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https://planetcalc.com/8287/?thanks=1 | 1,716,092,184,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057684.4/warc/CC-MAIN-20240519035827-20240519065827-00789.warc.gz | 413,045,795 | 10,201 | # Fraction to Percent and Percent to Fraction Calculator
This online calculator can calculate the percentage value for an entered fraction, or the fraction value for an entered percentage.
### This page exists due to the efforts of the following people:
#### Timur
Created: 2019-07-02 05:27:27, Last updated: 2023-04-13 03:11:53
You can use the calculators below to convert the fraction to the percent and vice versa.
### Fraction to Percent
To convert a fraction to a percentage, enter the numerator and denominator of the fraction. The calculator divides the numerator by the denominator, multiplies the result by 100 to obtain the percentage, and formats the result according to the desired precision.
#### Fraction to Percent
Percent
Digits after the decimal point: 2
### Percent to Fraction
Converting a percentage to a fraction is a bit more complicated if you want a simplified result. Let's take the example of 37.5%.
First, the calculator takes the numerator and denominator (37.5 and 100, respectively) and multiplies both by 10 to eliminate the decimal point and obtain whole numbers: 375 and 1000. Then, the calculator finds the greatest common divisor (GCD) of these two integers. Both integers are then divided by the GCD to simplify the fraction. Here's how to do it with our example:
The GCD of 375 and 1000 is 125. 375 divided by 125 equals 3, and 1000 divided by 125 equals 8. The resulting fraction is 3/8.
#### Percent to Fraction
Fraction
URL copied to clipboard
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PLANETCALC, Fraction to Percent and Percent to Fraction Calculator | 365 | 1,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-22 | latest | en | 0.840664 |
https://www.geeksforgeeks.org/digital-low-pass-butterworth-filter-in-python/ | 1,701,651,275,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00418.warc.gz | 861,098,673 | 52,676 | # Digital Low Pass Butterworth Filter in Python
In this article, we are going to discuss how to design a Digital Low Pass Butterworth Filter using Python. The Butterworth filter is a type of signal processing filter designed to have a frequency response as flat as possible in the pass band. Let us take the below specifications to design the filter and observe the Magnitude, Phase & Impulse Response of the Digital Butterworth Filter.
The specifications are as follows:
• Sampling rate of 40 kHz
• Pass band edge frequency of 4 kHz
• Stop band edge frequency of 8kHz
• Pass band ripple of 0.5 dB
• Minimum stop band attenuation of40 dB
We will plot the magnitude, phase, and impulse response of the filter.
### Step-by-step Approach:
Step 1: Importing all the necessary libraries.
## Python3
`# import required modules ` `import` `numpy as np ` `import` `matplotlib.pyplot as plt ` `from` `scipy ``import` `signal ` `import` `math`
Step 2: Define variables with the given specifications of the filter.
## Python3
`# Specifications of Filter ` ` ` ` ``# sampling frequency ` `f_sample ``=` `40000` ` ` `# pass band frequency ` `f_pass ``=` `4000` ` ` `# stop band frequency ` `f_stop ``=` `8000` ` ` `# pass band ripple ` `fs ``=` `0.5` ` ` `# pass band freq in radian ` `wp ``=` `f_pass``/``(f_sample``/``2``) ` ` ` `# stop band freq in radian ` `ws ``=` `f_stop``/``(f_sample``/``2``) ` ` ` `# Sampling Time ` `Td ``=` `1` ` ` ` ``# pass band ripple ` `g_pass ``=` `0.5` ` ` `# stop band attenuation ` `g_stop ``=` `40`
Step3: Building the filter using signal.buttord function.
## Python3
`# Conversion to prewrapped analog frequency ` `omega_p ``=` `(``2``/``Td)``*``np.tan(wp``/``2``) ` `omega_s ``=` `(``2``/``Td)``*``np.tan(ws``/``2``) ` ` ` ` ` `# Design of Filter using signal.buttord function ` `N, Wn ``=` `signal.buttord(omega_p, omega_s, g_pass, g_stop, analog``=``True``) ` ` ` ` ` `# Printing the values of order & cut-off frequency! ` `print``(``"Order of the Filter="``, N) ``# N is the order ` `# Wn is the cut-off freq of the filter ` `print``(``"Cut-off frequency= {:.3f} rad/s "``.``format``(Wn)) ` ` ` ` ` `# Conversion in Z-domain ` ` ` `# b is the numerator of the filter & a is the denominator ` `b, a ``=` `signal.butter(N, Wn, ``'low'``, ``True``) ` `z, p ``=` `signal.bilinear(b, a, fs) ` `# w is the freq in z-domain & h is the magnitude in z-domain ` `w, h ``=` `signal.freqz(z, p, ``512``)`
Output:
Step 4: Plotting the Magnitude Response.
## Python3
`# Magnitude Response ` `plt.semilogx(w, ``20``*``np.log10(``abs``(h))) ` `plt.xscale(``'log'``) ` `plt.title(``'Butterworth filter frequency response'``) ` `plt.xlabel(``'Frequency [Hz]'``) ` `plt.ylabel(``'Amplitude [dB]'``) ` `plt.margins(``0``, ``0.1``) ` `plt.grid(which``=``'both'``, axis``=``'both'``) ` `plt.axvline(``100``, color``=``'green'``) ` `plt.show()`
Output:
Step 5: Plotting the Impulse Response.
## Python3
`# Impulse Response ` `imp ``=` `signal.unit_impulse(``40``) ` `c, d ``=` `signal.butter(N, ``0.5``) ` `response ``=` `signal.lfilter(c, d, imp) ` ` ` `plt.stem(np.arange(``0``, ``40``), imp, use_line_collection``=``True``) ` `plt.stem(np.arange(``0``, ``40``), response, use_line_collection``=``True``) ` `plt.margins(``0``, ``0.1``) ` ` ` `plt.xlabel(``'Time [samples]'``) ` `plt.ylabel(``'Amplitude'``) ` `plt.grid(``True``) ` `plt.show()`
Output:
Step 6: Plotting the Phase Response.
## Python3
`# Phase Response ` `fig, ax1 ``=` `plt.subplots() ` ` ` `ax1.set_title(``'Digital filter frequency response'``) ` `ax1.set_ylabel(``'Angle(radians)'``, color``=``'g'``) ` `ax1.set_xlabel(``'Frequency [Hz]'``) ` ` ` `angles ``=` `np.unwrap(np.angle(h)) ` ` ` `ax1.plot(w``/``2``*``np.pi, angles, ``'g'``) ` `ax1.grid() ` `ax1.axis(``'tight'``) ` `plt.show() `
Output:
Below is the complete program based on the above approach:
## Python
`# import required modules ` `import` `numpy as np ` `import` `matplotlib.pyplot as plt ` `from` `scipy ``import` `signal ` `import` `math ` ` ` ` ` `# Specifications of Filter ` ` ` ` ``# sampling frequency ` `f_sample ``=` `40000` ` ` `# pass band frequency ` `f_pass ``=` `4000` ` ` `# stop band frequency ` `f_stop ``=` `8000` ` ` `# pass band ripple ` `fs ``=` `0.5` ` ` `# pass band freq in radian ` `wp ``=` `f_pass``/``(f_sample``/``2``) ` ` ` `# stop band freq in radian ` `ws ``=` `f_stop``/``(f_sample``/``2``) ` ` ` `# Sampling Time ` `Td ``=` `1` ` ` ` ``# pass band ripple ` `g_pass ``=` `0.5` ` ` `# stop band attenuation ` `g_stop ``=` `40` ` ` ` ` `# Conversion to prewrapped analog frequency ` `omega_p ``=` `(``2``/``Td)``*``np.tan(wp``/``2``) ` `omega_s ``=` `(``2``/``Td)``*``np.tan(ws``/``2``) ` ` ` ` ` `# Design of Filter using signal.buttord function ` `N, Wn ``=` `signal.buttord(omega_p, omega_s, g_pass, g_stop, analog``=``True``) ` ` ` ` ` `# Printing the values of order & cut-off frequency! ` `print``(``"Order of the Filter="``, N) ``# N is the order ` `# Wn is the cut-off freq of the filter ` `print``(``"Cut-off frequency= {:.3f} rad/s "``.``format``(Wn)) ` ` ` ` ` `# Conversion in Z-domain ` ` ` `# b is the numerator of the filter & a is the denominator ` `b, a ``=` `signal.butter(N, Wn, ``'low'``, ``True``) ` `z, p ``=` `signal.bilinear(b, a, fs) ` `# w is the freq in z-domain & h is the magnitude in z-domain ` `w, h ``=` `signal.freqz(z, p, ``512``) ` ` ` ` ` `# Magnitude Response ` `plt.semilogx(w, ``20``*``np.log10(``abs``(h))) ` `plt.xscale(``'log'``) ` `plt.title(``'Butterworth filter frequency response'``) ` `plt.xlabel(``'Frequency [Hz]'``) ` `plt.ylabel(``'Amplitude [dB]'``) ` `plt.margins(``0``, ``0.1``) ` `plt.grid(which``=``'both'``, axis``=``'both'``) ` `plt.axvline(``100``, color``=``'green'``) ` `plt.show() ` ` ` ` ` `# Impulse Response ` `imp ``=` `signal.unit_impulse(``40``) ` `c, d ``=` `signal.butter(N, ``0.5``) ` `response ``=` `signal.lfilter(c, d, imp) ` `plt.stem(np.arange(``0``, ``40``), imp, use_line_collection``=``True``) ` `plt.stem(np.arange(``0``, ``40``), response, use_line_collection``=``True``) ` `plt.margins(``0``, ``0.1``) ` `plt.xlabel(``'Time [samples]'``) ` `plt.ylabel(``'Amplitude'``) ` `plt.grid(``True``) ` `plt.show() ` ` ` ` ` `# Phase Response ` `fig, ax1 ``=` `plt.subplots() ` `ax1.set_title(``'Digital filter frequency response'``) ` `ax1.set_ylabel(``'Angle(radians)'``, color``=``'g'``) ` `ax1.set_xlabel(``'Frequency [Hz]'``) ` `angles ``=` `np.unwrap(np.angle(h)) ` `ax1.plot(w``/``2``*``np.pi, angles, ``'g'``) ` `ax1.grid() ` `ax1.axis(``'tight'``) ` `plt.show() `
Output:
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Next | 2,430 | 7,051 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2023-50 | latest | en | 0.707964 |
https://oeis.org/A029572 | 1,638,133,051,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358591.95/warc/CC-MAIN-20211128194436-20211128224436-00177.warc.gz | 511,938,727 | 3,824 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A029572 Number of permutations of an n-set containing a 5-cycle. 2
0, 0, 0, 0, 0, 24, 144, 1008, 8064, 72576, 653184, 7185024, 86220288, 1120863744, 15692092416, 237124952064, 3793999233024, 64497986961408, 1160963765305344, 22058311540801536, 441004037348818944, 9261084784325197824, 203743865255154352128, 4686108900868550098944 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,6 LINKS FORMULA a(n) = n! (1 - sum(k=0...floor(n/5), (-1)^k/(k!*5^k)). a(n)/n! is asymptotic to 1-e^(-1/5) = 1 - A092618. E.g.f.: (1-exp(-x^5/5))/(1-x) - Geoffrey Critzer, Jun 01 2013 MATHEMATICA nn = 20; a = Log[1/(1 - x)] - x^5/5; Range[0, nn]! CoefficientList[Series[1/(1 - x) - Exp[a], {x, 0, nn}], x] (* Geoffrey Critzer, Jun 01 2013 *) PROG (PARI) x='x+O('x^66); concat([0, 0, 0, 0, 0], Vec(serlaplace((1-exp(-x^5/5))/(1-x)))) \\ Joerg Arndt, Jun 01 2013 CROSSREFS Column k=5 of A293211. Sequence in context: A054118 A223373 A001342 * A129923 A064138 A158047 Adjacent sequences: A029569 A029570 A029571 * A029573 A029574 A029575 KEYWORD nonn AUTHOR STATUS approved
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Last modified November 28 13:47 EST 2021. Contains 349413 sequences. (Running on oeis4.) | 587 | 1,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.510056 |
https://domain.glass/search/?q=how%20much%20is%208ft%20in%20inches | 1,660,493,735,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00018.warc.gz | 220,010,830 | 6,502 | ## "how much is 8ft in inches"
Request time (0.033 seconds) [cached] - Completion Score 260000
how much is 3 ft 8 inches1 how much is 2 ft in inches0.49
10 results & 0 related queries
### How much is 8 ft in inches? - Answers
8 feet is 96 inches
### How much is 6 ft in inches? - Answers
There are 12 inches in & a foot, so 6ft will have 6 12=72 inches
Inch32.3 Foot (unit)25.4 Square foot3.1 Centimetre1 Cubic foot0.8 Imperial units0.7 Metre0.4 Steel0.3 Charles Dickens0.2 Lightning0.2 Arithmetic0.2 Length0.2 Megabyte0.2 Unit of measurement0.2 Pound (mass)0.2 Line (geometry)0.1 Yard0.1 Distance0.1 Hexagon0.1 Penny0.1
### How much is 5ft 8 inches into cm? - Answers
How much is 5ft 8 inches into cm? - Answers 72.72 centimeters.
Black Moon (group)12.5 Wiki (rapper)1.3 Q (magazine)0.9 Crystal Reed0.4 Soundpieces: Da Antidote0.3 Charles Dickens0.3 Juneteenth0.2 Twelve-inch single0.2 Ratking (group)0.2 Single (music)0.2 Multiply (Xzibit song)0.2 Singing0.2 Phonograph record0.1 2:540.1 Answer song0.1 Disclaimer (Seether album)0.1 RIAA certification0.1 Questions (Chris Brown song)0.1 Multiply (Jamie Lidell album)0.1 Multiply (ASAP Rocky song)0.1
### How much is 6 ft 4 in in inches? - Answers
How much is 6 ft 4 in in inches? - Answers In 8 6 4 order to answer this question, you must first know how many inches are in a foot the answer is If there are 12 inches Add 4 inches to 72 and you get 76 inches
### How much is 8 inches in cm? - Answers
There are 2.54 centimeters in & \$ one inch. So take your measurement in In this case, the answer is I G E 20.32 centimeters. Algebraic Steps / Dimensional Analysis Formula 8 in 2.54 cm 1 in =20.32 cm
Centimetre38.9 Inch32.9 Foot (unit)3.4 Measurement3 Unit of measurement2.3 Dimensional analysis2.3 Imperial units1.8 Reciprocal length0.9 Linearity0.8 Wavenumber0.8 Metric system0.7 Pyramid inch0.7 Multiplication0.4 Millimetre0.4 Length0.3 Cubic centimetre0.2 Calculator input methods0.2 Charles Dickens0.2 Lightning0.2 Ruler0.2
### What is 5 ft 8 inches in cm? - Answers
What is 5 ft 8 inches in cm? - Answers 172.72 cm
Inch26.6 Centimetre23.9 Foot (unit)10.9 Fraction (mathematics)0.5 Dimensional analysis0.4 Imperial units0.3 Metre0.3 Decimal0.2 Length0.2 Metric system0.2 Charles Dickens0.2 Lightning0.2 Cubic centimetre0.2 Megabyte0.2 Arithmetic0.2 Line (geometry)0.2 Unit of measurement0.2 80.2 Gigabyte0.1 Pound (mass)0.1
### How much is 3ft in inches? - Answers
How much is 3ft in inches? - Answers P N L 2012-02-03 00:36:47 Best Answer Copy 36 2012-02-03 00:36:47 This answer is 6 4 2: Add your answer: Earn 5 pts Q: much is 3ft in inches ? much is 3ft 2 inches in cm? much is 33 inches Math and Arithmetic Area Jobs & Education Brain Teasers and Logic Puzzles Units of Measure Length and Distance Building and Carpentry Soil Trending Questions Give me food and I will live give me water and I will die what am I? Asked By Wiki User Best foods for weight loss?
Wiki6.3 Mathematics2.6 User (computing)2.5 Inch2.1 Arithmetic2 Puzzle1.6 Weight loss1.3 Food1.2 Cut, copy, and paste0.9 Q0.9 Distance0.8 Measurement0.8 Binary number0.7 Centimetre0.7 Education0.7 Unit of measurement0.7 Water0.6 Die (integrated circuit)0.6 Multiplication0.6 Brain0.6
### How much is 2 feet in inches? - Answers
How much is 2 feet in inches? - Answers 24 inches equals 2 feet.
Inch28.5 Foot (unit)27.3 Metre1 Imperial units0.7 Square foot0.5 Square inch0.4 Centimetre0.3 Yard0.3 Charles Dickens0.3 Lightning0.3 Arithmetic0.3 Length0.3 Megabyte0.2 Unit of measurement0.2 Line (geometry)0.2 Distance0.2 Penny0.1 Mathematics0.1 Pixel0.1 Gigabyte0.1
### How much is 3 feet in inches? - Answers
How much is 3 feet in inches? - Answers Three 3 feet is 36 inches . There are 12 inches to a foot, and 3 x 12 = 36
Foot (unit)31.6 Inch29.8 Centimetre1.6 Imperial units1.6 Yard1 Metre0.8 Triangle0.4 Length0.3 Charles Dickens0.3 Lightning0.2 Arithmetic0.2 Distance0.2 Geometry0.2 Megabyte0.2 Pound (mass)0.2 Carburetor0.2 Unit of measurement0.2 Line (geometry)0.2 Penny0.1 Mathematics0.1 | 1,346 | 4,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.865541 |
https://wiki.services.openoffice.org/w/index.php?title=Documentation/OOo3_User_Guides/Calc_Guide/Solver&oldid=240643 | 1,586,237,450,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371665328.87/warc/CC-MAIN-20200407022841-20200407053341-00305.warc.gz | 711,216,380 | 8,496 | # Using the Solver
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Tools > Solver amounts to a more elaborate form of Goal Seek. The difference is that the Solver deals with equations with multiple unknown variables. It is specifically designed to minimize or maximize the result according to a set of rules that you define.
Each of these rules defines whether an argument in the formula should be greater than, lesser than, or equal to the figure you enter. If you want the argument to remain unchanged, you enter a rule that the cell that contains it should be equal to its current entry. For arguments that you would like to change, you need to add two rules to define a range of possible values: the limiting conditions. For example, you can set the constraint that one of the variables or cells must not be bigger than another variable, or not bigger than a given value. You can also define the constraint that one or more variables must be integers (values without decimals), or binary values (where only 0 and 1 are allowed).
Once you have finished setting up the rules, click the Solve button to begin the automatic process of adjusting values and calculating results. Depending on the complexity of the task, this may take some time..
## Solver example
Let's say you have \$10,000 that you want to invest in two mutual funds for one year. Fund X is a low risk fund with 8% interest rate and Fund Y is a higher risk fund with 12% interest rate. How much money should be invested in each fund to earn a total interest of \$1000?
To find the answer using Solver:
1. Enter labels and data:
• Row labels: Fund X, Fund Y, and total, in cells A2 thru A4.
• Column labels: interest earned, amount invested, interest rate, and time period, in cells B1 thru E1.
• Interest rates: 8 and 12, in cells D2 and D3.
• Time period: 1, in cells E2 and E3.
• Total amount invested: 10000, in cell C4.
2. Enter an arbitrary value (0 or leave blank) in cell C2 as amount invested in Fund X.
3. Enter formulas:
• In cell C3, enter the formula C4-C2 (total amount - amount invested in Fund X) as the amount invested in Fund Y.
• In cells B2 and B3, enter the formula for calculating the interest earned (see below).
• In cell B4, enter the formula B2+B3 as the total interest earned.
Example setup for solver
4. Choose Tools > Solver. The solver dialog opens.
The Solver dialog
5. Click in the Target cell field. In the sheet, click in the cell that contains the target value. In this example it is cell B4 containing total interest value.
6. Select Value of and enter 1000 in the field next to it. In this example, the target cell value is 1000 because your target is a total interest earned of \$1000. Select Maximum or Minimum if the target cell value needs to be one of those extremes.
7. Click in the By changing cells field and click on cell C2 in the sheet. In this example, you need to find the amount invested in Fund X (cell C2).
8. Enter limiting conditions for the variables by selecting the Cell reference, Operator and Value fields. In this example, the amount invested in Fund X (cell C2) should not be greater than the total amount available (cell C4) and should not be less than 0.
9. Click OK. A dialog appears informing you that the Solving successfully finished. Click Keep Result to enter the result in the cell with the variable value. The result is shown below.
Result of Solver operation
The default solver supports only linear equations. For nonlinear programming requirements, try the EuroOffice Solver or Sun’s Solver for Nonlinear Programming [Beta]. Both are available from the OpenOffice.org extensions repository. (For more about extensions, see Setting up and Customizing Calc | 842 | 3,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-16 | latest | en | 0.904454 |
https://amp.doubtnut.com/question-answer/in-triangle-a-b-c-tana-tanb-tanc6a-n-dtanatanb2-then-the-values-of-tana-tanb-tanc-are-respectively-a-22417 | 1,575,772,814,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00231.warc.gz | 272,156,495 | 19,413 | IIT-JEE
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Question From class 11 Chapter TRIGONOMETRIC FUNCTIONS
In triangle then the values of are, respectively (a) (b) (c) (d) none of these
If , show that :
2:26
If , prove that :
4:22
If (n being an integer). Show that .
2:19
If A+B=C ,then find tanA tanB tanC.
1:38
In a triangle ABC then
3:13
In a , prove that :
2:49
In an acute-angled triangle ABC,
3:25
Prove that a triangle ABC is equilateral if and only if
4:33
In any , prove that <br>
6:00
In a triangle . If then the least value of is
3:14
Prove that :
8:51
If in a triangle ABC, then prove that
5:14
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If A,B,C are the angles of a given triangle ABC . If cosA.cosB.cosC= and sinA.sinB.sinC= The value of is (A) (B) 6+sqrt(3) 6-sqrt(3)
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What are the signs that you are wasting your life ? | 747 | 2,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-51 | longest | en | 0.77975 |
https://help.libreoffice.org/6.1/en-GB/text/sbasic/shared/03030206.html | 1,643,467,236,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306181.43/warc/CC-MAIN-20220129122405-20220129152405-00587.warc.gz | 356,263,752 | 4,765 | # TimeValue Function
Calculates a serial time value from the specified hour, minute and second - parameters passed as strings - that represents the time in a single numeric value. This value can be used to calculate the difference between times.
## Syntax:
`TimeValue (Text As String)`
Date
## Parameters:
Text: Any string expression that contains the time that you want to calculate in the format "HH:MM:SS".
Use the TimeValue function to convert any time into a single value, so that you can calculate time differences.
This TimeValue function returns the type Variant with VarType 7 (Date) and stores this value internally as a double-precision number in the range 0 to 0.9999999999.
As opposed to the DateSerial or the DateValue function, where serial date values result in days relative to a fixed date, you can calculate with the values that are returned by the TimeValue function, but you cannot evaluate them.
In the TimeSerial function, you can pass individual parameters (hour, minute, second) as separate numeric expressions. For the TimeValue function, however, you can pass a string as a parameter containing the time.
## Error codes:
5 Invalid procedure call
13 Data type mismatch
## Example:
`Sub ExampleTimerValue`
`Dim daDT As Date`
`Dim a1, b1, c1, a2, b2, c2 As String`
` a1 = "start time"`
` b1 = "end time"`
` c1 = "total time"`
` a2 = "8:34"`
` b2 = "18:12"`
` daDT = TimeValue(b2) - TimeValue(a2)`
` c2 = a1 & ": " & a2 & chr(13)`
` c2 = c2 & b1 & ": " & b2 & chr(13)`
` c2 = c2 & c1 & ": " & trim(Str(Hour(daDT))) & ":" & trim(Str(Minute(daDT))) & ":" & trim(Str(Second(daDT)))`
` MsgBox c2`
`End Sub` | 451 | 1,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-05 | latest | en | 0.691007 |
https://documen.tv/question/a-1-500-kg-truck-has-a-net-force-of-4-200-n-acting-on-it-what-is-the-trucks-acceleration-24073191-63/ | 1,675,768,619,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00085.warc.gz | 221,853,809 | 19,768 | # A 1,500-kg truck has a net force of 4,200 N acting on it . What is the trucks’ acceleration
Question
A 1,500-kg truck has a net force of 4,200 N acting on it . What is the trucks’ acceleration
in progress 0
2 years 2021-07-20T21:47:48+00:00 1 Answers 38 views 0
## 2.8 m/s²
Explanation:
The acceleration of an object given it’s mass and the force acting on it can be found by using the formula
$$a = \frac{f}{m} \\$$
f is the force
m is the mass
From the question we have
$$a = \frac{4200}{1500} = \frac{42}{15} \\ = 2.8$$
We have the final answer as
### 2.8 m/s²
Hope this helps you | 206 | 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": 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} | 4.1875 | 4 | CC-MAIN-2023-06 | latest | en | 0.896577 |
https://www.ibpsguide.com/ibps-clerk-prelims-quantitative-aptitude-questions-day-15/ | 1,632,848,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00572.warc.gz | 845,469,382 | 72,257 | # IBPS Clerk Prelims 2018 – Quantitative Aptitude Questions Day-15
Dear Readers, Bank Exam Race for the Year 2018 is already started, To enrich your preparation here we have providing new series of Practice Questions on Quantitative Aptitude – Section. Candidates those who are preparing for IBPS Clerk Prelims 2018 Exams can practice these questions daily and make your preparation effective.
[WpProQuiz 4134]
Click here to view Quantitative Aptitude Questions in Hindi
Directions (Q. 1 – 5): What approximate value should come in place of question mark (?) in the following questions?
1) 26 % of 349.75 – 32 % of 599.27 =? – 157.94 – (27.21)2
a) 820
b) 568
c) 786
d) 634
e) 912
2) 5 2/3 of 299.78 + 3 1/8 of 429.67 =? % of 749.85
a) 384
b) 360
c) 448
d) 412
e) 490
3) (553.672 + 959.325 – 432.72) ÷ 14.85 =?
a) 72
b) 86
c) 65
d) 58
e) 95
4) [((149 ÷ 3) ÷ 9.97)2 + (64.13 × 9.05)1/2] = (?)2
a) 12
b) 18
c) 7
d) 23
e) 30
5) ? ÷ 3.9 + 72.03 × 15 = 4824.12 + 1598.23
a) 25682
b) 21368
c) 18924
d) 16780
e) 27950
Directions (Q. 6 – 10) Study the following information carefully and answer the given questions.
The following table shows the total number of tyres (In lakhs) manufactured by 5 different companies in a certain year.
6) Total number of tyres rejected by the company P is approximately what percentage more/less than the total number of tyres rejected by the company T?
a) 7 % less
b) 15 % more
c) 15 % less
d) 24 % more
e) 7 % more
7) Find the average number of tyres manufactured by all the given companies together?
a) 32.4 lakhs
b) 28.5 lakhs
c) 30.8 lakhs
d) 34.6 lakhs
e) None of these
8) Find the difference between the total number of tyres sold by the company Q to that of company S?
a) 792600
b) 756800
c) 833200
d) 809270
e) None of these
9) Find the ratio between the total number of tyres rejected by the company Q to that of R?
a) 72: 55
b) 53: 115
c) 84: 137
d) 65: 96
e) None of these
10) Total number of tyres sold by the company P is approximately what percentage of total number of tyres sold by the company R?
a) 60 %
b) 75 %
c) 50 %
d) 85 %
e) 105 %
Answers :
Direction (1-5) :
1) Answer: c)
26 % of 350 – 32 % of 600 = x – 158 – 272
(26/100)*350 – (32/100)*600 = x – 158 – 729
91 – 192 + 158 + 729 = x
X = 786
2) Answer: d)
6 of 300 + 3 of 430 = x% of 750
(300*6) + (3*430) = (x/100)*750
1800 + 1290 = 15x/2
(3090*2)/15 = x
X = 412
3) Answer: a)
(554 + 959 – 433)/15 = x
X = 1080/15 = 72
4) Answer: c)
[((150 ÷ 3) ÷ 10)2 + (64 × 9)1/2] = x2
(150/30)2 + (8*3) = x2
25 + 24 = x2
X2 = 49
X = 7
5) Answer: b)
x ÷ 4 + 72 × 15 = 4824 + 1598
(x/4) + 1080 = 6422
(x/4) = 5342
X = 5342*4 = 21368
Direction (6-10) :
6) Answer: e)
Total number of tyres rejected by the company P
= > 3200000*(2/100) = 64000
Total number of tyres rejected by the company T
= > 4000000*(1.5/100) = 60000
Required % = [(64000 – 60000)/60000]*100 = 7 % more
7) Answer: a)
The average number of tyres manufactured by all the given companies together
= > (32 + 26 + 48 + 16 + 40)/5
= > 162/5 = 32.4 lakhs
8) Answer: c)
The total number of tyres sold by the company Q
= >2600000*(95/100)*(84/100) = 2074800
The total number of tyres sold by the company S
= > 1600000*(97/100)*(80/100) = 1241600
Required difference = 2074800 – 1241600 = 833200
9) Answer: d)
The total number of tyres rejected by the company Q
= > 26*(5/100) = 1.3 lakhs
The total number of tyres rejected by the company R
= > 48*(4/100) = 1.92 lakhs
Required ratio = 1.3: 1.92 = 130: 192 = 65: 96
10) Answer: b)
Total number of tyres sold by the company P
= > 3200000*(98/100)*(78/100) = 2446080
Total number of tyres sold by the company R
= > 4800000*(96/100)*(72/100) = 3317760
Required % = (2446080/3317760)*100 = 73.726 = 75 %
Daily Practice Test Schedule | Good Luck
Topic Daily Publishing Time Daily News Papers & Editorials 8.00 AM Current Affairs Quiz 9.00 AM Current Affairs Quiz (Hindi) 9.30 AM IBPS Clerk Prelims – Reasoning 10.00 AM IBPS Clerk Prelims – Reasoning (Hindi) 10.30 AM IBPS Clerk Prelims – Quantitative Aptitude 11.00 AM IBPS Clerk Prelims – Quantitative Aptitude (Hindi) 11.30 AM Vocabulary (Based on The Hindu) 12.00 PM IBPS PO Prelims – English Language 1.00 PM SSC Practice Questions (Reasoning/Quantitative aptitude) 2.00 PM IBPS PO/Clerk – GK Questions 3.00 PM SSC Practice Questions (English/General Knowledge) 4.00 PM Daily Current Affairs Updates 5.00 PM IBPS PO Mains – Reasoning 6.00 PM IBPS PO Mains – Quantitative Aptitude 7.00 PM IBPS PO Mains – English Language 8.00 PM
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View all comments | 1,744 | 4,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2021-39 | latest | en | 0.80883 |
http://www.thenakedscientists.com/forum/index.php?topic=43551.0 | 1,477,378,285,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00316-ip-10-171-6-4.ec2.internal.warc.gz | 732,744,318 | 18,511 | # The Naked Scientists Forum
### Author Topic: problem with helicoids and pressure (Read 8586 times)
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### problem with helicoids and pressure
« on: 24/03/2012 13:54:02 »
Hi,
My son are looking for systems on internet for give energy free, I know it's impossible and until now I arrive to say where is the problem but I don't find for this one ? Can I expose the system ? It's not very complicated for explain, it's only 2 wood screws and differents pressures.
Regards
#### CliffordK
• Neilep Level Member
• Posts: 6321
• Thanked: 3 times
• Site Moderator
##### Re: problem with helicoids and pressure
« Reply #1 on: 24/03/2012 14:43:15 »
Hello,
You'll have to explain a bit better what the system you are discussing, and where energy is involved.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #2 on: 24/03/2012 16:04:20 »
Ok, thanks
The system is composed with 2 wood screws (helicoids). Each screw can only turn around its longitudinal axis (translation is not possible). Screws are put closed, the thread overlap other thread. All the system is under pressure P except one screw has around itself 3P and small volumes has pressure 2P. (Small volume=interface) Small volumes are where threads overlap. Screws turn at the same rotational speed in the same direction. All volumes are constant. I tested with 2 real wood screws, effectively screws turn, I see interface move up but like the interface is composed with 2 surfaces from threads and 2 gaskets's surfaces which are in the longitudinal axis, I see nothing lost energy.
First drawing: the principle, but green area can be very small like second drawing show. Third drawing show the volume of the interface composed with 2 parts of screw and 2 gaskets.
this for find the problem easily:
1/ Screws turn at the same speed in the same direction, so they don't move up/down, it's mechanical
2/ The interface with 2P pressure put a difference of pressure which want to turn screws in the same direction. My son see screws giving energy and me too !
3/ Interface has always the volume constant.
4/ Interface is moving in translation only in the longitudinal axis
5/ Interface is composed with 2 surfaces of helicoids and 2 surfaces of gaskets. Gaskets's surfaces are are parallel to longitudinal (or can be) so zero energy is needed for move in longitudinal axis the interface.
6/ All system can be put under P pressure like that red helicoid don't need something for retain pressure.
7/ Black helicoid is under pressure 3P, this pressure can be retain by a very thin film of pressure (with a hard material and gaskets for retain pressure). The hard material is moving when the interface is coming. Like the film can be very small, this need to take off very small pressure. The volume of the black helicoid with 3P pressure is constant because the interface is moving (need to take off and need to put 3P pressure). This need zero energy in theory for me.
Sure this system need gaskets, a lot ! and a special screw for retain 3P pressure. The film of pressure can be very thin, the side must move in real time with the moving of interfaces, difficult to do in practise but in theory this can't give more energy than received !!
I need help for explain where is the problem I don't see it :)
« Last Edit: 24/03/2012 16:35:41 by fgt55 »
#### RD
• Neilep Level Member
• Posts: 8088
• Thanked: 51 times
##### Re: problem with helicoids and pressure
« Reply #3 on: 24/03/2012 16:53:17 »
A mechanism like a lever or a screw can multiply force (pressure),
but does not multiply energy. Energy cannot be created or destroyed.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #4 on: 24/03/2012 16:57:43 »
I'm sure, you're right but I don't find where is exactly the problem can you explain ?
#### CliffordK
• Neilep Level Member
• Posts: 6321
• Thanked: 3 times
• Site Moderator
##### Re: problem with helicoids and pressure
« Reply #5 on: 24/03/2012 17:02:59 »
Obviously with gaskets, you get friction.
But, that aside, in a system with a pressure differential, you tend to get movement in a direction that relieves the pressure differential. If motion doesn't relieve the pressure differential, then it would be unlikely to turn.
So, ignoring friction, the pressure would tend to try to drive the 3P screw out of whatever is holding it. And, it would also tend to drive the 1P screw further into it. If it is designed so that the screws don't actually move, then the turning the screws won't actually relieve pressure, and there would be nothing to drive the movement forward.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #6 on: 24/03/2012 17:14:01 »
Yes, I'm ignoring friction.
If the screw move with longitudinal axis, screw can only turn, the axis can block the longitudinal translation I think (with bearing). Rotation is only the liberty for screw I don't know how one screw can go to another ?
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #7 on: 24/03/2012 18:28:07 »
What you have is esentially a nut and bolt.
You can multiply the force at the expense of motion, but, because of friction, you are unlikely to multiply the motion at the expense of force.
The reason nuts do not unscrew themselves is because of friction between the nut and the bolt, although, when there is vibration, or repeated heating and cooling, a nut can creep on a bolt and lose tension.
Friction is heat energy, so if there is any friction at all (and there always is) you will get less work (mechanical energy) out than you put in.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #8 on: 24/03/2012 18:43:49 »
Quote
What you have is esentially a nut and bolt.
Not really.
Have you read all my first message ? Have you understand all the message, if not don't hesitate to ask.
We can limit all friction but not to zero sure ! But friction can't cancel all the energy give by rotation unless you want it. Screws can't move in translation, so the interface can move up easily, gaskets and friction is only here because there is a difference of pressure.
A torque exist and rotate screws. The another thing to move is interface: 2 longitudinal surfaces, how the energy is lost ? I need an explanation for explain.
You're right friction is energy, so if screws give energy, if friction is energy where the energy is lost ? Do you think all energy give from screws are transform in heat ? Why, explain please.
Quote
is because of friction between the nut and the bolt
it's 2 wood screws like drawing show, so they can turn togeter easily.
« Last Edit: 24/03/2012 19:14:54 by fgt55 »
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #9 on: 25/03/2012 18:05:31 »
Hi,
I take time again to thinking about this problem and I don't know where friction is, if you could explain ? because when I take 2 wood screws in my hands they turn freely IF rotational speeds are the same.
For the interface, gaskets's surfaces can be hard surfaces except at junction where it's possible to use something like radial shaft seal technology for example, so the friction can be reduce I think. If not, can you explain please ?
Regards
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #10 on: 25/03/2012 18:34:26 »
Are you rotating both screws at the same time?
If you are, there never needs to be any contact between the screws so there is no force transmitted between them. If there is no force, there will be no friction.
You can get the same effect by rotating two meshed gears so that the teeth never actually make contact. Or you can put a nut on a bolt and rotate the bolt and allow the nut to rotate with the bolt.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #11 on: 25/03/2012 19:12:30 »
Yes, I'm rotating both screws at the same time. There is always a line in contact (I used wood screw), see the drawing. One screw never block other, sure the speed must be exaclty the same. A line is not a surface but the volume of the interface is something like the yellow part in last drawing: 2 surfaces from helicoids and 2 hard surfaces which is the really the interface, these surfaces are curve but in longitudinal axis. Maybe if you have 2 wood screws this can help me to understand.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #12 on: 25/03/2012 19:19:34 »
I add a drawing for show the real volume of interface. It's the other side of the last drawing. One screw is under P, other under 3P and interface under 2P.
« Last Edit: 25/03/2012 19:30:03 by fgt55 »
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #13 on: 25/03/2012 20:26:46 »
I really don't understand what the question is.
The two screws rotate and don't interfere with each other. A lot of things (like bearings for example) can have relative rotation without interfering each other. This isn't really any different.
Are you thinking this mechanism could be part of a pump or something?
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #14 on: 25/03/2012 21:30:24 »
Sorry. It's not a pump, all volumes are constant and pressure P, 2P, 3P never change. Red screw has around it P, black screw has around it 3P, the interface has inside 2P. Black screw see a differential pressure 3P/2P on a small part of a surface, this rotate with torque the black screw. Red screw see a differential 2P/P on a small part of a surface, this rotate (in the same direction than black screw) with torque the red screw. Interface move in longitudinal axis only but like surfaces are in longitudinal axis this don't need energy (2 surfaces of the interface is screws but screws don't move in longitudinal axis, but they turn). So if the friction for contain pressure can be small, where the energy is lost ?
« Last Edit: 25/03/2012 21:32:36 by fgt55 »
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #15 on: 25/03/2012 22:29:40 »
What you have is a type of helical gear.
http://en.wikipedia.org/wiki/Gear#Helical
However, the slope of the helix with two screws is so small that the mechanism will tend to wedge as soon as you use it to transmit torque from one screw to the other. If the screws had much steeper slopes, and you had multiple helixs on each screw, you would have two helical gears meshed, and you would be able to transmit torque.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #16 on: 25/03/2012 22:59:35 »
It's not one screw which transmit torque to another. It's the difference of pressure which create 2 torques, one for red screw and one for black screw. Screws are independant (but controlled for have same rotational speed). The slope is the slope of the pitch of the screw is not so small.
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #17 on: 25/03/2012 23:26:05 »
Yes, but you still have a helical gear : )
OK - I give up.
If both of the screws are being driven at the same speed, what does it matter whether the helixs contact each other or not, and what is the purpose of this machine?
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #18 on: 26/03/2012 00:03:00 »
My son has seen that idea on internet, I don't know where. The purpose is to give more energy than the system receive...I know it's not possible and often I find where is the problem on several systems. But here no ! Here I think the idea is to put 3 fixed pressures, 2 screws, gaskets, controlled system, etc. and recover energy from rotation of helixs (for me the translation of interface don't need energy), the rest of the system don't lost energy. I hope you don't give up ;)
We can imagine sensors with controlled system which driven at the same speed screws, it's only a technical problem (at least for me). The contact of helixs is not essential, it's the volume of interface which must be constant (for not loss energy I think). The volume of interface give a surface for each helix, the differential's pressure rotate screws and torque.
The volume of interface is not 0, see yellow part in reply #2
Maybe if you read the reply #2 you can find where an hypothesis is false ?
« Last Edit: 26/03/2012 00:09:36 by fgt55 »
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #19 on: 26/03/2012 01:03:01 »
You need to draw a box around the system and identify all the energy inputs and all the energy outputs. Total input must equal total output. If they are not equal you are overlooking something.
I can assure you that what you have is a variation of a helical gear, and there isn't a gearing system that is even 100% efficient let alone more than 100% efficient. I encourage you to study how gears work, and where they lose energy.
I'll be moving this topic to the "New Theories" forum in a little while.
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #20 on: 26/03/2012 08:24:03 »
Ok, thanks
Something important: 2 screws are motors, it's not one motor and one receptor, each give motor's torque.
I think 2 screws can be really independant, I don't see them like a gear when a screw will turn other, and maybe it's the problem. But I'm thinking about your message.
For now, maybe my misunderstood is I think the screw turn where there is a difference of pressure on a small part of surface (only one side), see drawing. Here there is ONLY one screw. One screw turn like that and give a torque ? Sure like that the sum of energy is 0 because I need to adjust the position of the red point.
« Last Edit: 26/03/2012 09:15:02 by fgt55 »
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #21 on: 26/03/2012 16:28:38 »
I think I understood your point of view, maybe you think one screw turn and give torque to the second screw ? Two Screws turns independantly and give a torque for turn 2 receptors. If you see 2 screws like motors, one screw don't give energy to other. Tell me if it's that ?
« Last Edit: 26/03/2012 17:43:10 by fgt55 »
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #22 on: 26/03/2012 19:10:10 »
Are you trying to make something like this?
http://en.wikipedia.org/wiki/Rotary_screw_compressor#Superchargers
#### fgt55
• Jr. Member
• Posts: 42
• LCF
##### Re: problem with helicoids and pressure
« Reply #23 on: 26/03/2012 19:30:18 »
It's not exatly that, more complicated and it's the contrary: pressure => torque
Please, take the reply #20, for you the screw turn and give torque ?
#### Geezer
• Neilep Level Member
• Posts: 8328
• "Vive la résistance!"
##### Re: problem with helicoids and pressure
« Reply #24 on: 27/03/2012 05:27:29 »
I still not sure I understand what it's supposed to do - probably a language thing.
Are you saying that fluid pressure contained by the green line will make the screws turn?
If that's it, the screws will not turn because no work is being done to make them turn. That would only happen if the volume of the 3P zone changed. The geometry of the screws ensures that the volume never changes.
If you contain high pressure gas in a cylinder with a piston, it only does work when the volume increases as the piston moves. If you look at the picture of the screw compressor I posted you can see that the helices have opposite screw directions so that work can be done on them to compress a gas. If the screws had the same direction they would rotate, but they would not compress anything.
Hope that's it!
#### The Naked Scientists Forum
##### Re: problem with helicoids and pressure
« Reply #24 on: 27/03/2012 05:27:29 » | 4,300 | 16,387 | {"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} | 2.65625 | 3 | CC-MAIN-2016-44 | longest | en | 0.930893 |
http://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=pounds | 1,653,093,407,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00623.warc.gz | 13,694,543 | 3,789 | Partner with ConvertIt.com
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 523 | 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} | 3.09375 | 3 | CC-MAIN-2022-21 | latest | en | 0.672464 |
https://www.888casino.com/blog/novelty-games/ultimate-texas-holdem-the-known-card | 1,632,169,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00237.warc.gz | 671,248,578 | 28,912 | Ultimate Texas Hold'em (UTH) is a rich source of advantage play opportunities. This post considers yet another possibility: when the player knows one of the two cards he will be dealt in advance of placing his initial bet. I often refer to this knowledge as "top carding," though the known card can be either player card. Knowledge of one player card can be gained, for example, by edge sorting, accidental exposure, card location, skilled cutting, shuffler issues, incidental marks on the cards and outright cheating. Whatever the means, the edge the player can get is extraordinary.
The work presented here is based on the combinatorial analysis I completed for the main game (see this post). No specialized computer analysis was required. After all, once the initial wager has been made, the hand will be played according to basic strategy. The only advantage the AP gets is whether his known card favors the player or the house.
The following table shows the edge, based on the known card:
Note that the average edge across all known cards is -2.1850%, the house edge (last row in table). If the known card is a Ten, Jack, Queen, King or Ace, then the AP has the edge, otherwise the house has the edge. The AP exploits this knowledge by making a large wager when he has the edge, otherwise he will either make a small wager or sit out the hand.
After I completed this work, I decided to check my results against those presented by James Grosjean in Exhibit CAA (page 362). I was surprised to discover that our results don't agree. For example, I give an edge of -56.97% if the known card is a Deuce. Grosjean gives an edge of -57.96% in this same situation.
Here are Grosjean's known card results:
Note that Grosjean'saverage edge across all known cards is -2.358%, which is NOT the house edge (last row in table). It follows that Grosjean's results are not correct. I do not mean this to be a gotcha against Grosjean -- I am simply making a methodological point. These differences reinforce theconjecture I've made in other posts that Grosjean's UTH results were obtained by simulation and not by exhaustive analysis. These differences are not consequential to the AP. (Note. As David Spence commented below, it appears that Grosjean's numbers may bebased on his basicstrategy with house edge 2.35%, and not computer-perfect basic strategy.)
The following table gives the statistics for known card play against UTH:
Knowledge of one card is a stronger opportunity than hole-carding. Even if the AP knows one dealer hole-card and the Turn/River hole-card, he "only" has an edge of about 25%. An average edge over 40% is a good day. A desirability indexof 50 puts this play in the stratosphere (Note: I used the overall standard deviation for UTH when I computed the DI, not the standard deviation when the hand is restricted to containing one of the cards, T, J, Q, K, A.).
The following procedures can be used to protect UTH against known-card advantage play:
• Include a turn before the deck is placed into the shuffler.
• Do not feed the deck from the previous round into the shuffler until all players have placed their wagers for the next round.
• Watch for a player at first base who has a large spread in his bets.
• Watch for team play to sort the cards and keep them sorted.
• Watch for signaling if the shuffler is located at third base.
• Watch for dirty and marked cards.
• Replace the cards if something just doesn't look right and hole-carding has been excluded. | 784 | 3,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2021-39 | longest | en | 0.968771 |
http://www.algebra.com/tutors/your-answers.mpl?userid=stanbon&from=20700 | 1,369,224,935,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701670866/warc/CC-MAIN-20130516105430-00085-ip-10-60-113-184.ec2.internal.warc.gz | 312,642,128 | 27,783 | Algebra -> Tutoring on algebra.com -> See tutors' answers! Log On
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54990..55019 , 55020..55049 , 55050..55079 , 55080..55109 , 55110..55139 , 55140..55169 , 55170..55199 , 55200..55229 , 55230..55259 , 55260..55289 , 55290..55319 , 55320..55349 , 55350..55379 , 55380..55409 , 55410..55439 , 55440..55469 , 55470..55499 , 55500..55529 , 55530..55559 , 55560..55589 , 55590..55619 , 55620..55649 , 55650..55679 , 55680..55709 , 55710..55739 , 55740..55769 , 55770..55799 , 55800..55829 , 55830..55859 , 55860..55889 , 55890..55919 , 55920..55949 , 55950..55979 , 55980..56009 , 56010..56039 , 56040..56069 , 56070..56099 , 56100..56129 , 56130..56159 , 56160..56189 , 56190..56219 , 56220..56249 , 56250..56279 , 56280..56309 , 56310..56339 , 56340..56369 , 56370..56399 , 56400..56429 , 56430..56459 , 56460..56489 , 56490..56519 , 56520..56549 , 56550..56579 , 56580..56609 , 56610..56639 , 56640..56669 , 56670..56699 , 56700..56729 , 56730..56759 , 56760..56789 , 56790..56819 , 56820..56849 , 56850..56879 , 56880..56909 , 56910..56939 , 56940..56969 , 56970..56999 , 57000..57029 , 57030..57059 , 57060..57089 , 57090..57119 , 57120..57149 , 57150..57179 , 57180..57209 , 57210..57239 , 57240..57269, >>Next
Rational-functions/389811: Given that x is an integer between -2 and 2, state the relation represented by the equation y=2-|x| by listing a set of ordered pairs. Then state whether the relation is a function Thank you so much!!!!!! God Bless and Happy New Year!!!!!!!!1 solutions Answer 276335 by stanbon(57308) on 2011-01-01 22:18:50 (Show Source): You can put this solution on YOUR website!Given that x is an integer between -2 and 2, state the relation represented by the equation y=2-|x| by listing a set of ordered pairs. Then state whether the relation is a function ----------- If x = -4, y = 2-4 = -2 If x = -3, y = 2-3 = -1 If x = -2, y = 2-2 = 0 If x = -1, y = 2-1 = 1 If x = 0, y = 2-0 = 2 --- Plot those points and connect them to get get: --- Yes it is a function. Cheers, Stan H.
Rational-functions/389809: how do you find the zeros in this equation? x^5-5x^4-3x^3+15x^2-4x+20=01 solutions Answer 276332 by stanbon(57308) on 2011-01-01 22:14:22 (Show Source): You can put this solution on YOUR website!find the zeros in this equation? x^5-5x^4-3x^3+15x^2-4x+20=0 --- I graphed it and found Real zeros at x = 2 and x = -2 and x = 5 ---- Use synthetic division to look for other zeros: 2)....1....-5....-3....15....-4....20 ........1.....-3....-9....-3....-10...|..0 ==== -2)......1......-5....1.....-5....|..0 ==== 5)........1.......0.....1.....|0 ---- Quotient: x^2+1 --- Complex zeros: x^2+1 = 0 (x+i)(x-i) = 0 x = -i or x = i ========================= Cheers, Stan H.
test/389807: What is the sulm of the polynomials 3a2b + 2a2b2 and -ab2 + a2b21 solutions Answer 276329 by stanbon(57308) on 2011-01-01 22:00:21 (Show Source): You can put this solution on YOUR website!What is the sum of the polynomials 3a^2b + 2a^2b^2 and -ab^2 + a^2b^2 ----- Add "like" terms to get: ---- (3a^2b) + (2a^2b^2+a^2b^2)-ab^2 --- = 3a^2b + 3a^2b^2 - ab^2 ========================= Cheers, Stan H. ============
Probability-and-statistics/389806: 4. What is the probability that a that a worker at Random drives more that 15 miles to work if the distribution of all worker’s mileage is a Normal Distribution that ranges from 4 miles to 26 miles? 1 solutions Answer 276328 by stanbon(57308) on 2011-01-01 21:56:39 (Show Source): You can put this solution on YOUR website!What is the probability that a worker at Random drives more that 15 miles to work if the distribution of all worker’s mileage is a Normal Distribution that ranges from 4 miles to 26 miles? ---- range = 6 sigma ---- 6 sigma = 26-4 = 22 --- sigma = 11/3 --- mean = 4+(22/2) = 15 ------------------------- P(x > 15) = 0.5000 ======================== Cheers, Stan H.
test/389808: Which of the following is a factor of the polynomial x2 - x - 20?1 solutions Answer 276326 by stanbon(57308) on 2011-01-01 21:52:20 (Show Source): You can put this solution on YOUR website!Which of the following is a factor of the polynomial x^2 - x - 20? ----- Factor form: (x-5)(x+4) =============================== Cheers, Stan H.
Triangles/389800: The sine ratio of any acute angle is equal to the cosine ratio of its complement ? true or false1 solutions Answer 276322 by stanbon(57308) on 2011-01-01 21:04:48 (Show Source): You can put this solution on YOUR website!true --- Cheers, Stan H.
Volume/389770: the sum of the digits of a two digit number is 11 and the difference of the digit is 5. what is the number1 solutions Answer 276308 by stanbon(57308) on 2011-01-01 18:46:11 (Show Source): You can put this solution on YOUR website! the sum of the digits of a two digit number is 11 and the difference of the digit is 5. what is the number ---- Let the number be 10t+u --- Equations: t + u = 11 t - u = 5 ----- 2t = 16 t = 8 --- Since t + u = 11, u = 3 ---------------- Ans: Number is 83 ====================== Cheers, Stan H.
Volume/389774: what is the height of a rectangular solid with a lenght of 7 feet width of 5 feet and a volume of 105 cubic feet1 solutions Answer 276306 by stanbon(57308) on 2011-01-01 18:36:54 (Show Source): You can put this solution on YOUR website!what is the height of a rectangular solid with a length of 7 feet, width of 5 feet and a volume of 105 cubic feet --------- Volume = height*width*length 105 sq.ft. = h*5'*7' ---- h = 105/35 --- height = 3 ft. ==================== Cheers, Stan H.
Equations/389776: The sum of two number is 33. Thier different is 7. what are the two number help please1 solutions Answer 276305 by stanbon(57308) on 2011-01-01 18:32:12 (Show Source): You can put this solution on YOUR website!The sum of two number is 33. Their difference is 7. what are the two numbers? ==================== Equations: x+y = 33 x-y = 7 --- Add and solve for "x": 2x = 40 x = 20 ---- Substitute and solve for "y": 20 + y = 33 y = 13 ----------------- Cheers, Stan H.
Money_Word_Problems/389761: four years ago, jonny invested \$4000 at 5.4% intrest compunded annually. five years ago, he invested \$3700 at 5.6% intrest compounded quartly. when are the two investments worth the same (or has the smaller one alwreday surprassed the greater one)?1 solutions Answer 276303 by stanbon(57308) on 2011-01-01 18:27:46 (Show Source): You can put this solution on YOUR website!four years ago, johnny invested \$4000 at 5.4% interest compounded annually. ---- five years ago, he invested \$3700 at 5.6% interest compounded quarterly. ==== when are the two investments worth the same (or has the smaller one already surpassed the greater one)? ------ Value of the \$3700 investment after one year. 3700(1+0.056) = \$3907 ------ Solve for "t": 3907(1+0.056)^t = 4000(1+0.054)^t [1.056/1.054]^t = 4000/3907 1.001898^t = 1.02375 ---- t = log(1.02375)/log(1.001898) --- t = 12.38 years ====================== Same value 8 1/3 years from now. ====================== Cheers, Stan H.
Matrices-and-determiminant/389758: What is Matrices, determinant, and the cramer rule? 1 solutions Answer 276299 by stanbon(57308) on 2011-01-01 17:55:50 (Show Source): You can put this solution on YOUR website!Please use Google or other search engine to find extensive discussion of these topics. ---- Cheers, Stan H. ====
Equations/389768: find the range,median and mode for 80,80,100,90,851 solutions Answer 276298 by stanbon(57308) on 2011-01-01 17:52:06 (Show Source): You can put this solution on YOUR website!find the range,median and mode for 80,80,100,90,85 ---- range = 100-80 = 20 ---- median = 85 ---- mode = 80 ==================== Cheers, Stan H.
Functions/389763: I am trying to solve by factoring, and I do not know if I am doing this correctly. Here is my work and answer: If f(x) = 20x and g(x) = x - 4, what is g(g(f(3))? f(3) = 20(3) g(60) = 60 - 4 g(56) = 56 - 4 = 52 [this is my answer, but I need someone to check the answer and walk me through the solution.] Thanks in advance for any assistance. 1 solutions Answer 276297 by stanbon(57308) on 2011-01-01 17:50:08 (Show Source): You can put this solution on YOUR website!If f(x) = 20x and g(x) = x - 4, what is g(g(f(3))? ------- g(g(f(3)) --- = g(g(60)) --- = g(56) ----- = 52 ============ Cheers, Stan H.
Evaluation_Word_Problems/389731: a scientist develops a formula which can turn a light bulb into a camel.working alone it takes the scientist 4 hours to create one dose of formula.the scientist assistant can make one dose in 12 hours. how long working together will it take the scientist and his assistant to create one dose of formula? please answer the question by using a table and the 5 step method. thank you.1 solutions Answer 276280 by stanbon(57308) on 2011-01-01 15:28:31 (Show Source): You can put this solution on YOUR website!a scientist develops a formula which can turn a light bulb into a camel. --------- working alone it takes the scientist 4 hours to create one dose of formula. --------- the scientist assistant can make one dose in 12 hours. --------- how long working together will it take the scientist and his assistant to create one dose of formula? --- Your able has three rows: scientist ; assistant ; together --- Your table has two coluns: time and rate ----- Scientist time = 4 hrs ; rate = 1/4 job/hr --- Assistant time = 12 hrs ; rate = 1/12 job/hr --- Together time = x hrs ; rate = 1/x job/hr ------------------------------------------- Equation: rate + rate = rate 1/4 + 1/12 = 1/x Multiply thru by 12x --- 3x + x = 12 4x = 12 x = 3 hrs (time to do the job together) ========= Cheers, Stan H.
Polynomials-and-rational-expressions/389745: What would the polynomial with real coefficients whose zeroes are -4, 1-i, and 1 + i be?1 solutions Answer 276277 by stanbon(57308) on 2011-01-01 15:20:08 (Show Source): You can put this solution on YOUR website!What would the polynomial with real coefficients whose zeroes are -4, 1-i, and 1 + i be? ---- f(x) = (x+4)(x-(1-i))(x-(1+i)) ---- f(x) = (x+4)((x-1)+i)((x-1)-i) ---- f(x) = (x+4)((x-1)^2+1) --- f(x) = (x+4)(x^2-2x+2) --- f(x) = x^3+2x^2-4x+8 =========================== Cheers, stan H.
Trigonometry-basics/389206: I really need to answer these trigonometry problems. 1. The measure of two of the angles of a triangle are 50 degrees and 55 degrees. If its longest side measures 17 cm, find the perimeter of the triangle. 2. The vertex angle of an isosceles triangle is 72 degrees and each of the equal sides is 10 cm. Find the perimeter and are of the triangle. 3. The diameter of a circle C is 34 cm. Two radii AC and BC from the angle of 96 degrees. Find the length of the cord AB. 1 solutions Answer 275829 by stanbon(57308) on 2010-12-29 17:08:11 (Show Source): You can put this solution on YOUR website!I really need to answer these trigonometry problems. 1. The measure of two of the angles of a triangle are 50 degrees and 55 degrees. If its longest side measures 17 cm, find the perimeter of the triangle. ---- The measure of the third angle is 180-(50+55) = 75 degrees ---- Find the length of the side opposite the 50 degree angle. Use the Law of Sines: a/sin(50) = 17/sin(75) --- a = sin(50)[17/sin(75)] ---- a = 13.48 ================== Find the length of the side opposite the 55 degree angle. Use the Law of Sines: b/sin(55) = [17/sin(75)] b = sin(55)[17/sin(75)] b = 14.42 ======================= Perimeter: 17 + 13.48 + 14.42 =================================== 2. The vertex angle of an isosceles triangle is 72 degrees and each of the equal sides is 10 cm. Find the perimeter and area of the triangle. --- Use the Law of Sines as in problem #1. =================================== 3. The diameter of a circle C is 34 cm. Two radii AC and BC form the angle of 96 degrees. Find the length of the cord AB. --- Perimeter = (pi)d = 34pi --- arc = (96/360)(34pi) = 28.48...cm Cheers, Stan H.
Triangles/389289: Find the value of x. A triangle. One exterior angle is 70 and the nonadjacent interior angles are X and 2X + 10.1 solutions Answer 275828 by stanbon(57308) on 2010-12-29 16:37:33 (Show Source): You can put this solution on YOUR website!A triangle. One exterior angle is 70 and the nonadjacent interior angles are X and 2X + 10. --- The sum of the exterior angles is 360 degrees. The exterior corresponding to "x" is 180-x The exterior correspondiing to "2x+10" = 180-(2x+10) = -2x+170 --- Solve: (180-x)+(-2x+170) + 70 = 360 -3x + 420 = 360 -3x = -60 x = 20 degrees ================= Cheers, Stan H. ============ ===========================
Equations/389243: hello, I really need help in solving these two equations. The question is as following: write the following as a single power, then evaluate. (-32)^3/5 x (-32)^4/5 / (-32)^2/5 and 4096^3/4 / 4096^2/3 x 4096^5/6 thanking you in advance!1 solutions Answer 275797 by stanbon(57308) on 2010-12-29 13:24:09 (Show Source): You can put this solution on YOUR website!write the following as a single power, then evaluate. (-32)^3/5 x (-32)^4/5 / (-32)^2/5 --- = (-32)^[3/5 + 4/5 - 2/5] --- = (-32)^[5/5] -- = -32 ================================== and 4096^3/4 / 4096^2/3 x 4096^5/6 --- = 4096^[3/4 + 2/3 + 5/6] --- = 4096^[(9+8+10)/12] --- = 4096^(27/12) --- = 2^27 --- Cheers, Stan H.
Exponents/389259: the model y = 17(1.9)x is classified as a. exponential growth b. exponential decay. The model y = 22(0.8)x is classified as a. exponential growth b. exponential decay.1 solutions Answer 275795 by stanbon(57308) on 2010-12-29 13:18:04 (Show Source): You can put this solution on YOUR website!duplicate
Exponents/389260: the model y = 17(1.9)x is classified as a. exponential growth b. exponential decay. The model y = 22(0.8)x is classified as a. exponential growth b. exponential decay.1 solutions Answer 275794 by stanbon(57308) on 2010-12-29 13:17:20 (Show Source): You can put this solution on YOUR website!the model y = 17(1.9)^x is classified as a. exponential growth b. exponential decay. Since 1.9 is greater than 1, it is exponential growth. --- The model y = 22(0.8)^x is classified as a. exponential growth b. exponential decay. Since 0.8 is greater than zero and less than one it is exponential decay. --- Cheers, Stan H. =========== | 17,372 | 41,157 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2013-20 | latest | en | 0.208096 |
https://economics.stackexchange.com/questions/32416/how-to-calculate-ppp-from-lcus | 1,716,465,930,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058625.16/warc/CC-MAIN-20240523111540-20240523141540-00519.warc.gz | 186,694,235 | 39,207 | # How to calculate PPP$from LCUs? Is there one single formula which allows to calculate the income of a person in PPP\$ from her income in LCUs (local currency units), given for example the PPP conversion factor, GDP (LCU per international $) as published by the World Bank. Assume for example an Indian farmer with a monthly income of ₹50,000, given India's PPP conversion factor of 18 and exchange rate ₹70 =$1. What is his monthly income in PPP\$and by which formula? Is other or further information needed? Vice versa: What is the farmer's monthly income in rupees when his income is known to be PPP\$30?
References:
It works like any other exchange rate:
• Divide ₹50,000 by 70 to get about US$714 / month at market exchange rates • Divide ₹50,000 by 18 to get about PPP$2778 / month at PPP rates (so not a poor farmer)
• Multiply USD$30 by 70 to get ₹2100 at market exchange rates • Multiply PPP$30 by 18 to get ₹540 at PPP rates
• You are right: ₹50,000 is more than the average Indian farmer earns, this is more like ₹10,000 or ₹20,000 (with sideline jobs). Oct 24, 2019 at 20:42
• Nevertheless, this doesn't work: According to Gapminder's Dollar Street, PPP$30 is a somehow not atypical income of poor farmers, but ₹540 per month is an unrealistically low income: not even the poorest farmers earn only such a small amount per month. Oct 24, 2019 at 20:47 • @Hans-PeterStricker Let's take the first example I came across from Gapminder of a fruit collector and family in West Bengal with a monthly income of PPP$29 per adult so presumably about ₹520. Now take the second case I found of a farmer and family near Chennai with a monthly income of PPP\$245 per adult so say about ₹4400 a month. Looking around their homes suggests different farmers can have very different incomes. Oct 24, 2019 at 23:16 | 453 | 1,814 | {"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} | 3.46875 | 3 | CC-MAIN-2024-22 | latest | en | 0.952375 |
http://catalog.flatworldknowledge.com/bookhub/reader/5228?e=hoyle-ch14_s06 | 1,386,615,139,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163995757/warc/CC-MAIN-20131204133315-00096-ip-10-33-133-15.ec2.internal.warc.gz | 27,005,523 | 13,323 | # Financial Accounting, v. 2.0
by Joe Hoyle and C.J. Skender
## 14.6 Bonds with Other Than Annual Interest Payments
### Learning Objectives
At the end of this section, students should be able to meet the following objectives:
1. Realize that cash interest payments are often made more frequently than once a year, such as each quarter or semiannually.
2. Determine the stated interest rate, the effective interest rate, and the number of time periods to be used in a present value computation when interest payments cover a period of time of less than a year.
3. Compute the stated cash interest payments and the effective interest rate when interest is paid on a bond more frequently than once each year.
4. Prepare journal entries for a bond when the interest payments are made for a period of time shorter than a year.
Question: In the previous examples, both the interest rates and payments always covered a full year. How is this process affected if interest payments are made at other time intervals such as each quarter or semiannually?
As an illustration, assume that on January 1, Year One, an entity issues term bonds with a face value of 500,000 that will come due in six years. Cash interest payments at a 6 percent annual rate are required by the contract. However, the actual disbursements are made every six months on June 30 and December 31. In setting a price for these bonds, the debtor and the creditor negotiate an effective interest rate of 8 percent per year. How is the price of a bond determined and the debt reported if interest payments occur more often than once each year? Answer: None of the five basic steps for issuing and reporting a bond is affected by a change in the frequency of interest payments. However, both the stated cash rate and the effective rate must be set to agree with the time interval between payment dates. The number of time periods used in the present value computation also varies based on the time that passes from one payment to the next. In this current example, interest is paid semiannually, so each time period is only six months in length. The stated cash interest rate to be paid during that period is 3 percent or 6/12 of the annual 6 percent rate listed in the bond contract. Similarly, the effective interest rate is 4 percent or 6/12 of the annual 8 percent negotiated rate. Both of these interest rates must align with the specific amount of time between payments. Over the six years until maturity, the bond is outstanding for twelve of these six-month periods of time. Thus, for this bond, the cash flows will be the interest payments followed by settlement of the face value. • Interest Payments:500,000 face value times 3 percent stated rate for a $15,000 interest payment every six months during these twelve periods. Equal payments are made at equal time intervals making this an annuity. Because payments are made at the end of each period, these payments constitute an ordinary annuity. • Face value:$500,000 is paid at the end of these same twelve periods. This cash payment is a single amount.
As indicated, the effective rate to be used in determining the present value of these cash payments is 4 percent per period or 6/12 times 8 percent.
### Present Value of Annuity Due of $1 http://www.principlesofaccounting.com/ART/fv.pv.tables/pvof1.htm ### Present Value of Ordinary Annuity of$1
http://www.principlesofaccounting.com/ART/fv.pv.tables/pvofordinaryannuity.htm
• The present value of $1 in twelve periods at an effective rate of 4 percent per period is$0.62460.
• The present value of an ordinary annuity of $1 for twelve periods at an effective rate of 4 percent per period is$9.38507.
• The present value of the face value cash payment is $500,000 times$0.62460 or $312,300. • The present value of the cash interest payments every six months is$15,000 times $9.38507 or$140,776 (rounded).
• The total present value of the cash flows established by this contract is $312,300 plus$140,776 or $453,076. As shown in Figure 14.24 "January 1, Year One—Issuance of$500,000 Bond with a 3 Percent Stated Rate to Yield Effective Rate of 4 Percent Semiannually", the bond is issued for this present value amount. With this payment, the agreed-upon effective rate of interest (8 percent for a full year or 4 percent for each six-month period) will be earned by the investor over the entire life of the bond.
Figure 14.24 January 1, Year One—Issuance of $500,000 Bond with a 3 Percent Stated Rate to Yield Effective Rate of 4 Percent Semiannually On June 30, Year One, the first$15,000 interest payment is made as reported in Figure 14.25 "June 30, Year One—Cash Interest Paid on Bond for Six-Month Period". However, the effective interest rate for that period is the principal of $453,076 times the six-month negotiated rate of 4 percent or$18,123 (rounded). Therefore, the interest to be compounded for this first six-month period is $3,123 ($18,123 interest less $15,000 payment). That is the amount of interest recognized but not yet paid that is added to the liability as shown in Figure 14.26 "June 30, Year One—Interest on Bond Adjusted to Effective Rate". Figure 14.25 June 30, Year One—Cash Interest Paid on Bond for Six-Month Period Figure 14.26 June 30, Year One—Interest on Bond Adjusted to Effective Rate The compound interest recorded previously raises the bond’s principal to$456,199 ($453,076 principal plus$3,123 in compound interest). The principal is gradually moving to the $500,000 face value. Another$15,000 in cash interest is paid on December 31, Year One (Figure 14.27 "December 31, Year One—Cash Interest Paid on Bond for Six-Month Period"). The effective interest for this second six-month period is $18,248 (rounded) or$456,199 times 4 percent interest. The compound interest recognized on December 31, Year One, is $3,248 ($18,248 less $15,000), a balance that is recognized by the entry presented in Figure 14.28 "December 31, Year One—Interest on Bond Adjusted to Effective Rate". Figure 14.27 December 31, Year One—Cash Interest Paid on Bond for Six-Month Period Figure 14.28 December 31, Year One—Interest on Bond Adjusted to Effective Rate The Year One income statement will report interest expense of$18,123 for the first six months and $18,248 for the second, giving a total for that year of$36,371. The second amount is larger than the first because of compounding.
The December 31, Year One, balance sheet reports the bond payable as a noncurrent liability of $459,447. That is the original principal (present value) of$453,076 plus compound interest of $3,123 (first six months) and$3,248 (second six months). Once again, interest each period has been adjusted from the cash rate stated in the bond contract to the effective rate negotiated by the two parties. Here, the annual rates had to be halved because payments were made semiannually.
### Test Yourself
Question:
Friday Corporation issues a two-year bond on January 1, Year One, with a $300,000 face value and a stated annual cash rate of 8 percent. The bond was sold to earn an effective annual rate of 12 percent. Interest payments are made quarterly beginning on April 1, Year One. The present value of$1 at a 3 percent rate in eight periods is $0.78941. The present value of$1 at a 12 percent rate in two periods is $0.79719. The present value of an ordinary annuity of$1 at a 3 percent rate for eight periods is $7.01969. The present value of an ordinary annuity of$1 at a 12 percent rate for two periods is $1.69005. What amount will Friday receive for this bond? 1.$276,055
2. $278,941 3.$284,785
4. $285,179 Answer: The correct answer is choice b:$278,941.
Explanation:
This bond pays interest of $6,000 every three months ($300,000 × 8 percent × 3/12) and $300,000 in two years. Based on quarterly payments, effective interest is 3 percent (12 percent × 3/12 year) and cash flows are for eight periods (every three months for two years). Present value of the interest is$6,000 × $7.01969, or$42,118. Present value of the face value is $300,000 ×$0.78941, or $236,823. Total present value (price of the bond) is$42,118 + $236,823, or$278,941.
### Key Takeaway
Bonds often pay interest more frequently than once a year—for example, at an interval such as every three months or six months. If the stated cash rate and the effective rate differ, determination of present value is still required to arrive at the principal amount to be paid when the bond is issued. However, the present value computation must be adjusted to reflect the change in the length of a time period. The amount of time between payments is considered one period. The stated cash interest rate, the effective rate negotiated by the parties, and the number of time periods until maturity are all set for that particular time. The actual accounting and reporting are not affected, merely the method by which the interest rates and the number of periods are calculated.
### Talking With a Real Investing Pro (Continued)
Following is a continuation of our interview with Kevin G. Burns.
Question: Assume that you are investigating two similar companies. You are thinking about recommending one of them to your clients as an investment possibility. The financial statements look much the same except that one of these companies has an especially low amount of noncurrent liabilities whereas the other has a noncurrent liability total that seems quite high. How does that difference impact your thinking? Long-term liabilities are a great way to gain financing because a company can make use of someone else’s money. However, debt does increase risk.
Kevin Burns: I have managed to do well now for many years by being a conservative investor. My preference is always for the company that is debt free or as close to debt free as possible. I do not like financial leverage, never have. I even paid off my own home mortgage more than ten years ago.
On the other hand, long-term liabilities have to be analyzed as they are so very common. Is any of the debt convertible into capital stock so that it could potentially dilute everyone’s ownership in the company? Is the company forced to pay a high rate of interest? Why was the debt issued? In other words, how did the company use all that cash that it received? As with virtually every area of financial reporting, you have to look behind the numbers to see what is actually happening. That is why transparency is important. If the debt was issued at a low interest rate in order to make a smart acquisition, I am impressed. If the debt has a high rate of interest and the money was not well used, that is not attractive to me at all.
### Video Clip
Professor Joe Hoyle talks about the five most important points in Chapter 14 "In a Set of Financial Statements, What Information Is Conveyed about Noncurrent Liabilities Such as Bonds?". | 2,498 | 10,854 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.970139 |
https://software.intel.com/en-us/forums/intel-math-kernel-library/topic/419414 | 1,480,972,961,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541839.36/warc/CC-MAIN-20161202170901-00484-ip-10-31-129-80.ec2.internal.warc.gz | 885,496,281 | 20,939 | # Use Which Routine for Sparse Matrix-Sparse Vector Multiplication
## Use Which Routine for Sparse Matrix-Sparse Vector Multiplication
I have a huge matrix with very few non-zero elements. Some of the columns and rows may also be completely zero. This matrix should be multiplied by a very long vector which has only few non-zero elements. I know that mkl_?cscmv performs the sparse matrix-vector multiplication but apparently only the matrix can be sparse. I am wondering if there is any MKL routine to calculate the production of such matrix and vector.
4 posts / 0 new
For more complete information about compiler optimizations, see our Optimization Notice.
You can have a look at mkl_?csrmm, mkl_?bsrmm, mkl_?cscmm, or mkl_?coomm and simply treat the vector as a sparse matrix.
The answer depends quite a bit on the representation that you have used for the sparse vector that you wish to multiply the sparse matrix into. Do you know the indices of the vector entries that are not equal to zero? Do you know the indices of the result vector that are not equal to zero?
Quote:
mecej4 wrote:
The answer depends quite a bit on the representation that you have used for the sparse vector that you wish to multiply the sparse matrix into. Do you know the indices of the vector entries that are not equal to zero? Do you know the indices of the result vector that are not equal to zero?
Each sparse vector is represented by <nz, nRows, ind, data>, where nz and nRows represent number of non-zero elements and number of rows, ind is an array representing index for non-zero elements, and data is an array representing value of non-zero elements. So, I do have the indices of input vector.
For the resulting vector, I know everything except data, i.e. I know what the indices for non-zero elements will be.
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# The Earth s rivers constantly carry dissolved salts into its
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The Earth s rivers constantly carry dissolved salts into its [#permalink]
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14 Aug 2008, 22:45
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The Earth’s rivers constantly carry dissolved salts into its oceans. Clearly, therefore, by taking the resulting increase in salt levels in the oceans over the past hundred years and then determining how many centuries of such increases it would have taken the oceans to reach current salt levels from a hypothetical initial salt-free state, the maximum age of the Earth’s oceans can be accurately estimated.
Which of the following is an assumption on which the argument depends?
(A) The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
(B) At any given time, all the Earth’s rivers have about the same salt levels.
(C) There are salts that leach into the Earth’s oceans directly from the ocean floor.
(D) There is no method superior to that based on salt levels for estimating the maximum age of the Earth’s oceans.
(E) None of the salts carried into the Earth’s oceans by rivers are used up by biological activity in the oceans.
[Reveal] Spoiler: OA
If you have any questions
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15 Aug 2008, 00:23
hibloom wrote:
The Earth’s rivers constantly carry dissolved salts into its oceans. Clearly, therefore, by taking the resulting increase in salt levels in the oceans over the past hundred years and then determining how many centuries of such increases it would have taken the oceans to reach current salt levels from a hypothetical initial salt-free state, the maximum age of the Earth’s oceans can be accurately estimated.
Which of the following is an assumption on which the argument depends?
* The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
* At any given time, all the Earth’s rivers have about the same salt levels.
* There are salts that leach into the Earth’s oceans directly from the ocean floor.
* There is no method superior to that based on salt levels for estimating the maximum age of the Earth’s oceans.
* None of the salts carried into the Earth’s oceans by rivers are used up by biological activity in the oceans.
Between A & E.
I am more inclined towards A though, but confusing.
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15 Aug 2008, 04:11
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ill say E. Without E, the authors method wont work since he relies on none of the salt being used up .... if it were, the age would be inaccurate
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15 Aug 2008, 06:06
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IMO E.
* The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
Deposit could be large but that never hampers the result
* At any given time, all the Earth’s rivers have about the same salt levels.
OOS
* There are salts that leach into the Earth’s oceans directly from the ocean floor.
OOS
* There is no method superior to that based on salt levels for estimating the maximum age of the Earth’s oceans.
OOS
* None of the salts carried into the Earth’s oceans by rivers are used up by biological activity in the oceans.
If biological process uses the same salt, concentration will definitely fall
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15 Aug 2008, 11:33
hey alpha y dont you share your line of reasoning
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15 Aug 2008, 17:17
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Conclusion: the maximum age of the Earth’s oceans = (Current Salt Level - hypothetical initial salt-free state) / (the oceans over the past hundred years)
Assumption: For every year there is constant salt deposit in the ocean. If not, this argument falls apart.
B
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15 Aug 2008, 18:24
Only B & E makes sense
B : only if the salt content carried by the river is constant, will the calculation of salt content in the ocean be possible.
E : this seems to more directly implied than the previous one.
Hence I would go with E
Whats the OA??
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15 Aug 2008, 18:37
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priyankur_saha@ml.com wrote:
IMO E.
* The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
Deposit could be large but that never hampers the result
If the quantities of dissolved salts deposited by rivers in the Earth’s oceans have been unusually large during the past hundred years , it will surely affect the result.
Consider the salts deposited in the past century is X.
The current salt level is say 5X
so as per the result ocean's age is 5 centuries.
But if X is unusually large and the deposited salt level of previous centuries were say only X/4 per century then the result will have a blunder!
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15 Aug 2008, 21:05
alpha_plus_gamma wrote:
priyankur_saha@ml.com wrote:
IMO E.
* The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
Deposit could be large but that never hampers the result
If the quantities of dissolved salts deposited by rivers in the Earth’s oceans have been unusually large during the past hundred years , it will surely affect the result.
Consider the salts deposited in the past century is X.
The current salt level is say 5X
so as per the result ocean's age is 5 centuries.
But if X is unusually large and the deposited salt level of previous centuries were say only X/4 per century then the result will have a blunder!
I'm confused between A and E, but I think A is the best. Partly I agree with alpha_plus_gamma, partly I think E is too an extreme assumption
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15 Aug 2008, 21:14
OA is A
nice explanation by alpha
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15 Aug 2008, 21:23
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between A and E...
one of the best ways for CR assumption questions
negate the assumption and the argument should fall apart.
negate E
Some of the salts carried into the Earth’s oceans by rivers are used up by biological activity in the oceans.
OK.. but if the portion of salt consumed by biological activities has remained constant over the years ... we can still predict the age of the oceans ... the argument doesnt fall apart
negate A
The quantities of dissolved salts deposited by rivers in the Earth’s oceans have been unusually large during the past hundred years.
it means that the rate has not been constant .... we cant predict the age .... agrument falls apart ...
right Option A
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05 Oct 2010, 04:08
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+1 for OA.
Have not been large is same as constant.. i.e. over the last 100 years the the amount of salt dissolving into sea is constant
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05 Oct 2010, 06:10
alpha_plus_gamma wrote:
priyankur_saha@ml.com wrote:
IMO E.
* The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
Deposit could be large but that never hampers the result
If the quantities of dissolved salts deposited by rivers in the Earth’s oceans have been unusually large during the past hundred years , it will surely affect the result.
Consider the salts deposited in the past century is X.
The current salt level is say 5X
so as per the result ocean's age is 5 centuries.
But if X is unusually large and the deposited salt level of previous centuries were say only X/4 per century then the result will have a blunder!
the problem says "...from a hypothetical initial salt-free state..."
in every century, there could be an increase and this increase should be roughly constant, then we could calculate the result. if portion of the salt was used in other way, ie. not calculated, then the result cannot be produced.
imo, E.
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05 Oct 2010, 07:29
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hibloom wrote:
Which of the following is an assumption on which the argument depends?
(A) The quantities of dissolved salts deposited by rivers in the Earth’s oceans have not been unusually large during the past hundred years.
(B) At any given time, all the Earth’s rivers have about the same salt levels.
(C) There are salts that leach into the Earth’s oceans directly from the ocean floor.
(D) There is no method superior to that based on salt levels for estimating the maximum age of the Earth’s oceans.
(E) None of the salts carried into the Earth’s oceans by rivers are used up by biological activity in the oceans.
i was a difficult choice between a and e.
I eliminated E with the foll. reasoning.
If at the beginning of the last century,the ocean contained x kgs of salt.
If in the last 100 years , 100 kgs of ice was dropped into the ocean by the rivers.
and suppose that in the past 100 yrs 5 kgs of salt was used up by the ocean because of its biological activity.
then 95 kgs was dropped in the last century.
So the level becomes x + 95
still we can find out the age of the ocean by taking 95 kgs as the average increase in 1 century.
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05 Oct 2010, 09:02
Got it right.Good One thanks.
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05 Oct 2010, 09:09
ankitranjan wrote:
Got it right.Good One thanks.
u can add value to this forum by explaining.This forum is for give as well as take.
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05 Oct 2010, 09:11
A
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05 Oct 2010, 18:24
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I initially had a strong urge to choose E after I narrowed down A and E.
However after some time thought option E could be broken but not A. Here is why:
Option E talks about "none of the salt getting used up". By taking the resulting increase in salt levels in the oceans over the past hundred years we could accommodate the the condition wherein some of the salt is being used up.
Hence option E is not the safe assumption to make. Hence A is the answer.
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05 Oct 2010, 19:10
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Has to be A as biological activity will happen every year.
So if rivers bring salt S and biological consumption is s then for a year the Salt increase would be S-s.
So even if we look at a long period every year s amount of salt gets consumed by biological consumption.
That way we can calculate the no of years based on X/(S-s) where X is total increase in the Salt content. So biological activity should not be a problem.
Thanks
Arun
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05 Oct 2010, 21:30
C and D are out .
From A, B and E
I chose E.
For A-it is not about large quantities or small quantities.
For B-We are concerned about the total salt content of all rivers to be same not independent rivers.
For E-it is a perfect answer.
Furthermore if you negate E.conclusion weakens that we may not get the true age.
Please let me know if i am correct
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Re: CR Salt content [#permalink] 05 Oct 2010, 21:30
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5 The Earth s rivers constantly carry dissolved salts into its 28 19 Jun 2008, 10:44
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Display posts from previous: Sort by | 3,941 | 15,036 | {"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} | 3.296875 | 3 | CC-MAIN-2017-04 | latest | en | 0.915472 |
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How can I analyze a cryptographic algorithm performance to make a table and charts comparing them. I have seen several papers but there's no paper explaining how it is exactly done.
As comparing data block size through time, etc...
Like key size, block size, power, throughtput, data length, processed bytes (MiB/s) and cycles per byte.
Example papers:
• Run under the same conditions? What are those papers? Also, see OpenSSL speed? Commented Dec 10, 2020 at 18:46
• @kelalaka Yes under the same condition. I've added in the post 3 paper examples. I haven't seen OpenSSL, actually I've been using Crypto++ Commented Dec 10, 2020 at 18:50
• If you read 1. paper The results of our implementations appear in Table 2. A Commented Dec 10, 2020 at 19:07
• @kelalaka Yes I know. The thing is, I want to test myself the algorithms Commented Dec 10, 2020 at 19:14
• @ModalNest Thank you! You helped me a lot! Commented Dec 11, 2020 at 14:44
It is not given because it is assumed that you either know it anyway, if you deal with cryptographic algorithms, or you don't need it, because its complex and time consuming and only a few people in the world need it, and can properly read the result.
To test throughput (Mb/sec, same as processed data) you need a fast device to write to, usually hdd itself adds too much delay to measure this correctly. You could use ram filesystem for it, and write there. You didn't mention the OS, so I will assume linux. You write a big file and measure the time it took
Knowing how big the file you've made and how much time it have taken, you can calculate the throughput, dividing file size by time taken. You could use writing to /dev/null, but its hard to predict how much optimization OS will do, so I would offer to avoid this.
Key size is given in algorithm initialization, you cant start without giving this information. If Crypto++ hides this information from you, try to search for something else. Or look for default values. For AES its 16 bytes.
Same for block size. Same link shows that block size is 16 bytes for AES too. As with the previous case you set it up yourself, and many algorithms accept different versions of key size and block size. Those are given at the start by the user. Or are defaults, that can be found about the particular implementation, algorithm + program that make it work.
Power is hard to estimate because efficiency of devices differ dramatically. Even if you measure it for your device, it will be different for another device. Simplest way is to write a large file in RAM, delete it, write again, in a loop for hours till your laptop battery drains out. Then check how big your battery was.
Or try checking the power consumption of your flat on a power meter, but this is even less precise. Proper method would use something like this
You could get a bit more precise result if you compare idle operating computer with computer that is doing your task. This way you automatically exclude all the parts that are not working for the cryptographic task, like a screen.
The cycles per byte is the most complex part. For this you actually need to read the code and see how many operations are done from start to finish, and then divide this amount of operations by the size of an output. If an algorithm took 17 operations like addition, xor and multiplication, and a device can do all of them in in one clock time, and output is 4 byte long, then the result is 17/4=4.25 cycles per byte. Keep in mind that modern CPU can operate on a very large words, 128bit (16byte), and that some operations can take a very long time, like exponentiation, that can take a 100 operations. You need to find the cost of each operation on a given machine. Usually it is 1 for addition, bit operations, multiplication, about 10 for division, sqrt, 100 for more complex operations like exponentiation, trigonometric functions for a modern CPU. Microcontrollers may only have fast addition. GPU may have fast division, but not bit operations. By fast I mean one cycle operation. So that in one clock step it will make one action like addition.
You may also be interested in checking how strong an algorithm is. This is done by reducing the key size and using a program that analyses the patterns in the output, TestU01 for example. Here is my favorite work on this topic, I like it for simplicity and visualisations mostly:
And this website in general. It allows to see how the cryptographic people think in general, what tools they use, what they value in an algorithm.
TLDR: you are asking about a lifetime worth of time spent collecting tools needed for a very complex task. I cant give you all of them. And there is no ready solution. Very few people need this. | 1,076 | 4,797 | {"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} | 2.6875 | 3 | CC-MAIN-2024-33 | latest | en | 0.939235 |
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LED strip lights do not cost a lot of electricity compared to traditional incandescent lights. Consumption is directly determined by the length of the strip light and its light density. A standard 5-meter strip will cost less than \$3 a year to run, on average.
## Why is my electric bill so high?
One of the main reasons your electric bill may be high is that you leave your appliances or electronics plugged in whether you’re using them or not. … The problem is, these devices are sitting idle, sucking electricity out of your home while waiting for a command from you, or waiting for a scheduled task to run. | 1,144 | 5,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-49 | latest | en | 0.926755 |
http://thebiglead.com/2014/04/15/your-odds-at-winning-100k-from-the-rams-less-than-1-in-100-trillion/ | 1,513,415,477,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948587496.62/warc/CC-MAIN-20171216084601-20171216110601-00180.warc.gz | 285,417,232 | 25,626 | # Your Odds at Winning \$100K from the Rams? Less Than 1 in 100 Trillion.
As you may have heard by now, the St. Louis Rams announced yesterday that they are offering \$100,000 to anybody who correctly forecasts their schedule for next year. We already know who their scheduled opponents are, but in order to win the bounty you need to properly prognosticate who and where they will play each week, the date of their bye, and which day each of their games falls on.
What are the odds of that?
Not bueno.
Noting that there are only byes from Week 4 through Week 12, if we work backwards, the chance of getting Week 17 right is 1/6 — we know they’ll play a division opponent. Then, the chance of getting Week 16 right, after assuming we’ve gotten Week 17 correct, is 1 in 15.
Keep replicating that methodology all the way back and we’re looking at about a .000000000000014% chance ( or 1.4 x 10^-14) at just getting the weeks of the games right. Putting that in word form, that’s about 1.4 in 100 Trillion chance, or a 1 in 71 Trillion chance of nailing all matchups, without guessing when the Rams will play on Monday and Thursday.
(Note for those that care: The odds first calculated the chances of getting weeks 17 to 13 right, starting with division games in week 17, then other non-bye weeks. That’s why it skips to weeks 3 to 1, before returning to the bye week possibilities, in the denominator).
We also have to account for whether games are scheduled on Thursdays or Mondays occasionally as well. Yeah, most games will be on Sunday, but the odds of getting all those right and nailing the primetime games on other days, too, brings us to worse than 1 in 10 quadrillion.
For comparison’s sake, take the odds of Powerball, where the odds of winning Wednesday’s \$110 million jackpot are 1 in 175,223,510 (the Rams contest is more than 50 million times tougher once you have to also guess the day of each game). Sure, there’s a \$2 cost, but would you rather part with that sum of money, or provide the Rams with your contact information?
So, this is pretty similar to the Warren Buffett NCAA Tournament bracket, where you’ll later get spammed in return for little more than an illusion of a payday. One major difference, however, is that NFL scheduling does not come down to the bounce of a ball after the contest closes; whereas no one really knows for sure who is going to win any given matchup, there is a group of people who have knowledge of the NFL schedule before it is released.
If somebody wins this Rams contest, with massive odds that make Powerball look like a safe investment, it would almost certainly be the result of a leak out of the league office or a hack by someone now motivated by a six figure payout. All told, that would be pretty hilarious.
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The story continues.
More NFL | 924 | 3,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-51 | latest | en | 0.933066 |
http://www.softwareandfinance.com/Turbo_C/Point_Inside_Circle.html | 1,696,440,985,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00796.warc.gz | 80,065,456 | 5,144 | # Turbo C - Check whether a point is inside or outside the Circle
Here is the Turbo C program to check whether a given point is inside or outside the Circle.
The distance between the center of the circle and the given point is calculated first. If the distance is less than or equal to the radius, then the point is inside the circle.
## Source Code
#include <stdio.h>
#include <math.h>
void main()
{
int nCountIntersections = 0;
float x, y, cx, cy, radius;
float distance;
printf("Program to find the given point inside or outside the circle:\n");
printf("Enter Center Point - X: ");
scanf("%f", &cx);
printf("Enter Center Point - Y: ");
scanf("%f", &cy);
printf("Enter Point - X: ");
scanf("%f", &x);
printf("Enter Point - Y: ");
scanf("%f", &y);
distance = sqrt( (double)(cx-x)*(cx-x) + (cy-y)*(cy-y));
printf("\nDistance between the point and center of the circle: %.4f", distance);
printf("\nGiven point is inside the circle");
else
printf("\nGiven point is outside the circle");
}
## Output
Program to find the given point inside or outside the circle:
Enter Center Point - X: 10
Enter Center Point - Y: 10
Enter Point - X: 12
Enter Point - Y: 4
Distance between the point and center of the circle: 6.32456
Given point is inside the circle | 324 | 1,278 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.712222 |
http://www.eecs.wsu.edu/~holder/courses/cse5311/lectures/l4/node26.html | 1,513,559,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948599549.81/warc/CC-MAIN-20171218005540-20171218031540-00586.warc.gz | 349,909,124 | 1,814 | Up: l4 Previous: Select2
# Analysis
T(n) + T(7n/10 + 6) + O(n)
To get T(7n/10 + 6), note that the smallest partition size would be
7n/10 + 6 denotes number of elements in larger partition
Assume T(n) cn for and 7n/10+6
T(n) = O(n)
T(n) c(n/5) + c(7n/10 + 6) + O(n)
= cn/5 + c7n/10 + c6 + O(n)
= 1/10(2cn + 7cn + 60c + O(n)) cn
2cn + 7cn + 60c + O(n) 10cn
60c + O(n) cn
c(n - 60) O(n)
c O(n) / (n - 60)
c is a valid constant for large enough n.
Stopping condition: T(n) = , if n 80
Up: l4 Previous: Select2 | 229 | 509 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-51 | latest | en | 0.594192 |
http://www.englishclub.com/esl-exams/ets-toeic-practice-3.htm | 1,448,741,805,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453805.6/warc/CC-MAIN-20151124205413-00277-ip-10-71-132-137.ec2.internal.warc.gz | 408,678,451 | 5,408 | # Part III: Short Conversations
In this part of the test you will listen to a short conversation between a man and a woman. After the conversation, you will answer three questions about the dialogue. There will be four possible answers for each question. Typical questions include, who, what, where, when, why, and how. You may also be asked to make an inference.
## Example 1:
First you will hear a short conversation:
1. What are the man and woman mainly discussing?
A) A vacation
B) A budget
C) A company policy
D) A conference
2. How is the woman traveling?
A) By plane
B) By bus
C) By taxi
D) By car
3. Why aren't the man and woman going together?
A) The woman needs to arrive earlier.
B) The man has to work overtime.
C) The woman dislikes air travel.
D) The man has to go to the bank first.
Explanation 1:
• Choice A is mentioned, but the man is asking if she needs to take a "vacation day". In Part III there is often one or two choices that are mentioned but are not correct.
• Choice B is related to saving money for the company, but this is not the main topic of the conversation. Be careful with main subject questions, because incorrect choices may be small details from the conversation.
• Choice C repeats the word "company," but no policy is mentioned. In Part III there is often one choice that includes a word from the conversation. You may have heard the word, but it is not the correct choice.
Explanation 2:
• Choice A is how the man is getting to the conference. In Part III there is often one or two choices that are mentioned but are not correct.
• Choice C is how the man is getting to the airport. "Taxi" is mentioned but is not correct.
• Choice D is not mentioned. In Part III there is often one choice that is not mentioned at all.
Explanation 3:
• Choice A confuses the idea of "leaving earlier" and "arriving earlier".
• Choice B repeats the word "overtime", but it was the woman who did overtime last week.
• Choice D uses the homonym bank, but in this conversation the term "bank" means to store up for later use, not a financial institution. In Part III there are often homonyms as distractors.
Transcript:
Man: Do you want to share a taxi to the airport? We can save on expenses that way, and as you know the company is trying to cut costs.
Woman: Actually I'm not flying. I'm going to the conference by bus. I have to leave tomorrow because it's going to take two days to get there.
Man: That's right. I forgot that you are afraid of flying. Are you taking a vacation day tomorrow?
Woman: Well, I worked some overtime last week, so I just banked it instead of wasting a holiday day.
## Example 2:
4. What does the man have to do today?
A) Visit his lawyer
B) Get a massage
C) Go to the doctor
D) Make an appointment
5. What can be inferred from the conversation?
A) The woman is the man's receptionist.
B) The lawyer works in the same building.
C) The woman has no deadlines today.
D)The man and woman have a meeting this afternoon.
6. What does the woman offer to do for the man?
B) Call his lawyer
C) Pick up the newspaper
D) Take notes at the meeting
Explanation 4:
• Choice A is who the man is expecting a call from. He needs to go there next week.
• Choice B confuses the sound "message" with "massage". Similar sound is a common distractor in Part III.
• Choice D is not correct because the man has already made the appointment. If the question was in the past, then it could be correct. Tense is a common distractor used in Part III.
Explanation 5:
• Choice A is incorrect because the woman doesn't normally answer the man's calls.
• Choice B is incorrect because the man mentions that he will have to go to a different location (downtown) to sign papers in front of his lawyer.
• Choice D is not correct because there is no mention of this. The man will be at the doctor's office.
Explanation 6:
• Choice B is the opposite of what the woman offers to do. She offers to take the man's calls, not make them.
• Choice C repeats the word "paper," but the conversation does not mention newspaper. Similar sound is a common distractor in Part III.
• Choice D confuses the idea of taking notes at a meeting and taking a message from a phone call.
Transcript:
Man: I have a doctor's appointment this afternoon. Are you going to be in the office, or do you have a meeting?
Woman: I'll be here. And, don't worry. I don't have much on for today, so I'll handle all of your calls.
Man: Thanks. I'm expecting a call from my lawyer. He's supposed to be sending me some changes to the contracts.
Woman: I'll make sure to take a detailed message if he calls. Is there anything you want to tell him?
Man: Well, you could remind him that I'm going to need to come downtown and sign a few papers in front of him. I'll have to set something up for next week.
Part IV > | 1,146 | 4,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-48 | latest | en | 0.95643 |
http://www.jiskha.com/display.cgi?id=1406826216 | 1,498,201,542,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320023.23/warc/CC-MAIN-20170623063716-20170623083716-00636.warc.gz | 561,212,468 | 3,731 | # maths
posted by on .
Zama had 140 packet of chips and each packet hada mass of 50g what is the mass of all packet of chips together ? Write your answer in kg.find the answer using cross multiply
• maths - ,
(140 * 50)/1000 = | 62 | 230 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.920675 |
http://www.loopandbreak.com/bit-operators-integer-only/ | 1,606,250,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177566.10/warc/CC-MAIN-20201124195123-20201124225123-00350.warc.gz | 131,387,031 | 13,343 | ### Bit Operators (Integer-only)
Python integers may be manipulated bitwise and the standard bit operations are supported: inversion, bitwise AND, OR, and exclusive OR (a.k.a. XOR), and left and right shifting. Here are some facts regarding the bit operators:
Negative numbers are treated as their 2’s complement value.
Left and right shifts of N bits are equivalent to multiplication and division by (2 ** N) without overflow checking.
For long integers, the bit operators use a “modified” form of 2’s complement, acting as if the sign bit were extended infinitely to the left.
The bit inversion operator ( ~ ) has the same precedence as the arithmetic unary operators, the highest of all bit operators. The bit shift operators ( << and >> ) come next, having a precedence one level below that of the standard plus and minus operators, and finally we have the bitwise AND, XOR, and OR operators (&, ^, | ), respectively. All of the bitwise operators are presented in the order of descending priority in following Table.
#### Integer Type Bitwise Operators
Bitwise Operator Function
~ num (unary) invert the bits of num, yielding -( num + 1)
num1 << num2 expr1 left shifted by expr2 bits
num1 >> num2 expr1 right shifted by expr2 bits
num1 & num2 expr1 bitwise AND with expr2
num1 ^ num2 expr1 bitwise XOR (exclusive OR) with expr2
num1 | num2 expr1 bitwise OR with expr2
We will now present some examples using the bit operators using 30 (011110), 45 (101101), and 60 (111100):
>>> 30 & 45
12
>>> 30 | 45
63
>>> 45 & 60
44
>>> 45 | 60
61
>>> ~30
-31
>>> ~45
-46
>>> 45 << 1
90
>>> 60 >> 2
15
>>> 30 ^ 45
51 | 418 | 1,616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2020-50 | longest | en | 0.865195 |
http://converter.ninja/length/yards-to-centimeters/434-yd-to-cm/ | 1,717,071,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00516.warc.gz | 7,624,967 | 5,367 | # 434 yards in centimeters
## Conversion
434 yards is equivalent to 39684.96 centimeters.[1]
## Conversion formula How to convert 434 yards to centimeters?
We know (by definition) that: $1\mathrm{yd}=91.44\mathrm{cm}$
We can set up a proportion to solve for the number of centimeters.
$1 yd 434 yd = 91.44 cm x cm$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{cm}=\frac{434\mathrm{yd}}{1\mathrm{yd}}*91.44\mathrm{cm}\to x\mathrm{cm}=39684.96\mathrm{cm}$
Conclusion: $434 yd = 39684.96 cm$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 centimeter is equal to 2.51984630953389e-05 times 434 yards.
It can also be expressed as: 434 yards is equal to $\frac{1}{\mathrm{2.51984630953389e-05}}$ centimeters.
## Approximation
An approximate numerical result would be: four hundred and thirty-four yards is about zero centimeters, or alternatively, a centimeter is about zero times four hundred and thirty-four yards.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 338 | 1,184 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.817737 |
http://vumultan.com/MTH202.aspx | 1,555,600,488,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517682.16/warc/CC-MAIN-20190418141430-20190418163430-00246.warc.gz | 185,218,201 | 8,026 | Mathematics MTH001 Elementary Mathematics MTH100 General Mathematics MTH101 Calculus And Analytical Geometry MTH201 Multivariable Calculus MTH202 Discrete Mathematics MTH301 Calculus II MTH302 Business Mathematics & Statistics MTH303 Mathematical Methods MTH401 Differential Equations MTH501 Linear Algebra MTH601 Operations Research MTH603 Numerical Analysis
MTH202 - Discrete Mathematics Course Page Mcqs Q & A Video Downloads
Course Category: Mathematics Course Level: Graduate Credit Hours: 3 Pre-requisites: N/A
# Course Synopsis
The course uses the patterns found in Sets, Relations and Functions. Mathematical ideas and concepts relevant to Logic, Combinatorics and Probability will be studied in this course. Matrices will be used to solve problems. The course also include: Mathematical Induction, Algorithm and Graph theory.
# Course Learning Outcomes
At the end of the course, you should be able to:
• Express statements with the precision of formal logic
• Analyze arguments to test their validity
• Apply the basic properties and operations related to sets
• Apply to sets the basic properties and operations related to relations and functions
• Define terms recursively
• To prove a given statement using mathematical induction
• To prove statements using direct and indirect methods
• To compute probability of simple and conditional events
• Identify and use formulae of Combinatorics in different problems
• Illustrate the basic definitions of graph theory and properties of graphs
# Course Contents
Logic, Sets & Operations on sets, Relations & Their Properties, Functions, Sequences & Series, Recurrence Relations, Mathematical Induction, Loop Invariants, Loop Invariants, Combinatorics, Probability, Graphs and Trees. | 355 | 1,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2019-18 | longest | en | 0.777977 |
http://www.physicsforums.com/showthread.php?t=623526 | 1,368,963,838,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368697442043/warc/CC-MAIN-20130516094402-00066-ip-10-60-113-184.ec2.internal.warc.gz | 630,016,276 | 10,107 | ## Stokes Theorem paraboloid intersecting with cylinder
1. The problem statement, all variables and given/known data
Use stokes theorem to elaluate to integral $\int\int_{s} curlF.dS$ where $F(x,y,z)= x^2 z^2 i + y^2 z^2 j + xyz k$ and s is the part of the paraboliod $z=x^2+ y^2$ that lies inside the cylinder $x^2 +y^2 =4$ and is orientated upwards
2. Relevant equations
3. The attempt at a solution
so i use Stokes theorem $\int\int_{s} curlF.dS = \oint_{c} F.dv$
so i want to get parametric equations for C and so i get x=2cost y=2sint and z=4 i came up with these as the boundary curve c is a circle of raadius 2 and my z value of 4 because that is where the parabaloid and cylinder intersect...am i right in my thinking here?
so then dx= -2sintdt ; dy=2costdt and dz=0
so then i get $\oint_{c} F.dv = \oint_{c} x^2 z^2 dx - y^2z^2 dy + xyzdz$
which becomes $\int^{2\pi}_{0} ((4cos^2 t )(16)(-2sint) - (4sin^2t)(16)(ccost)) dt$
$\int^{2\pi}_{0} ((-128cos^2 t )(sint) - (128sin^2 t)(cost)) dt$
am i working on the right lines here?
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
whoops hold up a sec...not finished putting up my attempt, im editing the post!
the origional post is complete now. anyone got any ideas?
## Stokes Theorem paraboloid intersecting with cylinder
anyone?
Recognitions: Gold Member I think the last part is wrong and it should be: ##\int^{2\pi}_{0} (-128cos^2 t sint + 128sin^2 tcost) .dt##
Recognitions:
Gold Member
Homework Help
Quote by gtfitzpatrick 1. The problem statement, all variables and given/known data Use stokes theorem to elaluate to integral $\int\int_{s} curlF.dS$ where $F(x,y,z)= x^2 z^2 i + y^2 z^2 j + xyz k$ and s is the part of the paraboliod $z=x^2+ y^2$ that lies inside the cylinder $x^2 +y^2 =4$ and is orientated upwards 2. Relevant equations 3. The attempt at a solution so i use Stokes theorem $\int\int_{s} curlF.dS = \oint_{c} F.dv$ so i want to get parametric equations for C and so i get x=2cost y=2sint and z=4 i came up with these as the boundary curve c is a circle of raadius 2 and my z value of 4 because that is where the parabaloid and cylinder intersect...am i right in my thinking here? so then dx= -2sintdt ; dy=2costdt and dz=0 so then i get $\oint_{c} F.dv = \oint_{c} x^2 z^2 dx - y^2z^2 dy + xyzdz$
Check that - sign.
which becomes $\int^{2\pi}_{0} ((4cos^2 t )(16)(-2sint) - (4sin^2t)(16)(ccost)) dt$ $\int^{2\pi}_{0} ((-128cos^2 t )(sint) - (128sin^2 t)(cost)) dt$ am i working on the right lines here?
Looks good except for that minus sign.
Recognitions: Gold Member The integration can be tricky but there is a simple trick to it. Let us know how it goes. However, i have a question myself, since i'm still learning. I suppose it is OK for me to ask here? What if the orientation of the surface S was downwards? Then, how would the calculations from post #1 change?
Hi LCkurtz and Sharks Thanks a million for comments. To integrate i used substitution u=cost du=-sintdt and similar for sin. as in $\int cos^2 t sint =\frac{1}{3} cos^3 t$ i think this works so i get $\frac{128}{3} (cos^3 t + sin^3 t)^{2\pi}_{0}$ =-$\frac{128}{3}$ i hope :)
Quote by sharks However, i have a question myself, since i'm still learning. I suppose it is OK for me to ask here? What if the orientation of the surface S was downwards? Then, how would the calculations from post #1 change?
i'm afraid i havent a clue, sorry!
Recognitions: Gold Member But isn't the whole integral this: ##\int^{2\pi}_{0} (-128cos^2 t sint + 128sin^2 tcost) .dt## Then, it would become: ##\int^{2\pi}_{0} (-128cos^2 t sint).dt + \int^{2\pi}_{0} (128sin^2 tcost).dt##
Recognitions:
Gold Member
Homework Help
Quote by sharks The integration can be tricky but there is a simple trick to it. Let us know how it goes. However, i have a question myself, since i'm still learning. I suppose it is OK for me to ask here? What if the orientation of the surface S was downwards? Then, how would the calculations from post #1 change?
I was tempted in my original post to ask the OP if he had thought about the orientation or whether he just took a guess. You use the right hand rule. In this case the normal was pointed upwards, so the right hand rule would give you counterclockwise in the xy plane as viewed from above which gives ##t:\, 0\rightarrow 2\pi##. For downward orientation you go the other way around the circle which could be ##t:\, 2\pi\rightarrow 0##. The effect is to change the sign of the answer.
Recognitions:
Gold Member
Quote by LCKurtz I was tempted in my original post to ask the OP if he had thought about the orientation or whether he just took a guess. You use the right hand rule. In this case the normal was pointed upwards, so the right hand rule would give you counterclockwise in the xy plane as viewed from above which gives ##t:\, 0\rightarrow 2\pi##. For downward orientation you go the other way around the circle which could be ##t:\, 2\pi\rightarrow 0##. The effect is to change the sign of the answer.
Thank you for this quick and accurate answer, LCKurtz.
Quote by sharks Then, it would become: ##\int^{2\pi}_{0} (-128cos^2 t sint).dt + \int^{2\pi}_{0} (128sin^2 tcost).dt##
$\int^{2\pi}_{0} (-128cos^2 t sint)dt + \int^{2\pi}_{0} (128sin^2 t cost)dt$
=128[$\frac{1}{3} cos^3 t ]^{2\pi}_{0} + 128[\frac{1}{3} sin^3 t ]^{2\pi}_{0}$
=$\frac{128}{3}[-1] = \frac{-128}{3}$
Quote by LCKurtz I was tempted in my original post to ask the OP if he had thought about the orientation or whether he just took a guess. You use the right hand rule. In this case the normal was pointed upwards, so the right hand rule would give you counterclockwise in the xy plane as viewed from above which gives ##t:\, 0\rightarrow 2\pi##. For downward orientation you go the other way around the circle which could be ##t:\, 2\pi\rightarrow 0##. The effect is to change the sign of the answer.
ahhh thanks a million i had wondered about that alright. that cleared it up. thanks a million.
Recognitions:
Gold Member
Quote by gtfitzpatrick $\int^{2\pi}_{0} (-128cos^2 t sint)dt + \int^{2\pi}_{0} (128sin^2 t cost)dt$ =128[$\frac{1}{3} cos^3 t ]^{2\pi}_{0} + 128[\frac{1}{3} sin^3 t ]^{2\pi}_{0}$ =$\frac{128}{3}[-1] = \frac{-128}{3}$
I don't think the evaluation part is correct.
Quote by gtfitzpatrick =$\frac{128}{3}[-1] = \frac{-128}{3}$
whoops should be $\frac{128}{3}[1-1] = 0$
Thanks a million sharks! your brill! | 2,005 | 6,511 | {"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} | 3.484375 | 3 | CC-MAIN-2013-20 | latest | en | 0.809854 |
https://studylib.net/doc/14894317/blc-190-name--worksheet-2-1 | 1,701,577,305,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00208.warc.gz | 607,111,135 | 12,420 | # BLC 190 Name: Worksheet 2 1
```BLC 190
Name:
Worksheet 2
1
1. Suppose that f (x) = x2 + 3x and g(x) = . Evaluate each of the following, simplifying
x
(a) f (a + b)
(b) g
(a)
b
(c)
f (a + h) − f (a)
h
(d)
g(a + h) − g(a)
h
2. Sketch the graph of y = (x − 1)2 − 4
&
%
\$
#
"
& % \$ # "
"
"
#
\$
%
&
"
#
\$
%
&
#
\$
%
&
3. Sketch the graph of y =
√
x+3−2
&
%
\$
#
"
& % \$ # "
"
#
\$
%
&
2
4. Find the equations for the lines through the given points.
(a) (−1, −3) and (1, 5)
(b) (−5, 8) and (1, −4)
(c) (1, 5) and (3, 5)
(d) (−1, 3) and (−1, 8)
3
5. Find the equation for the line with slope m = 3 through the point (−1, 5).
6. Suppose that f is a linear function and that f (2) = 3 and f (6) = 5.
(a) Find f (10).
(b) Find a so that f (a) = 20.
4
7. Consider the following function:
&
%
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#
"
& % \$ # "
"
"
#
\$
%
&
#
\$
%
Find an expression for this function.
8. At age 3, Jane is 36 inches tall. Starting at age 3, she grows 2 inches a year. How tall is
Jane at age n?
5
9. An oceanographer is taking undersea temperature readings using a thermistor temperature sensor attached to a Niskin bottle. At a depth of 300 meters, she measures a
temperature of 18◦ C. At a depth of 500 meters, she measures a temperature of 13◦ C.
(a) Assume that the temperature depends linearly on the the depth. Find a formula for
the temperature T at a depth of x meters.
(b) Use your answer to part (a) to estimate the temperature at a depth of 600 meters.
(c) Estimate the depth at which the water temperature is 16◦ C.
6
10. In 2005, there were 17.3 million students enrolled in college, and that number was increasing at a rate of 433,000 per year.
(a) Write an equation expressing the relationship between the number of college students
and the year.
(b) Estimate the number of college students in 2015.
7
``` | 711 | 1,933 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-50 | latest | en | 0.853461 |
https://www.lessonplanet.com/search?concept_ids%5B%5D=29885 | 1,718,313,653,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00514.warc.gz | 761,105,749 | 30,854 | Lesson Plan
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Your learners use many different number representations like positive and negative integers, exponents, radicals, and combinations of operations as they learn and strengthen their use of the order of operations. Thoughtfully organized to... | 1,703 | 7,903 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-26 | latest | en | 0.896092 |
https://www.studyadda.com/question-bank/factors-multiples_q39/4556/361147 | 1,576,384,952,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541301598.62/warc/CC-MAIN-20191215042926-20191215070926-00208.warc.gz | 871,953,459 | 19,868 | • question_answer A number N = 897324A64B is divided by both 8 and 9. Which of the following is the value of (A+B)? (i) 2 (ii) 11 (iii) 9 A) Either (i) or (ii)B) Either (ii) or (iii)C) Either (i) or (ii) or (iii) D) None of these
N = 897324A64B For N divisible by 8, last three digits should be divisible by 8. For N divisible by 9, sum of digits should be divisible by 9. However, 64B is divisible by 8 when B equals 0 and 8. Now, if $B=0,$then A should be 2. And if $B=8,$then A should be 3. Then, $(A+B)=2$and 11. | 203 | 553 | {"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} | 3.171875 | 3 | CC-MAIN-2019-51 | latest | en | 0.826426 |
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# 第三章 电解质溶液和离子平衡 - PowerPoint PPT Presentation
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### 第三章电解质溶液和离子平衡
3.1 一元弱酸、弱碱的离解平衡
• 3.1.1 离解常数
• 以一元弱酸HA为例，其离解平衡式及其标准离解常数为：
• HA H+ + A-
KaΘ = {c(H+)/ cΘ}·{c(A-)/ cΘ}
c(HA)/cΘ = c‘(H+) ·c’(A-)c‘(HA)若以BOH表示弱碱，则：BOH B+ + OH-KbΘ = {c(B+)/ cΘ}·{c(OH-)/ cΘ}c(BOH)/cΘ = c‘(B+) ·c’(OH-)c‘(BOH)KaΘ可用来表示酸的强弱.KaΘ越大，该弱酸的酸性越强，HA Ka HCOOH 1.77×10-4 HAC 1.75×10-5 HClO 2.80×10-8 HCN 6.20×10-10
3.1.2 离解度对于弱电解质来说,除了用离解常数表示电解质的强弱外,还可用离解度(α)来表示其离解的程度:α= 已离解的弱电解质浓度 ×100% 弱电解质的起始浓度 以一元弱酸HA为例,其离解度α, 离解常数KaΘ和浓度c之间的关系推导如下: HA H+ + A-
KaΘ = c‘(H+) ·c’(A-) = c‘α ·c’ α =c‘α 2c‘(HA) c‘(1-α) 1-α
1. 精确公式:也就是说, c‘α 2 + KaΘα - KaΘ = 0,
α =-KaΘ + √ (KaΘ)2 + 4c’KaΘ
2c’
c(H+) = cα = c·-KaΘ + √(KaΘ)2 + 4c’KaΘ
2c’
= －KaΘ + √(KaΘ)2 + 4c’KaΘ ·cΘ
2
2. 近似公式:
α = √KaΘ/c’c'(H+) =√KaΘ·c’
α = √KbΘ/c’c’(OH-) = √KbΘ·c’
3.1.3 一元弱酸、弱碱溶液中离子浓度及pH的计算
(2)如将此溶液稀释至0.010 mol·L-1, 求此时溶液的H+浓度、pH及离解度。
HA H+ + Ac-
KaΘ = c‘(H+) ·c’(Ac-) = x·x
c‘(HAc) 0.10 - x
∵c/ KaΘ = 0.10÷(1.75×10-5) >500, ∴可用近似公式
c'(H+) = √KaΘ·c’
X =c’(H+)= c’(Ac-)=√KaΘ·c’=√1.75×10-5×0.10
=1.3×10-3
pH = -lgc’(H+) =-lg 1.3×10-3 = 2.88
α = (1.3×10-3/0.10) ×100% = 1.3%
(2)∵c/ KaΘ = 0.010÷(1.75×10-5) >500,
∴仍可用近似公式
x = c’(H+) = √KaΘ·c’ =√1.75×10-5×0.010
= 4.2×10-4
pH = -lgc’(H+) = -lg 4.2×10-4 = 3.38
α = (4.2×10-4/0.010) ×100% = 4.2%
c(OH-) = 6.0×10-4 mol·L-1
NH3+ H2O NH4+ + OH-
KbΘ（NH3）= c'(NH4+) ·c’(OH-) = (6.0×10-4)2 = 1.8 ×10-5
c'(NH3) 0.020
α = c’(OH-) ×100% = 6.0×10-4×100% = 3.0%
c'(NH3) 0.020
3.2 多元弱酸的离解平衡
1．碳酸分两步离解：
Ka1Θ（H2CO3）= c'(H+) ·c’(HCO3-) = 4.4 ×10-7
c'(H2CO3)
Ka2Θ（H2CO3）= c'(H+) ·c’(CO32-) = 4.7 ×10-11
c'(HCO3-)
2．磷酸分三步离解：
Ka1Θ（H3PO4）= c'(H+) ·c’(H2PO4-) = 7.1 ×10-3
c'(H3PO4)
Ka2Θ（H3PO4）= c'(H+) ·c’(HPO42-) = 6.3 ×10-8
c'(H2PO4-)
Ka3Θ（H3PO4）= c'(H+) ·c’(PO43-) = 4.2 ×10-13
c'(HPO42-)
3．对于多元弱酸的分步离解，因为Ka1Θ >> Ka2Θ, 其氢离子浓度主要来自第一级电离,当求氢离子浓度时可当作一元弱酸来处理.
c’(H+)= c’(HCO3-) = √KaΘ·c’ =√4.4×10-7×0.040
= 1.3×10-4
Ka2Θ（H2CO3）=c'(H+) ·c’(CO32-)
c'(HCO3-)
=c’(CO32-) = 4.7 ×10-11 | 1,612 | 2,632 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-30 | latest | en | 0.33535 |
https://communities.sas.com/t5/SAS-Procedures/Using-the-FCMP-Procedure-to-Store-a-Formula-in-a-Function/td-p/106489?nobounce | 1,532,294,026,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00423.warc.gz | 620,387,942 | 26,986 | ## Using the FCMP Procedure to Store a Formula in a Function
Solved
Frequent Contributor
Posts: 90
# Using the FCMP Procedure to Store a Formula in a Function
Hello Everyone,
So I have the following program in my homework:
data test;
set orion.order_fact(keep=Employee_ID Quantity
Total_Retail_Price);
if Quantity>2 then Kick_Back_Amt=Quantity*Total_Retail_Price/5;
else Kick_Back_Amt=Quantity*Total_Retail_Price/10;
run;
proc print data=test(obs=5);
run;
I have to take this program and create a PROC FCMP where I have a formula stored that I can query later.
This is my program so far:
options cmplib=orion.functions;
proc fcmp outlib=work.functions.Marketing;
function kb(Quantity, Total_Retail_Price) \$50;
if Quantity>2 then return (Quantity*Total_Retail_Price/5);
else return (Quantity*Total_Retail_Price/10);
endsub;
run;
quit;
data kick_backs;
set orion.order_fact;
Kick_Back_Amt=kb(Quantity, Total_Retail_Price);
run;
proc print data=kick_backs (obs=5);
run;
However, when I try to access my formula, I get the following:
472 options cmplib=orion.functions;
473 proc fcmp outlib=work.functions.Marketing;
474 function kb(Quantity1, Total_Retail_Price) \$50;
475 if Quantity>2 then return (Quantity1*Total_Retail_Price/5);
NOTE: Numeric value converted to character.
476 else return (Quantity1*Total_Retail_Price/10);
NOTE: Numeric value converted to character.
477 endsub;
478 run;
NOTE: Function kb saved to work.functions.Marketing.
NOTE: PROCEDURE FCMP used (Total process time):
real time 0.20 seconds
cpu time 0.03 seconds
479 quit;
480
481 /*part c*/
482 data kick_backs;
483 set orion.order_fact;
484 Kick_Back_Amt=kb(Quantity, Total_Retail_Price);
484 Kick_Back_Amt=kb(Quantity, Total_Retail_Price);
--
68
ERROR 68-185: The function KB is unknown, or cannot be accessed.
485 run;
I am not sure where I am going wrong with my program, but if someone could take a look at it and let me know, it will be greatly appreciated.
Thanks!
Alisa
Accepted Solutions
Solution
05-07-2012 02:55 PM
Frequent Contributor
Posts: 90
## Re: Using the FCMP Procedure to Store a Formula in a Function
Never mind....just needed to change
options cmplib=orion.functions; to work.functions.
All Replies
Solution
05-07-2012 02:55 PM
Frequent Contributor
Posts: 90
## Re: Using the FCMP Procedure to Store a Formula in a Function
Never mind....just needed to change
options cmplib=orion.functions; to work.functions.
🔒 This topic is solved and locked. | 671 | 2,557 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-30 | latest | en | 0.649119 |
https://math.stackexchange.com/questions/2958334/proof-cauchy-schwarz-inequality | 1,563,819,610,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528208.76/warc/CC-MAIN-20190722180254-20190722202254-00507.warc.gz | 471,093,414 | 35,921 | # Proof Cauchy-Schwarz inequality
I want to proof $$|\langle x, y \rangle| \leq \|x\| \, \|y\|$$ for all $$x,y \in \mathbb{R}^n$$ or $$\mathbb{C}^n$$. I know there exist a ton of proofs for this inequality, but it want to proof it through a specific schematic.
1. We know that the length of a Vector is $$\geq 0$$, hence for an arbitrary $$t \in \mathbb{R}$$ it follows that $$0 \leq \| tx + y \|$$.
2. I think its possible to rearrange this inequality such that $$0 \leq at^2 + bt +c$$.
3. Know we can look at the discriminant $$b^2 -4ac$$ formulate a new inequality such that the Cauchy-Schwarz inequality follows.
• – Clement C. Oct 16 '18 at 19:02
Start by squaring your inequality in (1), giving $$0 \le \|tx+y\|^2 = \def\<#1>{\left<#1\right>}\ = t^2\|x\|^2 + 2t\ + \|y\|^2$$ The discriminant is therefore given by $$4\^2 - 4\|x\|^2\|y\|^2$$ As the discriminant cannot be positive (note that a non-negative quadratic real polynomial has at most one root), we have $$4\^2 - 4\|x\|^2\|y\|^2 \le 0 \iff \^2 \le \|x\|^2\|y\|^2.$$ Taking square roots gives Cauchy-Schwarz.
• Thank you! I have two questions. 1) For $\mathbb{R}$ i understand the rearrangement, but for $\mathbb{C}$ this is only sesquilinear, isn't it? Why is it also for $\mathbb{C}$ correct ? 2) Why exactly ist the discriminant not positiv? – faoeoe Oct 16 '18 at 19:36 | 467 | 1,340 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-30 | longest | en | 0.768664 |
http://support.sas.com/documentation/cdl/en/ormpug/66107/HTML/default/ormpug_lpsolver_examples06.htm | 1,558,522,989,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256797.20/warc/CC-MAIN-20190522103253-20190522125253-00045.warc.gz | 198,884,819 | 7,581 | ### Example 6.6 Migration to OPTMODEL: Generalized Networks
The following example shows how to use PROC OPTMODEL to solve the example Generalized Networks: Using the EXCESS= Option in Chapter 6: The NETFLOW Procedure in SAS/OR 12.3 User's Guide: Mathematical Programming Legacy Procedures. The input data sets are the same as in the PROC NETFLOW example.
```title 'Generalized Networks';
data garcs;
input _from_ \$ _to_ \$ _cost_ _mult_;
datalines;
s1 d1 1 .
s1 d2 8 .
s2 d1 4 2
s2 d2 2 2
s2 d3 1 2
s3 d2 5 0.5
s3 d3 4 0.5
;
```
```data gnodes;
input _node_ \$ _sd_ ;
datalines;
s1 5
s2 20
s3 10
d1 -5
d2 -10
d3 -20
;
```
The following PROC OPTMODEL statements read the data sets, build the linear programming model, solve the model, and output the optimal solution to a SAS data set called `GENETOUT`:
```proc optmodel;
set <str> NODES;
num _sd_ {NODES} init 0;
read data gnodes into NODES=[_node_] _sd_;
set <str,str> ARCS;
num _lo_ {ARCS} init 0;
num _capac_ {ARCS} init .;
num _cost_ {ARCS};
num _mult_ {ARCS} init 1;
read data garcs nomiss into ARCS=[_from_ _to_] _cost_ _mult_;
NODES = NODES union (union {<i,j> in ARCS} {i,j});
```
``` var Flow {<i,j> in ARCS} >= _lo_[i,j];
min obj = sum {<i,j> in ARCS} _cost_[i,j] * Flow[i,j];
con balance {i in NODES}: sum {<(i),j> in ARCS} Flow[i,j]
- sum {<j,(i)> in ARCS} _mult_[j,i] * Flow[j,i] = _sd_[i];
num infinity = min {r in {}} r;
/* change equality constraint to le constraint for supply nodes */
for {i in NODES: _sd_[i] > 0} balance[i].lb = -infinity;
solve;
num _supply_ {<i,j> in ARCS} = (if _sd_[i] ne 0 then _sd_[i] else .);
num _demand_ {<i,j> in ARCS} = (if _sd_[j] ne 0 then -_sd_[j] else .);
num _fcost_ {<i,j> in ARCS} = _cost_[i,j] * Flow[i,j].sol;
create data gnetout from [_from_ _to_]
_cost_ _capac_ _lo_ _mult_ _supply_ _demand_ _flow_=Flow _fcost_;
quit;
```
To solve a generalized network flow problem, the usual balance constraint is altered to include the arc multiplier _mult_[i,j] in the second sum. The balance constraint is initially declared as an equality, but to mimic the EXCESS=SUPPLY option in PROC NETFLOW, the sense of this constraint is changed to by relaxing the constraint’s lower bound for supply nodes. The output data set is displayed in Output 6.6.1.
Output 6.6.1: Optimal Solution with Excess Supply
Obs _from_ _to_ _cost_ _capac_ _lo_ _mult_ _supply_ _demand_ _flow_ _fcost_
1 s1 d1 1 . 0 1.0 5 5 5 5
2 s1 d2 8 . 0 1.0 5 10 0 0
3 s2 d1 4 . 0 2.0 20 5 0 0
4 s2 d2 2 . 0 2.0 20 10 5 10
5 s2 d3 1 . 0 2.0 20 20 10 10
6 s3 d2 5 . 0 0.5 10 10 0 0
7 s3 d3 4 . 0 0.5 10 20 0 0
The log is displayed in Output 6.6.2.
Output 6.6.2: OPTMODEL Log
NOTE: There were 6 observations read from the data set WORK.GNODES. NOTE: There were 7 observations read from the data set WORK.GARCS. NOTE: Problem generation will use 4 threads. NOTE: The problem has 7 variables (0 free, 0 fixed). NOTE: The problem has 6 linear constraints (3 LE, 3 EQ, 0 GE, 0 range). NOTE: The problem has 14 linear constraint coefficients. NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). NOTE: The OPTMODEL presolver is disabled for linear problems. NOTE: The LP presolver value AUTOMATIC is applied. NOTE: The LP presolver removed 2 variables and 2 constraints. NOTE: The LP presolver removed 4 constraint coefficients. NOTE: The presolved problem has 5 variables, 4 constraints, and 10 constraint coefficients. NOTE: The LP solver is called. NOTE: The Dual Simplex algorithm is used. Objective Phase Iteration Value Time D 1 1 0.000000E+00 0 D 2 2 1.500000E+01 0 D 2 4 2.500000E+01 0 NOTE: Optimal. NOTE: Objective = 25. NOTE: The Dual Simplex solve time is 0.05 seconds. NOTE: The data set WORK.GNETOUT has 7 observations and 10 variables.
Now consider the previous example but with a slight modification to the arc multipliers, as in the PROC NETFLOW example:
```data garcs1;
input _from_ \$ _to_ \$ _cost_ _mult_;
datalines;
s1 d1 1 0.5
s1 d2 8 0.5
s2 d1 4 .
s2 d2 2 .
s2 d3 1 .
s3 d2 5 0.5
s3 d3 4 0.5
;
```
The following PROC OPTMODEL statements are identical to the preceding example, except for the balance constraint. The balance constraint is still initially declared as an equality, but to mimic the PROC NETFLOW EXCESS=DEMAND option, the sense of this constraint is changed to by relaxing the constraint’s upper bound for demand nodes.
```proc optmodel;
set <str> NODES;
num _sd_ {NODES} init 0;
read data gnodes into NODES=[_node_] _sd_;
set <str,str> ARCS;
num _lo_ {ARCS} init 0;
num _capac_ {ARCS} init .;
num _cost_ {ARCS};
num _mult_ {ARCS} init 1;
read data garcs1 nomiss into ARCS=[_from_ _to_] _cost_ _mult_;
NODES = NODES union (union {<i,j> in ARCS} {i,j});
var Flow {<i,j> in ARCS} >= _lo_[i,j];
for {<i,j> in ARCS: _capac_[i,j] ne .} Flow[i,j].ub = _capac_[i,j];
min obj = sum {<i,j> in ARCS} _cost_[i,j] * Flow[i,j];
con balance {i in NODES}: sum {<(i),j> in ARCS} Flow[i,j]
- sum {<j,(i)> in ARCS} _mult_[j,i] * Flow[j,i] = _sd_[i];
num infinity = min {r in {}} r;
/* change equality constraint to ge constraint */
for {i in NODES: _sd_[i] < 0} balance[i].ub = infinity;
solve;
num _supply_ {<i,j> in ARCS} = (if _sd_[i] ne 0 then _sd_[i] else .);
num _demand_ {<i,j> in ARCS} = (if _sd_[j] ne 0 then -_sd_[j] else .);
num _fcost_ {<i,j> in ARCS} = _cost_[i,j] * Flow[i,j].sol;
create data gnetout1 from [_from_ _to_]
_cost_ _capac_ _lo_ _mult_ _supply_ _demand_ _flow_=Flow _fcost_;
quit;
```
The output data set is displayed in Output 6.6.3.
Output 6.6.3: Optimal Solution with Excess Demand
Obs _from_ _to_ _cost_ _capac_ _lo_ _mult_ _supply_ _demand_ _flow_ _fcost_
1 s1 d1 1 . 0 0.5 5 5 5 5
2 s1 d2 8 . 0 0.5 5 10 0 0
3 s2 d1 4 . 0 1.0 20 5 0 0
4 s2 d2 2 . 0 1.0 20 10 5 10
5 s2 d3 1 . 0 1.0 20 20 15 15
6 s3 d2 5 . 0 0.5 10 10 0 0
7 s3 d3 4 . 0 0.5 10 20 10 40
The log is displayed in Output 6.6.4.
Output 6.6.4: OPTMODEL Log
NOTE: There were 6 observations read from the data set WORK.GNODES. NOTE: There were 7 observations read from the data set WORK.GARCS1. NOTE: Problem generation will use 4 threads. NOTE: The problem has 7 variables (0 free, 0 fixed). NOTE: The problem has 6 linear constraints (0 LE, 3 EQ, 3 GE, 0 range). NOTE: The problem has 14 linear constraint coefficients. NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). NOTE: The OPTMODEL presolver is disabled for linear problems. NOTE: The LP presolver value AUTOMATIC is applied. NOTE: The LP presolver removed 2 variables and 2 constraints. NOTE: The LP presolver removed 4 constraint coefficients. NOTE: The presolved problem has 5 variables, 4 constraints, and 10 constraint coefficients. NOTE: The LP solver is called. NOTE: The Dual Simplex algorithm is used. Objective Phase Iteration Value Time D 1 1 0.000000E+00 0 D 2 2 4.997000E+01 0 D 2 5 7.000000E+01 0 NOTE: Optimal. NOTE: Objective = 70. NOTE: The Dual Simplex solve time is 0.05 seconds. NOTE: The data set WORK.GNETOUT1 has 7 observations and 10 variables. | 2,771 | 7,121 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-22 | latest | en | 0.596094 |
https://thisrunsforyou.com/qa/what-is-4-8-hours-in-hours-and-minutes.html | 1,606,835,628,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674594.59/warc/CC-MAIN-20201201135627-20201201165627-00513.warc.gz | 512,985,793 | 6,582 | # What Is 4.8 Hours In Hours And Minutes?
## What is 7.70 in hours and minutes?
7.70 hours in hours, minutes, and seconds 7.70 hours is 7 hours, 42 minutes and 0 seconds.
7.70 hours is also equivalent to 462 minutes and 0 seconds or 27720 seconds..
## How much is 4.8 hours in hours and minutes?
This conversion of 4.8 hours to minutes has been calculated by multiplying 4.8 hours by 60 and the result is 288 minutes.
## What is 3.7 hours in hours and minutes?
Three point seven hours is equal to three hours and forty-two minutes.
## Is .5 a half hour?
For example 15 minutes (¼ hour) equals . 25, 30 minutes (½ hour) equals . 5, etc.
## What is 7.75 hours in hours and minutes?
7:75 with the colon is 7 hours and 75 minutes.
## How much is 4.7 hours?
This conversion of 4.7 hours to minutes has been calculated by multiplying 4.7 hours by 60 and the result is 282 minutes.
## What does 0.75 mean in hours?
HoursMinutes0.65390.70420.75450.80486 more rows
## How many minutes is 5 hour?
Hours to Minutes Conversion TableHoursMinutes3 Hours180 Minutes4 Hours240 Minutes5 Hours300 Minutes6 Hours360 Minutes20 more rows
## What is 2.6 hours in hours and minutes?
Two point six hours is equal to two hours and thirty-six minutes.
## What is 3.75 hours in hours and minutes?
3.75 hours with the decimal point is 3.75 hours in terms of hours. 3:75 with the colon is 3 hours and 75 minutes.
## What is 4.9 hours and minutes?
This conversion of 4.9 hours to minutes has been calculated by multiplying 4.9 hours by 60 and the result is 294 minutes.
## What is 3.9 hours and minutes?
This conversion of 3.9 hours to minutes has been calculated by multiplying 3.9 hours by 60 and the result is 234 minutes.
## What is .8 of an hour?
Billing Increment Chart—Minutes to Tenths of an HourMinutesTime25-30.531-36.637-42.743-48.86 more rows
## What is 6.3 in hours and minutes?
This conversion of 6.3 hours to minutes has been calculated by multiplying 6.3 hours by 60 and the result is 378 minutes.
## What is 4.5 hours in hours and minutes?
4.5 h = 270 min….Conversion table: Hours to Minutes.HOURSMINUTES1=602=1203=1804=2405 more rows
## What is 6.75 hours in hours and minutes?
6.75 hours with the decimal point is 6.75 hours in terms of hours. 6:75 with the colon is 6 hours and 75 minutes. .
## Is 1.5 hours 1 hour and 30 minutes?
1.5 hours is therefore 1 hour and 30 minutes.
## Is 75 minutes an hour and 15 minutes?
75 minutes equals 1.25 hours. | 708 | 2,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.933558 |
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Members
7
1. ## Concerns about the geometry of the real number line
A real number line is a continuous line that extends infinitely in both directions. Each point on the line represents a real number. The real numbers on the line are arranged in sequential order. Any two real numbers on the line can be compared, and we can determine which number is greater or smaller than the other. The line can be divided into segments. Suppose we take a segment on the real number line, let us say between two integers, 1 and 2. The segment is finite but infinite in the number of real numbers it contains. There are countless real numbers between 1 and 2. However, there is no next number on this segment. By this, we mean that if you pick any number between 1 and 2, you can always find another number between them. There is no limit or endpoint to the number of real numbers between the two integers. Thus, there can never be a next number in this segment. The concept of the next number is not applicable to the real number line. The set of real numbers is complete, meaning that there is no need for any new number to fill in any gaps or provide solutions to any problems. Any two real numbers on the line are separated by an infinite number of other real numbers. This implies that there exists no next number or any number missing from the real number line. Each number on the line is unique and independent. Furthermore, the real number line is dense, which means that every point on the line can be approached arbitrarily close by a sequence of real numbers. This property further emphasizes that there is no next number on the real number line. For any point on the line, we can approach it arbitrarily close by finding real numbers that are infinitely close to the point. Since there exists no smallest positive number on the real number line, there is no next number. there is an end to the real number line segment or not? At first glance, it may seem that the real number line goes on forever without any end. This idea aligns with the concept of infinity, which is an unbounded quantity or magnitude that extends indefinitely without a limit or boundary. Mathematically, we can represent infinity by the symbol ∞, which denotes an infinitely large or small value that cannot be expressed or reached in the usual sense. In this sense, the real number line appears to be infinite both to the left and right of zero, with no end in sight. However, upon further analysis, we realize that there are limits to the real number line, albeit they may not be intuitive or straightforward. For instance, we can define bounds or intervals on the real number line that contain only a finite or countable number of real numbers. For example, we can define the interval [0,1], which contains all real numbers between zero and one, including both endpoints. In this case, the interval [0,1] is bounded, meaning it has an upper and lower limit, which are 1 and 0, respectively. Furthermore, there are situations where the real number line is incomplete or non-existent in certain spots. For instance, we can consider the imaginary or complex number system, which extends the real number line by introducing the imaginary unit i, such that i^2=-1. The complex number system includes both real and imaginary numbers, and it can be represented as a two-dimensional plane with a horizontal axis for the real part and a vertical axis for the imaginary part. However, there are points on the complex plane where the real part is zero, and only the imaginary part exists. Such points are called purely imaginary or vertical lines and are not part of the real number line. Therefore, the real number line is incomplete or does not exist in such cases. Another situation where the real number line segment is incomplete is in the case of limits or approaches to infinity. For example, we can consider the function f(x)=1/x, which approaches zero as x approaches infinity. In this case, the real number line seems to have an end or limit at zero, but we can still consider values greater than zero by taking the limit as x approaches infinity, which yields a value of zero. Therefore, the real number line segment has a limit or end but extends infinitely beyond it.
2. ## Concerns about the geometry of the real number line
The real numbers can have the" next " number? The answer to this question is a resounding no. Real numbers are already infinite, so it's impossible for them to have the "next" number - after all, they don't even know what that would be! It's like asking an endless ocean if it can contain one more drop of water; the answer will always be no! Intervals on the real number line are an important concept in math, but they don't have to be complicated! For example, if we look at a number line from 0 to 10, then (2, 8) is an interval that includes all numbers between 2 and 8 (but not including either of those two numbers). Similarly, [5.5 , 9] is another interval that contains all real numbers starting from 5.5 up until 9 - simple as can be! Intervals on the real number line are like a date night for math -- nothing too serious or committed, just a pleasant distraction from the day-to-day of dealing with numbers that just won't behave. They divide up the real number line into manageable chunks so there's no more guessing whether you should be adding, subtracting, multiplying or dividing. Plus it's got some pretty nifty consequences for graphing equations (no velocity limit here!). any more questions?
3. ## Concerns about the geometry of the real number line
Real numbers are the backbone of mathematics, and they always exist along the real number line. This line is made up of infinitely many segments that stretch from negative infinity to positive infinity. Each segment contains an infinite amount of numbers, making it impossible for us to ever run out! So no matter how much we explore math and its applications, real numbers will always be there waiting for us on the real number line. Taking the real numbers off the line doesn't mean they all disappear. They're still right there on the line. And the real numbers are, in actual fact, not the line. Yeah, they're all numbers, not geometric shape. You can imagine yourself deleting any real number on the line( or deleting anything in the universe ). But...don't let imaginations play tricks on you! No one in the universe can actually delete or remove any real number. Real numbers always exist in the mathematical realm of nature!
4. ## Philosophical Implications Of Infinite Parallel Multiverses
Parallel Multiverses = a theory, not reality No proof yet. No equations about that theory.
5. ## Consciousness Always Exists
If consciousness always exists, where is it while a patient is in coma? The patient can't experience things around him or her. He or she loses his or her consciousness. Where does consciousness go? Can you explain that?
6. ## Getting started
Philosophy or math is absolutely not a magic wand. That's why we need to study a variety of subjects in schools or colleges. Yeah, learn as if you were to live forever.
7. ## moment of inertia in round house kick (derivative or integral)
You don't need Calculus. Just basic math skills can solve the problem. You must like MuayThai so much, right? lol. Reference: How to calculate Moment of Inertia? - Formulas and Solved Examples – https://www.geeksforgeeks.org/how-to-calculate-moment-of-inertia/
× | 1,582 | 7,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2023-14 | latest | en | 0.932091 |
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# 3241 Lecture 4 - MAE 3241 AERODYNAMICS AND FLIGHT MECHANICS...
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MAE 3241: AERODYNAMICS AND FLIGHT MECHANICS Review: Bernoulli Equation and Examples Mechanical and Aerospace Engineering Department Florida Institute of Technology D. R. Kirk
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LECTURE OUTLINE Review of Euler’s Equation Euler’s equation for incompressible flow → Bernoulli’s Equation Review of Basic Aerodynamics How does an airfoil or wing generate lift? What are effects of viscosity? Why does an airfoil stall? Why are golf balls dimpled?
WHAT DOES EULER’S EQUATION TELL US? Euler’s Equation (Differential Equation) Relates changes in momentum to changes in force ( momentum equation ) Relates a change in pressure (dp) to a chance in velocity (dV) Assumptions we made: Steady flow Neglected friction (inviscid flow), body forces, and external forces dp and dV are of opposite sign IF dp increases dV decreases → flow slows down IF dp decreases dV increases → flow speeds up Valid for Incompressible and Compressible flows Valid for Irrotational and Rotational flows VdV dp ρ - =
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INVISCID FLOW ALONG STREAMLINES 0 2 2 0 0 2 1 2 2 1 2 2 1 2 1 = - + - = + = + V V p p VdV dp VdV dp V V p p ρ ρ ρ Relate p 1 and V 1 at point 1 to p 2 and V 2 at point 2 Integrate Euler’s equation from point 1 to point 2 taking ρ =constant
BERNOULLI’S EQUATION = + + = + 2 2 2 2 2 1 1 2 2 2 V p V p V p ρ ρ ρ If flow is irrotational p+½ ρ V 2 = constant everywhere Remember: Bernoulli’s equation holds only for inviscid (frictionless) and incompressible ( ρ =constant) flows Relates properties between different points along a streamline or entire flow field if irrotational For a compressible flow Euler’s equation must be used ( ρ is a variable) Both Euler’s and Bernoulli’s equations are expressions of F =m a expressed in a useful form for fluid flows and aerodynamics Constant along a streamline
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HOW DOES AN AIRFOIL GENERATE LIFT? Lift is mainly due to imbalance of pressure distribution over the top and bottom surfaces of airfoil If pressure is lower than pressure on bottom surface, lift is generated Why is pressure lower on top surface? We can understand answer from basic physics Continuity Newton’s 2 nd law
HOW DOES AN AIRFOIL GENERATE LIFT? 1. Flow velocity over the top of airfoil is faster than over bottom surface Streamtube A senses upper portion of airfoil as an obstruction Streamtube A is squashed to smaller cross-sectional area Mass continuity ρ AV=constant, velocity must increase Streamtube A is squashed most in nose region (ahead of maximum thickness) A B
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HOW DOES AN AIRFOIL GENERATE LIFT? 2. As velocity increases pressure decreases Incompressible: Bernoulli’s Equation Compressible: Euler’s Equation Called Bernoulli Effect 2. With lower pressure over upper surface and higher pressure over bottom surface, airfoil feels a net force in upward direction → Lift VdV dp V p ρ ρ - = = + constant 2 1 2 Most of lift is produced in first 20-30% of wing (just downstream of leading edge)
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Jill Tulane University ‘16, Course Hero Intern | 1,104 | 4,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-26 | latest | en | 0.801656 |
http://www.mytechinterviews.com/red-and-blue-marbles | 1,656,819,361,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00277.warc.gz | 93,227,288 | 15,753 | # Red and Blue Marbles
Question: You have 50 red marbles, 50 blue marbles and 2 jars. One of the jars is chosen at random and then one marble will be chosen from that jar at random. How would you maximize the chance of drawing a red marble? What is the probability of doing so? All 100 marbles should be placed in the jars.
Answer: Seems tricky at first right? Given that the number of red and blue marbles are the same, you would tend to think that the odds are 50-50. You would try different combinations, such as 25 of each colored marble in a jar or putting all red marbles in one jar and all the blue in the other. You would still end up with a chance of 50%.
So lets think of a better way to distribute the marbles. What if you put a single red marble in one jar and the rest of the marbles in the other jar? This way, you are guaranteed at least a 50% chance of getting a red marble (since one marble picked at random, doesn’t leave any room for choice). Now that you have 49 red marbles left in the other jar, you have a nearly even chance of picking a red marble (49 out of 99).
So let’s calculate the total probability.
P( red marble ) = P( Jar 1 ) * P( red marble in Jar 1 ) + P( Jar 2 ) * P( red marble in Jar 2 )
P( red marble ) = 0.5 * 1 + 0.5 * 49/99
P( red marble ) = 0.7474
Thus, we end up with ~75% chance of picking a red marble.
Have a better solution? Let us know through the comments section!
If you're looking for some serious preparation for your interviews, I'd recommend this book written by a lead Google interviewer. It has 189 programming questions and solutions:
## 8 Responses
1. Debra S says:
I agree, I put 1 marble in 1 jar. 49 marbles in the other. The Blue in the trash. The probability is 1.
2. I also came up with this answer intuitively. It’s a good solution but this definitely isn’t a proof it’s the best possible solution.
In order to prove this seems pretty complicated though.
This is what I did:
r1 is the number of red marbles in jar 1
b1 is the number of blue marbles in jar 1
we know r2 = 50-r1
and b2 = 50-b1
then by the regular conditional probability rules:
P(picking red) = 1/2 * r1/(b1+r1) + 1/2 * ((50-r1)/(100-b1-r1))
if we let x, y = r1, b1 respectively and we know they lie in a range of [0…50].
then z = P(picking red)
The equation for the probability is now:
def f(x,y):
return 1/2 * x/(y+x) + 1/2 * ((50-x)/(100-y-x));
plotting that in 3d shows you that there are indeed 2 maxima close to 0.75.
I used sage to do this, and used the ranges 0…49 and 1…50 since there are discontinuities when you divide by zero.
P = plot3d(f,(0,49),(1,50), adaptive=True, color=rainbow(60, ‘rgbtuple’));
here’s a pic:
http://bit.ly/qLHLuY
Now to we need to use calculus to prove that the maxima are inded global maxima and not local maxima. We can find them using the method of gradient descent/ascent. Where the path of greatest ascent/descent on the curve has a tangent vector at any point that coincides with the gradient vector at that point.
(i and j are the unit vectors along x and y.)
ie. its partial derivatives with respect to x and y must coincide with the above.
So thats as far as I got, I know that the method of gradient ascent/descent isn’t guaranteed to converge to global maxima/minima it might get stuck and local maxima/minima. But I think we can get around this by finding roots and using the second derivative test / etc.
Not 100% sure about that part though.
There must an easier way to prove this else why would they ask it? I mean otherwise you’re just submitting a clever hunch as your solution.
Let me know what you think.
Or if you can prove that they are global maxima/minima without waving your hands in the air.
Cheers.
3. Nitin says:
Damn! I came across this questions, and took a very long time to answer (with some hints from teh interviewer!)
wish I discovered this blog on the day it started! 🙂
4. P K says:
it’ a bit of trick question. it depends on how many times a marble is getting drawn.
5. Brian B. says:
I could physically feel mind mind wrench when I read this beautiful solution. Awesome!
6. Doggie says:
Hey John, smart answer but it clears says in the problem that “All 100 marbles should be placed in the jars”. I guess you have to think again now 🙂
7. jason says:
I believe this is the best result we can achieve
8. John S. says:
I would put one red marble in jar 1, 49 red marbles in jar 2, and the blue marbles in the trash. The probability of picking a red marble is now 1.
I don’t see anything in the problem statement that prevents this, so I win.
XHTML: These are some of the tags you can use: `<a href=""> <b> <blockquote> <code> <em> <i> <strike> <strong>` | 1,255 | 4,691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-27 | longest | en | 0.939155 |
https://www.etutorworld.com/math/pre-algebra-online-tutoring/ordering-of-rational-numbers.html | 1,604,170,591,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00324.warc.gz | 701,208,042 | 18,500 | Select Page
# Rational Numbers
Home | Online Math Tutoring | Pre-Algebra Tutoring | Ordering of Rational Numbers
rational number is one that is a part of a whole denoted as a fraction, decimal or a percentage. It is a number which cannot be expressed as a fraction of two integers (or we can say that it cannot be expressed as a ratio).
For instance, let’s consider the square root of 3. It is irrational and cannot be expressed by two integers.
It is an irrational number because it cannot be denoted as a fraction with just two integers.
#### Integers and Rational numbers
They are the numbers you usually count and they will continue upto infinity.
Whole numbers are all natural numbers including 0.
Example: 0, 1, 2, 3, 4…
Integers are all whole numbers and their negative sides too.
Example….. -6,-5,-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6…
Every integer is a rational number, as each integer n can be written in the form .
For example 5 = and so 5 is a rational number. However, numbers like ,,,, and – are also rational; since they are fractions whose numerator and denominator are integers.
1. Arrange the following integers in the ascending order:
1. -17, 1, 0, -15, 16,8, 33, 6
2. -44, -66, 0, 23, 41, 55, 15, 11, 10, -1, -2
1. Identify whether the following is rational or irrational number.
1. 0.5
2. 9.0
3. 5
1. -17, -15, 0, 1, 6, 8, 16, 33
2. -66, -44, -2, -1, 0, 10, 11, 15, 23, 41, 55.
1. Rational
2. Irrational
3. Rational
4. Rational
5. Rational
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Grammar | 593 | 1,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-45 | latest | en | 0.797657 |
https://discussions.unity.com/t/understanding-the-cost-of-different-functions/131816 | 1,716,363,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00346.warc.gz | 175,133,220 | 7,244 | # Understanding the cost of different functions
So this is something I’ve never been sure of, but since I have a concrete example right now, I figured I’d run it by you all so I can learn for the future.
Basically, I’m making a submarine game, and the submarine returns to a neutral position if no button is pressed. As it stands right now, I check every frame to see if a button is being pressed, and if it isn’t, I Lerp between the current position and the neutral position, like so:
``````if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
``````
However, this feels wasteful to me, as it will continue to Lerp even when it’s returned to a neutral position. I was considering adding an if statement to check to see if it’s at a neutral position, something like this:
``````if Mathf.Abs (currentX > .01f){
if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
}
``````
Alternately, I could do it:
``````if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
if Mathf.Abs (currentX > .01f){
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
}
``````
So I guess my question is, which of these 3 would be the fastest? Thanks!
The difference is quite insignificant with modern technology. And since the calculation is only executed once per frame and not many times for many objects, you could probably just skip this thought. But since you asked, and I was curious myself, I decided to run a test. Like most technical problems, the answer will be… it depends. Depends on the hardware, what value currentX is, if you’re pressing any keys, etc…
To run your own tests, get the profile script here: http://wiki.unity3d.com/index.php/Profiler
Now according to my super-scientific research, the results show that the 3rd method is the fastest on average.
no keys being pressed, currentX being greater than 0.01 for most of the time
• Method 1 took 0.003008700 seconds to complete over 1000 iterations, averaging 0.000003009 seconds per call
• Method 2 took 0.001624600 seconds to complete over 1000 iterations, averaging 0.000001625 seconds per call
• Method 3 took 0.000533800 seconds to complete over 1000 iterations, averaging 0.000000534 seconds per call
Holding up arrow key
• Method 1 took 0.003370300 seconds to complete over 1000 iterations, averaging 0.000003370 seconds per call
• Method 2 took 0.002678900 seconds to complete over 1000 iterations, averaging 0.000002679 seconds per call
• Method 3 took 0.001618500 seconds to complete over 1000 iterations, averaging 0.000001619 seconds per call
Now having said all that, I would personally choose the method that is easiest to read and maintain. Code readability is far superior than micro-optimizations. That doesn’t mean I completely disregard performance, but something as trivial as this seems to be a waste of time and energy that could be focused on other tasks, or be optimizing something that would have a noticeable effect on performance.
Here’s the test script I used:
``````using UnityEngine;
public class SomeProfileTest : MonoBehaviour {
float currentX = 1000f;
int iterations = 1000;
int elapsedIterations = 0;
// Update is called once per frame
void Update () {
float frameCurrentX = currentX;
Profile.StartProfile("Input Method 1");
if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
Profile.EndProfile("Input Method 1");
currentX = frameCurrentX;
Profile.StartProfile("Input Method 2");
if (Mathf.Abs (currentX) > .01f){
if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
}
Profile.EndProfile("Input Method 2");
currentX = frameCurrentX;
Profile.StartProfile("Input Method 3");
if (!Input.GetKey (KeyCode.UpArrow) && !Input.GetKey (KeyCode.DownArrow)) {
if (Mathf.Abs (currentX) > .01f){
currentX = Mathf.Lerp (currentX, 0, Time.deltaTime);
}
}
Profile.EndProfile("Input Method 3");
elapsedIterations++;
if (elapsedIterations >= iterations)
{
Application.Quit();
Debug.Break();
}
}
void OnApplicationQuit () {
Profile.PrintResults();
}
void OnGUI() {
GUI.Label(new Rect(0f, 0f, 100f, 100f), currentX.ToString());
}
}
``````
I don’t think the difference between those would really be noticable, so I wouldn’t bother with trying to figure out the most optimized method. Instead write the code in the way that you think makes the most sense.
If you want to check for a neutral position, then, imo, that’s the last one - I also suspect checking Mathf.Abs() takes a little more to check than 2 input booleans and that way lets you get out of the if statment before you check Mathf.Abs() - I’m not sure about that but I would be surprised if it was different.
I don’t think checking for it makes any real difference performance-wise though, Lerp should be a fairly cheap method.
In fact, not checking may be the best way performance-wise. I don’t know how Lerp compares to (x > Math.Abs), but in cases where you need to Lerp you’ll have to evaluate both. If you only very rarely DONT Lerp (wether this is the case depends on your game), constantly checking for a neutral position might be a net decrease in performance.
Not a definitive answer to your question as to what is fastest I suppose, but some things worth considering in any case. | 1,338 | 5,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-22 | latest | en | 0.908292 |
http://www.bbc.co.uk/blogs/blogradio4/posts/does_muslim_demographics_abuse_numbers?postId=84004342 | 1,397,840,689,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00560-ip-10-147-4-33.ec2.internal.warc.gz | 304,175,704 | 27,587 | ## Does 'Muslim Demographics' abuse numbers?
Friday 7 August 2009, 14:30
#### Tagged with:
On More or Less we patiently survey the statistical landscape. It's a kind of mathematical stakeout. When, finally, a number-abuser strays out into the open - we pounce.
Our victims are usually journalists or politicians. But of course now there are other channels open to the purveyor of rogue statistics.
A recent YouTube video, Muslim Demographics, uses data to portray the rapid Islamification of Europe and the United States. The claims it makes are rather startling.
But is a YouTube video fair game for More or Less? It's an interesting question - not least because we don't know who made the video. Or why (though one might speculate).
We decided to pounce. After all, the video's been played over 10 million times. That's a big hit. And it chimes with a thesis - the rise of 'Eurabia' - which has some traction elsewhere.
So how reliable are the statistics in the Muslim Demographics video? The short answer is: not very. But the long answer is more interesting, because the video is mix of the right, the wrong and the unknowable.
It's quite hard to dispute a figure for which there's no firm data either way. Take, for example, the video's claim that half of Dutch new-borns are Muslim. The Dutch cannot provide the relevant data because they don't collect it.
But Dutch statisticians estimate a Muslim population of 5 per cent of the total population. So to put it bluntly: could 5 per cent of Dutch women really be having 50 per cent of Dutch babies?
It sounds unlikely. But it's not an easy question to answer. If you want to see how we set about it, you might like to read this essay by my colleague Oliver Hawkins. Indeed, if you're into maths, we positively encourage you to; you might be able to suggest an even more elegant calculation.
The video is over seven minutes long, covering more ground than we could deal with on the radio. So we've made a video of our own - a more thorough analysis - and we've posted it on YouTube as a reply to the original video. It's embedded here, too.
If you like it, do pass it on.
Richard Knight is Series Editor of More or Less
• You can embed the More or Less video on your own web site: click the 'share' button at bottom right of the video and copy the embed code to your web page.
• The new series of More or Less starts today at 1330 on Radio 4. In the first programme Tim Harford investigates statistics which some claim reveal the 'Islamification' of Europe and checks whether the Home Office has been doing its sums properly. Do its claims about the DNA Database really add up?
• Richard Knight has written a longer piece about Muslim Demographics for the BBC News Magazine.
• rate this
0
• rate this
0
#### Comment number 2.
But surely this has only addressed the quantitative point, and not the qualitative argument, that it really wouldn't matter if 80% of Europeans were Muslim if Enlightenment values were to be preserved, civil liberties protected, human rights enshrined in law in perpetuity, nightclubs and bars remained open and the legal system based on total equality before the law, and not some shariah-hybrid nonsense.
I am fed up of listening to people say that we must be fully signed up to a 'tablets of stone' view of political correctness when some aspects of it, such as not being homophobic or islamophobic run diametrically opposite to each other and therefore make it impossible to square the circle. It is because few of us want to live in either a zionist Israel or Saudi Arabia that we choose to live the modern liberal democracy which the United Kingdom still purports to be.
• rate this
0
#### Comment number 3.
One point that hasn't been addressed is the assumption that people only have children with those in the same religon as themselves and that those children will also have the same religon. Belief in no religon is on the rise, but I don't think this is just down to atheists having lots of kids.
Its also clear that whoever made the original video dosen't have a clue. They managed to mix up Great Britain, the United Kingdom and the British Isles all at the same time.
To be honest, I dont know why more or less felt the need to cover this. Usually they deal with statistics that have been misleadingly manipulated, not just blatenly made up.
• rate this
0
#### Comment number 4.
Although I'm concerned about some aspects of rapidly changing Muslim demographics in Europe, I did find the original video a little hard to believe. I've always respected "More or Less", and they use more "proof" than the original, so I'm happy to accept their version.
However, turning to comment number 1 by jonmarriott, I'm more concerned with our education system to be honest! Jon - do you seriously not know the difference between race and religion?!? Criticism of Islam (a religious and political ideology open to all) cannot, by any definition, be mixed with the word "racism", which is discrimination against "any group of people who are defined by reference to their race, colour, nationality (including citizenship) or ethnic or national origin."
By falsely conflating the two, you are in effect belittling a powerful word and "diluting" the suffering of those who have suffered genuine and real racism based on skin colour, ethnic grouping etc. If you follow your logic fully, you yourself are being "racist" by suggesting only brown beardy people can be Muslim. You know it's not true, so please don't do it.
• rate this
0
#### Comment number 5.
The 'More or Less' response to the crude "Muslim Demographics" was balanced and helpful but it decided not to report the ONS data, from the UK, on what we know about the UK Muslim fertility rate and how it compares with other ethnic groups.
Here are two links which provide reasonably up-to-date official ONS data, relevant to this discussion:-
http://www.statistics.gov.uk/cci/nugget.asp?id=961
http://www.statistics.gov.uk/cci/nugget.asp?ID=951
• rate this
0
#### Comment number 6.
Thanks to U3280211 for the statistics links, and to alanparker for your point about conflating race and religion.
The original video is pretty unpleasant right-wing christian propaganda, and the More or Less reply knocks major holes in it. To an extent, however, the first few minutes seem to validate the original video’s assertion about fertility rates. The More or Less video casts doubt on the claim that muslim families are much larger than non-muslim ones, and states that immigrant family-sizes tend to move towards the norm of the indigenous population over time. Whilst the 8.1 figure for the average muslim family size is unsupported, this does not seem a strong rebuttal, and the ONS statistics linked to show that UK muslim families are almost double the size of most others.
The MoL video points out that figures given for immigration into Europe and Canada treat totals as if they were 100% muslim: the MoL video implies this is incorrect – I’m sure it is - but it would be more persuasive if accurate figures were provided.
The point of the MoL video was to highlight distortions and bad practice in the original, and it does this fairly and well. As the voice-over admits at the end, however, it doesn’t show that the original is completely false.
• rate this
0
#### Comment number 7.
The "More or Less" argument looks misleading to me. It says there is a high proportion of women of Turkish/Moroccan origin between 15 and 49 (61%/57% as against the mainstream 45%).
However, it would make a lot more sense to look at the number of Muslim women between 15 and 25. I think it possible (indeed likely) that there is a much greater population bulge in this very high-fertility age-group.
Realistically, of course, the original claim is worthless. However, it might well be true in many Netherland cities, the places which exert most influence on education and planning policies. I think the Dutch are right to be very worried. 5% of the population is the point at which Halal food (licensed by mosques ignorant of BSE restrictions) starts to take over in schools and hospitals.
• rate this
0
#### Comment number 8.
What a round-table discussion. This is why I love BBC blogs.
This entry is now closed for comments
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## Woman's Hour Power List 2014
Identifying the top ten game changers operating in the UK today.
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Woman's Hour Power List judges, 2014
Identifying the top ten game changers operating in the UK today.
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https://www.livemint.com/opinion/columns/keep-calm-and-be-greedy-sometimes-11623352131656.html | 1,632,528,999,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00664.warc.gz | 887,980,712 | 66,591 | OPEN APP
Home >Opinion >Columns >Keep calm and be greedy, sometimes
Two thought experiments to start this column, both of which you may have run into, or even used yourself.
First, recall your college days. In particular, imagine sitting down for a tough year-end mathematics exam. You’ve kept up in class, studied the subject well, but even so you expect the exam to push you. When you get the paper, you approach it the way your professor and parents and various other well-meaning folks have always advised: spend several minutes reading it through, then start by answering the question you recognize as the easiest. Done with that, you go on to the next-easiest, and so on.
The idea is to quickly lock in the marks you’re sure you can get. When that’s done to the extent possible, use the remaining time to solve questions that need more thought and effort and time, whose intricate calculations might induce mistakes. Why tackle the exam this way? Why not jump into the deep end right away, with the more difficult questions? Because you don’t want the difficult questions to take so much of your time that you have none left for the easy ones. That is, you don’t want to lose the easy marks that you can nail down with minimal effort. You might say, you’re greedy for them. Hold that thought.
Second, one rainy morning, you tell me I need to make a run to the nearby convenience store to stock up on food for Aziz the cat. As I grumble and then scrounge around for an umbrella, you produce a fistful of notes of several denominations. You hand them over with this stern instruction: use the minimum number of notes to pay for the cat food. On my way to the store, I marvel for a while at the eccentricity of this injunction. Why would you care how many notes I use, let alone insist on the fewest possible number? But then I start to wonder how I’m going to satisfy you.
So, let’s make this a little more real. Your fistful contains notes in these denominations: 500, 100, 50, 20 and 10. When I am done picking out Aziz’s food, the bill comes to 1,780. You’ve given me enough cash to pay, and enough notes of each denomination that there are several combinations of notes I can use. But there is that constraint you’ve handed me. How can I be sure of using the smallest number of notes?
It’s not that difficult, actually, as a little thought will tell you. We start with the note of the highest value that’s less than what we owe, and use as many of them as possible. In this case, that’s the 500 note. Three of them sum to 1,500 and, of course, we can’t use any more. We hand those over to the store-owner. What’s left of the bill is 280 (1,780 - 1,500 =280). Repeat: choose the note of the highest value that’s less than that amount, use as many as possible. That would be two 100 notes. Keep this going and we quickly have our solution: three 500s, two 100s, one 50, one 20 and one 10.
How do I know this is the correct solution? Well, the only way to use fewer notes is if some combination of notes can be replaced by one of a higher denomination. Not possible with the 500s, because I have no note worth more than that. Not possible with the 100s, because there are only two that I use to pay, and I need to have five to replace them with a 500 note. In general, not possible with any other combination of notes, because my selection of notes proceeded by switching to lower-denomination notes only when it wasn’t possible to use the higher denomination.
Intuitively at any rate, this probably makes sense. At every stage, I want to be sure to use higher-denomination notes first. You might say, at least in this task, where I must pay with as few notes as possible, that I’m greedy for the high-value notes.
“Greedy", again! As it happens, a whole class of mathematical algorithms is called “greedy". These two thought experiments capture the spirit of that greed. The point, every time you’re faced with a choice during the process, is to pick the option that you think will take you furthest right then towards the overall goal—score well in an exam, or use the fewest currency notes. That way, when the task is done, you will likely have reached that overall goal. Put it this way: while searching for an optimal solution to a problem, at each step the algorithm makes the optimal choice from those available.
Here’s a nicely mathematical example, taking off from last week’s column about series. First, an “arithmetic progression" (AP) is a sequence of three or more numbers that differ by the same amount. For example, 1, 3, 5, 7; or 101, 111, 121, 131, 141 and 151. Now you would agree that every increasing sequence of numbers doesn’t have to have an AP pop up somewhere along the way. Take the squares—1, 4, 9, 16, 25 ... - or the powers of 2 - 2, 4, 8, 16, 32 ...—clearly neither will contain an AP. You can think of other examples, too.
But consider a greedy algorithm that will produce a sequence of numbers, one that is guaranteed never to have an arithmetic progression. Put down 0 to begin with. The next entry in the sequence is selected greedily—the smallest number that will not immediately form an AP with numbers already in the series. The second entry is thus 1 (because anyway we can’t form an AP with just two numbers). The third cannot be 2, because (0, 1, 2) is an AP. So it is 3. The fourth is 4 because (0, 1, 4), (0, 3, 4) and (1, 3, 4) are all not APs. So, we have 0, 1, 3, 4. What’s the fifth in the sequence? It can’t be 5, because (1, 3, 5) is an AP. Not 6, because (0, 3, 6) is an AP. 7 gives us (1, 4, 7) and 8 (0, 4, 8), both APs; so neither of those can be next. In fact, it is 9. Staying greedy every time, we produce this sequence: 0, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36, 37, 39, 40 ... (Aside: if you were presented this sequence—even this many entries—and not the greedy algorithm that produced it, would you be able to work out the next entry? I seriously doubt I would).
As computer science students know well, greedy algorithms are interesting programming challenges in various familiar mathematics and computer science problems—finding the shortest path through a maze, for example, or compressing data for more efficient storage, or maybe even playing a game of chess. Though as CS students also know well, interesting programming challenges don’t necessarily solve problems: greed may not actually succeed. The famous travelling salesman problem is not so easily tackled, and what looks superficially like the best move in a given chess situation may well lead to disaster several moves down the road. Still, greedy algorithms are valuable tools in many problems. Maybe even in decisions about vaccinations? If they don’t always produce a correct solution, they may get us a workable solution. Sometimes, that’s good enough. Sometimes, you see, greed is good.
Once a computer scientist, Dilip D’Souza now lives in Mumbai and writes for his dinners. His Twitter handle is @DeathEndsFun | 1,701 | 6,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-39 | latest | en | 0.948045 |
https://physics.stackexchange.com/questions/135865/quantized-modes-of-em-field-in-a-cubic-cavity | 1,569,270,907,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514578201.99/warc/CC-MAIN-20190923193125-20190923215125-00339.warc.gz | 624,362,730 | 30,395 | # Quantized modes of EM field in a cubic cavity
I am trying to solve for Electric field modes in a cubic cavity with perfectly reflecting walls. I know that for the modes, I have the standard Helmholtz Equation $$(\nabla^2+k^2)E_i=0$$ And I want the field to vanish at the walls. I solved it out just like a 3D 'Box', so that $$E(n_x,n_y,n_z)=\sqrt{\frac{8}{V}} \sin \left( \frac{n_x\pi x}{L} \right) \sin \left( \frac{n_y\pi y}{L} \right) \sin \left( \frac{n_z\pi z}{L} \right) .$$
This solves the equation, but I am not sure if this is what is wanted for solving for the modes. Thanks for pointing me in the right direction.
• What you have found are by definition the electric field profiles of the modes of the cavity. Good job! I note that the dimensions of your formula don't make sense though. $1/\sqrt{\text{volume}}$ is not electric field. – DanielSank Sep 17 '14 at 0:46
• If this is a homework problem, please add the homework tag and include a verbatim transcription of what the assignment says. Since you're asking whether or not your work satisfies the problem, we need to know what the problem is ;) – DanielSank Sep 17 '14 at 0:48
• So if I were to instead say that it was some constant E_0 and then normalize this equation, I should get something that makes a little more sense? – yankeefan11 Sep 17 '14 at 1:01
• Presumably to get the equation you have, you write down a guess for $E$ and then plug it into Poisson's equation or something like that. So, when you write down your guess, just make sure it has the right dimensions. Putting a constant $E_0$ out front is exactly the right thing to do. Once you work through Poisson's equation (or whatever) you'll wind up with an actual expression for $E_0$ in terms of parameters of the problem and natural constants. – DanielSank Sep 17 '14 at 3:29
• The electric field is a vector. Your general solution is a function. Are you sure about the boundary conditions for the electric field? – suresh Sep 17 '14 at 12:19 | 540 | 1,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-39 | latest | en | 0.907501 |
https://www.chess.com/forum/view/game-analysis/how-to-figure-the-last-piece-standing-in-a-typical-exchange | 1,527,184,682,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866733.77/warc/CC-MAIN-20180524170605-20180524190605-00392.warc.gz | 719,592,562 | 13,346 | # How to figure the last piece standing in a typical exchange?
Yes...you have six pieces pointed and covering at the d6 ...and your opponent has seven pieces pointed at the d6 pawn...shoot the whole force or back off because it's not in your favor....it doesn't always play out the way you thought (using analysis board)...so whats the trick... if your not the last man standing ...is it a bad trade-down-exchange? What quick rule-of-thumb do the better players apply??
I'd look for away to divert the attention from the d6 to a square that favors me. A shoot out as you describe seems (to me) like a near empty board with my position a piece down. Not good.
I'd look for a fork or check or a way of forcing one of my opponents pieces to a weak square.
Is there a pin available?
A quiet move?
ednorton wrote:
I'd look for away to divert the attention from the d6 to a square that favors me. A shoot out as you describe seems (to me) like a near empty board with my position a piece down. Not good.
I'd look for a fork or check or a way of forcing one of my opponents pieces to a weak square.
Is there a pin available?
A quiet move?
I usually do...it's the exchange i get drawn into that doesn't go in my favor...i just wondered how others considered their particular choices... my real ambition is to shift focus from an overburdened piece...the exchanges seldom go to my favor....I with your advice on the quiet move approach.
I always run through the board position assuming everyone puts all their pieces in. Then, while doing that, I make sure he doesn't have any damaging intermediate moves. If the position is not favorable to me, I usually try to punish him somehow for using that tempo to capture whatever square the forces are all built up on. Like a gambit in the middle of a game, I guess.
And if that's not possible I do what ednorton suggests.
when there's tension in one square, usually one of the pieces attacking it is one that can hold the others (like a knight, which doesn't represent a direct attack). When the attack gets to the knight positioning, it's best to keep it defended & attack another flank. A couple of months ago I was in this position & instead of doing this, I kept attacking & in the end, lost pieces & positioning...
I guess, for a mere mortal...meaning myself....I try and simplify things by evaluating the exchange. If the exchange costs me a piece and a loss of tempo, well...I'm not gonna bite. If the exchange costs me a piece, but leaves me on the move...maybe there is a little light at the end of the tunnel.
What I have been learning as I have moved up the ranks from 1200 to 1400 is that at 1400, if I am a piece and a pawn down...my opponent is gonna toast me. Its just a matter of time. And I resign. (sometimes) | 638 | 2,781 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.966351 |
https://www.hetnieuwekroost.nl/Jul/1910_ball-mill-operating-weight-mean/ | 1,620,556,632,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00250.warc.gz | 830,173,962 | 7,060 | # ball mill operating weight mean
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Mill liners fill a basic role to protect your mills from the intense wear and tear that comes from grinding down hard raw materials. But even if they all do the same job not all mill liners are created equal. From the abrasion resistance of rubber to the impact resistance of steel different material parts offer different advantages. FL supplies the full range of mill liners options
• ### A Method to Determine the Ball Filling in Miduk Copper
Therefore ball abrasion rate in the mill determined by Equation 10. A𝑟= At F =4003321.8 131328 =30.48 g/ton (10) In above equations 𝑣m 3total mill volume (m ) 𝑁 number of balls which exist in mill 𝐴𝑏 each ball abrasion (g) 𝐴t total ball abrasion in the mill (g) 𝑣b each ball volume (m3) 𝑓b
• ### Optimum choice of the make-up ball sizes for maximum
If the ball wear rate is proportional to ball weight Δ = 1. If it is proportional to ball surface area Δ = 0. Austin and Klimpel s analysis of the experimental data showed inconsistencies in the data but Δ = 0 is a good approximation for ball milling.
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Mill balls weight M b kg . 21.125 . the Bond mill operating under the sam e conditions as those in the Bond is defined in a Bond ball mill on the samples of standard size − 3.327 0
• ### Ball Mill Operating principles components Uses
Jul 05 2020 · A ball mill also known as pebble mill or tumbling mill is a milling machine that consists of a hallow cylinder containing balls mounted on a metallic frame such that it can be rotated along its longitudinal axis. The balls which could be of different diameter occupy 3050 of the mill volume and its size depends on the feed and mill size.
• ### Improving Mill Shoe Bearing Reliability and Productivity
Cement Ball Mills. A cement ball mill is designed to grind clinker gypsum and for the drying of cement additives. Lighter mill weight (less energy to operate) Remember just because the oil can handle the operating temperature doesn t mean the bearing can. Other causes of high heat at the shoe bearings include Lack of oil flow. Oil
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you cannot use work index analysis for the same purpose on ball mill circuits because of the complex interactions between grinding and classification. In this module you will learn how to relate design and operating variables to ball mill circuit efficiency through functional performance analysis . This Introduction is seven pages long.
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• ### Ball Mill Design/Power Calculation
Dec 12 2016 · If P is less than 80 passing 70 microns power consumption will be. Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a
• ### Keeping an eye out for grinding mill dropped charge and
Nov 06 2019 · To help the grinding process water and steel balls are added to the process which then form the charge in the mill. Ball mill charge contains a significant amount of steel balls and can become extremely heavy. For example a large 24 foot (7.3 m) diameter ball mill charge typically weighs around 2 million lbs (907 tons).
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Working Principle Operation. The apparent difference in capacities between grinding mills (listed as being the same size) is due to the fact that there is no uniform method of designating the size of a mill for example a 5′ x 5′ Ball Mill has a working diameter of 5′ inside the liners and has 20 per cent more capacity than all other ball mills designated as 5′ x 5′ where the
• ### The grinding balls bulk weight in fully unloaded mill
Apr 11 2017 · The correct determination of the grinding balls bulk weight in mill allows accurately determination the mill balls feed weight. The mill balls feed weight is necessary for calculating the grinding media specific consumption and avoid mill overloading thereby eliminating the motor load increasing possibility.
• ### MODULE #5 FUNCTIONAL PERFOMANCE OF BALL MILLING
you cannot use work index analysis for the same purpose on ball mill circuits because of the complex interactions between grinding and classification. In this module you will learn how to relate design and operating variables to ball mill circuit efficiency through functional performance analysis . This Introduction is seven pages long.
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A 1.5 mio t/a cement plant is having a closed circuit ball mill for cement grinding The mill has been operating with satisfactory performance in-terms of system availability and output however power consumption was on higher side. 3.1 System Description Mill Rated capacity 150 t/h OPC at
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Effect of Surface Cleanliness on Mean Wear Rate 3 Hour Tests to production results ball size and mill operating conditions. of ore are the Bond strength test and the JK Drop Weight Test
• ### PROCESS DIAGNOSTIC STUDIES FOR CEMENT MILL
A 1.5 mio t/a cement plant is having a closed circuit ball mill for cement grinding The mill has been operating with satisfactory performance in-terms of system availability and output however power consumption was on higher side. 3.1 System Description Mill Rated capacity 150 t/h OPC at
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Mill liners fill a basic role to protect your mills from the intense wear and tear that comes from grinding down hard raw materials. But even if they all do the same job not all mill liners are created equal. From the abrasion resistance of rubber to the impact resistance of steel different material parts offer different advantages. FL supplies the full range of mill liners options
• ### (PDF) A comparison of wear rates of ball mill grinding media
Effect of Surface Cleanliness on Mean Wear Rate 3 Hour Tests to production results ball size and mill operating conditions. of ore are the Bond strength test and the JK Drop Weight Test
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The profile can be better customised to suit mill speed and filling and therefore optimise performance and it allows more material in the lifter for a given base width but the mill must only run in one direction. 7. High–low double wave ball mill linersThese are a refinement of the wave liner Figure 5.
• ### A Method to Determine the Ball Filling in Miduk Copper
Therefore ball abrasion rate in the mill determined by Equation 10. A𝑟= At F =4003321.8 131328 =30.48 g/ton (10) In above equations 𝑣m 3total mill volume (m ) 𝑁 number of balls which exist in mill 𝐴𝑏 each ball abrasion (g) 𝐴t total ball abrasion in the mill (g) 𝑣b each ball volume (m3) 𝑓b
• ### Improving Mill Shoe Bearing Reliability and Productivity
Cement Ball Mills. A cement ball mill is designed to grind clinker gypsum and for the drying of cement additives. Lighter mill weight (less energy to operate) Remember just because the oil can handle the operating temperature doesn t mean the bearing can. Other causes of high heat at the shoe bearings include Lack of oil flow. Oil
• ### Ball Mill Operating principles components Uses
Jul 05 2020 · A ball mill also known as pebble mill or tumbling mill is a milling machine that consists of a hallow cylinder containing balls mounted on a metallic frame such that it can be rotated along its longitudinal axis. The balls which could be of different diameter occupy 3050 of the mill volume and its size depends on the feed and mill size.
• ### Dealing With Scat in Mill ProcessingPumpEng Submersible
In a mining plant or processing mill they have a big grinding circuit for crushing raw material these are usually Ball or SAG mills. Inside these Sag/Ball mill are big steel balls that grind ore via an impact process. In simple terms these hard metal steel balls get to the top of the rotating chamber and fall down crushing the ore being
• ### A Method of C alculating Autogenous/ Semi-Autogenous
1.27 for the rod mill and 0.67 for the ball mill. The particle size distributions reported are shown in Figure 1. It follows that the No 1 rod mill/ball mill circuit was operating very effi ciently when comparing the combined operating work indices for both mills with the rod mill and ball mill laboratory work indices.
• ### Ball Mill Design/Power Calculation
Dec 12 2016 · If P is less than 80 passing 70 microns power consumption will be. Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a | 2,394 | 11,025 | {"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} | 2.796875 | 3 | CC-MAIN-2021-21 | latest | en | 0.842596 |
oviss.jp | 1,618,159,134,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00033.warc.gz | 552,306,765 | 19,930 | # Fourier Transforms Using Mathematica.ISBN9781510638556
## Fourier Transforms Using Mathematica
8,334（税込）
【在庫有り】
書名 Fourier Transforms Using Mathematica Mathematicaを使ったフーリエ変換 著者・編者 Goodman, J.W. 発行元 SPIE 発行年／月 2020年11月 装丁 ソフトカバー ページ数 110 ページ ISBN 978-1-5106-3855-6 発送予定 1-2営業日以内に発送致します
Description
The Fourier transform is a ubiquitous tool used in most areas of engineering and physical sciences. The purpose of this book is two-fold: (1) to introduce the reader to the properties of Fourier transforms and their uses, and (2) to introduce the reader to the program MathematicaR and demonstrate its use in Fourier analysis. Unlike many other introductory treatments of the Fourier transform, this treatment will focus from the start on both one-dimensional and two-dimensional transforms, the latter of which play an important role in optics and digital image processing, as well as in many other applications. It is hoped that by the time readers have completed this book, they will have a basic understanding of Fourier analysis and Mathematica.
The PDF is a Read Me First file with links to the Mathematica interactive book and to the free Wolfram Player for readers who do not have the full Mathematica program.
Contents:
1 Introduction
1.1 Why Mathematica?
1.2 What the Reader Should Know at the Start
1.3 Why Study the Fourier Transform?
2 Some Useful 1D and 2D Functions
2.1 User-Defined Names for Useful Functions
2.2 Dirac Delta Functions
2.3 The Comb Function
3 Definition of the Continuous Fourier Transform
3.1 The 1D Fourier Transform and Inverse Fourier Transform
3.2 The 2D Fourier Transform and Inverse Fourier Transform
3.3 Fourier Transform Operators in Mathematica
3.4 Transforms in-the-Limit
3.5 A Table of Some Frequently Encountered Fourier Transforms
4 Convolutions and Correlations
4.1 Convolution Integrals
4.2 The Central Limit Theorem
4.3 Correlation Integrals
5 Some Useful Properties of Fourier Transforms
5.1 Symmetry Properties of Fourier Transforms
5.2 Area and Moment Properties of 1D Fourier Transforms
5.3 Area and Moment Properties of 2D Fourier Transforms
5.4 Fourier Transform Theorems
5.5 The Projection-Slice Theorem
5.6 Widths in the x Domain and the u Domain
6 Fourier Transforms in Polar Coordinates
6.1 Using the 2D Fourier Transform for Circularly Symmetric Functions
6.2 The Zero-Order Hankel Transform
6.3 The Projection Transform Method
6.4 Polar-Coordinate Functions with a Simple Harmonic Phase
7 Linear Systems and Fourier Transforms
7.1 The Superposition Integral for Linear Systems
7.2 Invariant Linear Systems and the Convolution Integral
7.3 Transfer Functions of Linear Invariant Systems
7.4 Eigenfunctions of Linear Invariant Systems
8 Sampling and Interpolation
8.1 The Sampling Theory in One Dimension
8.2 The Sampling Theory in Two Dimensions
9 From Fourier Transforms to Fourier Series
9.1 Periodic Functions and Their Fourier Transforms
9.2 Example of a Complex Fourier Series
9.3 Mathematica Commands for Fourier Series
9.4 Other Types of Fouier Series
9.5 Circular Harmonic Expansions
10 The Discrete Fourier Transform
10.1 Sampling in Both Domains
10.2 Vectors and Matrices in Mathematica
10.3 The Discrete Fourier Transform (DFT)
10.4 The DFT and Mathematica
10.5 DFT Properties and Theorems
10.6 Discrete Convolutions and Correlations
10.7 The Fast Fourier Transform
11 The Fresnel Transform
11.1 Definition of the 1D Fresnel Transform
11.2 Approximations to the Bandwidth of the Interval-Limited Quadratic-Phase Exponential
11.3 Equivalent Bandwidth of the Interval-Limited Quadratic-Phase Exponentiald
11.4 The 2D Fresnel Transform
11.5 The Fresnel Transform of Circularly Symmetric Functions
11.6 Examples of Fresnel Transforms
11.7 The Frensel-Diffraction Transfer Function
11.8 The Discrete Frensel-Diffraction Integral
12 Fractional Fourier Transforms
12.1 Definition of the Fractional Fourier Transform
12.2 Mathematica Calculation of the Fractional Fourier Transform
12.3 Relationship between the Fractional Fourier Transform and the Fresnel-Diffraction Integral
13 Other Transforms Related to the Fourier Transform
13.1 The Abel Transform | 1,034 | 4,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-17 | latest | en | 0.711179 |
https://www.kyoto2.org/what-is-the-angle-of-rotation-of-a-pentagon/ | 1,726,621,075,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00058.warc.gz | 777,945,878 | 14,594 | # Kyoto2.org
Tricks and tips for everyone
# What is the angle of rotation of a pentagon?
## What is the angle of rotation of a pentagon?
The angle of rotation of an n sided regular polygon is given by 360∘n. A pentagon has 5 sides. ⟹Angle of rotation of a regular pentagon =360∘5=72∘. So, if you rotate the pentagon by an angle of 72∘, then it will look identical to the initial figure. Mathematics.
How do you calculate counterclockwise rotation?
Here are the rotation rules: 90° clockwise rotation: (x,y) becomes (y,-x) 90° counterclockwise rotation: (x,y) becomes (-y,x)
What is the rule for rotating counterclockwise by 180 degrees?
When rotating a point 180 degrees counterclockwise about the origin our point A(x,y) becomes A'(-x,-y). So all we do is make both x and y negative.
### What is the rule for a 270 degree counterclockwise rotation?
The rule for a rotation by 270° about the origin is (x,y)→(y,−x) .
What is angle of rotation of a polygon?
For rotation symmetry, the order of any regular polygon is the number of sides. The angle of rotation would be 360 degrees divided by that order. For example, an octagon is an 8-sided figure, so the order is eight. If you divide 360 by 8, you get 45, which means the octagon has an angle of rotation of 45 degrees.
Does a pentagon has a rotational symmetry?
A pentagon has 5 rotational symmetry as shown in figure has 5 point when rotated produces symmetrical image.
#### Is a 90 degree rotation clockwise or counterclockwise?
Since the rotation is 90 degrees, you will rotating the point in a clockwise direction.
What is the rule for a rotation of 90 degrees counterclockwise?
When we rotate a figure of 90 degrees counterclockwise, each point of the given figure has to be changed from (x, y) to (-y, x) and graph the rotated figure.
What is 90 degrees counterclockwise?
## What is the angle of rotation for this counterclockwise rotation about the origin?
270°
The point of rotation is the origin, draw lines joining one of the points, say X and it’s image to the origin. You can see that the lines form an angle of 270° , in the counterclockwise direction. Therefore, ΔX’Y’Z’ is obtained by rotating ΔXYZ counterclockwise by 270° about the origin.
How do you find the rotation of a polygon?
Coordinates of Rotation: The point (x,y) rotated an angle of θ counter-clockwise about the origin will land at point (x′,y′) where x′=xcos(θ)−ysin(θ) x ′ = x cos ( θ ) − y sin and y′=ycos(θ)+xsin(θ) y ′ = y cos ( θ ) + x sin .
How many order of rotation does a pentagon?
5
Therefore the order of rotational symmetry of a regular pentagon is 5. In fact, any manifestation of the regular pentagonal symmetry will have an order of 5.
### How many lines of rotational symmetry does a pentagon have?
5 lines of symmetry
Line symmetry in regular polygons A regular pentagon has 5 sides and 5 lines of symmetry. | 760 | 2,892 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.908665 |
https://ocw.mit.edu/courses/2-29-numerical-marine-hydrodynamics-13-024-spring-2003/pages/assignments/ | 1,713,909,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00803.warc.gz | 385,774,853 | 10,046 | 2.29 | Spring 2003 | Graduate
# Numerical Marine Hydrodynamics (13.024)
## Assignments
### Notes about Problem Sets 6 and 8
In problem sets 6 and 8, students write MATLAB® programs to solve two-dimensional boundary integral equations based on Green’s Theorem. In problem set 8, the inviscid streaming flow about an arbitrary two-dimensional object is calculated. The solution is done for a circular cylinder.
In problem set 8, the method is extended to the flow around a lift-generating airfoil with a wake across which there is a jump in the velocity potential. The two-dimensional Green function that is used is G = -ln r, where r is the distance between a “source point” and a “field point”.
The student is not expected to write an efficient MATLAB® m-file for computing the integral of the Green Function over a panel. Rather, that m-file is given to the students and it is called rank2d.m. This m-file computes the integral of the Green function, g, and of the normal derivative of the Green function, dg/dn, over a panel. This m-function works in local coordinates for which the “source panel” is approximated as a line on a local x-axis with the center of the line at the local origin. The “field point” is at (x,y) in local coordinates. The normal vector to the panel is in the positive local y-direction.
To use rank2d.m function, the panel length and the location of the field point in local coordinates must first be determined. This is done in the m-function “localize.m”, which should also be provided to the student who writes and used the remainder of the set of programs needed to complete the problem sets.
The m-functions, rank2d and localize are provided with problem set 6.
### Problem Sets
• Problem Set 1 (PDF)
• Problem Set 2 (PDF)
• Problem Set 3 (PDF)
• Problem Set 4 (PDF)
• Problem Set 5 (PDF)
• Problem Set 6 (PDF)
• Problem Set 7 (PDF)
• Problem Set 8 (PDF)
• Problem Set 9 (PDF)
• Problem Set 10 (PDF)
### Supporting Files
• 64a012.fin (FIN)
• LOCALIZE.M (M)
• RANK2D.M (M)
• wig4125.out (OUT)
• wigley5.out (OUT)
• wigley9.out (OUT)
## Course Info
Spring 2003
##### Learning Resource Types
Lecture Notes
Course Introduction
Problem Sets
Programming Assignments | 548 | 2,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-18 | latest | en | 0.857263 |
https://www.jiskha.com/questions/91279/Ascorbic-acid-is-a-weak-organic-acid-also-known-as-vitamin-C-A-student-prepares | 1,555,744,947,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578528702.42/warc/CC-MAIN-20190420060931-20190420082931-00477.warc.gz | 724,590,830 | 5,593 | # chemistry
Ascorbic acid is a weak organic acid also known as vitamin C. A student prepares a 0.20 mol/L aqueous solution of ascorbic acid and measures its pH as 2.40. Based on this evidence, what is the Ka of ascorbic acid?
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1. Based on what I can find on the internet, ascorbic acid is a monoprotic acid. If that is so, then let's call ascorbic acid HC.
Then HC ==> H^+ + C^-
Ka = ((H^+)(C^-)/(HC)
If pH = 2.40, then (H^+) = 0.00398/
(C^-) is the same.
(HC) = 0.2 = 0.00398
Plug in the Ka expression and solve for Ka. I get a little over 8 x 10^-5 which gives a pKa of 4.09 which isn't far from the pKa listed in my quant book of 4.17.
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posted by DrBob222
2. Thank you! Umm...Whats a quant book? A book read by extraterrestrials??:P
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3. Quantitative analysis chemistry book
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## Similar Questions
1. ### NEED HELP WITH CHEMISTRY HW
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6. A sample of a vitamin supplement was analyzed by titrating a 0.2894 g sample dissolved in water with 0.0215 M NaOH. A volume of 10.83 mL of the base was
asked by Anonymous on October 1, 2011
2. ### Chemistry
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6. A sample of a vitamin supplement was analyzed by titrating a 0.3252 g sample dissolved in water with 0.0284 M NaOH. A volume of 13.27 mL of the base was
asked by Anonymous on February 25, 2014
3. ### Chemistry
Ascorbic acid (vitamin C) is a diprotic acid having the formula H2C6H6O6. A sample of a vitamin supplement was analyzed by titrating a 0.3252 g sample dissolved in water with 0.0284 M NaOH. A volume of 13.27 mL of the base was
asked by PM on February 25, 2014
4. ### urgent i need help within an hour!
Ascorbic acid (vitamin C, C6H8O6) is a water-soluble vitamin. A solution containing 83.3 g of ascorbic acid dissolved in 210. g of water has a density of 1.23 g/mL at 55°C Calculate the molarity of ascorbic acid in this solution.
asked by Samantha on January 21, 2013
5. ### chemistry
How would you do this question: A student dissolved 5.0 g of vitamin C in 250 mL of water. The molar mass of ascorbic acid is 176 g/mol, and its Ka is 8.0 × 10−5. Calculate the pH of the solution. Note: Abbreviate the formula
asked by Q on June 7, 2014
6. ### chemistry
Titration of Vitamin C tablets Prior to titration, Vitamin C samples were dissolved in dilute sulfuric acid, treated with a 60mL portion of 0.3 M KIO3 and excess KI. Given equations: (1) ascorbic acid + I2 + 2H2O dehydroascorbic
7. ### chemistry
A student prepares a 0.20 mol/L aqueous solution of ascorbic acid and meausres its pH as 2.40. Based on theis evidence, what is the Ka of ascorbic acid? H+ = 0.00398 C- is the same (HC) = 0.2 = 0.00398 (0.00398)(0.00398) / 0.0199
asked by Dustin on April 7, 2008
8. ### chemisrty
Q1) A solution of ascorbic acid (C6H8O6, Formula mass = 176 g/mol) is made up by dissolving 80.5 g of ascorbic acid in 210 g of water, and has a density of 1.22 g/mL at 55 oC. Calculate the following: a) the mass percentage of
asked by lyan on January 16, 2012
9. ### Chemistry
I don't understand where to start first. Can you break it down for me? Ascorbic acid, also known as vitamin C, has a percentage composition of 40.9% C, 4.58% H, and 54.5% O. The molar mass of ascorbic acid is 176.1 g/mol.
asked by GABS on January 19, 2016
10. ### chemistry
A solution that contains 55.0 g of ascorbic acid (Vitamin C) in 250. g of water freezes at −2.34°C. Calculate the molar mass (in units of g/mol) of the ascorbic acid. Kf of water is 1.86°C/m.
asked by Diana on September 8, 2010
More Similar Questions | 1,242 | 3,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-18 | latest | en | 0.92752 |
https://www.numere-romane.ro/cum_se_scrie_numarul_arab_cu_numerale_romane.php?nr_arab=139000&nr_roman=(C)(X)(X)(X)M(X)&lang=en | 1,627,202,981,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151641.83/warc/CC-MAIN-20210725080735-20210725110735-00599.warc.gz | 954,006,843 | 9,663 | # Convert number: 139,000 in Roman numerals, how to write?
## Latest conversions of Arabic numbers to Roman numerals
139,000 = (C)(X)(X)(X)M(X) Jul 25 08:49 UTC (GMT) 53 = LIII Jul 25 08:49 UTC (GMT) 533,590 = (D)(X)(X)(X)MMMDXC Jul 25 08:49 UTC (GMT) 95,876 = (X)(C)(V)DCCCLXXVI Jul 25 08:49 UTC (GMT) 9,076 = M(X)LXXVI Jul 25 08:49 UTC (GMT) 500,369 = (D)CCCLXIX Jul 25 08:49 UTC (GMT) 965,990 = (C)(M)(L)(X)(V)CMXC Jul 25 08:49 UTC (GMT) 935 = CMXXXV Jul 25 08:49 UTC (GMT) 256,330 = (C)(C)(L)(V)MCCCXXX Jul 25 08:49 UTC (GMT) 870 = DCCCLXX Jul 25 08:49 UTC (GMT) 114,998 = (C)(X)M(V)CMXCVIII Jul 25 08:49 UTC (GMT) 200,251 = (C)(C)CCLI Jul 25 08:49 UTC (GMT) 745 = DCCXLV Jul 25 08:49 UTC (GMT) converted numbers, see more...
## The set of basic symbols of the Roman system of writing numerals
• ### (*) M = 1,000,000 or |M| = 1,000,000 (one million); see below why we prefer this notation: (M) = 1,000,000.
(*) These numbers were written with an overline (a bar above) or between two vertical lines. Instead, we prefer to write these larger numerals between brackets, ie: "(" and ")", because:
• 1) when compared to the overline - it is easier for the computer users to add brackets around a letter than to add the overline to it and
• 2) when compared to the vertical lines - it avoids any possible confusion between the vertical line "|" and the Roman numeral "I" (1).
(*) An overline (a bar over the symbol), two vertical lines or two brackets around the symbol indicate "1,000 times". See below...
Logic of the numerals written between brackets, ie: (L) = 50,000; the rule is that the initial numeral, in our case, L, was multiplied by 1,000: L = 50 => (L) = 50 × 1,000 = 50,000. Simple.
(*) At the beginning Romans did not use numbers larger than 3,999; as a result they had no symbols in their system for these larger numbers, they were added on later and for them various different notations were used, not necessarily the ones we've just seen above.
Thus, initially, the largest number that could be written using Roman numerals was:
• MMMCMXCIX = 3,999. | 686 | 2,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.90181 |
https://www.physicsforums.com/threads/radius-of-curvature-derivation-help.390494/ | 1,708,872,417,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00646.warc.gz | 974,161,468 | 16,662 | • Grand
In summary, the conversation is about understanding the derivation of the radius of curvature from a book. The equations for tangent and normal unit vectors are given, and the parametric equation of a circle is mentioned. The conversation also touches on a different way of expressing the concept of curvature.
Homework Statement
This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.
Homework Equations
Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:
$$\textbf{u}=\frac{d\textbf{r}}{ds}$$
and
$$\frac{d\textbf{u}}{ds}=k\textbf{n}$$
where k is called the curvature.
We can expand r(s) around a certain point:
$$\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...$$
using the equations for u and n:
$$\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...$$
which is the same as:
$$\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...$$
And from here they conclude that this is the equation of a circle with radius $$\frac{1}{k}$$, which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.
Thanks to advance to whoever is able to help.
The parametric equation of a circle is ( a+ Rcosu, b+Rsinu).
Grand said:
Homework Statement
This is not exactly a question, but I am trying to understand the derivation of radius of curvature from a boof I'm reading. I would be extremely grateful if someone is able to help me.
Homework Equations
Let u and n be the tangent and normal unit vectors respectively. If r(s) is the path function, we know that:
$$\textbf{u}=\frac{d\textbf{r}}{ds}$$
and
$$\frac{d\textbf{u}}{ds}=k\textbf{n}$$
where k is called the curvature.
We can expand r(s) around a certain point:
$$\textbf{\textbf{r(s)}}=\textbf{a}+s\frac{d\textbf{r}}{ds}+\frac{1}{2}s^{2}\frac{d^{2}\textbf{r}}{ds^{2}}+...$$
using the equations for u and n:
$$\textbf{r(s)}=\textbf{a}+s\textbf{u}+\frac{1}{2}s^{2}k\textbf{n}+...$$
which is the same as:
$$\textbf{r(s)}=\textbf{a}+\frac{sin(ks)}{k}\textbf{u}+\frac{1}{k}(1-cos(ks))\textbf{n}+...$$
And from here they conclude that this is the equation of a circle with radius $$\frac{1}{k}$$, which I don't quite understand. Tried to square it, extract sines and cosins, but still don't understand why this is a circle.
Thanks to advance to whoever is able to help.
I'm wondering if your latter equation is well derived. Did you take it from your book?
Yes, it is from the book and I was able to derive it by myself.
@Eynstone, that accounts only for the sin term, not for the -cos term
Oh wait! it's not expressed differently, it's just rearranged so that you can manage to get the radius of curvature, check that first website I handled, and enjoy up to point (31). ;)
What is the radius of curvature and why is it important in science?
The radius of curvature is a measure of how curved a curve or surface is at a given point. It is important in science because it helps us understand and analyze the shapes and movements of objects in the physical world.
How is the radius of curvature derived?
The radius of curvature is derived using the formula R = (1 + (dy/dx)^2)^3/2 / (d^2y/dx^2), where dy/dx represents the slope of the curve at a given point and d^2y/dx^2 represents the rate of change of slope.
What are the applications of the radius of curvature in different fields of science?
The radius of curvature has various applications in physics, engineering, and astronomy. It is used to analyze the shape of lenses and mirrors in optics, calculate the turning radius of objects in motion, and determine the curvature of space-time in general relativity.
Can the radius of curvature be negative?
Yes, the radius of curvature can be negative. A negative radius of curvature indicates that the curve or surface is concave, meaning it curves inward. A positive radius of curvature indicates a convex curve or surface, which curves outward.
Are there any limitations to using the radius of curvature as a measure of curvature?
While the radius of curvature is a useful measure of curvature, it is limited to measuring only the curvature at a single point on a curve or surface. It does not provide information about changes in curvature along the length of the curve. Additionally, it may not accurately represent the overall shape of complex curves or surfaces. | 1,256 | 4,626 | {"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": 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} | 3.953125 | 4 | CC-MAIN-2024-10 | latest | en | 0.881249 |
https://seostudiotools.com/probability-calculator | 1,726,294,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00142.warc.gz | 462,832,923 | 15,528 | # Probability Calculator
## Probability Calculator: How to Calculate the Probability
Master Probability Calculations: Learn how to calculate probabilities, including conditional and binomial, with Excel. Your guide to precise decision-making!
Knowing how to compute probabilities is a basic competency in the fields of statistics and probability. The ability to compute probabilities is very useful, whether you are a researcher, student, or someone who just wants to make well-informed judgments.
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Probability is essential to problem-solving and decision-making. Probability aids in result prediction and decision-making, from stock market projections to weather forecasts.
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How to Calculate Probability
The number of favorable outcomes must be counted and divided by the total number of potential outcomes in order to compute probability. Using this simple formula, you may determine the probability that an event will occur.
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Probability plays a major role in statistics analysis, conclusion-making, and prediction-making. Any statistical study requires an understanding of probability.
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The possibility of an event happening in light of the occurrence of another event is known as conditional probability. In cases when events are interconnected, it's a useful tool.
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Using Probability in Excel
Strong tools are available in Microsoft Excel for probability calculations. Discover how to use Excel to make complicated probability calculations simpler.
### Calculating the Probability of Multiple Events
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Success or failure are two conceivable outcomes that are dealt with by binomial probability. Discover how to compute binomial probability for a variety of uses.
### Practical Examples of Probability Calculations
We'll give examples of probability estimates from real-world scenarios to help you better understand the idea. Probability is present in everything, from medical diagnosis to gambling chances.
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In risk assessment, result prediction, and decision-making, probability plays a critical role.
### Can probability calculations be done in Microsoft Excel?
Yes, Excel has a number of tools that make it simple to compute probabilities.
### What is conditional probability, and how is it different from regular probability?
Conditional probability is more situation-specific and scenario-dependent since it takes into account occurrences contingent on the occurrence of another event.
### How can I apply probability in a practical scenario, such as in a business context?
Probability is useful for anticipating products, analyzing risks, and improving business plans. | 657 | 3,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-38 | latest | en | 0.94137 |
https://admin.clutchprep.com/physics/practice-problems/38549/a-circular-loop-of-wire-of-radius-l-is-in-a-uniform-magnetic-field-with-the-plan | 1,601,065,784,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228707.44/warc/CC-MAIN-20200925182046-20200925212046-00151.warc.gz | 235,119,008 | 25,389 | # Problem: A circular loop of wire of radius L is in a uniform magnetic field, with the plane of the loop perpendicular to the direction of the field. The magnetic field varies with time according to B(t) = a + bt, where a and b are constants. a) Calculate the magnetic flux through the loop at t = 0. b) Calculate the emf induced in the loop. c) If the resistnace of the loop is R, what is the induced current? d) At what rate is energy being delivered to the resistance of the loop?
###### Problem Details
A circular loop of wire of radius L is in a uniform magnetic field, with the plane of the loop perpendicular to the direction of the field. The magnetic field varies with time according to
B(t) = a + bt, where a and b are constants.
a) Calculate the magnetic flux through the loop at t = 0.
b) Calculate the emf induced in the loop.
c) If the resistnace of the loop is R, what is the induced current?
d) At what rate is energy being delivered to the resistance of the loop? | 235 | 988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-40 | latest | en | 0.921477 |
http://icpc.njust.edu.cn/Problem/Pku/2777/ | 1,604,064,788,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00205.warc.gz | 52,016,442 | 9,987 | # Count Color
Time Limit: 1000MS
Memory Limit: 65536K
## Description
Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board: 1. "C A B C" Color the board from segment A to segment B with color C. 2. "P A B" Output the number of different colors painted between segment A and segment B (including). In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.
## Input
First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.
## Output
Ouput results of the output operation in order, each line contains a number.
## Sample Input
2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2
## Sample Output
2
1
## Source
POJ Monthly--2006.03.26,dodo | 437 | 1,558 | {"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} | 2.765625 | 3 | CC-MAIN-2020-45 | latest | en | 0.922322 |
https://www.studyladder.com.au/games/activity/area-using-square-tiles-28409?backUrl=/games/mathematics/au-all-years/mathematics-area-646 | 1,561,068,423,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999273.79/warc/CC-MAIN-20190620210153-20190620232153-00465.warc.gz | 925,472,416 | 13,636 | ## Area using square tiles
Activity type: Interactive Activity
## Area using square tiles
Course
Mathematics
Year 4
Section
Area
Outcome
Area using square tiles
Activity Type
Interactive Activity
Activity ID
28409
## Testimonials
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## Australia – Australian Curriculum
• ##### Measurement and Geometry
• Using units of measurement
• ACMMG290 – Compare objects using familiar metric units of area and volume
• Shape
• ACMMG087 – Compare the areas of regular and irregular shapes by informal means
## New Zealand – National Standards
• ##### 4.GM – Geometry and measurement
• 4.GM.1 – Measure the lengths, areas, volumes or capacities, weights, and temperatures of objects and the duration of events, reading scales to the nearest whole number and applying addition, subtraction, and simple multiplication to standard units
## United Kingdom – National Curriculum
• ##### Year 4 programme of study
• KS2.Y4.M – Measurement
• Pupils should be taught to:
• KS2.Y4.M.3 – Find the area of rectilinear shapes by counting squares
## United States – Common Core State Standards
• ##### 3.MD – Measurement & Data
• Mathematics
• 3.MD.5 – Recognize area as an attribute of plane figures and understand concepts of area measurement.
• 3.MD.5.a – A square with side length 1 unit, called “a unit square,” is said to have “one square unit” of area, and can be used to measure area.
• 3.MD.7 – Relate area to the operations of multiplication and addition.
• 3.MD.7.a – Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths.
• 3.MD.7.c – Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning. | 484 | 2,044 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-26 | longest | en | 0.872154 |
https://getrevising.co.uk/revision-cards/waves-139 | 1,591,476,892,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348519531.94/warc/CC-MAIN-20200606190934-20200606220934-00168.warc.gz | 359,003,636 | 16,197 | # WAVES
?
• Created by: florrie
• Created on: 23-01-19 09:22
## WAVES DEFINITION
TRANSVERSE = the oscillations are perpendicular to the direction of energy transfer, not all trasnverse waves require a medium
LONGITUDINAL = the oscillations are parallel to the direction of energy transfer, require a medium to travel in
AMPLITUDE = is the maximum dispacement of a point on a wave away from its undisturbed position
WAVE LENGTH = distance from a point on one wave to the equivalent on the adjacent wave (measure from rarefraction to another rarefraction or compressions in longitudinal)
FREQUENCY = number of waves passing a point each second (Hz)
PERIOD = time in seconds for one wave to pass a point (s)
WAVESPEED = speed at which the waves moves through the medium
V = F X (LAMDA)
1 of 6
## ELECTROMAGNETIC WAVES
electromagnetic waves are trasverse (travel through a vacuum not a medium)
all waves travel at same speed through vacuum (3 x 10^8 m/s)
RED = LOWER FREQUENCY + LONG WAVELENGTH
VIOLET = HIGHER FREQUENCY + SHORT WAVELENGTH
radio --- micro --- infrared --- visible light --- ultraviolet --- x rays --- gamma rays
when electromagnetic waves are genrated or absorbed, changes take place in atoms or nuclei
absorbtion of waves can change energy levels, waves can be emitted and absorbed over a wide frequency range
ultraviolet = skin cancer, x-rays and gamma = ionising radiation, mutation of genes (cancer)
radio waves produced when electrons oscillate in electrical circuits, absorbed by aerials but then the electrons in aerial osicillate, creates alternating current with same frequency of RW
2 of 6
## REFRACTION
Waves can change direction when they change speed, moving from one medium to another
light to glass = slows down = bend towards the 'normal'
glass to air = speeds up (velocity) = bend away from the 'normal'
wavefront is an imaginary line that connects all the same points in a set of waves
when passing into glass the front of the wavefront slow down, get closer together = smaller wavelength = change direction to normal
when passing into glass AT NORMAL the whole wavefront slows down, no change in direction
3 of 6
## uses of em waves
RADIO - transmitt radio + terrestrial tv signals (freeview), used because they can trvel long distances before being absorbed, long wavelengths can also spread out between hills, reflect off a layer of charged particles in atmosphere allows us to send radio waves around Earth
MICRO - microwaves oven. communicate with satellites, pass through atmpsohere without being reflected or refracted
INFRARED - cook food in ovens because energy is easily absorbed, heat loss
VISIBLE LIGHT - fibre optics, transmit pulses of light which carry information, telephone and cable TV signals, short wavelength = lots of informations
ULTAVIOLET - energy efficent lightbulbs, contians more energy than viisble light, sun tanning
X RAYS - broken bones,very penetrative, pass through body tissue, absorbed by bones,medical treatment
GAMMA - detect cancers,very penetrative, pass through body tissue, medical treatment
4 of 6
## FORCES
Momentum (kg m/s) = mass x velocity
closed system = before momentum = after momentum
double velocity of a car = kinetic energy quadruples
BRAKING = friction between brake and wheel, themal energy in brakes, temp increases, a large braking force causes overheat and driver loose control
force = mass x acceleration (newtons 2nd law)
stopping distance = thinking distance + braking distance
thinking distance = distance travelled by the car during the driver's reaction time
5 of 6
## force2
1ST = stationary resultant force is 0 then it will be stationary, resultant force on moving object is 0 then object will continue moving with same velocity
2ND = acceleration is proportional to the resultant force and inversely proportional to the mass
inertia mass = a measure of how difficult it is to change the velocity of an object, ratio of the force needed to accelerate and object over the acceleration produced
3RD = whenever two objects interact the forces they exert on each other are equal and opposite
acceleration = change in velocity / time
total area under velocity time graph tell us the displacement
terminal velocity = maximum speed of an object, reached when the forces moving the object are balanced by its fricitonal forces (constant velocity)
6 of 6 | 977 | 4,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-24 | latest | en | 0.852683 |
https://www.coursehero.com/file/8935690/T-rSt-Ers-rBt-ErB-Cov-1T-Correlation-Coefficient-Correlation/ | 1,524,194,457,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937113.3/warc/CC-MAIN-20180420022906-20180420042906-00196.warc.gz | 759,811,625 | 26,024 | {[ promptMessage ]}
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Lecture2_StockPricing
# T rst ers rbt erb cov 1t correlation coefficient
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Unformatted text preview: T Covariance observations): observations): Cov (rs, rB) = 1/T ∑tt=1,..,T [ rSt – E(rs) ] [ rBt – E(rB) ] Cov =1,..,T Correlation Coefficient: Correlation ρ SB Cov(rS , rB ) = σ Sσ B Range of values for ρ: -1.0 < ρ < 1.0 Range ρ: 5-35 Covariance and Correlation Coefficient – probability-based Covariance (with probabilities of K scenarios): Cov (rs, rB) = ∑k=1,..,K pk [ rSk – E(rs) ] [ rBk – E(rB) ] Cov k=1,..,K Correlation Coefficient: Correlation ρ SB Cov(rS , rB ) = σ Sσ B Range of values for ρ: -1.0 < ρ < 1.0 Range ρ: 5-36 Example: Bond and Stock Returns (1) Expected returns Bond = 6% Stock = 10% Standard deviation Standard Bond = 12% Stock = 25% Bond Initial weights Bond = 50% Stock = 50% Correlation coefficient (Bonds and Stock) Correlation = 0, as a simplistic setting (will be changed later) later) 5-37 Example: Bond and Stock Returns (2) Expected return = 8% .5(0.06) + .5 (0.10) =8% Standard deviation = 13.87% [(.5)2 (.12)2 + (.5)2 (.25)2 + 2 (.5) (.12) (.5) (.25) (0)] ½ =13.87% Now, let the weight and correlation Now, change: “2 risky assets line” in “Lecture2_StockPricing” “Lecture2_StockPricing” 5-38 Figure: The risk-return trade-off of the portfolio of 2 risky assets 5-39 Exercise 3 1. Plot the risk-return relation for the 1. following two assets: following • Expected returns Bond = 5% Stock = 7% • Standard deviation Standard Bond = 8% Stock = 12% Bond • Correlation coefficient = -0.5 2. What’s the minimum volatility we can 2. reach using these two assets? Ans: 4.77% reach 5-40 More than 2 assets – we will discuss when we use Matlab 5-41 CAPM model and market risk 5-42 Specification of CAPM Model So, the relation between individual stock r i and the market portfolio rm can now be stated as: market can E(ri) - rf = α + β [E(rm) - rf ] or rit - rft = α + β [ rmt - rft ] + εit or – Note that, we need to consider the risk-free rate Note – β measures “market risk exposure” (“market beta”) of (“market stock i stock – α measures “abnormal returns”...
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https://www.askiitians.com/forums/Analytical-Geometry/what-is-the-equation-of-a-common-tangent-to-two-gi_202107.htm | 1,726,153,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651460.54/warc/CC-MAIN-20240912142729-20240912172729-00244.warc.gz | 604,990,192 | 43,059 | what is the equation of a common tangent to two given parabolas?
Arun
25750 Points
6 years ago
Hey Aman.
I think this will work.
Assume a line y = mx + c touches the two given curves.
Hence, when you take one of the curves, and substitute the value of y, you will get a quadratic equation in x which can have only one solution. Because this quadratic equation contains the terms m and c, you can use b^2 - 4ac = 0. This will get you one equation in m and c. Now , take the second curve and repeat the same thing. This gives you another equation in m and c.
Take these two equations and solve them. Depending on the combinations of the curves, it may be very simple or very lengthy.
You may get upto four tangents that way, assuming that the equations you get are quadratic in both m and c. | 192 | 796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-38 | latest | en | 0.93946 |
https://tombewley.com/posts/2020/05/weekly-readings-26/ | 1,685,287,056,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00560.warc.gz | 627,389,440 | 12,978 | Published:
## đ Papers
### Atkeson, Christopher G., and Stefan Schaal. Robot Learning From Demonstration, 1997.
This paper describes implementation of a model-based kind of imitation learning to perform the pendulum swing-up task by acceleration-based control of the horizontal position $x$ of a physical robot hand. The pendulum starts at angle $\theta=\pi$ and a successful swing-up moves it to $\theta=0$.
The approach makes use of a task model and a imitation-based cost function to be used by an optimal control planner. For the task model, the authors use a discrete-time parametric model of the angular velocity of an idealised pendulum attached to a horizontally-moving hand:
$\dot{\theta}_{k+1}=\left(1-\alpha_{1}\right) \dot{\theta}_{k}+\alpha_{2}\left(\sin \left(\theta_{k}\right)+\ddot{x}_{k} \cos \left(\theta_{k}\right) / g\right)$
The two parameters $\alpha_1$ and $\alpha_2$, corresponding to the viscous damping and $\Delta g/l$ respectively ($\Delta$ is the timestep), are learned from data by linear regression. An alternative nonparametric locally-weighted learning model, to predict $\dot{\theta}{k+1}$ as a function of $(\theta{k}, \dot{\theta}{k}, x{k}, \dot{x}{k}, \ddot{x}{k})$, is also learned.
These models are used by an optimal control planner to minimise a cost function that penalises deviations from a demonstration trajectory and squared acceleration:
$r\left(\mathbf{x}_{k}, \mathbf{u}_{k}, k\right)=\left(\mathbf{x}_{k}-\mathbf{x}_{k}^{\mathrm{d}}\right)^{\mathrm{T}}\left(\mathbf{x}_{k}-\mathbf{x}_{k}^{\mathrm{d}}\right)+\mathbf{u}_{k}^{\mathrm{T}} \mathbf{u}_{k}$
where the state is $\mathbf{x}=(\theta, \dot{\theta}, x, \dot{x})$, $\mathbf{x}^{\mathrm{d}}$ is the demonstrated motion, $k$ is the sample index, and the control is $\mathbf{u}=(\ddot{x})$.
With both parametric and non-parametric models, this approach successfully imitates human demonstrations with one back-and-forth motion of the pendulum, but it fails at the more difficult task of âpumpingâ the pendulum twice before swinging it up.
It is suggested that this is likely due to a mismatch between the model structure and the true system. To combat this, an additional free parameter is introduced: allowing the target angle used by the planner to vary away from $\theta=0$. Searching across a range of values identifies a value which allows the more complex task to be learned.
### Bhattacharyya, Raunak P., Derek J. Phillips, Changliu Liu, Jayesh K. Gupta, Katherine Driggs-Campbell, and Mykel J. Kochenderfer. âSimulating Emergent Properties of Human Driving Behavior Using Multi-Agent Reward Augmented Imitation Learningâ. In ArXiv:1903.05766 [Cs], 2019.
Both imitation learning and reinforcement learning are challenging and unstable in multi-agent domains where undesirable emergent properties may develop. The authors propose to tackle this problem by augmenting the generative adversarial imitation learning (GAIL) approach with prior knowledge in the form of hand-crafted rewards to disincentivise visiting a set of undesirable state-action pairs $U$.
GAIL uses an adversarial two-model setup, whereby a discriminator network $D_\psi$ is trained to output a high score given state-action pairs from a demonstration policy $\pi^\ast$, and a low score given pairs from a learner policy $\pi_\theta$.
Rather than applying the reward augmentation as a hard constraint (which makes optimisation much more difficult), it is added as a regularisation term to the GAIL objective:
$\underset{\theta}{\min}\ \underset{\psi}{\max}\ \mathbb{E}_{s,a\sim d_{\pi}^\ast}\left[D_{\psi}(s,a)\right]-\mathbb{E}_{s,a\sim d_{\pi_\theta}}\left[D_{\psi}(s,a)\right]+r \mathbb{E}_{\pi_{\theta}}\left[\mathbf{1}_{U}\right]$
This objective is solved in two iterative steps:
• Optimise the discriminator parameters $\psi$ using data from new rollouts.
• Use Trust Region Policy Optimisation to update the policy parameters $\theta$. Part of the loss comes from the discriminator, and part comes from the reward augmentation term.
In the context of a traffic simulator environment with real-world training data, one reward augmentation scheme (âbinaryâ) consists of a large penalty $R$ if a vehicle crashes or drives off the road, and a smaller one $R/2$ if it performs a hard brake. An alternative (âsmoothâ) uses a linearly-increasing penalty as a function of distance to the road edge or braking magnitude.
RAIL is compared with GAIL on various metrics: positional divergence of vehicles from the training data, error rate (collision, off-road, hard-brake), and frequency of emergent phenomena (lane changes, time gaps between vehicles) compared with the training data. During these evaluations, a certain proportion of the vehicles use the learned policy, and the rest follow the training data. RAIL is clearly seen to do better.
### Brennen, Andrea. âWhat Do People Really Want When They Say They Want âExplainable AIâ? We Asked 60 Stakeholders.â, 2020, 7.
Here the author summarises two key findings from stakeholder interviews.
Lack of consistent terminology:
• Identified synonyms for âExplainableâ included âAccountableâ, âFairâ, Reliableâ and âVerifiableâ. Some interviewees used terms interchangeably that others insisted were very different.
• Similarly, the two terms âInterpretabilityâ and âExplainabilityâ were used by some to mean the same thing, and others to describe two alternative approaches. What those alternatives actually were also varied!
• Finally, even âAIâ meant different things to different people. Some were interested only in deep learning models, while others extended their definition to incorporate nearly any automated manipulation of data.
• Miscommunication is exacerbated when words have both technical and nontechnical definitions, e.g. âFairnessâ.
Multiple motivations for explainability:
• Most people started with the notion of AI as a black box, and that the existence of a mismatch between what they wanted to understand and what they currently understood.
• But there are at least three very distinct variants:
• Debugging models: explainability as a tool to be used by engineers for improving performance.
• Identifying bias: explainability to identify problematic data or inferences, important from business and legal perspectives.
• Building trust: explainability to help end-users understand unfamiliar technologies and assuage their anxiety.
• With different explainees and different content requirements, it seems reasonable to expect weâll need different solutions.
Identifying these issues, which hinder discussion and mutual understanding, is the first important step toward mitigating them.
### Elton, Daniel C. âSelf-Explaining AI as an Alternative to Interpretable AIâ. ArXiv:2002.05149 [Cs, Stat], 24 April 2020.
The double descent phenomenon indicates that deep neural networks typically operate by interpolating between data points rather than by extracting a few high level rules as we might have hoped. This is a âdumbâ process: any apparent regularities that appear to be captured internally are properties of the data rather than a self-directed distillation process.
In addition to making failure likely outside the training distribution, this behaviour makes interpretation (which makes the implicit assumption that such latent rules exist) inherently hard. With this in mind, how can we ever hope to trust AI systems? Instead of post-hoc methods, Elton advocates for self-explaining AI models. In addition to the prediction itself, a self-explaining model would output:
• A human-understandable explanation;
• Confidence levels both both prediction and explanation.
Prediction and explanation branches from a shared set of primary layers. How can we trust that the explanation is actually relevant? One possible option is to measure the mutual information between the prediction $Y$ and the activation vector of the final shared latent layer $L$, and also between the explanation $E$ and $L$:
$\text{Relevance}(E,Y)\doteq\sum_{i=1}^{\vert L\vert}\text{MI}(L_i,Y)\text{MI}(L_i,E)$
A disadvantage of this is that mutual information is extremely expensive to compute!
A couple more points made during the discussion:
• The term âblack boxâ should not be used as often as it is, because given enough time and effort we can know exactly whatâs inside a neural network, and how it got there. The issue is that we donât have a good way of relating this knowledge to questions we actually care about.
• Given that deep neural networks seem to operate using dumb interpolation, there seems to be a direct conflict between the accuracy of an interpretation, and its relevance to the domain of application.
## đ Theses
### Ross, Stephane. âInteractive Learning for Sequential Decisions and Predictions.â (2013).
Learning actively through interaction is necessary to obtain good and robust predictors for sequential prediction tasks. No-regret online learning methods provide a useful class of algorithms to learn efficiently from these interactions, and provide good performance both theoretically and empirically.
This thesis introduces interactive learning methods that address the major limitation of behavioural cloning: the sequential nature of control problems breaks the i.i.d. assumption. Any open-loop error causes the test distribution to diverge from the training one in the closed-loop setting, and poor performance results.
The ideal solution is to train on the distribution that the learned predictor will see in closed-loop, but this is a chicken-and-egg problem! A common strategy to deal with such problems is to adopt an iterative training approach, and this is what is done here. Overall, the effect is to reduce imitation learning (and related problems) to online learning.
#### Learning Behaviour from Demonstrations
Here we start to tackle the problem of learning good control policies from demonstration. All analysis assumes an episodic MDP of fixed length $T$, but does generalise to discounted infinite-horizon tasks. Performance is quantified in terms of a non-negative cost to be minimised $C(s,a)\in[0,C_{\max}]$, rather than reward $R(s,a)$ which may be positive or negative. This simplifies much of the analysis.
Our ultimate goal is to minimise $J(\pi)$, the expected sum of costs accumulated by $\pi$ over $T$-timestep episodes under the induced state distribution $d_\pi$. Since this is challenging to do directly, we introduce a surrogate loss $\ell(\pi,\pi^\ast)$ that quantifies some measure of deviation from a high-performing expert policy $\pi^\ast$. In the special case when we are interested in the learnerâs ability to predict the expertâs actions, $C=\ell$.
Behavioural cloning is the most naive solution to this problem, which simply performs supervised learning on a training distribution generated by $\pi^\ast$. $\epsilon$ the expected loss under this distribution. It can be proven that if $\ell$ is the 0-1 loss (or an upper bound such as the hinge or logistic loss), then $J(\pi) \leq J\left(\pi^{*}\right)+C_{\max } T^{2} \epsilon$, meaning the closed-loop cost of $\pi$ compared with $\pi^\ast$ grows quadratically with $T$. This is much worse than the i.i.d. setting, where the cost of classifying $T$ samples grows linearly with $T$.
What follows is a selection of approaches for closing this gap.
##### Forward Training
As a simple starting point, we can imagine training a sequence of policies $\pi_1,âŚ,\pi_T$ in order. For each $\pi_t$, we sample multiple $t$-step trajectories, starting in the initial state distribution and continued by executing $\pi_1,âŚ,\pi_{t-1}$. This produces a distribution of states for time $t$. We query the expert for these states and use the results to train $\pi_t$.
By training sequentially in this way, each $\pi_t$ is trained under the distribution of states itâs going to encounter during closed-loop execution. If previous policies produce poor predictions, then $\pi^\ast$ will demonstrate the necessary recovery behaviours at future steps.
Iterative forward training attains closed-loop loss that grows linearly with $T$ (the best possible case), but it is impractical when $T$ is large (or the MDP is continuing), and inefficient in that the same or similar behaviours may have to be learned many times over for different timesteps.
##### Stochastic Mixing Training
The two approaches reviewed here achieve similar guarantees by adopting the same general philosophy of learning gradually.
SEARN works by starting from a policy $\pi_0=\pi^\ast$, i.e. one that just queries the expert. A new policy $\hat{\pi}_1$ is trained to minimise the surrogate loss $\ell$ on data from the execution of $\pi_0$. This is then stochastically mixed with $\pi_0$ to obtain the next policy
$\pi_1=(1-\alpha)\pi_0+\alpha\hat{\pi}_1$
which at every timestep $t$, follows $\hat{\pi}_1$ with probability $\alpha$, and follows $\pi_0$ otherwise. It keeps iterating in a similar fashion, so that
$\pi_{n}=(1-\alpha) \pi_{n-1}+\alpha \hat{\pi}_{n}=(1-\alpha)^{n} \pi_{0}+\alpha \sum_{i=1}^{n}(1-\alpha)^{n-i} \hat{\pi}_{i}$
After some large number of iterations $N$, SEARN terminates and returns a final policy $\pi_N$ that never has to query $\pi^\ast$ by renormalising:
$\pi_{N}=\frac{\alpha}{1-(1-\alpha)^{N}} \sum_{i=1}^{N}(1-\alpha)^{N-i} \hat{\pi}_{i}$
By default, SEARN involves simulating an entire trajectory to collect each training datapoint, which is often impractical. A variant, called SMILE, addresses this issue. It keeps the same update relationship between policies, mediated by $\alpha$, but at each step $n$ the training is simply to imitate the expert $\pi^\ast$ on the latest dataset. Hence each $T$-step trajectories yields $T$ datapoints for training instead of one. However, SMILE has less strong performance guarantees than SEARN or forward training.
Both SEARN and SMILE also have the disadvantage of requiring a stochastic policy which might be undesirable in practical (safety critical) applications.
##### Dataset Aggregation
This approach, the flagship of the thesis, is shortened to DAgger. It can learn deterministic policies, and is even simpler than the preceding methods.
DAgger starts by gathering a dataset $\mathcal{D}$ of expert-generated trajectories, trains a policy $\pi_2$ to imitate the expert on those data. Then at iteration $n$, it proceeds by collecting a dataset at each iteration under $\pi_n$, appends this to $\mathcal{D}$, and trains $\pi_{n+1}$ on the concatenation of all collected datasets. It is a kind of follow the leader algorithm in that at each iteration we find the policy that would give the best imitation in hindsight on all previous data.
This basic algorithm can be modified to use a mixture policy at each iteration $n$, that samples from either $\pi_n$ or $\pi^\ast$ according to a parameter $\beta_n$. In practice however, it is found that itâs often best to just set $\beta_1=1$ and $\beta_n=0$ for all $n\neq1$.
DAgger works with any âno-regretâ online learning algorithm.
A couple of questions:
• Why not just collect data everywhere, e.g. via a random or noisy policy? Because (1) itâs very data-inefficient, and (2) when the policy class $\Pi$ is not expressive enough to fully realise the target policy, we need to trade-off accuracy in different regions correctly fitting irrelevant data may actually harm performance.
• Why not just use the last collected dataset as in policy iteration? This tends to be very unstable and leads to oscillation between multiple mediocre policies. When we constantly append to $\mathcal{D}$, the fraction of new data decreases over time, so the policy will tend to change more and more slowly, and stabilise around some behaviour that produces good predictions under inputs that were collected by similar predictors. We get increasingly close to the âidealâ of matching training and test distributions.
#### Learning Behaviour using Cost Information
Now consider the case where an expert is present, and additional cost information is available. Intuitively, this should enable better performance!
##### Forward Training with Cost-to-Go
Instead of only observing the expertâs actions in states reached at time $t$ by execution of $\pi_1,âŚ,\pi_{t-1}$, we instead explore actions to perform at $t$, follow $\pi^\ast$ until the end of the episode, and observe the total cost $Q$ of this sequence. Under the assumption that $\pi^\ast$ is a decent policy and that $\Pi$ contains similar good policies, then these $Q$ values give us a rough estimate of what good policies in $\Pi$ will be able to achieve at future steps.
$\pi_t$ is trained to minimise cost on the collected dataset of $(s_t,a_t,Q)$ triples. If we are lucky enough to be able to sample all actions from each $s_t$, we have all the information we need for cost-sensitive classification learning. Otherwise, we can reduce the problem to a regression one (predicting cost for a state-action pair) or use importance weighting techniques.
To select exploring actions from an $s_t$, we can simply use a random strategy. Better results might be attained by casting this as a contextual bandit problem. Developing efficient algorithms for this problem is still an open problem.
In cases where the expert is much better than any policy in $\Pi$, then the cost estimates may be very optimistic, and not reflective of the true prospects of the learner. This will not lead to good performance.
##### DAgger with Cost-to-Go
A similar idea is used here. At iteration $n$, we collect each of $m$ sample as follows:
• Follow current policy $\pi_n$ from the start of the episode to a randomly-sampled time $t\in{1..T}$.
• Execute some exploration action $a_t$ in $s_t$.
• Follow $\pi^\ast$ from $t+1$ to $T$, observing the total cost $Q$ starting at $t$.
After adding the $m$ new samples to $\mathcal{D}$, we train $\pi_{n+1}$ as a cost-sensitive classifier.
By minimising cost, this algorithm is effectively a regret reduction of imitation learning, rather than an error reduction as obtained when minimising classification loss in conventional DAgger.
Utilising cost-to-go can often be expensive and impractical compared with simple imitation loss minimisation. A potential combination of the two approaches could be very effective.
## đď¸ Key Insights
• Atkeson and Schaalâs classic paper, oft-cited in the imitation learning literature, takes quite a heavily model-based approach compared with more recent works.
• That said, Bhattacharyya et alâs much newer RAIL model makes use of domain knowledge, in the form of hand-crafted penalties for known failure modes, added as an extra term to the GAIL objective.
• Brennenâs review lays bare the concerning lack of consistency or rigour in uses of words such as âinterpretabilityâ and âexplainabilityâ. With so many potential applications in mind, it might be that both terms are just too broad to be meaningful.
• Eltonâs proposal for self-explaining AI is to me less interesting than some of the high-level observations he makes along the way. The term âblack boxâ is really inappropriate when what is missing is not knowledge of a neural networkâs internal representations, but a way of relating them to issues we care about. And this seems doubly challenging if we believe the evidence that such representations encode mere interpolation rules rather than high-level regularities.
• Rossâ thesis gives compelling arguments for the critical importance of interaction when doing imitation learning in dynamic contexts. His DAgger algorithm is an easy-to-implement scheme for implementing this interaction.
Tags: | 4,580 | 19,762 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | longest | en | 0.727281 |
https://webapps.stackexchange.com/questions/51715/how-to-do-summax | 1,723,584,396,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00189.warc.gz | 478,313,191 | 42,138 | # How to do sum(max(...))?
Here is my raw data:
`````` Site IP Visits Age (in hours)
1 uu.com 1.2.3.4 10 3
2 vv.com 2.9.7.5 2 4
3 kk.com 2.9.7.5 8 24
4 uu.com 1.2.3.4 12 24
5 vv.com 2.9.7.5 10 24
6 kk.com 2.9.7.5 8 48
``````
`````` Col. A Col. B MAX of Visits MAX of Age
1 1.2.3.4 uu.com 12 24
2 2.9.7.5 kk.com 8 48
3 vv.com 10 24
``````
`````` Col. A Col. B MAX of Visits MAX of Age
1 1.2.3.4 uu.com 12 24
2 Total for 1.2.3.4 12 24
3 2.9.7.5 kk.com 8 48
4 vv.com 10 24
5 Total for 2.9.7.5 10 48
``````
This is not what I'm expecting because I want the SUM('Max of visits') as total. Clearly, this is the final pivot table I want:
`````` Col. A Col. B MAX of Visits MAX of Age
1 1.2.3.4 uu.com 12 24
2 Total for 1.2.3.4 12 24
3 2.9.7.5 kk.com 8 48
4 vv.com 10 24
5 Total for 2.9.7.5 18 48
``````
Notice on line `5` , column `'Max of Visits'`, I want 18 and not 10 as Google Spreadsheet gives me.
How can I acheive that ?
Perhaps you can use the following formula.
## Formula
``````=QUERY(DATA!C1:F7, "SELECT C, MAX(E), MAX(F) GROUP BY C PIVOT D")
``````
## Remark
In order to get the summation, I simply added a summation for the consecutive columns (see example).
## Example
I've created an example file for you: SUM and MAX in PIVOT TABLE
• Tks for your useful answer.Actually, I'd like to perform this: `SELECT D,C,SUM(MAX(E)) FROM <DATA> GROUP BY D,C` Commented Nov 15, 2013 at 22:37
• @Alex The <DATA> is contained in the SELECT statement....Quote from Google Query Language: `The "from" clause has been eliminated from the language.` Commented Nov 15, 2013 at 22:40
• You're right, but I have tried this `SELECT D,C,SUM(MAX(E)) GROUP BY D,C` and got `#VALUE` error. Commented Nov 15, 2013 at 22:50
• @Alex Haha, added a summation below the result (didn't you see that....haha). There's no other way. I will edit my answer. Commented Nov 15, 2013 at 22:52
• I did see the summation ;) Commented Nov 15, 2013 at 23:03
Try this:
``````=MAX(Start Cell:End Cell)
``````
Here is a demo: | 860 | 2,473 | {"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} | 2.96875 | 3 | CC-MAIN-2024-33 | latest | en | 0.802276 |
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Calculating the Put/Call Ratio for Delta Neutral ?
Discussion in 'Options' started by Aston01, Jun 5, 2012.
1. Aston01
I am trying to calculate a delta neutral position. Most of the time it seems the size of these positions are based on the number of shares someone is trying to hedge and the cost is whatever it ends up being to create the hedge. I am trying to calculate it for a trade the other way around, assuming I want to buy \$5k worth of calls and puts based on the delta and price of each, how many of each side I need to purchase to stay delta neutral and as close to \$5k as possible.
Assuming I was buying a \$5k combination of the following:
Call Option
Price - \$.59 each
Delta - .0759
Put Option
Price - \$.60 each
Delta - .0544
=5000*0.0759/(0.0759+0.0544)
which equals \$2,912.51
and
=5000*0.0544/(0.0759+0.0544)
which equals \$2,087.49
I end up with a Call Delta of 265.65 & Put Delta of 261.12 ... 4.53 Difference ...tolerable for my purposes.
My thought process was that if I could determine the relative difference between the 2 delta's I could determine how much of the \$5k needed to be allocated to each side. The above calculation achieved just that, but I am missing something it seems. When I put in a second set of values like below things didn't balance as effectively.
2nd Set of Values
Call Option
Price - \$.82 each
Delta - .1265
Put Option
Price - \$.50 each
Delta - .0587
=5000*0.1265/(0.1265+0.0587)
which equals \$3,415.22
and
=5000*0.0587/(0.1265+0.0587)
which equals \$1,584.77
I end up with a Call Delta of 240.35 & Put Delta of 399.16 ... -158.81 difference ... Not so good
Any ideas where I messed this one up ?
2. 1245
You made this very complicated. Delta is used to calculate your equivalent share position. Your using dollars in your calculation. If you only want to use options in a spread and not hedge with stock, try inverting the delta with a spread to achieve delta neutral.
In your example, I believe you're buying a combination of an OTM puts and an OTM calls. Call delta .0759, put delta .0544.
Calls 544
puts 759
or stay in that ratio, you'll be delta neutral. The option price will not matter in your example.
3. RPEX
I don't get it, what is the objective? Are you trying to hedge something?
4. heiasafari
I think you should be more specific in what you are trying to acheive... There are many option strategies that are delta neutral... Also who says you need to use both call and puts, you could use just 1 of the 2, again depending on your purpose... Also you can be delta neutral and long vol (like a calendar) or neutral and short vol (iron condor for example). Many ways to be delta neutral yet used in very different circonstances...
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# GENERALIZED PASCAL’S TRIANGLE AND METALLIC RATIOS
Independent Research Scholar, California Public Univeristy, USA
How to cite this article (APA): Sivaraman, R (2021). Generalized pascal’s triangle and metallic ratios. International Journal of Research - GRANTHAALAYAH, 9(7), 179. doi: 10.29121/granthaalayah.v9.i7.2021.4109
# Abstract
In this paper, I had demonstrated the way to determine the sequence of metallic ratios by generalizing the usual Pascal’s triangle. In doing so, I found several interesting properties that had been discussed in detail in this paper. I had proved four new results upon generalizing Pascal’s triangle. Thus, the primary aim of this paper is to connect the idea of Generalized Pascal’s triangle with that to the sequence of metallic ratios.
Keywords
Generalized Pascal's Triangle, Metallic Ratio of Order K, Binomial Coefficient, Hockey Stick Property, Binet's Formula Respresentation
## INTRODUCTION
Though the concept of Pascal’s triangle became notable among mathematical world through Pascal, the concept of constructing the triangle was well known to ancient Indian and Chinese mathematicians. Similarly, the sequence of metallic ratios has been used in almost all branches of Science and Engineering. In this paper, I had connected these two concepts and derived some interesting results related to them.
## DEFINITION
Let k be a positive integer. The sequence of Metallic ratios of order k is defined recursively by ${M}_{n+2}=k{M}_{n+1}+{M}_{n}\left(2.1\right),n\ge 1$ where ${M}_{0}=0,{M}_{1}=1,{M}_{2}=k$ The terms of the sequence defined by (2.1) are given by $0,1,k,{k}^{2}+1,{k}^{3}+2k,{k}^{4}+3{k}^{2}+1,{k}^{5}+4{k}^{3}+3k,{k}^{6}+5{k}^{4}+6{k}^{2}+1,...$ (2.2)
Notice that for k = 1, the above sequence is the usual Fibonacci sequence.
### METALLIC RATIOS OF ORDER K
Using the shift operator, the recurrence relation in (2.1), yield the quadratic equation ${m}^{2}-km-1=0$ The two real roots of this quadratic equation are given by $m=\frac{k±\sqrt{{k}^{2}+4}}{2}$ The positive value among these two roots is defined as the metallic ratio of order k denoted by ${\rho }_{k}$ Thus, ${\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2}$ (2.3) Since the sum of two roots is k, the other root is $k-{\rho }_{k}=\frac{k-\sqrt{{k}^{2}+4}}{2}$ (2.4)
## CONSTRUCTION OF GENERALIZED PASCAL’S TRIANGLE
Using the same rule of construction as that of usual Pascal’s triangle, I now construct a generalized Pascal’s triangle from which we can explore several properties. Let k be a positive integer. Consider a triangle displayed below.
We notice that each entry in the Generalized Pascal’s triangle of Figure 2 , is obtained using the recurrence relation defined in (2.1). In particular, if ${T}_{n,r}$ is the rth entry in $\eta$ th row, where
If k = 1, then (3.1) reduces to the definition of usual Pascal’s triangle and Figure 2 would then become the well known Pascal’s triangle. From Figure 2 , we notice that the rth entry in nth row where $0\le r\le n$ of the Generalized Pascal’s triangle in Figure 2 is given by where $\left(\begin{array}{c}n\\ r\end{array}\right)$ is the binomial coefficient.
### THEOREM 1
With respect to Generalized Pascal’s triangle of Figure 2 , we have the following properties
Proof:
The sum of all entries in nth row of Generalized Pascal’s triangle is given by
${\sum }_{r=0}^{n}{T}_{n,r}={\sum }_{r=}^{n}\left(\begin{array}{c}n\\ r\end{array}\right){k}^{n-r}={\sum }_{r=0}^{n}\left(\begin{array}{c}n\\ r\end{array}\right){k}^{n-r}{1}^{r}=\left(k+1{\right)}^{n}$ . This proves (3.3)
Using (3.2), we have
This proves (3.4)
Using (3.2) , we have
This proves (3.5) and hence completes the proof.
## METALLIC RATIOS FROM GENERALIZED PASCAL’S TRIANGLE
In this section, I demonstrate a method to obtain sequence of metallic ratios of order k as defined in (2.2) from the generalized Pascal’s triangle described in Figure 2 . First, we will rearrange the terms of the triangle in right triangle pattern as shown below.
If we add identical colored terms in the re-arranged triangle in North-East diagonal direction then we get
We notice that the terms of above sequence precisely forms the terms of sequence of metallic ratios of order k as defined in (2.2). Thus, the triangle in Figure 6 , generates sequence of metallic ratio of order k.
### THEOREM 2
The ratio of (n+1) th term to that of nth term of sequence of metallic ratios of order k converges to the number ${\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2}\left(4.1\right)$
Proof: From the recursive relation of metallic ratio of order k as defined in (2.1), notice that for some real numbers we get ${M}_{n}=\alpha {{\rho }_{k}}^{n}+\beta {\left(k-{\rho }_{k}\right)}^{n}\left(4.2\right)$ where ${\rho }_{k}$ and $k-{\rho }_{k}$ are the real numbers given by ${\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2},k-{\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2}$ Equation (4.2) is referred as Binet’s formula representation for sequence of metallic ratios of order k
We now notice that $-1 for all positive integers k. Hence ${\left(k-{\rho }_{k}\right)}^{n}\to 0\left(4.3\right)$ as $n\to \infty$ . Thus using (4.2) and (4.3), we obtain
$\underset{n\to \infty }{lim}\frac{{M}_{n+1}}{{M}_{n}}=\underset{n\to \infty }{lim}\frac{\alpha {{\rho }_{k}}^{n+1}+\beta {\left(k-{\rho }_{k}\right)}^{n+1}}{\alpha {{\rho }_{k}}^{n}+\beta {\left(k-{\rho }_{k}\right)}^{n}}=\underset{n\to \infty }{lim}\frac{\alpha {{\rho }_{k}}^{n+1}+0}{\alpha {{\rho }_{k}}^{n}+0}={\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2}$
This completes the proof.
### GOLDEN, SILVER AND BRONZE RATIOS
As mentioned in 2.2 of section 2, we notice that the Golden, Silver and Bronze ratios are special cases of ${\rho }_{k}$ k = 1, 2, 3. Hence if we consider the triangle in Figure 2 , with the values k = 1, 2 and 3 we can generate the terms of sequence of metallic ratios of orders 1, 2 and 3 respectively whose ratio of successive terms approach to ${\rho }_{1}=\frac{1+\sqrt{5}}{2},{\rho }_{2}=1+\sqrt{2},{\rho }_{3}=\frac{3+\sqrt{13}}{2}$ respectively. Thus we can produce number triangles which are generalized versions
## CONCLUSION
In this paper, by defining a generalized Pascal’s triangle, first, I had established three interesting properties in Theorem 1. The third property proved in (3.5) is the generalized Hockey stick property which works only if we could sum numbers in South – East diagonal direction unlike in usual Pascal’s in which Hockey stick property works in either South – East diagonal way or in South – West diagonal way owing to its symmetrical entries. In the generalized Pascal’s triangle discussed in this paper in Figure 2 , by (3.4), since the entries are not symmetrical with respect to the central vertical line, we get Hockey stick property only along South – East direction.
Further, I had obtained the terms of sequence of metallic ratios of order k through the generalized Pascal’s triangle in section 4. Finally, I had proved that the ratio of successive terms of sequence of metallic ratios is precisely the metallic ratio of order k given by ${\rho }_{k}$ . Thus, in this paper, I had established some new properties by generalizing the usual Pascal’s triangle and in doing so, I had obtained the terms of metallic ratio sequence. By considering k = 1, 2, 3 in Figure 2 , we can generate three new triangles from which produces three sequences as given in (2.2), whose ratio of successive terms approaches to Golden, Silver and Bronze ratios respectively, through the generalized Pascal’s triangle, we could generate sequences whose ratio of successive terms converges to the metallic ratio of order k given by ${\rho }_{k}=\frac{k+\sqrt{{k}^{2}+4}}{2}$ The ideas discussed in this paper will add more information to the study of metallic ratios and pave way for further explorations. | 2,227 | 7,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 26, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.888796 |
https://scoop.eduncle.com/uestion-if-the-demand-function-is-p-400-4x-what-will-be-maximum-tr-and-when-options-10-000-lnit-at-x | 1,610,841,499,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507971.27/warc/CC-MAIN-20210116225820-20210117015820-00018.warc.gz | 558,342,849 | 13,182 | UGC NET Follow
August 22, 2020 7:48 pm 30 pts
uestion If the demand function is p = 400-4x, what will be maximum TR and when ? Options : 10,000 Lnit at x= 50 unit 10,000 Taf x= 50 sa ET 1 20,000 unit 20,000 gT 2. 15,000 unit 15,000 gT 1,900 unit at x = 5 unit 41,900IR TAfar x= 5 zT E 4. | 124 | 287 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.605869 |
https://dsp.stackexchange.com/questions/72887/what-is-the-algorithm-to-generate-sine-waves-of-arbitrary-frequency-in-the-stft | 1,621,349,334,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00487.warc.gz | 212,067,549 | 38,388 | # What is the algorithm to generate sine waves of arbitrary frequency in the STFT domain?
I'd like to write a DSP algorithm to do additive synthesis using arbitrary sine waves with inverse rectangular FFTs.
This requires two things:
1. The ability to generate phase/amplitude lists that cause the IFFT to create sine waves that are not integer multiples of 1/(IFFT length)
Example: How do I generate 1000 Hz with 16 sample IFFTs at 44100 Hz?
1. The ability combine two frames pre-IFFT so they are added post-fft.
What are the necessary steps here?
• This can be done but it's tedious and inefficient. There are some extremely efficient oscillator algorithms that work in the time domain. What's wrong with using those ? – Hilmar Jan 30 at 0:18
• I'm working in wavetable synthesis and I'm sort of assuming that if I pile up enough oscilators in the fft domain it'll outperform a typical additive synth. Note: The oscillators here would be of stable frequency but not amplitude. – Audiomatt Jan 31 at 3:22
• what is the connection you have between wavetable synthesis and sinusoidal modeling with the STFT? i don't see them as directly related to each other. – robert bristow-johnson Feb 5 at 19:23
• My first choice would be to use a simple NCO (Numerically Controlled Oscillator) to generate any arbitrary sinewave with extremely high precision, fidelity, ability for nearly instantaneous frequency change and minimum resources. Have you ruled this out? dsp.stackexchange.com/questions/37803/… – Dan Boschen Mar 7 at 21:06 | 359 | 1,530 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-21 | latest | en | 0.905585 |
https://www.mail-archive.com/everything-list@googlegroups.com/msg41596.html | 1,532,025,112,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591150.71/warc/CC-MAIN-20180719164439-20180719184439-00327.warc.gz | 926,094,273 | 4,609 | # Re: Step 3
```
On 30 Oct 2013, at 10:31, LizR wrote:```
```
```
```On 30 October 2013 19:03, Jason Resch <jasonre...@gmail.com> wrote:
```
My point was only that the traditional notions of personal identity: saying this person is that one particular continuation of that biological organism, or of that one brain, do not work. They fail in cases of fusion, fission, duplication, radical change, amnesia, etc. and must be rejected in favor of more consistent definitions of personal identity.
```
```
That is exactly what comp does, and that is at least part of the point of the teleportation thought experiments. One of the results of comp is that personal identity is split into steps, normally called observer moments (the length of these moments isn't known),
```
```
Those are the computational steps ("3-OM"). Up to step 6, we can take the quasi-identity thesis, for the purpose of the reasoning (it does not really work with neither Newton, and is an open problem with QM, but with comp step seven shows a similar "problem").
```
```
and that personal survival from moment to moment is exactly the same as survival during a duplication experiment. In comp, at least, a person is a series of discrete states, a "Capsule theory" of memory and identity rather like the pigeonholes in Fred Hoyle's "October the First is too late".
```
```
Yes, but to belong to a computation, means that there is a universal machine which do the computation. It is a computable sequence, and computer science "laws" applies (it is not trivial). All states are related by universal machines, and the first person states, eventually, by infinities of universal machine.
```
```
If I send your Gödel number (a scanning of you) in an alien galaxy, they will only be able to reconstituted you if we succeed in betting on a universal interpreter. Universal code exists, but we need to bet on some inference inductive ability, and interest from the Alien.
```
```
So a person is not a series of discrete states, but a series of state computed by a universal machines. A series of discrete states can emulate a person, when it belongs to some genuine computations which involve some universal machine, in the 3p, and an infinity of universal machines in the 1p.
```
```
```
I'm sure Bruno will correct me if I have got anything wrong there.
```
```
```
Again, not wrong, but perhaps slightly unclear. (explaining the deaf short deaf dialog with Jason, perhaps).
```
```
All "pronouns" difficulties are cleared by taking literally the first person and third person notion defined in term of diaries taken or not taken in the teleportation boxes. By definition of comp, the doctor does not have to solve the "who am I " problem. Just to bet on the right genuine functionality level of the brain.
```
```
All this thread illustrates that the "personal identity" (Who am I) problem is orthogonal to the first person experience predictibility problem in self-duplication.
```
```
Comp makes it clear, that we survive in both city, from a 3p view, but that personally, we feel to survive in only one city. We can't predict which one in advance, as most copies will see that such a prediction has been shown wrong. John Clark and Chris Peck seems to avoid the last step of the experience: listen to the copy, or reading their personal (and unique for each copy) diaries.
```
If I am correct, both you and Jason have no problem with that, OK?
Bruno
PS I have to go. (Might answer other posts later. Sorry for the delays).
http://iridia.ulb.ac.be/~marchal/
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To unsubscribe from this group and stop receiving emails from it, send an email | 862 | 3,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-30 | latest | en | 0.949508 |
https://discuss.analyticsvidhya.com/t/p-value-and-the-corresponding-type-1-error/16741 | 1,550,722,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247499009.48/warc/CC-MAIN-20190221031117-20190221053117-00430.warc.gz | 532,496,092 | 3,561 | # P value and the corresponding type 1 error
#1
my null hypothesis: mu<=0
alternate: mu>5 based on this right tail test
but my t statistics comes as -0.388978596.
what will my p value be 0.35 or 0.65?
I get my p value as 0.65 then the type 1 error will be any value greater then 0.65? | 91 | 288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.786052 |
https://puzzling.stackexchange.com/tags/word-property/hot | 1,719,301,789,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865694.2/warc/CC-MAIN-20240625072502-20240625102502-00080.warc.gz | 429,005,719 | 21,794 | # Tag Info
Accepted
### Grandma likes coffee but not tea
Grandma likes Why?
• 21.1k
Accepted
### What is a Shy Word™?
A Shy Word™ is a word ... There is no CSV version, because ...
• 61.9k
Accepted
### After the funeral
So first is to note that I’d imagine that Julia got nothing because As it turns out, To Agnes, I bequeath 12,000 pounds. To Eli, I bequeath 201,000 pounds. To Eric, I bequeath 95,000 pounds. To ...
• 32.3k
Accepted
### What is a Jeeves Word™?
A Jeeves Word™ is one which The words are
• 70.9k
Accepted
### What is a God Word™?
A God Word™ is a word that...
• 8,093
### “I just can’t solve this easy puzzle!”
In this puzzle, all of the sums: However, try as I might (and having applied some pretty strict criteria) I just could not find another pair of numbers which could belong to this set - no matter ...
• 146k
Accepted
### What is a Croupier's Word™?
A Croupier's Word™ For example, TRASHED is one because For completeness,
• 15.4k
Accepted
• 1,481
Accepted
### What is a Racist Word™?
I think the answer is Racist Words™ Non-Racist Words™ They are called Racist™ because
• 137k
Accepted
### What makes an Interesting Puzzle™?
Subpuzzles A Fringe Word™ A Not Fringe Word™ A Leaping Set™ A Not Leaping Set™ A Persistent Phrase™ A Not Persistent Phrase™ A Square Number™ A Not Square Number™ A Double Jump Word™ A Not ...
Accepted
### What is a Peruvian Word™?
I think a Peruvian word is one for which The examples are
• 70.9k
Accepted
### What is a Bumpy Word™?
In a word, Then a Bumpy word is Example: Why "Bumpy"?
• 19.6k
Accepted
### What is an Organizable Word™?
An Organizable Word is Below is a list of the example words that shows how they are Organizable:
• 147k
Accepted
### What is a Ping Pong Word™?
A Ping Pong Word™ is a word that
• 13.6k
Accepted
### What is a Perfect Word™?
A perfect word is Example How I found my answer
• 1,130
Accepted
### What is a Green Word™?
I'm not sure if this is correct, but
• 6,875
Accepted
### Silly Sally likes wearing glasses but not spectacles
I think the idea here is that: BONUS:
• 15.2k
Accepted
### What is an Eternal Word™?
An Eternal Word™ is The name:
• 78.2k
Accepted
### What is a Casino Word™?
I think a Casino Word has the property that Examples
• 137k
Accepted
I believe...
• 6,469
Accepted
### Properties: Left of the colon
First puzzle Second puzzle Third puzzle
• 10.5k
Accepted
### What kind of car do I have?
I think you have a Reasoning
• 137k
Accepted
Koffka words
• 147k
Accepted
### What is a Romeo Word™?
I bet a Romeo Word™ is one which The table taught as: What's wrong with the title?
• 70.9k
Accepted
### What is a Bind Word™?
Bind Words™ are ones which are Examples are Double Bind Words™ are And they are called Bind words because
• 70.9k
Accepted
### What is a bi-directional narcissistic word?
I think that Bi-Directional Narcissistic words have the property that On the explanation of the word narcissistic
• 137k
Accepted
### What is a CirKle Word™?
It seems like all CirKle words Example Meaning of the name
• 137k
Accepted
### Can you "spot" the pattern?
The pattern for "DEFRAUD" would be: First, some observations: But ... But, but ... The patterns are: Or, to account for the visual tag:
• 61.9k
Accepted
### What is an Inflated Word™?
The answer is: And the extra inflated:
• 776 | 970 | 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} | 2.578125 | 3 | CC-MAIN-2024-26 | latest | en | 0.901823 |
https://redcrab-software.com/en/Calculator/Finance/Percentage | 1,606,665,822,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141201836.36/warc/CC-MAIN-20201129153900-20201129183900-00287.warc.gz | 443,842,491 | 4,917 | # Online Calculator - Percentage
Calculation of the percentage of the difference between two values
## Percentage online calculator
The percentage calculator calculates the percentage of a starting or base value and a percentage value.
For example, what is the percentage if the base amount is 200 and the percentage amount 30? The percentage is 15%.
Base Value
Future Value
Decimal places
### Result
Percentage
Surcharge or discount
## Description of the parameters
Basic Value
The starting or base value.
Future Value
The future value after surcharge or discount on the base value.
## Calculate percentage
An example is used to calculate the percentage by which an initial value of 850 must increase. to reach the percentage of $$1000$$.
The basic value is known $$G = 850$$ and the percent value $$W = 1000$$.
The formula is $$\displaystyle P=\frac{W·100}{G}=\frac{1000·100}{850}=117.65$$
The percent value of $$1000$$ is $$117.65%$$ of the start value of $$850$$. So it needs an increase of $$117.65 - 100 = 17.65\%$$ to reach the percent value of $$1000$$. | 262 | 1,080 | {"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} | 3.859375 | 4 | CC-MAIN-2020-50 | latest | en | 0.763217 |
https://www.jamiletheteacher.com/geometry/question-which-statement-is-assumed-to-be-true-in-euclidean-geometry.html | 1,660,053,733,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570977.50/warc/CC-MAIN-20220809124724-20220809154724-00175.warc.gz | 741,546,821 | 10,349 | ## Question: Which Statement Is Assumed To Be True In Euclidean Geometry?
In Euclidean geometry, the parallel line postulate holds true: Through a given point not on a line, there is one and only one line parallel to that line. Parallel lines lie in the same plane and never intersect in Euclidean geometry, even when they are infinitely long.
## Which of the following choices is a term that represents a statement that is assumed true in the geometric system?
Postulates are statements that require proof, while theorems cannot be proven. Both postulates and theorems do not require proof and are assumed to be true. There is no difference between postulates and theorems.
## What defines a Euclidean geometry?
Euclidean geometry, the study of plane and solid figures on the basis of axioms and theorems employed by the Greek mathematician Euclid (c. 300 bce). In its rough outline, Euclidean geometry is the plane and solid geometry commonly taught in secondary schools.
You might be interested: What Is The Best Compass For Geometry?
## What term is not described in Euclidean geometry?
The term line is not defined in Euclidean geometry. There are three words in geometry that are not properly defined. These words are point, plane and line and are referred to as the “three undefined terms of geometry”.
## Which of the following is assumed to be true without proof?
A postulate is a statement that is assumed to be true without a proof. It is considered to be a statement that is “obviously true”. Postulates may be used to prove theorems true. The term “axiom” may also be used to refer to a “background assumption”.
## Which term describes a geometric statement that is assumed to be true without proof theorem postulate definition conjecture?
postulate. a statement that describes a fundamental relationship between the basic terms of geometry. Postulates are accepted as true without proofs. proof.
## Is Euclidean geometry complete?
Euclidean geometry is a first-order theory. Although Hilbert thought Euclidean geometry could be put on a firmer foundation by rewriting it in terms of arithmetic, in fact Euclidean geometry is complete and consistent in a way that Godel’s theorem tells us arithmetic can never be.
## Which Euclidean geometry properties hold for the geometry?
The five axioms for Euclidean geometry are:
• Any two points can be joined by a straight line.
• Any straight line segment can be extended indefinitely in a straight line.
• Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.
• All right angles are congruent.
## Where is Euclidean geometry used?
An application of Euclidean solid geometry is the determination of packing arrangements, such as the problem of finding the most efficient packing of spheres in n dimensions. This problem has applications in error detection and correction.
You might be interested: Question: Mathematicians Who Have Contributed To Geometry?
## What do we mean by Euclidean geometry quizlet?
Set of all points, boundless and three dimensional. Collinear. Set of two points, that all lie on the same line. Non-Collinear.
## How do you differentiate Euclidean and non-Euclidean geometry?
While Euclidean geometry seeks to understand the geometry of flat, two-dimensional spaces, non-Euclidean geometry studies curved, rather than flat, surfaces. Although Euclidean geometry is useful in many fields, in some cases, non-Euclidean geometry may be more useful.
## What is undefined term in geometry?
Undefined Terms. In geometry, point, line, and plane are considered undefined terms because they are only explained using examples and descriptions. Name the points, Lines, & Planes. Collinear points are points. that lie on the same line.
## Is the term line defined in Euclidean geometry?
In Euclidean geometry. In modern geometry, a line is simply taken as an undefined object with properties given by axioms, but is sometimes defined as a set of points obeying a linear relationship when some other fundamental concept is left undefined. | 832 | 4,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-33 | latest | en | 0.94712 |
https://analyticsindiamag.com/should-you-love-or-be-scared-of-maths-required-for-data-science/?utm_source=CloudQuant&utm_campaign=nlp&utm_content=MachineLearning | 1,675,275,119,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499949.24/warc/CC-MAIN-20230201180036-20230201210036-00514.warc.gz | 122,227,921 | 42,575 | # Should You Love Or Be Scared Of Maths Required For Data science?
Data science is the future, everyone wants to learn this budding technology. Is everyone able to learn? The answer is “No”. Do you know the reason, it is none other than “mathematics”. What….did I say mathematics? Yes, Mathematics or simply math. While reading this, people who know what is data science or has worked in this field would be confused and would be asking how maths is responsible. Let me, rephrase my answer, “ Fear to Mathematics”.
Starting from our elementary education to our higher education we see students scared of mathematics or we can say students have a math phobia. Not sure who has created this buzz that data science requires a long list of math topics as a prerequisite.
It is not completely correct, elementary math is required but, as a beginner, you don’t need that much math for data science. Also, there is another side to data science and that is the practical side. For practical data science, a great deal of math is not required. Practical data science only requires skills to select the right tools. Being said that let’s understand how theoretical and practical data science differs.
## Difference between Theory and Practice
When we talk about data science, it is important to set our goal. What we want to achieve from learning data science? It is for academic learning or for practical purpose to build career on.
Why this goal setting is important because priorities and deliverables are different in both theoretical and practical data science. Learning data science for academic purposes is more for publishing research papers and push the field forward. While practical use of data science is to generate reports, build models and system software.
## Skills required for foundational Data science
As a beginner in data science, one will primarily work on foundational/fundamental skills of data science. These skills are required in each and every data science project.
What are these skills?
1. Data manipulation
2. Data Visualization
3. Data analysis, also know as EDA (Exploratory Data Analysis)
It is a known fact that in any data science project 75% of effort and time is spent doing these fundamental steps. These are the core skills required for success of any data science project. If any of the above step is missed out or is not carried out with utmost precision, final model might not be as good as it should be.
Coming back to the question, how much math is required for these core skills? – Very little.
So, by now you must be pretty much clear and convinced that math required for data science is not scary at all.
## What concepts of math are required for these foundation skills of data science?
You would be wondering that still some math is required , so what all topics are there.
Let me break the good news to you, you just require lower-level algebra and simple statistics. Don’t be astonished, it is a fact. Feeling delighted !!
Let me explain it in a bit more detail.
1. For getting the data, cleaning it and understanding it, doesn’t require any math. If new variable creation is required by deriving it’s value from already present variable. Then , it requires elementary math of addition, subtraction, multiplication and division. It may be required to manipulate the data by calculating mean, median or mode. It could be seen that none of the calculations here require complex maths.
For 95% of cases above method holds true but exceptions will always be there. Very rare cases would require complex computation. Again it is very rare.
• One major tasks in data science is data visualization. In this step graphs and plots are created to check patterns in data. This is like a subset of exploratory data analysis step. So, where math is required in this? One should be aware of which plot to create and which tool to use. So, problem solved, no math is required in this step as well.
One should know how to read plots and graphs whether it be scatter plot, histogram, line graph, point plot, etc. This is the only demand from this step.
• By now 75% work of any data science project is done. Now comes model creation using machine learning. Here again concept of theory and practical applicability of machine learning comes into picture. We have to create models for our business purpose and not to dig deep in model theoretically and publish a research paper. Conceptual understanding of machine learning models again don’t require math.
One should learn which model to used in which situation. How to interpret the model. How to check assumptions in model. How to make predictions from model. Have I mentioned math anywhere, no. So, in short no advance math is require to become a data scientist.
## Basic skills required to become a foundational data scientist
1. Selection of right tool
2. Understanding the syntax
3. Basic graphs and charts
4. Conceptual understanding of different machine learning models
5. Basic Algebra
6. Basic Statistics
This is mostly all you need to get started learning data science. Motivation and will to learn is utmost important.
But always remember saying by Galileo Galilei:
Mathematics is the language with which God has written the universe.”
## More Great AIM Stories
### TypeScript vs JavaScript: Who’s Winning The 10-year-long Battle?
Netali Agrawal is a part of the AIM Writers Programme. She is a Business Analyst who loves to explore new ideas in different industries through machine learning and artificial intelligence. She holds a bachelors degree in engineering along with post-graduation certification in business analytics and business intelligence. She is working with an MNC as a business analyst and leading a project for machine learning and artificial intelligence. Netali loves to write about analytics, machine learning and artificial intelligence. She loves to explore data and mould it in the best possible shape to get all possible insights from the data. She resides in Hyderabad, India. Linkedin Bio: www.linkedin.com/in/netali-agrawal-31192a71
## AIM Upcoming Events
Early Bird Passes expire on 3rd Feb
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16-17th Mar, 2023
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### Telegram group
Discover special offers, top stories, upcoming events, and more.
### Discord Server
Stay Connected with a larger ecosystem of data science and ML Professionals
### Kickstart your career in Data Science and Business Analytics with this program from Great Learning
The curriculum of the PGP in Data Science and Business Analytics: V.22 has been updated in consultation with industry experts, academicians and program alums.
### How to build healthcare predictive models using PyHealth?
PyHealth is a Python-based toolbox. As the name implies, this toolbox contains a variety of ML models and architecture algorithms for working with medical data and modeling.
### Explained: Prospective learning in AI
A paper published earlier this year argued that retrospective learning isn’t a good representation of true intelligence.
### AI in SEO is so evolved now, it’s pitting against itself
With the integration of AI into SEO, can brands overcome the strict and ever vigilant guidelines of SERPs?
### Council Post: Key things to remember while building data teams
The AI team consists of ‘an external team’ (a team external to the data team but part of the core AI team) that works closely with the data team and then there is the core data team itself.
### IBM launches new Mainframe model, aims to regain lost ground
Despite the cost-saving benefits and ease of sharing resources, only 25% of enterprise workloads have been moved to the cloud.
### How AI is used for the early detection of breast cancer
CNNs are efficient in detecting malignancies from scans.
### Google says no to FLoC, replaces it with Topic
Topic is a Privacy Sandbox proposal for internet-based advertising, which is replacing FLoC (Federated Learning of Cohorts).
### Learning Scala 101: The best books, videos and courses
Tagged as “the definitive book on Scala”, this book is co-authored by Martin Odersky, the designer of the Scala language.
### Meta AI proposes a new approach to improve object detection
Detic, like ViLD, uses CLIP embeddings as the classifier. | 1,716 | 8,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-06 | latest | en | 0.94191 |
http://syllabus.sic.shibaura-it.ac.jp/syllabus/2024/sys/146675.html?g=V00 | 1,721,888,890,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518579.47/warc/CC-MAIN-20240725053529-20240725083529-00845.warc.gz | 24,209,730 | 4,271 | Course title
V04104002
Calculus with Differential Equations
Course description
You will learn what a differential equation is and how to recognize some of the basic different types. You will learn how to apply some common techniques used to obtain general solutions of differential equations and how to fit initial or boundary conditions to obtain a unique solution. You will appreciate how differential equations arise in applications and you will gain some experience in applying your knowledge to model a number of engineering problems using differential equations.
Purpose of class
The purpose of this class is to learn how to recognize some of the basic different types of differential equations, to learn how to apply some common techniques used to obtain solutions of differential equations and to appreciate how differential equations arise in applications. This class also includes a review on the content learned in the class of differential equations at the time of first grade.
Goals and objectives
1. You can describe how to recognize some of the basic different types of differential equations
2. You can describe how to apply some common techniques used to obtain solutions of differential equations
3. You can describe how differential equations arise in applications
Relationship between 'Goals and Objectives' and 'Course Outcomes'
Mid-term exam Final exam Total.
1. 20% 20% 40%
2. 20% 20% 40%
3. 10% 10% 20%
Total. 50% 50% -
Language
English
Class schedule
Class schedule HW assignments (Including preparation and review of the class.) Amount of Time Required
1. Orientation, Case study of modelling Exercises for the models 190minutes
2. General solution Review general solutions 190minutes
3. Method of separation of variables Applying the method of separation of variables 190minutes
4. Inverse function Review inverse functions 190minutes
5. Exact equation Solving exact equations 190minutes
6. Finding the Integrating factor Solving equations via integrating factors 190minutes
7. Bernoulli differential equation Review Bernoulli differential equations 190minutes
8. Mid-term exam and review Review Sessions 1-7 190minutes
9. 2nd order constant coefficient equations Finding complementary functions 190minutes
10. 2nd order constant coefficient equations: exercise Finding complementary functions 190minutes
11. Particular integral Finding particular integrals 190minutes
12. General solution of inhomogeneous case Finding general solutions 190minutes
13. Applications Exercises for the models 190minutes
14. Final exam and review Review Sessions 9-13 190minutes
Total. - - 2660minutes
Evaluation method and criteria
Mid-term exam and Final exam. As a criterion, if you determine that you understand 60% of the content covered in the class, your final score will be 60 points.
Feedback on exams, assignments, etc.
ways of feedback specific contents about "Other"
The Others Provide feedback in class or via ScombZ/email, etc.
Textbooks and reference materials
No textbook
References (English in mathematics for Japanese students)
Prerequisites
Basic knowledge of calculus is required
Office hours and How to contact professors for questions
• Lunchtime on every Tuesday.
Regionally-oriented
Non-regionally-oriented course
Development of social and professional independence
• Course that cultivates an ability for utilizing knowledge
Active-learning course
More than one class is interactive
Course by professor with work experience
Work experience Work experience and relevance to the course content if applicable
N/A N/A
Education related SDGs:the Sustainable Development Goals
• 4.QUALITY EDUCATION
• 9.INDUSTRY, INNOVATION AND INFRASTRUCTURE | 752 | 3,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-30 | latest | en | 0.904014 |
https://www.jiskha.com/display.cgi?id=1314970453 | 1,516,476,115,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889681.68/warc/CC-MAIN-20180120182041-20180120202041-00092.warc.gz | 929,022,818 | 3,679 | # physics
posted by .
Consider a ball of mass m1 moving with a velocity u1 collides elastically with a ball of mass m2 at rest.Show that the maximum energy transfer takes place when m1=m2.
## Similar Questions
1. ### Physics
A softball of mass 0.200 kg that is moving with a speed of 8.2 m/s collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.9 m/s. 1. Calculate the velocity …
2. ### physics
A softball of mass 0.220 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward …
A softball of mass 0.220 kg that is moving with a speed of 6.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward …
4. ### physics
A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward …
5. ### Science Physics
A rubber ball of mass m1= 10kg is moving to the right with speed v1= 10m/s. it collides elastically with another ball of mass m2= 50kg, which is sitting at rest. m2 is larger than m1, what are the speeds of the balls, v1 and v2 after …
6. ### physics
a ball of mass m , moving with uniform speed, collides elastically with another stationary ball.the incident ball will lose maximum kinetic energy when the mass of the stationary ball is?
7. ### physics
A white billiard ball with mass mw = 1.32 kg is moving directly to the right with a speed of v = 2.91 m/s and collides elastically with a black billiard ball with the same mass mb = 1.32 kg that is initially at rest. The two collide …
8. ### PHYSICS!!
I've tried this problem again and again using conservation of momentum and energy but I am not having any luck: A ball of mass 0.300kg that is moving with a speed of 8.0m/s collides head-on and elastically with another ball initially …
9. ### physics
A ball of mass M1 moves with a velocity U1, collides perfectly elastically with a stationary ball of mass M2 . After the collision, M1 and M2 move with velocities V1 and V2 respectively. Show that : V1 =(M1-M2) / (M1 + M2 ) * U1 .
10. ### physics
A ball A of mass M collides elastically with another identical ball B at rest .Initially velocity of ball A is u m/s.After collision
More Similar Questions | 655 | 2,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-05 | latest | en | 0.910154 |
http://www.thelearningpoint.net/home/mathematics/compound-interest-part-1 | 1,519,439,632,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00490.warc.gz | 549,199,677 | 21,595 | The Learning Point > Mathematics >
### Compound Interest Part 1
Question 1: Suppose Ankit has 2300 Rs that he invests in an account that pays 7% interest compounded bi-annually. How much money does Ankit have at the end of 5 years? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 2300 Rs is the amount being invested or P. The interest rate r is 7% which must be changed into a decimal and becomes r = 0.07. The interest is compounded bi-annually, which tells us that n = 2. The money will stay in the account for 5 years so t = 5. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 2300 x ( 1 + 7/2)(2) x (5) = 3244.38 Rs. So after 5 years, the account is worth 3244.38 Rs. Question 1: Suppose Ankit has 2300 Rs that he invests in an account for 5 years at a certain rate compounded bi-annually. This becomes 3244.38 Rs at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (3244.38/2300)(1/(bi-annually x 57) - 1 = 7 %. Question 3: Suppose John has a principal P that he invests in an account that pays 1% interest compounded monthly. John has 1751.75 \$ at the end of 3 years. What was the starting principal P? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1700 \$ is the amount being invested or P. The interest rate r is 1% which must be changed into a decimal and becomes r = 0.01. The interest is compounded monthly, which tells us that n = 12. The money will stay in the account for 3 years so t = 3. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 1751.75/ ( 1 + 1/12)(12) x (3) = 1700 \$ So after 5 years, the account is worth 1751.75 \$. Question 4: Suppose Nina has 1500 Rs that she invests in an account that pays 15% interest compounded annually. In how many years will this become 2281.31 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1500 Rs is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded annually, which tells us that n = 1. The final amount is 2281.31 Rs. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (2281.31) - log (1500)) / (1* log(1 + 0.15/1)) => t = (LOG2281.31-LOG1500)/ (1* (1 + LOG0.15/1)) => t = 3 years So the time required is 3 years. Question 4: Suppose Nina has 1500 Rs that she invests in an account for 3 years at a certain rate compounded annually. This becomes 2281.31 Rs at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (2281.31/1500)(1/(annually x 315) - 1 = 15 %. Question 6: Suppose Mary has 2900 \$ that she invests in an account that pays 7% interest compounded quarterly. How much money does Mary have at the end of 3 years? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 2900 \$ is the amount being invested or P. The interest rate r is 7% which must be changed into a decimal and becomes r = 0.07. The interest is compounded quarterly, which tells us that n = 4. The money will stay in the account for 3 years so t = 3. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 2900 x ( 1 + 7/4)(4) x (3) = 3571.17 \$. So after 5 years, the account is worth 3571.17 \$. Question 7: Suppose Ankit has 100 Rs that he invests in an account that pays 15% interest compounded monthly. In how many years will this become 210.72 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 100 Rs is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded monthly, which tells us that n = 12. The final amount is 210.72 Rs. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (210.72) - log (100)) / (12* log(1 + 0.15/12)) => t = (LOG210.72-LOG100)/ (12* (1 + LOG0.15/12)) => t = 5 years So the time required is 5 years. Question 8: Suppose Martha has 900 £ that she invests in an account that pays 11% interest compounded annually. In how many years will this become 1108.89 £? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 900 £ is the amount being invested or P. The interest rate r is 11% which must be changed into a decimal and becomes r = 0.11. The interest is compounded annually, which tells us that n = 1. The final amount is 1108.89 £. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (1108.89) - log (900)) / (1* log(1 + 0.11/1)) => t = (LOG1108.89-LOG900)/ (1* (1 + LOG0.11/1)) => t = 2 years So the time required is 2 years. Question 9: Suppose John has 200 \$ that he invests in an account that pays 14% interest compounded bi-annually. In how many years will this become 262.16 \$? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 200 \$ is the amount being invested or P. The interest rate r is 14% which must be changed into a decimal and becomes r = 0.14. The interest is compounded bi-annually, which tells us that n = 2. The final amount is 262.16 \$. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (262.16) - log (200)) / (2* log(1 + 0.14/2)) => t = (LOG262.16-LOG200)/ (2* (1 + LOG0.14/2)) => t = 2 years So the time required is 2 years. Question 9: Suppose John has 200 \$ that he invests in an account for 2 years at a certain rate compounded bi-annually. This becomes 262.16 \$ at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (262.16/200)(1/(bi-annually x 214) - 1 = 14 %. Question 10: Suppose Nina has 1000 Rs that she invests in an account for 2 years at a certain rate compounded quarterly. This becomes 1104.49 Rs at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (1104.49/1000)(1/(quarterly x 25) - 1 = 5 %. Question 12: Suppose Mary has 1300 \$ that she invests in an account that pays 3% interest compounded annually. In how many years will this become 1420.55 \$? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1300 \$ is the amount being invested or P. The interest rate r is 3% which must be changed into a decimal and becomes r = 0.03. The interest is compounded annually, which tells us that n = 1. The final amount is 1420.55 \$. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (1420.55) - log (1300)) / (1* log(1 + 0.03/1)) => t = (LOG1420.55-LOG1300)/ (1* (1 + LOG0.03/1)) => t = 3 years So the time required is 3 years. Question 13: Suppose Ankit has 1600 Rs that he invests in an account that pays 4% interest compounded bi-annually. In how many years will this become 1731.89 Rs? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 1600 Rs is the amount being invested or P. The interest rate r is 4% which must be changed into a decimal and becomes r = 0.04. The interest is compounded bi-annually, which tells us that n = 2. The final amount is 1731.89 Rs. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (1731.89) - log (1600)) / (2* log(1 + 0.04/2)) => t = (LOG1731.89-LOG1600)/ (2* (1 + LOG0.04/2)) => t = 2 years So the time required is 2 years. Question 14: Suppose Martha has a principal P that she invests in an account that pays 15% interest compounded quarterly. Martha has 1252.89 £ at the end of 5 years. What was the starting principal P? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 600 £ is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded quarterly, which tells us that n = 4. The money will stay in the account for 5 years so t = 5. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 1252.89/ ( 1 + 15/4)(4) x (5) = 600 £ So after 5 years, the account is worth 1252.89 £. Question 15: Suppose John has a principal P that he invests in an account that pays 5% interest compounded monthly. John has 946.05 \$ at the end of 1 years. What was the starting principal P? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 900 \$ is the amount being invested or P. The interest rate r is 5% which must be changed into a decimal and becomes r = 0.05. The interest is compounded monthly, which tells us that n = 12. The money will stay in the account for 1 years so t = 1. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 946.05/ ( 1 + 5/12)(12) x (1) = 900 \$ So after 5 years, the account is worth 946.05 \$. Question 15: Suppose John has 900 \$ that he invests in an account for 1 years at a certain rate compounded monthly. This becomes 946.05 \$ at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (946.05/900)(1/(monthly x 15) - 1 = 5 %. Question 17: Suppose Alex has 400 £ that he invests in an account that pays 14% interest compounded bi-annually. In how many years will this become 786.86 £? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 400 £ is the amount being invested or P. The interest rate r is 14% which must be changed into a decimal and becomes r = 0.14. The interest is compounded bi-annually, which tells us that n = 2. The final amount is 786.86 £. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values to compute 't'. Taking logarithms: log A = log P + nt log (1 + r/n) => t = (log (786.86) - log (400)) / (2* log(1 + 0.14/2)) => t = (LOG786.86-LOG400)/ (2* (1 + LOG0.14/2)) => t = 5 years So the time required is 5 years. Question 18: Suppose Mary has 200 \$ that she invests in an account that pays 15% interest compounded quarterly. How much money does Mary have at the end of 3 years? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. The 200 \$ is the amount being invested or P. The interest rate r is 15% which must be changed into a decimal and becomes r = 0.15. The interest is compounded quarterly, which tells us that n = 4. The money will stay in the account for 3 years so t = 3. We have values for four of the variables. We can use this information to solve for A. Let's plug in the values. The required amount = 200 x ( 1 + 15/4)(4) x (3) = 311.09 \$. So after 5 years, the account is worth 311.09 \$. Question 18: Suppose Mary has 200 \$ that she invests in an account for 3 years at a certain rate compounded quarterly. This becomes 311.09 \$ at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (311.09/200)(1/(quarterly x 315) - 1 = 15 %. Question 19: Suppose Ankit has 2900 Rs that he invests in an account for 2 years at a certain rate compounded monthly. This becomes 3469.6 Rs at the end of the period. What was the Rate%? Let’s look at the compound interest formula and see how many values for the variables we are given in the problem. Let's plug in the values to compute the rate 'r' r = (A/P)(1/nt)-1 = (3469.6/2900)(1/(monthly x 29) - 1 = 9 %. | 3,955 | 13,558 | {"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} | 4.8125 | 5 | CC-MAIN-2018-09 | longest | en | 0.933616 |
http://math.stackexchange.com/questions/201052/find-the-coefficient-of-x3y2z3-in-the-expansion-2x3y-4zw9 | 1,467,032,899,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783396027.60/warc/CC-MAIN-20160624154956-00120-ip-10-164-35-72.ec2.internal.warc.gz | 196,661,613 | 18,365 | # Find the coefficient of $x^3y^2z^3$ in the expansion $(2x+3y-4z+w)^9$
The exercise says:
In the expansion $(2x+3y-4z+w)^9$, find the coefficient of $x^3y^2z^3$.
The formula to find the coefficient of $x_1^{r_1}x_2^{r^2}\dots x_k^{r_k}$ in $(x_1+x_2+\dots+x_k)^n$ is:
$$\frac{n!}{r_1!r_2!\dots r_k!}$$
Am I supposed to multiply the result determined by the above formula by $2$, $3$ and $4$? What to do when $x$ and $y$ has coefficients?
-
The coefficient of $x^3y^2z^3$ is zero. Do you mean the coefficient of $x^3y^3z^3$? – Michael Albanese Sep 23 '12 at 8:09
@MichaelAlbanese: Umm, no that's what I see in the book. What's wrong with zero? It's a nice number! – Gigili Sep 23 '12 at 8:15
Fair enough, it just makes the question much easier. – Michael Albanese Sep 23 '12 at 8:20
@MichaelAlbanese: Not everyone is as smart as you, and yes I Didn't notice it is zero and meant to ask the question for the general case. – Gigili Sep 23 '12 at 8:26
My comment wasn't intended to imply anything about my (or anyone else's) intelligence; if it came across as arrogant, I apologise but that was not my intention. I just thought that there may have been a typo (either yours or in the book). Given that you mentioned the multinomial coefficients, I assumed the exercise wanted you to use them, hence my question. – Michael Albanese Sep 23 '12 at 9:02
Every term in $(2x+3y-4z+w)^9$ is a product of nine factors, each of which is one of the four terms in parentheses. Thus, before you collect like terms each factor will have the form $(2x)^i(3y)^j(-4z)^kw^\ell$, where $i+j+k+\ell=9$. Since the exponents in $x^3y^2z^3$ add up to only $8$, not $9$, there is no such term in the product, and its coefficient is $0$.
If you actually meant the coefficient of $x^3y^3z^3$, each such term must arise as the product of three factors of $2x$, three of $3y$, and three of $-4z$, so it must be $(2x)^3(3y)^3(-4z)^3=2^33^3(-4)^3x^3y^3z^3$, with a coefficient of $2^33^3(-4)^3=-13824$. Your formulat tells you that there are
$$\frac{9!}{3!3!3!}=1680$$
such terms, so the total coefficient of $x^3y^3z^3$ is $-13824\cdot1680=-23~224~320$.
-
Sum of exponents in $x^3 y^2 z^3$ equals $8$ :) – M. Strochyk Sep 23 '12 at 8:17
@M.Strochyk: Tell my fingers to behave better! :-) (Thanks for catching it.) – Brian M. Scott Sep 23 '12 at 8:18
Surely the question (although perhaps badly worded) is asking for the term in the expansion which contains $x^3y^2z^3$ as written? This term will be of the form $nx^3y^2z^3w$ for some integer $n$, so its "coefficient" is $nw$. We can expand
$(2x+3y-4z+w)^9 = ((2x+3y-4z)+w)^9$
as
$(2x+3y-4z)^9 + 9w(2x+3y-4z)^8 + \ldots$,
so $n$ is equal to 9 times the coefficient of $x^3y^2z^3$ in the expansion of $(2x+3y-4z)^8$. From Brian M. Scott's answer, we see that $n$ is equal to $$9 \times \frac{8!}{3!2!3!}$$ which I will let you calculate yourself. Also note that this is equal to $$\frac{9!}{3!2!3!1!}$$ which is the coefficient of $x^3y^2z^3w$, as expected.
- | 1,070 | 2,994 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2016-26 | latest | en | 0.907804 |
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Equ. #9:
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# Math Choice Questions
Section N1
Level: Easy
Which of the
following would
you compare
to
Which of the
following would
you compare
to
Which of the
following would
you compare
to
Section N1
Level: Hard
Which of the
following would
you compare
to
What are all the
values of p, for
which
converges?
E There are no
values of p for
which the
Section 11.1
Level: Easy
Identify for the
sequence
2, 4, 6, 8, ...
Identify an for the
sequence
Section 11.1
Level: Hard
Identify an for the
sequence
0, 3, 8, 15, 24, ldots
Section 11.1
Level: Easy
The statement
means
that for each ε > 0
there exists an N
such that
Which of the
following
sequences has this
graph
Section111.eps
Identify the value of
the
Identify the value of
the
Identify the value of
the
Section 11.1
Level: Hard
Identify the value of
the
Consider the
following three
sequences:
A and
converge,
diverges
B and
converge,
diverges
C and
Section 11.2
Level: Easy
True or False: If
then
converges.
A True
B False
Determine the
general term for the
following series
D None of the
above
Section 11.2
Level: Hard
Which of the
following series are
geometric?
Section N2
Level: Easy
What is the value of
the series
C The series
doesn’t
converge
What is the formula
for
C The series
doesn’t
converge
Which of the
following is not an
example of
geometric growth?
A An endowment
B Earning annual
interest on an
account
C A population
doubling each
year
D All of the above
are examples of
Section N2
Level: Hard
Which of the
following is a
correct reindex of
Section N3
Level: Easy
For what values of
p does the series
converge?
Section N3
Level: Hard
True or False
A True
B False
Section N3
Level: Easy
Which series should
we use to decide if
converges
Which series should
we use to decide if
converges
Which series should
we use to decide if
converges | 748 | 2,394 | {"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} | 3.578125 | 4 | CC-MAIN-2017-51 | longest | en | 0.897058 |
https://www.jiskha.com/display.cgi?id=1359077935 | 1,531,868,595,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00552.warc.gz | 919,167,539 | 3,773 | # Math
posted by Mary
Adam drew two same size rectangles and divided them into the same number of eqal parts. He shaded 1/3 of one rectangle and 1/4 of the other rectangle. What is the least number of parts into which both rectangles could be divided?
1. Liam
12
2. John
12
3. 100_mathStudent
um I don't know lol I asked the question :P it's 12 lol I copied the others
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A large rectangle has an area of 28 square units. It has been divided into two smaller rectangles. One of the smaller rectangles has an area of 4Â square units. What are possible whole-number dimensions of the large rectangle?
9. ### math
Leah has two same size rectangles divided into the same number of Equal parts. One rectangle has 1/3 of the parts shaded and the other has 2/5 of the parts shaded. What is the least number of parts into which both rectangles can be …
10. ### Math(Connexus 6th)Unit 6
A room has 2 sections as shown in the picture below. Which expression uses the Distributive Property to find the total area of the room?
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Creating a macro - Writing a Script - Using the API
### [Solved] Function rnd()*20 = 20, sometimes
Hello everyone,
yesterday I tried to implement a dice-roller in basic and use it as a macro. That worked. Then I wanted to see how "random" the rnd() function actually is. To do that I created a function that generates a number of rolls (with a fixed die-size), writes the result into an array. Since I use a 20-sided die, I assumed that an array with 20 entries would suffice. Yet I was mistaken as I encountered out of bounds accesses. Looking deeper into this I found that my function would produce values in the range of (1:21) which was clearly not intended. However, I fail to see what this is caused by.
Code: Select all Expand viewCollapse view
`function fnf()RandomizeDim value as IntegerDim values(19) as LongFor I = 1 to 100000 value = CInt(Int(rnd()* 20)+1) values(value-1) = values(value-1) + 1Next Irstring = ""For I=0 to 19 rstring = rstring & CStr(I+1) & ":" & CStr(values(I)) & ", "Next I MsgBox rstringEnd function`
After testing I found that rnd()*20 results in values of 20, yet rnd() itself is never 1. The values which result in 21 seem random to me: I found a 0.98.. or a 0.67 as return value for rnd(). According to the doc, rnd() returns a Single which should be able to handle multiplication without an overflow. I assume this error is caused by some missing type conversion I simply did not do.
Last edited by BlueSkyBlackBird on Sat May 06, 2017 1:43 am, edited 3 times in total.
OpenOffice 3.1.3 Win7 x64
BlueSkyBlackBird
Posts: 3
Joined: Fri May 05, 2017 5:02 pm
### Re: Function rnd()*20 = 20, sometimes
yet rnd() itself is never 1
That is incorrect, rnd() does return the value 1, thus the 21 you are seeing in your results.
If your problem has been solved, please edit this topic's initial post and add "[Solved]" to the subject line
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UnklDonald418
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### Re: Function rnd()*20 = 20, sometimes
Again: Not according to doc. Also the 21 is produced by values other than a 1 for rnd().
OpenOffice 3.1.3 Win7 x64
BlueSkyBlackBird
Posts: 3
Joined: Fri May 05, 2017 5:02 pm
### Re: Function rnd()*20 = 20, sometimes
Not according to doc.
Not sure what doc you are referring to, but it appears to be wrong. The rnd() function returns both 0 and 1. The following code will verify that though rare they are both returned.
Code: Select all Expand viewCollapse view
`REM ***** BASIC ***** function fnf() Randomize Dim value as Integer Dim values(20) as Long Dim raw Dim I Dim test0 as integer Dim test1 as integer test0 = 0 : test1 = 0 For I = 1 to 100000 raw = rnd() if raw = 1 then test1 = test1 + 1 end if if raw = 0 then test0 = test0 + 1 end if value = CInt(Int(raw * 20) + 1) values(value-1) = values(value-1) + 1 Next I print test0 & " 0's returned : " & test1 & " 1's returned" rstring = "" For I=0 to 20 rstring = rstring & CStr(I+1) & ":" & CStr(values(I)) & ", " Next I MsgBox rstring End function`
If your problem has been solved, please edit this topic's initial post and add "[Solved]" to the subject line
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UnklDonald418
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### Re: Function rnd()*20 = 20, sometimes
It looks like you were right. Testing this made also clear what the problem was. The 21 was caused by the rnd() returning one. I also tested this using a variable called rnd, coming from a java background, I did not see a problem with that. However here It seems that one can call functions without using the braces. So what really happen is that every call of rnd as variable called rnd() instead and gave me a new value.
I thank you very much and I think I can change this to [solved].
OpenOffice 3.1.3 Win7 x64
BlueSkyBlackBird
Posts: 3
Joined: Fri May 05, 2017 5:02 pm
### Re: [Solved] Function rnd()*20 = 20, sometimes
Yes, function names are reserved in Basic. When there are no arguments being passed the braces are optional. Leaving them off doesn't help much with readability.
I did a little more testing and it looks like Int() and CInt() round anything below .05 to 0.
If your problem has been solved, please edit this topic's initial post and add "[Solved]" to the subject line
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UnklDonald418
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### Re: [Solved] Function rnd()*20 = 20, sometimes
You are not restricted to StarBasic. I recommend Python as macro language. The office suite comes with a Python runtime and a bridge between the Python language and the UNO API
Please, edit this topic's initial post and add "[Solved]" to the subject line if your problem has been solved.
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Location: Germany | 1,460 | 5,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | longest | en | 0.895678 |
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Andy
Didnt think that this should go in the theoretical physics section, but if it should feel free to move it.
We all know that a large gravitational field can bend light, but my question is would it be possible for a large enough magnetic field to bend light? I know that this doesn't happen around the Sun or at least i think it doesnt, but if it was more concentrated, if its possible for a magnetic field to be of a higher concentration, could it bend light?
I think that's long winded for what i want to ask, but you should get the idea.
No.
Gravitational fields bend space and time, so the
light rays passing through the bent area, bend with
it.
Magnetic and electrostatic fields, are supposed to be
completely different from gravitational fields, since
they don't bend space+time.
These instead operate by sending out virtual photons,
which ultimately lead to a force being exerted.
Homework Helper
It is also true that magnetic and electric fields add linearly without affecting one another. Since light is a wave in a magnetic-electric field, magnetic fields cannot affect light.
mmwave
Originally posted by HallsofIvy
It is also true that magnetic and electric fields add linearly without affecting one another. Since light is a wave in a magnetic-electric field, magnetic fields cannot affect light.
This is certainly true in a vacuum or a linear medium. In non-linear media two light waves can interact to produce new frequencies as in a Fabry-Perot cavity modulator.
Originally posted by Andy
Didnt think that this should go in the theoretical physics section, but if it should feel free to move it.
We all know that a large gravitational field can bend light, but my question is would it be possible for a large enough magnetic field to bend light? I know that this doesn't happen around the Sun or at least i think it doesnt, but if it was more concentrated, if its possible for a magnetic field to be of a higher concentration, could it bend light?
I think that's long winded for what i want to ask, but you should get the idea.
Since the field has energy and since energy has mass and since mass is the source of gravity then it follows that a magnetic field will general a gravitational field and such a field can delfect light.
Even light can generate a gravitational field and deflect light! :-)
Pete
McQueen
Originally posted by HallsofIvy
Since light is a wave in a magnetic-electric field, magnetic fields cannot affect light.
This is a wrong statement , magnetic fields do affect light , they change the spectral output of light .
Staff Emeritus
Gold Member
Originally posted by McQueen
This is a wrong statement , magnetic fields do affect light , they change the spectral output of light .
Well, light doesn't have any quality called 'spectral output' that I know of!
I think you're talking about the hyperfine splitting, and how magnetic fields can affect the light produced by atoms immersed in the field -- and this is correct.
However, once emitted by the atom, the photons themselves are not affected by the magnetic field in any way.
- Warren
Creator
Originally posted by HallsofIvy
It is also true that magnetic and electric fields add linearly without affecting one another. Since light is a wave in a magnetic-electric field, magnetic fields cannot affect light.
Well, as mmwave pointed out, this is only true in a vacuum. Magnetic field effects upon light are observed in solids and liquids, generally referred to as magneto-optic effects.
Generally a magnetic field aligned to the propogation axis of light causes left and right circularly polarized light to travel at different speeds in the substance, (i.e., the index of refraction for left and right circular polarization is changed).
The most common example is Faraday rotation, where, due to the above effect, the plane of polarization of linearly polarized light is rotated as it passes thru the material, the amount of rotation being proportional to the magnetic field and the path length through the material.
Many ingenious & useful devices have been developed as a result of these effects.
Creator
P.S. Magneto-optic effects also take place in plasmas, where, for ex., Faraday rotation is observed in polarized radio waves through the Earth's ionosphere.
Last edited:
Creator
Originally posted by mmwave
This is certainly true in a vacuum or a linear medium. In non-linear media two light waves can interact to produce new frequencies as in a Fabry-Perot cavity modulator.
Thanks mmwave.
Furthermore, (and somewhat contradicting my previous post), according to quantum electrodynamics two light waves DO interact even in the vacuum, violating the (classical) superposition principle.
This is a very small interaction but represents a non-linear departure from the classical predictions. It's a result of the virtual polarization of the vacuum which causes a non-vanishing photon scattering cross section. In QED photon-photon scattering is predicted, but very small.
ALSO, let it be known that photon interaction with the nuclear coulomb field is also shown to occur and has been measured exactly as predicted by QED; it is usually referred to as Delbruck scattering.
Creator
Creator
Originally posted by pmb
Since the field has energy and since energy has mass and since mass is the source of gravity then it follows that a magnetic field will general a gravitational field and such a field can delfect light.
Even light can generate a gravitational field and deflect light! :-)
pmb,
It may interest you to know (and I found quite interesting) that, using Quantum Field Theory, Tolman, Ehrenfest, & Podolsky found that two light beams propogating in the same direction do not interact gravitationally; BUT two beams going in opposite directions DO interact! Two different derivations are given in A. Zee's book entitled, Quantum Field Theory ...
However, I'm not sure that proof has ever been given of gravitating magnetic fields.
Creator
Originally posted by Creator
...using Quantum Field Theory, Tolman, Ehrenfest, & Podolsky found that two light beams propogating in the same direction do not interact gravitationally; BUT two beams going in opposite directions DO interact!
That's true only for low energy graviton exchange. If we include higher energy exchanges, we do see the expected gravitational interaction of photons, whatever their relative momentum.
Originally posted by Creator
However, I'm not sure that proof has ever been given of gravitating magnetic fields.
Magnetic and electric fields are different aspects of the electromagnetic field, which is light.
Last edited:
Creator
Originally posted by jeff
That's true only for low energy graviton exchange. ...
Yes, it was shown to be so in the 'weak field limit'. However I still find it interesting. I'm not quite sure what 'high energy exchange' would represent physically for two photon beams.
Magnetic and electric fields are different aspects of the electromagnetic field, which is light.
Yes, Jeff, I know, but that is still not considered explicit proof that magnetic fields gravitate.
Creator
Last edited:
Originally posted by Creator
...not quite sure what 'high energy exchange' would represent physically for two photon beams.
Then maybe you should find out.
Originally posted by Creator
...that [Magnetic and electric fields are different aspects of the electromagnetic field, which is light] is still not considered explicit proof that magnetic fields gravitate.
If you accept that magnetic fields are just photons, then how can you doubt that light gravitates? After all, light has energy, and all energy gravitates.
Originally posted by pmb
Since the field has energy and since energy has mass and since mass is the source of gravity then it follows that a magnetic field will general a gravitational field and such a field can delfect light.
Why do you insist on inserting this extra step in your reasoning by saying "energy has mass and mass generates gravity" when mass-energy equivalence allows you to dispense with this and just say that energy gravitates?
I looked at your site and you make a big deal out of the issue of whether or not the use of the term "relativistic mass" is legitimate. But really it's just a convention, some people like to use it - I'm one of them - and others think it's better not to, and I think they're being silly.
Originally posted by jeff
Why do you insist on inserting this extra step in your reasoning by saying "energy has mass and mass generates gravity" when mass-energy equivalence allows you to dispense with this and just say that energy gravitates?
I looked at your site and you make a big deal out of the issue of whether or not the use of the term "relativistic mass" is legitimate. But really it's just a convention, some people like to use it - I'm one of them - and others think it's better not to, and I think they're being silly.
So you have your opinions and I have mine. I thought you agreed to disagree? What is it that you want?
I don't make a big deal of "relativistic mass." It's simply something I'm studying. I use that website to record what I'm doing. The site is has a purpose which is not what you may think it is. I solidify concepts in my mind by putting them in writing. And I put them on the internet because it's easier to communicate with people I know in e-mail this way. I don't have to hope that they can read MS - Word etc. And since I discuss this a huge amount on the internet then it's a great way to discuss this since it's easier to see equations this way.
Regarding why I refer to mass instead of energy - I've explained that to you already. Was I unclear?
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1-Newbie
## How to set the condition for the solution of a 4th order equation?
Hi All, I have a 4th order equation for an unknown parameter. The equation includes some constants whose number depends on the external setting.
The parameter must be a positive real number. In the attachment, I listed two possible symbolic solutions k2_1 and k2_2.
How can I set the if else condition correctly to achieve the following requirement?
(1) if both k2_1 and k2_2 are real and positive, show both of them as the solution of k2
(2) if only one of k2_1 and k2_2 is real and positive show this real number
(3) if neither k2_1 or k2_2 can satisfy the positive real condition, show an error message
Thank you!
1 ACCEPTED SOLUTION
Accepted Solutions
24-Ruby V
(To:yhuang-3)
Maybe something like the attached.
As you can see, as so often the symbolics ignores some constraints given via "assume".
25 REPLIES 25
23-Emerald III
(To:yhuang-3)
Introduce a new variable, e.g. 'x' with as definition:
x=k2^2, then your equations turns into:
(4/9)*x^2 - constant7*x + constant5 = 0
You can use simple school mathematics to solve x.
Then k2 values are found by taking the positive or negative square root of x.
Success!
Luc
23-Emerald III
(To:yhuang-3)
There are only 4 possible solutions (because it is a 4th order problem)
They are found with:
Success!
Luc
23-Emerald III
(To:yhuang-3)
The other option is to ask Mathcad to solve the 4th order straight away:
Success!
Luc
24-Ruby V
(To:yhuang-3)
Maybe something like the attached.
As you can see, as so often the symbolics ignores some constraints given via "assume".
1-Newbie
(To:Werner_E)
Hi Werner, I have a similar but more complicated calculation.
Now I have three variables with several constant numbers (depending on external setting). How can I get the symbolic solution of three equations?
My variables are k2, QE1 and QE2. Are are real numbers, and k2 must be positive. Would you teach me how to solve it symbolically?
It seems like I am not allowed to upload another worksheet in this thread. If you want me to open a new thread to upload the worksheet, I can do that.
Thank you!
23-Emerald III
(To:yhuang-3)
You should not be limited in uploading worksheets. But the option to attach a worksheet is only available when you use the 'advanced editing' mode, or when you edit your entry (lower left corner: choose 'actions', the 'edit').
Luc
1-Newbie
(To:LucMeekes)
Thank you LucMeekes! I tried to solve the equations and Mathcad yields that the results are too large to display. Is there any way I can see the results individually? For example, can I see the symbolic results of k2, QE1 and QE2 separately? (the file is attached in this thread)
24-Ruby V
(To:yhuang-3)
Here is an approach using a function with your eight constants as arguments.
As you can see Mathcad is not able to solve the sixth order equation it arrives at.
Not even with specific values for your constants, let alone with symbolic constants.
So you could try your luck using a numerical solve block. But you would have to provide a suitable guess and also you would get just one solution for each variable. To get another solution you would have change your guess.
24-Ruby V
(To:yhuang-3)
As Luc already explained you sure are allowed to add more worksheets but this system makes it unnecessarily difficult to do so.
As an alternative to manually switching to the advanced editor you could send your message at first without attaching your file. Then, when you chose "Actions" at the lower left and then "Edit" you will end up automatically(?) in advance mode.
To solve a system of 3 equations in 3 variables symbolically you either pot the equations in a 3 x 1 vector and the use the symbolic "solve, var1, var2, var3" or you setup a solve block (Given ..... Find) and evaluate this block (the finat "find") symbolically.
1-Newbie
(To:Werner_E)
24-Ruby V
(To:yhuang-3)
I just was about to edit my previous post as I had just seen you already attached your file and used the vector method anyway.
It looks like the result of your symbolic eval is not a normal 1 x 3 matrix as whenever I try to assign it a variable I get the message "variable not defined" !?
1-Newbie
(To:Werner_E)
Thank you Werner! I have some update in my calculation. If we take a step back and we do not need to worry about my previously mentioned 3 unknown variable stuff, would you teach me how to solve this fourth order symbolic equation in mathcad? It seems like a direct solve cannot work in Mathcad. Thank you!
23-Emerald III
(To:yhuang-3)
It's no problem to solve that equation with Mathcad (11). Here's an extremely small part of the result:
For further information see here Cubic equation with complex coefficients. this is on a 3rd order polynomial, your 4th order gives much bigger results!
This is as far as I can zoom out (10 %), every vertical bar marks a page; and you're looking at the first fraction only, but it is clear that there are 4 solutions:
Even mathcad 15 should be able to do it but likely will not show you the answer, and instead tell you that you can use the answer in further calculations.
Success!
Luc
24-Ruby V
(To:LucMeekes)
> It's no problem to solve that equation with Mathcad (11).
It is in Mathcad 15 (and other version using MuPad as symbolic processor) as you can see in my reply
> Even mathcad 15 should be able to do it but likely will not show you the answer, and instead tell you that you can use the answer in further calculations.
True, but this can be quite tricky as my reply shows.
23-Emerald III
(To:yhuang-3)
And of course, matters are much simplified if you can give the values of your constants up front.
Example:
Success!
Luc
24-Ruby V
(To:yhuang-3)
The symbolics in Mathcad (= an old subset of muPad) can be quite cumbersome and sure is far below the capabilities of Mathematica or Maple (Maple is the symbolic engine in Mathcad 11 and so Luc is able to outperform us 😉
Here are some remarks:
1-Newbie
(To:Werner_E)
Hi Werner, thank you so much for this detailed calculation. One problem I got in editing your worksheet is the function "trim". My mathcad 14 cannot recognize this function. How can I use the equivalent function to replace "trim"? Thank you!
24-Ruby V
(To:yhuang-3)
"trim" is a routine from the "Data analysis extension pack". Thats one of four add-on packs which usually were included with Mathcad. As far as I remember they were not included with educational single user student licences.
trim simply removes all entries/rows) in the first argument vector(matrix) at the positions given by the second argument vector.
24-Ruby V
(To:Werner_E)
Here's a trim alternative
Its a quick hack - no error checking, etc.
1-Newbie
(To:Werner_E)
Thank you Werner. I am using a corporation license instead of the educational student license. It is strange that I do not have it. Do you have any suggestions to use the alternative method to achieve the same thing? For example, can we use "if" and "otherwise" to do this?
24-Ruby V
(To:yhuang-3)
I just posted a workaround / trim replacement one minute before your reply 😉
I am not sure but maybe you had to pay for those extension packs at the times of MC14. I am pretty sure that they were free addons which came along and got installed automatically with MC15 using a regular license (singe user or corporate float). But I saw a student license of MC15 though were the help files/e-books got installed but the license would lock the features.
24-Ruby V
(To:yhuang-3)
Of course you don't need to duplicate "trim" or use a replacement.
The filtering could be done in many different ways - here's one of them:
1-Newbie
(To:Werner_E)
Thank you Werner, that helps a lot!
I have an additional question. If for this 4th order equation, if we know there will be 2 real solutions and 2 conjugated complex solutions, is it easier to get the symbolic solutions of them? Or there is no difference from the general 4th order equation with all coefficients unknown?
24-Ruby V
(To:yhuang-3)
I guess that this knowledge does not help to find the symbolic solutions.
23-Emerald III
(To:yhuang-3)
Mathcad 11 will solve this system purely symbolically, but the result contains root functions:
This is the full solution:
When provided with actual values for the eight constants you might get:
In this case, only the first (top) two sets of answers satisfy your requirements (QEi are real and k2 is positive)
Success!
Luc
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1. Oct 2, 2011
### leroyjenkens
1. The problem statement, all variables and given/known data
A 110-m-wide river flows due east at a uniform speed of 3.3 m/s. A boat with a speed of 8.6 m/s relative to the water leaves the south bank pointed in a direction 37 degrees west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground? Give the direction as the angle of the velocity from due north, positive if to the east and negative if to the west. (c) How long does it take for the boat to cross the river?
2. Relevant equations
I was using the PA = PB + BA equation. Which is supposed to be set up like this: the velocity of something with respect to something else, equals, the velocity of something with respect to something else, plus, the velocity of something with respect to something else. Now I know how to do this problem, because it's simple enough to do without using the equation I just mentioned. But I was supposed to use this equation to do the problem.
What I did instead was I just thought about it logically. The river is moving east and the boat is moving in the westward direction, so I separated the boat's velocity into X and Y components and subtracted the X component from the velocity of the river and then solved from there.
But as far as the equation goes, how do I set up the equation? Which velocity goes on the left side of the equation and which velocities go on the right side? As far as I can tell, it makes a huge difference.
I tried using that equation to solve the problem, but the way I set up the equation, I ended up with an equation that would add the boat's velocity to the river's velocity, which would mean the boat would go faster in the west direction, even though the river is moving east. That would make no sense.
So anyone know how to set up that equation? In my book it's under the "Relative Motion in One direction" section, if that helps.
Thanks.
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Draft saved Draft deleted | 491 | 2,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-47 | longest | en | 0.965907 |
https://www.jiskha.com/display.cgi?id=1248407215 | 1,516,617,446,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00535.warc.gz | 929,119,866 | 4,599 | # Chem
posted by .
Methylaime, CH3NH2.
If a 0.100 mol/L solution of methyamine has a pH of 11.80, calculate the ionization constant for the weak base. What is an ionization constant. Is it just K?
This is what I did so far:
11.80 = -log (concentration of h3o+)
10^-11.80 = (concentration of h3o+)
1.58 x 10^.12 = (concentration of h3o+)
K = (1.58 x 10^-12)^2 / (0.1-1.58 x 10^-12)
Not too sure what to do from here :S
• Chem -
Think NH3 and yes, Ka is the ionization constant.
You know what to do with NH3.
NH3 + HOH ==> NH4^+ + OH^- so in an analogous fashion,
CH3NH2 (this is just NH3 with a CH3 in place of one of the H atoms. These are bases, too, and react the same way.
CH3NH2 + HOH ==> CH3NH3^+ + OH^-
So use an ICE chart to determine OH^-, CH3NH3^+ and CH3NH2 and calculate Ka.
I am getting something like 4 x 10^-4. Look up Kb for CH3NH2 in your text tables to see what it should be.
• Chem -
I don't understand how to figure out the equation. I thought it would be this:
CH3NH2 + HOH ==> CH3NH+ + H30
:S How do I know what the products will be?
• Chem -
You KNOW
NH3 + HOH ==> NH4^+ + OH^-. You KNOE NH3 is a weak base but it is a stronger base than H2O, therefore, the H^+ is pulled from the HOH to form NH4^+.
Amines are just substituted ammonia and there are number of them. When you see NH2 it is RNH2 or RNH or R3N so we are talking about three classes of amines where R may vary. In this case we have
CH3NH2 where CH3 is the R group added. And RNH2 + HOH ==> RNH3^+ + OH^-/
The reaction is strictly analogous to NH3. Do accordingly.
• Chem -
I understand now. Thank you :)
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# cell phone
0
397
1
+129
In 1985, there were 285 cell phone subscribers in the small town of Centerville. The number of subscribers increased by 75% per year after 1985. How many cell phone subscribers were in Centerville in 1994?
Apr 13, 2018
#1
+95985
+1
We have the following exponential function
N = 285 (1.75)x where N is the number of subscribers after x years
Since 1994 is 9 years after 1985, we have
N = 285 (1.75)9 ≈ 43,872 subscribers
( That's a pretty big increase !!! }
Apr 13, 2018 | 173 | 522 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-09 | longest | en | 0.861386 |
https://www.emathhelp.net/pt/pinches-imperial-to-gills-imperial/ | 1,652,773,033,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00705.warc.gz | 904,944,795 | 6,038 | # Pinches (imperial) to gills (imperial)
This free conversion calculator will convert pinches (imperial) to gills (imperial), i.e. pinch (imp) to gi (imp). Correct conversion between different measurement scales.
A fórmula é $V_{\text{gi (imp)}} = 0.005208333333333 V_{\text{pinch (imp)}}$, onde $V_{\text{pinch (imp)}} = 15$.
Portanto, $V_{\text{gi (imp)}} = 0.078125$.
Resposta: $15 \text{pinch (imp)} = 0.078125 \text{gi (imp)}$. | 154 | 436 | {"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} | 2.875 | 3 | CC-MAIN-2022-21 | latest | en | 0.275967 |
https://community.qlik.com/thread/103575 | 1,532,182,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592579.77/warc/CC-MAIN-20180721125703-20180721145703-00570.warc.gz | 616,940,237 | 24,518 | 11 Replies Latest reply: Jan 8, 2014 2:23 AM by Senthil Raja
# Difference between sum and expression Total in Straight table
Hi,
Can anyone please tell me the difference between sum and Expression total in the expression tab of Straight table?
• ###### Re: Difference between sum and expression Total in Straight table
Grand total of the expression. u can suppress the Grand total by selecting No Total Option.
• ###### Re: Re: Difference between sum and expression Total in Straight table
Hi Raja,
Thanks for the reply.
I want to know the difference between 'expression total' and 'sum', as highlighted in the below picture. The Problem is not about showing the total or hiding it...
• ###### Re: Difference between sum and expression Total in Straight table
Hi,
Sum - will give you the sum of all rows in the table
Expression Total - will give you the overall expression total.
Regards,
Jagan.
• ###### Re: Difference between sum and expression Total in Straight table
Do u mean it like, Expression total will return in the sum without considering what dimension we are using?
• ###### Re: Difference between sum and expression Total in Straight table
Hi Sundarakumar,
I lot cases the output of the both options is same. Except the some Situation Like, when you are having for one parent node as two child's (Value 100) the total of parent in Expression Total is 10 and sum total will be 20.
• ###### Re: Difference between sum and expression Total in Straight table
Thanks for the reply, sorry i dont understand what u r trying to explain.
Can u please elaborate?
• ###### Re: Difference between sum and expression Total in Straight table
Hi,
Sum - will give you the sum of all rows in the table
Expression Total - will give you the overall expression total.
For example:
Year A B
2013 50 100
2014 50 150
When you use A/B as expression and use Year as Dimension
Sum Mode - 83% (50 + 33) (50/100 + 50/150 = 50% + 33 % = 83%)
Expression Mode - 40% (100/250) (50 + 50) / (100 + 150)
Regards,
Jagan.
• ###### Re: Re: Difference between sum and expression Total in Straight table
Hi jagan,
Please find the attachment.
Y am i not getting the expression total in it? I have used what you have explained above in the app.
Can you please explain?
Thanks for the response.
-Sundar
• ###### Re: Difference between sum and expression Total in Straight table
Yes Jagan Exatly..Thanks
Based on the your needs you can set.
Fr example,
If you want max sale or min sales, avg sales, least year sales..
• ###### Re: Difference between sum and expression Total in Straight table
Hi Sundar,
PFA...
Reagrds,
raja
• ###### Re: Difference between sum and expression Total in Straight table
Difference is shown significantly when using distinct in expressions,
Expression Total Calculates the Expression regardless of all dimensions just similar as you write your expression in a text box.
Sum of Rows > Sums the values calculated by expression in all rows of the chart. | 696 | 3,031 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.836078 |
https://oeis.org/A107134 | 1,669,542,417,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710218.49/warc/CC-MAIN-20221127073607-20221127103607-00098.warc.gz | 488,014,569 | 4,302 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A107134 Primes of the form x^2+28y^2. 2
29, 37, 53, 109, 113, 137, 149, 193, 197, 233, 277, 281, 317, 337, 373, 389, 401, 421, 449, 457, 541, 557, 569, 613, 617, 641, 653, 673, 701, 709, 757, 809, 821, 877, 953, 977, 1009, 1033, 1061, 1093, 1117, 1129, 1201, 1213, 1229, 1289, 1297, 1373, 1381 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Discriminant=-112. See A107132 for more information. LINKS Vincenzo Librandi and Ray Chandler, Table of n, a(n) for n = 1..10000 [First 1000 terms from Vincenzo Librandi] N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references) FORMULA The primes are congruent to {1, 9, 25} (mod 28). - T. D. Noe, Apr 29 2008 MATHEMATICA QuadPrimes2[1, 0, 28, 10000] (* see A106856 *) PROG (Magma) [ p: p in PrimesUpTo(2000) | p mod 28 in {1, 9, 25} ]; // Vincenzo Librandi, Jul 23 2012 CROSSREFS Cf. A139643. Sequence in context: A355161 A031925 A325058 * A139851 A139895 A304688 Adjacent sequences: A107131 A107132 A107133 * A107135 A107136 A107137 KEYWORD nonn,easy AUTHOR T. D. Noe, May 13 2005 STATUS approved
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Last modified November 27 03:51 EST 2022. Contains 358362 sequences. (Running on oeis4.) | 667 | 1,896 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.717406 |
https://math.stackexchange.com/questions/1823013/coding-theory-linear-codes | 1,571,460,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986688826.38/warc/CC-MAIN-20191019040458-20191019063958-00049.warc.gz | 577,892,771 | 32,329 | # Coding theory - Linear codes
How can I prove the following :
1) Let C be a binary linear code in a binary vector space V. Let v be a vector not in C and consider the coset v+C. Show v+C is not a linear code.
2) suppose C is a linear code and that v+C is a coset of C. Suppose a and b are members of v+C. Prove a-b is a member of C.
What I know :
1) A coset of C is x + C = {x + y | y ∈ C}.
2) A linear code is a subspace of V(n, q) which is a union of disjoint cosets of C;
3) A linear code C is a code for which whenever x,y is in C then ax+by is in C for all a,b in Fq , that is C is a linear subspace of (Fq)^n
I understand that I would have to piece together the above three statements to prove both questions.
Any help or guidance is much appreciated.
• looks like homework – NaCl Jun 12 '16 at 9:39
• No, More like preparation for my final exam, it's a question that popped up in a past exam, something I don't want to regret not learning later on. – Keshav Chetty Jun 12 '16 at 9:41
For 1, ask yourself: can $0$ be in $v + V$, if $v \notin V$?
For 2, write $a = v + v_1$ for some $v_1 \in V$. Also $b = v + v_2$ for some $v_2 \in V$. Now what is $a - b$?
For 1, your task is simplified, since you are in the binary vector space domain. Now consider two words $v + k_1$ and $v + k_2$, where $k_1$ and $k_2$ are codewords in $C$. So the sum of the two words becomes $k_1 + k_2 = k'$ where $k' \in C$, and hence does not belong to $v + C$.
For 2, by similar reasoning that the difference of two codewords in $C$ is also a codeword, you can get that $a - b \in C$.
• Exploiting characteristic $2$ does not meaningfully simplify the argument, and somewhat obscures the essence: you could just as easily take the difference of the two words (which you already have done in part 2). – Erick Wong May 4 '17 at 18:48 | 566 | 1,827 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-43 | latest | en | 0.937504 |
https://www.scribd.com/document/293702390/ME-Problems-PC-Monopoly-1 | 1,553,383,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203093.63/warc/CC-MAIN-20190323221914-20190324003914-00351.warc.gz | 925,440,227 | 54,461 | You are on page 1of 7
# Problems on Perfect Competition & Monopoly
1. True and False questions. Indicate whether each of the following statements is true
or false and why.
(a) In long-run equilibrium, every firm in a perfectly competitive industry earns an
economic profit.
(b) Pure competition exists in a market when firms are price makers as opposed to
price takers.
(c) Downward-sloping industry demand curves characterize monopoly markets;
horizontal demand curves characterize perfectly competitive markets.
(d) Market structure describes the competitive environment in the market for any good
or service.
(e) A pure monopoly does not have to worry about suffering losses because it has
the power to set its price at any level it desires.
(f) Assuming a linear demand curve , a firm that wants to maximize its revenue will
charge a lower price than a firm that wants to maximize its profits.
(g) If P > AVC, a firm's total fixed cost will be greater than its loss.
(h) When a firm is able to set its price, its price will always be less than its MR.
2. A firm its output in a perfectly competitive market. The firm's total cost function is
given in the following schedule:
Output (units)
0
10
20
30
40
50
60
Total cost (\$)
50
120
170
210
260
330
430
The prevailing market price is \$7 per unit.
(a) What is the firm's profit maximizing output level?
3. A company operating in a perfectly competitive is faced with the following
total costs per ton (Q) of its output:
TC = \$500,000 + \$400Q + \$0.04Q2
(a) Calculate the industry price necessary to induce short-run firm supply of 5,000,
10,000, and 15,000 tons of output. Assume that MC > AVC at every point along
the firm's marginal cost curve and that total costs include a normal profit.
(b) Calculate short-run firm supply at industry prices of \$400, \$1,000, and \$2,000
per ton.
21 50. Q (1000) 1 2 3 4 5 6 7 8 9 10 Price 1650 1570 1490 1410 1330 1250 117 1090 1010 930 850 MR AVC AC MC 1570 1410 1250 1090 930 770 610 450 290 130 1281 1134 1009 906 825 766 729 714 721 750 2281 1634 1342. for which the cost of production is given below.67 871.00 48.78 43.70 30.10 40.90 33. What impact will this have on the firm's production levels and profit? Explain.19 46. What is the amount of profit (or loss) earned by the firm at the optimal level of production? (b) Suppose the firm does enter the market and that competition causes the price of the telephones to fall to \$35. what would be the profit maximizing output? What is the total profit (loss)? Should the firm produce at this price? 5.10 40.59 44.10 30.70 31. What would you advice this firm to do? 6.90 37.73 45.10 38.00 MC 30.10 30.36 45.90 51.10 42.96 44.00 AC 52. Do you think this company should enter the market? Explain.10 (a) Suppose the average wholesale price of this a wireless phone currently \$50. It estimates that if it were to begin production.70 36.90 38. Also given are quantities and prices that the firm believes it will be able to sell.40 37.74 44. A firm operating in perfectly competitive market (PCM) has the following total cost function: TC = 6000 + 400Q – 20Q2 + Q3 (a) What is the lowest (or minimum) price at which the firm will shut down in the short run? (b) If the market price is \$310. A manufacturer of electronics products is considering entering the telephone equipment business.50 37.60 37.70 57.86 839 832.00 39.11 850 1281 987 759 597 501 471 507 609 777 1011 .60 37.70 46.85 43.17 43. This same manufacturer of electronics products has developed a handheld computer. its short run cost function would be as follows: Q (1000) 9 10 11 12 13 14 15 16 17 18 19 20 AVC 41.4.40 39.33 1156 1025 932.10 39.
(d) price exceeds average revenue. (b) no barriers to entry or exit. (b) industry demand equals industry supply. A firm will earn normal profits when price: (a) equals average total cost. (c) increases in productivity that are successfully anticipated. (c) industry demand is less than industry supply. MCQs 7. (c) total variable cost exceeds total revenue. In long-run equilibrium. (b) total cost exceeds total revenue. (d) pure luck. . At the profit maximizing level of output for a monopolist: (a) P = AR and AR = AC (b) P = MC and MR > MC (c) P > MC and MR = MC (d) P = MR and AC = MC 9. (b) equals average variable cost. what amount would you recommend? Explain. monopoly prices are set a level where: (a) price exceeds marginal revenue. (c) equals marginal cost. (c) What arguments cam be made for charging a price lower than the profit maximizing level? If a lower is indeed established.(a) What price should the firm charge if it wants to maximize profit in the short run? (b) What arguments cam be made for charging a price higher than this price? If a higher price is indeed established. a perfectly competitive firm will shut down and produce nothing if: (a) excess profits equal zero. 10. 8. (d) complete knowledge of market price. Which of the following is not characteristic of perfect competition? (a) a differentiated product. 12. (d) exceeds minimum average total cost. (c) large number of buyers. 11. (d) the market price falls below the minimum average total cost. (b) decreases in cost that are successfully anticipated. what amount would you recommend? Explain. Above-normal profits in a perfectly competitive market are caused by: (a) increases in demand that are successfully anticipated. In the short run.
When the slope of the total revenue curve is equal to the slope of the total cost curve: (a) monopoly profit is maximized. Assume a profit maximizing firm's short-run cost is TC = 700 + 60Q. (b) continue operating in the short run even though it is losing money. When a firm has the power to establish its price: (a) P = MR. (b) P = MC. If its demand curve is P = 300 . The main difference between the price-quantity graph of a perfectly competitive firm and a monopoly is: (a) that the competitive firm's demand curve is horizontal. True & false . (d) that a monopoly does not incur increasing marginal cost. (b) normal. 16. (c) continue operating because it is earning an economic profit. (c) shut down in the long run. (d) Cannot be determined from the above information. (c) that a monopoly maximizes its profit when marginal revenue is greater than marginal cost. 17. it should: (a) shut down immediately. (c) P > MR. When a firm produces at the point where MR = MC. (c) above normal. 18. (b) that a monopoly always earns an economic profit while a competitive company always earns only normal profit. 14. the profit that it is earning is considered to be: (a) maximum. (b) the total revenue is maximum.15Q. (d) shut down if this loss exceeds fixed cost. If a perfectly competitive firm incurs an economic loss. 15. (d) not enough information is provided.13. (d) P < MR. ANSWERS: 1. (c) the marginal cost curve intersects the total average cost curve. (b) try to raise its price. what should it do in the short run? (a) shut down. (d) the total cost curve is at its minimum. while that of the monopoly is downward sloping.
and conditions of entry and exit. a firm will price its product at the point where MR=0. then the company will cover some of its fixed costs. can they stop the erosion in demand due to the one-hour photo developing machines and cameras that record images electronically on discs?) (f) True. 2. loss will be less than fixed cost (Contribution Margin is positive). this must be a lower price than the point where MR=MC. In long-run equilibrium. even if Polaroid continues to have a monopoly on cameras that use instant developing film. However. If P>AVC but P<AC. high cost producers will be forced to exit. Even a pure monopoly has to consider the possibility of demand falling below the level sufficient to earn a profit. At Q1 = 10 π = (P · Q) − TC = 7 (10) − 100 = −\$30 At Q2 = 50 π = 7 (50) − 310 = \$40 . Profit-maximizing output level (Q*) occurs where MR = MC. the firms that remain will continue to operate and earn a normal rate of return on investment. Such firms take industry prices as a given. you need to calculate MC: Output (units) 0 10 20 30 40 50 60 Total cost (\$) 50 120 170 210 260 330 430 MC -7 5 4 5 7 10 (a) P = MR = \$7 Profit-maximizing output level (Q*) occurs where MR = MC MC = \$7 at Q1 = 10 and Q2 = 50 units. (c) False. Following a decrease in industry prices. the degree to which products are similar or dissimilar. (e) False.(a) False. Market structure is typically characterized on the basis of four important industry characteristics: the number and size distribution of active buyers and sellers and potential entrants. Horizontal demand curves characterize perfectly competitive firms. (For example. (b) False. the amount and cost of information about product price and quality. thus. (g) True. Downward sloping demand curves follow from the law of diminishing marginal utility and characterize both perfectly competitive and monopoly market structures. In order to maximize revenue. Pure competition exists in a market when individual firms have no influence over price. Price will be more than MR. (h) False. every firm in a perfectly competitive industry earns zero excess profit. (d) True. By implication.
If P = \$50.5.600 (Note: Variable Cost = \$400Q + \$0.000) = 7. below which the firm should shutdown) (b) @ P =\$310.625 (substitute Q* =10.000: P = MC = \$400 + \$0. and AVC = \$400 + \$0. However.500 P = \$2.000: Q = -5. at P = \$400: Q = -5. b. Here: MC = ∂TC/∂Q = \$400 + \$0.000 + 12. 3. It might also want to set a higher price if it suspected that future competition would eventually force all competitors to lower their . at: Q = 5.08(10.Therefore Q* = Q2 = 50 units (b) Industry is not in long-run equilibrium since MC = MR = P > ATC (i.000) = \$800 Q = 10. or where MR = MC) @ Q* =10. However.200 Q = 15. Thus the firm should shut down.000 + 12.000) = 20.08Q = -400 + P Q = -5. it may want to consider charging a higher price if it wanted to position its product as a “premium” product.96 (rounded to \$100).000: Q = -5.36.e..5 in both TR & TC functions). this would cause it to lose \$136. above "normal" profits are being earned by one (or more) firms in the industry).72. by following the MR=MC rule). profit maximizing output Q* =10.) (b) When quantity is expressed as a function of price.5(\$2.5 units (where TR – TC is highest.08(5.000 4. The above price would enable a firm to earn a maximum amount of total profit in the short run.5(\$1.04Q2. so MC > AVC at each point along the firm's short-run supply curve. a sum greater than the implied fixed cost of \$99. (a) The marginal cost curve constitutes the short-run supply curve for firms in perfectly competitive industries provided price exceeds average variable cost. @ Q* =10. then the best that the firm could do by operating would be to produce 14 units (i. assuming P > AVC.000) = \$1.5. (a) Yes.000) = \$1..000: P = MC = \$400 + \$0. this firm makes a loss of \$5897. the price necessary to induce short-run firm supply of a given amount is found by setting P = MC.5P Therefore. The firm should continue production so long it covers its AVC (contribution margin is positive) 5.04Q.000: P = MC = \$400 + \$0.08(15.5(\$400) = 0 P = \$1.000 + 12. (b) If P = \$35.000 + 12. the firm's supply curve can be written: P = MC = \$400 + \$0. a. P* = \$1090. (a) P = \$300 ( = Minimum AVC. AVC = \$300. Because P = MR.e. the firm should produce 18 units and earn an economic profit of \$108.08Q Therefore. 6.5 < P = \$310.08Q 0.
if the firm charged \$850. d 14.d 15.price. c 12. As can be seen in the numerical example. we can only say that the firm would set a higher price if it gives greater priority to goals mentioned above. c. a 11. The firm would want to consider setting a price lower than \$1090 if it wanted to increase its revenue (i. its total revenue would be \$850.a . a 13. There may be other reasons for lowering the price..c 16.e. It may also choose to be an aggressive price-cutter in an oligopolistic market. In fact.c 17. Without more specific data about these other considerations. For example. As a generalization.000 (as compared to \$763. market share). a 8. c 9. d 10. it would be difficult to suggest a specific price that is higher than \$1090.a 18. it could continue lowering its price in order to increase its revenue up to the point at which MR=0 (not shown in the table). the firm may wish to use the strategy of “learning curve pricing” (see Chapter 8).000 at the price of \$1090). MCQS 7. | 3,359 | 12,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.889407 |
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• 1.
### 16. A sound wave is an example of a(n) ______________ wave.
• A.
Transverse
• B.
Longitudinal
• C.
Crest-and-trough
• D.
Electromagnetic
B. Longitudinal
Explanation
A sound wave is an example of a longitudinal wave because it consists of compressions and rarefactions that travel in the same direction as the wave itself. In a longitudinal wave, the particles of the medium vibrate parallel to the direction of the wave propagation. This is in contrast to a transverse wave, where the particles vibrate perpendicular to the direction of the wave. A crest-and-trough wave refers to a type of transverse wave, and electromagnetic waves are a different type of wave altogether, characterized by the oscillation of electric and magnetic fields.
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• 2.
### 17. A sound wave consists of a series of
• A.
Compressions and rarefactions.
• B.
Longitudes and latitudes.
• C.
Hills and valleys.
• D.
Perpendicular vibrations
A. Compressions and rarefactions.
Explanation
A sound wave is a longitudinal wave, meaning it travels by compressing and rarefying the medium it passes through. In compressions, the particles are pushed closer together, resulting in a region of high pressure. In rarefactions, the particles are spread out, creating a region of low pressure. These alternating compressions and rarefactions form the series of disturbances that propagate as a sound wave.
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• 3.
### 18. The human perception of pitch depends on a sound’s
• A.
Velocity.
• B.
Wavelength.
• C.
Frequency.
• D.
Amplitude
C. Frequency.
Explanation
The human perception of pitch refers to our ability to perceive high or low frequencies of sound. Pitch is directly related to the frequency of a sound wave, which is the number of cycles or vibrations per second. Higher frequencies are perceived as higher pitches, while lower frequencies are perceived as lower pitches. Therefore, the correct answer is frequency.
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• 4.
### 19. When you hear the sound from a vehicle that is moving toward you, the pitch is higher than it would be if the vehicle were stationary. The pitch sounds higher because the
• A.
Sound waves arrive more frequently.
• B.
Sound from the approaching vehicle travels faster.
• C.
Wavelength of the sound waves becomes greater.
• D.
Amplitude of the sound waves increases.
A. Sound waves arrive more frequently.
Explanation
When a vehicle is moving towards you, the sound waves it produces are compressed, causing them to arrive at your ears more frequently. This results in a higher pitch because the frequency of the sound waves determines the perceived pitch.
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• 5.
### 20. The highness or lowness of a sound is perceived as
• A.
Compression.
• B.
Ultrasound.
• C.
Wavelength.
• D.
Pitch.
D. Pitch.
Explanation
The highness or lowness of a sound is perceived as pitch. Pitch refers to the frequency of a sound wave, with higher frequencies being perceived as higher pitches and lower frequencies being perceived as lower pitches. Compression, ultrasound, and wavelength are not directly related to the perception of the highness or lowness of a sound.
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Related Topics | 854 | 3,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-10 | latest | en | 0.884184 |
https://serc.carleton.edu/sp/ssac_home/general/examples/17797.html | 1,508,503,206,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00318.warc.gz | 863,562,658 | 8,147 | # What Time Did The Potato Die?
This material is replicated on a number of sites as part of the SERC Pedagogic Service Project
#### Summary
In this Spreadsheets Across the Curriculum activity, students use "crime scene" data of body temperature vs. time to determine the time of death of a potato crime victim. Students will learn to create a spreadsheet, make graphs, add trend lines, and work with linear and exponential curves in an effort to graphically estimate and algebraically calculate the potato's time of death. The module introduces the mathematical concepts of linear and exponential graphing, unit conversions, and trend lines.
## Learning Goals
Students will:
• Learn how to create and work with a spreadsheet in Excel.
• Learn how to create a graph within Excel.
• Gain experience with adding and working with trend lines.
• Gain experience with converting units while solving a problem.
• Gain experience with interpreting graphs and choosing the best graph to represent a given set of data.
In the process, the students will:
• Model an exponential-decay phenomenon to solve a problem.
## Context for Use
This activity was designed for lower-division college students taking forensic science. Student math levels were expected to be diverse with some students not having math beyond high school precalculus. The module could be used in an advanced high school class.
## Description and Teaching Materials
PowerPoint SSAC2005:HV8079.RS1.1_student (PowerPoint 403kB May27 10)
The module is a PowerPoint presentation with embedded spreadsheets.
If the embedded spreadsheets are not visible, save the PowerPoint file to disk and open it from there.
This PowerPoint file is the student version of the module. An instructor version is available by request. The instructor version includes the completed spreadsheet. Send your request to Len Vacher (vacher@usf.edu) by filling out and submitting the Instructor Module Request Form.
## Teaching Notes and Tips
This module best works as an in-class assignment in a computer lab where students each have access to a computer. I schedule a two-hour time block for this module. If you have a lot of students to assist, it is quite effective to have those students very familiar with Excel help others.
## Assessment
The last page of the module has a homework assignment asking students to determine the time of death of a zucchini crime victim. | 482 | 2,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-43 | latest | en | 0.870486 |
https://www.thestudentroom.co.uk/showthread.php?page=7&t=4122915 | 1,519,130,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812938.85/warc/CC-MAIN-20180220110011-20180220130011-00783.warc.gz | 911,793,152 | 42,767 | You are Here: Home >< Maths
# OCR (Not MEI) S1 Wednesday 8th June 2016 watch
1. (Original post by catwat99)
How?? I did (3x2x1x2)/7P4
Isn't it because of the repetitions it's over 210
2. (Original post by Breynolds2671999)
I thought i did it wrong, can't believe i actually got it right
thats was i did!! omd i thought i was wrong, does anyone remember wat answer came out?
3. (Original post by RochL)
S2g im only person who got 2/35 for ABBA 😂😂 what was the actual answer?
I got that too
4. for the barbara question, did anyone else get 1/210?
5. (Original post by catwat99)
How?? I did (3x2x1x2)/7P4
i did same as you but over 7c4 as those were the total ways in which the 4 cards could be selected...... i thought
6. For the second part of the 1st question, could you use binomial ?
Posted from TSR Mobile
7. (Original post by Jamesk123)
I got 2/35
I was unsure but asked my teacher who got 1/70 same as me and so did one of my friends but there also seem to be a lot of other answers out there so I am unsure
8. so for the ABBA question, wat was the actual answer??
9. (Original post by student199919)
for the barbara question, did anyone else get 1/210?
No i got 1/210 but there were multiple a, b and r's so i did 1/210 x 3!2!2!, probably got it wrong but it is only 1 mark
10. (Original post by student199919)
for the barbara question, did anyone else get 1/210?
Yes
11. who thinks this years paper was harder than last years?
12. (Original post by Ella rivers1)
For the second part of the 1st question, could you use binomial ?
Posted from TSR Mobile
Yeah i think you were meant to, i did anyway
13. How do you do the standard deviation question and what was the answer
14. (Original post by Ozil5)
Yeah i think you were meant to, i did anyway
wat was the question?
15. Also did anyone get 146.5 for the mean of the apples?
16. (Original post by StruanCaughey)
I was unsure but asked my teacher who got 1/70 same as me and so did one of my friends but there also seem to be a lot of other answers out there so I am unsure
I got 7C4 as the total amount =35
And then 2! X 2! = 4 so 4/35
17. How did you work out the box plot question?
18. For the question with 9<X<20 I did P(X<20) - P(X<10), which is the same as (1-P(X>19)) - (1- P(X>9)) then just work it out ...
Is that right?
19. (Original post by kassy12324)
How did you work out the box plot question?
149-140
20. (Original post by Breynolds2671999)
No i got 1/210 but there were multiple a, b and r's so i did 1/210 x 3!2!2!, probably got it wrong but it is only 1 mark
I did exactly the same and I don't see why it's wrong?
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Let there be a pack of $100$ cards numbered $1$ to $100$. The $i^{th}$ card states: "There are at most $i - 1$ true cards in this pack". Then how many cards of the pack contain TRUE statements?
1. $0$
2. $1$
3. $100$
4. $50$
5. None of the above.
edited | 307 views
Option D should be the correct one.
that is $50$ cards of the pack contain true statements.
Why?
Because if the statement written on card number $x$ is true then all the statements written in card numbers $x + 1$ to $100$ must be true.
For Example if Card number $3$ is true, then according to the statement written in this card, "There are at most $3 - 1(=2)$ true cards in the pack".
This implies number of true cards must be less than $3$.
Now the statement on card number $4$ will imply that the number of true cards must be less than $4$.
similarly the statement on card number $100$ will imply that the number of true cards must be less than $100$.
So if statement written on card number $3$ is true then all the statements written on the card numbers $4$ to $100$ will vacuously be true.
But now the the number of true cards will be $98$ (from card number $3$ to $100$) hence the statement on the card number $3$ must be false.
Clearly this is inconsistent so card number $3$ can not be a true card.
Conclusion:
If card number $x$ is a true card then:
1. There are at least $(100 - x) + 1$ true cards. and
2.Total Number of true cards must be less then or equal to $x - 1$
(where $x$ belongs to Integers between $1$ and $100$).
For any value of $x \leq 50$ both of the above statements can not be true simultaneously, so none of the cards from $1$ to $50$ is a true card.
For any value of $x \geq 51$ both of the above statements can be true at the same time, so all of the cards from $51$ to $100$ must be true cards & the total number of true cards will be $50$.
It can be observed that comparing one of the boundary cases of each of the above two statements will give us of the boundary cases of our answer.
that is, on solving:
$(100 - x) + 1 = x - 1$
we get $x = 51$ which indeed is our smallest true card.
by Boss (14.3k points)
selected by
0
wow !! | 593 | 2,164 | {"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} | 4.09375 | 4 | CC-MAIN-2020-05 | latest | en | 0.890788 |
https://gulfnews.com/games/play/play-sudoku---a-numbers-game-anyone-can-enjoy-1.1621226497793 | 1,623,795,413,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00501.warc.gz | 271,726,348 | 17,043 | Welcome to Sudoku.
This puzzle does involve numbers, but you definitely don’t have to be a maths whiz to solve it. All you really need to do is think logically and be unafraid to use trial and error to explore different numerical placements.
Let’s consider the grid – it’s a 9x9 square containing 81 cells. The grid is subdivided into nine 3x3 blocks. Some of the cells already have numbers filled in and these are called givens. Your goal is to complete the whole grid using the numbers 1 to 9 so that each row, column and block contains each number only once.
Seems simple enough, right? Except, mathematicians have calculated that there are approximately 6.671×1021 or 6,670,903,752,021,072,936,960 different possible combinations for Sudoku puzzles. That’s a number with seven commas! I had to Google what such an astronomical number was called, since it’s not something you encounter every day. The answer is – 6 sextillion. That’s the number of possible Sudoku grids one can create.
Don’t panic just yet. This is good news for puzzle solvers like you. It means you don’t have to ever work on the same Sudoku puzzle twice.
It also means hours of relaxing fun that you can indulge in on gulfnews.com, while you pass time on the Metro, play with friends or and family, have a leisurely weekend breakfast, or take a quick break at work.
Remember, Sudoku isn’t a guessing game. Take your time, use logic, and revel in the accomplishment of completing one of these tricky puzzles.
To enjoy the game even more, invite a friend using the ‘Play Together’ button and solve the puzzle together.
Let us know if you found the Sudoku puzzles challenging or entertaining at games@gulfnews.com. | 382 | 1,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.942951 |
https://www.eevblog.com/forum/chat/web-building-software-wordpress-or-joomla/?prev_next=next;PHPSESSID=gvr77mjsq0a4gv9oa71n04mjl3 | 1,675,443,943,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500058.1/warc/CC-MAIN-20230203154140-20230203184140-00189.warc.gz | 751,856,733 | 13,260 | ### Author Topic: How useful is triangle wave in function generator ? (Read 10881 times)
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#### xani
• Frequent Contributor
• Posts: 400
##### How useful is triangle wave in function generator ?
« on: April 10, 2010, 03:14:51 pm »
Im thinking about buying this: http://translate.google.pl/translate?hl=pl&sl=pl&tl=en&u=http%3A%2F%2Fwww.ambm.pl%2Findex.php%2Furzadzenia-kontrolno-pomiarowe%2Fdd1a.html for my mini lab but it doesn't generate triange wave, just sinus and square and im wondering if i should ge thos or "normal" func. generator. How useful is triangl wave out ?
#### alm
• Guest
##### Re: How useful is triangle wave in function generator ?
« Reply #1 on: April 10, 2010, 10:41:18 pm »
It's useful for testing amplifiers, it's much easier to see distortion and clipping than with sine waves. It's also useful for testing comparators.
This seems quite different from a regular function generator. A function generator can usually generate sine/square/triangle waves from 1Hz to a few MHz, with amplitudes up to 20V in a high-Z load. Most offer symmetry adjustment (adjust duty cycle for square waves and makes triangle waves into ramps). The frequency stability is usually lousy, especially for the non-DDS ones.
This seems more targeted at the RF segment, more like a signal generator with a square wave thrown in. The frequency range is much higher, the amplitude is lower (max 2Vp-p in high-Z load). Max 5mV offset. Probably fairly good stability since it's DDS. The spectral purity at 1Vp-p, 1MHz looks quite good (all harmonics < -57dB or so). The square wave is pretty fast (<8ns rise/fall time), but has a fixed output amplitude. Not very useful for eg. 3.3V logic.
It may be a fine unit for the intended use, I have no experience with this product, but the specs seem pretty good. It seems less versatile than a normal function generator (which is the jack-of-all-trades of signal sources), as long as you don't need the extra range, stability or purity.
#### xani
• Frequent Contributor
• Posts: 400
##### Re: How useful is triangle wave in function generator ?
« Reply #2 on: April 11, 2010, 12:18:38 am »
Im wondering what i should get because for similiar price i can get typical 2-3MHz func. gen. For now i plan to use it for audio/SMPS (and i plan building tesla coil ^^ ) but after seein EEVBlog Instek review i didnt like that "wobbling" in wave frequency and thats why im wondering if i should get that DDS gen.
#### Kiriakos-GR
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##### Re: How useful is triangle wave in function generator ?
« Reply #3 on: April 11, 2010, 02:18:58 am »
Well there is another alternative .. way .
All that you need are one generator , in KIT with one TL082 ... ( that you have to assemble by your self ) , and one Frequency counter ..
I have make this setup for my self , but I own one good frequency counter ,
the LEADER LDC-831 .
As alternative of the dedicated frequency counter , you can use an multimeter with such ability , it would possibly cost less.
Still , the only issue with this setup , its that you have to tune it , its time that you like to use it,
by using both devices simultaneously .
Just an idea ..
« Last Edit: April 11, 2010, 02:59:05 am by Kiriakos-GR »
#### xani
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• Posts: 400
##### Re: How useful is triangle wave in function generator ?
« Reply #4 on: April 11, 2010, 08:51:29 am »
I can always use scope for that
Well i have ICL8038 so i could make like 100kHz func. gen with it but its kinda low compared to cheapest func. gen. and i'd still have to make all that additional circuits (frequency sweep, DC offset etc.) to make it comparable to even cheapest func. gen.
I guess I'll go with normal func. generator for now, i dont need such high freq generator for now and things like frequency sweep sounds quite useful for what i'll be doing
#### Kiriakos-GR
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##### Re: How useful is triangle wave in function generator ?
« Reply #5 on: April 11, 2010, 11:30:58 am »
Well , I have buy this gen. about 15 years back , from a Greek company called as " Smart Kit " ,
it has lots of Kits .
I just visited their page , and found the link ..
So have a look, to see what I am talking about.
http://www.smartkit.gr/details2.php?lang=1&wh=6&searchttile=1008&thepid=119&lang=2
There are four potentiometers on it ...
The first adjusts the frequency ,
the other three adjust the outputs , of its signal .
It has three separate outputs .
sinusoidal - triangular - square .
About the range it is 20 Hz and 20 KHz , with three scales selector .
Because of this thread , I took in my hands , so to test it , I do not use it much this days .
And found the frequency potentiometer , to be out of shape ...
I will probably exchanged , with a modern "10 turns" one , so to be easier to get tuned , with better accuracy .
Speaking about cost , even today it cost as 22 EUR or about 30\$ .
#### xani
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##### Re: How useful is triangle wave in function generator ?
« Reply #6 on: April 11, 2010, 01:28:33 pm »
i got one of similiar kits (it was 10Hz-100khz function generator) but in the end it was qute buggy schematic + it had just one "cover-it-all" range so even with 10 turn pot. it was hard to get some low freq wave out of it.
I was thinking about getting one of those 4-in-1 things like for example http://www.alibaba.com/product-gs/240859288/UNIVERSAL_SYSTEMS_DF6911.html as i need some decent power supply and 2nd multimeter and im wondering if that wouldn't be better instead of dedicated generator
#### hans
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##### Re: How useful is triangle wave in function generator ?
« Reply #7 on: April 11, 2010, 01:55:47 pm »
A triangle wave is very useful use in Telecommunications, analog encoding of audio (PWM signals, for example in class D amplifiers). Missing out on that for a DDS generator is quite a bad thing I suppose.
Though you can easily create a little triangle generator if you know the frequency with an opamp, resistor and capacitor.
You would need to calculate the R and C for the specific frequency you would be working on, because it's related to tau (tau = R*C). Just take a capacitor for you frequency range and a 10k pot so you can shift it a bit around. The input is just a basic 0V to 10V square in this case.
So; having it built is very useful, but if you only use it so rarely (as it seems, because you're asking us how useful it is ) you could get away with building a circuit up for it everytime. In the end you can't keep your devices connected to a function generator all the time, so building up a square to triangle wave is something you would have to do anyway.
#### xani
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##### Re: How useful is triangle wave in function generator ?
« Reply #8 on: April 11, 2010, 02:20:32 pm »
yeah, but then atm i dont need high freq. generator either, so i was wondering what's better, to have more availble freq or be able to generate triange + frequency sweep + wave width regulation, now im thinking that "real" function generator would be more useful ;]
#### Kiriakos-GR
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##### Re: How useful is triangle wave in function generator ?
« Reply #9 on: April 11, 2010, 03:21:11 pm »
i got one of similar kits (it was 10Hz-100khz function generator) but in the end it was qute buggy schematic + it had just one "cover-it-all" range so even with 10 turn pot. it was hard to get some low freq wave out of it.
My generator has an three range selector !!
About you last question ... " usability "
Well , now that we have set all the options, of tools , on table ..
I will say that in my generator , even if it has three outputs on the PCB , my box has only one
The other two are connected with thin air ...
I work with repairs of sound equipment , mixers and amplifiers , for home use and large dancing halls .
Primarily what is needed are the acoustic range , 80Hz - 9kHz .
For acoustic frequency's testing , you need other the sound generator , one fasma-scope.
( well my English does not helping me here , but I am trying )
fasma-scope or fasmatoscope ... its a graphical interface , equal of what we see in some cheap, and high end EQ devices ( Equalizers ) , that they have one little screen in the center , with led bars , one for every "base" frequency ..
Kenwood had build one fantastic EQ , with 16+16 "base" frequency's and a huge "spectrum analyzer" display, on it .
With all this tools , I was able to check the total frequency response , of the device that I was testing.
So , the importance of the generator it self , has a little value , if there is no " fasma-scope or fasmatoscope or spectrum analyzer with display , at the end of the chain .
Ok ... theoretically , you can insert an single signal as 1kHz as example, and measure the output in mV ,with a simple multimeter , but this are the theory ... in praxis you need " real " tools = more advanced .
In SUM , you will need just one simple sound generator.
« Last Edit: April 11, 2010, 03:57:47 pm by Kiriakos-GR »
#### jimmc
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##### Re: How useful is triangle wave in function generator ?
« Reply #10 on: April 11, 2010, 04:04:28 pm »
If you want an audio signal generator and spectrum analyser then consider using a PC sound card.
I posted this a few days ago:
I came across this website a while ago...
http://www.sillanumsoft.org/ZRLC.htm
I've used older versions the software for some time and the Spectrum Analyser works well,
the latest version has the facility to measure R, L & C with a simple external circuit (two OP-Amps and a few resistors + decoupling!).
Jim
Can also generate two channels of sine, square, triangle, noise etc. with frequency sweep if required.
#### xani
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• Posts: 400
##### Re: How useful is triangle wave in function generator ?
« Reply #11 on: April 11, 2010, 09:29:11 pm »
My scope has FFT and yeah u can always use soundcard. But i wanna play with things like SMPS and Tesla coils.
Smf | 2,687 | 10,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.868669 |
https://learn.careers360.com/ncert/question-compute-the-bulk-modulus-of-water-from-the-following-data-initial-volume-1000-litre-pressure-increase-1000-atm-1-atm-1013-105-pa-final-volume-1005-litre-compare-the-bulk-modulus-of-water-with-that-of-air-at-constant-temperature-expl/ | 1,718,827,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861832.94/warc/CC-MAIN-20240619185738-20240619215738-00327.warc.gz | 304,813,214 | 35,838 | #### Q12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm ( ), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Pressure Increase, P = 100.0 atm
Initial Volume = 100.0 l
Final volume = 100.5 l
Change in Volume = 0.5 l
Let the Bulk Modulus of water be B
The bulk modulus of air is
The Ratio of the Bulk Modulus of water to that of air is
This ratio is large as for the same pressure difference the strain will be much larger in the air than that in water. | 166 | 641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.895701 |
https://buildingspeed.org/blog/2012/06/17/why-200-mph-laps-at-michigan-are-not-like-200-mph-laps-at-daytona-or-talladega/ | 1,568,911,229,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00071.warc.gz | 399,738,211 | 22,810 | The Science of Fast
# Why 200 mph Laps at Michigan are not like 200-mph Laps at Daytona or Talladega
That fact that people are even talking about restrictor plates for Cup racing at Michigan International Speedway indicates a lack of understanding of the issues that give rise to concerns about cars getting airborne. I touched on the difference between average and instantaneous quantities last week with the pit road speeding issue at Pocono. Instantaneous speed is the speed you are going at some particular instant. A radar gun measures instantaneous speed.
That fact that people are even talking about restrictor plates for Cup racing at Michigan International Speedway indicates a lack of understanding of the issues that give rise to concerns about cars getting airborne.
I touched on the difference between average and instantaneous quantities last week with the pit road speeding issue at PoconoInstantaneous speed is the speed you are going at some particular instant. A radar gun measures instantaneous speed.
Average speed takes into account that speed varies over time and gives you one number that represents your speed over a number of different points or times. Speeding loops measure average speed.
For example, let’s say I look at my speedometer five times over the course of an hour. I note that I’m going 55 mph, 62 mph, 67 mph, 52 mph and 64 mph. Adding those numbers up comes to 300 mph. Since there were five measurements, divide by five and my average speed is 60 mph. The measurements could vary much more widely and still produce the same average. Any five speeds that add up to 300 mph would give you an average speed of 60 mph.
## Michigan
I’ve diagrammed to the right a rough sketch of speed vs. time (or distance). A diagram of Michigan International Speedway is in the upper left-hand corner of the picture, with corners and straights labeled. Those positions correspond to the positions shown in the graph.
The maximum speeds of 218 mph being reported happen generally toward the end of the straightaways. You’re accelerating all the way down the straight, reaching maximum speed just before you have to brake to enter the turn. You slow down going into the turn and speed up coming out of it. (I drew a symmetric graph only because I have limited time and even more limited drawing skills.)
Pre new left-side tire, cars were reaching a maximum of 218 mph with average lap speeds a little more than 200 mph. This tells you there must be a significant number of places on the track where cars are going slower than 200 mph. Those places would be the turns, where they are slowing down to a little over 190 mph (Thank you @chrisneville84 and @DRodmanNASCAR.) The solid line in my graph shows the instantaneous speed, while the average speed is shown as a dashed line.
### Does that mean we don’t have to worry about lift-off at Michigan?
As I mentioned in my last post, the big concern is that the car becomes unstable against lift off when it a) reaches a high enough speed and b) spins. Cars are more likely to spin in the turns, but the cars aren’t going 218 mph in the turns at Michigan – they’re going much more slowly.
The issue remains that two cars could hit on the straightaway, spinning one or both cars. This concern is heightened because of the possibility that a spinning car can hook a tire on the asphalt/grass transition in the frontstretch.
A corresponding picture of speed vs. time (or distance) at a plate track would look like the picture to the left. Cars are flat out all the way around the track. The speed doesn’t change very much through an entire lap. At Daytona, they are taking the corners near 200 mph. That’s a far different situation than at Michigan, where the speeds vary considerably around the track. The propensity for getting that perfect storm of speed and angle of the car with respect to the air increases under those conditions.
## Why isn’t this a concern for Nationwide?
We’ve heard all week that the Nationwide cars are running wide open all the way around the track, just like they do at Daytona. Their average lap times, however, are in the 190 mph range, which puts the safety concern squarely in the dangers of pack racing rather than cars going airborne.
## Other Michigan Stories
The myth of the 200 mph lift-off speed
Michigan: Don’t Believe the speeds
## 3 thoughts on “Why 200 mph Laps at Michigan are not like 200-mph Laps at Daytona or Talladega”
1. The last few airborne cars at Talladega or Daytona have happened on the backstretch or the tri-oval, not in the corners. That is good information in your explanation, though. It is very different, but the danger for liftoff is still high if a car gets turned there. Let’s hope it doesn’t happen!
2. Greg says:
I feel like the last few airborne cars at Talledega or Daytona had a “Wing” on the back, and as soon as that thing was pointing forward (in the direction of travel) the car took off (literally).
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,098 | 5,068 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-39 | latest | en | 0.926157 |
http://www.mathworks.com/help/dsp/ref/median.html?requestedDomain=www.mathworks.com&nocookie=true | 1,511,490,824,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00589.warc.gz | 447,113,313 | 18,266 | # Documentation
### This is machine translation
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# Median
Median value of input
• Library:
• DSP System Toolbox / Statistics
## Description
The Median block computes the median of each row or column of the input, or along vectors of a specified dimension of the input. It can also compute the median of the entire input. You can specify the dimension using the Find the median value over parameter. While computing the median, the block first sorts the input values. If the number of values is odd, the median is the middle value. If the number of values is even, the median is the average of the two middle values. To sort the data, you can specify the Sort algorithm parameter as either ```Quick sort``` or `Insertion sort`. The block sorts complex inputs according to their magnitude.
## Ports
### Input
expand all
The block accepts real-valued or complex-valued multichannel and multidimensional inputs. The input data type must be double precision, single precision, integer, or fixed point, with power-of-two slope and zero bias.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `uint8` | `uint16` | `uint32` | `fixed point`
Complex Number Support: Yes
### Output
expand all
The data type of the output matches the data type of the input.
The block computes the median value in each row or column of the input, or along vectors of a specified dimension of the input. It can also compute the median of the entire input. Each element in the output array `y` is the median value of the corresponding column, row, or entire input. The output array `y` depends on the setting of the Find the median value over parameter. Consider a three-dimensional input signal of size M-by-N-by-P. When you set Find the median value over to:
• `Entire input` — The output at each sample time is a scalar that contains the median value of the M-by-N-by-P input matrix.
• `Each row` — The output at each sample time consists of an M-by-1-by-P array, where each element contains the median value of each vector over the second dimension of the input. For an M-by-N matrix input, the output is an M-by-1 column vector.
• `Each column` — The output at each sample time consists of a 1-by-N-by-P array, where each element contains the median value of each vector over the first dimension of the input. For an M-by-N matrix input, the output at each sample time is a 1-by-N row vector.
In this mode, the block treats length-M unoriented vector inputs as M-by-1 column vectors.
• `Specified dimension` — The output at each sample time depends on the value of the Dimension parameter. If you set the Dimension to `1`, the output is the same as when you select `Each column`. If you set the Dimension to `2`, the output is the same as when you select `Each row`. If you set the Dimension to `3`, the output at each sample time is an M-by-N matrix containing the median value of each vector over the third dimension of the input.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `uint8` | `uint16` | `uint32` | `fixed point`
Complex Number Support: Yes
## Parameters
expand all
### Main Tab
Specify the sorting algorithm as either `Quick sort` or ```Insertion sort```.
• `Each column` — The block outputs the median value over each column.
• `Each row` — The block outputs the median value over each row.
• `Entire input` — The block outputs the median value over the entire input.
• `Specified dimension` — The block outputs the median value over the dimension specified in the Dimension parameter.
Specify the dimension (one-based value) of the input signal over which the block computes the median. The value of this parameter must be greater than 0 and less than the number of dimensions in the input signal.
#### Dependencies
To enable this parameter, set Find the median value over to `Specified dimension`.
### Note
To use these parameters, the data input must be fixed point. For all other inputs, the parameters on the Data Types tab are ignored.
Specify the rounding mode for fixed-point operations as one of the following:
• `Floor`
• `Ceiling`
• `Convergent`
• `Nearest`
• `Round`
• `Simplest`
• `Zero`
For more details, see rounding mode.
When you select this parameter, the block saturates the result of its fixed-point operation. When you clear this parameter, the block wraps the result of its fixed-point operation. For details on `saturate` and `wrap`, see overflow mode for fixed-point operations.
### Note
The Rounding mode and Saturate on integer overflow parameters have no effect on numeric results when all these conditions are met:
• Product output data type is ```Inherit: Inherit via internal rule```.
• Accumulator data type is ```Inherit: Inherit via internal rule```.
With these data type settings, the block operates in full-precision mode.
Product output specifies the data type of the output of a product operation in the Median block. For more information, see Fixed Point and Multiplication Data Types.
• `Inherit: Same as input` — The block specifies the product output data type to be the same as the input data type.
• `fixdt([],16,0)` — The block specifies an autosigned, binary-point, scaled, fixed-point data type with a word length of 16 bits and a fraction length of 0.
Alternatively, you can set the Product output data type by using the Data Type Assistant. To use the assistant, click the button.
For more information on the data type assistant, see Specify Data Types Using Data Type Assistant (Simulink).
Accumulator specifies the data type of the output of an accumulation operation in the Median block. For more details, see Fixed Point.
You can set this parameter to:
• `Inherit: Same as product output` — The block specifies the accumulator data type to be the same as the product output data type.
• `Inherit: Same as input` — The block specifies the accumulator data type to be the same as the input data type.
• `fixdt([],16,0)` — The block specifies an autosigned, binary-point, scaled, fixed-point data type with a word length of 16 bits and a fraction length of 0.
Alternatively, you can set the Accumulator data type by using the Data Type Assistant. To use the assistant, click the button.
For more information on the data type assistant, see Specify Data Types Using Data Type Assistant (Simulink).
Output specifies the data type of the output of the Median block. For more details, see Fixed Point.
You can set this parameter to:
• `Inherit: Same as accumulator` — The block specifies the output data type to be the same as the accumulator data type.
• `Inherit: Same as input` — The block specifies the output data type to be the same as the input data type.
• `Inherit: Same as product output` — The block specifies the output data type to be the same as the product output data type.
• `fixdt([],16,0)` — The block specifies an autosigned, binary-point, scaled, fixed-point data type with a word length of 16 bits and a fraction length of 0.
Alternatively, you can set the Output data type by using the Data Type Assistant. To use the assistant, click the button.
For more information on the data type assistant, see Specify Data Types Using Data Type Assistant (Simulink).
Specify the minimum value that the block can output. The default value is `[]` (unspecified). Simulink® uses this value to perform:
• Simulation range checking. See Signal Ranges (Simulink).
• Automatic scaling of fixed-point data types.
Specify the maximum value that the block can output. The default value is `[]` (unspecified). Simulink uses this value to perform:
• Simulation range checking. See Signal Ranges (Simulink).
• Automatic scaling of fixed-point data types.
Select this parameter to prevent the fixed-point tools from overriding the data types you specify on the block.
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https://www.coursehero.com/file/p1s49g1/Check-the-links-in-the-step-by-step-above-for-methods-to-read-lines-from-the/ | 1,555,772,177,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529839.0/warc/CC-MAIN-20190420140859-20190420162029-00033.warc.gz | 642,806,248 | 77,583 | Project7_making_change
# Check the links in the step by step above for methods
• Notes
• 3
This preview shows page 3 out of 3 pages.
Check the links in the step-by-step above for methods to read lines from the keyboard in Processing. You might find some use for buttons as well, as in this example: Remember to watch for boundary conditions – these are the limiting or extreme cases that your logic might otherwise miss – for example: price and amount paid exactly equal, either price or amount paid greater than 50, or less than 0, and so on. Catch these with if statements, issue an informative error message and stop processing or try to recover if you like. [And here's a BIG hint – you can avoid a lot of possible pitfalls if you work everything out in pennies (i.e. \$5.67 = 567 cents) and use integer math instead of floating point.] Scoring Please submit your source code and sample output for several values of price and amount paid. Extra Credit Add code to allow your program to deal with needing to make change when your till is out of a particular denomination. In other words, I buy something for \$5.13 with a \$20, but I have no \$10's in the till. (So I would use two \$5's). And so forth. It might be overkill for this problem, but you could handle this by throwing an OutOfDenomination ( int faceValue ) exception in some makeChange method, and then catching it and retrying with the next lower value, and so forth. Just a thought. A good job on this part could double the points this exercise is worth, because I think you'll get a concrete example of the programmer's adage -- “Error handling is half the work”.
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• Spring '08
• GeraldReed
• \$5, Meyer, \$5.13
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# chemistry edexcel synoptic watch
1. Well, there aren't really any big equations you need to know..
moles = mass / RMM
moles = (vol * conc.)/1000 (as you're normally given conc.s in moldm-3 and volumes in cm3)
Relative atomic mass = ((isotope mass * % abundance) + (isotope mass * % abundance)) / ∑%abundances
That's about it.. at an extreme stretch maybe you would have to know the Avagadro constant is 6.02 * 1023, but I don't think so.
2. Dus ne1 have ne idea ov wat kinda questions gona *** up..!!!.....??
3. 2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l)
so ratio of NaOH : H2SO4 is 2:1
4. (Original post by k0rrupter)
sum1 help me out here,
NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2H2O(l)
HOW is the mole ratio of NAOH2SO4 2:1 ?
You've balanced everything except the sodium hydroxide. You need 2 of them
6. i thought the equation was:
NaOH(aq) + H2SO4(aq) -----> NaHSO4(aq) + H2O(l)
7. (Original post by ram)
i thought the equation was:
NaOH(aq) + H2SO4(aq) -----> NaHSO4(aq) + H2O(l)
.
yes this is correct also
8. Has anyone looked at the sample of the new synoptic layout? If not then check it out http://www.edexcel.org.uk/VirtualCon...June_20052.pdf
Is it just me that finds it really difficult/bordering on impossible? and do you think that the questions tomorrow will be on similar topics?
9. do u have answers for that paper??
might print it of and try it lol
- yeah i guess im gonna have to write tiny to fit it in
10. (Original post by ram)
i thought the equation was:
NaOH(aq) + H2SO4(aq) -----> NaHSO4(aq) + H2O(l)
in alkaline conditions the sulphuric acid donates both protons to the solution
11. no, answers aren't on the website, which makes it worse as i can't even see how i should answer the questions
12. i hv tried doing unit 6 past papers, finding them VERY hard, does anyone know what the grade boundaries are like?
13. probably 40 / 50 for A?
14. Jan 05 Unit 6B paper anyone plz?
15. Ok, on the examiners report for June 03, if you did coursework and synoptic, the raw marks are 82/100 ! But then I guess people who do coursework would just keep doing it again and again until they virtually get 100%. For the practical and synoptic, its 74/100.
16. (Original post by kpg)
From the past papers I've done I'd suggest we've gotta be pretty clued up on
-Calculations...of any kind!
-Eθ calculations, equations, feasibility etc.
-All organic reactions
-All organic mechanisms (though there were 3 in the unit 5)
-Rates of reactions
-Equilibria + calculations
-Shapes of molecules
-Acid/Base theory
-Transition metals
-Extraction of aluminium
-Intermolecular forces esp. hydrogen bonding and solubility of organic molecules
Oh and reactions of hydrogen halides with sulphuric acid has come up twice so thats something to keep in mind.
does ne1 know wot the three mechanisms were in the unit 5 - i didnt do it so i dont know - thanks
17. Free-radical polymerisation, electrophilic addition of hydrogen bromide to alkene and Sn2 nucleophilic addition of OH- to a halogenoalkane I believe.
..What actually is the reaction between hydrogen halides and sulphuric acid?
18. Hydrogen Halides and CONC Sulphuric Acid
Chloride - steamy fumes of HCL
Bromides - steamy fumes of HBr, Br2 and SO2 gas.
Iodide - Iodine Vapour
HI is a powerful reducing agent, reduces H2S04 and itself is oxidised to I2
HBr is just powerful enough to reduce H2S04 to SO2 and its oxidised to Br2
HCl is a weak reducing agent so doesn't reduce H2S04...
19. I just did this question from Jun 2001 I think, spent ages on it working out oxidation numbers because Id forogtten it.. thankfully there was a clue it was about oxidation in it
20. (Original post by Whatsername)
But then I guess people who do coursework would just keep doing it again and again until they virtually get 100%.
Very true! I <3 the way you can do that with the coursework!
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3. Bishop Grosseteste University | 1,258 | 4,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2018-34 | latest | en | 0.897432 |
https://www.studymode.com/essays/Boiling-Water-Condensation-And-Dissolving-63605728.html | 1,579,384,313,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593937.27/warc/CC-MAIN-20200118193018-20200118221018-00430.warc.gz | 1,120,288,905 | 25,071 | Boiling Water, Condensation and Dissolving.
Topics: Water, Solid, Vapor Pages: 3 (973 words) Published: September 25, 2008
Assignment One
Eglstgeest wrote, “The purpose of teachers’ questions should be to promote children’s activity and reasoning” (pp 41) which is why it is up to the teachers of today to establish an understanding of the different sorts of questions that can be used to get different sorts of responses from the children that would initiate participation in science activities on everyday things that seem ‘natural’ to us. (Benbow, A. & Mebly, C. 2002). The focus of this essay is to explain the existing matter in solids, liquid and gaseous states applied to the concept cartoons which are then compared to a child’s conception of the following scientific investigations; sugar added to a cup of tea, when water is boiling, and the reasons behind condensation on the outside of a glass filled with water and ice cubes. I am then to compared the 'correct' answers to my child’s answers and explore where the child is coming from and why they believe their answer is correct.
The child I interviewed is 10 year old Amy. She is currently in year 5. She isn’t like most kids, as she attended a Korean public school in South Korea from years 1 to 3. Her background made it difficult for her to understand most of the scientific words as she probably hasn’t heard them before. She had to learn how to read and write English at the age of 9 while everyone else in her class was far ahead of her.
The particle theory of matter is the answer to many questions about our everyday life that everyday people barely ever think of questioning. “It explains a whole range of phenomena that you encounter in your daily life” (James, M. pp18). Matter is made up of many small particles. These particles differ in size depending on the substance and their speed varies on the temperature of the substance.
Boiling water
Water boiling is the cause of particles moving around each other rapidly as the liquid heats up which enables the particles near the surface to escape into the air creating gaseous...
References: Benbow, A. & Mebly, C. (2002) Planning a Science Investigation in Science Education for Elementary Teachers. Wadsworth/Thomson Learning: California. Pp. 189-196
Eglstgeest, J. (1985) The right question at the right time. In Harlen, W. Primary science: Taking the plunge. Pp. 36-45
Gallagher, R. & Ingram, P. (1989) Chemistry. Oxford University Press, Oxford (‘Particles in solids, liquids in gases’. Pp. 10-11; ‘A closer look at gases’, pp. 10-13)
James, M. (etal). (1999) Matter, the kinetic theory of matter. Chemical Connections 1. Jacaranda. National library of Australia. Pp. 18-20 | 634 | 2,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-05 | latest | en | 0.966182 |
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