering that what are the proper commands to use in maple for solving the system as fast as possible. Take a look at the files in the download link if you want to see the system of linear algebraic equations.
http://pc.cd/h79
Please provide me any suggesitons that you may think will be helpful like using other sofwares that are good in doing this work such as MATLAB or something else.
## Updating Maple Integration Database...
Maple 18
I have an integral that maple can not solve but I can solve it by hand. How can I add this to maple integration database?
f:=int(r^2*BesselJ(0,a*r)*BesselI(1,b*r),r) | 442 | 1,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-26 | latest | en | 0.917628 |
https://mybiasedcoin.blogspot.com/2008/08/on-simulations.html?showComment=1218326280000 | 1,726,824,845,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00413.warc.gz | 367,308,040 | 18,594 | ## Wednesday, August 06, 2008
### On Simulations
I've been coding up some simulations for some Allerton papers that are due all too soon. Of late I've depended far too much on my (now former) student Adam Kirsch to take care of doing our simulations, but he's graduated, and they're needed, so off I go. (Adam's graduating is all to the good, but clearly, I'll be missing him, especially when it comes time to write simulations.)
I'm always amazed at how averse theory people seem to be to doing simulations. I find them useful for generating ideas and thinking about problems in the early stages -- cutting off wrong directions and giving insight into the right ones. If you don't like doing simulations for such purposes, because it doesn't work for you, or you're clever enough to not need data, I have no issue with that -- people work differently.
But I also use simulations as a way of checking my work. If I have a theorem that says that a random process will behave a certain way, and it's possible to code a simulation of the process, I'll check my theorem with code. If the theory and the code don't match up, my assumption is that something is wrong somewhere, and the result is not ready until the two match or I know why they don't. Surprisingly, I think it's about 50-50 as to which I end up finding is wrong, the code or the theorem. (By the same token, if I don't have a theorem, and there's more than one way to simulate a process, I'll code multiple simulations, and make sure they match!)
Of course not all results can be checked by coding something up -- but many can. Particularly in the study of random processes, which is my area. And it's clear to me that many researchers don't check by coding -- because I (or students working with me) have several times found mistakes by doing a simple implementation and finding that we get different numbers out than the paper gives. Generally the mistakes aren't "fatal" -- usually a constant is off somewhere, and often eventually the O notation will take care of it -- but of course it is grating as a reader when something in a paper is plain wrong and you're left to figure out why. When someone doesn't do a check-by-code, I must admit, it lowers my trust of and my overall opinion of the person's work. Sure, people make mistakes (myself included) -- but if you're ignoring a straightforward approach for checking your work, that doesn't inspire confidence.
I imagine some theory people are so out of practice coding they "can't" do a simulation. (But hey, that's not really an excuse, that's what grad students are for...) And others probably just consider it a waste of time. If you really are good enough not to need to check your work this way, more power to you. Me, I'll get back to the (admitted drudgery of) coding my things up and seeing if they work the way I think they should....
Anonymous said...
Full quotation. Simulations help
understanding the problem and moreover to prove you are at least
not wrong when you write theorems.
As a side note, since people publishes
theorems, I think they shall
also publish the source code
(e.g. on their webpage).
The purpose is twofold: to help
others take confidence with your
theoretical frameworks, saving
time for future workup, and to
keep alive the confrontation between theory and practice.
Mark Wilson said...
Worst-case bounds for algorithm performance can often be shown up as pretty flabby by some basic simulation, but that doesn't seem to be done by most authors.
I agree that when theoretical results are corroborated by computation, the source code and results should be made available publicly and referenced in the paper. Perhaps there should be some sort of policy by journals or SIGACT, for example, on this.
Unknown said...
"Worst-case bounds can often be shown up as pretty flabby..."
My experience is somewhat different. As a practioner who uses theoretical work, the typical problem is that the simulation is much more optimistic than reality. My favorite rule is from colleagues who do switch design -- "the Devil gets to pick your input patterns" and the challenge, of course, is to figure out what the Devil would pick (usually it is the output of a nasty, hard to simulate or analyze, multi-state process). And it locks you right against the worst-case bound.
Anonymous said...
My cynical opinion is that the arguments in this thread beautifully explain why most theorists don't run any simulations:
(1) You might find theoretical mistakes. Why take that chance? In a field where conferences don't publish retractions, there's no shame in making a mistake in a conference paper. If nobody catches the mistake, then so much the better. If somebody catches a fatal mistake, then at least you got one extra conference paper out of it. (Non-experts will never know, and experts probably won't really hold it against you, since anybody could make a mistake.) If somebody catches a correctable mistake, then you silently fix it in the journal version (if any) while vaguely thanking the person who caught it for "helpful conversations" in the acknowledgments.
Overall, if you don't want people to make mistakes, then you have to penalize them for it. There's enough strategic behavior on the part of students seeking jobs that you can't just count on them to do the right thing. (Of course, most people try to do the right thing, but even a minority can have a big effect on the field.)
(2) Mark Wilson's comment about flabbiness is quite right. If you prove a quadratic bound while simulations suggest the algorithm really runs in n log n time, then your result looks a little less impressive. It's rare for the theoretical analysis to be quite as strong as the simulation results, so including simulations in your paper is practically an advertisement for the weaknesses of your analysis.
Anonymous said...
Regarding (2): this should be
exactly the point of distinguishing
theory and practice and a step
towards living with the fact that
no big-Oh notation is as good
as it seems. Everybody lives with
it outside the theoretician community, why shouldn't the community itself.
Anonymous said...
Isn't (2) really a positive thing? If you can prove a quadratic bound, but simulations show n log n, you can say that we show a worst case n^2 bound, but the algorithm is even better in practice! It is only an advertisement for your algorithm.
Anonymous said...
^^
Haha, exactly! Thats what I thought too when I read it :)
Michael Mitzenmacher said...
I think it's a very interesting open question whether a paper that had results of the form: "We design an algorithm with O(n^2) performance. We do not have a corresponding lower bound. Experimental results suggest the performance is O(n log n) on many cases of interest." would be considered "desirable' at a high-level (FOCS/STOC) theory conference. I'd like to think so -- I certainly think it wouldn't be an issue for many more practically oriented theory conferences. But I wouldn't be surprised to see a review for such a paper that said something like "Remove the experiments. They're a waste of space and add nothing to the results," or even "While the O(n^2) result is nice, the authors themselves seem to suggest that the right answer might be O(n \log n). They should resolve this issue by either proving this bound or providing a lower bound before publication."
The problem is I don't think the community in general appreciates simulation work, and there's not really a standard as to when such work is desirable or acceptable as an adjunct to the theory. In networking, it's more clear -- such simulations are almost by default required (indeed, to the extreme of whether they actually add meaningfully to the results or not).
I'm enjoying the debate and hope for more comments on this.
Anonymous said...
Michael, I fail to see the debate that you are enjoying. There's no theorist is in this "debate", just some confused youngsters who like to bash what they think is "theory."
Your position on this is sufficiently well known in the theory circles, to the point that everyone is bored. It is also not a constructive position. You need to have a lot more standing in theory before your somewhat radical positions will be treated seriously. Say, if someone like Feige was telling us to include more experiments in STOC/FOCS papers, then there would be a debate.
PS: This comment is not meant as a personal attack, though I can see how it lends itself to such an interpretation. I'm just saying that any realistic attempt to change a field begins by convincing those guys that you can be one of the best by their own rules, and only then proposing new rules.
Michael Mitzenmacher said...
Anon: I suppose I believe it's more effective to persuade by argument than by force of personality. While I would agree with you on Feige's stature in the theory community, I doubt a pronouncement by him would be any more (or less) effective than a pronouncement by me about what FOCS/STOC should be like. Convincing the people who serve on PCs what is acceptable is the main way to enact change at the conference level -- and there I expect younger people will be either to persuade, again by argument and (where possible) by example.
Anonymous said...
"I doubt a pronouncement by [Feige] would be any more (or less) effective than a pronouncement by me about what FOCS/STOC should be like."
I'm not a great fan of proof-by-reputation either, but I think you are suffering from a bit of Harvarditis here ;)
Anonymous said...
Michael, I agree with you in general and it certainly makes sense to perform and/or include simulations for certain, specific types of results. However, I would guess that a large majority of theory papers are either not amenable to simulations at all (e.g., lower bounds, crypto, coding theory, complexity, learning theory) or would not benefit from having experimental results included (e.g., algorithms papers where the improvement doesn't show up until impractically large instance size).
In fact, here is an experiment: what fraction of papers in the last few STOC/FOCS proceedings would you say would be helped by simulations? (One can ask the same question for SODA, where I expect the percentage would be larger but still not a majority.)
Michael Mitzenmacher said...
Anon 12: I would certainly agree that some, or perhaps even many, theory papers are not suitable for simulations; but I'm skeptical it's a large majority. Of course, obviously I'm biased as I work in areas where simulations, I think, should be more common.
As for your experiment, I'm not following FOCS/STOC enough to comment on the fraction that could benefit from simulation. But based on your comment I went to the FOCS 2008 home page to take a look, since I feel you've "called me out". The first paper that caught my interest, k-wise independent graphs by Alon and Nussboim, would seem to be improved by simulation. Indeed, in their abstract they claim as a motivation for looking at k-wise independent graphs that handling large G(N,p) graphs becomes infeasible and "cheaper" graphs such as k-wise independence graphs must be used instead. (I'm essentially quoting them; please see their abstract.) It would have been nice to see some evidence to back this statement up somewhere in the paper, by simulation. Simulations might also have been useful to test the accuracy of their statement and check when the asymptotics kick in as well.
Going down further in the list, I saw the Succincter paper by Patrascu, a paper I had looked at before since it deals with compression, an area I like. (Indeed, Mihai, if you're reading this, you should cite my Compressed Bloom Filter paper in the journal version, where I essentially ask for the solution your paper provides...) This is a paper that (to me) screams for implementation results. I've put it on my personal short list of "good implementation projects for undergrads"; I'm curious as to how well the methods work in practice, and the paper, naturally, gives no real clue.
So of the papers I've looked at, which are in areas I'm interested in, I think 2 for 2 could benefit from some simulation work.
To be clear, I'm not saying these are bad papers in any way. Indeed, the point is that they're good papers -- they interest me. I think they could be better papers -- both more interesting, and more useful to more people -- if they had some additional simulation work as well.
Anonymous said...
I'm just saying that any realistic attempt to change a field begins by convincing those guys that you can be one of the best by their own rules, and only then proposing new rules.
This sort of attitude is exactly why so many things in academia that should change don't. Aside from giving a good rationalization for not listening to anybody, it's simply a false observation about change. Worse yet, it's false in both directions: this strategy is neither sufficient nor necessary for effecting change.
Establishing an incredible reputation doesn't help much in changing what a field values (witness Madhu Sudan and his aliens paper). If Feige announced that FOCS/STOC papers should feature more simulations, then a few people might listen to him, but most would ignore him, and graduate students would go around talking about how poor Uri was getting soft in his old age.
In the other direction, systemic change doesn't generally come from brilliant scientists convincing their colleagues through pure reason. Instead, it comes from political machinations. For example, the prevalence of open access publishing in biomedical research didn't come about by personally convincing scientists that it was ethically or scientifically necessary. Rather, it came about because NIH started requiring it for all grant-supported research, after a big fight.
The upshot is that if you want to change the way CS theory is done, it won't be easy to do it by convincing individual theorists that there's a better path (regardless of whether you are right, or for that matter whether you are Uri Feige). It may be better to focus on convincing hiring committees, or tenure committees, or NSF panels, that their standards and goals should shift a little. Wherever the funding and jobs go, the research community will follow.
There's no theorist is in this "debate", just some confused youngsters who like to bash what they think is "theory."
Actually, I'm commenter #5, and I'm a theorist, as well as no longer so young. I freely admit I'm no Uri Feige, though (I think it's safe to say the few Uri Feige-level theorists don't waste their time commenting anonymously on blogs).
Anonymous said...
While I agree with Michael, I'd like to point out that in fields where people do run experiments, you get terrible mistakes also. And they are as baffling (or more baffling) than when no experiment was done.
Someone implements some algorithm and says that some measured constant is 1.222. You implement the same algorithm, and get a constant of 1.444. Where is the catch? If you cannot compare the implementations, it is very difficult to even publish a note saying there is a contradiction. You can try emailing the authors, but good luck getting access to their implementation! (There were studies done on this topic and most authors will refuse to provide you with their implementations!)
So, in my opinion, it is not enough to run experiments, you must also publish your source code. If you made a mistake, make sure everyone can see it. That is real science!
As for implementation being a waste of time... Knuth was not afraid to get his hands dirty and even maintain the code he published.
Finally, the idea that only people at the top of the food chain, the very people who benefit the most from the current system, can or should change the system is wrong in so many ways... that's called cultivating biases. That's also a fallacy because changes almost never occur because an old man decided to reform the system.
Anonymous said...
"Convincing the people who serve on PCs what is acceptable is the main way to enact change at the conference level..."
In my experience in ACM and IEEE, this approach sometimes works, but is a hard slog and often runs into institutional resistance.
The better solution is to create your own conference with like-minded fellows. If the paper produces wonderful papers and exciting results, it can change the field. I've seen it happen at least 4 times over the past 15 years in data comm research. (I'll note the hazard here is that the conference almost booms -- is big enough to be financially viable but never quite achieves enough oomph to become one of the top 3 or 4 conferences in the field -- and then it lingers on the calendar for years). | 3,570 | 16,779 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-38 | latest | en | 0.974645 |
https://web2.0calc.com/questions/quadrilaterals_1 | 1,521,753,923,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648003.58/warc/CC-MAIN-20180322205902-20180322225902-00069.warc.gz | 744,521,225 | 6,090 | +0
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The angles of a quadrilateral are \$x\$, \$5x + 15^\circ \$, \$3x - 25^\circ\$, and \$4x - 20^\circ \$. Find the measure of the largest angle of the quadrilateral.
eileenthecoolbean Aug 7, 2017
#1
+6604
+3
The sum of the angles in a quadrilateral = 360° So.....
x + 5x + 15 + 3x - 25 + 4x - 20 = 360
Combine like terms.
13x - 30 = 360
Add 30 to both sides of the equation.
13x = 390
Divide both sides of the equation by 13 .
x = 30
Since x is positive, the largest angle must be 5x + 15 .
( It has the largest coefficient of x and all the others either add nothing or subtract something. )
And..... if x = 30, 5x + 15 = 5(30) + 15 = 165°
hectictar Aug 7, 2017
Sort:
#1
+6604
+3
The sum of the angles in a quadrilateral = 360° So.....
x + 5x + 15 + 3x - 25 + 4x - 20 = 360
Combine like terms.
13x - 30 = 360
Add 30 to both sides of the equation.
13x = 390
Divide both sides of the equation by 13 .
x = 30
Since x is positive, the largest angle must be 5x + 15 .
( It has the largest coefficient of x and all the others either add nothing or subtract something. )
And..... if x = 30, 5x + 15 = 5(30) + 15 = 165°
hectictar Aug 7, 2017
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# Proportion and Quartiles
Please see the attached file for the questions.
1. A set of data whose histogram is bell shaped yields a sample mean and standard deviation of 50 and 4, respectively. Approximately what proportion of observations are:
a) Between 46 and 54?
b) Between 42 and 58?
c) Between 38 and 62?
2. Calculate the first, second, and third quartiles for the following sample.
5 8 2 9 5 3 7 4 2 7 4 10 4 3 5
3. There is a garbage crisis in North America - too much garbage and no place to put it. As a consequence, the idea of recycling has become quite popular. A waste-management company in a large city is willing to begin recycling newspapers, aluminum cans, and plastic containers. However, it is profitable to do so only if a sufficiently large proportion of households are willing to participate. In this city, 1 million households are potential recyclers. After some analysis it was determined that, for every 1,000 households that participate in the program, the contribution to profit is \$500. It was also discovered that fixed costs are \$55,000 per year. It is believed that 50,000, 100,000, 200,000, or 300,000 households will participate with probabilities of .5, .3, .1, and .1, respectively. A preliminary survey was performed where 25 households were asked whether they would be willing to be part of this recycling program.
Suppose only 3 of the 25 respond favorably. Incorporate this information into a decision-making process to decide whether the waste-management company should proceed with the recycling venture.
4. Suppose the following statistics were calculated from data gathered from a randomized block experiment with k = 4 and b = 10:
SS(total) = 1,210 SST = 275 SSB = 625
(a) Can we conclude from these statistics that the treatment means differ? (use α =.01)
(b) Can we conclude from these statistics that the block means differ? (Use α = .01)
5. A random sample of 50 observations yielded the following frequencies for the standardized intervals:
Interval Frequency
Z ≤ - 1 6
-1 < Z ≤ 0 27
0 < Z ≤ 1 14
Z >1 3
Can we infer that the data are not normal? Use α = .10.
#### Solution Summary
This solution is comprised of a detailed explanation for different questions on proportions, frequencies, chi-square test. All the formulas are given with respect to different questions such as mean, standard deviation, z score, probability calculation using standard normal distribution, quartile, median and chi-square test. Full explanantion is given for every question.
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https://www.physicsforums.com/threads/understanding-the-circuit-diagram-for-an-amp-hour-meter-a-guide-for-beginners.72777/ | 1,713,486,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817249.26/warc/CC-MAIN-20240418222029-20240419012029-00888.warc.gz | 832,615,755 | 27,169 | # Understanding the Circuit Diagram for an Amp Hour Meter: A Guide for Beginners
• mrjeffy321
In summary: I don't know what it does, or where it goes, or why it's neede.So I am just going to guess at it and hope for the best?In summary, the circuit diagram shows a way to measure how much an amplifier has been used by connecting it to a resistor and measuring the voltage across it. There are several leads that need to be connected to a power supply, and an optional resistor is included. The diagram also includes a button which is used to reset the chip. The screen for the device needs to be connected to an LCD, and the device itself needs to be solder to a IC "seat".
mrjeffy321
This thread is a continuation of my previous thread, except this one is primarily focusing on the contruction of an amp hour meter and programming it, rather than the gathering of parts for it. (since the parts have just arrived, very slow mail order)
I am getting very confused when looking at the circuit diagram (provided by faust9), I don't really know what is going on here, I am just going off faith that it will work.
For the purposes of my questions, I will divide up the diagram into 3 sections,
Section 1 (from the far left side to R5)
-W4, and W5 are "connectors of some sort, earlier given as bannana plugs. this is where my circuit that I wish to measure is connected up correct, accorss the . 1 ohm ressitor? that would put the amp hour meter in parallel with my circuit, so far so good if I am understanding.
Section 2 (from R5 to where R3 reconnects to PB3)
-The triangle shaped symbol with all the numbers is the Op Amp Right?, Which is really shaped like a recangular chip with no sort of numbers on it coresponding to the diagram, luck me.
-Pointing up and down from the Op Amp, from #7 and #4 are arrowsm one is +5 v, which I assume I need to connect to my separate input powersupply to supply it with 5 volts, but what about the other arrow, does that goto a ground?
-by R7, which is an optional ressitor which I was going to leave out, there is an arrow pointing in both directions, with a -5v, so I assume that this needs to be connected with the -5 volts from my power supply, what is the arrow pointing in the other direction? what do I do if I choose to leave out R7, or even if I keep it in, how is it connected?
Section 3 (from where R3 reconnects to PB3 to the right side of the page).
-I am still unlear on why the "I" (labeled SW) symbol is needed or what exactly I am looking to put in there. I know it is a Push Button Normally open switch, but why is it neede, what does it do? and the arrow comming out to its left, is that ground?
-Then there are several more +5v leads that I need to connect to my input 5 volt power supply.
-Down at the bottom, on the ship, #4, there is a ground arrow, but I don't even know where I will connect this to a ground, the only ground I ever use is a screw in a plate of a light switch cover on the wall. but I don't think this is what you were meaning.
-Then there are the W s, which are for in circuit programming I think. What if I decide I don't want to do any of that, I will just program it correctly out of circuit, then once I get it right, solder it in then and only then. couldn't I leave those out then?
-Then it connects to the LCD screen, I am sure I'll have problems when I get to that stage.
I am getting really confuses since I don't know what I am doing here, help if very much appreciated.
1. No, the R0 resistor goes in series with the circuit. It "converts" the current going through it into a voltage for the differential amplifier (741) to measure.
2. Yes, its an op-amp, yes its 5V, yes, to ground.
-R7 is a variable resistor across pins 1 and 5 of the 741 and with the resistor wiper at -5V. It is used to offset the op-amp. Depending on how big the offset needs to be you could leave the resistor out, as long as you put something in the program code that will subtract the offset after the chip has done its ADC. This could however lead to continualy re-programming the device to zero the meter.
3. The "I" is a switch which is used to reset the chip, by taking pin 1 low. It's used to reset the the display. The arrow is connected to ground.
-Pin4 of the chip goes to the same ground as the others.
-I would highly recommend that you solder a IC "seat" and not the chip itself. Then you can program, slot the chip in, try it out, unplug the chip to reprogram, put into try again etc, etc. without all the soldering in between. It will also prevent you heating the IC up too much if you solder it directly.
Do you have any deatils on the LCD screen you intend using?
updated for clarity, grammer and cos I misinterpreted the function
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Thanks, that brought me a little closer to understanding all this.
All this stuff that apperantly came so easy to you, is all new to me.
what I meant by the series/parallel above was the my original circuit would be in series with the R0 resistor, but the amp hour meter circuit would be in parallel with my original circuit. so the amp hour meter won't really affect my original circuit.
As for my LCD screen, I got a slightly different one than I said I was going to in the previous thread, I found one cheaper somewhere else.
I don't know how to descibe it to well, it is a 2x16 character display and has a 8-bit LCD controller (HD44780 or equiv).
it has 16 holes along the bottom (and a couple smaller holes).
Here is the info I have on it:
1: Gnd
2: Vdd (power)
3: Vee (contrast adjust) works well near Gnd
4: RS
5: RW
6: E - Clock line
7-14: Data0-Data7
Interfacing to the display simply involves sending a 8-bit command and display data.
As for the reset button, I will get some, excuse the term, cheapo button that is normally open and I push it to close the circuit.
For the ground, I don't know what I will do, as I said before, wheneber I see a ground in a circuit, I would always use the screen on a light swirch cover (or a water focet/pipe), but I doubt this is what is meant? Will just leaving it as a wire attached to nothing work in this case, if the ground isn't that necesary in this case. What do "normal" people's circuits use for a ground in this kind of case?
So if I chose to leave off the optional [varaible] resistor, R7, how would I connect the -5v to the opAmp? would I connect it to both pins 1 and 5?
Sorry about the ground. In this application ground is not earthing but simply the 0V line of your power supply.
If you don't use the optional resistor leave the pins unconnected. But if you have a -5V supply available then I would definitely make use of it.
Can't find much on your LCD display
LCD stuff:
http://www.doc.ic.ac.uk/~ih/doc/lcd/
Ever seen those LCD displays on computer cases?
http://www.beyondlogic.org/parlcd/parlcd.htm
As for the ground, in electronics, ground can be something other than zero(floating ground) or Earth as mentioned by Delta. Ground is simply a reference that your circuit is built around. Your powersupply to your circuit (your wall wart or batteries) should be used as the reference where the ground is the negative battery terminal or the negative terminal of your wall-wart connector(somoe of these are center (+) while others are center (-)---make sure you verify the polarity if you use one of these).
Delta covered the rest it seems
Oh, those 'W's are used for something called in-circuit programming. You don't need this functionality but it makes life easier. They are a little incomplete in that I didn't show isolation circuits (something like a 10K resistor between the LCD and the W) because these are not needed. They are just a covenience. AVR datasheets cover ICSP pretty well. Also, LCDs read 8 bits ususaaly. A lot of LCD controllers have a provision for two four bit transmission but if yours doesn't working around that isn't really that tough. You'll need to find the datasheet to verify your controllers exact functionality.
Last edited by a moderator:
I hate to keep asking what might seem like the same questions over and over again, but it is helping.
this "wall wart", it is some type of external power supply to the amp hour meter circuit, for example a battery or AC/DC converter plugged into the wall?
This power supply's output should be fairly close to 5v, but it isn't that critical that I get it right on, because I have the voltage regulator to "regulate" it.
Now if I may turn your attention to the PS circuit (figure 5).
The input on the left, would that not be the + side of the power supply of whatever I find to power it (battery, ...), and the output, the - side?
The rectangle in the middle, labeled, "MC78XX/LM78XX", is that the voltage regulator? I now have both my capacitors (1 .1 uF tantalum and a 1 uF electrolytic capacitor <- the closest one I could find to .33 uF). Then at the bottom middle, you see a 3 pronged device, what is that? could it be the voltage regulator insted and the "MC78XX..." is the battery?
OK, so now where ever I see a ground or -5v in the circuit, I will connect it up to the negative side of the power supply, and where ever I see a +5v, I will connect it up to the positive side of the power supply.
I did end up getting some of those IC socket/"Seat" things to colder into the circuit rather than the real chips (OpAmp and Microcontroller), so they will make life a bit easier.
http://catb.org/~esr/jargon/html/W/wall-wart.html
http://encyclobeamia.solarbotics.net/articles/wall_wart.html
The answer to your P/S question is yes. The left pin (looking down at the regulator with the metal tab away from you--the metal tab extends up above the plastic and is embedded in the back of the 7805 as well) is the input power pin from your external source. The middle pin is connected to the ground or (-) terminal of your source as well as the ground of all of the components you use. The right pin is the 5V output.
The rectangle in the middle is the voltage regulator and the three pronged device is know as chassis ground.
Next, yes. Like delta said if you don't want to offset your OpAmp then leave those pins unconnected (floating) and connect the 5V to all of the appropriate sources as well as all of the GNDS.
Additionally, the power to the 7805 can be fairly high (up to 18V for an 05 if I recall correctly) so as long as your external P/S is a little over 5V all the way up to 18V then you should be OK.
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OK, I think I have assembled the circuit correctly (I don't know for sure, since I can't test it, I can't test it until I program it).
the basic idea for how this will work is, I will be continually (as fast as the chip will let me), be checking the amps in the circuit, then from that, get the amp hours. I will add that current cycles amp hours to the running total, and display it on the LCD screen. Since the chip runs at 20 MHz (so claims the manufacturer), that it will have 20 million (20,000,000) cycles per second, we will also assume a constant current in between cycles, after all how much can it change in 1/20 millionth of a second.
Check the amps, multiply the number of amps by 1.38888889 × 10-11 (the number of hours in each cycles), and then add that to the total.
Now obviously, that is going to give us a very small number, a very very small number indeed, also we dont really need that accurate of a measurement, so we might want to use a counter variable to count off every 20,000 cycles or so, and then take a measurement, insted of every single cycle.
Sound good? we will still get a measurement taken every milisecond (1/1000), and that would mean that we would then multiply the amp ready by 2.77777778 × 10-07, giving us a much higher (and more significant) number.
Also, maybe insted of displaying the running total of amp hours, we can insted update it only every second, so only update the LCD display every 20,000,000th cycle.
And then later, if I discover that I want to have "super duper" acuracy, it won't be all that hard to change.
I think that if I write this code in C, it will one, be a lot easier for me to write, understand, and maintain in the furture, as opposed to writting it in assembly.
The only problem is, since this is specialized code here (they don't teach this stuff in any learn to program books, nor do you run into it on an everyday basic), I don't have a clue on how to do the most critical elements of the program, read the amps and display the data on the LCD.
Wanting to do this in C rather than assembly makes it even harder to find any sample source code to use.
Any ideas?
mrjeffy321 said:
OK, I think I have assembled the circuit correctly (I don't know for sure, since I can't test it, I can't test it until I program it).
the basic idea for how this will work is, I will be continually (as fast as the chip will let me), be checking the amps in the circuit, then from that, get the amp hours. I will add that current cycles amp hours to the running total, and display it on the LCD screen. Since the chip runs at 20 MHz (so claims the manufacturer), that it will have 20 million (20,000,000) cycles per second, we will also assume a constant current in between cycles, after all how much can it change in 1/20 millionth of a second.
Check the amps, multiply the number of amps by 1.38888889 × 10-11 (the number of hours in each cycles), and then add that to the total.
Now obviously, that is going to give us a very small number, a very very small number indeed, also we dont really need that accurate of a measurement, so we might want to use a counter variable to count off every 20,000 cycles or so, and then take a measurement, insted of every single cycle.
Sound good? we will still get a measurement taken every milisecond (1/1000), and that would mean that we would then multiply the amp ready by 2.77777778 × 10-07, giving us a much higher (and more significant) number.
Also, maybe insted of displaying the running total of amp hours, we can insted update it only every second, so only update the LCD display every 20,000,000th cycle.
And then later, if I discover that I want to have "super duper" acuracy, it won't be all that hard to change.
I think that if I write this code in C, it will one, be a lot easier for me to write, understand, and maintain in the furture, as opposed to writting it in assembly.
The only problem is, since this is specialized code here (they don't teach this stuff in any learn to program books, nor do you run into it on an everyday basic), I don't have a clue on how to do the most critical elements of the program, read the amps and display the data on the LCD.
Wanting to do this in C rather than assembly makes it even harder to find any sample source code to use.
Any ideas?
I would suggest coding in asm because your chip only has 1Kx8 of flash memory to work with. You can do it in C; however, writing code in C as efficient as asm code takes as muck time and effort to do as writing in asm to begin with(it can be done though).
Now, your chip comes with a default clock speed of 9.6Mhz---you don't need to change that because you are still sampeling the ADC at a rate far greater than one second intervals.
Here's the basic idea of what you want to do:
If you do things in this order, then life should be a little easier when you start debugging your program:
1) set up all of your ports---you need to tell the MCU which ports are input and which are output ports.
2) Set up your ADC---you need to determine the ADC prescaler, left justify (tthe ADC is 10 bits, the MCU is an 8 bit processor---doing it this way will make things easier and you'll only loose resolution at the very bottom end of your scale: outside of your defined operating conditions I might add) set the voltage reference to Vcc, which channel is read, and what not.
4) start your main program. Write everything you want to do in little functions i.e. send data to LCD and call these little functions as needed. It's a good idea to send data to the lcd every second so your LCD doesn't turn into a blur of numbers. You will have to write a timer function to do this though. The timer on the MCU is an 8 bit timer thus it counts from 0 to 255---every clock cycle. When the clock overflows (goes from 255 to 0 again) an interrupt is triggered (this is a good place to use interrupts). Write an interrupt routine to clear the overflow bit, and increment an a bit register. You'll probably have to chain a few registers together to get your one second delay between LCD updates. You'll be well served to set your timer prescaler to 1024 which means you'll have to count off about 9570 clock pulses between LCD writes or 2^14 bits--two registers. You'll keep incrementing one register when the timer interrupt occurs. Check the carru flag of the sreg. If the carry flag is set then increment the upper reg. check to see if the 5th (5:0) bit is set. If set send data to LCD and clear both high and low registers. If cleared then start process over again. You can do this by decrementing as well (there's actually a good reason to use decrementing for this vice incrementing but it elludes me at the moment).
Make and ACD read function--move contents of ACDH to a reg and perform the needed math on that. reset adc start process over again.
If you want to keep a running total of amps consumed then you will need to determine the largest time needed(days, weeks, months, years...) and set aside enough registers to store your results.If you wanted to keep a days worth of data for instance then you'd have 24(3600)(5)=432000 which equates to 19 bits of information (3 registers).
Your going to have to write the whole program out then once you have everything working you'll need to determine a trimmer variable to correct the time. Basically, make your circuit, connect it to a test circuit with a know current and let things run for a time (the longer the better not to exceed you max time though)---say an hour. Your MCU will say (XXXX) while the real amps/hour may only be XXX. Use this to develope a trimmer value which you can add to a register or use as a multiplier to correct the MCU to real life.
Well, I know I missed a few things, but the above should help you get going. Good luck and don't let asm scare ya. It's easy if you think things out step by step including "where do I want this value to be stored". C takes care of the where question usually in asm you have to handle it.
OK, that was way over my head. I understood some of it, and the concept of most of it, but knowing how to actually do that is a different story.
Also,
I don't understand how it will be able to calculate the current from just reading the voltage across the .1 ohm ressitor. If you use Ohms law (V = RI), so the current = V/R, or V/.1 in this case, but the .1 resistor isn't the only resistor in the circuit to be measured, how will the amp hour meter know this? There is something missing in my mind on how it works. And we haven't even gotten into the specifics of doing the actual programming yet.
Machine code is by no means my specialty, but I can tell you a little about the .1 ohm resistor concern. Any current going through your main load HAS to also go through the .1 ohm resistor. Using ohms law you can compute the current through it. There is NO MORE to this. Current in .1 ohm = current in load = current in the wires carrying the current to the load. Series circuit, current is the same.
I am afraid I don't understant.
Lets say we have a 5 volt power source for the mail circuit.
connect this power source to a circuit with say... a 10 ohm resistor, which is them connected in series with the .1 ohm resistor in the amp hour meter circuit, for a combined ressitance of 10.1 ohms.
I = V / R
We know the potential difference/voltage is 5 volts.
We also know that the real combined resistance is 10.1 ohms.
So the real current is 5 / 10.1 = .495 amps.
But for the hour meter, all it knows if that it is measuring the voltage across a .1 ohm resistor, so
I = 5 / .1 = 50 amps!
or is this where the trimmer would normally come in, or in my case, the corretion calculation in code?
Don't forget you also have an opamp (10x gain IIR) and you'll need to multiply each calculation by the time required to complete one sample. The opamp will be easy to factor in as will the timeing. once you have your code you'll then be able to figure out the average time to complete a cycle. You'll end up with some correction factor which consists of a time factor and the gain factor multipied by the adc result(it's a good thing avr's have a floating point because this would be a bear to do as an integer). Keep a running total of the results and you're good to go.
mrjeffy321 said:
I am afraid I don't understant.
Lets say we have a 5 volt power source for the mail circuit.
connect this power source to a circuit with say... a 10 ohm resistor, which is them connected in series with the .1 ohm resistor in the amp hour meter circuit, for a combined ressitance of 10.1 ohms.
I = V / R
We know the potential difference/voltage is 5 volts.
We also know that the real combined resistance is 10.1 ohms.
So the real current is 5 / 10.1 = .495 amps.
But for the hour meter, all it knows if that it is measuring the voltage across a .1 ohm resistor, so
I = 5 / .1 = 50 amps!
or is this where the trimmer would normally come in, or in my case, the corretion calculation in code?
You're measuring the voltage dropped across the 0.1 ohm resistor only. The 10 ohm resistor will drop most of the voltage while the 0.1: $$V_{0.1}=\frac{V_{source}}{R_{10}+R_{0.1}}*R_{0.1}}=0.049$$
Since you know ohm's law and you are measuring the voltage dropped by the resistor and you know the value of the resistor then you know two of the three variables. Here: let's say the voltage on the output of the opamp is 3 volts with a 10x gain. That tellsyou there is a .3 volt drop across the resistor and 0.2volts/o.1ohms=I through resistror=3 amps. Now if it takes 100 clock cycles to figure this out (AVR studio will tell you this) and one clock cycle is 103 nanoseconds then you'd have (103 nanoseconds)*100*3/3600=8.58333333 × 10-09 amp-hours for that one conversion. Just keep adding the conversions to get get the running total.
Last edited:
I was discussing this project and the circuit diagram with some other people and they expressed some concerns about why an OpAmp was choosen as "the shunt resistor instead of the differential input of the ATtiny13?". Also, there was a commend made that the 741 [OpAmp] isn't in common use now-a-days, why choose it over say the build in feature of the chip?
I'm sure there's loads of missed stuff in here, as only a sample of the circuit is available.
However a look at the ATtiny13 datasheet seems to show that the chip only gives a logic output if the analogous signal on one pin (Ain0) is greater on the other (AIN1). Using the chips ADC directly isn't much use if we are measuring incircuit current, unless one end of the shunt is at 0V line with the ATtiny13. Therefore an intermittant comparator is being used.
## 1. What is an amp hour meter?
An amp hour meter is a device that measures the amount of electrical charge that has been consumed or discharged from a battery over time. It is typically used in applications where the battery's capacity and usage need to be monitored, such as in electric vehicles or renewable energy systems.
## 2. How does an amp hour meter work?
An amp hour meter works by measuring the current flowing in and out of the battery and integrating it over time. It uses a shunt resistor to measure the current and a microcontroller to calculate and display the amp hour reading. Some meters also have additional features such as temperature compensation and state of charge indicators.
## 3. Can an amp hour meter be used for any type of battery?
Yes, an amp hour meter can be used for any type of battery as long as it is within the meter's voltage and current rating. However, different types of batteries may require different calibration settings due to their different discharge characteristics.
## 4. What are the benefits of using an amp hour meter?
An amp hour meter provides accurate and real-time monitoring of a battery's usage, which can help prolong its lifespan and prevent overcharging or undercharging. It also allows for better management of energy usage and can provide insights into the performance of a battery over time.
## 5. How do I install an amp hour meter?
The installation process of an amp hour meter may vary depending on the specific model and application. However, it typically involves connecting the meter's shunt resistor in series with the battery's positive terminal and wiring the meter's display to a power source. It is recommended to follow the manufacturer's instructions for proper installation and calibration.
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931 | 6,138 | 25,319 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-18 | latest | en | 0.969405 |
http://www.chegg.com/homework-help/questions-and-answers/consider-the-solid-inside-the-surface-x2y2z2-9-and-outside-the-surface-x2y2z2-1use-spheric-q3312732 | 1,369,249,109,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702298845/warc/CC-MAIN-20130516110458-00044-ip-10-60-113-184.ec2.internal.warc.gz | 391,223,525 | 7,993 | ## A quick description of your question...
Consider the solid inside the surface x^2+y^2+z^2 = 9 and outside the surface x^2+y^2+z^2 = 1.Use spherical coordinates to write the integral to calculate the volume of the solid, then calculate the integral. | 63 | 252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.665911 |
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# Factorize (i) (x^4+3x^2+4) <br> (ii) (x^4+5x^2+9)
Question from Class 9 Chapter Factorisation Of Polynomials
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Solution :
We have <br> `(i) (x^4+3x^2+4)=(x^4+4x^2+4)-x^2` <br> `=(x^2+2)^2-x^2` <br> `=(x^2+2-x)(x^2+2+x)` <br> `=(x^2-x+2)(x^2+x+2)`. <br> ` therefore (x^4+3x^2+4)=(x^2-x+2)(x^2+x+2)`. <br> `(ii) (x^4+5x^2+9)=(x^4+6x^2+9)-x^2` <br> `=(x^2+3)^2-x^2` <br> `=(x^2+3-x)(x^2+3+x)` <br> `=(x^2-x+3)(x^2+x+3)`. <br> `therefore (x^4+5x^2+9)=(x^2-x+3)(x^2+x+3)`.
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http://thetopsites.net/article/52637589.shtml | 1,611,821,665,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704839214.97/warc/CC-MAIN-20210128071759-20210128101759-00715.warc.gz | 98,942,606 | 6,893 | ## Haskell:Check if first element in a list is a duplicate
how to check if a list is empty in haskell
haskell remove first element from list
haskell index of element in list
erlang remove duplicates from list
ocaml remove duplicates from list
I am writing a recursive function called `threeN` that takes a list with one element and adds a new number depending on if it's even or odd. If it's even divide it by 2, and if it's odd multiply it by three and subtract 1. I'm having trouble with my base case which will check the list to see if an element is already contained in the list.
It should be like this:
```prelude> threeN [9]
[14,5,10,20,7,14,28,56,19,38,13,26,9]
```
```prelude> threeN [9]
[9]
```
This is my code so far:
```threeN :: [Integer] -> [Integer]
threeN [] = []
threeN [n]
| n `elem` [n] = [n]
| n `mod` 2 ==0 = threeN[n `div` 2] ++ [n]
| n `mod` 2 ==1 = threeN[(3*n)-1] ++ [n]
```
This is supposed to be done with basic functions, so I can't use very advanced ones to solve the issue, which is why I'm having trouble.
`threeN` consumes a list: `xs`, starting with a single element in it. It produces a new element: `x'` based on the value of the head: `x` of `xs`; it prepends `x'` to `xs` if `x'` hasn't occurred in `xs`.
```threeN :: [Int] -> [Int]
threeN [] = []
threeN l@(x:xs)
| x' `elem` l = l -- base case
| otherwise = threeN (x':l)
where
x' = if even x then x `div` 2 else x * 3 - 1
```
Haskell function that tests if a list has repeated (duplicate) elements , This means that you'll have to keep up with a list of elements that you've already visited so you can check this. So first, write a function that Comparing the length of the list against the length of the set has the disadvantage that it needs to traverse the entire list rather than short-circuiting on the first duplicate. If you want it to work (fsvo work) on infinite lists, then this is not an option. A Set is still a useful data structure for this,
Determining whether a list contains duplicates - haskell, Hi /r/Haskell, long-time lurker and fan of Haskell, first time actually writing non-toy I have a list of Ord a, and would like to "efficiently" determine whether or not it duplicates hasDuplicatesWith seen (x:xs) = -- If we have seen the current item In case no such element exists you have to print -1. If there are multiple elements in a which are repeated at least k times, then print these elements ordered by their first occurrence in the list. So I wrote a few different functions to help with this. count which counts the number of occurrences of an element in a list
Why the signature is `[Integer] -> [Integer]`? the input is actually just a number. The following code works.
```threeN :: Integer -> [Integer]
threeN n = threeN' n []
where threeN' n acc
| n `elem` acc = n:acc
| even n = threeN' (n `div` 2) (n:acc)
| odd n = threeN' (3 * n - 1) (n:acc)
```
If you are force to use `[Integer]` as input signature:
```threeN :: [Integer] -> [Integer]
threeN [n] = threeN' n []
where threeN' n acc
| n `elem` acc = n:acc
| even n = threeN' (n `div` 2) (n:acc)
| odd n = threeN' (3 * n - 1) (n:acc)
```
But I think It does not make sense.
Regards!
Data.List.Unique, Safe Haskell, Safe. Language, Haskell2010. Data.List.Unique. Description. Library provides the functions to find unique and duplicate elements in the list the first - all elements with removed duplicates (like sortUniq but the result is not isUnique function is to check whether the given element is unique in the list or not. Check if list contains a value x. list is an iterable finite container. List::Util 'first' Find first index of an element in list;
You could use `head` and `tail` with `elem` to test whether the first element already exists in the list. Note however that `head` and `tail` are unsafe functions. They will crash if given an empty list.
```threeN :: [Integer] -> [Integer]
threeN ns | n `elem` tail ns = ns
| even n = threeN ([n `div` 2]++ns)
| odd n = threeN ([3*n-1]++ns)
where
```
Also if you do not want the repeat number to be in the output, then have the first guard just equal `tail ns` instead of `ns`. There is probably a more efficient algorithm to generate these lists, but this just modifies what you've provided.
Filter Duplicate Elements in Haskell, If you write uniq as a right fold, you don't need to pass an accumulator through, and the list comes out in the right order: uniq :: Eq a => [a] -> [a] uniq [] = [] uniq Make a new list containing just the first N elements from an existing list. take n xs. Split a list into two smaller lists (at the Nth position). splitAt n xs (Returns a tuple of two lists.) Delete the just Nth element of a list. This is tricky. AFAIK, there is no built-in function that does this.
Find the first duplicated element, A List with a duplicate element, with 0 or more elements before, between, and Haskell, 35 bytes xŒQ¬\$Ḣ Input: array M \$ Operate on M ŒQ Distinct sieve - Returns a boolean mask where an index is truthy for the first occurrence of an element Test suite. Remove from Q the first appearance of every element in Q, then Documentation. complex :: Eq a => [a] -> ([a], [a], [a]) Source # complex function is a complex investigation of the list. It returns triple: the first - all elements with removed duplicates (like sortUniq but the result is not sorted) the second - the elements that are repeated at least once in the list
[PDF] Solutions to Exercises, Provide a function to check if a character is alphanumeric, that is lower case, upper case or numeric. the element duplicated. The answer is n elements from a list and that it is also possible to drop the first m elements from a list. Of course, it If the first list contains duplicates, so will the result. >>> [1,2,2,3,4] `intersect` [6,4,4,2] [2,2,4] It is a special case of intersectBy, which allows the programmer to supply their own equality test. If the element is found in both the first and the second list, the element from the first list will be used.
List.Extra - list-extra 8.2.4, Find the first minimum element in a list using a comparison function Indicate if list has duplicate values when supplied function are applyed on each values. The function foldl1 took in 7.0.0 was b -> a -> b consistent with the Haskell A function is applied to each element of the list and then the equality check is To check if a list contains any duplicate element follow the following steps, Add the contents of list in a set. As set contains only unique elements, so no duplicates will be added to the set. Compare the size of set and list.
• Your code matches the line `9 `elem` [9] = [9]` so it just returns `[9]`
• Why is the second to last element in the example list `26` instead of `28`? `9` is odd, so `9*3+1=28`? Am I missing something? | 1,824 | 6,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-04 | latest | en | 0.885802 |
https://www.aqua-calc.com/what-is/density/milligram-per-us-tablespoon | 1,723,596,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00409.warc.gz | 513,027,576 | 22,370 | # What is a milligram per US tablespoon (unit)
## The milligram per US tablespoon is a unit of measurement of density
The milligram per US tablespoon density measurement unit is used to measure volume in US tablespoons in order to estimate weight or mass in milligrams
• What is densityInstant conversionsConversion tables
• 1 mg/tbsp = 0.0676280454 µg/mm³
• 1 mg/tbsp = 67.6280454 µg/cm³
• 1 mg/tbsp = 67 628.0454 µg/dm³
• 1 mg/tbsp = 67 628 045.4 µg/m³
• 1 mg/tbsp = 67.6280454 µg/ml
• 1 mg/tbsp = 67 628.0454 µg/l
• 1 mg/tbsp = 338.140227 µg/metric tsp
• 1 mg/tbsp = 1 014.42068 µg/metric tbsp
• 1 mg/tbsp = 16 907.0114 µg/metric c
• 1 mg/tbsp = 1 108.22511 µg/in³
• 1 mg/tbsp = 1 915 012.99 µg/ft³
• 1 mg/tbsp = 51 705 350.6 µg/yd³
• 1 mg/tbsp = 333.333333 µg/tsp
• 1 mg/tbsp = 1 000 µg/tbsp
• 1 mg/tbsp = 2 000 µg/fl.oz
• 1 mg/tbsp = 16 000 µg/US c
• 1 mg/tbsp = 32 000 µg/pt
• 1 mg/tbsp = 64 000 µg/US qt
• 1 mg/tbsp = 256 000 µg/US gal
• 1 mg/tbsp = 6.76280454×10-5 mg/mm³
• 1 mg/tbsp = 0.0676280454 mg/cm³
• 1 mg/tbsp = 67.6280454 mg/dm³
• 1 mg/tbsp = 67 628.0454 mg/m³
• 1 mg/tbsp = 0.0676280454 mg/ml
• 1 mg/tbsp = 67.6280454 mg/l
• 1 mg/tbsp = 0.338140227 mg/metric tsp
• 1 mg/tbsp = 1.01442068 mg/metric tbsp
• 1 mg/tbsp = 16.9070114 mg/metric c
• 1 mg/tbsp = 1.10822511 mg/in³
• 1 mg/tbsp = 1 915.01299 mg/ft³
• 1 mg/tbsp = 51 705.3507 mg/yd³
• 1 mg/tbsp = 0.333333333 mg/tsp
• 1 mg/tbsp = 2 mg/fl.oz
• 1 mg/tbsp = 16 mg/US c
• 1 mg/tbsp = 32 mg/pt
• 1 mg/tbsp = 64 mg/US qt
• 1 mg/tbsp = 256 mg/US gal
• 1 mg/tbsp = 6.76280454×10-8 g/mm³
• 1 mg/tbsp = 6.76280454×10-5 g/cm³
• 1 mg/tbsp = 0.0676280454 g/dm³
• 1 mg/tbsp = 67.6280454 g/m³
• 1 mg/tbsp = 6.76280454×10-5 g/ml
• 1 mg/tbsp = 0.0676280454 g/l
• 1 mg/tbsp = 0.000338140227 g/metric tsp
• 1 mg/tbsp = 0.001014420681 g/metric tbsp
• 1 mg/tbsp = 0.01690701135 g/metric c
• 1 mg/tbsp = 0.00110822511 g/in³
• 1 mg/tbsp = 1.91501299 g/ft³
• 1 mg/tbsp = 51.7053507 g/yd³
• 1 mg/tbsp = 0.000333333333 g/tsp
• 1 mg/tbsp = 0.001 g/tbsp
• 1 mg/tbsp = 0.002 g/fl.oz
• 1 mg/tbsp = 0.016 g/US c
• 1 mg/tbsp = 0.032 g/pt
• 1 mg/tbsp = 0.064 g/US qt
• 1 mg/tbsp = 0.256 g/US gal
• 1 mg/tbsp = 6.76280454×10-11 kg/mm³
• 1 mg/tbsp = 6.76280454×10-8 kg/cm³
• 1 mg/tbsp = 6.76280454×10-5 kg/dm³
• 1 mg/tbsp = 0.0676280454 kg/m³
• 1 mg/tbsp = 6.76280454×10-8 kg/ml
• 1 mg/tbsp = 6.76280454×10-5 kg/l
• 1 mg/tbsp = 3.38140227×10-7 kg/metric tsp
• 1 mg/tbsp = 1.014420681×10-6 kg/metric tbsp
• 1 mg/tbsp = 1.690701135×10-5 kg/metric c
• 1 mg/tbsp = 1.108225108×10-6 kg/in³
• 1 mg/tbsp = 0.00191501299 kg/ft³
• 1 mg/tbsp = 0.05170535073 kg/yd³
• 1 mg/tbsp = 3.33333333×10-7 kg/tsp
• 1 mg/tbsp = 1.0×10-6 kg/tbsp
• 1 mg/tbsp = 2.0×10-6 kg/fl.oz
• 1 mg/tbsp = 1.6×10-5 kg/US c
• 1 mg/tbsp = 3.2×10-5 kg/pt
• 1 mg/tbsp = 6.400000007×10-5 kg/US qt
• 1 mg/tbsp = 0.0002560000003 kg/US gal
• 1 mg/tbsp = 6.76280454×10-14 t/mm³
• 1 mg/tbsp = 6.76280454×10-11 t/cm³
• 1 mg/tbsp = 6.76280454×10-8 t/dm³
• 1 mg/tbsp = 6.76280454×10-5 t/m³
• 1 mg/tbsp = 6.76280454×10-11 t/ml
• 1 mg/tbsp = 6.76280454×10-8 t/l
• 1 mg/tbsp = 3.38140227×10-10 t/metric tsp
• 1 mg/tbsp = 1.014420681×10-9 t/metric tbsp
• 1 mg/tbsp = 1.690701135×10-8 t/metric c
• 1 mg/tbsp = 1.108225108×10-9 t/in³
• 1 mg/tbsp = 1.91501299×10-6 t/ft³
• 1 mg/tbsp = 5.170535073×10-5 t/yd³
• 1 mg/tbsp = 3.33333333×10-10 t/tsp
• 1 mg/tbsp = 1.0×10-9 t/tbsp
• 1 mg/tbsp = 2.0×10-9 t/fl.oz
• 1 mg/tbsp = 1.6×10-8 t/US c
• 1 mg/tbsp = 3.2×10-8 t/pt
• 1 mg/tbsp = 6.400000007×10-8 t/US qt
• 1 mg/tbsp = 2.560000003×10-7 t/US gal
• 1 mg/tbsp = 2.385509097×10-9 oz/mm³
• 1 mg/tbsp = 2.385509097×10-6 oz/cm³
• 1 mg/tbsp = 0.0023855091 oz/dm³
• 1 mg/tbsp = 2.3855091 oz/m³
• 1 mg/tbsp = 2.385509097×10-6 oz/ml
• 1 mg/tbsp = 0.0023855091 oz/l
• 1 mg/tbsp = 1.19275455×10-5 oz/metric tsp
• 1 mg/tbsp = 3.57826365×10-5 oz/metric tbsp
• 1 mg/tbsp = 0.000596377275 oz/metric c
• 1 mg/tbsp = 3.909149024×10-5 oz/in³
• 1 mg/tbsp = 0.0675500952 oz/ft³
• 1 mg/tbsp = 1.82385257 oz/yd³
• 1 mg/tbsp = 1.175798731×10-5 oz/tsp
• 1 mg/tbsp = 3.527396192×10-5 oz/tbsp
• 1 mg/tbsp = 7.05479239×10-5 oz/fl.oz
• 1 mg/tbsp = 0.0005643833904 oz/US c
• 1 mg/tbsp = 0.00112876678 oz/pt
• 1 mg/tbsp = 0.00225753356 oz/US qt
• 1 mg/tbsp = 0.00903013426 oz/US gal
• 1 mg/tbsp = 1.490943188×10-10 lb/mm³
• 1 mg/tbsp = 1.490943188×10-7 lb/cm³
• 1 mg/tbsp = 0.0001490943188 lb/dm³
• 1 mg/tbsp = 0.149094319 lb/m³
• 1 mg/tbsp = 1.490943188×10-7 lb/ml
• 1 mg/tbsp = 0.0001490943188 lb/l
• 1 mg/tbsp = 7.45471594×10-7 lb/metric tsp
• 1 mg/tbsp = 2.236414782×10-6 lb/metric tbsp
• 1 mg/tbsp = 3.72735797×10-5 lb/metric c
• 1 mg/tbsp = 2.44321814×10-6 lb/in³
• 1 mg/tbsp = 0.00422188095 lb/ft³
• 1 mg/tbsp = 0.1139907856 lb/yd³
• 1 mg/tbsp = 7.34874207×10-7 lb/tsp
• 1 mg/tbsp = 2.20462262×10-6 lb/tbsp
• 1 mg/tbsp = 4.409245244×10-6 lb/fl.oz
• 1 mg/tbsp = 3.52739619×10-5 lb/US c
• 1 mg/tbsp = 7.05479239×10-5 lb/pt
• 1 mg/tbsp = 0.0001410958478 lb/US qt
• 1 mg/tbsp = 0.0005643833912 lb/US gal
• 1 mg/tbsp = 1.043660231×10-6 gr/mm³
• 1 mg/tbsp = 0.001043660231 gr/cm³
• 1 mg/tbsp = 1.04366023 gr/dm³
• 1 mg/tbsp = 1 043.66023 gr/m³
• 1 mg/tbsp = 0.001043660231 gr/ml
• 1 mg/tbsp = 1.04366023 gr/l
• 1 mg/tbsp = 0.005218301157 gr/metric tsp
• 1 mg/tbsp = 0.01565490345 gr/metric tbsp
• 1 mg/tbsp = 0.2609150586 gr/metric c
• 1 mg/tbsp = 0.01710252703 gr/in³
• 1 mg/tbsp = 29.5531667 gr/ft³
• 1 mg/tbsp = 797.935501 gr/yd³
• 1 mg/tbsp = 0.005144119446 gr/US tsp
• 1 mg/tbsp = 0.01543235835 gr/US tbsp
• 1 mg/tbsp = 0.03086471671 gr/fl.oz
• 1 mg/tbsp = 0.2469177336 gr/US c
• 1 mg/tbsp = 0.4938354673 gr/pt
• 1 mg/tbsp = 0.9876709346 gr/US qt
• 1 mg/tbsp = 3.95068374 gr/US gal
• 1 mg/tbsp = 4.633993094×10-12 sl/mm³
• 1 mg/tbsp = 4.633993094×10-9 sl/cm³
• 1 mg/tbsp = 4.633993094×10-6 sl/dm³
• 1 mg/tbsp = 0.0046339931 sl/m³
• 1 mg/tbsp = 4.633993094×10-9 sl/ml
• 1 mg/tbsp = 4.633993094×10-6 sl/l
• 1 mg/tbsp = 2.316996547×10-8 sl/metric tsp
• 1 mg/tbsp = 6.950989641×10-8 sl/metric tbsp
• 1 mg/tbsp = 1.158498274×10-6 sl/metric c
• 1 mg/tbsp = 7.59375414×10-8 sl/in³
• 1 mg/tbsp = 0.000131220072 sl/ft³
• 1 mg/tbsp = 0.003542941944 sl/yd³
• 1 mg/tbsp = 2.28405886×10-8 sl/tsp
• 1 mg/tbsp = 6.85217659×10-8 sl/tbsp
• 1 mg/tbsp = 1.370435321×10-7 sl/fl.oz
• 1 mg/tbsp = 1.09634825×10-6 sl/US c
• 1 mg/tbsp = 2.19269651×10-6 sl/pt
• 1 mg/tbsp = 4.385393026×10-6 sl/US qt
• 1 mg/tbsp = 1.75415721×10-5 sl/US gal
• 1 mg/tbsp = 7.45471594×10-14 short tn/mm³
• 1 mg/tbsp = 7.45471594×10-11 short tn/cm³
• 1 mg/tbsp = 7.45471594×10-8 short tn/dm³
• 1 mg/tbsp = 7.45471595×10-5 short tn/m³
• 1 mg/tbsp = 7.45471594×10-11 short tn/ml
• 1 mg/tbsp = 7.45471594×10-8 short tn/l
• 1 mg/tbsp = 3.72735797×10-10 short tn/metric tsp
• 1 mg/tbsp = 1.118207391×10-9 short tn/metric tbsp
• 1 mg/tbsp = 1.863678985×10-8 short tn/metric c
• 1 mg/tbsp = 1.22160907×10-9 short tn/in³
• 1 mg/tbsp = 2.110940475×10-6 short tn/ft³
• 1 mg/tbsp = 5.69953928×10-5 short tn/yd³
• 1 mg/tbsp = 3.674371035×10-10 short tn/US tsp
• 1 mg/tbsp = 1.10231131×10-9 short tn/US tbsp
• 1 mg/tbsp = 2.204622622×10-9 short tn/fl.oz
• 1 mg/tbsp = 1.763698095×10-8 short tn/US c
• 1 mg/tbsp = 3.527396195×10-8 short tn/pt
• 1 mg/tbsp = 7.05479239×10-8 short tn/US qt
• 1 mg/tbsp = 2.821916956×10-7 short tn/US gal
• 1 mg/tbsp = 6.655996375×10-14 long tn/mm³
• 1 mg/tbsp = 6.655996375×10-11 long tn/cm³
• 1 mg/tbsp = 6.655996375×10-8 long tn/dm³
• 1 mg/tbsp = 6.655996384×10-5 long tn/m³
• 1 mg/tbsp = 6.655996375×10-11 long tn/ml
• 1 mg/tbsp = 6.655996375×10-8 long tn/l
• 1 mg/tbsp = 3.3279981875×10-10 long tn/metric tsp
• 1 mg/tbsp = 9.9839945625×10-10 long tn/metric tbsp
• 1 mg/tbsp = 1.663999094×10-8 long tn/metric c
• 1 mg/tbsp = 1.090722384×10-9 long tn/in³
• 1 mg/tbsp = 1.884768281×10-6 long tn/ft³
• 1 mg/tbsp = 5.088874357×10-5 long tn/yd³
• 1 mg/tbsp = 3.28068842411×10-10 long tn/US tsp
• 1 mg/tbsp = 9.84206526786×10-10 long tn/US tbsp
• 1 mg/tbsp = 1.968413055×10-9 long tn/fl.oz
• 1 mg/tbsp = 1.574730442×10-8 long tn/US c
• 1 mg/tbsp = 3.149460888×10-8 long tn/pt
• 1 mg/tbsp = 6.298921777×10-8 long tn/US qt
• 1 mg/tbsp = 2.519568711×10-7 long tn/US gal
• 1 mg/tbsp = 1.06495942×10-11 st/mm³
• 1 mg/tbsp = 1.06495942×10-8 st/cm³
• 1 mg/tbsp = 1.06495942×10-5 st/dm³
• 1 mg/tbsp = 0.01064959421 st/m³
• 1 mg/tbsp = 1.06495942×10-8 st/ml
• 1 mg/tbsp = 1.06495942×10-5 st/l
• 1 mg/tbsp = 5.3247971×10-8 st/metric tsp
• 1 mg/tbsp = 1.59743913×10-7 st/metric tbsp
• 1 mg/tbsp = 2.66239855×10-6 st/metric c
• 1 mg/tbsp = 1.745155814×10-7 st/in³
• 1 mg/tbsp = 0.000301562925 st/ft³
• 1 mg/tbsp = 0.008142198971 st/yd³
• 1 mg/tbsp = 5.249101479×10-8 st/US tsp
• 1 mg/tbsp = 1.574730443×10-7 st/US tbsp
• 1 mg/tbsp = 3.149460889×10-7 st/fl.oz
• 1 mg/tbsp = 2.519568707×10-6 st/US c
• 1 mg/tbsp = 5.039137421×10-6 st/pt
• 1 mg/tbsp = 1.007827484×10-5 st/US qt
• 1 mg/tbsp = 4.031309937×10-5 st/US gal
• 1 mg/tbsp = 2.174292148×10-9 oz t/mm³
• 1 mg/tbsp = 2.174292148×10-6 oz t/cm³
• 1 mg/tbsp = 0.002174292146 oz t/dm³
• 1 mg/tbsp = 2.17429215 oz t/m³
• 1 mg/tbsp = 2.174292148×10-6 oz t/ml
• 1 mg/tbsp = 0.002174292146 oz t/l
• 1 mg/tbsp = 1.087146074×10-5 oz t/metric tsp
• 1 mg/tbsp = 3.261438219×10-5 oz t/metric tbsp
• 1 mg/tbsp = 0.0005435730388 oz t/metric c
• 1 mg/tbsp = 3.563026465×10-5 oz t/in³
• 1 mg/tbsp = 0.06156909729 oz t/ft³
• 1 mg/tbsp = 1.66236563 oz t/yd³
• 1 mg/tbsp = 1.071691551×10-5 oz t/US tsp
• 1 mg/tbsp = 3.215074656×10-5 oz t/US tbsp
• 1 mg/tbsp = 6.430149315×10-5 oz t/fl.oz
• 1 mg/tbsp = 0.000514411945 oz t/US c
• 1 mg/tbsp = 0.00102882389 oz t/pt
• 1 mg/tbsp = 0.00205764778 oz t/US qt
• 1 mg/tbsp = 0.008230591125 oz t/US gal
• 1 mg/tbsp = 1.81191012326×10-10 troy/mm³
• 1 mg/tbsp = 1.811910123×10-7 troy/cm³
• 1 mg/tbsp = 0.0001811910122 troy/dm³
• 1 mg/tbsp = 0.1811910122 troy/m³
• 1 mg/tbsp = 1.811910123×10-7 troy/ml
• 1 mg/tbsp = 0.0001811910122 troy/l
• 1 mg/tbsp = 9.05955062×10-7 troy/metric tsp
• 1 mg/tbsp = 2.717865182×10-6 troy/metric tbsp
• 1 mg/tbsp = 4.529775323×10-5 troy/metric c
• 1 mg/tbsp = 2.96918872×10-6 troy/in³
• 1 mg/tbsp = 0.005130758108 troy/ft³
• 1 mg/tbsp = 0.1385304689 troy/yd³
• 1 mg/tbsp = 8.930762927×10-7 troy/US tsp
• 1 mg/tbsp = 2.67922888×10-6 troy/US tbsp
• 1 mg/tbsp = 5.358457762×10-6 troy/fl.oz
• 1 mg/tbsp = 4.286766208×10-5 troy/US c
• 1 mg/tbsp = 8.573532418×10-5 troy/pt
• 1 mg/tbsp = 0.0001714706484 troy/US qt
• 1 mg/tbsp = 0.0006858825938 troy/US gal
• 1 mg/tbsp = 4.348584296×10-8 dwt/mm³
• 1 mg/tbsp = 4.348584296×10-5 dwt/cm³
• 1 mg/tbsp = 0.04348584292 dwt/dm³
• 1 mg/tbsp = 43.4858429 dwt/m³
• 1 mg/tbsp = 4.348584296×10-5 dwt/ml
• 1 mg/tbsp = 0.04348584292 dwt/l
• 1 mg/tbsp = 0.0002174292149 dwt/metric tsp
• 1 mg/tbsp = 0.0006522876438 dwt/metric tbsp
• 1 mg/tbsp = 0.01087146078 dwt/metric c
• 1 mg/tbsp = 0.0007126052929 dwt/in³
• 1 mg/tbsp = 1.23138195 dwt/ft³
• 1 mg/tbsp = 33.2473125 dwt/yd³
• 1 mg/tbsp = 0.0002143383103 dwt/US tsp
• 1 mg/tbsp = 0.0006430149313 dwt/US tbsp
• 1 mg/tbsp = 0.001286029863 dwt/fl.oz
• 1 mg/tbsp = 0.0102882389 dwt/US c
• 1 mg/tbsp = 0.0205764778 dwt/pt
• 1 mg/tbsp = 0.04115295561 dwt/US qt
• 1 mg/tbsp = 0.1646118225 dwt/US gal
#### Foods, Nutrients and Calories
CHOCOLATE ASSORTMENT SNOWFLAKES, UPC: 040000413318 contain(s) 526 calories per 100 grams (≈3.53 ounces) [ price ]
209328 foods that contain Total lipid (fat). List of these foods starting with the highest contents of Total lipid (fat) and the lowest contents of Total lipid (fat)
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Crystal River weighs 1 521.75 kg/m³ (94.99975 lb/ft³) with specific gravity of 1.52175 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Gold(III) fluoride [AuF3] weighs 6 750 kg/m³ (421.38873 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 15W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
A centimeter per second squared (cm/s²) is a derived metric SI (System International) measurement unit of acceleration
Energy is the ability to do work, and it comes in different forms: heat (thermal), light (radiant), motion (kinetic), electrical, chemical, nuclear and gravitational.
t/ft³ to mg/US c conversion table, t/ft³ to mg/US c unit converter or convert between all units of density measurement.
#### Calculators
Estimate the total volume of blood, blood plasma and blood cells | 6,199 | 12,646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2024-33 | latest | en | 0.279231 |
http://www.framsticks.com/trac/framsticks/browser/cpp/f0-fuzzy/neuroimpl-fuzzy.h?rev=109 | 1,695,790,403,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510259.52/warc/CC-MAIN-20230927035329-20230927065329-00652.warc.gz | 60,306,827 | 5,385 | # source:cpp/f0-fuzzy/neuroimpl-fuzzy.h@109
Last change on this file since 109 was 2, checked in by Maciej Komosinski, 14 years ago
added f0-fuzzy (the Fuzzy neuron and its f0-based mutation and crossover)
File size: 2.9 KB
Line
1#ifndef _NEUROIMPLFUZZY_H_
2#define _NEUROIMPLFUZZY_H_
3
4#include <stdlib.h>
5#include <math.h>
6
7#include "neuroimpl.h"
8#include "sstring.h"
9
10extern ParamEntry NI_FuzzyNeuro_tab[];
11
12/** Does the fuzzyfication process (of inputs) and defuzzyfication proces (of outpurs) - represents fuzzy rules
13*/
14class NI_FuzzyNeuro : public NeuroImpl
15{
16private:
17
18 double *fuzzySets; /// list of four digits which represents fuzzy sets: [0]-l, [1]-m, [2]-n, [3]-r, ... fuzzySet[4*i] = left, fuzzySet[4*i + 1] = midleft, fuzzySet[4*i + 2] = midright, fuzzySet[4*i + 3] = right
19
20 /** Determines, which fuzzy set is connected with each input of neuron. For instance third rule:
21 * 'IF input3 = fuzzy set #3 AND input5 = fuzzy set #1 then output2 = fuzzy set #6 AND output7 = fuzzy set #5'
22 * the variables shoul have values as shown below:
23 * RulesDef[4]=2; RulesDef[5]=2; //rule 3: 2 inputs, 2 outputs
24 * Rules[2][0]=3, Rules[2][1]=3, Rules[2][2]=5, Rules[2][3]=1, Rules[2][4]=2, Rules[2][5]=6, Rules[2][6]=7, Rules[2][3]=5
25 */
26 int *rulesDef; ///list of rules definitions: nr of inputs in rule 1, nr of outputs in rule 1, ... and so on for each rule
27 int **rules; ///list of rules body: input nr, fuzzy set nr, ... , output nr, fuzzy set nr, ... and so on for each rule
28
29 /**
30 * Sets defuzzyfication parameters: determines - for each rule - cut level <0;1> (minimum membership function of current rule).
31 * In fact, defuzzParam remembers the values from 'first layer' - fuzzyfication layer (see neuron at documentation)
32 * i.e. rule 1: defuzzParam[0] = 0.3522
33 */
34 double *defuzzParam; /// i.e.: defuzParam[5] = 0.455 means that rule #6 has got a minimum membership function (of given inputs set for this rule) at value 0.455 (it's cut level)
35
36protected:
37
38 ///Fuzzy functions
39 double TrapeziumFuzz(int which_fuzzy_set, double input_val);
40 int Fuzzyfication();
41 int Defuzzyfication();
42 int GetFuzzySetParam(int set_nr, double &left, double &midleft, double &midright, double &right);
43
44public:
45
46 int fuzzySetsNr; /// number of fuzzy sets
47 int rulesNr; ///number of rules
48 SString fuzzySetString; /// strings containing all fuzzy sets given in f0
49 SString fuzzyRulesString; /// strings containing all fuzzy rules given in f0
50
51 NI_FuzzyNeuro() {paramentries=NI_FuzzyNeuro_tab; fuzzySets=defuzzParam=NULL; rulesDef=NULL; rules=NULL;}
52 ~NI_FuzzyNeuro();
53 NeuroImpl* makeNew() { return new NI_FuzzyNeuro(); };
54 void go();
55 int lateinit();
56 /** Function build model based on given genotype and conts number of neurons connected with fuzzy neuro,
57 also checks number of fuzzy neuron inputs.
58 \param genotype genotype to be scanned
59 \param inputs number of fuzzy neuron inputs
60 \param output number of fuzzy neuron outputs (= number of neurons connected to fuzzy neuron)
61 @return success or failure
62 **/
63 static int countOuts(const Model *m, const Neuro *fuzzy);
64
65};
66
67#endif
Note: See TracBrowser for help on using the repository browser. | 1,033 | 3,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.613973 |
http://www.interweave.com/article/knitting/converting-stitch-patterns-for-working-in-the-round/ | 1,490,787,227,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190295.4/warc/CC-MAIN-20170322212950-00073-ip-10-233-31-227.ec2.internal.warc.gz | 569,947,513 | 20,450 | # Knitting in the Round: Converting Knitting Stitch Patterns Like a Pro
Just a few of the knitting stitches you’ll find in our Knit Stitch Dictionary.
Traditionally, knitting stitches in a stitch dictionary are presented for knitting flat (back and forth in rows). If you want to use one of the stitches for something knitted in the round (a sock, say, or a hat), then you have to do a little bit of conversion magic with knitting stitch patterns.
## Here are the basic conversion steps for knitting stitch patterns:
1. Pick an appropriate stitch pattern.
Some patterns are easy to convert from rows to rounds; some can be mind-bendingly difficult. Before you get your heart set on a particular stitch pattern for use in a cute baby hat, check to see if it looks like it will convert easily to working in the round. There are two things you want to look for: First, a pattern where the wrong-side rows contain only purl stitches or knit stitches, and second, a pattern where the number of edge stitches is the same on all rows. (Remember that edge stitches are the ones outside the repeat section.)
2. When figuring out how many stitches to cast on: Drop the “balancing” stitches from your calculations.
In other words: Drop the Y number in the “multiple of X stitches plus Y” notation discussed above. Cast on only the “multiple of X” number for your in-the-round pattern.
3. When you are knitting: Work from asterisk to semi-colon only.
In other words, everything between those two punctuation marks is your stitch repeat, and you will knit just those stitches around and around your “tube.”
4. Convert the wrong-side rows to right-side rounds.
You’re always on the right side when you knit in the round, right? So there are no “wrong-side rows,” technically. There are two steps to getting the wrong-side right in your circular knitting. First, all purl stitches become knit stitches and all knit stitches become purl stitches. Second, the “wrong-side” instructions are read backwards! An example will help here: Say that the wrong-side ROW instructions tell you to work “p3, k1” across. For circular knitting, you would work “p1, k3” around. So: Purls became knits, knits became purls, and you work the repeat in reverse. This can be a bit confusing, so WRITE OUT the wrong-side instructions in the new circular “language” in order to help yourself keep it all “straight”!
## If in doubt, chart it out.
The steps given above for converting flat stitch patterns to in-the-round stitch patterns will work well for symmetrical patterns that are simple in design. If you have your heart set on a complex or asymmetrical pattern, then graph paper, pencils, swatching, and patience are your best friends. Or, for the computer-savvy individuals, try using a spreadsheet program, setting a narrow column width and using your own set of symbols for the stitches to see how things line up. I do this with all my own designs, and although the results are not publishable, my home-grown charts are a huge help in my knitting.
Here are a couple of resources on knitting stitch patterns:
The Knit Stitch Dictionary, by Debbie Tomkies—This is an excellent reference guide for 250 of the most popular, fresh, and customizable knit stitches with both written and charted instructional examples.
17 Free Knit Stitches: A Guide to Knitting Stitches—We’ve gathered some of our favorite knitting stitch patterns to share with you in this free eBook. With 17 techniques in all, you’ll discover knitting stitches for beginners to tackle, plus more intermediate and complex designs such as the lattice pattern, herringbone and many more! | 780 | 3,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-13 | latest | en | 0.898936 |
https://fishinginthefootsteps.com/best-629 | 1,670,493,455,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00713.warc.gz | 290,158,171 | 5,664 | # Free math app that shows work
Free math app that shows work can be found online or in math books. We will give you answers to homework.
## The Best Free math app that shows work
Free math app that shows work is a software program that helps students solve math problems. Once you have a good understanding of the basics, you can start to work on more complicated problems. If you're still having trouble, there are plenty of resources available online and in math textbooks
By inputting the letters that you have into the cheat, you can quickly find words that fit the puzzle. This can be a great way to save time and frustration when you are trying to complete a difficult word search.
There are a number of free online tutor chat services available for math. These services are a great resource for students who need extra help with their math homework or just want to brush up on their skills. The tutors on these chat services are usually math experts who can help students with any questions they may have.
There are many ways to solve word problems, but one effective method is to use an equation. First, identify the information given in the problem and what you are looking for. Then, set up an equation using that information. Finally, solve the equation to find the answer. Using an equation to solve word problems can be tricky, but with practice it can become easier. There are also many online tools that can help, like a word problem equation solver.
There are a few different methods that can be used to solve multi step equations. The most common method is to use the distributive property to simplify the equation and then solve it by using the order of operations. Another method is to use inverse operations to solve the equation.
If you're struggling to solve a word problem, try using a calculator. First, read the problem and identify what information you need to solve it. Then, use the calculator to input that information and solve the problem. Be sure to show your work so you can check your work later.
## We solve all types of math troubles
It's honestly a life saver. Not only it solves and explain it also have features like graphs. I hope to see it solve word problems in the future. I'm expecting the updates to be awesome.
Petra Watson
It's honestly a very good app. Right now, in the time of quarantine it helps me learn the steps to solving my math problems. I think that I'd have an even better experience if I could buy plus. I'd recommend it to any student who needs help.
Philippina Jackson
Solve Math-solving Findanswer Solve math problems free Problem solver app Solve equation on interval calculator Solve math problems step by step Take photo of math problem | 549 | 2,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-49 | latest | en | 0.956026 |
https://blog.beerriot.com/2017/03/04/ | 1,500,972,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00406.warc.gz | 634,526,372 | 37,359 | Archive for March 4th, 2017|Daily archive page
Beer IoT (Part 8)
Strike up the band, it’s time for the eighth, and final, installment of this fermentation instrumentation series. In part four, I placed several different sensors in several different carboys of beer beginning fermentation. In parts six and seven, I analyzed a week of data from two of the sensors. This post will cover the third sensor, a floating accelerometer.
The ADXL345 provides three readings for each sample: one for each axis in 3D space. I’m using the chip in “2g 10-bit” mode, which means each axis will report a number from -512 (2g negative acceleration) to +512 (2g positive acceleration). In this case, the only acceleration I want to measure is gravity, so I should see only values between -256 (1g negative acceleration; “this axis is pointing straight down”) and 256 (1g positive acceleration; “this axis is pointing straight up”). Using a bit of trigonometry, I should be able to figure out the angle at which the sensor is tilted.
My float is sort of a rounded rectangular prism. I’ve oriented the sensor such that the y-axis is in line with the long axis of the prism, with the positive end pointing toward the end I expect to float. The x-axis is horizontal across the short axis of the prism, and the z-axis is pointed “up”. The expectation is: y will start about zero, or slightly positive, as high buoyancy keeps the float “flat”; x will start about zero as well, because any dip should be along y; and z will start near max, almost straight up. As the beer ferments, reduced buoyancy should cause one end to dip, causing y to increase (because it will point upward more steeply), and z to decrease (because it will move off of straight upward), with x staying the same (because the rotation should be around that axis). So what actually happened?
Pictures may be worth a thousand words, but I’m pretty sure this one just says, “Not that.” We have both x and z increasing, and y is doing … I’m not even sure. Let’s see if there is anything to salvage.
Let’s check an assumption first. I’m expecting to only see acceleration from gravity here, so the total acceleration should always be 1g (plus or minus some measurement noise). We can check that with a bit of Pythagorus: the square root of the sum of the squares of the readings should be a constant 256.
Except for the sudden change in the middle, it’s a variance of about 1, which is 0.0039g for this chip. It’s interesting that it’s only 252 max, and I have no idea what that sudden shift is (it seems correlated with a sudden shift on the z axis, but nothing on the other axes), but it does look like we’re measuring approximately a constant 0.97-0.98g force.
The increasing x may a bit of a red-herring. It just means that the tube is “rolling” (turning around its long axis). This is why z is increasing as well: x returning to horizontal around the unchanging y axis means that z is returning to vertical. There is a chance that the float is rolling instead of tipping as buoyancy changes. This might be worth returning to later, but let’s see if we can save y first.
Despite the fact that our sanity check showed that we’re reading constant gravity as we expect, and therefor all axes agree, we could use Pythagorus again to compute what the value of the y reading should have been, given x, z, and our expected force:
The synthesized y reading is in blue, while the actual y reading is in red. This graph used 256 for the expected gravity. Let’s instead use the 253/250 mix we saw before, which will also account for that unexplained shift in z:
Many features are similar between these plots, but we appear to have exaggerated a somewhat steady descent in y during the period that x and z where steadily climbing (Feb 19 through 21). I expected y to start around zero and become more negative over time. Starting above zero, and decreasing anyway just means that the sensor was tilted away from the expected sinking angle to start. Interestingly, y moving from slightly up to closer to horizontal will also have the effect of bringing z up closer to vertical, just like x moving from negative toward zero did.
If the rolling is not the result of buoyancy change, then this change in y alone leaves us with a change from early Feb 19 mid afternoon Feb 21 of either 33-22 (computed, blue line) or 10-5 (observed, red line). asin(33/256) = 7.41º, asin(22/256) = 4.93º; asin(10/256) = 2.24º, asign(5/256) = 1.12º. Using 252 instead of 256 only alters the result by 0.1º. So, a change of at most 2.5º, and at least 1º. A bit of a narrow bad, if you ask me.
If the sensor shifted during placement, the x axis might be measuring pitch instead of roll. But if even if not, what if the fermentation primarily produced rolling instead if pitching? The x reading swings from -115 to -100. That’s 26.7º to 23º, meaning a change of 4.3º. That’s more, is about all that can be said about that.
If we take both roll and pitch together, we can just consider the change in z, but we also have to ignore the sudden shift near the end of the time range we were looking at. That gives 223 to 231, or 60.59º to 64.46º. Still just 4º change.
If I return to the design of the float, its weight of the float is 1.8oz. So, to float in fresh water, it will have to displace 1.8oz, or 3.24 cubic inches. The float is 4 inches long, by 1 3/16 inches wide, by 7/8 inches tall. A rough estimate places that at 4.15625 cubic inches. The rounded edges are tricky, though. Measuring via displacement shows it’s actually about 3.5 cubic inches. So, what I have is a float that is only just barely floating in water. That was the aim, and what was observed, but good to see the math line up.
If the float has to displace 3.24 cu.in. of water at specific gravity 1.000, then at our starting gravity of 1.040 it only has to displace 3.12 cu.in. of unfermented beer, and 3.22 cu.in. of beer at the finishing gravity of 1.0075. So we’re looking at a change of 0.1 cubic inches of displacement.
I found the center of mass to be about 3/8 inch closer to the end that is expected to sink. That’s not a huge margin of influence, but since the float will be almost entirely submerged anyway, it’s probably enough. The heavy end also happens to have less volume, due to the curvature, so it should have to sink more to displace the same amount.
Calculus is probably the correct way to solve this problem. 18.01 was a long time ago, though, so I’m hoping I can fudge it. Instead of trying to figure out how far this float should have tipped, let’s figure out if a 4º pitch could have changed the displacement 0.1 cu.in.
If we’re already mostly-submerged, we’re looking near an edge that is not 1 3/16 inches wide, but instead closer to 0.75 in (due to curvature). If the float were flat (which the y and z axis readings mostly suggest), 0.75 * 3.75 in would be above water 0.036 in. If we pitched that 4º, we would lose 0.12 cubic inches into the water:
Height lost at 4º over 3.75″: (sin(4/180)*3.75) = 0.08332647479 inches
Volume of 0.083 x 3.75 x 0.75 ” triangular prism: 0.75 * (0.083 * 3.75)/2 = 0.117 cubic inches
So 4º could actually the correct change for these parameters! This does require that the x and z axes were reading pitch, and not roll, though. A 4º roll with these parameters is only a change of about 0.02 cubic inches.
A final check: how well does the shape of this data match the shape of the BeerBug’s?
Comparing the Z axis, it looks like the story is similar to the pressure sensor: the change plateaus at about the same point as the BeerBug, signaling the end of primary fermentation. If the 4º change was measuring the correct thing, then math would have told me the correct final gravity, but it would have been much easier to have developed a calibration table with known angles of specific gravities beforehand.
That wraps up this experiment. I’ve added this data to the gist containing the other sensor data. What’s next for this beer sensing story … ?
For the BeerBug, it may be worth continuing to use their service. The device works when the service works, as shown in this data. If anything, I think I’d work on snooping its communication, so I could tee it off to my own storage, in case their service goes down again.
For the pressure sensor, it’s mostly about a new housing, and then calibration to that housing. It needs something that is both heavy enough to sink, and also flexible enough to compress. If both of those are taken care of, it seems like calibrating to known specific gravities may actually provide decent data.
For the tilt sensor, it’s also about a new housing. The weight distribution needs to be far more unbalanced, to ensure a larger change in angle. Something narrower, so that more sinking is required to balance displacement, would work to. If those can be taken care of, then calibration may make this as good as other options.
Both the pressure sensor and the tilt sensor would also benefit from getting the Helium Atom and battery onboard. Current results are probably affected by the cable running out of the carboy. For now, this would require fermenting in something with a wider neck, since the development board is too wide to fit in the carboy. That’s easy to do, and would avoid me having to design my own printed circuitry.
What was really amazing to me is how easy this sort of thing has become. A week after I got hardware, I put it into service. I2C is a nice standard communication protocol. Lua is a quick language to pick up. The Helium chip, library, and service work very smoothly. Between the dev kit and the sensors, I’m over \$100, but less than \$200 into this exploration. I can see why people are excited about IoT these days – it’s easy to get started, and fun to participate.
But for now, there are two cases of beer to sample in a couple of weeks, and they’re stacked under earlier brews, so I won’t have any more fermentation to measure for a while. I’m setting up one of my Atoms to monitor the temperature in the conditioning closet. I wonder what I should start measuring with the other.
Beer IoT (Part 7)
Keeping on the hop, it’s time for part seven of this fermentation instrumentation series. In part four, I placed a few different sensors in some actively fermenting beer to gather data. In the previous post, I looked at data from a commercial sensor. Now it’s time to examine the data from my experimental pressure sensor.
I have two atmospheric pressure sensors collecting data while this beer ferments. One is outside the carboy, while the other is in a non-rigid (i.e. squeezable) container near the bottom of the inside of the carboy. The idea is that as the beer ferments, it will become less dense (because alcohol is less dense than sugar), and thus the same volume of liquid will put less pressure on the sensor.
Let’s start with predictions. I took the long way around and made a table that says that if my sensor is four inches below the surface of the water (it is), it should see about nine millibars of pressure more than just sitting in the air, regardless of specific gravity. This was the long way around, because it turns out “water-inch” is a known pressure unit (equal to 2.49mbar).
Here is what my sensors measured:
Blue: External Pressure, Red: Internal Pressure
Unfortunately, there are two problems, relative to my predictions. The first is that the curves are not 9mbar apart. The second is that the red one is the internal sensor, consistently reading lower than the external sensor. Before I put the sensor in the carboy, I saw about the same difference. It’s possible that the missing 9 mbar is due to the fact that the sensor housing is not laying on the bottom of the carboy, as in the picture at the top of this post, but is instead resting with one end higher than the other. That means the pressure on it is not uniform, and I could be losing all of the additional pressure to the top end (which also happens to be the most flexible end). This might be enough to declare the experiment invalid, but let’s continue looking at what I have anyway.
What is more interesting from the earlier table is the difference we’re supposed to see as the beer ferments. I measured an original gravity (OG) of 1.040 when I pitched the yeast, and a final gravity (FG) of 1.00075 when I bottled (there are some sugars the yeast won’t eat, so we don’t reach 1.000). The predicted difference in pressure is just over 0.3 mbar. The pressure varies by over 15 mbar just due to the weather, though, so how can we tell? By subtracting the external, weather-only pressure from the internal, weather+beer pressure:
Hooray! We do see relative pressure change in the carboy. It’s noisy, but I think we have to compare the top-ish of the hump with the resting level of the plateau on the right. Why not compare the start point at the left with the resting point at the right? The climb on the left is likely one of two things: something similar to what the BeerBug sees, as discussed in part six (i.e. initial oxygen consumption, carbon dioxide production, or yeast proliferation), or the sensor moving. In either case, it does take the yeast 12-24 hours to really get working, and that’s about where the climb levels off, so that’s where the conversion of the sugar really starts.
Drawing some lines across the difference graph, I find the “max” pressure to be about -1.65, and the “end” pressure about -2.05, a difference of 0.4 mbar. That’s 30% off of the hypothesis. This doesn’t seem like a bad error (for a first attempt), but I’m skeptical that we saw this much pressure change without seeing the entire pressure difference (the missing 9 mbar).
So, let’s see if we can answer some unknowns. Before removing the sensor from the carboy, I attempted to figure out if the pressure leveled off higher due to pressure from the airlock. There is about a half inch of water that has to be moved out of the way, which would be 1.2 mbar. That’s nearly double the difference between the start pressure and the resting pressure, but to check, I let the gas out a couple of times between 5pm and 7pm in this graph:
There is no dip in pressure, just added noise from me jostling the cable. The resting pressure change is not from pressurizing the airlock. Follow-up experiments will be necessary to determine if it has anything to do with the specific mixture of absorbed gases, or the presence of yeast cells, I think for now the most likely answer is that the sensor moved.
I also tried to find the missing 9 mbar. After removing the sensor from the carboy, I put it in another container under about 4 inches of water, fully horizontal this time.
That graph starts with the sensor in the open air. It looks like I found about 6 mbar once I got the sensor truly in the bottom of the container. I think the last 3 mbar can probably be attributed to the rigidity of the container – it probably couldn’t deform further.
Perhaps most importantly, how does this compare to the BeerBug’s specific gravity curve? If I do just a little scaling and shifting, I can lay the curves on top of each other:
Blue: Differential Pressure, scaled and shifted; Red: Specific Gravity as measured by BeerBug
The two carboys do contain different strains of yeast (the original reason for using three separate carboys), and the BeerBug’s carboy started noticeably faster. So, the different in the start of the curves is to be expected. The head in both, and the bubbling out of the airlocks of both did seem to reduce at about the same time, so the simultaneous arrival at finishing plateau is expected as well.
Overall results are, unfortunately, inconclusive. It looks like the end of fermentation was signaled correctly. I would not have been able to predict the finishing gravity, though. There is enough here to warrant future experiments, I think. This was something of an opportunistic test. I was brewing these three batches anyway, so why not try the sensors? Something with more control (i.e. taking fermentation out of the equation) should be illuminating.
If you want to explore this data yourself, I’ve posted the data in CSV format in a gist.
Stay tuned for the final episode analyzing the accelerometer data soon!
Update: accelerometer data is live in part eight. | 3,846 | 16,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-30 | longest | en | 0.931538 |
https://125fps.com/casino/question-do-u-win-with-2-numbers-on-powerball.html | 1,642,930,100,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00069.warc.gz | 146,194,077 | 17,995 | # Question: Do u win with 2 numbers on Powerball?
Contents
## What does 2 Powerball numbers pay?
The table below demonstrates how the Power Play option affects each prize.
Numbers Matched Prize Prize With Multiplier
10x
3 \$7 \$70
2 + Powerball \$7 \$70
1 + Powerball \$4 \$40
## How many numbers do you need to win on the Powerball?
Then, the winning Powerball number is randomly selected from a pool of 20 balls numbered 1 to 20. To win Division 1, you need all 7 matching numbers to win in Powerball plus the Powerball number. You must match at least 2 numbers plus the Powerball number to win the lowest Powerball Division prize, Division 9.
## Do you win anything on the lottery with 2 numbers?
Yes, if you match two Mega Millions numbers during the drawing, this means that you will walk away with some money, but you won’t win the jackpot. However, you will only win some money if you matched a yellow ball and a white ball. Matching two white balls won’t get you any winnings.
## Is 3 numbers a prize in Powerball?
In this quick and easy guide to playing Powerball, we’ll take you through the different Powerball Division prizes, what the Powerball winning combinations are, and how much you could stand to win.
Jackpot Prize Levels.
Weeks without jackpot Jackpot Prize
1 \$3 Million
2 \$8 Million
3 \$20 Million
4 \$40 Million
THIS IS INTERESTING: What dice do you need for dice stacking?
## How do Powerball payouts work?
Lottery winners can collect their prize as an annuity or as a lump-sum. … A lump-sum payout distributes the full amount of after-tax winnings at once. Powerball and Mega Millions offer winners a single lump sum or 30 annuity payments over 29 years.
## Is 2 numbers on Mega Millions worth anything?
2 numbers plus the Mega Ball – \$10
If your ticket matches two numbers and the Mega Ball you’ll get \$10 but the odds to get there begin to skyrocket to 1 in 693. | 454 | 1,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2022-05 | latest | en | 0.877149 |
https://gbee.edu.vn/is-good-1350-sat-score-colleges-you-can-get-into-cg4wlf5v/ | 1,718,839,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00414.warc.gz | 235,456,341 | 17,866 | Are you a student who wants to know ‘is a good 1350 SAT score?’ Then this article is for you. Although there are various parameters for evaluating an admission application, SAT is important. In addition, all the extremely selective colleges have very high average SAT. Therefore, if you want to get into any of these colleges, you need to be aware of important factors for admission.
Table of Contents
## What is the SAT Score, and How is it Calculated?
SAT is a standardized test conducted to check students’ knowledge and learning before college application. There are two main sections of this test, namely
• Evidence-based reading and writing
• Math
It is a multiple-choice question based test and is conducted in offline mode. The SAT exam pattern is as follows:
Section Total time provided Total number of questions Full marks Math 88 minutes 58 800 Evidence based reading and writing (65+35) = 100 minutes 96 800
For each correct answer, an SAT taker gets one mark. There is no negative marking for wrong or unattempted questions. Thus, the raw score of a candidate is calculated on the number of correct answers. Then, the raw marks are further converted into scaled marks through equating. Finally, equating converts the raw scores according to the difficulty level of questions.
Afterward, to get the section score of evidence-based writing and reading, the test score is multiplied by 10. On the other hand, the section score of the Math test is multiplied by 20. The total section score of Math and evidence-based reading and writing gives the SAT score.
## What Does it Mean to have a 1350 SAT Score?
Students who achieve a 1350 SAT score often want to know is 1350 a good SAT score? Your SAT score includes scores of both the sections of SAT, i.e., Math and evidence-based reading-writing.
The answer to the question- ‘is 1350 a good score?’ lies with the college you want to apply to. Different colleges have their average SAT score requirements. Generally, the most competitive colleges have a higher average SAT score.
However, the national average SAT score is around 1051. This means that any SAT score higher than 1051 is considered optimum for getting admission in most colleges. Thus, as per the national average SAT score, the answer to your question – ‘is 1350 a good score?’ would be a very good score. Still, you need to search for the average SAT score of the colleges in your priority list to get a better idea of ‘is 1350 a good SAT score or not’.
### Importance of SAT Score in the Admission Application
Universities consider your SAT score one of the most important factors while reviewing admission applications. Applicants who submit their SAT scores greatly benefit their competitors who don’t consider submitting them.
There are Two Primary Reasons Why SAT Score is so Important for College Admission:
• Applicants to a college come from different schools and have taken different courses, participated in extracurricular activities, etc. The SAT score provides a common base for comparing all the applicants.
The SAT score of an applicant also reveals the accuracy of the school transcript and GPA. On many occasions, the SAT score of applicants helps the admission officers in determining if you are competent enough for the college or not.
Moreover, a high SAT score can make up for the low GPA of an applicant. For example, suppose the applicant has a perfect 1600 SAT score with a 3.1 GPA. In that case, the admission committee might ignore your GPA and judge your application according to the SAT score.
• The SAT score is also crucial in determining the rank of a university. If you have a 1350 SAT score and get admitted to a university, your score will be recorded in their SAT statistics. Every year the universities release the data of all the admitted students. According to this data, the public determines the competitiveness of the school. The higher the average score, the more competitive that school is perceived.
As the national average SAT is estimated at around 1051, students who want to know- ‘is 1350 a good SAT score?’, can rest assured that it is a great score. With a 1350 SAT score in your application, you can get admission to some of the most selective colleges.
### What is a 1350 SAT Score Percentile?
According to the term, an SAT percentile facilitates the comparison of students according to their SAT scores. For example, suppose you have secured 1000 SAT scores in both sections of the SAT. This score is calculated as 50 percentile. You have done better than 50% of the SAT test-takers.
Similarly, if the SAT percentile is 80, the candidate is under the top 80% of the SAT test-takers. Therefore students who want to know ‘is 1350 a good SAT score?’ should first figure out the 1350 SAT score percentile.
The Following Table Provides a Better Understanding of your Question ‘is 1350 a good SAT score’ by Demonstrating the 1350 SAT Score Percentile.
SAT score Vs. SAT percentile score
SAT Score SAT Percentile 1600 Above 99% 1550 99% 1510 99% 1450 98% 1400 95% 1350 90% 1230 80% 1170 71% 1110 61% 1060 51% 1000 40% 950 31% 890 21% 810 11% 750 5% 680 Below 1%
The above table shows that a 1350 SAT score percentile puts students in the top 90% of the test takers. This is an excellent score. Most selective universities have their average SAT at this score. Thus, if you think, ‘is 1350 a good SAT score‘, you would be happy to know that this score can help you earn an edge among your competitors.
Is 1350 a Good SAT Score for Admission to Most Selective Colleges?
There are multiple with average SAT scores at or below 1350. Students who have a 1350 SAT score can successfully pass the admission process in these colleges. Moreover, a high SAT score can compensate for a low GPA.
However, some extremely selective colleges have their average SAT score above 1350. Thus, applicants need to have a high GPA and ACT scores to secure a seat in these colleges.
### Steps to Improve your SAT Score
The answer to your question, ‘is 1350 a good SAT score?’, lies in the list of your preferred colleges. Suppose the colleges you want to attend have their average SAT above 1350. In that case, it will be better to improve your grades before starting the application process.
For improving your SAT score, you can do the following:
• You should prepare more efficiently and take a re-test. Following a proper study plan can be very beneficial for achieving your target without getting distracted. For this, you can seek help from your tutor.
• Consider superscoring your SAT results. Many colleges accept the best scores of each section from all the tests taken by you.
• Keep yourself motivated throughout the preparation because sometimes it gets tough to follow a routine.
### Highly Selective Colleges for Applicants with a 1350 SAT Score
Although an SAT score of 1350 is way higher than the national average, there are some colleges where students can find tough competitions for admission. Some of the extremely competitive colleges are:
University Location Average SAT Average ACT University of Michigan Ann Arbor 1435 33 Boston College Chestnut Hill 1420 33 Rensselaer Polytechnic Institute Troy 1409 31 New York University New York 1440 32 Emory University Altana 1435 33 Boston University Boston 1420 32 Villanova University Villanova 1395 33
### Moderately Selective Colleges for Applicants with a 1350 SAT Score
Numerous colleges have an average SAT score of around 1350. If you are an applicant with SAT score of 1350, you will have a great chance of securing a seat in the following colleges:
University Location Average SAT Average ACT University of California, Santa Barbara Santa Barbara 1355 29 University of Texas at Dallas Richardson 1350 30 Stony Brook University Stony Brook 1335 29 Lehigh University Betlehem 1365 31 University of Texas at Austin Austin 1355 30 University of Washington Seattle 1340 30 California Polytechnic State University, San Luis Obispo San Luis Obispo 1335 29
### Safe Colleges for Applicants with a 1350 SAT Score
Applicants with a 1350 SAT score are highly competitive for the following colleges:
University Location Average SAT Average ACT University of Massachusetts Amherst Amherst 1290 29 Syracuse University Syracuse 1275 28 Penn State University Park University Park 1265 28 Rochester Institute of Technology Rochester 1300 30 Drexel University Philadelphia 1290 28 United States Military Academy West Point 1270 28 University of Cincinnati Cincinnati 1265 27
#### Final Thought:
We hope that you have got a better idea about the competitiveness of a 1350 SAT score. If you have time in your hands, you must take a re-test to improve your grades for extremely competitive universities. However, there are many top-ranked colleges where you can secure a seat with a 1350 SAT score.
### 1. Is 1350 SAT score good?
Ans) It depends on the university you want to go to. The average SAT score requirements vary by college. Around 1051 is the average SAT score across the country. This indicates that the ideal SAT score for admission to the majority of universities is 1051 or higher.
2. Which highly selective institutions accept applicants with SAT scores of 1350?
• University of Michigan
• Boston College
• Rensselaer Polytechnic Institute
• New York University
• Emory University
• Boston University
• Villanova University
Comments:
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You are watching: Is Good 1350 SAT Score? Colleges You Can Get Into. Info created by GBee English Center selection and synthesis along with other related topics. | 2,347 | 10,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-26 | latest | en | 0.946108 |
https://www.physicsforums.com/threads/simple-but-poorly-worded-unit-conversion-question.332750/ | 1,582,292,435,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145529.37/warc/CC-MAIN-20200221111140-20200221141140-00338.warc.gz | 818,933,140 | 19,106 | # Simple but poorly worded unit conversion question
## Homework Statement
Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 54 centuries, what is the total (in hours) of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day, etc.)?
2. Relevant equations
Sum of an arithmetic sequence (see below)
## The Attempt at a Solution
I've seen this question "answered" a number of times via a google search (for example, https://www.physicsforums.com/showthread.php?t=208510"), but nobody seems to ask the more fundamental question: What are they even asking? Most attempts at answering the question use an arithmetic sequence. For example, if the day is 1ms longer at the end of each century, then at the first day it was 1/36525 ms (365.25 days/year, 100 years/century) longer than at the beginning of the first century, and at the second day it was 2/36525 ms longer than at the beginning of the first century, and more generally at the n'th day it was n/36525 ms longer than at the beginning of the first century. So the sum of all these gains is (1 + 2 + 3 + ... + k) / 36525, where k is the total number of days in 54 centuries.
I have three problems with all of these solutions.
First of all, it makes no sense. The answer you get is meaningless in the context of physics and has no physical interpretation that I can come up with. This is normal in a math book to have problems with no physical interpretation, but I'm suspicious of seeing a physics problem like this.
Second, I'm not sure what this formula is that people are using for the sum of an arithmetic sequence. In the link above, for example, someone claims that the sum is (n/2) · [2·a1 + (n-1)·d]. What are a1 and d? And what happened to n*(n+1) / 2?
Third, the wording of the question says "what is the total (in hours) of the daily increases in time". The daily increases. The increases between each day. The increases between each day are constant, 1/36525 ms. So the sum of the daily increases for a century should be 1ms, and the sum of the daily increases for 54 centuries should be 54 ms, or 1.5 × 10^-5 hours.
Anyone have any insight on what the heck this question is talking about?
Thanks
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Second, I'm not sure what this formula is that people are using for the sum of an arithmetic sequence. In the link above, for example, someone claims that the sum is (n/2) · [2·a1 + (n-1)·d]. What are a1 and d? And what happened to n*(n+1) / 2?
A term in an arithmetic sequence is given by $$u_{n} = a + (n-1) d$$.
a is the constant base term, while d is the common difference between consecutive terms.
The $$\frac{n(n+1)}{2}$$ that you cite is only true for the sum 1+2+3+...n, which would constitute an arithmetic sequence $$u_{n} = 1 + (n-1) (1) = n$$. Substituting a and d = 1 into the general formula thus yields $$\frac{n(n+1)}{2}$$.
A term in an arithmetic sequence is given by $$u_{n} = a + (n-1) d$$.
a is the constant base term, while d is the common difference between consecutive terms.
The $$\frac{n(n+1)}{2}$$ that you cite is only true for the sum 1+2+3+...n, which would constitute an arithmetic sequence $$u_{n} = 1 + (n-1) (1) = n$$. Substituting a and d = 1 into the general formula thus yields $$\frac{n(n+1)}{2}$$.
Ahh, I'm always bad at remembering formulas. I tend to just stick with a single formula whenever possible and then just massage everything into the right format to make it work.
Thanks, now just looking for any suggestions on the problem itself
Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 54 centuries, what is the total (in hours) of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day, etc.)?
As I read the question I just see it as your 3rd point. The sum of the gain of each of the days is simply 54ms. As the question gives you the increase per century independent of the length of the day I don't see why you are involving sumations at all. The increase per century is 1ms independent of the lenght of the day during the century as I see it.
ideasrule
Homework Helper
Anyone have any insight on what the heck this question is talking about?
I'm pretty sure I know what it means.
Suppose you have a cesium atomic clock that can keep time much better than Earth's rotation. (Actually, you don't have to suppose; they exist.) After 54 centuries, the clocks will say that the time is such-and-such. Because Earth's rotation has been slowing down, astronomical observations will indicate a different time. The clocks might say that it's noon in England, for example, even though the Sun has yet to rise. The question asks: what is the difference between the clock's time and the time determined by Earth's rotation?
This question is extremely meaningful because it turns out that the slowing of Earth's rotation is more than detectable with atomic clocks. Every few years, a "leap second" is added to the time to compensate for decreases in the length of the day. Many effects contribute to the decrease: the most significant are tidal interactions with the Moon and glacier rebound, but other effects are of a similar order of magnitude. That's why leap seconds don't get inserted at regular intervals: because Earth's rotation is not slowing at a constant rate. However, for this question, you have to assume that it is; you'll be able to get a reasonably accurate answer because the extremely regular tides generated by the Moon dominate other factors.
If you're interested, you can read http://tycho.usno.navy.mil/leapsec.html [Broken] for a more detailed explanation of the leap seconds that inspired your question.
About the question itself: using the arithmetic series formula is only an approximation, but it's more than accurate enough considering the uncertainty in the rate that Earth's rotation slows. What you're assuming is that Earth rotates at exactly the same rate for one revolution, then instantly changes its rotation speed and rotates exactly 1 revolution at the new speed, then repeats the process. In reality, rotation speed is slowing constantly and not just at year-end, and to rigorously account for that you'd have to use integration. For this problem, integration happens to be easier. Suppose you have a small period of time dt. After time t, dt has lengthened by ktdt, where k is 1 ms/century. Its length would then be (1+kt)dt. Integrating that gives t+kt2/2, so kt2/2 would be the time difference between real time and solar time.
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As I read the question I just see it as your 3rd point. The sum of the gain of each of the days is simply 54ms. As the question gives you the increase per century independent of the length of the day I don't see why you are involving sumations at all. The increase per century is 1ms independent of the lenght of the day during the century as I see it.
I actually agree with you, but myself and a friend both solved the problem independently, and I did it the 54ms way, and he did it the summation way. Neither of us could understand why the other was doing it the way they did, so we searched on google and found many "solutions" to the problem. Almost all used the summation way, which is really strange to me. I guess this was why I got my undergrad degree in math, so I dont' have to worry about lack of precision in the wording of problems :) | 1,870 | 7,706 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-10 | latest | en | 0.955712 |
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