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http://blog.symbolab.com/2018/ | 1,708,910,511,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474649.44/warc/CC-MAIN-20240225234904-20240226024904-00202.warc.gz | 6,282,225 | 20,655 | ## Tuesday, December 4, 2018
### Advanced Math Solutions - Matrix Inverse Calculator, Matrices
In this blog post, I will talk about how to get the inverse of a matrix. Please refresh yourself on RREF before continuing, if you haven’t mastered this topic yet.
What is the inverse of a matrix?
The inverse of a matrix is like the reciprocal of a number. The inverse of an n\times n matrix A is A^{-1},an n\times n matrix, such that A⋅A^{-1}=I, where I is the identity matrix. Just like how the product of a number and its reciprocal equals 1, (\frac{1}{n}⋅n=1), the product of a matrix and its inverse equals the identity matrix.
An identity matrix is an n\times n matrix, where the main diagonal of the matrix is all 1s and everywhere else in the matrix is 0s.
Note: You can only calculate the inverse of a square (n\times n) matrix.
There are a couple of methods used to find the inverse of a matrix. In this blog post, we will go over the method that Symbolab uses, which is one of the most common methods.
How do you calculate the inverse of a matrix?
1. Augment the n\times n matrix A with the n\times n identity matrix to create matrix (A|I)
• Augmenting a matrix means to create a matrix by appending the columns of two matrices
• Augmenting the matrix allows you to perform the same elementary row operations on both sides of the matrix
2. Put the matrix on the left hand side of the augmented matrix in RREF
• The matrix on the left hand side will be converted to the identity matrix
• Whatever elementary row operations you do to the left matrix will be done to the matrix on the right
3. The inverse matrix, A^{-1}, is to the right of the augmented matrix
Doesn’t sound too complicated right? As long as you’ve master putting matrices in RREF, this should be a piece of cake. Let’s see some examples!
First example
1. Augment the matrix with the identity matrix
2. Put the matrix on the left in RREF
R_2-R_1→R_2
\frac{1}{3} R_2→R_2
R_1-R_2→R_1
\frac{1}{2} R_1→R_1
3. The inverse matrix is on the right of the augmented matrix
You can verify that this is the inverse of the matrix by multiplying the inverse of the matrix and the matrix together (see here). If the product equals the identity matrix, then it is the inverse.
Next example (click here):
1. Augment the matrix with the identity matrix
2. Put the matrix on the left in RREF
Since we’ve already gone over how to put a matrix in RREF in a previous blog post and in the first example, we won’t go over how to do this.
3. The inverse matrix is on the right of the augmented matrix
Last example (click here)
1. Augment the matrix with the identity matrix
2. Put the matrix on the left in RREF
3. The inverse matrix is on the right of the augmented matrix
As you can see, as long as you know how to put a matrix in RREF, finding the inverse of a matrix doesn’t require too much additional work.
It is important to make sure that you double check your answer by verifying the product of the matrix and its inverse is the identity matrix because it is very easy to make mistake while adding and subtracting rows.
For more help or practice on this topic, check out Symbolab’s Practice.
Next blog post, I’ll talk about what the determinant of a matrix is and how to calculate it.
Until next time,
Leah
## Tuesday, October 23, 2018
### Advanced Math Solutions - Matrix Gauss Jordan Reduction Calculator, Gauss Jordan Elimination
In our previous blog posts, we talked about Row Echelon Form (Gaussian Elimination). If you haven’t, please look over it before continuing with this blog post.
In this blog post, we’ll talk about another advanced matrix topic that uses the same concepts, Gauss Jordan Elimination.
The Gauss Jordan Elimination is a method of putting a matrix in row reduced echelon form (RREF), using elementary row operations, in order to solve systems of equations, calculate rank, calculate the inverse of matrix, and calculate the determinant of a matrix (we will cover this in the next few blog posts).
RREF is when a matrix qualifies for the following four characteristics:
• Each non-zero row has 1, called a leading 1, as their first non-zero entry
• Each column with a leading 1 has zeros in every other entry
• As you move down the rows, the leading 1 moves to the right
• All zero rows are at the bottom
Another thing to note: Unlike matrices in REF, matrices in RREF are unique.
Here are examples of matrices in RREF:
Here are examples of matrices that aren’t in RREF:
You’ll use the same elementary row operations that you use to put a matrix in REF to put a matrix in RREF.
Here are guidelines on how to put a matrix in RREF:
1. Put the matrix in REF
2. If there are nonzero entries in the column of the leading coefficient in the first row, make them 0 by using the elementary row operations
3. If the leading coefficient in the first row is not a 1, make it a 1 by multiply the row by the reciprocal (this turns the leading coefficient into the leading 1)
4. Repeat steps 2-3 and replace “first” with “second”, then “third”, and so on
Gauss Jordan Elimination can be tricky the first few times, so I will walk you through 3 examples.
First example (click here):
1. Put the matrix in REF
2. Get rid of the nonzero entry in the column of the leading coefficient in Row 2: R_1-5R_2→R_1
3. Turn the leading coefficient into a 1 in Row 1: \frac{1}{5} R_1→R_1
4. Turn the leading coefficient into a 1 in Row 2: \frac{1}{2} R_2→R_2
Example 2 (click here):
1. Put the matrix in REF
2. Turn the leading coefficient in Row 1 into 1: \frac{1}{3} R_1→R_1
3. Turn the leading coefficient in Row 2 into 1: \frac{1}{8} R_2→R_2
In this problem there was no leading coefficient in Row 3, so we didn’t have to get rid of the entries in column 3.
Example 3 (click here):
1. Put the matrix in REF
2. Get rid of the nonzero entry in the column of the leading coefficient in Row 2: R_1-3R_2→R_1
3. Turn the leading coefficient in Row 1 into 1: \frac{1}{3} R_1→R_1
4. Turn the leading coefficient in Row 2 into 1: 3R_2→R_2
5. Turn the leading coefficient in Row 3 into 1: \frac{1}{2} R_3→R_3
Putting a matrix in RREF can require a lot of steps, so it is important to document what you are doing in each row. This will help you look over your work and remember what you did. Just like putting a matrix in REF, practice will help you get better at putting a matrix in RREF.
For more help or practice on this topic, visit Symbolab’s Practice.
Until next time,
Leah.
## Tuesday, October 9, 2018
### Advanced Math Solutions - Matrix Row Echelon Calculator, Gaussian Elimination (Row Echelon Form)
In our previous blog posts, we talked about the Matrix basics. Now, we are ready to talk about a more advanced matrix topic, Gaussian Elimination (also known as row echelon form).
The Gaussian Elimination, is a method of putting a matrix in row echelon form (REF), using elementary row operations.
REF is when a matrix qualifies for the following two characteristics:
• Each nonzero row has a leading coefficient (the first nonzero entry) that is to the right of the leading coefficient of the row above it
• There can’t be any nonzero entries below the leading coefficient in the leading coefficient’s column
• All zero rows are at the bottom
Note: A matrix in REF is not unique, so you may have a slightly different solution.
Here are examples of REF:
Here are examples of matrices that aren’t in REF:
The elementary row operations you’ll use to put your matrix in REF are:
• Switch any two rows
• Multiply each entry in a row by a non-zero constant
• Replace a row by the sum/difference of the row itself and another row, where it’s entries are multiplied by a non-zero constant
Here is a guideline on how to put a matrix in REF:
1. Move all zero rows to the bottom
2. Begin at the first row
3. If the first entry is a zero, switch the row with a row below it that has non-zero entry in the first column
4. If there are nonzero entries below the leading coefficient of the first row in the same column, cancel the entries by subtracting multiples of the the first row to the other rows (this will result in a zero entry)
5. Repeat steps 2 - 4 and replace “first” with “second”, then “third”, and so on until you can’t do anything more
This topic can be hard to understand at first, so let’s see some examples to better understand.
First example (click here):
1. Cancel the leading coefficient in Row 2: R_2-\frac{1}{3} R_1→ R_2
2. Simplify Row 2
We got rid of (canceled out) the 2 in Row 2 because there can’t be any entries below the leading coefficient in Row 1. We were able to do this using the elementary row operations. Multiplying Row 1 by ⅓, in order to turn 6 into 2, and then subtracting Row 2 by ⅓ Row 1.
Next example (click here):
1. Cancel the leading coefficient in Row 2: R_2-\frac{1}{2} R_1 →R_2
2. Cancel the leading coefficient in Row 3: R_3-(-\frac{3}{4} R_2) →R_3
Last example (click here):
1. Cancel the leading coefficient in Row 2: R_2-\frac{1}{2} R_1 →R_2
2. Cancel the leading coefficient in Row 3: R_3-\frac{1}{2} R_1→R_3
3. Since the second entry is a zero, switch the row with a row below it that has non-zero entry in the second column: R_3↔R_2
Putting matrices in REF can be tricky at first, but once you’ve done a handful of practice problems, it will come to you much easier. Check out Symbolab’s Row Echelon Calculator to help you better understand this topic.
Until next time,
Leah | 2,447 | 9,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-10 | latest | en | 0.849189 |
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# Example # 24.1: Find the magnetic field 5 cm from a long wire carrying 12 A....
Example # 24.1: Find the magnetic field 5 cm from a long wire carrying 12 A. Example # 24.2: Find the magnetic field inside a solenoid with 15,000 turns that is 20 cm long and carries 6.8 A of current. Example # 24.3: Find the force on a proton moving to the right (horizontally) at 330,000 m/s where there is an upward magnetic field of 0.0058 T. Example # 24.4: Find the speed of electrons moving horizontally in the same field as problem 24.3 if they move in a circle of radius 12.0 cm.
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Example # 24.3: Find the force on a proton moving to the right (horizontally) at 330,000 m/s where there is an upward magnetic field of 0.0058 T. Example # 24.4: Find the speed of electrons moving horizontally in the same field as problem 24.3 if they move in a circle of radius 12.0 cm.
• ### Example # 24.3: Find the force on a proton moving to the right (horizontally) at 330,000...
Example # 24.3: Find the force on a proton moving to the right (horizontally) at 330,000 m/s where there is an upward magnetic field of 0.0058 T. Example # 24.4: Find the speed of electrons moving horizontally in the same field as problem 24.3 if they move in a circle of radius 12.0 cm.
• ### Example # 24.1: Find the magnetic field 5 cm from a long wire carrying 12 A.
Example # 24.1: Find the magnetic field 5 cm from a long wire carrying 12 A.
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http://hpgeneralstudies.com/solved-paper-jbt-tet-2018-part-2-hp-bose/ | 1,575,691,236,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540495263.57/warc/CC-MAIN-20191207032404-20191207060404-00232.warc.gz | 73,498,025 | 14,338 | # Solved Paper JBT TET 2018 [Part-2] – HP BOSE
By | October 1, 2018
51. If the circumference of a 3cm, then its area (in cm2)
1. 9/4
2. 4/9
3. 9 /4
4. 3 /2
52. 210 – 1 is divisible by
1. 2
2. 3
3. 4
4. 10
53. In a right-angled triangle, the sum of two acute angles is
1. 45 °
2. 80°
3. 90°
4. None of these
54. The highest common factor of
1. 1
2. 12
3. 4/5
4. 1/60
55. The value of 42-2+120-20*0.6*10/(28-7*2)/(2000*0.3)
1. 600
2. 560
3. 614
4. 484
56. If 603*28=63*4*…….., then the number in the blank space is
1. 67
2. 21
3. 28
4. 63
57. The number 0.318564318564318564…….. is
1. A natural number
2. An integer
3. A rational number
4. An irrational number
58. Which of the following is an irrational number?
1. 15
2. 01516
3. 1516
4. 5015001500015…..
59. If x+1/x=5, then x3+1/x3 is equal to
1. 125
2. 100
3. 110
4. 25
60. If triangle PQR=EFD, then ED =
1. PQ
2. QR
3. PR
4. None of these
61 to 63. Choose the word which is the most opposite in meaning to the word printed in.
61. PRESENCE
1. Absence
2. Attendance
3. Absent
4. None of these
62. ACTIVE
1. Smart
2. Dull
3. Intelligent
4. Lazy
1. Human
2. Man
3. Children
4. Child
64 and 65. Choose the correct form of voice.
64. Ram was eating apples
1. Apples are being eaten by ram
2. Apples are eaten by ram
3. Apples were being eaten by Ram
4. Apples was eaten by ram
1. A newspaper was read by him.
2. A newspaper is read by him.
3. A newspaper is being read by him.
4. A newspaper was being read by him.
Choose the correct option of an indirect form of speech.
66. The teacher said, “The earth is round.”
1. The teacher said that the earth is round.
2. The teacher said that the earth was round.
3. The teacher told that the earth is round.
4. The teacher said earth is round.
67 to 69. Choose the correct synonyms for the words in bold letters.
67. AWFUL
1. Terrible
2. Ugly
3. Beautiful
4. Smart
68. ASSIST
1. Improve
2. Bear
3. Resist
4. Help
69. ACCOMPLISH
1. Complete
2. Start
3. Achieve
4. End
70 to 73. Identify the appropriate alternative from the given options for filling in the blanks.
70. You ……… obey your parents.
1. Should
2. Would
3. Could
4. May
71. It …………. rain tomorrow.
1. Could
2. Would
3. May
4. Might
72. We …. all die.
1. May
2. Must
3. Can
4. Could
73. You …… have my book, I have already it.
1. May
2. Must
3. Can
4. Could
74 to 77. Fill in the blanks with appropriate conjunction for the given questions.
74. I did not come……… you did not call me.
1. Because
2. Or
3. Either
4. Nor
75. ………… I am poor, ……. I am honest.
1. As – as
2. So – as
3. If – yet
4. So – as
76. The plural of ‘Radio’ is
4. None of these
77. The plural of ‘Sheep’ is
1. Sheep
2. Ship
3. Sheeps
4. Ships
78 and 79. Pick up the correctly spelt word.
78.
1. Harrassment
2. Harasment
3. Harrasment
4. Harassment
79.
1. Knowledge
2. Knowlege
3. Knowledg
4. None of these
80 and 81. Choose the correct option of an indirect form of speech.
80. I said, “I am quite well.”
1. I said that I am quite well
2. I said that I was quite well
3. I told that I was quite well
4. None of these
81. He said, “Mohan had gone.”
1. He said that Mohan has gone.
2. He said Mohan gone.
3. He said that Mohan have gone.
4. He said that Mohan is gone.
82 to 84. Choose the correct article from the given options.
82. Copper is …….. useful metal
1. A
2. An
3. The
4. None of these
83. Honest men speak ……. truth.
1. A
2. The
3. An
4. None of these
84. He is ….. honour to his profession
1. An
2. A
3. The
4. None of these
85 to 87. Choose the most suitable models from the given choices for blank spaces.
85. ………….. you have a very happy married life
1. Might
2. Can
3. Will
4. May
1. Will
2. Shall
3. Must
4. Would
87. When the eat is away, the mice …. play
1. Shall
2. Will
3. Can
4. Could
88 to 90. Choose the most appropriate preposition out of the given choices.
88. He is not aware …. His mistakes
1. With
2. At
3. To
4. Of
89. He was charged …. theft
1. With
2. Of
3. At
4. To
90. He is not eligible … this post
1. Of
2. For
3. With
4. To
91. यह महिला क्या कर रही है मे यह है?
1. सर्ब्नाम
2. बिशेषण
3. संज्ञा
4. सार्बनमीक
92. द्विकर्मक क्रिया युक्त वाक्य है?
1. नमिता हंसती है
2. बह रामायण पड़ता है
3. धोबी कपडे धोता है
4. किरायदार को मकान दिखाओ
93. आंध्र का बिलोम शब्द?
1. गीला
2. नम
3. सूखा
4. शुषक
94. शुद्ध शब्द छाँटिए ।
1. अनुशंसा
2. निरोग
3. आशीर्वाद
4. श्रीमति
95. मछली का पर्यायवाची नहीं है?
1. मीनाक्षी
2. मीन
3. शफरी
4. मकर
96. महत्व का बिग्रह है?
1. महा+तव
2. महा+त्तव
3. महत + त्व
4. मह+ तत्व
97. आगे नाथ न पीछे पगहा” कहव का सही अर्थ है
1. अबसे अलग
2. सबके साथ
3. निपुत्र
4. परम स्वतन्त्र
98. मनु+अंतर की संधि है?
1. मन्वन्तर
2. मनुन्तर
3. मनोन्तर
4. मनुवन्तर
99. शुद्ध शब्द छाँटिए?
1. गत्यवरोध
2. गतिवरोध
3. गत्यवरोध
4. गत्यावरोध
100. निम्न मैं से “तद्भव” शब्द छाँटिए
1. दुग्ध
2. कर्म
3. गेहूं
4. नृत्य | 1,939 | 4,863 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2019-51 | latest | en | 0.913655 |
https://rosettacode.org/wiki/Fractal_tree?oldid=358445 | 1,708,829,547,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00540.warc.gz | 505,238,721 | 100,391 | # Fractal tree
Fractal tree
You are encouraged to solve this task according to the task description, using any language you may know.
Generate and draw a fractal tree.
1. Draw the trunk
2. At the end of the trunk, split by some angle and draw two branches
3. Repeat at the end of each branch until a sufficient level of branching is reached
## 11l
Translation of: Nim
-V
Width = 1000
Height = 1000
TrunkLength = 400
ScaleFactor = 0.6
StartingAngle = 1.5 * math:pi
DeltaAngle = 0.2 * math:pi
F drawTree(outfile, Float x, Float y; len, theta) -> N
I len >= 1
V x2 = x + len * cos(theta)
V y2 = y + len * sin(theta)
outfile.write("<line x1='#.6' y1='#.6' x2='#.6' y2='#.6' style='stroke:white;stroke-width:1'/>\n".format(x, y, x2, y2))
drawTree(outfile, x2, y2, len * ScaleFactor, theta + DeltaAngle)
drawTree(outfile, x2, y2, len * ScaleFactor, theta - DeltaAngle)
V outsvg = File(‘tree.svg’, ‘w’)
outsvg.write(|‘<?xml version='1.0' encoding='utf-8' standalone='no'?>
<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN' 'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'>
<svg width='100%%' height='100%%' version='1.1' xmlns='http://www.w3.org/2000/svg'>
<rect width="100%" height="100%" fill="black"/>
’)
drawTree(outsvg, 0.5 * Width, Height, TrunkLength, StartingAngle)
outsvg.write("</svg>\n")
## Action!
Action! language does not support recursion. Therefore an iterative approach with a stack has been proposed.
DEFINE MAXSIZE="12"
INT ARRAY SinTab=[
0 4 9 13 18 22 27 31 36 40 44 49 53 58 62 66 71 75 79 83
88 92 96 100 104 108 112 116 120 124 128 132 136 139 143
147 150 154 158 161 165 168 171 175 178 181 184 187 190
193 196 199 202 204 207 210 212 215 217 219 222 224 226
228 230 232 234 236 237 239 241 242 243 245 246 247 248
249 250 251 252 253 254 254 255 255 255 256 256 256 256]
INT ARRAY xStack(MAXSIZE),yStack(MAXSIZE),angleStack(MAXSIZE)
BYTE ARRAY lenStack(MAXSIZE),dirStack(MAXSIZE)
BYTE stacksize=[0]
INT FUNC Sin(INT a)
WHILE a<0 DO a==+360 OD
WHILE a>360 DO a==-360 OD
IF a<=90 THEN
RETURN (SinTab(a))
ELSEIF a<=180 THEN
RETURN (SinTab(180-a))
ELSEIF a<=270 THEN
RETURN (-SinTab(a-180))
ELSE
RETURN (-SinTab(360-a))
FI
RETURN (0)
INT FUNC Cos(INT a)
RETURN (Sin(a-90))
BYTE FUNC IsEmpty()
IF stacksize=0 THEN
RETURN (1)
FI
RETURN (0)
BYTE FUNC IsFull()
IF stacksize=MAXSIZE THEN
RETURN (1)
FI
RETURN (0)
PROC Push(INT x,y,angle BYTE len,dir)
IF IsFull() THEN Break() FI
xStack(stacksize)=x yStack(stacksize)=y
angleStack(stacksize)=angle lenStack(stacksize)=len
dirStack(stacksize)=dir
stacksize==+1
RETURN
PROC Pop(INT POINTER x,y,angle BYTE POINTER len,dir)
IF IsEmpty() THEN Break() FI
stacksize==-1
x^=xStack(stacksize) y^=yStack(stacksize)
angle^=angleStack(stacksize) len^=lenStack(stacksize)
dir^=dirStack(stacksize)
RETURN
PROC DrawTree(INT x,y,len,angle,leftAngle,rightAngle)
BYTE depth,dir
Plot(x,y)
x==+Cos(angle)*len/256
y==-Sin(angle)*len/256
DrawTo(x,y)
Push(x,y,angle,len,0)
WHILE IsEmpty()=0
DO
Pop(@x,@y,@angle,@len,@dir)
IF dir<2 THEN
Push(x,y,angle,len,dir+1)
IF dir=0 THEN
angle==-leftAngle
ELSE
angle==+rightAngle
FI
len=13*len/16
Plot(x,y)
x==+Cos(angle)*len/256
y==-Sin(angle)*len/256
DrawTo(x,y)
IF IsFull()=0 THEN
Push(x,y,angle,len,0)
FI
FI
OD
RETURN
PROC Main()
BYTE CH=$02FC,COLOR1=$02C5,COLOR2=$02C6 Graphics(8+16) Color=1 COLOR1=$BA
COLOR2=$B2 DrawTree(140,191,40,110,35,15) DO UNTIL CH#$FF OD
CH=$FF RETURN Output: ## Ada Library: SDLAda with Ada.Numerics.Elementary_Functions; with SDL.Video.Windows.Makers; with SDL.Video.Renderers.Makers; with SDL.Video.Rectangles; with SDL.Events.Events; procedure Fractal_Tree is Width : constant := 600; Height : constant := 600; Level : constant := 13; Length : constant := 130.0; X_Start : constant := 475.0; Y_Start : constant := 580.0; A_Start : constant := -1.54; Angle_1 : constant := 0.10; Angle_2 : constant := 0.35; C_1 : constant := 0.71; C_2 : constant := 0.87; Window : SDL.Video.Windows.Window; Renderer : SDL.Video.Renderers.Renderer; Event : SDL.Events.Events.Events; procedure Draw_Tree (Level : in Natural; Length : in Float; Angle : in Float; X, Y : in Float) is use SDL; use Ada.Numerics.Elementary_Functions; Pi : constant := Ada.Numerics.Pi; X_2 : constant Float := X + Length * Cos (Angle, 2.0 * Pi); Y_2 : constant Float := Y + Length * Sin (Angle, 2.0 * Pi); Line : constant SDL.Video.Rectangles.Line_Segment := ((C.int (X), C.int (Y)), (C.int (X_2), C.int (Y_2))); begin if Level > 0 then Renderer.Set_Draw_Colour (Colour => (0, 220, 0, 255)); Renderer.Draw (Line => Line); Draw_Tree (Level - 1, C_1 * Length, Angle + Angle_1, X_2, Y_2); Draw_Tree (Level - 1, C_2 * Length, Angle - Angle_2, X_2, Y_2); end if; end Draw_Tree; procedure Wait is use type SDL.Events.Event_Types; begin loop while SDL.Events.Events.Poll (Event) loop if Event.Common.Event_Type = SDL.Events.Quit then return; end if; end loop; delay 0.100; end loop; end Wait; begin if not SDL.Initialise (Flags => SDL.Enable_Screen) then return; end if; SDL.Video.Windows.Makers.Create (Win => Window, Title => "Fractal tree", Position => SDL.Natural_Coordinates'(X => 10, Y => 10), Size => SDL.Positive_Sizes'(Width, Height), Flags => 0); SDL.Video.Renderers.Makers.Create (Renderer, Window.Get_Surface); Renderer.Set_Draw_Colour ((0, 0, 0, 255)); Renderer.Fill (Rectangle => (0, 0, Width, Height)); Draw_Tree (Level, Length, A_Start, X_Start, Y_Start); Window.Update_Surface; Wait; Window.Finalize; SDL.Finalise; end Fractal_Tree; ## Amazing Hopper Translation of: BBCBasic "Una suma que no quería salir, pero ya salió" :D /* Execute with:$ hopper jm/tree.jambo -x -o bin/tree
$rxvt -g 280x250 -fn "xft:FantasqueSansMono-Regular:pixelsize=1" -e ./bin/tree */ #include <jambo.h> Main Set '25, 0.76, 160, 100, 10' Init 'Spread, Scale, SizeX, SizeY, Depth' Color back '22', Cls Color back '15' Set '{SizeX} Mul by (2), -30, Div(SizeY,2), 90, Depth' Gosub 'Branch' Pause End Subrutines Define 'Branch, x1, y1, size, angle, depth' x2=0, y2=0 Let ( x2 := #(x1 + size * cos(d2r(angle))) ) Let ( y2 := #(y1 + size * sin(d2r(angle))) ) Draw a line ( #(180-y1), #(180-x1), #(180-y2), #(180-x2)) If ( #( depth > 0) ) Set (x2, y2, {size} Mul by 'Scale', {angle} Minus 'Spread',\ Minus one(depth)) Gosub 'Branch' Set (x2, y2, {size} Mul by 'Scale', {angle} Plus 'Spread',\ Minus one(depth)) Gosub 'Branch' End If Return ## Arturo width: 1000 height: 1000 trunkLength: 400 scaleFactor: 0.6 startingAngle: 1.5 * pi deltaAngle: 0.2 * pi drawTree: function [out x y len theta][ if len < 1 -> return null x2: x + len * cos theta y2: y + len * sin theta 'out ++ ~"<line x1='|x|' y1='|y|' x2='|x2|' y2='|y2|' style='stroke: white; stroke-width:1'/>\n" drawTree out x2 y2 len*scaleFactor theta+deltaAngle drawTree out x2 y2 len*scaleFactor theta-deltaAngle ] svg: { <?xml version='1.0' encoding='utf-8' standalone='no'?> <!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN' 'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'> <svg width='100%%' height='100%%' version='1.1' xmlns='http://www.w3.org/2000/svg'> <rect width="100%" height="100%" fill="black"/> } drawTree svg 0.5*width height trunkLength startingAngle 'svg ++ "</svg>" write "fractal.svg" svg Output: ## AutoHotkey Image - Link, since uploads seem to be disabled currently. Library: GDIP #SingleInstance, Force #NoEnv SetBatchLines, -1 ; Uncomment if Gdip.ahk is not in your standard library ; #Include, Gdip.ahk FileOut := A_Desktop "\MyNewFile.png" TreeColor := 0xff0066ff ; ARGB TrunkWidth := 10 ; Pixels TrunkLength := 80 ; Pixels Angle := 60 ; Degrees ImageWidth := 670 ; Pixels ImageHeight := 450 ; Pixels Branches := 13 Decrease := 0.81 Angle := (Angle * 0.01745329252) / 2 , Points := {} , Points[1, "Angle"] := 0 , Points[1, "X"] := ImageWidth // 2 , Points[1, "Y"] := ImageHeight - TrunkLength if (!pToken := Gdip_Startup()) { MsgBox, 48, Gdiplus error!, Gdiplus failed to start. Please ensure you have Gdiplus on your system. ExitApp } OnExit, Exit pBitmap := Gdip_CreateBitmap(ImageWidth, ImageHeight) , G := Gdip_GraphicsFromImage(pBitmap) , Gdip_SetSmoothingMode(G, 4) , pBrush := Gdip_BrushCreateSolid(0xff000000) , Gdip_FillRectangle(G, pBrush, -5, -5, ImageWidth + 10, ImageHeight + 10) , Gdip_DeleteBrush(pBrush) , pPen := Gdip_CreatePen(TreeColor, TrunkWidth/Decrease) , Gdip_DrawLine(G, pPen, Points.1.X, Points.1.Y, Points.1.X, ImageHeight) , Gdip_DeletePen(pPen) Loop, % Branches { NewPoints := {} pPen := Gdip_CreatePen(TreeColor, TrunkWidth) for Each, Point in Points { N1 := A_Index * 2 , N2 := (A_Index * 2) + 1 , NewPoints[N1, "X"] := Point.X + (TrunkLength * Sin(NewPoints[N1, "Angle"] := Point.Angle - Angle)) , NewPoints[N1, "Y"] := Point.Y - (TrunkLength * Cos(NewPoints[N1].Angle)) , NewPoints[N2, "X"] := Point.X + (TrunkLength * Sin(NewPoints[N2, "Angle"] := Point.Angle + Angle)) , NewPoints[N2, "Y"] := Point.Y - (TrunkLength * Cos(NewPoints[N2].Angle)) , Gdip_DrawLine(G, pPen, Point.X, Point.Y, NewPoints[N1].X, NewPoints[N1].Y) , Gdip_DrawLine(G, pPen, Point.X, Point.Y, NewPoints[N2].X, NewPoints[N2].Y) } TrunkWidth *= Decrease , TrunkLength *= Decrease , Points := NewPoints , Gdip_DeletePen(pPen) } Gdip_SaveBitmapToFile(pBitmap, FileOut) , Gdip_DisposeImage(pBitmap) , Gdip_DeleteGraphics(G) Run, % FileOut Exit: Gdip_Shutdown(pToken) ExitApp ## BASIC ### BASIC256 graphsize 300,300 level = 12 : len =63 # initial values x = 230: y = 285 rotation = pi/2 A1 = pi/27 : A2 = pi/8 # constants which determine shape C1 = 0.7 : C2 = 0.85 dim xs(level+1) : dim ys(level+1) # stacks fastgraphics color black rect 0,0,graphwidth,graphheight refresh color green gosub tree refresh imgsave "Fractal_tree_BASIC-256.png", "PNG" end tree: xs[level] = x : ys[level] = y gosub putline if level>0 then level = level - 1 len = len*C1 rotation = rotation - A1 gosub tree len = len/C1*C2 rotation = rotation + A1 + A2 gosub tree rotation = rotation - A2 len = len/C2 level = level + 1 end if x = xs[level] : y = ys[level] return putline: yn = -sin(rotation)*len + y xn = cos(rotation)*len + x line x,y,xn,yn x = xn : y = yn return ### Run BASIC 'Fractal Tree - for Run Basic - 29 Apr 2018 'from BASIC256 - http://rosettacode.org/wiki/Fractal_tree#BASIC256 'copy this text and go to http://www.runbasic.com WindowWidth = 500 'Run Basic max size 800 x 600 WindowHeight = 350 c = 255 '255 for white '0 for black graphic #w, WindowWidth, WindowHeight #w cls("black") 'black background color #w color(c,c,c) 'changes color to white level = 10 ' initial values leng = 50 x = 230: y = 325 ' initial values x = 230: y = 285 pi = 3.1415 rotation = 3.1415/2 'A1 = pi/27 : A2 = pi/8 ' constants which determine shape 'C1 = 0.7 : C2 = 0.85 ' tree is drifted left A1 = pi/9 : A2 = pi/9 ' constants which determine shape C1 = 0.85 : C2 = 0.85 ' Symmetrical Tree dim xs(level+1) : dim ys(level+1) ' stacks print : print "Welcome to the Run BASIC Fractal Tree Program" #w color("green") 'color green gosub [tree] render #w ' imgsave "Fractal_tree_BASIC-256.png", "PNG" Print "Thank you and goodbye" end [tree] xs(level) = x : ys(level) = y gosub [putline] if level>0 then level = level - 1 leng = leng*C1 rotation = rotation - A1 gosub [tree] leng = leng/C1*C2 rotation = rotation + A1 + A2 gosub [tree] rotation = rotation - A2 leng = leng/C2 level = level + 1 end if x = xs(level) : y = ys(level) return [putline] yn = -1*sin(rotation)*leng + y xn = cos(rotation)*leng + x #w line(x,y,xn,yn) x = xn : y = yn return 'end of code End ### BBC BASIC Output: Spread = 25 Scale = 0.76 SizeX% = 400 SizeY% = 300 Depth% = 10 VDU 23,22,SizeX%;SizeY%;8,16,16,128 PROCbranch(SizeX%, 0, SizeY%/2, 90, Depth%) END DEF PROCbranch(x1, y1, size, angle, depth%) LOCAL x2, y2 x2 = x1 + size * COSRAD(angle) y2 = y1 + size * SINRAD(angle) VDU 23,23,depth%;0;0;0; LINE x1, y1, x2, y2 IF depth% > 0 THEN PROCbranch(x2, y2, size * Scale, angle - Spread, depth% - 1) PROCbranch(x2, y2, size * Scale, angle + Spread, depth% - 1) ENDIF ENDPROC ### IS-BASIC 100 PROGRAM "Tree.bas" 110 OPTION ANGLE DEGREES 120 GRAPHICS HIRES 2 130 SET PALETTE 0,170 140 PLOT 640,10;ANGLE 90; 150 CALL TREE(200) 160 DEF TREE(N) 170 IF N<24 THEN EXIT DEF 180 PLOT FORWARD N;RIGHT 25; 190 CALL TREE(N*.75) 200 PLOT LEFT 50; 210 CALL TREE(N*.75) 220 PLOT RIGHT 25,BACK N, 230 END DEF ## C Library: SDL Library: SGE or Library: cairo #include <SDL/SDL.h> #ifdef WITH_CAIRO #include <cairo.h> #else #include <SDL/sge.h> #endif #include <cairo.h> #include <stdlib.h> #include <time.h> #include <math.h> #ifdef WITH_CAIRO #define PI 3.1415926535 #endif #define SIZE 800 // determines size of window #define SCALE 5 // determines how quickly branches shrink (higher value means faster shrinking) #define BRANCHES 14 // number of branches #define ROTATION_SCALE 0.75 // determines how slowly the angle between branches shrinks (higher value means slower shrinking) #define INITIAL_LENGTH 50 // length of first branch double rand_fl(){ return (double)rand() / (double)RAND_MAX; } void draw_tree(SDL_Surface * surface, double offsetx, double offsety, double directionx, double directiony, double size, double rotation, int depth) { #ifdef WITH_CAIRO cairo_surface_t *surf = cairo_image_surface_create_for_data( surface->pixels, CAIRO_FORMAT_RGB24, surface->w, surface->h, surface->pitch ); cairo_t *ct = cairo_create(surf); cairo_set_line_width(ct, 1); cairo_set_source_rgba(ct, 0,0,0,1); cairo_move_to(ct, (int)offsetx, (int)offsety); cairo_line_to(ct, (int)(offsetx + directionx * size), (int)(offsety + directiony * size)); cairo_stroke(ct); #else sge_AALine(surface, (int)offsetx, (int)offsety, (int)(offsetx + directionx * size), (int)(offsety + directiony * size), SDL_MapRGB(surface->format, 0, 0, 0)); #endif if (depth > 0){ // draw left branch draw_tree(surface, offsetx + directionx * size, offsety + directiony * size, directionx * cos(rotation) + directiony * sin(rotation), directionx * -sin(rotation) + directiony * cos(rotation), size * rand_fl() / SCALE + size * (SCALE - 1) / SCALE, rotation * ROTATION_SCALE, depth - 1); // draw right branch draw_tree(surface, offsetx + directionx * size, offsety + directiony * size, directionx * cos(-rotation) + directiony * sin(-rotation), directionx * -sin(-rotation) + directiony * cos(-rotation), size * rand_fl() / SCALE + size * (SCALE - 1) / SCALE, rotation * ROTATION_SCALE, depth - 1); } } void render(SDL_Surface * surface){ SDL_FillRect(surface, NULL, SDL_MapRGB(surface->format, 255, 255, 255)); draw_tree(surface, surface->w / 2.0, surface->h - 10.0, 0.0, -1.0, INITIAL_LENGTH, PI / 8, BRANCHES); SDL_UpdateRect(surface, 0, 0, 0, 0); } int main(){ SDL_Surface * screen; SDL_Event evt; SDL_Init(SDL_INIT_VIDEO); srand((unsigned)time(NULL)); screen = SDL_SetVideoMode(SIZE, SIZE, 32, SDL_HWSURFACE); render(screen); while(1){ if (SDL_PollEvent(&evt)){ if(evt.type == SDL_QUIT) break; } SDL_Delay(1); } SDL_Quit(); return 0; } ## C++ #include <windows.h> #include <string> #include <math.h> //-------------------------------------------------------------------------------------------------- using namespace std; //-------------------------------------------------------------------------------------------------- const float PI = 3.1415926536f; //-------------------------------------------------------------------------------------------------- class myBitmap { public: myBitmap() : pen( NULL ) {} ~myBitmap() { DeleteObject( pen ); DeleteDC( hdc ); DeleteObject( bmp ); } bool create( int w, int h ) { BITMAPINFO bi; void *pBits; ZeroMemory( &bi, sizeof( bi ) ); bi.bmiHeader.biSize = sizeof( bi.bmiHeader ); bi.bmiHeader.biBitCount = sizeof( DWORD ) * 8; bi.bmiHeader.biCompression = BI_RGB; bi.bmiHeader.biPlanes = 1; bi.bmiHeader.biWidth = w; bi.bmiHeader.biHeight = -h; HDC dc = GetDC( GetConsoleWindow() ); bmp = CreateDIBSection( dc, &bi, DIB_RGB_COLORS, &pBits, NULL, 0 ); if( !bmp ) return false; hdc = CreateCompatibleDC( dc ); SelectObject( hdc, bmp ); ReleaseDC( GetConsoleWindow(), dc ); width = w; height = h; return true; } void setPenColor( DWORD clr ) { if( pen ) DeleteObject( pen ); pen = CreatePen( PS_SOLID, 1, clr ); SelectObject( hdc, pen ); } void saveBitmap( string path ) { BITMAPFILEHEADER fileheader; BITMAPINFO infoheader; BITMAP bitmap; DWORD* dwpBits; DWORD wb; HANDLE file; GetObject( bmp, sizeof( bitmap ), &bitmap ); dwpBits = new DWORD[bitmap.bmWidth * bitmap.bmHeight]; ZeroMemory( dwpBits, bitmap.bmWidth * bitmap.bmHeight * sizeof( DWORD ) ); ZeroMemory( &infoheader, sizeof( BITMAPINFO ) ); ZeroMemory( &fileheader, sizeof( BITMAPFILEHEADER ) ); infoheader.bmiHeader.biBitCount = sizeof( DWORD ) * 8; infoheader.bmiHeader.biCompression = BI_RGB; infoheader.bmiHeader.biPlanes = 1; infoheader.bmiHeader.biSize = sizeof( infoheader.bmiHeader ); infoheader.bmiHeader.biHeight = bitmap.bmHeight; infoheader.bmiHeader.biWidth = bitmap.bmWidth; infoheader.bmiHeader.biSizeImage = bitmap.bmWidth * bitmap.bmHeight * sizeof( DWORD ); fileheader.bfType = 0x4D42; fileheader.bfOffBits = sizeof( infoheader.bmiHeader ) + sizeof( BITMAPFILEHEADER ); fileheader.bfSize = fileheader.bfOffBits + infoheader.bmiHeader.biSizeImage; GetDIBits( hdc, bmp, 0, height, ( LPVOID )dwpBits, &infoheader, DIB_RGB_COLORS ); file = CreateFile( path.c_str(), GENERIC_WRITE, 0, NULL, CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, NULL ); WriteFile( file, &fileheader, sizeof( BITMAPFILEHEADER ), &wb, NULL ); WriteFile( file, &infoheader.bmiHeader, sizeof( infoheader.bmiHeader ), &wb, NULL ); WriteFile( file, dwpBits, bitmap.bmWidth * bitmap.bmHeight * 4, &wb, NULL ); CloseHandle( file ); delete [] dwpBits; } HDC getDC() { return hdc; } int getWidth() { return width; } int getHeight() { return height; } private: HBITMAP bmp; HDC hdc; HPEN pen; int width, height; }; //-------------------------------------------------------------------------------------------------- class vector2 { public: vector2() { x = y = 0; } vector2( int a, int b ) { x = a; y = b; } void set( int a, int b ) { x = a; y = b; } void rotate( float angle_r ) { float _x = static_cast<float>( x ), _y = static_cast<float>( y ), s = sinf( angle_r ), c = cosf( angle_r ), a = _x * c - _y * s, b = _x * s + _y * c; x = static_cast<int>( a ); y = static_cast<int>( b ); } int x, y; }; //-------------------------------------------------------------------------------------------------- class fractalTree { public: fractalTree() { _ang = DegToRadian( 24.0f ); } float DegToRadian( float degree ) { return degree * ( PI / 180.0f ); } void create( myBitmap* bmp ) { _bmp = bmp; float line_len = 130.0f; vector2 sp( _bmp->getWidth() / 2, _bmp->getHeight() - 1 ); MoveToEx( _bmp->getDC(), sp.x, sp.y, NULL ); sp.y -= static_cast<int>( line_len ); LineTo( _bmp->getDC(), sp.x, sp.y); drawRL( &sp, line_len, 0, true ); drawRL( &sp, line_len, 0, false ); } private: void drawRL( vector2* sp, float line_len, float a, bool rg ) { line_len *= .75f; if( line_len < 2.0f ) return; MoveToEx( _bmp->getDC(), sp->x, sp->y, NULL ); vector2 r( 0, static_cast<int>( line_len ) ); if( rg ) a -= _ang; else a += _ang; r.rotate( a ); r.x += sp->x; r.y = sp->y - r.y; LineTo( _bmp->getDC(), r.x, r.y ); drawRL( &r, line_len, a, true ); drawRL( &r, line_len, a, false ); } myBitmap* _bmp; float _ang; }; //-------------------------------------------------------------------------------------------------- int main( int argc, char* argv[] ) { ShowWindow( GetConsoleWindow(), SW_MAXIMIZE ); myBitmap bmp; bmp.create( 640, 512 ); bmp.setPenColor( RGB( 255, 255, 0 ) ); fractalTree tree; tree.create( &bmp ); BitBlt( GetDC( GetConsoleWindow() ), 0, 20, 648, 512, bmp.getDC(), 0, 0, SRCCOPY ); bmp.saveBitmap( "f://rc//fracTree.bmp" ); system( "pause" ); return 0; } //-------------------------------------------------------------------------------------------------- ## Ceylon Translation of: Java Library: Swing Library: AWT Be sure to import java.desktop and ceylon.numeric in your module.ceylon file. import javax.swing { JFrame { exitOnClose } } import java.awt { Color { white, black }, Graphics } import ceylon.numeric.float { cos, toRadians, sin } shared void run() { value fractalTree = object extends JFrame("fractal tree") { background = black; setBounds(100, 100, 800, 600); resizable = false; defaultCloseOperation = exitOnClose; shared actual void paint(Graphics g) { void drawTree(Integer x1, Integer y1, Float angle, Integer depth) { if (depth <= 0) { return; } value x2 = x1 + (cos(toRadians(angle)) * depth * 10.0).integer; value y2 = y1 + (sin(toRadians(angle)) * depth * 10.0).integer; g.drawLine(x1, y1, x2, y2); drawTree(x2, y2, angle - 20, depth - 1); drawTree(x2, y2, angle + 20, depth - 1); } g.color = white; drawTree(400, 500, -90.0, 9); } }; fractalTree.visible = true; } ## Clojure Translation of: Java Library: Swing Library: AWT (import '[java.awt Color Graphics] 'javax.swing.JFrame) (defn deg-to-radian [deg] (* deg Math/PI 1/180)) (defn cos-deg [angle] (Math/cos (deg-to-radian angle))) (defn sin-deg [angle] (Math/sin (deg-to-radian angle))) (defn draw-tree [^Graphics g, x y angle depth] (when (pos? depth) (let [x2 (+ x (int (* depth 10 (cos-deg angle)))) y2 (+ y (int (* depth 10 (sin-deg angle))))] (.drawLine g x y x2 y2) (draw-tree g x2 y2 (- angle 20) (dec depth)) (recur g x2 y2 (+ angle 20) (dec depth))))) (defn fractal-tree [depth] (doto (proxy [JFrame] [] (paint [g] (.setColor g Color/BLACK) (draw-tree g 400 500 -90 depth))) (.setBounds 100 100 800 600) (.setResizable false) (.setDefaultCloseOperation JFrame/DISPOSE_ON_CLOSE) (.show))) (fractal-tree 9) ## Common Lisp Translation of: Clojure ;; (require :lispbuilder-sdl) (defun deg-to-radian (deg) "converts degrees to radians" (* deg pi 1/180)) (defun cos-deg (angle) "returns cosin of the angle expressed in degress" (cos (deg-to-radian angle))) (defun sin-deg (angle) "returns sin of the angle expressed in degress" (sin (deg-to-radian angle))) (defun draw-tree (surface x y angle depth) "draws a branch of the tree on the sdl-surface" (when (plusp depth) (let ((x2 (+ x (round (* depth 10 (cos-deg angle))))) (y2 (+ y (round (* depth 10 (sin-deg angle)))))) (sdl:draw-line-* x y x2 y2 :surface surface :color sdl:*green*) (draw-tree surface x2 y2 (- angle 20) (1- depth)) (draw-tree surface x2 y2 (+ angle 20) (1- depth))))) (defun fractal-tree (depth) "shows a window with a fractal tree" (sdl:with-init () (sdl:window 800 600 :title-caption "fractal-tree") (sdl:clear-display sdl:*black*) (draw-tree sdl:*default-surface* 400 500 -90 depth) (sdl:update-display) (sdl:with-events () (:video-expose-event () (sdl:update-display)) (:quit-event () t)))) (fractal-tree 9) ## D ### SVG Version Translation of: Raku import std.stdio, std.math; enum width = 1000, height = 1000; // Image dimension. enum length = 400; // Trunk size. enum scale = 6.0 / 10; // Branch scale relative to trunk. void tree(in double x, in double y, in double length, in double angle) { if (length < 1) return; immutable x2 = x + length * angle.cos; immutable y2 = y + length * angle.sin; writefln("<line x1='%f' y1='%f' x2='%f' y2='%f' " ~ "style='stroke:black;stroke-width:1'/>", x, y, x2, y2); tree(x2, y2, length * scale, angle + PI / 5); tree(x2, y2, length * scale, angle - PI / 5); } void main() { "<svg width='100%' height='100%' version='1.1' xmlns='http://www.w3.org/2000/svg'>".writeln; tree(width / 2.0, height, length, 3 * PI / 2); "</svg>".writeln; } ### Turtle Version This uses the turtle module from the Dragon Curve task, and the module from the Grayscale Image task. Translation of: Logo import grayscale_image, turtle; void tree(Color)(Image!Color img, ref Turtle t, in uint depth, in real step, in real scale, in real angle) { if (depth == 0) return; t.forward(img, step); t.right(angle); img.tree(t, depth - 1, step * scale, scale, angle); t.left(2 * angle); img.tree(t, depth - 1, step * scale, scale, angle); t.right(angle); t.forward(img, -step); } void main() { auto img = new Image!Gray(330, 300); auto t = Turtle(165, 270, -90); img.tree(t, 10, 80, 0.7, 30); img.savePGM("fractal_tree.pgm"); } ### Alternative version Translation of: Java Using DFL. import dfl.all; import std.math; class FractalTree: Form { private immutable DEG_TO_RAD = PI / 180.0; this() { width = 600; height = 500; text = "Fractal Tree"; backColor = Color(0xFF, 0xFF, 0xFF); startPosition = FormStartPosition.CENTER_SCREEN; formBorderStyle = FormBorderStyle.FIXED_DIALOG; maximizeBox = false; } private void drawTree(Graphics g, Pen p, int x1, int y1, double angle, int depth) { if (depth == 0) return; int x2 = x1 + cast(int) (cos(angle * DEG_TO_RAD) * depth * 10.0); int y2 = y1 + cast(int) (sin(angle * DEG_TO_RAD) * depth * 10.0); g.drawLine(p, x1, y1, x2, y2); drawTree(g, p, x2, y2, angle - 20, depth - 1); drawTree(g, p, x2, y2, angle + 20, depth - 1); } protected override void onPaint(PaintEventArgs ea){ super.onPaint(ea); Pen p = new Pen(Color(0, 0xAA, 0)); drawTree(ea.graphics, p, 300, 450, -90, 9); } } int main() { int result = 0; try { Application.run(new FractalTree); } catch(Exception e) { msgBox(e.msg, "Fatal Error", MsgBoxButtons.OK, MsgBoxIcon.ERROR); result = 1; } return result; } ## EasyLang # Fractal tree # color 555 proc tree x y deg n . . if n > 0 linewidth n * 0.4 move x y x += cos deg * n * 1.3 * (randomf + 0.5) y += sin deg * n * 1.3 * (randomf + 0.5) line x y tree x y deg - 20 n - 1 tree x y deg + 20 n - 1 . . timer 0 on timer clear tree 50 10 90 10 timer 2 . ## Evaldraw Evaldraw version creates a 3D tree with a camera rotating around the tree. static ratio = .75; static branchlength = 60; static max_branches = 4; struct vec3{x,y,z;}; () { t=klock(); srand(t * 1); zero1 = .5+.5*cos(t); maxbranches = int( 1+1 + zero1*5); cls(0); clz(1e32); distcam = -70; camrot = .5 * t; ca=distcam * cos(camrot); sa=distcam * sin(camrot); setcam(sa,-50,ca,camrot,0); angle = 2*pi / 8; branchlen = 10+50 * zero1; tree(maxbranches, 0, branchlen, 0,0,0, pi / 2, 0, angle); moveto(0,0); printf("N=%g, frame=%5.0f, cam:%3.0f", maxbranches, numframes, camrot / pi * 180); printf("\n%gx%g",xres,yres); sleep(16); } tree(mb, n, blen, x,y,z, ang_yx, ang_yz, angle) { n++; if( n> mb ) return; len = blen / n * ratio; c = 64 + 128 * n/7; setcol(100,c,38); dx=0; dy=0; dz=0; double mat[9]; vec3 axis = {0,0,1}; ang2mat(ang_yz, ang_yx, mat); transformPoint(axis,mat); dx=axis.x; dy=-axis.y; dz=axis.z; ox = x; oy = y; oz = z; x += len * dx; y += -len * dy; z += len * dz; rd = 8 / n; rd2 = 7 / (n+1); drawcone(ox,oy,oz,rd,x,y,z,rd2,DRAWCONE_FLAT + DRAWCONE_NOPHONG); nextangle = /*(-.5+1*rnd*pi) * */angle; tree(mb, n, blen, x, y, z, ang_yx - angle, ang_yz, nextangle); tree(mb, n, blen, x, y, z, ang_yx + angle, ang_yz, nextangle); tree(mb, n, blen, x, y, z, ang_yx, ang_yz - angle, nextangle); tree(mb, n, blen, x, y, z, ang_yx, ang_yz + angle, nextangle); } ang2mat(hang,vang,mat[9]) { mat[6] = cos(vang)*sin(hang); mat[0] = cos(hang); mat[7] = sin(vang); mat[1] = 0; mat[8] = cos(vang)*cos(hang); mat[2] =-sin(hang); mat[3] = mat[7]*mat[2] - mat[8]*mat[1]; mat[4] = mat[8]*mat[0] - mat[6]*mat[2]; mat[5] = mat[6]*mat[1] - mat[7]*mat[0]; } transformPoint(vec3 v, thisRot[9]) { NewX = v.x * thisRot[0] + v.y * thisRot[1] + v.z * thisRot[2]; NewY = v.x * thisRot[3] + v.y * thisRot[4] + v.z * thisRot[5]; NewZ = v.x * thisRot[6] + v.y * thisRot[7] + v.z * thisRot[8]; v.x=newx; v.y=newy; v.z=newz; } ## Delphi Works with: Delphi version 6.0 procedure DrawTree(Image: TImage; X1, Y1: integer; Angle: double; Depth: integer); var X2,Y2: integer; begin if Depth = 0 then exit; X2:=trunc(X1 + cos(DegToRad(Angle)) * Depth * 5); Y2:=trunc(Y1 + sin(DegToRad(Angle)) * Depth * 5); Image.Canvas.Pen.Color:=ColorMap47[MulDiv(High(ColorMap47),Depth,11)]; Image.Canvas.Pen.Width:=MulDiv(Depth,5,10); Image.Canvas.MoveTo(X1,Y1); Image.Canvas.LineTo(X2,Y2); DrawTree(Image, X2, Y2, Angle - 10, Depth - 1); DrawTree(Image, X2, Y2, Angle + 35, Depth - 1); end; procedure ShowFactalTree(Image: TImage); begin ClearImage(Image,clBlack); DrawTree(Image, 250, 350, -90, 11); Image.Invalidate; end; Output: Elapsed Time: 31.913 ms. ## F# Translation of: Raku let (cos, sin, pi) = System.Math.Cos, System.Math.Sin, System.Math.PI let (width, height) = 1000., 1000. // image dimension let scale = 6./10. // branch scale relative to trunk let length = 400. // trunk size let rec tree x y length angle = if length >= 1. then let (x2, y2) = x + length * (cos angle), y + length * (sin angle) printfn "<line x1='%f' y1='%f' x2='%f' y2='%f' style='stroke:rgb(0,0,0);stroke-width:1'/>" x y x2 y2 tree x2 y2 (length*scale) (angle + pi/5.) tree x2 y2 (length*scale) (angle - pi/5.) printfn "<?xml version='1.0' encoding='utf-8' standalone='no'?> <!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN' 'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'> <svg width='100%%' height='100%%' version='1.1' xmlns='http://www.w3.org/2000/svg'>" tree (width/2.) height length (3.*pi/2.) printfn "</svg>" ## Fantom using fwt using gfx class FractalCanvas : Canvas { new make () : super() {} Void drawTree (Graphics g, Int x1, Int y1, Int angle, Int depth) { if (depth == 0) return Int x2 := x1 + (angle.toFloat.toRadians.cos * depth * 10.0).toInt; Int y2 := y1 + (angle.toFloat.toRadians.sin * depth * 10.0).toInt; g.drawLine(x1, y1, x2, y2); drawTree(g, x2, y2, angle - 20, depth - 1); drawTree(g, x2, y2, angle + 20, depth - 1); } override Void onPaint (Graphics g) { drawTree (g, 400, 500, -90, 9) } } class FractalTree { public static Void main () { Window { title = "Fractal Tree" size = Size(800, 600) FractalCanvas(), }.open } } ## FreeBASIC Translation of: BBC BASIC ' version 17-03-2017 ' compile with: fbc -s gui Const As Double deg2rad = Atn(1) / 45 Dim Shared As Double scale = 0.76 Dim Shared As Double spread = 25 * deg2rad ' convert degree's to rad's Sub branch(x1 As ULong, y1 As ULong, size As ULong, angle As Double, depth As ULong) Dim As ULong x2, y2 x2 = x1 + size * Cos(angle) y2 = y1 + size * Sin(angle) Line (x1,y1) - (x2,y2), 2 ' palette color green If depth > 0 Then branch(x2, y2, size * scale, angle - spread, depth -1) branch(x2, y2, size * scale, angle + spread, depth -1) End If End Sub ' ------=< MAIN >=----- Dim As Double angle = -90 * deg2rad ' make sure that the tree grows up Dim As ULong SizeX = 800 Dim As ULong SizeY = SizeX * 3 \ 4 Dim As Double size = SizeY \ 4 Dim As ULong depth = 11 ScreenRes SizeX, SizeY, 8 WindowTitle ("Fractal Tree") branch(SizeX\2, SizeY, size, angle, depth) ' empty keyboard buffer While InKey <> "" : Wend windowtitle ("Fractal Tree, hit any key to end program") Sleep End ## Frege Works with: Frege version 3.23.888-g4e22ab6 module FractalTree where import Java.IO import Prelude.Math data AffineTransform = native java.awt.geom.AffineTransform where native new :: () -> STMutable s AffineTransform native clone :: Mutable s AffineTransform -> STMutable s AffineTransform native rotate :: Mutable s AffineTransform -> Double -> ST s () native scale :: Mutable s AffineTransform -> Double -> Double -> ST s () native translate :: Mutable s AffineTransform -> Double -> Double -> ST s () data BufferedImage = native java.awt.image.BufferedImage where pure native type_3byte_bgr "java.awt.image.BufferedImage.TYPE_3BYTE_BGR" :: Int native new :: Int -> Int -> Int -> STMutable s BufferedImage native createGraphics :: Mutable s BufferedImage -> STMutable s Graphics2D data Color = pure native java.awt.Color where pure native black "java.awt.Color.black" :: Color pure native green "java.awt.Color.green" :: Color pure native white "java.awt.Color.white" :: Color pure native new :: Int -> Color data BasicStroke = pure native java.awt.BasicStroke where pure native new :: Float -> BasicStroke data RenderingHints = native java.awt.RenderingHints where pure native key_antialiasing "java.awt.RenderingHints.KEY_ANTIALIASING" :: RenderingHints_Key pure native value_antialias_on "java.awt.RenderingHints.VALUE_ANTIALIAS_ON" :: Object data RenderingHints_Key = pure native java.awt.RenderingHints.Key data Graphics2D = native java.awt.Graphics2D where native drawLine :: Mutable s Graphics2D -> Int -> Int -> Int -> Int -> ST s () native drawOval :: Mutable s Graphics2D -> Int -> Int -> Int -> Int -> ST s () native fillRect :: Mutable s Graphics2D -> Int -> Int -> Int -> Int -> ST s () native setColor :: Mutable s Graphics2D -> Color -> ST s () native setRenderingHint :: Mutable s Graphics2D -> RenderingHints_Key -> Object -> ST s () native setStroke :: Mutable s Graphics2D -> BasicStroke -> ST s () native setTransform :: Mutable s Graphics2D -> Mutable s AffineTransform -> ST s () data ImageIO = mutable native javax.imageio.ImageIO where native write "javax.imageio.ImageIO.write" :: MutableIO BufferedImage -> String -> MutableIO File -> IO Bool throws IOException drawTree :: Mutable s Graphics2D -> Mutable s AffineTransform -> Int -> ST s () drawTree g t i = do let len = 10 -- ratio of length to thickness shrink = 0.75 angle = 0.3 -- radians i' = i - 1 g.setTransform t g.drawLine 0 0 0 len when (i' > 0)$ do
t.translate 0 (fromIntegral len)
t.scale shrink shrink
rt <- t.clone
t.rotate angle
rt.rotate (-angle)
drawTree g t i'
drawTree g rt i'
main = do
let width = 900
height = 800
initScale = 20
halfWidth = fromIntegral width / 2
buffy <- BufferedImage.new width height BufferedImage.type_3byte_bgr
g <- buffy.createGraphics
g.setRenderingHint RenderingHints.key_antialiasing RenderingHints.value_antialias_on
g.setColor Color.black
g.fillRect 0 0 width height
g.setColor Color.green
t <- AffineTransform.new ()
t.translate halfWidth (fromIntegral height)
t.scale initScale initScale
t.rotate pi
drawTree g t 16
f <- File.new "FractalTreeFrege.png"
void $ImageIO.write buffy "png" f Output is here due to Is file uploading blocked forever? ## Frink // Draw Fractal Tree in Frink // Define the tree function FractalTree[x1, y1, angleval, lengthval, graphicsobject] := { if lengthval > 1 { // Define current line end points (x2 and y2) x2 = x1 + ((cos[angleval degrees]) * lengthval) y2 = y1 + ((sin[angleval degrees]) * lengthval) // Draw line - notice that graphicsobject is the graphics object passed into the function. graphicsobject.line[x1,y1,x2,y2] // Calculate branches. You can change the lengthval multiplier factor and angleval summand to create different trees FractalTree[x2, y2, angleval - 20, lengthval * 0.7, graphicsobject] FractalTree[x2, y2, angleval + 20, lengthval * 0.7, graphicsobject] } } // Create graphics object g = new graphics // Start the recursive function. In Frink, a -90° angle moves from the bottom of the screen to the top. FractalTree[0, 0, -90, 30, g] // Show the final tree g.show[] ## FutureBasic _window = 1 _wndWidth = 680 void local fn BuildWindow window _window, @"Fractal Tree", ( 0, 0, _wndWidth, 600 ) WindowSetBackgroundColor( _window, fn ColorBlack ) WindowSubclassContentView( _window ) end fn local fn PlotFractalTree( x1 as double, y1 as double, size as long, angle as double, spread as long, depth as long, scale as double ) double x2, y2 pen 1.0, fn ColorGreen, NSLineCapStyleSquare // Convert angle to radians x2 = x1 + size * cos(angle * pi / 180) y2 = y1 + size * sin(angle * pi / 180) line x1, y1, x2, y2 if ( depth > 0 ) fn PlotFractalTree( x2, y2, size * scale, angle - spread, spread, depth - 1, scale ) fn PlotFractalTree( x2, y2, size * scale, angle + spread, spread, depth - 1, scale ) end if end fn void local fn DoDialog( ev as long, tag as long ) select ( tag ) case _windowContentViewTag double spread = ( 80.0 / (_wndWidth / 2 ) ) * 90 fn PlotFractalTree( _wndWidth / 2, 550, 140, -90, spread, 10, 0.75 ) end select select ( ev ) case _windowWillClose : end end select end fn on dialog fn DoDialog fn BuildWindow HandleEvents ## Go package main // Files required to build supporting package raster are found in: // * Bitmap // * Grayscale image // * Xiaolin Wu's line algorithm // * Write a PPM file import ( "math" "raster" ) const ( width = 400 height = 300 depth = 8 angle = 12 length = 50 frac = .8 ) func main() { g := raster.NewGrmap(width, height) ftree(g, width/2, height*9/10, length, 0, depth) g.Bitmap().WritePpmFile("ftree.ppm") } func ftree(g *raster.Grmap, x, y, distance, direction float64, depth int) { x2 := x + distance*math.Sin(direction*math.Pi/180) y2 := y - distance*math.Cos(direction*math.Pi/180) g.AaLine(x, y, x2, y2) if depth > 0 { ftree(g, x2, y2, distance*frac, direction-angle, depth-1) ftree(g, x2, y2, distance*frac, direction+angle, depth-1) } } ## Haskell An elegant yet universal monoidal solution. Library: Gloss import Graphics.Gloss type Model = [Picture -> Picture] fractal :: Int -> Model -> Picture -> Picture fractal n model pict = pictures$ take n $iterate (mconcat model) pict tree1 _ = fractal 10 branches$ Line [(0,0),(0,100)]
where branches = [ Translate 0 100 . Scale 0.75 0.75 . Rotate 30
, Translate 0 100 . Scale 0.5 0.5 . Rotate (-30) ]
main = animate (InWindow "Tree" (800, 800) (0, 0)) white $tree1 . (* 60) The solution gives rise to a variety of fractal geometric structures. Each one can be used by substituting tree1 in the main function by the desired one. --animated tree tree2 t = fractal 8 branches$ Line [(0,0),(0,100)]
where branches = [ Translate 0 100 . Scale 0.75 0.75 . Rotate t
, Translate 0 100 . Scale 0.6 0.6 . Rotate 0
, Translate 0 100 . Scale 0.5 0.5 . Rotate (-2*t) ]
--animated fractal clock
circles t = fractal 10 model $Circle 100 where model = [ Translate 0 50 . Scale 0.5 0.5 . Rotate t , Translate 0 (-50) . Scale 0.5 0.5 . Rotate (-2*t) ] --Pythagoras tree pithagor _ = fractal 10 model$ rectangleWire 100 100
where model = [ Translate 50 100 . Scale s s . Rotate 45
, Translate (-50) 100 . Scale s s . Rotate (-45)]
s = 1/sqrt 2
--Sierpinski pentagon
pentaflake _ = fractal 5 model $pentagon where model = map copy [0,72..288] copy a = Scale s s . Rotate a . Translate 0 x pentagon = Line [ (sin a, cos a) | a <- [0,2*pi/5..2*pi] ] x = 2*cos(pi/5) s = 1/(1+x) Alternative solution Using the method of the J contribution. Library: HGL import Graphics.HGL.Window import Graphics.HGL.Run import Control.Arrow import Control.Monad import Data.List enumBase :: Int -> Int -> [[Int]] enumBase n = mapM (enumFromTo 0). replicate n. pred psPlus (a,b) (p,q) = (a+p, b+q) toInt :: Double -> Int toInt = fromIntegral.round intPoint = toInt *** toInt pts n = map (map (intPoint.psPlus (100,0)). ((0,300):). scanl1 psPlus. ((r,300):). zipWith (\h a -> (h*cos a, h*sin a)) rs) hs where [r,h,sr,sh] = [50, pi/5, 0.9, 0.75] rs = take n$ map (r*) $iterate(*sr) sr lhs = map (map (((-1)**).fromIntegral))$ enumBase n 2
rhs = take n $map (h*)$ iterate(*sh) 1
hs = map (scanl1 (+). zipWith (*)rhs) lhs
fractalTree :: Int -> IO ()
fractalTree n =
runWindow "Fractal Tree" (500,600)
(\w -> setGraphic w (overGraphics ( map polyline $pts (n-1))) >> getKey w) main = fractalTree 10 ## Icon and Unicon procedure main() WOpen("size=800,600", "bg=black", "fg=white") | stop("*** cannot open window") drawtree(400,500,-90,9) WDone() end link WOpen procedure drawtree(x,y,angle,depth) if depth > 0 then { x2 := integer(x + cos(dtor(angle)) * depth * 10) y2 := integer(y + sin(dtor(angle)) * depth * 10) DrawLine(x,y,x2,y2) drawtree(x2,y2,angle-20, depth-1) drawtree(x2,y2,angle+20, depth-1) } return end Translation of: Java ## J require'gl2' coinsert'jgl2' L0=: 50 NB. initial length A0=: 1r8p1 NB. initial angle: pi divided by 8 dL=: 0.9 NB. shrink factor for length dA=: 0.75 NB. shrink factor for angle N=: 14 NB. number of branches L=: L0*dL^1+i.N NB. lengths of line segments NB. relative angles of successive line segments A=: A0*(dA^i.N) +/\@:*("1) _1 ^ #:i.2 ^ N NB. end points for each line segment P=: 0 0+/\@,"2 +.*.inv (L0,0),"2 L,"0"1 A wd {{)n pc P closeok; setp wh 480 640; cc C isidraw flush; pshow; }} gllines <.(10 + ,/"2 P-"1<./,/P) See the talk page for some implementation notes. ## Java Library: Swing Library: AWT import java.awt.Color; import java.awt.Graphics; import javax.swing.JFrame; public class FractalTree extends JFrame { public FractalTree() { super("Fractal Tree"); setBounds(100, 100, 800, 600); setResizable(false); setDefaultCloseOperation(EXIT_ON_CLOSE); } private void drawTree(Graphics g, int x1, int y1, double angle, int depth) { if (depth == 0) return; int x2 = x1 + (int) (Math.cos(Math.toRadians(angle)) * depth * 10.0); int y2 = y1 + (int) (Math.sin(Math.toRadians(angle)) * depth * 10.0); g.drawLine(x1, y1, x2, y2); drawTree(g, x2, y2, angle - 20, depth - 1); drawTree(g, x2, y2, angle + 20, depth - 1); } @Override public void paint(Graphics g) { g.setColor(Color.BLACK); drawTree(g, 400, 500, -90, 9); } public static void main(String[] args) { new FractalTree().setVisible(true); } } ## JavaScript Implementation using HTML5 canvas element to draw tree structure. <html> <body> <canvas id="canvas" width="600" height="500"></canvas> <script type="text/javascript"> var elem = document.getElementById('canvas'); var context = elem.getContext('2d'); context.fillStyle = '#C0C0C0'; context.lineWidth = 1; var deg_to_rad = Math.PI / 180.0; var depth = 9; function drawLine(x1, y1, x2, y2, brightness){ context.moveTo(x1, y1); context.lineTo(x2, y2); } function drawTree(x1, y1, angle, depth){ if (depth !== 0){ var x2 = x1 + (Math.cos(angle * deg_to_rad) * depth * 10.0); var y2 = y1 + (Math.sin(angle * deg_to_rad) * depth * 10.0); drawLine(x1, y1, x2, y2, depth); drawTree(x2, y2, angle - 20, depth - 1); drawTree(x2, y2, angle + 20, depth - 1); } } context.beginPath(); drawTree(300, 500, -90, depth); context.closePath(); context.stroke(); </script> </body> </html> ## jq The following generates SVG, which can be viewed by following the link below. # width and height define the outer dimensions; # len defines the trunk size; # scale defines the branch length relative to the trunk; def main(width; height; len; scale): def PI: (1|atan)*4; def precision(n): def pow(k): . as$in | reduce range(0;k) as $i (1; .*$in);
if . < 0 then - (-. | precision(n))
else
(10|pow(n)) as $power | (. * 10 *$power) | floor as $x | ($x % 10) as $r | ((if$r < 5 then $x else$x + 5 end) / 10 | floor) / $power end; def p2: precision(2); def tree(x; y; len; angle): if len < 1 then empty else (x + len * (angle|cos)) as$x2
| (y + len * (angle|sin)) as $y2 | (if len < 10 then 1 else 2 end) as$swidth
| (if len < 10 then "blue" else "black" end) as $stroke | "<line x1='\(x|p2)' y1='\(y|p2)' x2='\($x2|p2)' y2='\($y2|p2)' style='stroke:\($stroke); stroke-width:\($swidth)'/>", tree($x2; $y2; len * scale; angle + PI / 5), tree($x2; $y2; len * scale; angle - PI / 5) end ; "<svg width='100%' height='100%' version='1.1' xmlns='http://www.w3.org/2000/svg'>", tree(width / 2; height; len; 3 * PI / 2), "</svg>" ; main(1000; 1000; 400; 6/10) Output:$ jq -r -n -r -f Fractal_tree_svg.jq > Fractal_tree.svg
## Julia
Translation of: F#
const width = height = 1000.0
const trunklength = 400.0
const scalefactor = 0.6
const startingangle = 1.5 * pi
const deltaangle = 0.2 * pi
function tree(fh, x, y, len, theta)
if len >= 1.0
x2 = x + len * cos(theta)
y2 = y + len * sin(theta)
write(fh, "<line x1='$x' y1='$y' x2='$x2' y2='$y2' style='stroke:rgb(0,0,0);stroke-width:1'/>\n")
tree(fh, x2, y2, len * scalefactor, theta + deltaangle)
tree(fh, x2, y2, len * scalefactor, theta - deltaangle)
end
end
outsvg = open("tree.svg", "w")
write(outsvg,
"""<?xml version='1.0' encoding='utf-8' standalone='no'?>
<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN'
'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'>
<svg width='100%%' height='100%%' version='1.1'
xmlns='http://www.w3.org/2000/svg'>\n""")
tree(outsvg, 0.5 * width, height, trunklength, startingangle)
write(outsvg, "</svg>\n") # view file tree.svg in browser
## Kotlin
Translation of: Java
// version 1.1.2
import java.awt.Color
import java.awt.Graphics
import javax.swing.JFrame
class FractalTree : JFrame("Fractal Tree") {
init {
background = Color.black
setBounds(100, 100, 800, 600)
isResizable = false
defaultCloseOperation = EXIT_ON_CLOSE
}
private fun drawTree(g: Graphics, x1: Int, y1: Int, angle: Double, depth: Int) {
if (depth == 0) return
val x2 = x1 + (Math.cos(Math.toRadians(angle)) * depth * 10.0).toInt()
val y2 = y1 + (Math.sin(Math.toRadians(angle)) * depth * 10.0).toInt()
g.drawLine(x1, y1, x2, y2)
drawTree(g, x2, y2, angle - 20, depth - 1)
drawTree(g, x2, y2, angle + 20, depth - 1)
}
override fun paint(g: Graphics) {
g.color = Color.white
drawTree(g, 400, 500, -90.0, 9)
}
}
fun main(args: Array<String>) {
FractalTree().isVisible = true
}
## Lambdatalk
1) defining the function tree:
{def tree
{lambda {:e // last branch length
:s // trunks length
:k // ratio between two following branches
:a // rotate left
:b} // rotate right
{if {< :s :e}
then
else M:s T:a
{tree :e {* :k :s} :k :a :b}
T-{+ :a :b}
{tree :e {* :k :s} :k :a :b}
T:b M-:s }}}
2) Calling this function generates a sequence of commands mooving a pen:
• Tθ rotates the drawing direction "θ" degrees from the previous one
• and Md draws a segment "d" pixels in this direction.
{def T {tree 1 190 {/ 2 3} 15 45}}
and produces 40995 words beginning with:
M190 T15 M126.66666666666666 T15 M84.44444444444443 T15 M56.29629629629628 T15 M37.53086419753085 T15 M25.020576131687235 T15
M16.680384087791488 T15 M11.120256058527659 T15 M7.413504039018439 T15 M4.942336026012292 T15 M3.2948906840081946 ...
3) These words are sent to a the turtle lambdatalk primitive
which is a graphic device translating the sequence of Md and Tθ
into a sequence of SVG points x0 y0 x1 y1 ... xn yn
which will feed the points attribute of a polyline SVG element:
{svg {@ width="580px" height="580px" style="box-shadow:0 0 8px #000;"}
{polyline
{@ points="{turtle 230 570 180 {T}}"
fill="transparent" stroke="#fff" stroke-width="1"
}}}
This is an abstract of the output:
<svg width="580px" height="580px" style="box-shadow:0 0 8px #000;">
<polyline points="230 580 230 380 195 251 151 174 109 132 75 113 49 106 32 106 21 109 ...
... 413 286 324 286 230 380 230 580 "
fill="transparent" stroke="#888" stroke-width="1">
</polyline>
</svg>
The complete ouput can be seen displayed in http://lambdaway.free.fr/lambdawalks/?view=fractal_tree
## Liberty BASIC
LB includes Logo-type turtle commands, so can be drawn that way as well as that shown here.
NoMainWin
sw = 640 : sh = 480
WindowWidth = sw+8 : WindowHeight = sh+31
UpperLeftX = (DisplayWidth -sw)/2
UpperLeftY = (DisplayHeight-sh)/2
Open"Fractal Tree" For Graphics_nf_nsb As #g
#g "Down; Color darkgreen; TrapClose halt"
h$= "#g" 'initial assignments initAngle = Acs(-1)*1.5 'radian equivalent of 270 degrees theta = 29 * (Acs(-1)/180) 'convert 29 degrees to radians length = 110 'length in pixels depth = 25 'max recursion depth 'draw the tree Call tree h$, 320, 470, initAngle, theta, length, depth
#g "Flush; when leftButtonDown halt" 'L-click to exit
Wait
Sub halt handle$Close #handle$
End
End Sub
Sub tree h$, x, y, initAngle, theta, length, depth Scan newX = Cos(initAngle) * length + x newY = Sin(initAngle) * length + y #h$ "Line ";x;" ";y;" ";newX;" ";newY
length = length * .78
depth = depth - 1
If depth > 0 Then
Call tree h$, newX, newY, initAngle-theta, theta, length, depth Call tree h$, newX, newY, initAngle+theta, theta, length, depth
End If
End Sub
## Lingo
----------------------------------------
-- Creates an image of a fractal tree
-- @param {integer} width
-- @param {integer} height
-- @param {integer} fractalDepth
-- @param {integer|float} initSize
-- @param {float} [scaleFactor=1.0]
-- @return {image}
----------------------------------------
on fractalTree (width, height, fractalDepth, initSize, spreadAngle, scaleFactor)
if voidP(scaleFactor) then scaleFactor = 1.0
img = image(width, height, 24)
img.fill(img.rect, rgb(0,0,0))
_drawTree(img, width/2, height, -PI/2, fractalDepth, initSize, spreadAngle, scaleFactor)
return img
end
on _drawTree (img, x1, y1, angle, depth, size, spreadAngle, scaleFactor)
if (depth) then
x2 = x1 + cos(angle)*depth*size
y2 = y1 + sin(angle)*depth*size
img.draw(x1, y1, x2, y2, [#color:rgb(255,255,255)])
end if
end
Usage:
fractalDepth = 10
initSize = 7.0
scaleFactor = 0.95
img = fractalTree(480, 380, fractalDepth, initSize, spreadAngle, scaleFactor)
## Logo
to tree :depth :length :scale :angle
if :depth=0 [stop]
setpensize round :depth/2
forward :length
right :angle
tree :depth-1 :length*:scale :scale :angle
left 2*:angle
tree :depth-1 :length*:scale :scale :angle
right :angle
back :length
end
clearscreen
tree 10 80 0.7 30
## Lua
### Bitmap
Needs LÖVE 2D Engine
g, angle = love.graphics, 26 * math.pi / 180
wid, hei = g.getWidth(), g.getHeight()
function rotate( x, y, a )
local s, c = math.sin( a ), math.cos( a )
local a, b = x * c - y * s, x * s + y * c
return a, b
end
function branches( a, b, len, ang, dir )
len = len * .76
if len < 5 then return end
g.setColor( len * 16, 255 - 2 * len , 0 )
if dir > 0 then ang = ang - angle
else ang = ang + angle
end
local vx, vy = rotate( 0, len, ang )
vx = a + vx; vy = b - vy
g.line( a, b, vx, vy )
branches( vx, vy, len, ang, 1 )
branches( vx, vy, len, ang, 0 )
end
function createTree()
local lineLen = 127
local a, b = wid / 2, hei - lineLen
g.setColor( 160, 40 , 0 )
g.line( wid / 2, hei, a, b )
branches( a, b, lineLen, 0, 1 )
branches( a, b, lineLen, 0, 0 )
end
canvas = g.newCanvas( wid, hei )
g.setCanvas( canvas )
createTree()
g.setCanvas()
end
function love.draw()
g.draw( canvas )
end
### ASCII
Using the Bitmap class and text renderer from here, then extending...
function Bitmap:tree(x, y, angle, depth, forkfn, lengfn)
if depth <= 0 then return end
local fork, leng = forkfn(), lengfn()
local x2 = x + depth * leng * math.cos(angle)
local y2 = y - depth * leng * math.sin(angle)
self:line(math.floor(x), math.floor(y), math.floor(x2), math.floor(y2))
self:tree(x2, y2, angle+fork, depth-1, forkfn, lengfn)
self:tree(x2, y2, angle-fork, depth-1, forkfn, lengfn)
end
bitmap = Bitmap(128*3,128)
bitmap:tree( 64, 120, math.pi/2, 8, function() return 0.3 end, function() return 3 end)
bitmap:tree(192, 120, math.pi/2, 8, function() return 0.6 end, function() return 2.5 end)
bitmap:tree(320, 120, math.pi/2, 8, function() return 0.2+math.random()*0.3 end, function() return 2.0+math.random()*2.0 end)
bitmap:render({[0x000000]='.', [0xFFFFFFFF]='█'})
Output:
Shown at 25% scale:
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## Mathematica / Wolfram Language
fractalTree[
pt : {_, _}, \[Theta]orient_: \[Pi]/2, \[Theta]sep_: \[Pi]/9,
depth_Integer: 9] := Module[{pt2},
If[depth == 0, Return[]];
pt2 = pt + {Cos[\[Theta]orient], Sin[\[Theta]orient]}*depth;
DeleteCases[
Flatten@{
Line[{pt, pt2}],
fractalTree[pt2, \[Theta]orient - \[Theta]sep, \[Theta]sep,
depth - 1],
fractalTree[pt2, \[Theta]orient + \[Theta]sep, \[Theta]sep,
depth - 1]
},
Null
]
]
Graphics[fractalTree[{0, 0}, \[Pi]/2, \[Pi]/9]]
## MiniScript
This GUI implementation is for use with Mini Micro.
drawTree = function(x1, y1, angle, depth)
fork_angle = 20
base_len = 9
if depth > 0 then
radians = angle * pi / 180
x2 = x1 + cos(radians) * depth * base_len
y2 = y1 + sin(radians) * depth * base_len
gfx.line x1, y1, x2, y2, "#008000"
drawTree x2, y2, angle - fork_angle, depth - 1
drawTree x2, y2, angle + fork_angle, depth - 1
end if
end function
clear
gfx.clear "#87CEEB"
drawTree 480, 10, 90, 11
img = gfx.getImage(0, 0, 960, 640)
file.saveImage "/usr/fractalTree.png", img
## NetRexx
Translation of: Java
Library: Swing
Library: AWT
/* NetRexx */
options replace format comments java crossref symbols binary
import java.awt.Color
import java.awt.Graphics
import javax.swing.JFrame
class RFractalTree public extends JFrame
properties constant
isTrue = (1 == 1)
isFalse = \isTrue
-- ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
method RFractalTree() public
super('Fractal Tree')
setBounds(100, 100, 800, 600)
setResizable(isFalse)
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE)
return
-- ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
method drawTree(g = Graphics, x1 = int, y1 = int, angle = double, depth = int) private
if depth \= 0 then do
x2 = x1 + (int Math.cos(Math.toRadians(angle)) * depth * 10.0)
y2 = y1 + (int Math.sin(Math.toRadians(angle)) * depth * 10.0)
g.drawLine(x1, y1, x2, y2)
drawTree(g, x2, y2, angle - 20, depth - 1)
drawTree(g, x2, y2, angle + 20, depth - 1)
end
return
-- ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
method paint(g = Graphics) public
g.setColor(Color.BLACK)
drawTree(g, 400, 500, -90, 9)
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[])public static
RFractalTree().setVisible(isTrue)
return
## Nim
Translation of: Julia
import math
import strformat
const
Width = 1000
Height = 1000
TrunkLength = 400
ScaleFactor = 0.6
StartingAngle = 1.5 * PI
DeltaAngle = 0.2 * PI
proc drawTree(outfile: File; x, y, len, theta: float) =
if len >= 1:
let x2 = x + len * cos(theta)
let y2 = y + len * sin(theta)
outfile.write(
fmt"<line x1='{x}' y1='{y}' x2='{x2}' y2='{y2}' style='stroke:white;stroke-width:1'/>\n")
outfile.drawTree(x2, y2, len * ScaleFactor, theta + DeltaAngle)
outFile.drawTree(x2, y2, len * ScaleFactor, theta - DeltaAngle)
let outsvg = open("tree.svg", fmWrite)
outsvg.write("""<?xml version='1.0' encoding='utf-8' standalone='no'?>
<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN'
'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'>
<svg width='100%%' height='100%%' version='1.1'
xmlns='http://www.w3.org/2000/svg'>\n
<rect width="100%" height="100%" fill="black"/>\n""")
outsvg.drawTree(0.5 * Width, Height, TrunkLength, StartingAngle)
outsvg.write("</svg>\n") # View file tree.svg in browser.
## OCaml
Library: ocaml-cairo
#directory "+cairo"
let img_name = "/tmp/fractree.png"
let width = 480
let height = 640
let level = 9
let line_width = 4.0
let color = (1.0, 0.5, 0.0)
let pi = 4.0 *. atan 1.0
let angle_split = pi *. 0.12
let angle_rand = pi *. 0.12
let () =
Random.self_init();
let surf = Cairo.image_surface_create Cairo.FORMAT_RGB24 ~width ~height in
let ctx = Cairo.create surf in
Cairo.set_antialias ctx Cairo.ANTIALIAS_SUBPIXEL;
Cairo.set_line_cap ctx Cairo.LINE_CAP_ROUND;
let draw_line (x,y) (dx,dy) =
Cairo.move_to ctx x (float height -. y);
Cairo.line_to ctx dx (float height -. dy);
Cairo.stroke ctx;
in
let set_color (r,g,b) v =
Cairo.set_source_rgb ctx ~red:(r *. v) ~green:(g *. v) ~blue:(b *. v);
in
let trans_pos (x,y) len angle =
let _x = cos angle
and _y = sin angle in
(x +. (_x *. len),
y +. (_y *. len))
in
let rec loop ~level ~pos ~line_width ~line_len
~angle ~angle_split ~angle_rand ~intc =
if level > 0 then begin
(* draw the current segment *)
Cairo.set_line_width ctx line_width;
set_color color intc;
let pos_to = trans_pos pos line_len angle in
draw_line pos pos_to;
(* evolution of the parameters *)
let line_width = line_width *. 0.8
and line_len = line_len *. 0.62
and angle_split = angle_split *. 1.02
and angle_rand = angle_rand *. 1.02
and intc = intc *. 0.9
in
let next_loop =
loop ~level:(pred level) ~pos:pos_to ~intc
~line_width ~line_len ~angle_split ~angle_rand
in
(* split *)
let angle_left = angle +. angle_split +. Random.float angle_rand
and angle_right = angle -. angle_split -. Random.float angle_rand
in
next_loop ~angle:angle_left;
next_loop ~angle:angle_right
end
in
let pos = (float width *. 0.5, float height *. 0.1)
and line_len = float height *. 0.3
in
loop ~level ~pos ~angle:(pi /. 2.0)
~angle_split ~angle_rand
~line_width ~line_len ~intc:1.0;
Cairo_png.surface_write_to_file surf img_name
(*Cairo_png.surface_write_to_channel surf stdout*)
## PARI/GP
This version with recursion, in general, is a translation of JavaScript version. Some tweaks and options were added to make it reusable and outputting different size of a tree.
Translation of: JavaScript
Works with: PARI/GP version 2.7.4 and above
\\ Fractal tree (w/recursion)
\\ 4/10/16 aev
plotline(x1,y1,x2,y2)={plotmove(0, x1,y1);plotrline(0,x2-x1,y2-y1);}
plottree(x,y,a,d)={
my(x2,y2,d2r=Pi/180.0,a1=a*d2r,d1);
if(d<=0, return(););
if(d>0, d1=d*10.0;
x2=x+cos(a1)*d1;
y2=y+sin(a1)*d1;
plotline(x,y,x2,y2);
plottree(x2,y2,a-20,d-1);
plottree(x2,y2,a+20,d-1),
return();
);
}
FractalTree(depth,size)={
my(dx=1,dy=0,ttlb="Fractal Tree, depth ",ttl=Str(ttlb,depth));
print1(" *** ",ttl); print(", size ",size);
plotinit(0);
plotcolor(0,6); \\green
plotscale(0, -size,size, 0,size);
plotmove(0, 0,0);
plottree(0,0,90,depth);
plotdraw([0,size,size]);
}
{\\ Executing:
FractalTree(9,500); \\FracTree1.png
FractalTree(12,1100); \\FracTree2.png
FractalTree(15,1500); \\FracTree3.png
}
Output:
*** Fractal Tree, depth 9, size 500
*** last result computed in 140 ms.
*** Fractal Tree, depth 12, size 1100
*** last result computed in 236 ms.
*** Fractal Tree, depth 15, size 1500
*** last result computed in 1,095 ms
## Perl
using the GD::Simple module.
use GD::Simple;
my ($width,$height) = (1000,1000); # image dimension
my $scale = 6/10; # branch scale relative to trunk my$length = 400; # trunk size
my $img = GD::Simple->new($width,$height);$img->fgcolor('black');
$img->penSize(1,1); tree($width/2, $height,$length, 270);
print $img->png; sub tree { my ($x, $y,$len, $angle) = @_; return if$len < 1;
$img->moveTo($x,$y);$img->angle($angle);$img->line($len); ($x, $y) =$img->curPos();
tree($x,$y, $len*$scale, $angle+35); tree($x, $y,$len*$scale,$angle-35);
}
## Phix
Translation of: XPL0
Library: Phix/pGUI
Library: Phix/online
You can run this online here.
--
-- demo\rosetta\FractalTree.exw
-- ============================
--
with javascript_semantics
include pGUI.e
Ihandle dlg, canvas
cdCanvas cddbuffer, cdcanvas
procedure drawTree(integer level, atom angle, len, integer x, y)
integer xn = x + floor(len*cos(angle)),
yn = y + floor(len*sin(angle)),
red = 255-level*8,
green = level*12+100
cdCanvasSetForeground(cddbuffer, red*#10000 + green*#100)
cdCanvasSetLineWidth(cddbuffer,floor(5-level/3))
cdCanvasLine(cddbuffer, x, 480-y, xn, 480-yn)
if level<12 then
drawTree(level+1, angle-0.4, len*0.8, xn, yn) --left
drawTree(level+1, angle+0.1, len*0.8, xn, yn) --right
end if
end procedure
function redraw_cb(Ihandle /*ih*/, integer /*posx*/, /*posy*/)
cdCanvasActivate(cddbuffer)
cdCanvasClear(cddbuffer)
drawTree(0, -PI/2.0, 80.0, 360, 460)
cdCanvasFlush(cddbuffer)
return IUP_DEFAULT
end function
function map_cb(Ihandle ih)
cdcanvas = cdCreateCanvas(CD_IUP, ih)
cddbuffer = cdCreateCanvas(CD_DBUFFER, cdcanvas)
cdCanvasSetBackground(cddbuffer, CD_PARCHMENT)
return IUP_DEFAULT
end function
procedure main()
IupOpen()
canvas = IupCanvas("RASTERSIZE=640x480")
IupSetCallbacks(canvas, {"MAP_CB", Icallback("map_cb"),
"ACTION", Icallback("redraw_cb")})
dlg = IupDialog(canvas,"RESIZE=NO")
IupSetAttribute(dlg, "TITLE", "Fractal Tree")
IupShow(dlg)
if platform()!=JS then
IupMainLoop()
IupClose()
end if
end procedure
main()
## PHP
Image is created with GD module. Code adapted from the JavaScript version.
<?php
$width = 512;$height = 512;
$img = imagecreatetruecolor($width,$height);$bg = imagecolorallocate($img,255,255,255); imagefilledrectangle($img, 0, 0, $width,$width, $bg);$depth = 8;
function drawTree($x1,$y1, $angle,$depth){
global $img; if ($depth != 0){
$x2 =$x1 + (int)(cos(deg2rad($angle)) *$depth * 10.0);
$y2 =$y1 + (int)(sin(deg2rad($angle)) *$depth * 10.0);
imageline($img,$x1, $y1,$x2, $y2, imagecolorallocate($img,0,0,0));
drawTree($x2,$y2, $angle - 20,$depth - 1);
drawTree($x2,$y2, $angle + 20,$depth - 1);
}
}
drawTree($width/2,$height, -90, $depth); imagepng($img);
imagedestroy($img); ?> ## PicoLisp This uses the 'brez' line drawing function from Bitmap/Bresenham's line algorithm#PicoLisp. (load "@lib/math.l") (de fractalTree (Img X Y A D) (unless (=0 D) (let (R (*/ A pi 180.0) DX (*/ (cos R) D 0.2) DY (*/ (sin R) D 0.2)) (brez Img X Y DX DY) (fractalTree Img (+ X DX) (+ Y DY) (+ A 30.0) (dec D)) (fractalTree Img (+ X DX) (+ Y DY) (- A 30.0) (dec D)) ) ) ) (let Img (make (do 300 (link (need 400 0)))) # Create image 400 x 300 (fractalTree Img 200 300 -90.0 10) # Draw tree (out "img.pbm" # Write to bitmap file (prinl "P1") (prinl 400 " " 300) (mapc prinl Img) ) ) ## Plain English To run: Start up. Clear the screen to the lightest blue color. Pick a brownish color. Put the screen's bottom minus 1/2 inch into the context's spot's y coord. Draw a tree given 3 inches. Refresh the screen. Wait for the escape key. Shut down. To draw a tree given a size: If the size is less than 1/32 inch, exit. Put the size divided by 1/4 inch into the pen size. If the size is less than 1/4 inch, pick a greenish color. Remember where we are. Stroke the size. Turn left 1/16 of the way. Draw another tree given the size times 2/3. Turn right 1/16 of the way. Turn right 1/16 of the way. Draw a third tree given the size times 2/3. Turn left 1/16 of the way. Go back to where we were. Output: ## PL/pgSQL Works with: Postgres This piece of code generates the coordinates of each branch, builds a version in the standardized geometry representation format: WKT. A temporary table contains the results: coordinates and WKT representation of each branch. In a query (Postgres + postgis function), we can draw a unique geometry that can be displayed in a tool like QGis or DBeaver database manager for example. The query exploits the notion of CTE and its recursive form. drop table if exists my_temp_tree_table; do $$declare _length numeric := 1; -- a little random _random_length_reduction_max numeric := 0.6; _fork_angle numeric := pi()/12; -- a little random _random_angle numeric := pi()/12; _depth numeric := 9 ; begin create temporary table my_temp_tree_table as WITH RECURSIVE branch(azimuth, x1, y1, x2, y2, len, n) AS ( VALUES (pi()/2, 0.0, 0.0, 0.0, _length, _length, _depth) UNION all select azimuth+a, x2, y2, round((x2+cos(azimuth+a)*len)::numeric, 2), round((y2+sin(azimuth+a)*len)::numeric, 2), (len*(_random_length_reduction_max+(random()*(1-_random_length_reduction_max))))::numeric, n-1 FROM branch cross join ( select ((-_fork_angle)+(_random_angle)*(random()-0.5)) a union select ((_fork_angle)+(_random_angle)*(random()-0.5)) a2 ) a WHERE n > 0 ) select x1, y1, x2, y2, 'LINESTRING('||x1||' '||y1||','||x2||' '||y2||')' as wkt from branch ; end$$ ; -- coordinates and WKT select * from my_temp_tree_table; -- binary version (postgis) of each branch select ST_GeomFromEWKT('SRID=4326;'||wkt) geom from my_temp_tree_table; -- a unique geometry select st_union(ST_GeomFromEWKT('SRID=4326;'||wkt)) geom from my_temp_tree_table; Output: coordinates and WKT x1 |y1 |x2 |y2 |wkt | -----+----+-----+----+---------------------------------+ 0.0| 0.0| 0.0| 1|LINESTRING(0.0 0.0,0.0 1) | 0.0| 1| 0.15|1.99|LINESTRING(0.0 1,0.15 1.99) | 0.0| 1|-0.29|1.96|LINESTRING(0.0 1,-0.29 1.96) | 0.15|1.99| 0.36|2.68|LINESTRING(0.15 1.99,0.36 2.68) | 0.15|1.99| 0.05|2.70|LINESTRING(0.15 1.99,0.05 2.70) | ... ### a simple unparameterized version, without randomness WITH RECURSIVE noeuds(azimuth, x0, y0, x, y, len, n) AS ( VALUES (pi()/2, 0::real, 0::real, 0::real, 10::real, 10::real, 9::int) UNION all select azimuth+a, x, y, (x+cos(azimuth+a)*len)::real, (y+sin(azimuth+a)*len)::real, (len/2)::real, n-1 FROM noeuds cross join (select (-pi()/7)::real a union select (pi()/7)::real a2) a WHERE n > 0 ) , branche as ( select '('||x0||' '||y0||','||x||' '||y||')' b from noeuds ) select ST_GeomFromEWKT('SRID=4326;MULTILINESTRING('||string_agg(b, ',')||')') tree from branche ## PostScript %!PS %%BoundingBox: 0 0 300 300 %%EndComments /origstate save def /ld {load def} bind def /m /moveto ld /g /setgray ld /t /translate ld /r /rotate ld /l /lineto ld /rl /rlineto ld /s /scale ld %%EndProlog /PerturbateAngle {} def /PerturbateLength {} def % ** To add perturbations, define properly PerturbateAngle and PerturbateLength, e.g. % /PerturbateAngle {realtime 20 mod realtime 2 mod 1 eq {add} {sub} ifelse} def % /PerturbateLength {realtime 10 mod 100 div realtime 2 mod 1 eq {add} {sub} ifelse} def /fractree { % [INITLENGTH, SPLIT, SFACTOR, BRANCHES] dup 3 get 0 gt { 0 0 m dup 0 get 0 exch l gsave dup 0 get 0 exch t dup 1 get PerturbateAngle r dup 2 get dup PerturbateLength s dup aload pop 1 sub 4 array astore fractree stroke grestore gsave dup 0 get 0 exch t dup 1 get neg PerturbateAngle r dup 2 get dup PerturbateLength s dup aload pop 1 sub 4 array astore fractree stroke grestore } if pop } def % /BRANCHES 14 def /INITLENGTH 50 def /SPLIT 35 def /SFACTOR .75 def % % BB check %0 0 m 300 0 rl 0 300 rl -300 0 rl closepath stroke % 0 g 150 0 t [INITLENGTH SPLIT SFACTOR BRANCHES] fractree stroke % showpage origstate restore %%EOF Shorter version: %!PS-Adobe-3.0 %%BoundingBox: 0 0 300 300 /!0 { dup 1 sub dup 0 gt } def /trunk { 0 0 moveto 0 60 translate 0 0 lineto stroke } def /branch { gsave scale rotate dup d exch sub d div setgray tree grestore } def /L { 30 .8 .8 branch } def /M {-10 .7 .7 branch } def /R {-35 .7 .7 branch } def /tree { trunk !0 { L M R } if pop } def /d 10 def 5 setlinewidth 1 setlinecap 170 20 translate d tree pop %%EOF ## POV-Ray #include "colors.inc" #include "transforms.inc" #declare CamLoc = <0, 5, 0>; #declare CamLook = <0,0,0>; camera { location CamLoc look_at CamLook rotate y*90 } light_source { CamLoc color White } #declare Init_Height = 10; #declare Spread_Ang = 35; #declare Branches = 14; #declare Scaling_Factor = 0.75; #macro Stick(P0, P1) cylinder { P0, P1, 0.02 texture { pigment { Green } } } #end #macro FractalTree(O, D, S, R, B) #if (B > 0) Stick(O, O+D*S) FractalTree(O+D*S, vtransform(D, transform{rotate y*R}), S*Scaling_Factor, R, B-1) FractalTree(O+D*S, vtransform(D, transform{rotate -y*R}), S*Scaling_Factor, R, B-1) #end #end union { FractalTree(<-2,0,0>, <1,0,0>, 1, Spread_Ang, Branches) } ## Prolog SWI-Prolog has a graphic interface : XPCE. fractal :- new(D, window('Fractal')), send(D, size, size(800, 600)), drawTree(D, 400, 500, -90, 9), send(D, open). drawTree(_D, _X, _Y, _Angle, 0). drawTree(D, X1, Y1, Angle, Depth) :- X2 is X1 + cos(Angle * pi / 180.0) * Depth * 10.0, Y2 is Y1 + sin(Angle * pi / 180.0) * Depth * 10.0, new(Line, line(X1, Y1, X2, Y2, none)), send(D, display, Line), A1 is Angle - 30, A2 is Angle + 30, De is Depth - 1, drawTree(D, X2, Y2, A1, De), drawTree(D, X2, Y2, A2, De). ## PureBasic #Spread_Ang = 35 #Scaling_Factor = 0.75 #Deg_to_Rad = #PI / 180 #SizeH = 500 #SizeV = 375 #Init_Size = 100 Procedure drawTree(x1, y1, Size, theta, depth) Protected x2 = x1 + Cos(theta * #Deg_to_Rad) * Size, y2 = y1 + Sin(theta * #Deg_to_Rad) * Size LineXY(x1, y1, x2, y2, RGB(255, 255, 255)) If depth <= 0 ProcedureReturn EndIf ;draw left branch drawTree(x2, y2, Size * #Scaling_Factor, theta - #Spread_Ang, depth - 1) ;draw right branch drawTree(x2, y2, Size * #Scaling_Factor, theta + #Spread_Ang, depth - 1) EndProcedure OpenWindow(0, 0, 0, #SizeH, #SizeV, "Fractal Tree", #PB_Window_SystemMenu) Define fractal = CreateImage(#PB_Any, #SizeH, #SizeV, 32) ImageGadget(0, 0, 0, 0, 0, ImageID(fractal)) If StartDrawing(ImageOutput(fractal)) drawTree(#SizeH / 2, #SizeV, #Init_Size, -90, 9) StopDrawing() SetGadgetState(0, ImageID(fractal)) EndIf Repeat: Until WaitWindowEvent(10) = #PB_Event_CloseWindow ## Processing #### Using rotation void setup() { size(600, 600); background(0); stroke(255); drawTree(300, 550, 9); } void drawTree(float x, float y, int depth) { float forkAngle = radians(20); float baseLen = 10.0; if (depth > 0) { pushMatrix(); translate(x, y - baseLen * depth); line(0, baseLen * depth, 0, 0); rotate(forkAngle); drawTree(0, 0, depth - 1); rotate(2 * -forkAngle); drawTree(0, 0, depth - 1); popMatrix(); } } #### Calculating coordinates Translation of: Python void setup() { size(600, 600); background(0); stroke(255); drawTree(300, 550, -90, 9); } void drawTree(float x1, float y1, float angle, int depth) { float forkAngle = 20; float baseLen = 10.0; if (depth > 0) { float x2 = x1 + cos(radians(angle)) * depth * baseLen; float y2 = y1 + sin(radians(angle)) * depth * baseLen; line(x1, y1, x2, y2); drawTree(x2, y2, angle - forkAngle, depth - 1); drawTree(x2, y2, angle + forkAngle, depth - 1); } } ### Processing Python mode #### Using rotation Translation of: Processing def setup(): size(600, 600) background(0) stroke(255) drawTree(300, 550, 9) def drawTree(x, y, depth): fork_ang = radians(20) base_len = 10 if depth > 0: pushMatrix() translate(x, y - baseLen * depth) line(0, baseLen * depth, 0, 0) rotate(fork_ang) drawTree(0, 0, depth - 1) rotate(2 * -fork_ang) drawTree(0, 0, depth - 1) popMatrix() #### Calculating coordinates Translation of: Python def setup(): size(600, 600) background(0) stroke(255) drawTree(300, 550, -90, 9) def drawTree(x1, y1, angle, depth): fork_angle = 20 base_len = 10.0 if depth > 0: x2 = x1 + cos(radians(angle)) * depth * base_len y2 = y1 + sin(radians(angle)) * depth * base_len line(x1, y1, x2, y2) drawTree(x2, y2, angle - fork_angle, depth - 1) drawTree(x2, y2, angle + fork_angle, depth - 1) ## Python Library: pygame import pygame, math pygame.init() window = pygame.display.set_mode((600, 600)) pygame.display.set_caption("Fractal Tree") screen = pygame.display.get_surface() def drawTree(x1, y1, angle, depth): fork_angle = 20 base_len = 10.0 if depth > 0: x2 = x1 + int(math.cos(math.radians(angle)) * depth * base_len) y2 = y1 + int(math.sin(math.radians(angle)) * depth * base_len) pygame.draw.line(screen, (255,255,255), (x1, y1), (x2, y2), 2) drawTree(x2, y2, angle - fork_angle, depth - 1) drawTree(x2, y2, angle + fork_angle, depth - 1) def input(event): if event.type == pygame.QUIT: exit(0) drawTree(300, 550, -90, 9) pygame.display.flip() while True: input(pygame.event.wait()) ## QB64 _Title "Fractal Tree" Const sw% = 640 Const sh% = 480 Screen _NewImage(sw, sh, 8) Cls , 15: Color 2 Call tree(sw \ 2, sh - 10, _Pi * 1.5, _Pi / 180 * 29, 112, 15) Sleep System Sub tree (x As Integer, y As Integer, initAngle As Double, theta As Double, length As Double, depth As Integer) Dim As Integer iL, newX, newY, iX, iY, iD iL = length: iX = x: iY = y: iD = depth newX = Cos(initAngle) * length + iX newY = Sin(initAngle) * length + iY Line (iX, iY)-(newX, newY) iL = length * .78 iD = iD - 1 If iD > 0 Then Call tree(newX, newY, initAngle - theta, theta, iL, iD) Call tree(newX, newY, initAngle + theta, theta, iL, iD) End If End Sub ## Quackery [$ "turtleduck.qky" loadfile ] now!
[ [ 1 1
30 times
[ tuck + ]
swap join ] constant
do ] is phi ( --> n/d )
[ 2dup 5 1 v< iff
2drop done
2dup 5 1 v/
proper 2drop wide
2dup walk
1 5 turn
2dup phi v/
2dup recurse
-2 5 turn
recurse
1 5 turn
-v fly ] is tree ( n/d --> )
turtle
20 frames
-1 4 turn
-450 1 fly
500 1 tree
1 frames
Output:
## R
Translation of: PARI/GP
Works with: R version 3.3.3 and above
## Recursive FT plotting
plotftree <- function(x, y, a, d, c) {
x2=y2=0; d2r=pi/180.0; a1 <- a*d2r; d1=0;
if(d<=0) {return()}
if(d>0)
{ d1=d*10.0;
x2=x+cos(a1)*d1;
y2=y+sin(a1)*d1;
segments(x*c, y*c, x2*c, y2*c, col='darkgreen');
plotftree(x2,y2,a-20,d-1,c);
plotftree(x2,y2,a+20,d-1,c);
#return(2);
}
}
## Plotting Fractal Tree. aev 3/27/17
## ord - order/depth, c - scale, xsh - x-shift, fn - file name,
## ttl - plot title.
pFractalTree <- function(ord, c=1, xsh=0, fn="", ttl="") {
cat(" *** START FRT:", date(), "\n");
m=640;
if(fn=="") {pf=paste0("FRTR", ord, ".png")} else {pf=paste0(fn, ".png")};
if(ttl=="") {ttl=paste0("Fractal tree, order - ", ord)};
cat(" *** Plot file -", pf, "title:", ttl, "\n");
##plot(NA, xlim=c(0,m), ylim=c(-m,0), xlab="", ylab="", main=ttl);
plot(NA, xlim=c(0,m), ylim=c(0,m), xlab="", ylab="", main=ttl);
plotftree(m/2+xsh,100,90,ord,c);
dev.copy(png, filename=pf, width=m, height=m);
dev.off(); graphics.off();
cat(" *** END FRT:",date(),"\n");
}
## Executing:
pFractalTree(9);
pFractalTree(12,0.6,210);
pFractalTree(15,0.35,600);
Output:
> pFractalTree(9);
*** START FRT: Tue Mar 28 16:49:49 2017
*** Plot file - FRTR9.png title: Fractal tree, order - 9
*** END FRT: Tue Mar 28 16:49:50 2017
> pFractalTree(12,0.6,210);
*** START FRT: Tue Mar 28 17:32:15 2017
*** Plot file - FRTR12.png title: Fractal tree, order - 12
*** END FRT: Tue Mar 28 17:32:16 2017
> pFractalTree(15,0.35,600);
*** START FRT: Tue Mar 28 17:38:34 2017
*** Plot file - FRTR15.png title: Fractal tree, order - 15
*** END FRT: Tue Mar 28 17:38:41 2017
## Racket
#lang racket
(require graphics/turtles)
(define (tree n)
(when (> n 1)
(draw (/ n 2))
(tprompt (split* (turn 60) (turn -60))
(tree (/ n 2)))
(draw (/ n 2))
(turn 5)
(tree (- n 1))))
(turtles #t) (move 100) (turn 90) (move -200)
(tree 35)
(save-turtle-bitmap "tree.png" 'png)
## Raku
(formerly Perl 6) Image is created in SVG format.
my ($width,$height) = (1000,1000); # image dimension
my $scale = 6/10; # branch scale relative to trunk my$length = 400; # trunk size
say "<?xml version='1.0' encoding='utf-8' standalone='no'?>
<!DOCTYPE svg PUBLIC '-//W3C//DTD SVG 1.1//EN'
'http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd'>
<svg width='100%' height='100%' version='1.1'
xmlns='http://www.w3.org/2000/svg'>";
tree($width/2,$height, $length, 3*pi/2); say "</svg>"; multi tree($, $,$length where { $length < 1},$) {}
multi tree($x,$y, $length,$angle)
{
my ($x2,$y2) = ( $x +$length * $angle.cos,$y + $length *$angle.sin);
say "<line x1='$x' y1='$y' x2='$x2' y2='$y2' style='stroke:rgb(0,0,0);stroke-width:1'/>";
tree($x2,$y2, $length*$scale, $angle + pi/5); tree($x2, $y2,$length*$scale,$angle - pi/5);
}
## Red
Red [Needs: 'View]
color: brown
width: 9
view/tight/options/flags/no-wait [ ; click image to grow tree
img: image 1097x617 draw [
pen brown line-width 9 line 500x600 500x500] [grow]
] [offset: 0x0] [no-border]
ends: reduce [500x500 pi * 3 / 2] ; list of terminal nodes
da: pi * 30 / 180 ; angle of branches in radians
ea: pi * 5 / 180 ; offset added to angle to break symmetry
l: 200 ; branches initial lenght
scale: 0.7 ; branches scale factor
grow: does [ ; grows branches
l: l * scale
color: 2 * color + leaf / 3
width: width - 1
newends: copy []
foreach [p a] ends [
a1: a + da - ea
p1: p + as-pair l * cos a1 l * sin a1
a2: a - da - ea
p2: p + as-pair l * cos a2 l * sin a2
append img/draw compose/deep [
pen (color) line-width (width) line (p1) (p) (p2)]
append newends reduce [p1 a1 p2 a2]
]
ends: newends
]
Output:
## Ring
load "guilib.ring"
new qapp
{
win1 = new qwidget() {
setwindowtitle("drawing using qpainter")
setgeometry(100,100,500,500)
label1 = new qlabel(win1) {
setgeometry(10,10,400,400)
settext("")
}
draw()
show()
}
exec()
}
func draw
p1 = new qpicture()
color = new qcolor() {
setrgb(0,0,255,255)
}
pen = new qpen() {
setcolor(color)
setwidth(1)
}
new qpainter() {
begin(p1)
setpen(pen)
sizex = 400
sizey = 200
depth = 10
tree(self, sizex, 0, sizey/2, 90, depth)
endpaint()
}
label1 { setpicture(p1) show() }
func tree myObj, x1, y1, size, angle, depth
myObj{
scale = 0.76
x2 = x1 + size * cos(angle)
y2 = y1 + size * sin(angle)
drawline(x1, y1, x2, y2)
if depth > 0
tree(self, x2, y2, size * scale, angle - spread, depth - 1)
tree(self, x2, y2, size * scale, angle + spread, depth - 1) ok}
Output:
## Ruby
Library: Shoes
Shoes.app(:title => "Fractal Tree", :width => 600, :height => 600) do
background "#fff"
stroke "#000"
def drawTree(x1, y1, angle, depth)
if depth != 0
x2 = x1 + (Math.cos(angle * @deg_to_rad) * depth * 10.0).to_i
y2 = y1 + (Math.sin(angle * @deg_to_rad) * depth * 10.0).to_i
line x1, y1, x2, y2
drawTree(x2, y2, angle - 20, depth - 1)
drawTree(x2, y2, angle + 20, depth - 1)
end
end
drawTree(300,550,-90,9)
end
## Rust
Library: Piston
//Cargo deps :
// piston = "0.35.0"
// piston2d-graphics = "0.23.0"
// piston2d-opengl_graphics = "0.49.0"
// pistoncore-glutin_window = "0.42.0"
extern crate piston;
extern crate graphics;
extern crate opengl_graphics;
extern crate glutin_window;
use piston::window::WindowSettings;
use piston::event_loop::{Events, EventSettings};
use piston::input::RenderEvent;
use glutin_window::GlutinWindow as Window;
use opengl_graphics::{GlGraphics, OpenGL};
use graphics::{clear, line, Context};
const ANG: f64 = 20.0;
const COLOR: [f32; 4] = [1.0, 0.0, 0.5, 1.0];
const LINE_THICKNESS: f64 = 5.0;
const DEPTH: u32 = 11;
fn main() {
let mut window: Window = WindowSettings::new("Fractal Tree", [1024, 768])
.opengl(OpenGL::V3_2)
.exit_on_esc(true)
.build()
.unwrap();
let mut gl = GlGraphics::new(OpenGL::V3_2);
let mut events = Events::new(EventSettings::new());
while let Some(e) = events.next(&mut window) {
if let Some(args) = e.render_args() {
gl.draw(args.viewport(), |c, g| {
clear([1.0, 1.0, 1.0, 1.0], g);
draw_fractal_tree(512.0, 700.0, 0.0, DEPTH, c, g);
});
}
}
}
fn draw_fractal_tree(x1: f64, y1: f64, angle: f64, depth: u32, c: Context, g: &mut GlGraphics) {
let x2 = x1 + angle.to_radians().sin() * depth as f64 * 10.0;
let y2 = y1 - angle.to_radians().cos() * depth as f64 * 10.0;
line(
COLOR,
LINE_THICKNESS * depth as f64 * 0.2,
[x1, y1, x2, y2],
c.transform,
g,
);
if depth > 0 {
draw_fractal_tree(x2, y2, angle - ANG, depth - 1, c, g);
draw_fractal_tree(x2, y2, angle + ANG, depth - 1, c, g);
}
}
## Scala
Adapted from the Java version. Screenshot below.
import swing._
import java.awt.{RenderingHints, BasicStroke, Color}
object FractalTree extends SimpleSwingApplication {
val DEPTH = 9
def top = new MainFrame {
contents = new Panel {
preferredSize = new Dimension(600, 500)
override def paintComponent(g: Graphics2D) {
draw(300, 460, -90, DEPTH)
def draw(x1: Int, y1: Int, angle: Double, depth: Int) {
if (depth > 0) {
val x2 = x1 + (math.cos(angle.toRadians) * depth * 10).toInt
val y2 = y1 + (math.sin(angle.toRadians) * depth * 10).toInt
g.setColor(Color.getHSBColor(0.25f - depth * 0.125f / DEPTH, 0.9f, 0.6f))
g.setStroke(new BasicStroke(depth))
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON)
g.drawLine(x1, y1, x2, y2)
draw(x2, y2, angle - 20, depth - 1)
draw(x2, y2, angle + 20, depth - 1)
}
}
}
}
}
}
## Scheme
The tree is created as a list of line segments, which can then be drawn on a required device. For this program, the tree is output to an eps file.
(import (scheme base)
(scheme file)
(scheme inexact)
(scheme write))
(define *scale* 10) ; controls overall size of tree
(define *split* 20) ; controls angle of split (in degrees)
;; construct lines for tree as list of 5-tuples (x1 y1 x2 y2 depth)
;; - x1 y1 is start point
;; - angle of this line, in radians
;; - depth, depth within tree (controls length of line)
(define (create-tree x1 y1 angle depth)
(let ((pi 3.14159265358979323846264338327950288419716939937510582097))
(* d pi 1/180)))
;
(if (zero? depth)
'()
(let ((x2 (+ x1 (* (cos (degrees->radians angle)) depth *scale*)))
(y2 (+ y1 (* (sin (degrees->radians angle)) depth *scale*))))
(append (list (map truncate (list x1 y1 x2 y2 depth)))
(create-tree x2 y2 (- angle *split*) (- depth 1))
(create-tree x2 y2 (+ angle *split*) (- depth 1))))))
;; output the tree to an eps file
(define (output-tree-as-eps filename tree)
(when (file-exists? filename) (delete-file filename))
(with-output-to-file
filename
(lambda ()
(display "%!PS-Adobe-3.0 EPSF-3.0\n%%BoundingBox: 0 0 800 800\n")
;; add each line - sets linewidth based on depth in tree
(for-each (lambda (line)
(display
(string-append "newpath\n"
(number->string (list-ref line 0)) " "
(number->string (list-ref line 1)) " "
"moveto\n"
(number->string (list-ref line 2)) " "
(number->string (list-ref line 3)) " "
"lineto\n"
(number->string (truncate (/ (list-ref line 4) 2)))
" setlinewidth\n"
"stroke\n"
)))
tree)
(display "\n%%EOF"))))
(output-tree-as-eps "fractal.eps" (create-tree 400 200 90 9))
## Scilab
### L-System approach
This script uses complex numbers to represent (x,y) coordinates: real part as x position, and imaginary part as y position. The tree is generated using an L-system approach, and the lines are then drawn by interpreting the resulting sentence. The output is plotted onto graphic window.
trunk = 1; //trunk length
ratio = 0.8; //size ratio between two consecutive branches
depth = 9; //final number of branch levels
orign = 0; //origin of the tree (should be complex)
angle = 45*%pi/180; //angle between two branches [rad]
trunk_angle = 90*%pi/180; //angle between trunk and X-axis [rad]
right_angle = angle/2; //angles to the right or to the left
left_angle = 0.8*angle; //can be set independently or
//as function of 'angle'
//L-system definition:
//Alphabet: FBD[]+-
//F: go forward B: go backwards
//[: start new branch ]: end current branch
//+: branch to the right -: branch to the left
//D: double line (forward then backward)
//Axiom: D
//Rule: D -> F[+D-D]B
//L-system sentence generation
sentence = 'D'
rule = 'F[+D-D]B';
for i=1:depth
sentence = strsubst(sentence,'D',rule);
end
sentence = strsplit(sentence)';
//Empty tree
tree_size = 1.0...
+ length(find(sentence=='F'|sentence=='B'))...
+ 2 * length(find(sentence=='D'));
tree=zeros(tree_size,1);
//Drawing the tree
branch_level = 0;
curr_angle = trunk_angle;
curr_pos = 1;
for ind = 1:size(sentence,'c')
charac = sentence(ind);
select charac
case 'F' then //Draw line forward
tree(curr_pos+1) = tree(curr_pos)...
+ trunk * ratio^branch_level * exp(curr_angle*%i);
curr_pos = curr_pos + 1;
case 'B' then //Draw line backwards
tree(curr_pos+1) = tree(curr_pos)...
+ trunk * ratio^branch_level * exp((%pi+curr_angle)*%i);
curr_pos = curr_pos + 1;
case '[' then //New branch
branch_level = branch_level + 1;
case '+' then //Turn right
curr_angle = curr_angle - right_angle;
case '-' then //Turn left
curr_angle = curr_angle + right_angle + left_angle;
case ']' then //End of branch
branch_level = branch_level - 1;
curr_angle = curr_angle - left_angle;
case 'D' then //Double line
tree(curr_pos+1) = tree(curr_pos)...
+ trunk * ratio^branch_level * exp(curr_angle*%i);
tree(curr_pos+2) = tree(curr_pos+1)...
+ trunk * ratio^branch_level * exp((%pi+curr_angle)*%i);
curr_pos = curr_pos + 2;
end
end
scf(); clf();
xname('Fractal tree: '+string(depth)+' levels')
plot2d(real(tree),imag(tree),14);
set(gca(),'isoview','on');
set(gca(),'axes_visible',['off','off','off']);
### Recursive approach
Translation of: PHP
width = 512;
height = 512;
img=scf();
set(img,'figure_size',[width,height]);
function drawTree(x1, y1, angle, depth)
if depth ~= 0 then
x2 = x1 + cos(angle * %pi/180) * depth * 10;
y2 = y1 + sin(angle * %pi/180) * depth * 10;
plot2d([x1 x2],[y1 y2],14);
drawTree(x2, y2, angle - 20, depth - 1);
drawTree(x2, y2, angle + 20, depth - 1);
end
endfunction
drawTree(width/2,height,90,10);
set(gca(),'isoview','on');
$include "seed7_05.s7i"; include "float.s7i"; include "math.s7i"; include "draw.s7i"; include "keybd.s7i"; const float: DEG_TO_RAD is PI / 180.0; const proc: drawTree (in integer: x1, in integer: y1, in float: angle, in integer: depth) is func local var integer: x2 is 0; var integer: y2 is 0; begin if depth <> 0 then x2 := x1 + trunc(cos(angle * DEG_TO_RAD) * flt(depth * 10)); y2 := y1 + trunc(sin(angle * DEG_TO_RAD) * flt(depth * 10)); lineTo(x1, y1, x2, y2, white); drawTree(x2, y2, angle - 20.0, depth - 1); drawTree(x2, y2, angle + 20.0, depth - 1); end if; end func; const proc: main is func begin screen(600, 500); clear(curr_win, black); KEYBOARD := GRAPH_KEYBOARD; drawTree(300, 470, -90.0, 9); ignore(getc(KEYBOARD)); end func; Original source: [2] ## Sidef Translation of: Perl func tree(img, x, y, scale=6/10, len=400, angle=270) { len < 1 && return() img.moveTo(x, y) img.angle(angle) img.line(len) var (x1, y1) = img.curPos tree(img, x1, y1, scale, len*scale, angle+35) tree(img, x1, y1, scale, len*scale, angle-35) } require('GD::Simple') var (width=1000, height=1000) var img = %s|GD::Simple|.new(width, height) img.fgcolor('black') img.penSize(1, 1) tree(img, width/2, height) File('tree.png').write(img.png, :raw) ## Smalltalk This example is coded for Squeak Smalltalk. Object subclass: #FractalTree instanceVariableNames: '' classVariableNames: '' poolDictionaries: '' category: 'RosettaCode' Methods for FractalTree class: tree: aPoint length: aLength angle: anAngle | p a | (aLength > 10) ifTrue: [ p := Pen new. p up. p goto: aPoint. p turn: anAngle. p down. 5 timesRepeat: [ p go: aLength / 5. p turn: 5. ]. a := anAngle - 30. 3 timesRepeat: [ self tree: p location length: aLength * 0.7 angle: a. a := a + 30. ] ]. draw Display restoreAfter: [ Display fillWhite. self tree: 700@700 length: 200 angle: 0. ] Now open a new Workspace and enter: FractalTree new draw. ## SVG In the same style as Dragon curve#SVG. SVG has no parameterized definitions, so the recursion must be unrolled. <?xml version="1.0" standalone="yes"?> <!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20010904//EN" "http://www.w3.org/TR/2001/REC-SVG-20010904/DTD/svg10.dtd"> <svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="400" height="320"> <style type="text/css"><![CDATA[ line { stroke: black; stroke-width: .05; } circle { fill: black; } ]]></style> <defs> <g id="stem"> <line x1="0" y1="0" x2="0" y2="-1"/> </g> <g id="l0"><use xlink:href="#stem"/></g> <!-- These are identical except for the id and href. --> <g id="l1"> <use xlink:href="#l0" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l0" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l2"> <use xlink:href="#l1" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l1" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l3"> <use xlink:href="#l2" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l2" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l4"> <use xlink:href="#l3" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l3" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l5"> <use xlink:href="#l4" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l4" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l6"> <use xlink:href="#l5" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l5" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l7"> <use xlink:href="#l6" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l6" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l8"> <use xlink:href="#l7" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l7" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> <g id="l9"> <use xlink:href="#l8" transform="translate(0, -1) rotate(-35) scale(.7)"/> <use xlink:href="#l8" transform="translate(0, -1) rotate(+35) scale(.7)"/> <use xlink:href="#stem"/></g> </defs> <g transform="translate(200, 320) scale(100)"> <use xlink:href="#l9"/> </g> </svg> ## Swift Image - Link, since uploads seem to be disabled currently. In a playground: extension CGFloat { func degrees_to_radians() -> CGFloat { return CGFloat(M_PI) * self / 180.0 } } extension Double { func degrees_to_radians() -> Double { return Double(M_PI) * self / 180.0 } } class Tree: UIView { func drawTree(x1: CGFloat, y1: CGFloat, angle: CGFloat, depth:Int){ if depth == 0 { return } let ang = angle.degrees_to_radians() let x2:CGFloat = x1 + ( cos(ang) as CGFloat) * CGFloat(depth) * (self.frame.width / 60) let y2:CGFloat = y1 + ( sin(ang) as CGFloat) * CGFloat(depth) * (self.frame.width / 60) let line = drawLine(x1, y1: y1, x2: x2, y2: y2) line.stroke() drawTree(x2, y1: y2, angle: angle - 20, depth: depth - 1) drawTree(x2, y1: y2, angle: angle + 20, depth: depth - 1) } func drawLine(x1:CGFloat, y1:CGFloat, x2:CGFloat, y2:CGFloat) -> UIBezierPath { let path = UIBezierPath() path.moveToPoint(CGPoint(x: x1,y: y1)) path.addLineToPoint(CGPoint(x: x2,y: y2)) path.lineWidth = 1 return path } override func drawRect(rect: CGRect) { let color = UIColor(red: 1.0, green: 0.0, blue: 0.0, alpha: 1.0) color.set() drawTree(self.frame.width / 2 , y1: self.frame.height * 0.8, angle: -90 , depth: 9 ) } } let tree = Tree(frame: CGRectMake(0, 0, 300, 300)) tree ## Standard ML Works with PolyML open XWindows; open Motif; fun toI {x=x,y=y} = {x=Real.toInt IEEEReal.TO_NEAREST x,y=Real.toInt IEEEReal.TO_NEAREST y} ; fun drawOnTop win usegc ht hs {x=l1,y=l2} {x=r1,y=r2} = let val xy = {x=l1 - ht * (l2-r2) , y = l2 - ht * (r1-l1) } val zt = {x=r1 - ht * (l2-r2) , y= r2 - ht * (r1-l1) } val ab = {x= ( (#x xy + #x zt) + hs * (#y zt - #y xy ) )/2.0 , y = ( (#y zt + #y xy) - hs * (#x zt - #x xy )) /2.0 } in if abs (l1 - #x xy ) < 0.9 andalso abs (l2 - #y xy ) < 0.9 then XFlush (XtDisplay win) else (XFillPolygon (XtWindow win) usegc [ (XPoint o toI) {x=l1,y=l2}, (XPoint o toI ) xy , (XPoint o toI ) ab , (XPoint o toI ) zt , (XPoint o toI ) {x=r1,y=r2} ] Convex CoordModeOrigin ; drawOnTop win usegc (0.87*ht) hs xy ab ; drawOnTop win usegc (0.93*ht) hs ab zt ) end ; val demoWindow = fn () => let val shell = XtAppInitialise "" "tree" "top" [] [ XmNwidth 800, XmNheight 650] ; val main = XmCreateMainWindow shell "main" [ XmNmappedWhenManaged true ] ; val canvas = XmCreateDrawingArea main "drawarea" [ XmNwidth 800, XmNheight 650] ; val usegc = DefaultGC (XtDisplay canvas) ; in XtSetCallbacks canvas [ (XmNexposeCallback , (fn (w,c,t) => ( drawOnTop canvas usegc 8.0 0.85 {x=385.0,y=645.0} {x=415.0,y=645.0} ; t) ) ) ] XmNarmCallback ; XtManageChild canvas ; XtManageChild main ; XtRealizeWidget shell end ; demoWindow (); ## Tcl Library: Tk package require Tk set SIZE 800 set SCALE 4.0 set BRANCHES 14 set ROTATION_SCALE 0.85 set INITIAL_LENGTH 50.0 proc draw_tree {w x y dx dy size theta depth} { global SCALE ROTATION_SCALE$w create line $x$y [expr {$x +$dx*$size}] [expr {$y + $dy*$size}]
if {[incr depth -1] >= 0} {
set x [expr {$x +$dx*$size}] set y [expr {$y + $dy*$size}]
set ntheta [expr {$theta *$ROTATION_SCALE}]
# Draw left branch
draw_tree $w$x $y \ [expr {$dx*cos($theta) +$dy*sin($theta)}] \ [expr {$dy*cos($theta) -$dx*sin($theta)}] \ [expr {$size * (rand() + $SCALE - 1) /$SCALE}] $ntheta$depth
# Draw right branch
draw_tree $w$x $y \ [expr {$dx*cos(-$theta) +$dy*sin(-$theta)}] \ [expr {$dy*cos(-$theta) -$dx*sin(-$theta)}] \ [expr {$size * (rand() + $SCALE - 1) /$SCALE}] $ntheta$depth
}
}
pack [canvas .c -width $SIZE -height$SIZE]
draw_tree .c [expr {$SIZE/2}] [expr {$SIZE-10}] 0.0 -1.0 $INITIAL_LENGTH \ [expr {3.1415927 / 8}]$BRANCHES
## TUSCRIPT
Image is created in SVG-format
$$MODE TUSCRIPT dest="fracaltree.svg" ERROR/STOP CREATE (dest,fdf-o,-std-) ACCESS d: WRITE/ERASE/RECORDS/UTF8 dest s,text MODE DATA$$ header=*
<?xml version="1.0" standalone="yes"?>
<!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20010904//EN"
"http://www.w3.org/TR/2001/REC-SVG-20010904/DTD/svg10.dtd">
<svg xmlns="http://www.w3.org/2000/svg"
width="400" height="320">
<style type="text/css"><![CDATA[
line { stroke: brown; stroke-width: .05; }
]]></style>
$$WRITE/NEXT d header$$ defsbeg=*
<defs>
<g id="stem"> <line x1="0" y1="0" x2="0" y2="-1"/> </g>
$$WRITE/NEXT d defsbeg$$ LOOP n=10,21
$$id=n+1,lastnr=VALUE(n)$$ g=*
<g id="{id}"> <use xlink:href="#{n}" transform="translate(0, -1) rotate(-35) scale(.7)"/>
$$WRITE/NEXT d g$$ ENDLOOP
$$defsend = * </defs> <g transform="translate(200, 320) scale(100)"> <use xlink:href="#{lastnr}"/> </g>$$ MODE TUSCRIPT
WRITE/NEXT d defsend
WRITE/NEXT d "</svg>"
ENDACCESS d
## TypeScript
Translation of: JavaScript
// Set up canvas for drawing
var canvas: HTMLCanvasElement = document.createElement('canvas')
canvas.width = 600
canvas.height = 500
document.body.appendChild(canvas)
var ctx: CanvasRenderingContext2D = canvas.getContext('2d')
ctx.fillStyle = '#000'
ctx.lineWidth = 1
// constants
const degToRad: number = Math.PI / 180.0
const totalDepth: number = 9
/** Helper function that draws a line on the canvas */
function drawLine(x1: number, y1: number, x2: number, y2: number): void {
ctx.moveTo(x1, y1)
ctx.lineTo(x2, y2)
}
/** Draws a branch at the given point and angle and then calls itself twice */
function drawTree(x1: number, y1: number, angle: number, depth: number): void {
if (depth !== 0) {
let x2: number = x1 + (Math.cos(angle * degToRad) * depth * 10.0)
let y2: number = y1 + (Math.sin(angle * degToRad) * depth * 10.0)
drawLine(x1, y1, x2, y2)
drawTree(x2, y2, angle - 20, depth - 1)
drawTree(x2, y2, angle + 20, depth - 1)
}
}
// actual drawing of tree
ctx.beginPath()
drawTree(300, 500, -90, totalDepth)
ctx.closePath()
ctx.stroke()
## Wren
Translation of: Kotlin
Library: DOME
import "graphics" for Canvas, Color
import "dome" for Window
import "math" for Math
var Radians = Fn.new { |d| d * Num.pi / 180 }
class FractalTree {
construct new(width, height) {
Window.title = "Fractal Tree"
Window.resize(width, height)
Canvas.resize(width, height)
_fore = Color.white
}
init() {
drawTree(400, 500, -90, 9)
}
drawTree(x1, y1, angle, depth) {
if (depth == 0) return
var x2 = x1 + (Math.cos(r) * depth * 10).truncate
var y2 = y1 + (Math.sin(r) * depth * 10).truncate
Canvas.line(x1, y1, x2, y2, _fore)
drawTree(x2, y2, angle - 20, depth - 1)
drawTree(x2, y2, angle + 20, depth - 1)
}
update() {}
draw(alpha) {}
}
var Game = FractalTree.new(800, 600)
## XPL0
include c:\cxpl\codes;
proc DrawBranch(Lev, Dir, Len, X, Y);
int Lev; real Dir, Len; int X, Y;
int Red, Grn;
[Move(X, Y);
X:= X + fix(Len*Cos(Dir));
Y:= Y + fix(Len*Sin(Dir));
Red:= 255-Lev*8; Grn:= Lev*12+100;
Line(X, Y, Red<<16+Grn<<8);
if Lev < 12 then \limit level of recursion
[DrawBranch(Lev+1, Dir-0.4, Len*0.8, X, Y); \left
DrawBranch(Lev+1, Dir+0.1, Len*0.8, X, Y); \right
];
];
## zkl
Uses the PPM class from http://rosettacode.org/wiki/Bitmap/Bresenham%27s_line_algorithm#zkl
Translation of: BBC BASIC
Translation of: XPL0
fcn fractalTree(){
scale:=0.76;
sizeX:=400; sizeY:=300;
bitmap:=PPM(sizeX*2,sizeY*2,0xFF|FF|FF);
branch:='wrap(x1,y1,size,angle,depth){
x2:=x1 - size*ar.cos();
y2:=y1 + size*ar.sin();
color:=(0xff-depth*8).shiftLeft(16) + (depth*12+100).shiftLeft(8);
bitmap.line(x1,y1, x2,y2, color);
if(depth){
self.fcn(x2,y2,scale*size,angle - 30,depth - 1,vm.pasteArgs(5));
self.fcn(x2,y2,scale*size,angle + 8, depth - 1,vm.pasteArgs(5));
}
};
branch(sizeX,0,sizeY/2,90.0,10);
bitmap.write(File("foo.ppm","wb"));
}();
The funkyness (pasteArgs) in the recursion (self.fcn) is due to the closure ('wrap): the closed over args are stashed in the arglist, they need to be added to the parameters when recursing.
## ZX Spectrum Basic
Translation of: BASIC256
10 LET level=12: LET long=45
20 LET x=127: LET y=0
30 LET rotation=PI/2
40 LET a1=PI/9: LET a2=PI/9
50 LET c1=0.75: LET c2=0.75
60 DIM x(level): DIM y(level)
70 BORDER 0: PAPER 0: INK 4: CLS
80 GO SUB 100
90 STOP
100 REM Tree
110 LET x(level)=x: LET y(level)=y
120 GO SUB 1000
130 IF level=1 THEN GO TO 240
140 LET level=level-1
150 LET long=long*c1
160 LET rotation=rotation-a1
170 GO SUB 100
180 LET long=long/c1*c2
190 LET rotation=rotation+a1+a2
200 GO SUB 100
210 LET rotation=rotation-a2
220 LET long=long/c2
230 LET level=level+1
240 LET x=x(level): LET y=y(level)
250 RETURN
1000 REM Draw
1010 LET yn=-SIN rotation*long+y
1020 LET xn=COS rotation*long+x
1030 PLOT x,y: DRAW xn-x,y-yn
1040 LET x=xn: LET y=yn
1050 RETURN | 38,420 | 144,599 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-10 | latest | en | 0.511473 |
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# LRO Educator Resource Kit
Each lesson or activity in this toolkit is related to NASA's Lunar Reconnaissance Orbiter (LRO). The toolkit is designed so that each lesson can be done independently, or combined and taught in a sequence. The Teacher Implementation Guide provides... (View More)
# Heavyweight Champion: Jupiter
Learners will weigh themselves on scales modified to represent their weights on other worlds to explore the concept of gravity and its relationship to weight. They consider how their weights would be the highest of all the planets while standing on... (View More)
Keywords: Jupiter; Mass; Weight
# What Is Energy?
In this introductory textbook chapter, energy is defined and its sources, forms, and flow are introduced. Students are introduced to the first law of thermodynamics, conservation of energy. The resource includes links to current news articles, and a... (View More)
Audience: High school
Materials Cost: Free
# Enceladus, I Barely Knew You
This is a lesson that applys occultations to Saturn's Moon Enceladus. Learners will establish whether Saturn’s small moon, Enceladus, has an atmosphere, whether that atmosphere is over the entire planet, and what creates Saturn’s E-ring. The... (View More)
# Designing a Spectroscopy Mission
This is a math-science integrated unit about spectrographs. Learners will find and calculate the angle that light is transmitted through a holographic diffraction grating using trigonometry. After finding this angle, the students will build their... (View More)
Audience: High school
# Marvelous Martian Mineralogy
This is a lesson about determining planetary composition. Learners will use a reflectometer to determine which minerals are present (from a set of knowns) in a sample of Mars soil simulant. Requires the use of ALTA II spectrometers (which may be... (View More)
# Signal-to-Noise Ratio (New Horizons To Pluto and Beyond)
This is an activity about the Signal-to-Noise Ratio. Learners will engage with a hands-on activity and an online interactive to understand the terms signal and noise as they relate to spacecraft communication; quantify noise using a given dataset;... (View More)
Audience: Middle school
Materials Cost: Free
# Using Spectral Data to Explore Saturn and Titan
Learners will compare known elemental spectra with spectra of Titan and Saturn’s rings from a spectrometer aboard the NASA Cassini spacecraft. They identify the elements visible in the planetary and lunar spectra. The activity is part of Project... (View More)
# A Spectral Mystery
Learners will use a spectrograph to gather data about light sources. Using the data they’ve collected, students are able to make comparisons between different light sources and make conjectures about the composition of a mystery light source. The... (View More)
# Goldilocks and the Three Planets
Learners will interpret spectral graphs to determine the atmospheric composition of Earth, Venus, and Mars, and then mathematically compare the amount of the greenhouse gas, CO2, on these planets. Students will brainstorm what things, along with... (View More)
«Previous Page123 Next Page» | 752 | 3,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-22 | latest | en | 0.87305 |
https://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-the-number-of-weekdays-in-a-month.aspx | 1,540,118,134,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00358.warc.gz | 824,945,321 | 6,186 | # Thinking outside the box
Patron Saint of Lost Yaks
## How to calculate the number of weekdays in a month
Here http://weblogs.sqlteam.com/peterl/archive/2009/07/12/How-to-calculate-the-number-of-weekdays-for-any-given.aspx you can find a specialized version that calculates number of weekdays for any given period.
CREATE FUNCTION dbo.fnMonthWeekDays
(
@Year SMALLINT,
@Month TINYINT
)
RETURNS TINYINT
AS
BEGIN
RETURN (
SELECT 20 + SUM(1 - DATEDIFF(DAY, '17530101', dt) % 7 / 5)
FROM (
SELECT DATEADD(MONTH, 12 * @Year - 22800 + @Month, -1) AS dt
WHERE ISDATE(10000 * @Year + 100 * @Month + 31) = 1
UNION ALL
SELECT DATEADD(MONTH, 12 * @Year - 22800 + @Month, -2)
WHERE ISDATE(10000 * @Year + 100 * @Month + 30) = 1
UNION ALL
SELECT CASE ISDATE(10000 * @Year + 100 * @Month + 29)
WHEN 1 THEN DATEADD(MONTH, 12 * @Year - 22800 + @Month, -3)
ELSE '18991231'
END
) AS d
WHERE @Year BETWEEN 1753 AND 9999
AND @Month BETWEEN 1 AND 12
)
END
Print | posted on Sunday, July 12, 2009 1:50 PM | Filed Under [ SQL Server 2008 Algorithms SQL Server 2005 SQL Server 2000 ]
## #re: How to calculate the number of weekdays in a month
Can u please explain the logic ?
10/22/2009 3:12 PM | Sowmya
## #re: How to calculate the number of weekdays in a month
The first 28 days of any month always has 20 weekdays, no matter what day of week the month starts with.
So, you you only have to check day 29, 30, and 31 for weekday.
10/23/2009 9:41 AM | Peso
Comments have been closed on this topic. | 497 | 1,496 | {"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} | 3.5625 | 4 | CC-MAIN-2018-43 | latest | en | 0.656106 |
https://currentschoolnews.com/us/how-to-open-a-locker/ | 1,701,577,138,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00828.warc.gz | 227,939,557 | 18,927 | # How to Open a Locker (Checkout these Easy Steps)
To some people, how to open a locker might seem weird, that’s because they feel it is easy to do so. But do you know there are different locks in different lockers? Yes! You may not know how to open some complicated ones, but trust me, this article will give you a headlight through easy steps.
## How to Open a Locker
Most combination locks will have three digits, ranging from 0 to 39. If your lock is integrated into the locker.
However, these numbers may be higher, sometimes even reaching 99.
When opening a locker, note that each number must be turned and positioned such that it lines up with the indicator arrow on top of the lock.
Follow these steps to open a locker.
#### Step 1: Releasing the Lock
‣ First, release the lock.
‣ Turn the dial on the locker lock three times to the right.
‣ You must clear the combination again if you believe you entered it incorrectly before attempting again.
#### Step 2: Entering the Combination
‣ Right, left, right, enter the combination for your locker.
‣ It’s time to input your combination now that the lock has been opened.
‣ Remember to turn “right-left-right” as a fast reminder!
First Number: After opening the lock, turn the dial counterclockwise (to the right) until you reach your first number
Second Number: Now turn the dial in the opposite direction (to the left), stopping at your second number.
Third Number: Turn the dial in a clockwise direction (to the right) and stop at the last combo number.
#### Step 3: Unlock the Lock
‣ Pull down on the lock body after entering your last digit to unlock the lock.
‣ You can now open the lock by opening the door if the lock is a built-in feature of the locker.
### How Can You Pick Lock a School Locker With a Paperclip?
To do this, you’ll need:
‣ A pair of metal paper clips
‣ A pair of pliers
Follow these steps to pick a lock with a paper clip
‣ Insert a tension wrench into the keyhole’s bottom.
The lock should be turned in the same direction as the wrench.
‣ Insert your pick into the keyhole’s upper portion and its rear, then swiftly withdraw it while jiggling the pick upward.
Maintain tension on the tension wrench while you’re doing this. The lock picking won’t function properly if you don’t.
‣ Inside the lock, find the pins. The majority of American locks feature at least five pins that must be inserted in order to unlock the lock.
When you insert your lock pick, you will feel the pins against it. You will know where to depress them based on this.
‣ Push the pins down. As you press the pins down, make sure to rotate your tension wrench.
When you move the pins to their unlocked position, you should feel a slight give or possibly hear a tiny clicking sound.
‣ Move picks around until each pin releases. Jiggle the pick while increasing the tension with the tension wrench until each pin releases.
Make sure to turn the tension wrench to unlock the lock when you hear a snap or click.
#### Does a School Have a Right to Search School Lockers?
In general, a search may be conducted without notifying the kid or the parent of any object that is thought to be school property.
Even though you use it, the school administration might authorize the police to search your locker or desk.
Any such policies that schools may have must be made available to students upon request.
#### Can a School Empty Out Your Locker and Throw Everything Away?
Schools have the right to empty lockers and dispose of their contents if a student does not retrieve their possessions after class concludes.
#### Is it True that Teachers Keep a Spare Key for School Lockers?
The locker system is run by the assistant principal and head custodian. They would be in possession of an extra key.
Some systems allow students to purchase combination locks.
The assistant principal must see them use the combination (or teacher, who will pass the code on to the AP).
That is useful if the AP has to access a student’s locker or if a student consistently forgets their code.
The lock-cutters used to open any lock for which they don’t know the combination are the system’s “spare key.”
The administrator or the principal unquestionably holds the master key if there are lockers with turnkeys.
#### Did People Always Put Locks on their School Lockers?
Given that they are termed “lockers,” which require a lock in order to function as intended, people have always done this.
Locks are usually already installed on the lockers at some high schools where you kept your books, supplies, etc.
If you’re referring to lockers in the gym, whether or not locks were used depends on the person or what they brought to the class.
#### Why Do High School Lockers Have Vents?
The majority of lockers, not only those in high schools, have vents. These vents are designed to let exercise gear and shoes dry out.
These vents allow fresh air to flow into the locker, allowing items like sweaty clothing to dry off and preventing unwanted odors from developing.
#### Why Do Most High Schools Have Lockers?
The locker is essentially a storage space where students may keep their books and other belongings so they don’t have to carry bulky bags to school every day.
And yet, lockers are rapidly fading from use in many schools.
#### Why Don’t Schools in Britain Have Lockers?
Probably because they do not need them.
Since bags are not permitted in classrooms, American schools frequently use lockers so that students have a place to store all of their belongings.
In Britain, you just carry your bag from class to class, so there isn’t a need for a place to store things. In the UK, it isn’t a thing.
They also don’t have a lot of time between classes for getting things out of your locker because you are expected to travel directly from one class to the next.
As a result, you shouldn’t be using a locker on the opposite side of the building. Additionally, the cost of space is also significantly higher in the UK.
The property costs a fortune to purchase, and the building itself is nearly twice as expensive as it is in the US.
That makes schools smaller; narrower hallways are one of the simplest ways to conserve space.
That indicates that there just isn’t enough room to provide lockers for each student.
If they do, they will usually be saved for the seniors who have more load to carry.
We hope you now understand how to open a locker by yourself. If this article was helpful to you, you can share it with others who need it.
CSN Team. | 1,402 | 6,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-50 | latest | en | 0.921073 |
https://www.daniellitt.com/blog/2017/11/2/a-non-prorepresentable-deformation-functor | 1,511,489,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807056.67/warc/CC-MAIN-20171124012912-20171124032912-00383.warc.gz | 789,344,031 | 16,486 | # A non-prorepresentable deformation functor
I've been teaching a topics course on deformation theory this semester; so far we've covered quite a few interesting things -- the Tian-Todorov theorem and Green-Lazarsfeld's generic vanishing theorem, for example. I just wanted to use this post to record an easy and fun example that I wasn't able to find in the literature.
Question: What is a good example of a smooth projective variety $$X$$ over a field $$k$$ such that the deformation function $$\text{Def}_X$$ is not pro-representable? Here $$\text{Def}_X: \text{Art}/k\to \text{Sets}$$ is the functor sending a local Artin $$k$$-algebra $$A$$, with residue field $$k$$, to the set of isomorphism classes of flat $$A$$-schemes $$Y$$ equipped with an isomorphism $$\phi: Y_k\overset{\sim}{\to} X$$.
Answer: Let $$X=\text{Bl}_Z(\mathbb{P}^2)$$, where $$Z$$ is a set of $$10$$ points lying on a line.
It's not totally obvious, of course, that this works. The idea of the proof follows. Given a variety $$X$$, we let $$\text{DefAut}_X: \text{Art}/k\to \text{Sets}$$ be the deformation functor sending a local Artin $$k$$-algebra $$A$$, with residue field $$k$$, to the set of isomorphism classes of triples $$Y, \psi, \phi$$, where $$Y$$ is a flat $$A$$-scheme, $$\phi: Y_k\overset{\sim}{\to}X$$ is an isomorphism of the special fiber of $$Y$$ with $$X$$, and $$\psi: Y\overset{\sim}{\to} Y$$ is an automorphism of $$Y$$ which is the identity on $$Y_k$$.
Proposition. Let $$X$$ be a proper variety. The functor $$\text{Def}_X$$ is pro-representable if and only if the natural map $$\text{DefAut}_X\to \text{Def}_X$$ is formally smooth. (Here the natural map simply forgets the automorphism $$\psi$$.)
This is somewhat standard; it's an easy application of Schlessinger's criteria, for example. One way of saying this is that given a small extension $$0\to I\to B\to A\to 0,$$ and a deformation $$Y$$ of $$X$$ to $$B$$, there's a long exact sequence of sets $$0\to H^0(X, T_X)\otimes I\to \text{Aut}^0(Y)\to \text{Aut}^0(Y_A)\to H^1(X, T_X)\otimes I\to \text{Def}_X(B)\to \text{Def}_X(A)\to H^2(X, T_X)\otimes I.$$
Here $$\text{Aut}^0(Y)$$ is the set of automorphisms of $$Y$$ fixing $$Y_k$$. The fact that this is a long exact sequence means that the first four non-zero terms are an exact sequence of groups; that $$H^1(X, T_X)\otimes I$$ acts transitively on the fibers of the map $$\text{Def}_X(B)\to \text{Def}_X(A)$$, with kernel exactly the image of $$\text{Aut}^0(Y_A)$$; and that an element of $$\text{Def}_X(A)$$ lifts to an element of $$\text{Def}_X(B)$$ if and only if it maps to zero in $$H^2(X, T_X)\otimes I$$. Now formal smoothness of the map $$\text{DefAut}_X\to \text{Def}_X$$ is the same as surjectivity of the map $$\text{Aut}^0(Y)\to \text{Aut}^0(Y_A)$$ for all small extensions and all choices of $$Y$$. This is the same as faithfulness of the action of $$H^1(X, T_X)\otimes I$$ on the fibers of the map $$\text{Def}_X(B)\to \text{Def}_X(A)$$, which is a restatement of Schlessinger's pro-representability criterion in our setting.
So why does our example $$X$$ -- namely, $$\mathbb{P}^2$$ blown up at a bunch of points all lying on a line $$\ell$$ -- have non-prorepresentable deformation functor? The point is that infinitesimal automorphisms of $$X$$ are precisely infinitesimal automorphisms of $$\mathbb{P}^2$$ fixing the line $$\ell$$. But if one moves the points to general position, the infinitesimal automorphism group becomes trivial. So the map $$\text{DefAut}_X\to \text{Def}_X$$ is far from smooth. Pretty easy and cool.
Is there a reference with examples like these? I'm sure they're well-known, but I wasn't able to find any smooth proper examples in the literature. | 1,151 | 3,726 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2017-47 | longest | en | 0.791935 |
https://1-texasholdempoker.com/lottery/you-asked-has-anyone-won-the-lottery-using-the-lost-numbers.html | 1,643,439,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00044.warc.gz | 130,242,928 | 19,450 | # You asked: Has anyone won the lottery using the lost numbers?
Contents
## Has anyone won the lottery using the numbers from Lost?
Take for instance the 9,078 Lost fans who used the show’s mysterious sequence of figures (4, 8, 15, 16, 23 and 42) for lottery entries in New York’s recent Mega Millions draw. …
## How many times have the numbers from Lost won the lottery?
Sure – if you’re not stuck on the ‘Lost’ island. The winning numbers are 4, 8, 15, 25, 47. and the Mega Ball number is 42. Some of these numbers were used by Hugo ‘Hurley’ Reyes in the ABC drama, Lost in winning his jackpot, but as Losties know, the numbers control much, much more than just money.
## How many people use the Lost numbers?
Of course, there’s no way to tell just how many of those people were playing all of Hurley’s numbers. But “Lost” executive producer Damon Lindelof celebrated by tweeting his own figures. “9,078 people played Hurley’s numbers in the MegaMillions tonight, each winning \$150,” he wrote, “#THATSSORAVEN.”
## Is it possible to predict winning lottery numbers?
Keep in mind that every number in the lottery has an equal probability of being chosen as the winning number. There is no software that could predict the exact winning number. But you can increase your chance of not splitting the jackpot price if you use the not so common number.
## Did Hurley really win the lottery?
Flashbacks. In September 2004, Hurley wins the lottery, and over the ensuing weeks, everyone around him seems to suffer increasingly bad luck.
## How did Hurley win the lottery?
On the show, fan favorite Hurley won the lottery by using the numbers 4 8 15 16 23 42. The numbers tortured him, though, showing up everywhere on the island. … The winning numbers for the Tuesday night Mega Millions lottery, with its gigantic jackpot, were: 4 8 15 25 47 42.
## Why are the numbers cursed Lost?
The numbers 4, 8, 15, 16, 23 and 42 frequently recurred in Lost. Each corresponded with one of the final candidates to replace Jacob as protector of the Island. The numbers also formed the coefficients in an equation that predicted mankind’s extinction.
## Where did Hurley get the numbers?
The series of numbers first appeared in the show on the winning lotto ticket owned by Hurley (Jorge Garcia), though he was convinced that the numbers were cursed and brought him bad luck wherever he went.
## Is it worth to watch Lost?
The series is never as incredible as it was in the first season, and there are chunks of the second and third seasons that could be cut out without losing anything, in terms of the overall plot and character arcs. But, overall, it is definitely worth watching; in fact, I’d recommend watching it twice.
IT IS SURPRISING: When did Wesley Snipes say Always bet on black?
## When did Lost come out?
Lost is an American drama television series that originally aired on ABC from September 22, 2004, to May 23, 2010, over six seasons, comprising a total of 121 episodes.
## What is in the hatch in Lost?
Jack and Locke discover that the hatch is a 1970’s station built to control an electromagnetic anomaly on the island and learn that it was built by the Dharma Initiative, a scientific research team. Michael, Sawyer, and Jin find out that they are being held captive by the survivors of Flight 815’s tail section.
## Which lottery is easiest to win?
What is the easiest lottery to win the jackpot on? Believe it or not, the easiest to win is the Polish Mini Lotto. The odds of winning the jackpot prize is a cool 850,668 to one.
## Can you give family money if you win the lottery?
Each person can give away, during life or at death, a certain amount of property before the tax kicks in. … So by claiming the lottery winnings as a family partnership, a winner can claim that they are not making a taxable gift, because it was a family investment. This could save millions in gift taxes.
## What numbers are most likely to win the lottery?
The Most Common Primary Numbers
The most common primary Powerball numbers are 10, 42, 39, 28, 22, 23, 32, 16, 41, and 26. | 957 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-05 | latest | en | 0.954451 |
https://anhngq.wordpress.com/2019/04/ | 1,653,007,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00102.warc.gz | 145,314,594 | 25,311 | # Ngô Quốc Anh
## April 14, 2019
### Extending functions between metric spaces: Continuity, uniform continuity, and uniform equicontinuity
Filed under: Giải Tích 3, Giải tích 8 (MA5206) — Tags: — Ngô Quốc Anh @ 15:02
This topic concerns a very classical question: extend of a function $f : X \to Y$ between two metric spaces to obtain a new function $\widetilde f : \overline X \to Y$ enjoying certain properties. I am interested in the following three properties:
• Continuity,
• Uniformly continuity,
• Pointwise equi-continuity, and
• Uniformly equi-continuity.
Throughout this topic, by $X$ and $Y$ we mean metric spaces with metrics $d_X$ and $d_Y$ respectively.
CONTINUITY IS NOT ENOUGH. Let us consider the first situation where the given function $f : X \to Y$ is only assumed to be continuous. In this scenario, there is no hope that we can extend such a continuous function $f$ to obtain a new continuous function $\widetilde f : \overline X \to Y$. The following counter-example demonstrates this:
Let $X = [0,\frac 12 ) \cup (\frac 12, 1]$ and let $f$ be any continuous function on $X$ such that there is a positive gap between $f(\frac 12+)$ and $f(\frac12-)$. For example, we can choose
$\displaystyle f(x)=\begin{cases}x^2&\text{ if } x<\frac 12,\\x^3 & \text{ if } x>\frac 12.\end{cases}$
Since $f$ is monotone increasing, we clearly have
$\displaystyle f(\frac12-)-f(\frac 12+)=\frac18.$
Hence any extension $\widetilde f$ of $f$ cannot be continuous because $\widetilde f$ will be discontinuous at $x =\frac 12$. Thus, we have just shown that continuity is not enough. For this reason, we require $f$ to be uniformly continuous. | 473 | 1,658 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-21 | latest | en | 0.783718 |
https://puzzling.stackexchange.com/questions/9000/an-abundance-of-one-line-rebus-puzzles | 1,708,970,693,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00199.warc.gz | 479,299,246 | 42,021 | # An abundance of one-line Rebus puzzles
Solve the following ten one-line Rebus puzzles:
1. AGE BUT
2. marri Ag e
3. wWoOlOfL
4. "24 Hours"
5. TRADI$~$ TION
7. WHEATHER
8. SGEG
9. TOSYOBSYOBWN
10. ALL world
• Awesome puzzle! Keep making more rebus puzzles like these. Love it! Feb 16, 2015 at 8:28
I am going to combine all answers so that it is easy for everyone to see. All credit goes to Hellion, Gamow and Vikram.
1. AGE BUT
Age before beauty (B-U-T)
2. marri Ag e
Ag=silver with marriage = Silver wedding anniversary
3. wWoOlOfL
Wolf in WOOL = Wolf in sheep's clothing
4. "24 Hours"
Multiple possibilities:
Call it a day
Quote of the day
7. WHEATHER
8. SGEG
Permutation of EGGS = Scrambled eggs
9. TOSYOBSYOBWN
BOYS BOYS backwards, inside TOWN = The Boys are Back in Town
10. ALL world
It's a small world after all!
• Hello! As this is a compilation of existing answers, I've converted it to Community Wiki. CW posts are intended to be collaborated on by all users of the site, and don't earn their owners reputation. Thanks for posting this!
– user20
Feb 17, 2015 at 5:40
1:
Age before Beauty ("B, U, T")
7: (just a guess here)
10:
It's a small world after all.
• 1 and 10: great solutions! For 7, I think it should rather be "spell of bad weather". Feb 15, 2015 at 16:22
• @Gamow, I think you've got it. Changed my answer to suit. (with one slight alteration.) Feb 15, 2015 at 17:01
The answers for 2, 3, 8, and 9:
2:
"Ag=silver with marriage" = "Silver wedding anniversary"
3:
"wolf in WOOL" = "Wolf in sheep's clothing"
8:
"SGEG = permutation of EGGS" = "Scrambled eggs"
9:
"SYOBSYOB = BOYS BOYS backwards, inside TOWN" = "The Boys are Back in Town"
and the answer for 7 in collaboration with Hellion
7:
5:
2>
uneven marriage
4>
All day packed
5>
6> | 572 | 1,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-10 | latest | en | 0.90103 |
https://www.toktol.com/notes/context/1765/physics/gravitational-potential-and-gravitational-potential-energy/gravitational-potential-and-gravitational-force | 1,513,450,081,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588420.68/warc/CC-MAIN-20171216181940-20171216203940-00771.warc.gz | 838,945,325 | 13,247 | Use adaptive quiz-based learning to study this topic faster and more effectively.
# Gravitational potential and gravitational force
The gravitational force is the negative of the gradient (or the derivative) of the gravitational potential energy.
The gradient tells you by how much the gravitational potential energy changes per unit of distance that you move towards or away from the gravitational force.
The proof for the equation for the potential around a unit mass $M$ is as follows:
The potential is the work done by a gravitational field on a unit mass in bringing the mass from infinity to its current position $R$.
If the unit mass is at distance $S$ from the planet (the source of the gravitational field), the work done to move it infinitesimally from $S$ to $S+dS$ is $F(S) dS = -GM/S^{2}$ (the force is opposite to the movement).
Integrating between the current position $R$ to infinity and using the fact that the integral of $S^{-2}$ is $-S^{-1}$, we get:$$V =\int_R^\infty F(S) dS= -\int_R^\infty \frac{GM}{S^2}\, dS = \left[\frac{GM}{S}\right]_R^\infty = -\frac{GM}{R}$$ | 279 | 1,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2017-51 | latest | en | 0.884953 |
http://www.solutioninn.com/projects/crtique_analysis | 1,495,676,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607960.64/warc/CC-MAIN-20170525010046-20170525030046-00055.warc.gz | 665,383,851 | 7,942 | # crtique / analysis
## Project Description:
i've already completed the required research at attachemnt, however i'm having difficulity with answering the required word portion of part 1 part ii. i think the dupont analysis would be 14-18 using the values from 3-7. the analysis values are tabulated using the formulas in the pdf. the common size analysis is 22-26 using values from 3,9-12. the analysis values are obtained by dividing each value from 3,9-12 by the total revenue value in line 7 and then multiplying times 100 to get a %. but again, i'm sure how to apply this. please see the below and attached. unit ii problem solving part 1 for this task you will conduct a comparative dupont analysis of two companies. using a search engine, find one large corporation included in the s&p 500. then, find one of its largest competitors. go to the investor relations portion of each corporation's homepage and find their most recent annual report. calculate a complete dupont analysis calculating the roe, roa, the profit margin, total asset turnover and equity multiplier. critique the differences between the two corporations in approximately 100 words. part 2 using the most recent income statements (annual) for the two corporations from part 1 of the task, calculate a common size analysis using a spreadsheet. discuss the differences in the two corporations in approximately 75 words. your answer can be completed below your spreadsheet analysis. submit part 1 and part 2 together in one document.
supporting files attached:
Skills Required:
Project Stats:
Price Type: Negotiable
Expired
Total Proposals: 2
1 Current viewersl
28 Total views
Project posted by:
### Proposals
Proposals Reputation Price offered | 371 | 1,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-22 | longest | en | 0.909101 |
http://mathhelpforum.com/calculus/103709-general-flux-vector-print.html | 1,508,481,121,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823731.36/warc/CC-MAIN-20171020044747-20171020064747-00852.warc.gz | 213,725,474 | 3,253 | # General Flux Vector
• Sep 22nd 2009, 06:15 AM
Maccaman
General Flux Vector
Let $\rho = \rho_0 + \rho_1 \sin(\omega t)$ be the density of oxygen in a spherical shell with inner radius a and outer radius b.
Find the total amount of oxygen, Q(t) in this shell for any time t.
Find the most general flux vector of the form
$\vec{J}(x,y,x,t) = f(r,t)(\vec{r}) \ \mbox{where} \ r = \sqrt{x^2 +y^2 + z^2}$ and $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$ such that
$\frac{\delta \rho}{\delta t} + \ \mbox{div} \ \vec{J} = 0$
If at any time and anywhere on the the outer surface of the shell $(r=b)$ the flux vector satisfies $\vec{J}$. $\vec{r} = \frac{\omega}{3b} \sin (\omega t)$ then
$f(r,t) = \frac{\omega}{3r^3} (\rho_1 (b^3 - r^3) \cos (\omega t) + \sin (\omega t))$
Show that the total flux into the spherical shell per unit time is equal to
$\frac{d Q(t)}{dt}$
• Oct 3rd 2009, 11:14 PM
Maccaman
Quote:
Originally Posted by Maccaman
Let $\rho = \rho_0 + \rho_1 \sin(\omega t)$ be the density of oxygen in a spherical shell with inner radius a and outer radius b.
Find the total amount of oxygen, Q(t) in this shell for any time t.
[/tex]
This could be wrong but here is what I have got:
$Q(t) = \iiint\limits_V \ \rho \ dv = \rho_0 + \rho_1 \sin (\omega t) \iiint\limits_V \ \rho \ dv = (\rho_0 + \rho_1 \sin (\omega t)) \frac{4}{3}\pi (b-a)^3$
Is that correct? If not, could somebody please show me where I went wrong?
Thanks
• Oct 4th 2009, 06:55 AM
Jester
Quote:
Originally Posted by Maccaman
This could be wrong but here is what I have got:
$Q(t) = \iiint\limits_V \ \rho \ dv = \rho_0 + \rho_1 \sin (\omega t) \iiint\limits_V \ \rho \ dv = (\rho_0 + \rho_1 \sin (\omega t)) \frac{4}{3}\pi (b-a)^3$
Is that correct? If not, could somebody please show me where I went wrong?
Thanks
I think it should be
$Q(t) = \left( \rho_0 + \rho_1 \sin (\omega t)\right) \frac{4}{3}\pi (b^3-a^3)$ | 681 | 1,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-43 | longest | en | 0.681872 |
https://community.tableau.com/thread/268851 | 1,547,940,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583684033.26/warc/CC-MAIN-20190119221320-20190120003320-00502.warc.gz | 474,250,152 | 26,366 | 2 Replies Latest reply on May 8, 2018 11:50 AM by Javier Gironella
Count+1 previous step based on Status
I am pretty sure there is a simple way to solve this, but I can't seem to find it.
I have a workbook with a column that indicates which step is each one of our projects in, so we can keep track of them.
They are asking me to build a visualization in which they want to assume that if a project has reached "Step 10", it also counts as that project having gone through Steps 9, 8, 7, 6 ,5 ,4, etc.; so a project in Step 10 must also add +1 to the count of projects in Step 9, 8, 7, 6, 5, 4 etc.
My data is not structured that way, it only presents the furthest step that the project has reached, so I need to find some workaround for this.
I have tried using calculated fields with =COUNT(IF()) but when structuring them visually I do not get the expected result.
Does anyone have any ideas how to solve this?
• 1. Re: Count+1 previous step based on Status
HI Javier
This might work?
Thanks,
Shin
1 of 1 people found this helpful
• 2. Re: Count+1 previous step based on Status
Hi Shinichiro,
Your formula is exactly what I was looking for, thank you very much!
I would just like to clarify a few points in case anyone stumbles upon this issue and finds this answer:
1. Because this is a WINDOW_SUM, for the calculation to work correctly, the steps must be precisely ordered in the visualization from Step 1 to Step N.
2. If you need to calculate percentage of projects starting from base 100% (Step 1 is 100% of total projects and decreases every step), you can modify Shinichiro's calculation to the following:
WINDOW_SUM(COUNTD([Id]),0,LAST())/WINDOW_SUM(COUNTD([Id]),FIRST(),LAST())
3. If you need to order your steps in rows, from top to bottom (as in Shinichiro's example, Step 1 is the top row and Step N is bottom row), you need to compute using Table (down). If you presented your steps as columns (Step 1 is the first column, Step N is the last column), compute using Table (across).
I hope anyone who sees this finds Shinichiro Murakami's answer as valuable as I did, thanks for your help again! | 538 | 2,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.950909 |
https://hncppf.com/baccarat-gambling-part-1-a-fun-simple-game-you-can-win-easily/ | 1,717,026,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00597.warc.gz | 249,925,544 | 21,961 | Wager choice is defined as the process of selecting the winning decision (player or banker) in baccarat or (pass or do not pass) in craps. There are several approaches preferred by skilled players. Some like to follow the shoe in baccarat or the roll in craps. By this we indicate, the player wagers the like the coming before decision. One more preferred technique of selecting which side to bet on is the decision before last. Pattern recognition entriques lots of players as they look for engaging patterns. Streak play has actually sponded several systems in the baccarat video game.
Baccarat appears to hold out the most wish to the mathematically inclined. Given that the game is played out of a shoe that is generally not reshuffled until a baccarat game has actually been concluded. The mathematics types declare there is reliance in the baccarat video game. Reliance implies as the very first fifty percent of the baccarat footwear is played it needed establishes the composition of the second half of the baccarat shoe. For that reason, we need to have the ability to establish some edge over your home in the second component of the shoe since we currently recognize which cards have actually been played in the very first fifty percent of the shoe (obviously we would certainly have to keep track somehow to understand this). Card checking spearheaded and also made use of successfully in blackjack was tried in baccarat. The outcome was a little favorable however very frustrating as the player benefit was very really slight. My advise on baccarat card counting; do not squander your time, it will not work.
Various other methods of removing the “dependency” from the baccarat video game have been explored.
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Some players take a special method based upon the continuity of numbers in an unfavorable expectation video game like the baccarat video game or craps video game. The continuity of numbers concept behind this approach verifies out whenever we analyze the data on the major structure computer. This is not to state the idea will certainly provide you the foolproof gaming system. Below is the keynote. Allow’s claim we mosted likely to the casino site with a team of scientists and we tape-recorded every roll of the dice at the craps table 24 hours per day 7 days a week for 6 months. That would amount to countless rolls of the dice. Simply for discussion, allow’s claim 6 months of continual recording stands for 650,000 rolls or decisions. Currently, allow’s claim a private scientist lives to be 102 years of ages and for 80 years of his life he mosted likely to the gambling enterprise on a daily basis of his life. Daily he tape-recorded 10 mins of craps rolls at the craps table (not also at the very same table everyday!). Let’s claim for our little discussion, that the 80 years of 10 minutes a day created the amount of decisions, the very same 650,000 as the team of 24/7 scientists did in 6 months.
Let us take both sets of data and also become part of our computer system. The computer will certainly analyse the information against the mathematical design of the video game. We’ll evaluate the pattern circulation, the incident of numbers and also all of the percents. We’ll compute the casino site’s advantage over the player (i.e. – pass line 1.42%, do not pass 1.40%, and so on). The computer churns for a couple of minutes and spits out the two reports; Record 1- the “24/7 for 6 months information” and also Record 2- the “10 minute per day for 80 years information”. What do you think the differnce will be? ABSOLUTELY NOTHING !! That’s right, the stats as well as numbers will certainly coincide.
Well, what can we gain from this? As gamers we play just one huge craps game and one large baccarat video game (if those are your video games) in our life. You form a continuity of numbers that over the lifetime of your play will always attempt to adapt the morphology of the game or the mathematical distribution. At any kind of given time different aspects (patterns as well as numbers) will get on a variance from their typical mathematical circulation in the game. As an example, let us claim in our lifetime game we have actually played 1,000,000 hands of baccarat. We know since we tape-recorded everybody of those suckers on scorecards and also entered them into our computer system after every play session. [the pros call this treatment “cataloging their game”]
Every night after entering his data, the player cranks up his trusty computer analysis program. A report is derived; he finds out for instance, that patterns of seven bank decisions in row are 6 basic inconsistencies behind in incident in “his game”. The professional gamer has a desirable approach for playing the inconsistencies in his game. He understands the discrepancy IS mosting likely to come back to equipartition at some point. IT ALWAYS DOES, eventually! Don’t go to the online casino yet. The big question is the length of time lacking will the pattern stay before beginning to find back into the regular circulation version of the video game?
Baccarat Gambling – Part 1; A Fun, Simple Game You Can Win Easily! | 1,268 | 5,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.981004 |
http://www.algebra.com/algebra/homework/percentage/percentage.faq.question.226128.html | 1,368,906,765,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382764/warc/CC-MAIN-20130516092622-00055-ip-10-60-113-184.ec2.internal.warc.gz | 318,788,047 | 4,429 | # SOLUTION: a rake is priced \$9.00 it is on sale for 30% off. how much will you save?
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Algebra: Percentage and Pie Charts Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on percentage Question 226128: a rake is priced \$9.00 it is on sale for 30% off. how much will you save?Answer by Isabellaksmith(3) (Show Source): You can put this solution on YOUR website!\$2.70 \$9.00 times .30 ( 30 % changed to a decimal ) | 217 | 817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2013-20 | latest | en | 0.883513 |
https://www.compadre.org/stpbook/statistical-mechanics-1/ | 1,701,956,368,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100674.56/warc/CC-MAIN-20231207121942-20231207151942-00646.warc.gz | 805,256,729 | 4,589 | # Statistical and Thermal Physics 2E: Statistical Mechanics I
## Chapter 4: Methodology
We combine ideas from thermodynamics and probability theory to introduce statistical mechanics, which is the study of how microscopic behavior leads to macroscopic behavior in systems of many constituents. There only are a few models, such as the ideal gas whose macroscopic averages can be calculated analytically starting from the microscopic model.
In this chapter we use the computer to illustrate the counting of microstates and the behavior of some many particle systems.
## Chapter 5: Magnetic Systems
We apply the tools of statistical mechanics to magnetic systems. The most important and simplest model containing interactions, is the Ising model defined by the energy of a microstate $$E = -J \sum_{\lt ij \gt} s_i s_j - H \sum_i s_i,$$ where $J$ is the energy of interaction and is positive for a ferromagnetic interaction, $s_i = \pm 1$ is the spin of a spin 1/2 particle such as an electron, $\lt ij \gt$ denotes nearest-neighbor pairs of spins located on a lattice, and $H$ is the applied external magnetic field.
In addition to studying the Ising model and its variations, we consider various techniques such as transfer matrices, mean-field theories, and computer simulations, which have become essential for our understanding of magnetic systems.
## Chapter 6: Many-Particle Systems
We apply the techniques of statistical mechanics to many particle systems. Our focus will be on systems where we can treat the particles as noninteracting. The first half of the chapter discusses classical and semiclassical systems, and the second half discusses systems where quantum mechanics is essential.
The key results of classical systems are the Maxwell velocity and speed distributions and the equipartition theorem, which are valid for gases, liquids, and solids.
Quantum systems that are discussed include blackbody radiation (considered as a collection of photons), the Debye model of solids (considering quantized sound waves or normal modes as phonons), Fermi systems such as electrons in metals, neutron stars, and white dwarfs, and Bose gases which can condense into a single ground state at low enough temperatures. To study these systems we use the grand canonical ensemble to treat the indistinguishability of the particles, and introduce the concepts of single particle density of states and the mean occupation of each single particle state (distribution function) at a temperature $T$ and chemical potential $\mu$.
Overview TOC
OSP Projects: | 517 | 2,563 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.109375 | 3 | CC-MAIN-2023-50 | latest | en | 0.932629 |
https://codedump.io/share/FBpg615m48ls/1/int-and-char-scala-interpolation-generics-and-upper-bounds | 1,480,965,384,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541783.81/warc/CC-MAIN-20161202170901-00244-ip-10-31-129-80.ec2.internal.warc.gz | 832,398,156 | 9,226 | jalv1039 - 24 days ago 8
Scala Question
# Int and Char scala interpolation (generics and upper bounds)
I am playing with generics and upper bounds in scala and I am facing the following problem:
Let's say I have the following function (where I only expect either Int or Char):
``````def foo[T](tuple: Tuple2[T, T]) = tuple match {
case Tuple2(x: Int, y: Int) => (x to y).toArray
case Tuple2(x: Char, y: Char) => (x to y).toArray
}
``````
I would expect to have a better and compressed form like:
``````def foo2[T >: Int with Char <: AnyVal](tuple: (T, T)) = (tuple._1 to tuple._2).toArray
``````
But this is definitely not working.
Any ideas?
It would be easier to simply call `NumericRange.inclusive` yourself, rather than try to sort out the implicits and type constraints that would be needed to use the `to` syntax. If we require an implicit `Integral`, we'll know how to construct the range with a step of one (provided by the `Integral` instance). We'll also need a `ClassTag[A]` to create the `Array` generically:
``````import scala.reflect.ClassTag
import scala.collection.immutable.NumericRange
def foo2[A >: Int with Char : ClassTag](tuple: (A, A))(implicit int: Integral[A]) =
NumericRange.inclusive(tuple._1, tuple._2, int.one).toArray
scala> foo2(('a', 'z'))
res13: Array[Char] = Array(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z)
scala> foo2((1, 10))
res14: Array[Int] = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
``````
Others fail (if you want them to with the type constraint):
``````scala> foo2((1L, 10L))
<console>:19: error: could not find implicit value for parameter int: Integral[AnyVal]
foo2((1L, 10L))
^
`````` | 526 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-50 | longest | en | 0.742486 |
http://cvxr.com/cvx/examples/circuit_design/html/LC_osc_design.html | 1,653,622,816,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00762.warc.gz | 20,383,710 | 3,484 | ```% Boyd, Kim, and Mohan, "Geometric programming and its
% applications to EDA Problems", (DATE Tutorial 2005), pp.102-113.
% Original code by S. Mohan
% Written for CVX by Almir Mutapcic 02/08/06
%
% Designs an LC oscillator consisting of a loop inductor, varactors
% for fine tuning, binary weighted switching capacitors for coarse
% tuning, cross coupled NMOS transistors, and tail current source.
% The optimal LC oscillator design iwith minimum power consumption,
% and limits on phase noise, area, etc... can be formulated as a GP:
%
% minimize P
% s.t. N <= Nmax, A <= Amax, l >= lmin, etc.
%
% where optimization variables are loop inductor dimensions D,W,
% size of varactor Vc, size of switching caps Csw, width and length
% of transistors Wnmos, Lnmos, bias current Ibias, etc.
%********************************************************************
% problem data
%********************************************************************
Vdd = 1.2; % voltage
CL = 0.2e-12; % load capcitance
F = 5e9; % operating frequency in Hz
omega = 2*pi*F; % operating freq. in radians
FOff = 6e5; % offset frequency for phase noise calculation
LoopGainSpec = 2.0; % loop gain spec
Vbias = 0.2; % non ideality of current mirror
% tuning specs
T = 0.1; % +/- tuning range as a normalized value
CvarRatio = 3; % maximum to minimum value of CVar
CswBits = 3;
CswSegs = 2^(CswBits);
CvarCswLSBOverlap = 2;
%********************************************************************
% optimization of LC oscillator circuit (with tradeoff curve)
%********************************************************************
% varying phase noise parameter for the tradeoff curve
powers = [];
for PNSpec=0.7e-12:0.2e-12:1e-11
fprintf(' PNSpec = %5.2f dBc/Hz: ', 10*log10(PNSpec) );
cvx_begin gp quiet
% optimization variables
variable D; % diameter of loop inductor
variable W; % width of loop inductor
variable SRF; % self resonance frequency
variable l; % length of CMOS transistor
variable w; % width of CMOS transistor
variable Imax; % maximum current through CMOS transistor
variable VOsc; % differential voltage amplitude
variable CT; % total capacitance of oscillator
variable Csw; % maximum switching capacitance
variable Cvar; % minimum variable capacitance
variable IBias; % bias current
variable CMaxFreq; % capacitor max frequency
% minimize power = Vdd*IBias;
minimize( Vdd*IBias )
subject to
%*******************************************%
% loop inductor definitions and constraints %
%*******************************************%
SRFSpec = 3*F;
omegaSRF = 2*pi*SRF;
% inductance
L = 2.1e-06*D^(1.28)*(W)^(-0.25)*(F)^(-0.01);
% series resistance
R = 0.1*D/W+3e-6*D*W^(-0.84)*F^(0.5)+5e-9*D*W^(-0.76)*F^(0.75)+0.02*D*W*F;
% effective capacitance
C = 1e-11*D+5e-6*D*W;
% area, tank conductance, and inverse quality factor
Area = (D+W)^2;
G = R/(omega*L)^2;
invQ = R/(omega*L);
% loop constraints
Area <= 0.25e-6;
W <= 30e-6;
5e-6 <= W;
10*W <= D;
D <= 100*W;
SRFSpec <= SRF;
omegaSRF^2*L*C <= 1;
%****************************************%
% transistor definitions and constraints %
%****************************************%
GM = 6e-3*(w/l)^0.6*(Imax/2)^(0.4);
GD = 4e-10*(w/l)^0.4*(Imax/2)^(0.6)*1/l;
Vgs = 0.34+1e-8/l+800*(Imax*l/(2*w))^0.7;
Cgs = 1e-2*w*l;
Cgd = 1e-9*w;
Cdb = 1e-9*w;
% transistor constraints
2e-6 <= w;
0.13e-6 <= l;
l <= 1e-6;
%***************************************************%
% overall LC oscillator definitions and constraints %
%***************************************************%
invVOsc = (G+GD)/IBias;
% phase noise
kT4 = 4*1.38e-23*300;
kT4G = 2*kT4;
LoopCurrentNoise = kT4*G;
TransistorCurrentNoise = 0.5*kT4G*GM;
PN = 1/(160*(FOff*VOsc*CT)^2)*(LoopCurrentNoise+TransistorCurrentNoise);
% capacitance
Cfix = C+0.5*(CL+Cgs+Cdb+4*Cgd); % fixed capacitance
CDiffMaxFreq = Cfix+0.5*Cvar;
invLoopGain = (G+0.5*GD)/(0.5*GM);
% LC oscillator constraints
PN <= PNSpec;
omega^2*L*CT == 1;
omega^2*(1+T)^2*L*CMaxFreq == 1;
4*T/((1-T^2)^2)*CT <= Csw*(1+CvarCswLSBOverlap/CswSegs);
Csw*CvarCswLSBOverlap/CswSegs <= 0.5*Cvar*(CvarRatio-1);
CDiffMaxFreq <= CMaxFreq;
VOsc+2*Vbias <= 2*Vdd;
VOsc*invVOsc <= 1;
invLoopGain*LoopGainSpec <= 1; % loop gain spec
Vbias+Vgs+IBias/2*R/2 <= Vdd; % bias constraint spec
Imax == IBias;
cvx_end
fprintf('min_power = %3.2f mW\n', cvx_optval/1e-3);
powers = [powers cvx_optval];
end
PNSpec = 0.7e-12:0.2e-12:1e-11;
plot(10*log10(PNSpec),powers/1e-3);
xlabel('Phase Noise (dBc/Hz)')
ylabel('Power (mW)')
```
```Generating the optimal tradeoff curve...
PNSpec = -121.55 dBc/Hz: min_power = 19.00 mW
PNSpec = -120.46 dBc/Hz: min_power = 11.41 mW
PNSpec = -119.59 dBc/Hz: min_power = 8.35 mW
PNSpec = -118.86 dBc/Hz: min_power = 6.44 mW
PNSpec = -118.24 dBc/Hz: min_power = 5.64 mW
PNSpec = -117.70 dBc/Hz: min_power = 5.30 mW
PNSpec = -117.21 dBc/Hz: min_power = 5.02 mW
PNSpec = -116.78 dBc/Hz: min_power = 4.77 mW
PNSpec = -116.38 dBc/Hz: min_power = 4.56 mW
PNSpec = -116.02 dBc/Hz: min_power = 4.38 mW
PNSpec = -115.69 dBc/Hz: min_power = 4.21 mW
PNSpec = -115.38 dBc/Hz: min_power = 4.07 mW
PNSpec = -115.09 dBc/Hz: min_power = 3.93 mW
PNSpec = -114.81 dBc/Hz: min_power = 3.81 mW
PNSpec = -114.56 dBc/Hz: min_power = 3.70 mW
PNSpec = -114.32 dBc/Hz: min_power = 3.60 mW
PNSpec = -114.09 dBc/Hz: min_power = 3.51 mW
PNSpec = -113.87 dBc/Hz: min_power = 3.42 mW
PNSpec = -113.67 dBc/Hz: min_power = 3.34 mW
PNSpec = -113.47 dBc/Hz: min_power = 3.27 mW
PNSpec = -113.28 dBc/Hz: min_power = 3.20 mW
PNSpec = -113.10 dBc/Hz: min_power = 3.13 mW
PNSpec = -112.92 dBc/Hz: min_power = 3.07 mW
PNSpec = -112.76 dBc/Hz: min_power = 3.01 mW
PNSpec = -112.60 dBc/Hz: min_power = 2.96 mW
PNSpec = -112.44 dBc/Hz: min_power = 2.91 mW
PNSpec = -112.29 dBc/Hz: min_power = 2.86 mW
PNSpec = -112.15 dBc/Hz: min_power = 2.81 mW
PNSpec = -112.01 dBc/Hz: min_power = 2.77 mW
PNSpec = -111.87 dBc/Hz: min_power = 2.72 mW
PNSpec = -111.74 dBc/Hz: min_power = 2.68 mW
PNSpec = -111.61 dBc/Hz: min_power = 2.64 mW
PNSpec = -111.49 dBc/Hz: min_power = 2.61 mW
PNSpec = -111.37 dBc/Hz: min_power = 2.57 mW
PNSpec = -111.25 dBc/Hz: min_power = 2.54 mW
PNSpec = -111.14 dBc/Hz: min_power = 2.50 mW
PNSpec = -111.02 dBc/Hz: min_power = 2.47 mW
PNSpec = -110.92 dBc/Hz: min_power = 2.44 mW
PNSpec = -110.81 dBc/Hz: min_power = 2.41 mW
PNSpec = -110.71 dBc/Hz: min_power = 2.38 mW
PNSpec = -110.60 dBc/Hz: min_power = 2.36 mW
PNSpec = -110.51 dBc/Hz: min_power = 2.33 mW
PNSpec = -110.41 dBc/Hz: min_power = 2.31 mW
PNSpec = -110.32 dBc/Hz: min_power = 2.28 mW
PNSpec = -110.22 dBc/Hz: min_power = 2.26 mW
PNSpec = -110.13 dBc/Hz: min_power = 2.23 mW
PNSpec = -110.04 dBc/Hz: min_power = 2.21 mW | 2,595 | 6,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | latest | en | 0.603798 |
https://www.teacherspayteachers.com/Product/Pre-Algebra-Spider-Web-Equations-and-Coordinates-4064315 | 1,537,933,736,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00200.warc.gz | 872,607,390 | 20,252 | # Pre Algebra Spider Web (Equations and Coordinates)
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One-Step Equations and Coordinate Graphing
Four Quadrants (Grid 15 x 20 Portrait)
This equation solving / coordinate graphing project is fun for the student, makes a great bulletin board display and is ideal for reviewing coordinate graphing skills and equation solving skills.
By solving equations, students will create pairs of Cartesian coordinates to create the diagonal lines of the spider web. After they finish with that, they will plot the remainder of the web using a list of coordinate points.
This activity can be a class project or something to be worked on independently when time allows. It also works as an extra credit assignment or when there is a substitute teacher.
This lesson includes:
• 24 equations to solve
• Blank answer sheet for the student’s calculations
• Answer key for the equations with solutions
• Coordinate graph paper for plotting (with a light, medium and dark grid)
• Coordinate list of points to plot
• Sample of the completed project in color
Also included is:
• Student Directions for Plotting Coordinates
• Directions for Plotting Coordinates with Fractions and Decimals
• Directions for Plotting Coordinates in Quadrant I
• Directions for Plotting Coordinates in All Quadrants
• Details for Plotting Coordinates with Fractions in Quadrant I
• Details for Plotting Coordinates with Fractions in All Quadrants.
If you like these projects please let us know. There are many projects, activities and math labs covering the entire school year and some projects are cross curricular (Sports, Patriotism, Seasons, Holidays, History, and Science). Lessons begin at 2nd grade and are also appropriate for the Middle School and some for High School students. These projects have been used successfully at both the Elementary and Middle School levels.
http://www.teacherspayteachers.com/Store/Anthony-And-Linda-Iorlano
Anthony Iorlano and Linda Iorlano
NCTM Standards
Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
Common Core State Standards
CCSS.Math.Content.5.G.A.1 Use a pair of perpendicular number lines, called axes, to define a coordinate system, with the intersection of the lines (the origin) arranged to coincide with the 0 on each line and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Understand that the first number indicates how far to travel from the origin in the direction of one axis, and the second number indicates how far to travel in the direction of the second axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate).
CCSS.MATH.CONTENT.7.EE.B.4.A
Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers.
CCSS.MATH.CONTENT.8.EE.C.7.B
Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.
Keywords: Coordinate Drawing, Coordinate Drafting, Coordinate Graphing, Abscissa, Ordinate, Ordered Pair, Spider, Spider Web, Halloween, Pre-Algebra, PreAlgebra, Pre Algebra, Equations, One-step
Total Pages
17 pages
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Teaching Duration
N/A
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 759 | 3,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-39 | latest | en | 0.902488 |
http://www.solutioninn.com/mathematics-statistics | 1,484,719,941,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00110-ip-10-171-10-70.ec2.internal.warc.gz | 682,832,433 | 8,149 | Showing 1 to 20 of 73447 Questions
• a. Suppose that \$68,000 is to be allocated for advertising, research, and investment in the ratio 8:6:3. How much money will be allocated for each
b. Computer Warehouse sells batteries (\$2) and small boxes of mechanical pencils (\$6). In July, total sales were \$1056. Customers bought 5 times as many batteries as boxes of mechanical pencils. How many of each did Computer Warehouse sell?
• A quality characteristic of interest for a coffee-bag-filling process is the weight of the coffee in the individual bags. If the bags are under filled, two problems arise. First, customers may not be able to brew the coffee to be as strong as they wish. Second, the company may be in violation of the truth-in-labeling laws. In this example, the label weight on the package indicates that, on average, there are 5.45 grams of coffee in a bag. If the average amount of coffee in a bag exceeds the label weight, the company is giving away product. Getting an exact amount of coffee in a bag is problematic because of variation in the temperature and humidity inside the factory, differences in the density of the coffee, and the extremely fast filling operation of the machine (approximately 170 bags a minute). The following table provides the weight in grams of a sample of 50 bags produced in one hour by a single machine:
a. Compute the arithmetic mean and median.
b. Compute the first quartile and third quartile.
c. Compute the range, interquartile range, variance, standard deviation, and coefficient of variation.
d. Interpret the measures of central tendency within the context of this problem. Why should the company producing the coffee bags be concerned about the central tendency?
e. Interpret the measures of variation within the context of this problem. Why should the company producing the coffee bags be concerned about variation?
• An apple juice bottling company maintains records concerning the number of unacceptable bottles of juice obtained from the filling and capping machines. Based on past data, the probability that a bottle came from machine I and was nonconforming is 0.05 and the probability that a bottle came from machine II and was nonconforming is 0.075. These probabilities represent the probability of one bottle out of the total sample having the specified characteristics. Half the bottles are filled on machine I and the other half are filled on machine II.
a. If a filled bottle of juice is selected at random, what is the probability that it is a nonconforming bottle?
b. If a filled bottle of juice is selected at random, what is the probability that it was filled on machine II?
c. If a filled bottle of juice is selected at random, what is the probability that it was filled on machine I and is a conforming bottle?
• According to Investment Digest ("Diversification and the Risk/Reward Relationship", Winter 1994, 1-3), the mean of the annual return for common stocks from 1926 to 1992 was 15.4%, and the standard deviation of the annual return was 24.5%. During the same 67-year time span, the mean of the annual return for long-term government bonds was 5.5%, and the standard deviation was 6.0%. The article claims that the distributions of annual returns for both common stocks and long-term government bonds are bell-shaped and approximately symmetric. Assume that these distributions are distributed as normal random variables with the means and standard deviations given previously.
a. Find the probability that the return for common stocks will be greater than 0%.
b. Find the probability that the return for common stocks will be less than 20%.
• Compute a 95% confidence interval for the population mean, based on the sample 10, 12, 13, 14, 15, 16, and 49. Change the number from 49 to 16 and recalculate the confidence interval. Using the results, describe the effect of an outlier or extreme value on the confidence interval
• The director of admissions at the University of Maryland, University College is concerned about the high cost of textbooks for the students each semester. A sample of 25 students enrolled in the university indicates that X (bar) = \$315.4 and s = \$43.20.
a. Using the 0.10 level of significance, is there evidence that the population mean is above \$300?
b. What is your answer in (a) if s = \$75 and the 0.05 level of significance is used?
c. What is your answer in (a) if X (bar) = \$305.11 and s = \$43.20?
d. Based on the information in part (a), what decision should the director make about the books used for the courses if the goal is to keep the cost below \$300?
• A large candy manufacturer is concerned that the mean weight of their bag of Gooey Sour Worms is not greater than 7.3 ounces. It can be assumed that the population standard deviation is .5 ounces based on past experience. A sample of 169 gummy worms is selected and the sample mean is 7.35 ounces. Using a level of significance of .10, is there evidence that the population mean weight of the candy bars is greater than 7.3? Fully explain your answer.
• Each of the possible five outcomes of a random experiment is equally likely. The sample space is {a, b, c, d, e}.
Let A denote the event {a, b}, and let B denote the event {c, d, e}. Determine the following:
(a) P (A) (b) P (B)
(c) P (A’) (d) P (A U B)
(e) P (A ∩ B)
• The sample space of a random experiment is {a, b, c, d, e} with probabilities 0.1, 0.1, 0.2, 0.4, and 0.2, respectively. Let A denote the event {a, b, c}, and let B denote the event
{c, d, e}. Determine the following:
(a) P (A) (b) P (B)
(c) P (A’) (d) P (A U B)
(e) P (A ∩ B)
• A part selected for testing is equally likely to have been produced on any one of six cutting tools.
(a) What is the sample space?
(b) What is the probability that the part is from tool 1?
(c) What is the probability that the part is from tool 3 or tool 5?
(d) What is the probability that the part is not from tool 4?
• An injection-molded part is equally likely to be obtained from any one of the eight cavities on a mold.
(a) What is the sample space?
(b) What is the probability a part is from cavity 1 or 2?
(c) What is the probability that a part is neither from cavity 3 nor 4?
• A sample space contains 20 equally likely outcomes.
If the probability of event A is 0.3, how many outcomes are in event A?
• Orders for a computer are summarized by the optional features that are requested as follows:
Proportion of orders
No optional features 0.3
One optional feature 0.5
More than one optional feature 0.2
(a) What is the probability that an order requests at least one optional feature?
(b) What is the probability that an order does not request more than one optional feature?
• If the last digit of a weight measurement is equally likely to be any of the digits 0 through 9,
(a) What is the probability that the last digit is 0?
(b) What is the probability that the last digit is greater than or equal to 5?
• A sample preparation for a chemical measurement is completed correctly by 25% of the lab technicians, completed with a minor error by 70%, and completed with a major error by 5%.
(a) If a technician is selected randomly to complete the preparation, what is the probability it is completed without error?
(b) What is the probability that it is completed with either a minor or a major error
• A credit card contains 16 digits between 0 and 9. However, only 100 million numbers are valid. If a number is entered randomly, what is the probability that it is a valid number?
• Suppose your vehicle is licensed in a state that issues license plates that consist of three digits (between 0 and 9) followed by three letters (between A and Z). If a license number is selected randomly, what is the probability that yours is the one selected
• A message can follow different paths through servers on a network. The senders message can go to one of five servers for the first step, each of them can send to five servers at the second step, each of which can send to four servers at the third step, and then the message goes to the recipients server.
(a) How many paths are possible?
(b) If all paths are equally likely, what is the probability that a message passes through the first of four servers at the third step?
• Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 100 disks are summarized as follows:
Shock resistance
High Low
Scratch high 70 9
Resistance low 16 5
Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance.
If a disk is selected at random, determine the following probabilities:
(a) P (A) (b) P (A)
(c) P (A`) (d) P (A ∩ B)
(e) (A U B) (f) (A` U B)
• Samples of a Fast aluminum part are classified on the basis of surface finish (in micro inches) and edge finish. The results of 100 parts are summarized as follows:
Edge finish
Excellent Good
Surface excellent 80 2
Finish good 10 8
Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent length. If a part is selected at random,
determine the following probabilities:
(a) P (A) (b) P (A)
(c) P (A`) (d) P (A ∩ B)
(e) (A U B) (f) (A` U B)
| 2,200 | 9,217 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | longest | en | 0.93054 |
https://www.bhavinionline.com/2016/02/crack-the-code-if-rose-is-coded-as-682-what-is-code-for-search/ | 1,601,090,075,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00787.warc.gz | 737,900,305 | 18,738 | Crack The Code: If ROSE is coded as 682; What is Code for Search
From the hints given, crack the code for the given word.
If ROSE is coded as 6821,
CHAIR is coded as 73456 and
PREACH is coded as 961473,
What will be the code for SEARCH?
If you observe closely you will notice that each of the words are coded with the same number;
For example R in Rose and R in Chair are both coded as 6.
Similarly you have to find codes for all the letters in the word SEARCH.
The First two letters are in ROSE;
i.e SE = 21
The Next two letters can be found out from PREACH and CHAIR;
i.e. AR = 46
Last 2 letters can be found from PREACH;
i.e. CH = 73
So the code for SEARCH would be;
SEARCH = 214673
One Reply to “Crack The Code: If ROSE is coded as 682; What is Code for Search”
1. Looks like a substitution code with the following mapping:
ESHAIRCOP
0123456789
This means “SEARCH” is encoded as “214673”
Many permutations of all 9 digits encode interesting messages: 356124789, 538217649, … | 275 | 999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2020-40 | longest | en | 0.926981 |
http://www.chegg.com/homework-help/questions-and-answers/pilot-small-plane-encounters-shifting-winds-flies-260km-northeast-450-km-due-north-point-f-q639955 | 1,438,146,874,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986022.41/warc/CC-MAIN-20150728002306-00030-ip-10-236-191-2.ec2.internal.warc.gz | 378,152,716 | 12,086 | A pilot in a small plane encounters shifting winds. He flies 26.0km northeast, then 45.0 km due north. From this point, he flies anadditional distance in an unknown direction, only to find himselfat a small airstrip that his map shows to be 70.0 km directly northof his starting point.
Part A
what was the length of the third leg of his trip? (km)
Part B
what was the direction of the third leg of his trip?
θ = ? o north of west | 111 | 433 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2015-32 | latest | en | 0.986615 |
http://statkat.com/stattest.php?t=13&t2=4&t3=13&t4=12 | 1,555,871,317,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578532050.7/warc/CC-MAIN-20190421180010-20190421202010-00310.warc.gz | 171,832,647 | 9,174 | # Regression (OLS) - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Regression (OLS)
Chi-squared test for the relationship between two categorical variables
Regression (OLS)
Two way ANOVA
Independent variablesIndependent /column variableIndependent variablesIndependent variables
One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesOne categorical with $I$ independent groups ($I \geqslant 2$)One or more quantitative of interval or ratio level and/or one or more categorical with independent groups, transformed into code variablesTwo categorical, the first with $I$ independent groups and the second with $J$ independent groups ($I \geqslant 2$, $J \geqslant 2$)
Dependent variableDependent /row variableDependent variableDependent variable
One quantitative of interval or ratio levelOne categorical with $J$ independent groups ($J \geqslant 2$)One quantitative of interval or ratio levelOne quantitative of interval or ratio level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
$F$ test for the complete regression model:
• $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
or equivalenty
• The variance explained by all the independent variables together (the complete model) is 0 in the population: $\rho^2 = 0$
$t$ test for individual regression coefficient $\beta_k$:
• $\beta_k = 0$
in the regression equation $\mu_y = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$
• There is no association between the row and column variable
More precise statement:
• If there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
The distribution of the dependent variable is the same in each of the $I$ populations
• If there is one random sample of size $N$ from the total population:
The row and column variables are independent
$F$ test for the complete regression model:
• $\beta_1 = \beta_2 = \ldots = \beta_K = 0$
or equivalenty
• The variance explained by all the independent variables together (the complete model) is 0 in the population: $\rho^2 = 0$
$t$ test for individual regression coefficient $\beta_k$:
• $\beta_k = 0$
in the regression equation $\mu_y = \beta_0 + \beta_1 \times x_1 + \beta_2 \times x_2 + \ldots + \beta_K \times x_K$
ANOVA $F$ tests:
• For main and interaction effects together (model): no main effects and interaction effect
• For independent variable A: no main effect for A
• For independent variable B: no main effect for B
• For the interaction term: no interaction effect between A and B
We could also perform $t$ tests for specific contrasts and multiple comparisons, just like we did with one way ANOVA. However, this is more advanced stuff.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
$F$ test for the complete regression model:
• not all population regression coefficients are 0
or equivalenty
• The variance explained by all the independent variables together (the complete model) is larger than 0 in the population: $\rho^2 > 0$
$t$ test for individual $\beta_k$:
• Two sided: $\beta_k \neq 0$
• Right sided: $\beta_k > 0$
• Left sided: $\beta_k < 0$
• There is an association between the row and column variable
More precise statement:
• If there are $I$ independent random samples of size $n_i$ from each of $I$ populations, defined by the independent variable:
The distribution of the dependent variable is not the same in all of the $I$ populations
• If there is one random sample of size $N$ from the total population:
The row and column variables are dependent
$F$ test for the complete regression model:
• not all population regression coefficients are 0
or equivalenty
• The variance explained by all the independent variables together (the complete model) is larger than 0 in the population: $\rho^2 > 0$
$t$ test for individual $\beta_k$:
• Two sided: $\beta_k \neq 0$
• Right sided: $\beta_k > 0$
• Left sided: $\beta_k < 0$
ANOVA $F$ tests:
• For main and interaction effects together (model): there is a main effect for A, and/or for B, and/or an interaction effect
• For independent variable A: there is a main effect for A
• For independent variable B: there is a main effect for B
• For the interaction term: there is an interaction effect between A and B
AssumptionsAssumptionsAssumptionsAssumptions
• In the population, the residuals are normally distributed at each combination of values of the independent variables
• In the population, the standard deviation $\sigma$ of the residuals is the same for each combination of values of the independent variables (homoscedasticity)
• In the population, the relationship between the independent variables and the mean of the dependent variable $\mu_y$ is linear. If this linearity assumption holds, the mean of the residuals is 0 for each combination of values of the independent variables
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
• Sample size is large enough for $X^2$ to be approximately chi-squared distributed under the null hypothesis. Rule of thumb:
• 2 $\times$ 2 table: all four expected cell counts are 5 or more
• Larger than 2 $\times$ 2 tables: average of the expected cell counts is 5 or more, smallest expected cell count is 1 or more
• There are $I$ independent simple random samples from each of $I$ populations defined by the independent variable, or there is one simple random sample from the total population
• In the population, the residuals are normally distributed at each combination of values of the independent variables
• In the population, the standard deviation $\sigma$ of the residuals is the same for each combination of values of the independent variables (homoscedasticity)
• In the population, the relationship between the independent variables and the mean of the dependent variable $\mu_y$ is linear. If this linearity assumption holds, the mean of the residuals is 0 for each combination of values of the independent variables
• The residuals are independent of one another
• Variables are measured without error
Also pay attention to:
• Multicollinearity
• Outliers
• Within each of the $I \times J$ populations, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the $I \times J$ populations
• For each of the $I \times J$ groups, the sample is an independent and simple random sample from the population defined by that group. That is, within and between groups, observations are independent of one another
• Equal sample sizes for each group make the interpretation of the ANOVA output easier (unequal sample sizes result in overlap in the sum of squares; this is advanced stuff)
Test statisticTest statisticTest statisticTest statistic
$F$ test for the complete regression model:
• \begin{aligned}[t] F &= \dfrac{\sum (\hat{y}_j - \bar{y})^2 / K}{\sum (y_j - \hat{y}_j)^2 / (N - K - 1)}\\ &= \dfrac{\mbox{sum of squares model} / \mbox{degrees of freedom model}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square model}}{\mbox{mean square error}} \end{aligned}
where $\hat{y}_j$ is the predicted score on the dependent variable $y$ of subject $j$, $\bar{y}$ is the mean of $y$, $y_j$ is the score on $y$ of subject $j$, $N$ is the total sample size, and $K$ is the number of independent variables
$t$ test for individual $\beta_k$:
• $t = \dfrac{b_k}{SE_{b_k}}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$, with $s$ the sample standard deviation of the residuals, $x_j$ the score of subject $j$ on the independent variable $x$, and $\bar{x}$ the mean of $x$. For models with more than one independent variable, computing $SE_{b_k}$ becomes complicated
Note 1: mean square model is also known as mean square regression; mean square error is also known as mean square residual
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
$X^2 = \sum{\frac{(\mbox{observed cell count} - \mbox{expected cell count})^2}{\mbox{expected cell count}}}$
where for each cell, the expected cell count = $\dfrac{\mbox{row total} \times \mbox{column total}}{\mbox{total sample size}}$, the observed cell count is the observed sample count in that same cell, and the sum is over all $I \times J$ cells
$F$ test for the complete regression model:
• \begin{aligned}[t] F &= \dfrac{\sum (\hat{y}_j - \bar{y})^2 / K}{\sum (y_j - \hat{y}_j)^2 / (N - K - 1)}\\ &= \dfrac{\mbox{sum of squares model} / \mbox{degrees of freedom model}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square model}}{\mbox{mean square error}} \end{aligned}
where $\hat{y}_j$ is the predicted score on the dependent variable $y$ of subject $j$, $\bar{y}$ is the mean of $y$, $y_j$ is the score on $y$ of subject $j$, $N$ is the total sample size, and $K$ is the number of independent variables
$t$ test for individual $\beta_k$:
• $t = \dfrac{b_k}{SE_{b_k}}$
• If only one independent variable:
$SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}}$, with $s$ the sample standard deviation of the residuals, $x_j$ the score of subject $j$ on the independent variable $x$, and $\bar{x}$ the mean of $x$. For models with more than one independent variable, computing $SE_{b_k}$ becomes complicated
Note 1: mean square model is also known as mean square regression; mean square error is also known as mean square residual
Note 2: if only one independent variable ($K = 1$), the $F$ test for the complete regression model is equivalent to the two sided $t$ test for $\beta_1$
For main and interaction effects together (model):
• $F = \dfrac{\mbox{mean square model}}{\mbox{mean square error}}$
For independent variable A:
• $F = \dfrac{\mbox{mean square A}}{\mbox{mean square error}}$
For independent variable B:
• $F = \dfrac{\mbox{mean square B}}{\mbox{mean square error}}$
For the interaction term:
• $F = \dfrac{\mbox{mean square interaction}}{\mbox{mean square error}}$
Note: mean square error is also known as mean square residual or mean square within
Sample standard deviation of the residuals $s$n.a.Sample standard deviation of the residuals $s$Pooled standard deviation
\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}-\begin{aligned} s &= \sqrt{\dfrac{\sum (y_j - \hat{y}_j)^2}{N - K - 1}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} \begin{aligned} s_p &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - (I \times J)}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned} Sampling distribution of F and of t if H0 were trueSampling distribution of X^2 if H0 were trueSampling distribution of F and of t if H0 were trueSampling distribution of F if H0 were true Sampling distribution of F: • F distribution with K (df model, numerator) and N - K - 1 (df error, denominator) degrees of freedom Sampling distribution of t: • t distribution with N - K - 1 (df error) degrees of freedom Approximately a chi-squared distribution with (I - 1) \times (J - 1) degrees of freedomSampling distribution of F: • F distribution with K (df model, numerator) and N - K - 1 (df error, denominator) degrees of freedom Sampling distribution of t: • t distribution with N - K - 1 (df error) degrees of freedom For main and interaction effects together (model): • F distribution with (I - 1) + (J - 1) + (I - 1) \times (J - 1) (df model, numerator) and N - (I \times J) (df error, denominator) degrees of freedom For independent variable A: • F distribution with I - 1 (df A, numerator) and N - (I \times J) (df error, denominator) degrees of freedom For independent variable B: • F distribution with J - 1 (df B, numerator) and N - (I \times J) (df error, denominator) degrees of freedom For the interaction term: • F distribution with (I - 1) \times (J - 1) (df interaction, numerator) and N - (I \times J) (df error, denominator) degrees of freedom Here N is the total sample size Significant?Significant?Significant?Significant? F test: • Check if F observed in sample is equal to or larger than critical value F^* or • Find p value corresponding to observed F and check if it is equal to or smaller than \alpha t Test two sided: t Test right sided: t Test left sided: • Check if X^2 observed in sample is equal to or larger than critical value X^{2*} or • Find p value corresponding to observed X^2 and check if it is equal to or smaller than \alpha F test: • Check if F observed in sample is equal to or larger than critical value F^* or • Find p value corresponding to observed F and check if it is equal to or smaller than \alpha t Test two sided: t Test right sided: t Test left sided: • Check if F observed in sample is equal to or larger than critical value F^* or • Find p value corresponding to observed F and check if it is equal to or smaller than \alpha C\% confidence interval for \beta_k and for \mu_y; C\% prediction interval for y_{new}n.a.C\% confidence interval for \beta_k and for \mu_y; C\% prediction interval for y_{new}n.a. Confidence interval for \beta_k: • b_k \pm t^* \times SE_{b_k} • If only one independent variable: SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}} Confidence interval for \mu_y, the population mean of y given the values on the independent variables: • \hat{y} \pm t^* \times SE_{\hat{y}} • If only one independent variable: SE_{\hat{y}} = s \sqrt{\dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} Prediction interval for y_{new}, the score on y of a future respondent: • \hat{y} \pm t^* \times SE_{y_{new}} • If only one independent variable: SE_{y_{new}} = s \sqrt{1 + \dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} In all formulas, the critical value t^* is the value under the t_{N - K - 1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). -Confidence interval for \beta_k: • b_k \pm t^* \times SE_{b_k} • If only one independent variable: SE_{b_1} = \dfrac{\sqrt{\sum (y_j - \hat{y}_j)^2 / (N - 2)}}{\sqrt{\sum (x_j - \bar{x})^2}} = \dfrac{s}{\sqrt{\sum (x_j - \bar{x})^2}} Confidence interval for \mu_y, the population mean of y given the values on the independent variables: • \hat{y} \pm t^* \times SE_{\hat{y}} • If only one independent variable: SE_{\hat{y}} = s \sqrt{\dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} Prediction interval for y_{new}, the score on y of a future respondent: • \hat{y} \pm t^* \times SE_{y_{new}} • If only one independent variable: SE_{y_{new}} = s \sqrt{1 + \dfrac{1}{N} + \dfrac{(x^* - \bar{x})^2}{\sum (x_j - \bar{x})^2}} In all formulas, the critical value t^* is the value under the t_{N - K - 1} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). - Effect sizen.a.Effect sizeEffect size Complete model: • Proportion variance explained R^2: Proportion variance of the dependent variable y explained by the sample regression equation (the independent variables): \begin{align} R^2 &= \dfrac{\sum (\hat{y}_j - \bar{y})^2}{\sum (y_j - \bar{y})^2}\\ &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}\\ &= 1 - \dfrac{\mbox{sum of squares error}}{\mbox{sum of squares total}}\\ &= r(y, \hat{y})^2 \end{align} $$R^2 is the proportion variance explained in the sample by the sample regression equation. It is a positively biased estimate of the proportion variance explained in the population by the population regression equation, \rho^2. If there is only one independent variable, R^2 = r^2: the correlation between the independent variable x and dependent variable y squared. • Wherry's R^2 / shrunken R^2: Corrects for the positive bias in R^2 and is equal to$$R^2_W = 1 - \frac{N - 1}{N - K - 1}(1 - R^2)$$R^2_W is a less biased estimate than R^2 of the proportion variance explained in the population by the population regression equation, \rho^2 • Stein's R^2: Estimates the proportion of variance in y that we expect the current sample regression equation to explain in a different sample drawn from the same population. It is equal to$$R^2_S = 1 - \frac{(N - 1)(N - 2)(N + 1)}{(N - K - 1)(N - K - 2)(N)}(1 - R^2)$$Per independent variable: • Correlation squared r^2_k: the proportion of the total variance in the dependent variable y that is explained by the independent variable x_k, not corrected for the other independent variables in the model • Semi-partial correlation squared sr^2_k: the proportion of the total variance in the dependent variable y that is uniquely explained by the independent variable x_k, beyond the part that is already explained by the other independent variables in the model • Partial correlation squared pr^2_k: the proportion of the variance in the dependent variable y not explained by the other independent variables, that is uniquely explained by the independent variable x_k -Complete model: • Proportion variance explained R^2: Proportion variance of the dependent variable y explained by the sample regression equation (the independent variables):$$ \begin{align} R^2 &= \dfrac{\sum (\hat{y}_j - \bar{y})^2}{\sum (y_j - \bar{y})^2}\\ &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}}\\ &= 1 - \dfrac{\mbox{sum of squares error}}{\mbox{sum of squares total}}\\ &= r(y, \hat{y})^2 \end{align} $$R^2 is the proportion variance explained in the sample by the sample regression equation. It is a positively biased estimate of the proportion variance explained in the population by the population regression equation, \rho^2. If there is only one independent variable, R^2 = r^2: the correlation between the independent variable x and dependent variable y squared. • Wherry's R^2 / shrunken R^2: Corrects for the positive bias in R^2 and is equal to$$R^2_W = 1 - \frac{N - 1}{N - K - 1}(1 - R^2)$$R^2_W is a less biased estimate than R^2 of the proportion variance explained in the population by the population regression equation, \rho^2 • Stein's R^2: Estimates the proportion of variance in y that we expect the current sample regression equation to explain in a different sample drawn from the same population. It is equal to$$R^2_S = 1 - \frac{(N - 1)(N - 2)(N + 1)}{(N - K - 1)(N - K - 2)(N)}(1 - R^2)$$Per independent variable: • Correlation squared r^2_k: the proportion of the total variance in the dependent variable y that is explained by the independent variable x_k, not corrected for the other independent variables in the model • Semi-partial correlation squared sr^2_k: the proportion of the total variance in the dependent variable y that is uniquely explained by the independent variable x_k, beyond the part that is already explained by the other independent variables in the model • Partial correlation squared pr^2_k: the proportion of the variance in the dependent variable y not explained by the other independent variables, that is uniquely explained by the independent variable x_k • Proportion variance explained R^2: Proportion variance of the dependent variable y explained by the independent variables and the interaction effect together:$$ \begin{align} R^2 &= \dfrac{\mbox{sum of squares model}}{\mbox{sum of squares total}} \end{align} $$R^2 is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population. • Proportion variance explained \eta^2: Proportion variance of the dependent variable y explained by an independent variable or interaction effect:$$ \begin{align} \eta^2_A &= \dfrac{\mbox{sum of squares A}}{\mbox{sum of squares total}}\\ \\ \eta^2_B &= \dfrac{\mbox{sum of squares B}}{\mbox{sum of squares total}}\\ \\ \eta^2_{int} &= \dfrac{\mbox{sum of squares int}}{\mbox{sum of squares total}} \end{align} $$\eta^2 is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population. • Proportion variance explained \omega^2: Corrects for the positive bias in \eta^2 and is equal to:$$ \begin{align} \omega^2_A &= \dfrac{\mbox{sum of squares A} - \mbox{degrees of freedom A} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_B &= \dfrac{\mbox{sum of squares B} - \mbox{degrees of freedom B} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \\ \omega^2_{int} &= \dfrac{\mbox{sum of squares int} - \mbox{degrees of freedom int} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}\\ \end{align} $$\omega^2 is a better estimate of the explained variance in the population than \eta^2. Only for balanced designs (equal sample sizes). • Proportion variance explained \eta^2_{partial}:$$ \begin{align} \eta^2_{partial\,A} &= \frac{\mbox{sum of squares A}}{\mbox{sum of squares A} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,B} &= \frac{\mbox{sum of squares B}}{\mbox{sum of squares B} + \mbox{sum of squares error}}\\ \\ \eta^2_{partial\,int} &= \frac{\mbox{sum of squares int}}{\mbox{sum of squares int} + \mbox{sum of squares error}} \end{align}$ANOVA tablen.a.ANOVA tableANOVA table - n.a.n.a.n.a.Equivalent to ---OLS regression with two, categorical independent variables and the interaction term, transformed into$(I - 1)$+$(J - 1)$+$(I - 1) \times (J - 1)$code variables. Example contextExample contextExample contextExample context Can mental health be predicted from fysical health, economic class, and gender?Is there an association between economic class and gender? Is the distribution of economic class different between men and women?Can mental health be predicted from fysical health, economic class, and gender?Is the average mental health score different between people from a low, moderate, and high economic class? And is the average mental health score different between men and women? And is there an interaction effect between economic class and gender? SPSSSPSSSPSSSPSS Analyze > Regression > Linear... • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Independent(s) Analyze > Descriptive Statistics > Crosstabs... • Put one of your two categorical variables in the box below Row(s), and the other categorical variable in the box below Column(s) • Click the Statistics... button, and click on the square in front of Chi-square • Continue and click OK Analyze > Regression > Linear... • Put your dependent variable in the box below Dependent and your independent (predictor) variables in the box below Independent(s) Analyze > General Linear Model > Univariate... • Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factor(s) JamoviJamoviJamoviJamovi Regression > Linear Regression • Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates • If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well • Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes' Frequencies > Independent Samples -$\chi^2\$ test of association
• Put one of your two categorical variables in the box below Rows, and the other categorical variable in the box below Columns
Regression > Linear Regression
• Put your dependent variable in the box below Dependent Variable and your independent variables of interval/ratio level in the box below Covariates
• If you also have code (dummy) variables as independent variables, you can put these in the box below Covariates as well
• Instead of transforming your categorical independent variable(s) into code variables, you can also put the untransformed categorical independent variables in the box below Factors. Jamovi will then make the code variables for you 'behind the scenes'
ANOVA > ANOVA
• Put your dependent (quantitative) variable in the box below Dependent Variable and your two independent (grouping) variables in the box below Fixed Factors
Practice questionsPractice questionsPractice questionsPractice questions | 6,714 | 25,355 | {"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": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2019-18 | longest | en | 0.737999 |
http://mathhelpforum.com/pre-calculus/47117-express-function.html | 1,480,856,721,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541321.31/warc/CC-MAIN-20161202170901-00485-ip-10-31-129-80.ec2.internal.warc.gz | 171,097,772 | 11,350 | 1. ## Express the function...
Express the function in the form f g. (Use only elementary functions, but not the identity function, g(x) = x.)
I am totally lost on what to do.
When it gives mt F(x) and tells me to find g(x) I can do fine... but this.... this is weird.
AND this one confounds me as well
Express the function in the form f g. (Use only elementary functions, but not the identity function, g(x) = x.)
2. Originally Posted by leftyguitarjoe
Express the function in the form f g. (Use only elementary functions, but not the identity function, g(x) = x.)
I am totally lost on what to do.
Then it gives mt F(x) and tells me to find g(x) I can do fine... but this.... this is weird.
my first question may seem silly to you, but consider it seriously: do you understand what $f \circ g$ means.
i do not like problems like this, because you usually have several answers that are correct. but i bet this is the answer they are looking for:
$f(x) = \sqrt[3]{x}$ and $g(x) = \frac x{4 + x}$
now see if you can tell why that answer makes sense, after considering the question i asked you earlier
AND this one confounds me as well
Express the function in the form f compose g. (Use only elementary functions, but not the identity function, g(x) = x.)
if you think you got it, try this one yourself and tell us your answer
3. Yes, I am familiar with f(g)) and such.
Thanks for the help. I'll go work out the other one and report back.
4. ok, so the other one should be:
f(x) =
and
g(x) =
EDIT
err I mean to make those "t"'s not "x"'s
but ya catch my drift.
5. Originally Posted by leftyguitarjoe
ok, so the other one should be:
f(x) =
and
g(x) =
yup
6. gah.
I think too hard. thats my problem haha.
7. Originally Posted by leftyguitarjoe
gah.
I think too hard. thats my problem haha.
that's a lot of people's problem
good luck | 497 | 1,854 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2016-50 | longest | en | 0.937943 |
https://cracku.in/30-among-six-people-p-q-r-s-t-and-v-each-lives-on-a-d-x-ibps-po-2012 | 1,721,288,708,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00483.warc.gz | 174,786,166 | 23,599 | Instructions
Directions : Each of the questions below consists of a question and three statements numbered I, II and III given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read all the three statements and give answer-
Question 30
# Among six people P, Q, R, S, T and V each lives on a different floor of a six-storey building having six floors numbered one to six (the ground floor is numbered 1, the floor above it, number 2 and so on and the topmost floor is numbered 6. Who lives on the topmost floor? I) There is only one floor between the floors on which R and Q live. P lives on an even numbered floor. II) T does not live on an even numbered floor. Q lives on an even numbered floor. Q does not live on the topmost floor. III) S lives on an odd numbered floor. There are two floors between the floors on which S and P live. T lives on a floor immediately above R's floor.
Solution
Based on information in the paragraph, the person in the highest floor lives on 6 (even number). P, Q and R live in even numbered floors. T lives above R =. So, R is not the answer. Q is not in the top floor. So, P is in the top floor. Information from all three statements is used to arrive at this solution. | 298 | 1,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.955652 |
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Is There Really a Strictly Conserved Stress-Energy Tensor in GR?
Quote by DaleSpam I actually disagree here. I don't think that it is even approximately true, at least not globally. If it were approximately true then that would mean that you could get a GW which was spherical plus some small higher order terms. However, the spherical and dipole terms are identically 0. You can write a realistic GW as a quadrupole term plus some small higher order terms, but no lower than quadrupole.
If you're going to actually try to model the wave fields directly as spherical harmonics, yes, I agree; the l = 0 and l = 1 terms are identically zero.
However, I can see doing a geometric optics approximation where we model the GWs, globally, as expanding spherical wavefronts of "graviton pulses", similar to the way "photon" wavefronts are modeled as spherical in SR as an approximation, even though they're really not (the lowest-order EM radiation is dipole so the spherical term is 0 for that as well). Of course the GW wavelength has to be much, much smaller than the size of the spheres for this to work, i.e., the GWs have to be high frequency. I suspect that the paper Q-reeus linked to about HFGWs was doing something along those lines. But that is still only an approximation.
Furthermore, it's a useless approximation for trying to decide if GWs carry energy, because a "yes" answer to that question is built into the geometric optics approximation in the first place. That approximation assumes that the "gravitons" are massless particles carrying some finite amount of energy and momentum (if they carried zero energy and momentum they would have infinite wavelength, which obviously violates the small-wavelength assumption).
Quote by DaleSpam Of course, you can do local approximations, but then there is no advantage to expanding as local spherical waves rather than local plane waves. The symmetry argument doesn't apply, and you add needless complication to your approximation terms without adding accuracy.
Exactly.
Quote by PeterDonis So you believe you can exhibit a quadrupole wave which has spherical symmetry? This should be interesting. I eagerly await the new thread.
Sorry to have to say both yourself and DaleSpam are attacking a straw man in posts #83-86. Did I not make it clear in #78 I was talking about spherical wavefronts? You are both sadly uninformed about common terminology here. Spherical wavefront (often just the term 'spherical wave' is used - without confusion by those in the know) simply means that at large r (i.e. well into radiation zone) wavefronts of constant phase have spherical symmetry. And that much I clarified for you in #80 - so you are both without excuse for attacking this straw man of your own creation. From http://en.wikipedia.org/wiki/Antenna...#Compact_range
The CATR uses a source antenna which radiates a spherical wavefront and one or more secondary reflectors to collimate the radiated spherical wavefront into a planar wavefront within the desired test zone.
See also (perhaps you should inform author of gross ignorance in using the term 'spherical wave' in respect of antenna radiation! What an ignoramus!)
Also try
Although the wave emitted by the oscillating dipole is a spherical wave, it does not have the same intensity in all directions.
(between (4.16) and (4.16')) Gees - yet another ignoramus! Must be crawling with em out there.
Get used to it folks - spherical wave simply refers to phase of wavefront, and need have no bearing on angular dependence of field strength or direction - savvy?! I never once used the term spherically uniform field or monopole field or monopole moment - that all came from inside your heads.
Now, assuming your bonfire for the straw man has burnt out, listen up. Have been feeling my way on this issue - beginning with #76. Some statements made in #78 I now see are wrong (phase differential bit and what flowed from that), but stand by the overall thrust. It needs considerably more refinement and better presentation, and that I intend to do, but hands are tied up at the moment elsewhere. Sufficient to say I'm now sure GW's are a phantom. One more thing. Since you and DaleSpam have not heeded my request to leave it all for now - you might as well make good on that undertaking to provide reference material for Feynman's 'strange' sticky-bead argument that had stick pointing along propagation axis. Have you found one yet?
Mentor
Quote by Q-reeus Get used to it folks - spherical wave simply refers to phase of wavefront, and need have no bearing on angular dependence of field strength or direction - savvy?! I never once used the term spherically uniform field or monopole field or monopole moment - that all came from inside your heads.
Oh, you are correct. I was indeed thinking you were referring to complete spherical symmetry in every aspect of the wave instead of simply a spherical phase distribution. I stand corrected.
Of course, since you didn't intend to imply anything about anything other than the phase then your argument becomes a non sequiter. A GW has more than just phase, so these other components need not be spherically symmetric as your argument requires:
Quote by Q-reeus Now please explain how each and every sticky bead decides which way to move in this situation - east or west. I'll save you the trouble - by symmetry there can be no such motions.
I was mistakenly thinking that you were making a valid argument from an incorrect premise, when you were actually making an invalid argument from a correct premise.
Quote by DaleSpam Oh, you are correct. I was indeed thinking you were referring to complete spherical symmetry in every aspect of the wave instead of simply a spherical phase distribution. I stand corrected.
Thanks for at least admitting that - there's hope yet for you DS.
Of course, since you didn't intend to imply anything about anything other than the phase then your argument becomes a non sequiter.
Actually it's that statement that is the non sequiter - more below.
A GW has more than just phase, so these other components need not be spherically symmetric as your argument requires:
Wrong on last bit. My particular argument you presumably are referring to - in last section of #78, does not at all require spherical symmetry of field - merely axial symmetry in equatorial plane. And that was correctly applied.
Q-reeus: "Now please explain how each and every sticky bead decides which way to move in this situation - east or west. I'll save you the trouble - by symmetry there can be no such motions." I was mistakenly thinking that you were making a valid argument from an incorrect premise, when you were actually making an invalid argument from a correct premise.
Actually it is you hastily making an invalid judgement. What I wrote there is just basic fact and cannot be sensibly denied. Maybe you simply have not grasped what was being said there. Perhaps you have the basic geometry confused. Are you cognizant of the arrangement: A very large circular hoop encircling at large r an axial quadrupole oscillator, with latter's axis of oscillation normal to plane of hoop? Hoop lying in equatorial plane of oscillator. I certainly described it plainly enough as such, but it never fails to amaze me how readily some folks can still misinterpret. If you did understand arrangement, how can you criticize the bit you quoted? It necessarily is true.
Mentor This thread, like the ones before it, has degenerated to the personal. It's closed. Q-reeus, please do not start another one. | 1,616 | 7,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.973702 |
http://math.stackexchange.com/questions/155017/partial-derivation-lim-x-y-to0-0-fracx2-sin2y2x2y/155019 | 1,469,351,926,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823989.0/warc/CC-MAIN-20160723071023-00318-ip-10-185-27-174.ec2.internal.warc.gz | 163,131,325 | 16,962 | # Partial Derivation: $\lim_{(x,y)\to(0,0)}\frac{x^2+\sin^2y}{2x^2+y}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+\sin^2y}{2x^2+y}$$
Above problem is in my textbook. it's different from others because I don't know how to process with trigometric element: in this example is $\sin^2y$
Thanks :)
-
## 1 Answer
HINT: Don’t let the trig function throw you. What happens if you approach the origin along the $x$-axis, with $y=0$ What happens if you approach the origin along the $y$-axis, with $x=0$? (You could also see what happens when you approach along the line $y=x$.)
By the way, a useful thing to remember is that when $x$ is very close to $0$, both $\sin x$ and $\tan x$ are very very close to $x$. Thus, to get a quick idea of what’s probably going on near the origin with this function, temporarily replace $\sin^2 y$ by $y^2$: it’s a good approximation when $y$ is close to $0$.
- | 277 | 884 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 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} | 3.65625 | 4 | CC-MAIN-2016-30 | latest | en | 0.925332 |
https://www.learncram.com/dav-solutions/dav-class-6-maths-chapter-15-worksheet-2/ | 1,716,961,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00862.warc.gz | 723,115,328 | 13,531 | # DAV Class 6 Maths Chapter 15 Worksheet 2 Solutions
The DAV Class 6 Maths Book Solutions and DAV Class 6 Maths Chapter 15 Worksheet 2 Solutions of Perimeter and Area offer comprehensive answers to textbook questions.
## DAV Class 6 Maths Ch 15 WS 2 Solutions
Question 1.
The area of a plastic sheet is 300 sq cm and its length is 30 cm. What is the breadth of the plastic sheet?
Area of the plastic sheet = 300 sq. cm
Its length = 30 cm
Breadth of the plastic sheet = $$\frac{\text { Area }}{\text { Length }}$$
= $$\frac{300}{30}$$
= 10 cm
Hence the required breadth = 10 cm.
Question 2.
The area of a football ground is 2900 m2. If its breadth is 29 m, find the length of the ground.
The area of the football ground
= 2900 m2
Length of the ground = $$\frac{\text { Area }}{\text { Breadth }}$$
= $$\frac{2900}{29}$$
= 100 m
Hence the required length = 100 m.
Question 3.
The perimeter of a square lawn is 72 m. Find the length of the side of the square lawn.
Perimeter of the square lawn = 72 m
Side = $$\frac{1}{4}$$ × Perimeter
= $$\frac{1}{4}$$ × 72 m
= 18 m
Hence the required length of side = 18 m.
Question 4.
A rectangular garden is 230 m long and its area is 13800 m2. Find the width of the garden.
Area of the rectangular garden = 13800 m2
Its length = 230 m
The width of the garden = $$\frac{\text { Area }}{\text { Length }}$$
= $$\frac{13800}{230}$$
= 60 m
Hence the required width = 60 m.
Question 5.
Neha covered 95.52 m by running around a square garden. Find the length of each side of the garden.
Perimeter of a square garden = 95.52 m
Side = $$\frac{\text { Perimeter }}{4}$$
= $$\frac{95.52}{4}$$
= 23.88 m
Hence the side of the square garden – 23.88 m
### DAV Class 6 Maths Chapter 15 Value Based Questions
Question 1.
On the occasion of Republic Day, a school organised a poster making competition for the students of Class VI. The topic was ‘National Integration’. Each child was given a square sheet of side 60 cm. They had to make a border of width 2 cm using the coloured tape and the remaining portion had to be divided into four equal portions to showT main teachings of different religions.
(a) Find the length of the coloured tape needed for the border.
Side of square sheet = 60 cm
The coloured taped needed for the border = Perimeter of the square
= 4 × side
= 4 × 60
= 240 cm
(b) What will be the side of the square into which the sheet has been divided?
The side of the square (ABCD) into which the sheet has been divided
= $$\frac{60-2 \times 2}{2}=\frac{60-4}{2}=\frac{56}{2}$$ = 28 cm
(c) What do you learn from the given situation?
To inculcate team work among children.
Question 2.
There was a deserted land near Rajesh’s house. The colony people use to throw garbage on that land. Rajesh along with his friends contacted the people of the colony and they all decided to develop the land into a park with a walking track where the children can play and elders can walk.
(a) Find the cost for levelling the park including the track at the rate of ₹ 15 per square metre. | 864 | 3,023 | {"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} | 4.84375 | 5 | CC-MAIN-2024-22 | latest | en | 0.846695 |
https://numberworld.info/2120001020 | 1,660,849,251,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00486.warc.gz | 392,016,604 | 4,006 | Number 2120001020
Properties of number 2120001020
Cross Sum:
Factorization:
2 * 2 * 5 * 137 * 773723
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
7e5ca5fc
Base 32:
1v5p9fs
sin(2120001020)
-0.93624132309549
cos(2120001020)
-0.35135763109459
tan(2120001020)
2.6646392172522
ln(2120001020)
21.474682406762
lg(2120001020)
9.3263360698817
sqrt(2120001020)
46043.468809376
Square(2120001020)
Number Look Up
Look Up
2120001020 (two billion one hundred twenty million one thousand twenty) is a impressive number. The cross sum of 2120001020 is 8. If you factorisate the figure 2120001020 you will get these result 2 * 2 * 5 * 137 * 773723. 2120001020 has 24 divisors ( 1, 2, 4, 5, 10, 20, 137, 274, 548, 685, 1370, 2740, 773723, 1547446, 3094892, 3868615, 7737230, 15474460, 106000051, 212000102, 424000204, 530000255, 1060000510, 2120001020 ) whith a sum of 4484504304. The figure 2120001020 is not a prime number. 2120001020 is not a fibonacci number. The number 2120001020 is not a Bell Number. 2120001020 is not a Catalan Number. The convertion of 2120001020 to base 2 (Binary) is 1111110010111001010010111111100. The convertion of 2120001020 to base 3 (Ternary) is 12110202011012200022. The convertion of 2120001020 to base 4 (Quaternary) is 1332113022113330. The convertion of 2120001020 to base 5 (Quintal) is 13320210013040. The convertion of 2120001020 to base 8 (Octal) is 17627122774. The convertion of 2120001020 to base 16 (Hexadecimal) is 7e5ca5fc. The convertion of 2120001020 to base 32 is 1v5p9fs. The sine of the figure 2120001020 is -0.93624132309549. The cosine of 2120001020 is -0.35135763109459. The tangent of the figure 2120001020 is 2.6646392172522. The square root of 2120001020 is 46043.468809376.
If you square 2120001020 you will get the following result 4494404324801040400. The natural logarithm of 2120001020 is 21.474682406762 and the decimal logarithm is 9.3263360698817. that 2120001020 is very great figure! | 789 | 2,087 | {"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} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.645565 |
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M313Lec11 - Math 313 Lecture #11 § 2.3: Cramer’s Rule...
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Unformatted text preview: Math 313 Lecture #11 § 2.3: Cramer’s Rule The Adjoint of a Matrix. In addition to defining and computing the determinant of a matrix, the cofactors of a matrix also play a role in another formula for the inverse of an invertible matrix. The adjoint of an n × n matrix A is the n × n matrix adj( A ) whose ( i, j ) entry is the cofactor A ji of A : adj A = A 11 A 21 . . . A n 1 A 12 A 22 . . . A n 2 . . . . . . . . . . . . A 1 n A 2 n . . . A nn . Notice that adj( A ) is the transpose of the matrix whose ( i, j ) entry is A ij . To see how this is related to the inverse of A when A is invertible, we need to recall the “technical lemma” from last time: a i 1 A j 1 + a i 2 A j 2 + · · · + a in A jn = det( A ) if i = j, if i 6 = j. The ( i, j ) entry of the product A (adj A ) is n X k =1 a ik A jk = det( A ) if i = j, if i 6 = j. Therefore each diagonal entry of A (adj A ) is det( A ) and the entries off the diagonal are zero: A (adj A ) = det(...
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M313Lec11 - Math 313 Lecture #11 § 2.3: Cramer’s Rule...
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Ask a homework question - tutors are online | 506 | 1,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2017-17 | longest | en | 0.825475 |
http://www.reddit.com/r/haskell/comments/2a5n8i/which_refinement_types_arent_structural/ | 1,408,797,527,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500826016.5/warc/CC-MAIN-20140820021346-00284-ip-10-180-136-8.ec2.internal.warc.gz | 575,133,999 | 12,916 | [–] 4 points5 points (0 children)
Like gridaphobe says, the "subtraction" is more like the relative complement of sets than it is like subtraction on numbers. While there are a few notions of negative types, they don't arise from refinement so far as I've seen.
As for non-structuralizable refinements, I think you'll have to be more specific about what you want. Refinement types à la ∑-types are perfectly structural. It seems like you want to adjust the "data" side of the ∑-type so as to eliminate the need for the "proof" side, but this requires clarification about what precisely counts as "data" (and therefore acceptable to keep around) vs what counts as "proof" (and therefore must be eliminated).
Broadly speaking, we can view refinements of `A` as the pair `(B,b)` where `B` is a type and `b : B -> A` is a monomorphism. This is somewhat evil since there will be many such pairs which are isomorphic, so really we should think about isomorphism classes of such monomorphisms. For the refinement type `{x:A | P x}` we can choose `B = ∑ A P` and `b = fst`. This suggests that as an alternative to defining what you mean by "structure", you could try looking for models of type theory which (a) don't have pullbacks, or which (b) don't have subobject classifiers.
[–] 0 points1 point (2 children)
What do you mean by "strictly negative" types?
[–][S] 1 point2 points (1 child)
I'm not really sure, this is just the sort of intuition I have, coming from the idea of "virtual species", where we could have the quotient type (Bool,x)/2, isomorphic to x, but we could also have xn/n!, isomorphic to an unordered bag, for the first case we can just cancel the 2 with bool, to get a concrete x, but in the second case, there's no non-quotiented type algebraic type for bags, so it's "virtual". In the refinement type case, if you think of predicates as "subtracting out the values that don't hold", we could "subtract out too much", so that we can only describe it as a type with a predicate. I have no idea if such a concept makes sense though.
[–] 6 points7 points (0 children)
I think your intuition about subtraction is a good place to start, but it's more like set subtraction than integer subtraction. The number of inhabitants of a type is always going to be non-negative. This does mean however that you can end up with types for which you cannot construct any values, a simple example is
``````{v:Int | v == 0 && v == 1}
``````
which can be read as the set of Ints `v` that are equal to both 0 and 1. Of course, this set is empty, so we would say the type has no inhabitants. Note that you can also define uninhabited types in Haskell, e.g.
``````data Empty
`````` | 683 | 2,691 | {"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} | 2.9375 | 3 | CC-MAIN-2014-35 | latest | en | 0.958375 |
https://phys.libretexts.org/Bookshelves/Quantum_Mechanics/Introductory_Quantum_Mechanics_(Fitzpatrick)/11%3A_Time-Independent_Perturbation_Theory/11.09%3A_Zeeman_Effect | 1,725,872,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00169.warc.gz | 429,900,798 | 30,979 | # 11.9: Zeeman Effect
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Consider a hydrogen atom placed in a uniform $$z$$-directed external magnetic field of magnitude $$|{\bf B}|$$. The modification to the Hamiltonian of the system is $H_1 = -\mu\cdot{\bf B},$ where $\mu = - \frac{e}{2\,m_e}\,({\bf L} + 2\,{\bf S})$ is the total electron magnetic moment, including both orbital and spin contributions. [See Equations ([e10.57])–([e10.59]).] Thus, $H_1 = \frac{e\,B}{2\,m_e}\,(L_z+ 2\,S_z).$
Suppose that the applied magnetic field is much weaker than the atom’s internal magnetic field, ([e12.124]). Because the magnitude of the internal field is about 25 tesla, this is a fairly reasonable assumption. In this situation, we can treat $$H_1$$ as a small perturbation acting on the simultaneous eigenstates of the unperturbed Hamiltonian and the fine structure Hamiltonian. Of course, these states are the simultaneous eigenstates of $$L^2$$, $$S^2$$, $$J^{\,2}$$, and $$J_z$$. (See the previous section.) Hence, from standard perturbation theory, the first-order energy-shift induced by a weak external magnetic field is \begin{aligned} {\mit\Delta} E_{l,1/2;j,m_j} &= \langle l,1/2;j,m_j|H_1|l,1/2;j,m_j\rangle\nonumber\\[0.5ex] &= \frac{e\,B}{2\,m_e}\,\left(m_j\,\hbar + \langle l,1/2;j,m_j|S_z|l,1/2;j,m_j\rangle\right),\end{aligned} because $$J_z=L_z+S_z$$. Now, according to Equations ([e11.47]) and ([e11.48]),
$\label{e12.143} \psi^{(2)}_{j,m_j} = \left(\frac{j+m_j}{2\,l+1}\right)^{1/2}\psi^{(1)}_{m_j-1/2,1/2} + \left(\frac{j-m_j}{2\,l+1}\right)^{1/2}\,\psi^{(1)}_{m_j+1/2,-1/2}$ when $$j=l+1/2$$, and $\psi^{(2)}_{j,m_j} = \left(\frac{j+1-m_j}{2\,l+1}\right)^{1/2}\psi^{(1)}_{m_j-1/2,1/2} - \left(\frac{j+1+m_j}{2\,l+1}\right)^{1/2}\,\psi^{(1)}_{m_j+1/2,-1/2}$ when $$j=l-1/2$$. Here, the $$\psi^{(1)}_{m,m_s}$$ are the simultaneous eigenstates of $$L^2$$, $$S^2$$, $$L_z$$, and $$S_z$$, whereas the $$\psi^{(2)}_{j,m_j}$$ are the simultaneous eigenstates of $$L^2$$, $$S^2$$, $$J^{\,2}$$, and $$J_z$$. In particular, $\label{e12.145} S_z\,\psi^{(1)}_{m,\pm 1/2} = \pm \frac{\hbar}{2}\,\psi^{(1)}_{m,\pm 1/2}.$ It follows from Equations ([e12.143])–([e12.145]), and the orthormality of the $$\psi^{(1)}$$, that $\langle l,1/2;j,m_j|S_z|l,1/2;j,m_j\rangle = \pm \frac{m_j\,\hbar}{2\,l+1}$ when $$j=l\pm 1/2$$. Thus, the induced energy-shift when a hydrogen atom is placed in an external magnetic field—which is known as the Zeeman effect —becomes $\label{e12.147} {\mit\Delta} E_{l,1/2;j,m_j} = \mu_B\,B\,m_j\left(1\pm \frac{1}{2\,l+1}\right)$ where the $$\pm$$ signs correspond to $$j=l\pm 1/2$$. Here, $\mu_B = \frac{e\,\hbar}{2\,m_e} = 5.788\times 10^{-5}\,{\rm eV/T}$ is known as the Bohr magnetron. Of course, the quantum number $$m_j$$ takes values differing by unity in the range $$-j$$ to $$j$$. It, thus, follows from Equation ([e12.147]) that the Zeeman effect splits degenerate states characterized by $$j=l+1/2$$ into $$2\,j+1$$ equally spaced states of interstate spacing $\label{e12.149} {\mit\Delta} E_{j=l+1/2} = \mu_B\,B\left(\frac{2\,l+2}{2\,l+1}\right).$ Likewise, the Zeeman effect splits degenerate states characterized by $$j=l-1/2$$ into $$2\,j+1$$ equally spaced states of interstate spacing $\label{e12.150} {\mit\Delta} E_{j=l-1/2} = \mu_B\,B\left(\frac{2\,l}{2\,l+1}\right).$
In conclusion, in the presence of a weak external magnetic field, the two degenerate $$1S_{1/2}$$ states of the hydrogen atom are split by $$2\,\mu_B\,B$$. Likewise, the four degenerate $$2S_{1/2}$$ and $$2P_{1/2}$$ states are split by $$(2/3)\,\mu_B\,B$$, whereas the four degenerate $$2P_{3/2}$$ states are split by $$(4/3)\,\mu_B\,B$$. This is illustrated in Figure [fzee]. Note, finally, that because the $$\psi^{(2)}_{l,m_j}$$ are not simultaneous eigenstates of the unperturbed and perturbing Hamiltonians, Equations ([e12.149]) and ([e12.150]) can only be regarded as the expectation values of the magnetic-field induced energy-shifts. However, as long as the external magnetic field is much weaker than the internal magnetic field, these expectation values are almost identical to the actual measured values of the energy-shifts.
Figure 24: The Zeeman effect for the and states of a hydrogen atom. Here, $$$$\epsilon=\mu_{B} B$$$$. Not to scale.
• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
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This page titled 11.9: Zeeman Effect is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick. | 3,506 | 9,250 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-38 | latest | en | 0.194723 |
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by Activities for Learning
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Super Corners is a game of strategy, adding numbers so that the total is a multiple of 5 (5, 10, 15, and 20) while matching colors. Variations are also available for multiplies of 4 (4, 8, 12, 16, and 20) and multiples of 3 (3, 6, 9, 12, 15, and 18). Fifty cards are in each game and final scores are stored.
Version
1.1
Rating
NA
Size
59Mb
Genre
Education
Last updated
April 18, 2018
Release date
April 13, 2018
App Store Description
Super Corners is a game of strategy, adding numbers so that the total is a multiple of 5 (5, 10, 15, and 20) while matching colors. Variations are also available for multiplies of 4 (4, 8, 12, 16, and 20) and multiples of 3 (3, 6, 9, 12, 15, and 18). Fifty cards are in each game and final scores are stored.
Cards can be played at any position, provided the two numbers touching have the same color and add up to the chosen multiple.
When playing with the multiples of 5, first consider the green 4 on the top. What is needed to make 10, 15, or 20? Look at the five cards below; the only possibility is the green 6. Next, look at the red 10. There is one option with the red 5. The blue 3 has the option of the blue 7 and the black 5 has no options.
Click the chosen card and the spot on the board to place the card. The light green squares indicate available places to play a card.
When two cards form a corner, each touching numbers must match colors and add up to the multiple. Look at the last card on the right and the corner. The black 5s make 10 and the blue 7 and 8 make 15. Score will be 25, however because this is a CORNER, the score of 25 is multiplied by TWO for a total of 50 points.
When three cards create a BAY and a card is placed, the total of the three pairs is multiplied by THREE. Four cards create a WINDOW and the total of the four pairs is multiplied by FOUR.
Disclaimer:
AppAdvice does not own this application and only provides images and links contained in the iTunes Search API, to help our users find the best apps to download. If you are the developer of this app and would like your information removed, please send a request to [email protected] and your information will be removed. | 598 | 2,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-22 | longest | en | 0.925066 |
http://ask.learncbse.in/t/guess-the-value-of-the-limit-of-f-x-as-x-approaches-3-correct-to-six-decimal-places-if-an-answer-does-not-exist-enter-dne/32828 | 1,537,650,172,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158691.56/warc/CC-MAIN-20180922201637-20180922222037-00531.warc.gz | 19,617,567 | 2,891 | # Guess the value of the limit of f(x) as x approaches 3, correct to six decimal places. (If an answer does not exist, enter DNE.)
#1
Guess the value of the limit of f(x) as x approaches 3, correct to six decimal places. (If an answer does not exist, enter DNE.)
lim
x→3
x2 − 3x
x2 − x − 6 | 93 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-39 | latest | en | 0.844898 |
http://www.sciforums.com/showthread.php?113621-The-Impossibility-of-Knowing-Your-Own-Future/page7&p=2937812 | 1,394,527,026,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011160422/warc/CC-MAIN-20140305091920-00032-ip-10-183-142-35.ec2.internal.warc.gz | 517,711,337 | 24,387 | 1. Originally Posted by Neverfly
This is confusing me. My perspective is that any message need only be written ONCE. You would not have messages within messages simply because only one message would be produced.
That message being the final determinant result.
Anything else before that is just factoring.
But the message is part of the future, therefor we can't write the message only once. We would have to write the message and account for the events that he message itself has every moment onward. We can't write a message about the future in such a way that the message itself isn't part of it.
I may have created confusion here.
My point on the Forth Dimensional perspective was not relevant to the topic at hand directly.
It was an exercise to show how your mind works and figures things out and how it can fall short on topics of which it is not accustomed to thinking a certain way.
WE ALL suffer from this, self included, and my example could be applied to me as much as you.
Ok, fair enough.
This is why we are bouncing off of eachother, both of our minds work this way and the physics works its own way.
This is fundamentally inaccurate to determinism.
In determinism, the future is "Known" even if 'known' is the wrong applicable word. Determinant is the right word.
This being the case, what you just described is in error.
The future is 'known' but not written as a message that can be read by us.
The message recorded would show only the path of determinant trajectories.
So if I measure the path of billiard ball A and determine it's future trajectory (In a deterministic fashion, covering all factors that can influence the outcome) and then record that on paper- You can call that a prediction.
I then pick up the stick and call my shot- and shoot.
The prediction WILL come true.
Yes, I'm with you on this. As long as you shoot exactly as you predicted then the effects are going to be what you predicted (assuming that you followed each trajectory without error and also made no errors in the prediction).
In your example, you say you can choose to change the outcome by knowing the prediction.
But here's the clincher;
That projection, that trajectory- would not have been predicted.
The physical process that work in your brain are determinant as well. They would have been factors in the creation of that piece of paper with the message.
Had you been so motivated to await the prediction to be printed on paper so that you could choose against it, the prediction would have been altered to show whatever inevitable result would happen.
The original prediction in the thought experiment would never have existed.
I know, IF a prediction could be made THEN it would be accurate, but my argument is that the prediction can't be made in the first place.
We could predict all the events in another solar system, but as soon as we give them the message, then that message has to be predicted as well. If we didn't give them the message then everything would play out as predicted, but as soon as the prediction has any influence on their future then it can't be predicted because we would have to have the full message before it was completed (as the message is only a part of the future).
At this point you say a prediction cannot be reached, simply because any prediction finally printed out on paper would affect the next outcome.
Yes, so we would need to factor in the prediction before it was fully made. The prediction that we predicted would only be as completed as it is at the moment we need it. This is a real consequence that can't be escaped. We would have to have a full prediction in order to accurately describe the prediction itself.
And at this point I say that it can be reached just fine- even if our minds cannot fathom the events- because it would have determined what the result would be, based on our brains as a factor- the trajectories as a factor- and the prediction would be perhaps one were you could not alter events or decided not to alter events-- essentially, whatever the outcome of it all would be.
We can complicate things and talk about trajectories and other such things, but knowledge about if the message can be completed doesn't have to include all those trajectories, we need only one example where such a message can't be created and that is that it can't predict itself and how it influences the events. It could never be reached as the fully predicted message would have to be done in order to describe itself.
Yes but still not relevant to Determinism because in determinism, the future is "known" (actually determinant) all the way up til the end of time. Whatever that end may be.
Determinism doesn't say that we can know the future, it simply says that there is a future which is already determinant - it doesn't say that we can actually know it, because of that it is still relevant to determinism and doesn't contradict it.
Yes, my only point was that the names cannot be treated as the physical.
If the names are accurate to the physical then it can describe the physical. Otherwise the physical couldn't be described at all if we never would be allowed to use names for physical entities.
Yay! I have the last word (for a while at least)!
2. Originally Posted by Cyperium
But the message is part of the future, therefor we can't write the message only once. We would have to write the message and account for the events that he message itself has every moment onward. We can't write a message about the future in such a way that the message itself isn't part of it.
Why would a message have to be written down in this manner? It makes no sense.
Now, I apologize. The distraction of the Zimmerman thread has had me off kilter for a while (Squirrel and Yazata, if you would like to see me eat some tasty crow and tolerate HEAVY ABUSE, convoluted intellectual dishonesty and misquoting... read that thread- you'll love it)
I really haven't had time to think about this topic and really work over how I understand it as opposed to your view. Give me some time to catch up.
So... gonna try here:
If the determinant result is written on a message - then the prediction has been concluded.
The factors the play into the result do not need to be known by us, to be written down as successive predictions within predictions or messages within messages.
The process of measuring the trajectories would yield the inevitable result- not POSSIBLE results, depending on how you react.
Since the future is determinant (Or can be termed loosely as 'known') before this, then it does not matter how you react- only an act of Free Will can change it.
Originally Posted by Cyperium
Yes, I'm with you on this. As long as you shoot exactly as you predicted then the effects are going to be what you predicted (assuming that you followed each trajectory without error and also made no errors in the prediction).
This assumes I was not told the outcome of the prediction.
Rather, not IF I shoot...- I WILL shoot exactly as predicted. I have neither choice nor say in the matter. To do otherwise must be an Act Of Free Will.
But, Cyperium, It doesn't matter either way, man!
You seem to be thinking that our trajectories are determinant ONLY because we do not know what they are. That if we did know, we could somehow alter the physics of the Universe.
We cannot alter the physics of the Universe. We can only abide by our properties.
So if we assume I am told the prediction, I would not shoot the same way you say...
Well, I agree a different prediction would be reached- the prediction in the first scenario would never have been created, considered- it simply wouldn't have happened.
That the prediction would happen was a factor in the process of making since the origin of time. The factors of me trying to change it- also determinant from the origin of time.
They would not have Suddenly Appeared In The Universe when you saw the message, as you appear to be thinking.
The ONLY message that would exist would be what the result was of those factors. One message.
One prediction.
Only an act of Free Will could change it.
3. Originally Posted by Neverfly
(Squirrel and Yazata, if you would like to see me eat some tasty crow and tolerate HEAVY ABUSE, convoluted intellectual dishonesty and misquoting... read that thread- you'll love it)
I don't care..
(tried to elaborate and make this long winded..but this is what it boils down to, so i'll just leave it at that..)
anyway..
I think Issac Asimov tried to address this in his foundation series (I read them all)
he posited Psycohistory, where the action of society as a whole can be predicted, but the individual cannot be predicted (the Mule)
which asks if free will is just limited to the individual.
my point with this post is shouldn't we distinguish in the argument between the individual and the society??
4. Prediction: I will return to this post at 12:35pm today and place a small-case "o" into the following space between the parentheses: (o)
5. Post #124 has successfully refuted the OP's claims.
next.
6. Originally Posted by keith1
Post #124 has successfully refuted the OP's claims.
next.
um..
Last edited by keith1; Today at 01:35 PM
7. Originally Posted by NMSquirrel
um..
A page snapshot from my timezone (Pacific U.S.) is easily documented if required. It's all a part of the prediction strategy. The prediction will hold up as valid, and the OP remains refuted.
next.
8. Originally Posted by NMSquirrel
I don't care..
(tried to elaborate and make this long winded..but this is what it boils down to, so i'll just leave it at that..)
After all your whining and complaining- You don't care. LOL you shoulda just said that the first time instead of whining and complaining.
Originally Posted by NMSquirrel
I think Issac Asimov tried to address this in his foundation series (I read them all)
he posited Psycohistory, where the action of society as a whole can be predicted, but the individual cannot be predicted (the Mule)
which asks if free will is just limited to the individual.
my point with this post is shouldn't we distinguish in the argument between the individual and the society??
Fictitious nonsense and utterly irrelevant to the topic. I have read all the foundation series too.
ETA SPOILERS
And incidentally, the reason Psychohistory could not predict small motions was because the method of measurement relied on group activities and trends as a whole. It was not determined, it was statistical, based on averages.
"The Mule" disrupted that because he was not a part of the Human Society. He was a member of Gaia, which was an isolated SuperOrganism planet experiment of robot Daneel Olivaw. Gaia was unknown about by Hari Seldon. Daneel Olivaw also took great pains to keep the Gaia experiment separate from the psychohistory experiment.
It all stemmed from Asimovs three laws- Olivaw and another robot formulated a "Forth law" that demanded that humanity could not come to harm, by action or inaction. Even from itself. This inspired them to form the two experiments in order to influence events directly themselves to protect humanity as a whole.
As you can see, even in fiction, none of it related to actual predictions.
Originally Posted by NMSquirrel
um..
Says 1:35 on mine, Squirrel... Different Time Zones.
9. Originally Posted by Neverfly
...Says 1:35 on mine, Squirrel... Different Time Zones.
It helped that there was a time stamp on the prediction initiating phase. And a time stamp at the conclusion, and one at the verifying station(s) (Squirrel/Never).
Interesting symptomatic condition of dealing with prediction and time frames?
I should think so.
But there were interesting hang-ups at my end at prediction initiation (PI), which I will divulge, so as other minds can find some erroneous condition, which might salvage the OP's claims:
The PI time stamp remains the same (1:26pmPST). The message was altered and resent (at 12:35PST, instead of the originally planned 12:30PST), to compensate for the fact that the "last edited" (LE) time stamp fails, if the time interval between "message sent" and "first opportunity to engage LE time stamp" is not long enough. However, I do not believe this act changed the results of refuting the OP.
--The short time interval lessened the complexity level, even though the simplification makes the basic tenets more easy to examine.
--Some room for ease in the technical equipment advantage.
--One should have an onsite observer to witness the "o" being place at the promised time. This is a minor issue and one not questioned by the station observers, as the LE time stamp does show the required "o" in place at that time, as promised in the prediction eventually observed.
10. Using analogies from the movie "Back to the Future":
The similarity of the "o" to the lightening
and the time 12:35 PM PST to the clock,
Where I was given more play in adjusting the event parameters (space and time coordinates), to meet the intended event prediction.
11. Originally Posted by Neverfly
Why would a message have to be written down in this manner? It makes no sense.
Because your future has the message in it. How could you know your future without the message? How could the prediction account for your knowledge of the future (that is something that will have to be accounted for as part of the prediction), if the prediction doesn't account for the message?
Now, I apologize. The distraction of the Zimmerman thread has had me off kilter for a while (Squirrel and Yazata, if you would like to see me eat some tasty crow and tolerate HEAVY ABUSE, convoluted intellectual dishonesty and misquoting... read that thread- you'll love it)
I really haven't had time to think about this topic and really work over how I understand it as opposed to your view. Give me some time to catch up.
Sure.
So... gonna try here:
If the determinant result is written on a message - then the prediction has been concluded.
But how could it be concluded when the result has to cover the message as it is within your future? Everything must be accounted for in order for it to be a true prediction, we can't leave the message out of the prediction as the message influences you to a great deal.
The factors the play into the result do not need to be known by us, to be written down as successive predictions within predictions or messages within messages.
Then you must agree that we can only predict a future that doesn't itself have the prediction within it. To be frank, you say that the prediction itself doesn't need to be in the message, but the prediction exists in the future as the full prediction, while before we got to the full prediction we would have just parts of it. Because of that the prediction will not be able to predict your reaction to the message, since it doesn't have the full message when it gets to that point.
The process of measuring the trajectories would yield the inevitable result- not POSSIBLE results, depending on how you react.
Since the future is determinant (Or can be termed loosely as 'known') before this, then it does not matter how you react- only an act of Free Will can change it.
All trajectories that lead to the message itself will also have to be measured, already we have a message within the message. Every act of measuring, in the present, will also have to be considered in order to account for anything in the future. It is a futile attempt.
To have a true prediction of the future, we can't only look at the variables outside of the measuring process, we would have to account also for the measuring. This clearly shows a infinite loop, the message already has to account for the message even at this stage.
This assumes I was not told the outcome of the prediction.
Rather, not IF I shoot...- I WILL shoot exactly as predicted. I have neither choice nor say in the matter. To do otherwise must be an Act Of Free Will.
But, Cyperium, It doesn't matter either way, man!
You seem to be thinking that our trajectories are determinant ONLY because we do not know what they are. That if we did know, we could somehow alter the physics of the Universe.
We cannot alter the physics of the Universe. We can only abide by our properties.
So if we assume I am told the prediction, I would not shoot the same way you say...
Well, I agree a different prediction would be reached- the prediction in the first scenario would never have been created, considered- it simply wouldn't have happened.
That the prediction would happen was a factor in the process of making since the origin of time. The factors of me trying to change it- also determinant from the origin of time.
They would not have Suddenly Appeared In The Universe when you saw the message, as you appear to be thinking.
The ONLY message that would exist would be what the result was of those factors. One message.
One prediction.
Only an act of Free Will could change it.
I don't think that, but you conclude that the message can even be made, while I conclude that it can't be made, if there is no message then there is no knowledge about it to consider.
Originally Posted by keith1
Post #124 has successfully refuted the OP's claims.
next.
This is knowing the past, not the future, only when you succeeded in posting that specific time did your prediction come true. Before that anything could have happened to prevent you from posting exactly that time. You could say that it was indeed a true prediction - but we would have to take your word for it, saying that it was a true prediction is only true when it actually happened as you predicted, before that we would have to take it on faith that you were actually able to make a solid prediction before it happened.
12. Originally Posted by Cyperium
...This is knowing the past, not the future, only when you succeeded in posting that specific time did your prediction come true. Before that anything could have happened to prevent you from posting exactly that time....
It is the quality of the prediction that seals the level of it's successful fate.
In a qualitative sense, the odds were very good for repetitive success with that particular prediction. If I added time, those odds would diminish. Who knows if there will be an internet in 100 years, to support the transportation of my lengthy prediction. I would have to increase the predictions odds of success by careful construct.
...You could say that it was indeed a true prediction - but we would have to take your word for it, saying that it was a true prediction is only true when it actually happened as you predicted, before that we would have to take it on faith that you were actually able to make a solid prediction before it happened...
You have my word, many time checks, and a correct and solid result. You cannot ask for more, and not look impossible to please.
You will need to choose less "viscosity" in your sentences that oil your "appearance of wisdom" around areas you are plainly at a loss for "validity".
It has been a slippery walk with you, starting with the OP's first sentence.
Maybe we should restart there:
"...The Impossibility of Knowing Your Own Future...Because by knowing it you will not be able to keep to it, it would be like balancing on a infinitely thin line...."
I demand a more comprehensible rewrite.
13. Originally Posted by Neverfly
After all your whining and complaining- You don't care. LOL you shoulda just said that the first time instead of whining and complaining.
let me clarify..
I don't care that you "eat some tasty crow and tolerate HEAVY ABUSE, convoluted intellectual dishonesty and misquoting..."
so your point of "you'll love it" is irrelevant..
Fictitious nonsense and utterly irrelevant to the topic.
just because you say so?(you expect me to just take your word for it?)
you have claimed that the mind can be determined..so now you argue against a possible means?
this is an intelligent exercise,to which a conclusion can never be ascertained.
plus where is you claims of fictitious nonsense to the square circle and the immovable rock?
how does that go? without any proof to the contrary,then any hypothesis can be possible
(I know..i seriously misquoted that one...can someone correct it?)
14. Originally Posted by keith1
It is the quality of the prediction that seals the level of it's successful fate.
In a qualitative sense, the odds were very good for repetitive success with that particular prediction. If I added time, those odds would diminish. Who knows if there will be an internet in 100 years, to support the transportation of my lengthy prediction. I would have to increase the predictions odds of success by careful construct.
True, but it wouldn't apply as knowledge in the strictest sense. You believed that you were going to be able to make that post at exactly that time. When I say "know the future" then I'm using "know" in the strictest sense, not just being "very certain" of something. I mean you could have chosen to break the prediction instead, would it then count as knowledge? Some event could have taken place when you decided to post it so that you lost your connection or something, did you in fact know you were going to post it at exactly that time? Or were you just very certain of it?
You have my word, many time checks, and a correct and solid result. You cannot ask for more, and not look impossible to please.
What does "and not look impossible to please" mean? It wasn't impossible to break your prediction, in fact it could have been easily done just choosing not to follow through with it.
Let's say that you predict each footstep you take in the next day, how easy would it be to follow that prediction?
You will need to choose less "viscosity" in your sentences that oil your "appearance of wisdom" around areas you are plainly at a loss for "validity".
It has been a slippery walk with you, starting with the OP's first sentence.
Maybe we should restart there:
I demand a more comprehensible rewrite.
I don't see the trouble with that quote. To know is to know, if you don't know then you don't. This thread assumes that you would know the future, not just make a guess. But if you have trouble with it yet, then go ask at the linguistic forum instead, I have no wish to discuss the fluidity of words with you as it is straying the topic.
15. Originally Posted by Cyperium
All trajectories that lead to the message itself will also have to be measured, already we have a message within the message. Every act of measuring, in the present, will also have to be considered in order to account for anything in the future. It is a futile attempt.
None of what you just described creates a futile attempt.
Let's quote myself and address this one bit:
That the prediction would happen was a factor in the process of making since the origin of time. The factors of me trying to change it- also determinant from the origin of time.
They would not have Suddenly Appeared In The Universe when you saw the message, as you appear to be thinking.
The ONLY message that would exist would be what the result was of those factors. One message.
One prediction.
Only an act of Free Will could change it.
If the determinant factors have been bouncing toward the outcome of the measurements being made by a human and the determining factors bouncing toward the resultant prediction since the origin of time; please explain as best you can how it is that a message can ALTER trajectories already in motion?
Can the "message" add or remove causality?
Or is the "message" non causa pro causa?
Off topic Whining of my own:
Originally Posted by NMSquirrel
let me clarify..
I don't care that you "eat some tasty crow and tolerate HEAVY ABUSE, convoluted intellectual dishonesty and misquoting..."
so your point of "you'll love it" is irrelevant..
You cared a great deal when you were demanding that I consider I can be wrong.
I told it's possible and I've had to eat my share of crow.
You cared a great deal when you claimed I was so heavy handed.
And I said there's been far worse on the forums...
So I show you that I, Too, must deal with "Heavy handed" replies and I have shown what you asked- I DO sometimes eat crow.
But since it wasn't you- you do not care.
I think this is very telling about all that whining you and Yazata did, trying to explain to me how I am supposed to TREAT YOU.
So here's the deal, Squirrel, I conceded to make an effort to not stomp on your poor little toes. Consider that rescinded. Buck up with real science if you're going to participate or take the heavy handedness without crying like a baby demanding to be pampered.
Back on topic, again...
Originally Posted by NMSquirrel
just because you say so?(you expect me to just take your word for it?)
you have claimed that the mind can be determined..so now you argue against a possible means?
this is an intelligent exercise,to which a conclusion can never be ascertained.
plus where is you claims of fictitious nonsense to the square circle and the immovable rock?
how does that go? without any proof to the contrary,then any hypothesis can be possible
(I know..i seriously misquoted that one...can someone correct it?)
More gobbledogook and total nonsense. I spelled out what's in the fiction clearly in the post that you quoted.
Post an intelligent response, one well thought out and researched with scientific papers and not works of complete fiction and your liable to get a better reply.
16. Some event could have taken place when you decided to post it so that you lost your connection or something, did you in fact know you were going to post it at exactly that time? Or were you just very certain of it?
Yeah, I was certain of it, because of experience of doing it many times before, and expected no other outcome. Knowing by rote.
Let's say that you predict each footstep you take in the next day, how easy would it be to follow that prediction?
It would take planning it out ahead of time. Another aspect of tightening the parameters of the prediction to ensure a desired outcome.
Such acts are the mainstay of Guiness Book of Record feats. Difficult, but doable.
I didn't mean to knock your communication skills.
I like the subject. That's why I was persistent in understanding your intended gist.
I can know the present, which is the past's future.
It is futile to try to adjust the future without the technology to approach the issue.
I feel sorry for the last generation having to set in horse wagons and know rockets were coming into the conversation.
I feel sorry for our generation in the same way.
17. Originally Posted by Neverfly
None of what you just described creates a futile attempt.
Let's quote myself and address this one bit:
If the determinant factors have been bouncing toward the outcome of the measurements being made by a human and the determining factors bouncing toward the resultant prediction since the origin of time; please explain as best you can how it is that a message can ALTER trajectories already in motion?
Can the "message" add or remove causality?
Or is the "message" non causa pro causa?
It isn't that it removes causiality, it's that it has to account for its own causiality - in the present - when the message is made. I don't think that there is a system to measure in such a way that the measurement doesn't need to be accounted for. This is about the ability to create the message in the first place, not if causiality can be broken (which it obviously can't in a causually determined universe).
Originally Posted by keith1
Yeah, I was certain of it, because of experience of doing it many times before, and expected no other outcome. Knowing by rote.
Yes, but that is insufficient knowledge. There hasn't been any confusion as to whether or not it was actual knowledge about the future. Everyone that has responded has been clear that it was knowledge and not educated guessing - it is a requirement for the idea.
It would take planning it out ahead of time. Another aspect of tightening the parameters of the prediction to ensure a desired outcome.
Such acts are the mainstay of Guiness Book of Record feats. Difficult, but doable.
I didn't mean to knock your communication skills.
I like the subject. That's why I was persistent in understanding your intended gist.
I can know the present, which is the past's future.
It is futile to try to adjust the future without the technology to approach the issue.
I feel sorry for the last generation having to set in horse wagons and know rockets were coming into the conversation.
I feel sorry for our generation in the same way.
I could probably be more clear at some points, I didn't think it was necessary and having to think about the possible 'other' meanings of each word or concept I use would make the posts very hard to write. I'm kind of convenient in that way.
You have a good point, we don't have the technology and because of that we wouldn't know how the present could be measured in such a way that we would know the future without being able to change it because of our knowledge of it. I have found, though, through some simple thought experiments that it should be impossible. Each measurement of the present, is a event that has to be measured itself. In order to accurately account for all the events, we would have to account also for us measuring them and also plotting them into a medium that can be read later.
Another, easier way to see it is that if we have a message that contains the information about the future, IF that future has the message in it then the prediction can't account for the message itself, as the future has the full message, and the prediction hasn't got to the full message yet (as it is the message).
18. cyberium..
it seems you are talking Schrodinger..the act of measuring changes the outcome..
19. Originally Posted by Cyperium
It isn't that it removes causiality, it's that it has to account for its own causiality - in the present - when the message is made. I don't think that there is a system to measure in such a way that the measurement doesn't need to be accounted for. This is about the ability to create the message in the first place, not if causiality can be broken (which it obviously can't in a causually determined universe).
OK, agreed then, we cannot break causality.
No causality is added nor removed.
So how is there a paradox?
No one said the measurement doesn't need to be accounted for- in fact, I've been saying the exact opposite all throughout- Why are you suddenly claiming the opposite of what I have been saying? It's falling under ALL the factors. I've said and you have quoted me, the message is accounted for. It must be - it cannot NOT be accounted for.
About the only thing you've said that is unrelated is your concept of time. Which is irrelevant in a determinant universe.
You have not answered the question- not explained what can alter a set trajectory. You have not described a system in which causality can be altered in any way.
20. Originally Posted by NMSquirrel
cyberium..
it seems you are talking Schrodinger..the act of measuring changes the outcome..
It seems that way, but I haven't looked into that area, my idea assumes that all events can be measured. But that is also what causes the infinite loop when producing the message.
If certain interpretations of quantum mechanics are true then we don't have a predetermined universe, or at least not all events can actually be measured (like we can't measure both position and momentum at 100% accuracy).
Originally Posted by Neverfly
OK, agreed then, we cannot break causality.
No causality is added nor removed.
So how is there a paradox?
Because the message can't exist in the first place. If it would then that would create a paradox (for all the reasons I've stated). We would have a infinite message (as the message is in the future and would have to account for itself), also I still think that it is impossible to know everything about the future without being able to change at least what we are capable of changing. There is no limit on the message itself on what we can know about it. This is actually due to the inability of the prediction to account for the message and how that itself can have effects on the future. So it's not far off to show that the message can't be done to show that the knowledge of the contents of the message would enable us to change the future - it is simply one of the effects that the prediction can't account for as it doesn't have the full message before it is completed, because of that it can't account for the knowledge we would have about the message either.
No one said the measurement doesn't need to be accounted for- in fact, I've been saying the exact opposite all throughout- Why are you suddenly claiming the opposite of what I have been saying? It's falling under ALL the factors. I've said and you have quoted me, the message is accounted for. It must be - it cannot NOT be accounted for.
If there is a impossibility to account for the message in the first place then that certainly has to be considered. But taken that the message could in fact exist, then we get the paradox that we know the future but are unable to change even those things that should be easy to change. How can we have a entire future written down and easy to read where no event could be easy to change because of our knowledge of it? If not anything else we would feel very restricted. But it boils down to the fact that the message can't account for itself. There's no reason to assume the impossible.
About the only thing you've said that is unrelated is your concept of time. Which is irrelevant in a determinant universe.
You have not answered the question- not explained what can alter a set trajectory. You have not described a system in which causality can be altered in any way.
I haven't described it because that system would be impossible - if the message can't exist, then there would be no system to be able to hold it. It would be a infinite message where any knowledge of the message and the consequences thereof would fail as the prediction got there (because that is when it has to account for itself).
So in other words; the answer is that the system that could alter the future would have the message within itself in infinite regress, it couldn't be read as it couldn't escape the moment we read it (as that moment is when the message would have to be accounted for). If we were to just skip the message and the information of us reading it then we would have to skip a infinity of pages.
However, we can read about the past, what has happened since the message was made and until the moment we read the message, but when we get to the place in the message where we start reading the message (or when the message was handed to us) then it would be just a repetition of the message up until that point (because the message could be made so that the message itself wasn't accounted for, it could describe a system where the message itself has no influence on the future). Because of this we can't know our own future but there is a possibility that we can know the future of other people, or even the entire universe, as long as we don't know anything that the message itself could influence, because then we would have the infinite regress.
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# 20 mg equals how many grams
• 1/50 grams
the mass 1/50 grams
• tk10publ tk10canl
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• convert 20miligram to gram | 452 | 1,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2014-15 | longest | en | 0.845917 |
https://lists.dlitz.net/pipermail/pycrypto/2012q4/000635.html | 1,660,626,200,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00435.warc.gz | 358,160,909 | 3,231 | # [pycrypto] Confused about some code in PubKey/RSA/_slowmath.py
Lorenz Quack don at amberfisharts.com
Tue Nov 6 14:59:44 EST 2012
```Hi Shoufu,
first things first:
A) Cryptography is *very* hard to get right even if use a crypto library. So if
you ask these questions out of curiosity or educational purposes that is fine but
please don't use your own crypto in production code where you really need security!
B) I'm myself not a cryptography expert so take my answers with a grain of salt.
First some general explanations that might be helpful.
For answers to your actual questions scroll down to the "---" mark.
From a mathematical standpoint encryption and decryption are the same operation.
Namely raise a number (this is either your plain text "m" or your cipher text "c")
to an exponent modulo a large number (usually called "n"). The only difference is
what you take as the exponent. So in RSA you have two different exponents a private
one (let us call it "d") and a public one (let us call it "e"). combined with "n"
these are in essence your private and public key. So now we have one mathematical
operation but four variables ("m", "c", "d", and "e") giving you these combinations:
1) m**e mod n --> encryption
2) m**d mod n --> sign
3) c**e mod n --> verify
4) c**d mod n --> decryption
---
so to answer your first question:
"encrypting" using your private key is called signing. Think of it this way. if you
"encrypt" with your private key everybody would be able to decrypt it because what
could be done with your public key which is ... well *public*
as for the second question:
this question seems to stem from the same misconception as the first one.
encryption happens with the *public* key and decryption with the *private* key.
Everybody (i.e. the public) is allowed to send you encrypted messages but only you
should be able to decrypt them in private!
I hope that answers your questions.
Sincerely yours,
Lorenz
On 11/05/2012 06:26 AM, Shoufu Luo wrote:
> Hi all,
>
> I'm trying to encrypt a message with my private key and release the encrypted to
> others who will use my public key to decrypt. But, I failed.
>
> I was confused by the follwing code from PubKey/RSA/_slowmath.py of pyCrypto-2.6.
> If anyone can give any clues to answer the following questions, I will appreciate.
>
> 1. Theoretically, if I encrypt date using private key, I can decrypt the encrypted
> data using public key, and vice versa. Why the key must be a private key in
> decryption function, line 51-52? Can't I use private key to decrypt?
>
> 2. _sign() should be the signature process using private key to encrypt a piece of
> data, why it is trying to decrypt at line 70, and it should be decryption in
> '_verify', but why it is _encrypt()?
>
> 49 def _decrypt(self, c):
> 50 # compute c**d (mod n)
> 51 if not self.has_private():
> 52 raise TypeError("No private key")
> 53 if (hasattr(self,'p') and hasattr(self,'q') and hasattr(self,'u')):
> 54 m1 = pow(c, self.d % (self.p-1), self.p)
> 55 m2 = pow(c, self.d % (self.q-1), self.q)
> 56 h = m2 - m1
> 57 if (h<0):
> 58 h = h + self.q
> 59 h = h*self.u % self.q
> 60 return h*self.p+m1
> 61 return pow(c, self.d, self.n)
> 62
> 63 def _encrypt(self, m):
> 64 # compute m**d (mod n)
> 65 return pow(m, self.e, self.n)
> 66
> 67 def _sign(self, m): # alias for _decrypt
> 68 if not self.has_private():
> 69 raise TypeError("No private key")
> 70 return self._decrypt(m)
> 71
> 72 def _verify(self, m, sig):
> 73 return self._encrypt(sig) == m
>
>
> Thanks,
> Shoufu
>
```
More information about the pycrypto mailing list | 1,050 | 3,855 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.926062 |
https://www.ah-studio.com/231852-simplest-form-definition-algebra-things-that-make-you-love-and-hate-simplest-form-definition-algebra | 1,627,402,219,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153392.43/warc/CC-MAIN-20210727135323-20210727165323-00223.warc.gz | 639,225,226 | 5,800 | # Simplest Form Definition Algebra Things That Make You Love And Hate Simplest Form Definition Algebra
The activity should assignment as follows:
The activity fib will acquire an altercation of breadth and back it is invoked, it will acknowledgment the breadth of the Fibonacci Arrangement based on the length. For example, if I alarm fib(5), I appetite to acknowledgment the aboriginal 5 numbers in the Fibonacci Sequence: [0, 1, 1, 2, 3]
I started by creating a capricious to abundance the abject Fibonacci Sequence. It will be an arrangement of [0, 1] because the Fibonacci Arrangement consistently activate with 0 and 1.
Then, I created a abject case to acknowledgment the arrangement if true.
From there, I appetite to retrieve the aftermost two numbers in the arrangement to be acclimated to advance to the abutting number.
I acclimated the allotment adjustment and assigned it to a new variable.
Next, I sum the slices and advance it into the arrangement array.
Finally, I alarm the activity afresh to see if it hits the abject case.
We forgot about some bend cases. What if the activity is alleged and the altercation is beneath than 2? It would account an absolute bend so we charge to add checks.
We appetite the altercation breadth to be greater than 2:
Test it out on CodePen or JSFiddle to see it in action, I additionally added a few logs so you can see the results:
Is there a bigger way to do this? Probably! So I’m activity to accumulate practicing!
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Fractions – LA 8 + | simplest form definition algebra | 917 | 4,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-31 | latest | en | 0.895292 |
https://www.physicsforums.com/threads/parallel-flow-pipe-network-help.613775/ | 1,611,346,490,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531335.42/warc/CC-MAIN-20210122175527-20210122205527-00282.warc.gz | 944,522,831 | 14,364 | # Parallel flow pipe network help
Let's say I have a simple parallel flow problem as shown in the attachment. The flow is pressurized and the downstream of the return goes back to a piping system that is negligible. It does not vent to atmosphere.
I've tried reading up on the subject, but I got stuck at a certain point. At point A, I know that the flow going in must equal the flow going out to the two pipes. Therefore, it will look like Q = Q1+Q2. Expanding Q1 and Q2 gives me V*A for each respective pipe.
I also know that the head loss of Pipe 1 must equal the head loss of Pipe 2. Expanding those equations, I find that I am faced with an unknown, which is the friction factor for each pipe. The friction factor can be explicitly calculated using the Swamee and Jain formula, but is dependent on the Reynolds Number, which is dependent on the flow velocity, which is ultimately dependent on the flow rate of each of the pipes. After getting the friction factor, everything else is easy. Simply plug the velocities from the head loss equations back into the Q formula shown above and solve for the unknown.
I know one can attain the correct friction factor value by many iterations, but is there a way without doing iterations by hand or by Excel and not fancy software?
Also, how would I find the total pressure loss of the system?
Thanks!
#### Attachments
• 29.4 KB Views: 465 | 309 | 1,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-04 | latest | en | 0.951237 |
http://nrich.maths.org/6242/solution | 1,503,427,683,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112539.18/warc/CC-MAIN-20170822181825-20170822201825-00139.warc.gz | 319,271,015 | 4,865 | ### Hockey
After some matches were played, most of the information in the table containing the results of the games was accidentally deleted. What was the score in each match played?
### Friday 13th
This month there is a Friday the thirteenth and this year there are three. Can you explain why every year must contain at least one Friday the thirteenth?
### Christmas Trees
Christmas trees are planted in a rectangular array of 10 rows and 12 columns. The farmer chooses the shortest tree in each of the columns... the tallest tree from each of the rows ... Which is the taller tree, A or B?
# Rendezvous
##### Stage: 3 Short Challenge Level:
If the friends meet in Bergerac, the total distance travelled will be $$87+79+61+58 \; \text{km} = 285\; \text{km}$$
If the friends meet in Bordeux, the total distance travelled will be $$87+47+31+84 \; \text{km} = 249\; \text{km}$$
If the friends meet in Langon, the total distance travelled will be $$79+47+54+37 \; \text{km}=217\; \text{km}$$
If the friends meet in Libourne, the total distance travelled will be $$61+31+54+65 \; \text{km}=211\; \text{km}$$
If the friends meet in Marmande, the total distance travelled will be$$58+84+37+65 \; \text{km}=244\; \text{km}$$
So the friends should meet in Libourne.
This problem is taken from the UKMT Mathematical Challenges. | 365 | 1,326 | {"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} | 3.78125 | 4 | CC-MAIN-2017-34 | latest | en | 0.812582 |
https://www.pcreview.co.uk/threads/embarrassing-question-about-lookup.3643570/ | 1,723,295,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00136.warc.gz | 725,363,963 | 13,601 | E
#### Emil
I am attempting to create what should be a simple formula. I am sure there
are better ways to do this but I am somewhat new so just going for what I
know how to use. I have attempted VLOOKUP, but I get a bit confused on this.
D5 is a drop down menu which is the name. (This Cell never moves).
c5,c6,c7, etc… is a numerical value I would like my new target cell to show.
I only have 27 names.
Here’s my formula
=LOOKUP(D5,{"name1","name2","name3","name4","name5","name6","name7","name8","name9","name10","name11","name12","name13","name14","name15","name16","name17","name18","name19","name20","name21","name22","name23","name24","name25","name26","name27"},{"c5","c6","c7","c8","c9","c10","c11","c12","c13","c14","c15","c16","c17","c18","c19","c20","c21","c22","c23","c24","c25","c26","c27","c28","c29","c30","c31"})
I think this formula is perfect except instead of showing “c5â€, etc… I want
it show the value of C5.
This should work. It is a formula array so you have to type Shft-Cntl Enter
to get the square brackets
{=OFFSET(C5:C31,MATCH(TRUE,("name"&ROW(A1:A27)=D5),0)-1,0)}
The line below produces an array {"name1","name2,...,"name27"}
("name"&ROW(A1:A27)
The Match returns the index number of the item that matches D5
Then Offset get the item in C5:C31 using the index returned in the Match
function.
=LOOKUP(D5,{"name1","name2","name3"},C5:C7)
Extend it to 27 names!
But I suggest to store the 27 names in a cell range, say B5:B31,
then the formula is:
=LOOKUP(D5,B5:B31,C5:C31)
Regards,
Stefi
„Emil†ezt Ãrta:
Stefi:
Wow, that was soo easy! Worked liked a Charm** I can't thank you enough
- I have literally spent hours attempting to figure this out!
Also thank you Joel – I have a lot to learn and you have given me some
direction on what other functions I should know.
You are welcome! Thanks for the feedback!
Stefi
„Emil†ezt Ãrta: | 613 | 1,895 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-33 | latest | en | 0.8832 |
http://slidegur.com/doc/1290481/p1-2.1-energy-transfers-and-efficiency-powerpoint | 1,582,029,518,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143695.67/warc/CC-MAIN-20200218120100-20200218150100-00282.warc.gz | 131,518,354 | 17,136 | ### P1 2.1 Energy transfers and efficiency powerpoint
```P1.2.1 Energy transfers and efficiency
GCSE items that can be used to define successful outcomes
for peer and self assessment activities
•Assessment of practical work
Using Exam pro items to support successful
outcomes
Learners will be able to test their progress against learning
outcomes using questions taken from past AQA GCSE
questions.
Version 2.0
Q1 Foundation AQA-PH1FP-QP-Jun12 Q1
The appliances shown on the next slide transfer electrical
energy to other types of energy.
The vacuum cleaner is designed to transfer electrical
energy to kinetic energy.
a)Name the three other appliances which are also
designed to transfer electrical energy to kinetic energy.
Version 2.0
Version 2.0
Q1 cont.
b) Which two of the following statements are true?
•A Appliances only transfer part of the energy usefully.
•B The energy transferred by appliances will be destroyed.
•C The energy transferred by appliances makes the
surroundings warmer.
•D The energy output from an appliance is bigger than the
energy input.
Version 2.0
Q1 mark scheme
a) Washing machine, fan, drill
1 mark for each 3
b) A C
1 mark for each
Version 2.0
2
Q2 Foundation Exampro QM94R505
The diagram shows what happens to each 100 joules of energy from crude oil when it is
used as petrol in a car.
The widths of the arrows show exactly how much energy is transferred in each particular
way.
(a) Complete the diagram by adding the correct energy value alongside each
arrow. (3)
Version 2.0
(b) Calculate how efficient the car engine is at transferring
the energy from petrol into useful movement.
(2)
(c) Two students are discussing the diagram.
The first says that none of the energy released from the
crude oil is really lost.
The other says that all of the energy released from the crude
oil is really lost.
What do you think?
(4)
Version 2.0
Q2 mark scheme
(a) A = B = 20,
C = 60
3
(b) evidence of or 20%/0.2 gains 1 mark
but 25%/0.25 or figure consistent with B in part (a) gains 2 marks
2
(c) ideas that
“lost” energy is (also) transferred (in ways we don’t want)
total amount of energy (before and after any transfer) remains the same
energy becomes (progressively) more spread out/less easy to transfer again
this applies to all the energy/even to the energy usefully transferred
each for 1 mark
Version 2.0
Q3 higher AQA-PH1HP-QP-Jan12 Q 6
The table on the next slide gives data about two types of low
energy bulb.
(a) Both types of bulb produce the same useful power output.
(a) (i) Calculate the useful power output of the CFL.
Use the correct equation from the Physics Equations Sheet.
Useful power output = ................................................. W
(2 marks)
Version 2.0
Version 2.0
Q3 contd.
(a) (ii) Calculate the efficiency of the LED bulb.
Use the correct equation from the Physics Equations Sheet.
Efficiency = ......................................................
(1 mark)
Version 2.0
Q3 contd.
6 (b) Sketch and label a Sankey diagram for the CFL (2 marks)
6 (c) LED bulbs are expensive. This is because of the large
number of individual electronic LED chips needed to produce
sufficient light from each bulb.
6 (c) (i) Use the data in the table to evaluate the costeffectiveness of an LED bulb compared to a CFL. (2 marks)
6 (c) (ii) Scientists are developing brighter and more efficient
LED chips than those currently used in LED bulbs.
Suggest one benefit of developing brighter and more efficient
LED chips.
(1 mark)
Version 2.0
Q3 markscheme
Version 2.0 | 887 | 3,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-10 | latest | en | 0.92257 |
https://nikkiestutorials.com/2020/01/14/class-11-limits-exercise-1a-question-number-7-to-12-anjali-maths/ | 1,579,915,122,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250628549.43/warc/CC-MAIN-20200125011232-20200125040232-00400.warc.gz | 570,892,633 | 9,180 | # Class 11/Limits /Exercise 1A Question number 7 to 12/Anjali Maths
#rsaggarwalsolution #class11solution #setssolution #exercise1cSolution #exercise1dsolution #completesolutioninhindiandenglish
#notutionrequire #mathsonlinelearning #mathstrick #mathssolution #exampreparation #simplewayosoltionc #calculus #limits:
Limits Chapter 1A Question numer 1,2,3,4,5,and 6 Lnk is below:
https://youtu.be/D2A3EKBT_9c
What is Intersection Set:
https://youtu.be/JYpEo58Y3Zs
What is a union set:
https://youtu.be/u8ODpdOXA-s
Exercise 1C sets:
Question number1:
https://youtu.be/a3mCBLr2OME
Question number 2:
https://youtu.be/_ru0zXHmuOU
Question number 3:
https://youtu.be/O-kY3TkFjWs
Question number 4:
https://youtu.be/SiZWSYzbLNI
Question Number 5:
https://youtu.be/7uQTk0ACfto
Question Number 6:
https://youtu.be/c5YbcLpWXIE
Question Number 7
:https://youtu.be/-ztbUn7q2jg
Question Number8:
https://youtu.be/5tKkScrUFJo
Question Number 9:
https://youtu.be/Y7KDKJjouag
Question Number 10:
https://youtu.be/jOUZpE-L9X4
Question Number 11:
https://youtu.be/NGuzVgucW2o
Question Number 12:
https://youtu.be/LfazLvgb61A
Question Number 13:
https://youtu.be/d7kcHo3zSS8
question number 14
https://youtu.be/MS9YuB7cT2A
Question Number 15
https://youtu.be/elglcf8Si2M
Exercise 1B SOLUTION
exercise 1 B question number 1:
https://youtu.be/NUK8LKHxYKY
exercise 1 B question number 2:
https://youtu.be/woQomzwAfjs
Exercise 1B Question Number 3
https://youtu.be/oTo14QT1ZZY
Exercise 1B Question Number 4:
https://youtu.be/JR7uWj0j3rY
Exercise1B
https://youtu.be/h7n7PxTyOWw
Sets Exercise 1A/Question 1- 9 solution:
Question Number 9: https://youtu.be/4T8WjO4gpzw
Question Number 8: https://youtu.be/AHbxQIDMMCI
Question number 7: https://youtu.be/qBIOlp5uFiY
Question number 6: https://youtu.be/tXU9660kej0
Question number 5 : https://youtu.be/aHAki1vG1Dk
Question number 4 : https://youtu.be/93J2mV0vd-Y
Question number 3 : https://youtu.be/UZeWgk4JYXw
Question number 2: https://youtu.be/bikkTVo7jzg
Question number 1: https://youtu.be/JRlfY_sPyEQ
Class 11, complete solution, class 11 maths, tutorial, education,
the solution, simple, trick, easy way, rs Aggarwal solution, learn maths, rs Aggarwal solution, sets, tutorial sets | 789 | 2,253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.257603 |
http://www.mrscienceshow.com/2014/03/cricket-is-matter-of-life-and-death.html | 1,675,194,500,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00193.warc.gz | 74,334,872 | 23,471 | ## Tuesday, 18 March 2014
### Cricket is a matter of life and death
A guest post by Bernard Kachoyan
Ever thought of batting as a life and death struggle against hostile forces? It always seemed that way when I batted. Well you might be more accurate than you think.
The experience of a batsman can be described as a microcosm of life: when you go out to bat you are “born”, when you get out you “die”. But what happens when you are Not Out (NO)? More subtly, when you are Not Out you simply leave the sample pool, that is you live for a while then you stop being measured. In the parlance of statistics, this becomes “censored” data. In medical research the “born” moment is equivalent to when a patient is first being monitored (e.g. survival times of cancer patients after diagnosis). The question in medicine becomes, what is the “survival function”, the probability that a patient survives for X years after the start of observation? And how does the life expectancy curve of one population differ from another, in particular are people treated in a particular way different to a control group).
These type of problems are commonly addressed using Kaplan-Meier (KM) estimators. In economics, it can be used to measure the length of time people remain unemployed after a job loss. In engineering, it can be used to measure the time until failure of machine parts. Here we will apply those ideas to batting in cricket.
An important property of the KM estimate is that it is non-parametric in the sense that it does not assume any type of Normal distribution in the data, something which is patently untrue for this type of data. It also only uses the data itself to generate a survival curve (the term given to the survival function after it is drawn on a chart) and associated confidence limits. Hence the KM survival curve may look odd in that it declines in a series of steps at the observation times and the function between sampled observations is constant. However, when a large enough sample is taken, the KM approaches the true survival function for that population.
An important advantage of the KM method is that it can take into account censored data, particularly censoring if a patient withdraws from a study, i.e. is lost from the sample before the final outcome is observed. This makes it perfect for dealing with the NOs as described above.
When referring to batsmen, “death” means getting out, being “censored” means completing the innings before getting out (remaining NOT OUT) and “time” means number of runs scored (tj = scoring j runs). The idea of the KP estimator is pretty simple.
1. The conditional probability that an individual dies in the time interval from ti to ti+1, given survival up to time ti is estimated as di/ni where di is the number who die at time ti, and ni is the number alive just before time ti, including those who will die at time ti
2. Then the conditional probability that an individual survives beyond ti+1 is (ni – di)/ ni
3. When there is no censoring, ni is just the number of survivors just prior to time ti. With censoring, ni is the number of survivors minus the number of losses (censored cases). It is only those surviving cases that are still being observed (have not yet been censored) that are "at risk" of an observed death
4. The KP estimator of the survivor function at time t for tj ≤ t ≤ tj+1 is then formally:
Such KM curves have attractive properties, which perhaps explain their popularity in medical research for over half a century. They are fairly easy to calculate and they provide a visual depiction of all of the raw data—including the times of actual failure, yet still give a sense of the underlying probability model.
Let’s now apply the KM estimator to some cricket statistics. In this case I have arbitrarily chosen the batting statistics of Steve Waugh, Sachin Tendulkar (up to 2010 to keep roughly the same number of innings as Waugh) and Don Bradman. Without the consideration of the censored data (the Not Outs), then the curve simply reverts to the percentage of scores less than or equal to a certain number of runs - the value on the x axis. This is shown in Figure 1. Bradman of course is still clearly in a class of his own.
If we now properly include the NOs in the formulation we get survival curves as shown in Figure 2. I have omitted the Tendulkar curves here for clarity. As expected, the survival rates go up as the NOs do not indicate a true “death”. In Steve Waugh’s case, the increase is noticeable (I didn’t say “significant”!) since he has a large number of NOs compared to most batsmen within his number of test innings.
This is shown more starkly in Figure 3, where I have plotted both the censored and uncensored curves for Waugh and Tendulkar. I have plotted them on a logarithmic scale to highlight differences. It can be seen that Waugh’s censored survival curve (cf the raw curve) tracks Tedulkar’s very closely until a score of about 100. This reflects the large number of Waugh’s NOT OUTS (43 vs 29 in roughly the same number of innings, 260 vs 278). The diversity of the curves after that not only reflects the propensity of Tendulkar to go on to big scores, but also that a large number of Tendulkar’s not outs were after he had already scored a century (15 vs 2 for Waugh).
Figure 1 and Figure 2
Figure 3
The basic KM methodology has been around since the 1950s and of course has been extended in various ways by professional statisticians and alternative methods proposed. But their simplicity means it is still widely used.
There are several drawbacks, some of which can be seen in Figures 1-3. Firstly, the vertical drop at specific times is drawn from the data, and should not be seen as indicating particular “danger times”. This is particularly evident at larger scores where the naturally small sample size means that three are fewer data points (i.e. scores where a batsman actually gets out). So some sort of smoothing of the curve is thus necessary to provide an estimate of the true underlying functional dependency.
This reduction on the sample at large values also means the effect of each individual failure on the size of the step-down increases.
Another drawback of the KM method is that the estimate of the probability of surviving each “danger time” depends only on the number of patients at risk at that time. So if there are censored values the actual time between the last failure and the time of censoring is not considered.
It is natural at this point to question the underlying assumption of the KM method that the patients (i.e. innings) are independent. Is it common to talk in cricket about form slumps or purple patches. This can be examined statistically by considering the autocorrelation function of the scores, shown in Figure 4 assuming stationarity, where Waugh has been omitted for clarity. The figure clearly shown no evidence for time/innings correlation and although strictly speaking un-correlation does not imply true independence, it is evidence that the innings can be considered independent for the purposes of this analysis.
Figure 4
The question naturally arises is whether we can say anything statistically about whether the difference between survival curves is significant (cf treated vs control groups in medicine). Confidence intervals can be placed on the derived curves using the so-called Greenwood formula, dating back to the 1920s, or its more modern variations. These will suffer the drawback of being less accurate in the tail of the curves, where by definition the sample size is smallest. Not only will the formulas return a greater error because of that, the validity per se comes into question as the expressions rely on a normal approximation (through the central limit theorem), hence can only be considered valid for remaining innings bigger than say 20 or so.
Unfortunately, as we have seen above it is in the tails of the curve where the distinctions between very good and great batsman are often found.
Similarly a number of ways of comparing curves exist in the statistical literature, such as Kolmogorov–Smirnov test, the Log-rank test or the Cox proportional hazards test. These can rapidly become very mathematically complicated, especially if we want to try and distinguish one part of the curve specifically (say the high end).
Although I haven’t done the hard yards in this article, my intuition tells me we might be hard pressed to prove statistically significant differences between the Waugh and Tendulkar corrected survival curves. This is the drawback of applying statistical tests into areas where their applicability is not clear.
In any case, it can be seen that batting can most certainly be considered a true life and death struggle.
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9. Ever since the announcement of the International League T20 has been made by the ECB, there had been a huge interest by international investors to jump on board and purchase the franchise rights. https://cricketcorner.net/ilt20-live-streaming-tv-channels-list/ | 2,647 | 12,736 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | longest | en | 0.943707 |
http://dict.cn/parametric%20line | 1,548,333,421,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584547882.77/warc/CC-MAIN-20190124121622-20190124143622-00586.warc.gz | 55,296,324 | 7,606 | # parametric line
• 参量曲线, 坐标线
## parametric line的用法和样例：
### 例句
1. This paper presets a new fault location method using one terminal voltage current and another terminal current (or voltage) based on distributed parameters line model.
以分布参数作为高压输电线路的模型，提出了利用线路一端电压电流和另一端电流（或电压）的测距方法。
2. MA Wenqi,LANG Yansheng,MA Zhaoyan.Algorithm of fault simulation and fault-location for distributed parameter line model.Proceedings of the EPSA,1996,8(3):7-16.
[7]马文骐;郎燕生;马昭彦.;分布参数输电线路故障模拟及测距
3. The vertical line meets the horizontal one here.
垂直线与水平线在此相交。
4. One line cuts another at right angles.
一条线与另一条线成直角相交。
5. The printers have missed out a whole line here.
排版工人在这里漏掉了一整行。
6. Participation in tender is in our line of business.
参加投标属于我们的业务范围。 | 269 | 738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-04 | latest | en | 0.389543 |
http://mathoverflow.net/questions/122512/is-this-surface-diffeomorphic-to-a-spheres2/122517 | 1,466,806,201,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391519.2/warc/CC-MAIN-20160624154951-00198-ip-10-164-35-72.ec2.internal.warc.gz | 195,252,464 | 13,486 | Is this surface diffeomorphic to a sphere(S^2)? [closed]
let $f:R^3 -> R \ \ \ \ \ \ \ \ f(x,y,z)=x^4 + y^6 +z^8 \\$
$M = f^{-1}(1)$
Is M is diffeomorphic to a sphere $S^2$ ?
I tried to solve this problem, but I realized that I have no tools to solve it.
Constant rank theorem tells me M is a smooth 2 dimensional manifold, but does not tell me how it looks like.
And more generally, when $N = \{ x,y,z \in R^3 | ax^n + by^m + cz^l = 1\}$ is diffeomorphic to a sphere? What tools can I use to solve this problem?
-
closed as off topic by Chris Gerig, Benoît Kloeckner, Mariano Suárez-Alvarez♦, Martin Brandenburg, Willie WongFeb 21 '13 at 12:09
Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.
math.stackexchange.com is a better place for these questions. – Mariano Suárez-Alvarez Feb 21 '13 at 8:51
It is. The homeomorphism from $X$ to the unit sphere is $$x\mapsto \mathrm{sign}(x)\cdot x^2,\ y\mapsto y^3,\ z\mapsto\mathrm{sign}(z)\cdot z^4.$$
Here is a plot of the surface $x^4+y^6+z^8 = 1$. It is approximately a cube with opposite corners $(-1,-1,-1)$ and $(1,1,1)$. | 428 | 1,428 | {"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} | 3.1875 | 3 | CC-MAIN-2016-26 | latest | en | 0.905138 |
https://accounting.uworld.com/cpa-review/lc/accounting-dictionary/term/coefficient-of-determination/ | 1,725,987,899,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00372.warc.gz | 68,111,732 | 50,691 | ## The Coefficient of Determination or r squared, tells you how much of the variance in one variable is due to changes in another variable.
If the coefficient of determination is 47% it means 47% of the change in one variable is due to change in another variable. | 58 | 264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-38 | latest | en | 0.931796 |
https://www.coursehero.com/file/p3l0g2qg/This-example-illustrates-an-application-of-the-general-theory-of-second-best/ | 1,596,995,434,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738562.5/warc/CC-MAIN-20200809162458-20200809192458-00199.warc.gz | 631,470,546 | 165,820 | # This example illustrates an application of the
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This example illustrates an application of the general theory of second best . Essentially, the theory states that when two opposing factors contribute to efficiency losses, they can offset one another’s distortions.
51 Corrective Subsidies: A Means of Internalizing Positive Corrective Subsidies: A Means of Internalizing Positive Externalities Externalities A corrective subsidy is similar in concept to a corrective tax. Figure 3.6 shows how a corrective subsidy for inoculations can result in the efficient output of this good. The competitive market equilibrium output would be 10 million inoculations per year at the competitive market price of \$25 per inoculation. This is inefficient because the marginal social benefit ( MPB i + MEB ) at that level of consumption exceeds the marginal social cost. A corrective subsidy is a payment made by government to either buyers or sellers of a good so that the price paid by consumers is reduced. The payment must equal the marginal external benefit of the good or service. In this case, \$20 is the marginal external benefit associated with each person inoculated.
52 Corrective Subsidies: A Means of Internalizing Positive Corrective Subsidies: A Means of Internalizing Positive Externalities Externalities Suppose the government announces that it will pay each person inoculated a subsidy of \$20. This subsidy adds \$20 to the marginal private benefit of each inoculation. The demand curve for inoculations shifts upward from D=MPB i to D′ = MPB i + \$20. As the demand for inoculations increases, the market equilibrium moves from point U to point V in Figure 3.6. At that point, the market price of an inoculation increases to \$30 to cover increased marginal costs of production. However, the net price after receiving the subsidy declines for consumers. The net price is now \$30 - \$20 = \$10 per inoculation. This reduction in the net price to consumers increases the quantity demanded to 12 million per year, the efficient output.
53 Corrective Subsidies: A Means of Internalizing Positive Corrective Subsidies: A Means of Internalizing Positive Externalities Externalities The effect of the subsidy is to increase the benefit of inoculations accruing to those other than the buyers or the sellers of inoculations from \$200 million per year to \$240 million per year (\$20 per person inoculated multiplied by 12 million inoculations per year). The government accomplishes this by making a total of \$240 million in subsidy payments to the 12 million people inoculated each year. This is represented by the area RVXY in Figure 3.6. The subsidy is paid from tax revenues.
54 Subsidy Payments Figure 3.6 A Corrective Subsidy 0 Price, Benefit, and Cost (Dollars) Inoculations per Year (Millions) Y 10 X D' = MPB i + \$20 = MSB D = MPB i S = MSC 25 10 U 30 12 V R 45 Z
55 Corrective Subsidies: A Means of Internalizing Positive Corrective Subsidies: A Means of Internalizing Positive Externalities Externalities Examples of corrective subsidies include the provision of certain government services at levels below the marginal cost of such services. For example,
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• Externality, MEC
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Jill Tulane University ‘16, Course Hero Intern | 919 | 4,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-34 | latest | en | 0.89979 |
http://cboard.cprogramming.com/c-programming/143693-function-days-date-december-31st-printable-thread.html | 1,454,985,437,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701155060.45/warc/CC-MAIN-20160205193915-00215-ip-10-236-182-209.ec2.internal.warc.gz | 33,385,368 | 3,761 | # function days from date to december 31st
• 11-25-2011
cameuth
function days from date to december 31st
Code:
```#include <stdio.h> #include <time.h> int days, month, year, days_1, month_1, year_1, leapy, ndays, ddays, days_total, days_beforeleap; int daysinmonth(int month) /*there is a problem in this statement, your printf is reading out ridiculous numbers for days, don't forget to ask*/ { switch(month) { case 1: case 3: case 5: case 7: case 8: case 10: case 12: return 31; case 2: return 28; case 4: case 6: case 9: case 11: return 30; default: return -1; } } int leap(int year) { if(year%400 ==0 || (year%100 != 0 && year%4 == 0)) { leapy=1; } else { leapy=0; } } int daystoyear(month, month_1, year, year_1) { int i=0, days_toend, years_todays; days_toend=daysinmonth(month)-days; while(month<month_1) { daysinmonth(month); month++; } years_todays=365*(year_1-year); days_beforeleap=days_toend+years_todays; return days_beforeleap; } daysinyear(int days, int month, int days_1, int month_1) { int i=0; days_total = daysinmonth(month) - days; for(i = month+1;i < month_1 - 1;++i); days_total += daysinmonth(i); days_total += days_1; return days_total; } int main(void) { int days_1, month_1, year_1; char a, b, a1, b1; printf("intput a date in the follwing format:\n"); printf("mmddyyyy: in other words the two digit day,two digit month,\n"); printf("and four digit year, with no slashes or heifens anywhere.\n"); scanf("%d%c%d%c%d", &month, &a, &days, &b, &year); if(month<0 || month>12 || days<0 || days>31 || year<0) { printf("data not valid"); } printf("enter the later date in the same format:\n"); scanf("%d%c%d%c%d", &month_1, &a1, &days_1, &b1, &year_1); daystoyear(month, month_1, year, year_1); printf("total days=%d",days_beforeleap); }```
the function days in year is supposed to read the number of days from a date to the end of the year. I don't think the while loop is working but I don't know why. Is it a recursive function problem?
• 11-25-2011
BillyTKid
Code:
```for(i = month+1;i < month_1 - 1;++i); /* <<-- here is your newbie mistake */ days_total += daysinmonth(i); days_total += days_1; return days_total; }```
• 11-25-2011
cameuth
I think that for loop is what causes my date totals to be 60 and 61 for 1/1/1 to 1/1/1 and 1/1/1 to 1/1/2. I've toyed with it a bit, but I can't seem to find the exact logic problem.
Am I overthinking it? would this work?
Code:
`for(i=month;i<month_1;++i);`
• 11-25-2011
CommonTater
Delete line 60 and see what happens...
• 11-25-2011
cameuth
AHA! the semicolon! Thanks, and hello again tater, always a pleasure!
• 11-25-2011
cameuth
ok, it now works for dates in the same month(and year of course), but the values come up about two months short for long amounts of time. For example, 1/1/1 to 12/31/1 comes out to 305 days. Here's the new function:
Code:
```daysinyear(int days, int month, int days_1, int month_1) { int i; for(i = month+1;i < month_1 - 1;++i) days_total += daysinmonth(i); days_total+=days_1; days_total=days_total-1; return days_total; }```
• 11-25-2011
cameuth
Oh, and I added line 7 because the values in the same month were 1 day off, but I'm not sure how wise that was now... | 1,102 | 3,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-07 | latest | en | 0.247274 |
https://www.altsci.com/3/s/Every-lattice-element-of-the-structure-is-in-its-p | 1,624,463,300,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00433.warc.gz | 594,275,860 | 7,281 | Page "Absolute zero" ¶ 8
from Wikipedia
## Some Related Sentences
Every and lattice
Every Boolean algebra is a Heyting algebra when a → b is defined as usual as ¬ a ∨ b, as is every complete distributive lattice when a → b is taken to be the supremum of the set of all c for which a ∧ c ≤ b. The open sets of a topological space form a complete distributive lattice and hence a Heyting algebra.
* Every totally ordered set that is a bounded lattice is also a Heyting algebra, where is equal to when, and 1 otherwise.
* Every Boolean algebra is a distributive lattice.
* Every Heyting algebra is a distributive lattice.
* Every totally ordered set is a distributive lattice with max as join and min as meet.
Every complete lattice is also a bounded lattice, which is to say that it has a greatest and least element.
Every lattice in can be generated from a basis for the vector space by forming all linear combinations with integer coefficients.
* Every substructure is the union of its finitely generated substructures ; hence Sub ( A ) is an algebraic lattice.
Also, a kind of converse holds: Every algebraic lattice is isomorphic to Sub ( A ) for some algebra A.
Every complemented distributive lattice has a unique orthocomplementation and is in fact a Boolean algebra.
Every distributive lattice is modular.
Every non-modular lattice contains a copy of N < sub > 5 </ sub > as a sublattice.
Every quasinormal subgroup is a modular subgroup, that is, a modular element in the lattice of subgroups.
Every and element
Every such subset has a smallest element, so to specify our choice function we can simply say that it maps each set to the least element of that set.
** Zorn's lemma: Every non-empty partially ordered set in which every chain ( i. e. totally ordered subset ) has an upper bound contains at least one maximal element.
** Tukey's lemma: Every non-empty collection of finite character has a maximal element with respect to inclusion.
: Every non-empty set A contains an element B which is disjoint from A.
Every Boolean algebra ( A, ∧, ∨) gives rise to a ring ( A, +, ·) by defining a + b := ( a ∧ ¬ b ) ∨ ( b ∧ ¬ a ) = ( a ∨ b ) ∧ ¬( a ∧ b ) ( this operation is called symmetric difference in the case of sets and XOR in the case of logic ) and a · b := a ∧ b. The zero element of this ring coincides with the 0 of the Boolean algebra ; the multiplicative identity element of the ring is the 1 of the Boolean algebra.
Every element of is a member of the equivalence class.
Every repetition of insertion sort removes an element from the input data, inserting it into the correct position in the already-sorted list, until no input elements remain.
Every singleton is a terminal object, with the functions mapping all elements of the source sets to the single target element as morphisms.
Every simple module is cyclic, that is it is generated by one element.
Every element s, except a possible greatest element, has a unique successor ( next element ), namely the least element of the subset of all elements greater than s. Every subset which has an upper bound has a least upper bound.
Every element has a successor ( there is no largest element ).
* Every non-empty set of left ideals of R, partially ordered by inclusion, has a maximal element with respect to set inclusion.
Every time a pixel on a triangle is rendered, the corresponding texel ( or texture element ) in the texture must be found.
Every non-inner automorphism yields a non-trivial element of Out ( G ), but different non-inner automorphisms may yield the same element of Out ( G ).
Every and structure
Every finite tree structure has a member that has no superior.
Every congruence relation has a corresponding quotient structure, whose elements are the equivalence classes ( or congruence classes ) for the relation.
Every group has a presentation, and in fact many different presentations ; a presentation is often the most compact way of describing the structure of the group.
Every algebraic structure has its own notion of homomorphism, namely any function compatible with the operation ( s ) defining the structure.
Every solid has its own characteristic energy band structure.
Every chapter contains a comprehensive compilation all of the published examples of the reaction organized in tables according to the structure of the starting material.
Every Riemann surface is a two-dimensional real analytic manifold ( i. e., a surface ), but it contains more structure ( specifically a complex structure ) which is needed for the unambiguous definition of holomorphic functions.
: Every oriented prime closed 3-manifold can be cut along tori, so that the interior of each of the resulting manifolds has a geometric structure with finite volume.
* " What Every Programmer Should Know About Memory " by Ulrich Drepper — explains the structure of modern memory subsystems and suggests how to utilize them efficiently
Every H * is very special in structure: it is pure-injective ( also called algebraically compact ), which says more or less that solving equations in H * is relatively straightforward.
Every structure is associated with a certain quantity of energy, which determines the stability of the molecule or ion ( the lower energy, the greater stability ).
Every Hermitian manifold is a complex manifold which comes naturally equipped with a Hermitian form and an integrable, almost complex structure.
Every animal cell is enclosed in a plasma membrane, which has the structure of a lipid bilayer with many types of large molecules embedded in it.
Every structure from the World's Columbian Exposition was long ago destroyed by fire, demolished or moved elsewhere, except for the old Palace of Fine Arts, now the Museum of Science and Industry, The Palace of Fine Arts, the only fireproof building at the fair, fell into disrepair and was rehabilitated with a \$ 5 million grant in 1930 from Julius Rosenwald ( President of Sears, Roebuck and Co .).
Every five years, starting the day after Labor Day, the New Jersey Department of Environmental Protection ’ s ( DEP ) water level management plan allows the lake to be lowered five feet to allow for inspection of the dam structure, property owners ’ repairs to lakeshore structures, aquatic vegetation control, and silt and drainage material removal from areas around the lake.
Every timber structure on the site had been burnt, the charcoal being the only organic matter that survived the acid soils.
Every server uses its own query logic and structure.
Every ( normal ) Boolean algebra with operators can be represented as a field of sets on a relational structure in the sense that it is isomorphic to the complex algebra corresponding to the field.
Every compact smooth manifold of dimension 2n, which has only handles of index ≤ n, has a Stein structure provided n > 2, and when n = 2 the same holds provided the 2-handles are attached with certain framings ( framing less than the Thurston-Bennequin framing ).
Every window was lit, to lurid effect, as if the interior of the structure might be on fire.
Every Menards store shares a common structure.
Every structure shown here was built during this period, along with tens of others
0.646 seconds. | 1,503 | 7,205 | {"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-2021-25 | latest | en | 0.915692 |
https://mathematica.stackexchange.com/questions/146618/center-of-mass-of-2d-region/146619 | 1,721,820,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518277.99/warc/CC-MAIN-20240724110315-20240724140315-00886.warc.gz | 355,327,961 | 41,186 | # Center of mass of 2D region
The problem states:
Find the center of mass of a thin plate covering the region between the x-axis and the curve $$y=20/x^2, 5 \leq x \leq8$$ if the plate's density at a point (x,y) is $\delta(x)=2x^2$.
If the density were uniform I would find the center of mass using RegionCentroid:
In[1]:= reg = ImplicitRegion[{5 <= x <= 8, 0 <= y <= 20/x^2}, {x, y}]; RegionCentroid[reg] Out[1]= {40/3 Log[8/5], 43/160}
Is there an "easy" way like this to compute the center of mass when the density function is given? I'm trying to avoid setting up integrals manually.
It seems to me that what I need is Geometric Centroid. "http://mathworld.wolfram.com/GeometricCentroid.html" says:
The centroid is center of mass of a two-dimensional planar lamina or a three-dimensional solid. The mass of a lamina with surface density function $\sigma(x,y)$..."*. "The geometric centroid of a region can be computed in the Wolfram Language using Centroid[reg].
However it doesn't work for me. Does this function still exist in Mathematica 11 or do I need to use Needs?
Since it is a thin plate and so essentially 2D, you can use the density as the third dimension.
reg = ImplicitRegion[{5 <= x <= 8, 0 <= y <= 20/x^2, 0 <= z <= 2 x^2}, {x, y, z}];
RegionCentroid[reg]//Most
out = {13/2, 1/4}
• Nice! I think this solution is valid for 3 dimension regions by adding a density function as the 4th dimension. Am I right? Commented May 22, 2017 at 16:44
• I believe so... Commented May 22, 2017 at 16:55
You can use region integrals:
reg=ImplicitRegion[{5<=x<=8,0<=y<=20/x^2},{x,y}];
δ[x_,y_]:=2x^2
Integrate[{x,y} δ[x,y], {x,y} ∈ reg] / Integrate[δ[x,y], {x,y} ∈ reg]
{13/2, 1/4} | 537 | 1,698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-30 | latest | en | 0.827868 |
https://studysoup.com/tsg/965524/linear-algebra-with-applications-8-edition-chapter-3-1-problem-4 | 1,618,640,768,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00601.warc.gz | 634,778,536 | 12,761 | ×
Get Full Access to Linear Algebra With Applications - 8 Edition - Chapter 3.1 - Problem 4
Get Full Access to Linear Algebra With Applications - 8 Edition - Chapter 3.1 - Problem 4
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# Let A = [ -0 -6 1 Find the following minors and cofactors of A.(a) M13 and C13 (b) M22
ISBN: 9781449679545 435
## Solution for problem 4 Chapter 3.1
Linear Algebra with Applications | 8th Edition
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4
Problem 4
Let A = [ -0 -6 1 Find the following minors and cofactors of A.(a) M13 and C13 (b) M22 and C22(c) M31 and C31 (d) M33 and C33
Step-by-Step Solution:
Step 1 of 3
MATH 1220 Notes for Week #14 18 April 2016 ● Find lim sin(x. x→0 tan(x) ○ lim sin(x= lim ssin(x)lim sin(x)cos(= limcos(x) = 1 x→0 tan(x) x→0 cos(x) x→0 sin(x) x→0 ○ Recognize that the limit of the ratios of these functions near x = 0 is 1 because that exhibit the same characteristics (slope of 1 in this case) near x = 0 . ● Find the 3rd degree Taylor polynomial to f(x) = tan(x) near x = 0 ○ (0) f (0) = tan(0) = 0 y 0 0 (1) cos (0)+sin (0) 1 f (0) = (tan(0)) ′ = cos (0) = cos (0)= y 1 x 1!
Step 2 of 3
Step 3 of 3
##### ISBN: 9781449679545
This textbook survival guide was created for the textbook: Linear Algebra with Applications, edition: 8. The full step-by-step solution to problem: 4 from chapter: 3.1 was answered by , our top Math solution expert on 03/15/18, 05:22PM. Since the solution to 4 from 3.1 chapter was answered, more than 240 students have viewed the full step-by-step answer. Linear Algebra with Applications was written by and is associated to the ISBN: 9781449679545. The answer to “Let A = [ -0 -6 1 Find the following minors and cofactors of A.(a) M13 and C13 (b) M22 and C22(c) M31 and C31 (d) M33 and C33” is broken down into a number of easy to follow steps, and 29 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 56 chapters, and 1286 solutions.
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# Thread: Checking for a whole number
1. ## Checking for a whole number
Is there anyway I can check if a number is a whole number?
I am retrieving images from a database and on every third image I want to start a new row in the table displaying them. I have set \$i to be 0 and increased it with every reiteration of the loop. After having increased it I want to check if \$i divided by 3 produces a whole number or not.
How do I do this?
Thanks
WillowFae
• If you want the whole number every time:
round(\$variable, 0);
1.3423
would return
1
IF, you just want to see if its a full number or has a .
\$blah = explode("\.", \$variable);
if \$blah[1] == "" then
the number is whole
else then
the number is not whole
end if
or something
• if(eregi("\.", \$i))
{
echo "not a whole number";
}
else
{
echo "is a whole number";
}
if \$i has a "." it will not be considered a whole number.
• or you can just use
<?php
round(\$i);
?>
which will round the number to a whole number.
• Ooh, thanks for those. Just what I wanted - the checking option, not the rounding up one.
• Originally posted by coffeedemon
if \$i has a "." it will not be considered a whole number. [/B]
Ahh, but mathematically speaking, a whole number is...
"a positive integer", so
-1 although it would pass your code, is not a whole number. Neither is 0. So you would need to do an additional test to see if the number is > 0. If so, then it's whole.
Just getting picky
• ## Validate Whole Number in PHP
Hi Friends,
There is a Better and Faster Method than preg_match for whole number
validation. that is : ctype_digit().... It take only String parameter , so make
our input as String First like eg:
\$var = (string)\$value;
then You can Check like this:
if (ctype_digit(\$var) ){
print "whole number\n";
}else{
print "not whole number\n";
}
ok Be happy , if any clarification Revert to me
anes.pa@gmail.com
Bye Anes
• ...the hell? This is a 7 year old post!
•
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• | 578 | 2,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-49 | longest | en | 0.849507 |
https://economics.stackexchange.com/questions/11685/stata-estimation-of-own-price-and-cross-price-elasticities-for-logit-nested-lo | 1,582,282,648,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145500.90/warc/CC-MAIN-20200221080411-20200221110411-00294.warc.gz | 351,666,529 | 30,516 | # STATA: Estimation of Own-Price and Cross-Price Elasticities for Logit, Nested Logit, and BLP Models
I am working with some data on automobile sales and characteristics in a small country. It is an unbalanced panel dataset taken over 6 years and over 5 market segments. I am trying to construct some discrete-choice demand models using Stata, using two OLS models (one where Price comes in linearly, one where Price is logged) and a nested logit. I will do my BLP instruments model next.
The OLS models are: δj=ln P(j)-ln P(0)=βxj-αpj+εj
and
δj=ln P(j)-ln P(0)=βxj-αln(pj)+εj
The nested logit is as follows:
δj=ln P(j)-ln P(0)=βxj-αpj+σ[ln P(j)- ln P(g)]+ξj for j∈g.
Where P(j) and P(0) represent, respectively, the probability of buying good j and the probability of buying an outside good. Xj is a vector of the characteristics of each good j (in this case, I am using CC/weight, Maximum Speed/100 and Size = Length*Width/10000), and pj is the price of good j.
The code I have used to define my variables and then runmy regressions is as follows:
gen lnPj=ln(Sj)
gen lnPg=ln(Sg1)
gen deltaj=ln(Sj/S0)
gen MSP100=0.01*MSpeed
gen LW10000= Length*Width*.0001
gen diffPjPg= lnPj-lnPg
reg deltaj CCW MSP100 LW10 P, robust
reg deltaj CCW MSP100 LW10000 lnP, robust
reg deltaj CCW MSP100 LW10 lnP diffPjPg, robust
So far, I have computed the own-price elasticities with the code:
gen OPE_PLogit=-.4894868*P*(1-Sj) for the first model and
gen OPE_lnPLogit=-1.959741*P*(1-Sj) for the second model
where Sj is the market share of that particular car for that particular year. I do know that in Stata, generally, one computes the elasticity using the -margins- command.
Let kk be any good but jj, α be the coefficient on price, η be the price elasticity, Sj be the own-market share, Sg be the share for a particular class of cars, e.g. Sedans, Sk be the share for all cars within a group except car j, and S0 be the overall share for cars other than J. The question is this: given that I have directly specified the regression equation and obtained good regression coefficients, is it still appropriate to use the formula η=αPj⋅(1−Sj)η=αPj⋅(1−Sj) normally used to compute the own-price elasticity and η=αPk⋅Skη=αPk⋅Sk (.) for the cross-price elasticity if I am using an unbalanced panel dataset? I think I recall my professor saying that an unbalanced panel dataset will adversely affect the elasticities, but I'm not sure how to adjust for this. | 722 | 2,453 | {"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} | 2.75 | 3 | CC-MAIN-2020-10 | latest | en | 0.865478 |
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• 0
# Sharing a sandwhich
## Question
Given globs of ham, bread, and cheese (in any shape), placed any way you like, Prove that with a knife there exists a way to bisect each of the ham, bread, and cheese. In other words, show that you can share it with a friend so that each of you can have the exact same amount of the three globs of food.
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You mean show that for any shape there exists a direction such that if you slice through it you get two equal halves?
Well then just put the shape at point (0,0,0) and imagine slicing it parallel to the yz axis at point x, let v1(x) be the volume of the first half and v2(x) the volume of the second, if x is at the far left then v1(x)=0 and v2(x)=1, if x is at the far right then v1(x)=1 and v2(x)=0, all that's left to show is that v2 and v2 are continous and then simply apply the mean value theorem and you know there exists some x where v1(x)=v2(x)=0.6
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Spoiler
Identify the centerpoint of each of the 3 items and then slice a plane that passes through all 3 centerpoints. Since a line through the center point of each item bisects that item, you've cut the assemblage in half as well.
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https://pdffox.com/nvp-what-is-it-for-pdf-free.html | 1,718,995,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00538.warc.gz | 389,130,871 | 8,655 | Don't fear change. The surprise is the only way to new discoveries. Be playful! Gordana Biernat
#### Idea Transcript
White Paper
NVP, what is it for?
In this paper, Simon Robinson, Excel Product Manager, explains the term NVP and how it is used in infrastructure data cabling. NVP, what is it for? NVP, yet another acronym to understand in infrastructure cabling… NVP is a property of copper data cabling that is, hopefully, stated on specification sheets and required to carry out testing, but why? Firstly, NVP stands for It is the speed at which the Nominal Velocity of Propagation. It is the speed data signals travel down at which the data signals the cable expressed as a travel down the cable percentage of the speed of expressed as a percentage light in a vacuum. of the speed of light in a vacuum. So a typical NVP for an unscreened cable may be 69% (sometimes expressed as 0.69). The speed of light in a vacuum (designated by the letter c) is a constant and is 299 792 458 ms-1 (metres per second) – far easier to quote the percentage than use the value. What is NVP used for? Part of the requirement of testing is to record the length of a copper data cable. This can be done manually by reading the cable sheath, and hoping that the metre marks are reasonably accurate and then recording it. This is not very easy once the cable is installed and would also require manual input in the test results. So by knowing the speed that the signal is transmitted down the cable it is a simple matter of the test equipment using: Distance=Speed×Time The testers measure the delay in the signal (time) and use the speed (NVP x c). Which length is used? There are in fact different lengths when looking at a twisted pair cable. l Sheath Length l Copper Length The Sheath Length is the physical length and most cables, including Excel, have metre marks to aid this measurement. Copper Length is the actual length of the copper conductors. Data cables are constructed of twisted pair cables. Each of the four pairs has a different twist rate, and so is of different length.
Different twist rates are used on the pairs to improve the performance and reduce cross talk issues. These different lengths accounts for the delay skew (see below). When the length of the cable is referred to in cable or installation design it is the sheath length that is the measurement.
Attenuation (Insertion Loss) is directly proportional to the length of the installed cable. The longer the cable the more the signal is attenuated or lost. Why is the length important? Attenuation (Insertion Loss) is directly proportional to the length of the installed cable. The longer the cable the more the signal is attenuated or lost. The components for infrastructure cabling systems up to and including Class FA are designed to ensure that the insertion loss does not exceed certain values when the channel length is at a maximum of 100 m. By specifying these limits, applications can be developed knowing that they will work on compliant Generic Cabling systems. As mentioned above, the twist rates of each pair, is different within the sheath. The different lengths of the pairs means that the delay (the time taken for the signal to be carried down the cable) will be different across the pairs. The difference in delay between the longest and shortest pair is the delay skew. It is important that the delay skew is within limits to ensure that the applications, that use multiple pairs, will function correctly.
continued overleaf
How is NVP calculated?
Finding the NVP
The standards use the delay measured at 10 MHz of the shortest pair on 100 m (sheath length) of the cable.
The NVP of data cables are stated on specification sheets and often stored in the test equipment as well under the manufacturers name – for example Excel. These are, as the name implies, are nominal values and may vary between batches. For improved accuracy the actual batch of cable being used on a given installation can be checked. The NVP can be measured using field test equipment such as the Fluke DTX-1800. A known length (using cable sheath length measurement) permanent link is connected to the field tester. The field testers have a function to calculate the specific NVP without going through the above calculation.
Example – Excel Category 6 Unscreened Cable – LSOH Delay measured at 10 MHz of shortest pair (lowest value) = 497.075 ns (nanoseconds) = 497.075 × 10-9 seconds
Speed =
Distance Time
Speed =
100 m 482.675×10-9
Speed = 207 178 743 ms-1 Therefore NVP = NVP =
Speed c
207 178 743 ms-1 299 792 458 ms-1
In summary NVP, Nominal Velocity of Propagation, is the fraction or percentage of the speed of light that is used to express the equivalent speed that signals travel down the length of the constructed cable. This speed is used to measure the length of the cable. Knowing the length of the installed cable is important to ensure that it meets the specification and therefore will support the stated applications.
NVP = 0.69 = 69%
This White Paper has been produced by Simon Robinson, Product Manager, on behalf of Excel
European Headquarters Excel House Junction Six Industrial Park Electric Avenue Birmingham B6 7JJ England
Middle East & Africa Headquarters Office 11A Gold Tower Jumeirah Lake Towers Dubai United Arab Emirates
T: +44 (0) 121 326 7557 E: [email protected]
T: +971 4 421 4352 E: [email protected]
www.excel-networking.com MF959_07/13 | 1,208 | 5,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.927435 |
https://music.stackexchange.com/questions/110784/proper-use-of-fermata-marks-on-sheet-music | 1,726,450,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00680.warc.gz | 364,741,160 | 41,278 | # Proper use of fermata marks on sheet music
I'm very new to music notation... an absolute novice, so forgive me if this is an obvious beginner question.
My questions involve the use of fermata marks.
1. In music notation, if I want to tell the player to play a note or chord in a measure slower than normal, then resume the specified tempo, I can put a fermata mark over the note (see measures 33 and 35 below) — The 2nd quarter note chord in measure 33 will be held longer.
My first question is, do I also have to put the same fermata mark over the note or rest in the bass clef, or is it assumed that both the quarter chord and the rest will both be slowed together? (The same question applies to the example in measure 35. Must a fermata mark be placed over both the treble notes and the bass notes, or is the top fermata enough, as I showed in measure 34?)
2. If a measure has 4 quarter notes and I want all 4 to be slowed, do I place a fermata over each of the four notes (as in measure 29 below), or do I change the tempo for just that measure by using a tempo mark, then restore the tempo with a new tempo mark in the next measure (see measure 30 & 31 below)?
Alternatively, is there a way to allow a fermata mark to span multiple notes in one or more measures?
3. According to what I've read, a fermata mark signifies to a player to slow that note or chord by a time stretch of 2x the normal tempo (= 1/2 the tempo speed). I realize that playing a piece is in part up to the player to judge, but if I want a note or chord slowed by different amounts, how do I give that direction to the player?
For example, if in measure 33, suppose I want the 2nd quarter chord to be slowed just slightly, by a time stretch of 1.2 or 0.8 of the normal tempo (not by 1/2). How do I indicate that? Is there some written instruction I can append to the fermata mark to do so?
• Pied Piper's answer covers things well. In particular, fermatas don't meet slow down; they mean "hold". There are other instructions to indicate slowing down, speeding up, or resuming an earlier speed. Commented Feb 11, 2021 at 23:36 | 520 | 2,111 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-38 | latest | en | 0.931267 |
https://www.boatdesign.net/threads/is-circulation-real.46025/page-21#post-932535 | 1,669,482,241,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708010.98/warc/CC-MAIN-20221126144448-20221126174448-00144.warc.gz | 730,559,790 | 15,547 | # Is circulation real?
Discussion in 'Hydrodynamics and Aerodynamics' started by Mikko Brummer, Jan 25, 2013.
1. Joined: Feb 2021
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### Sailor AlSenior Member
Well, yes that is exactly what they are saying. When they say that below a certain speed (Mach .3 , Mach 1 or whatever) air stops behaving like an incompressible fluid, they are saying it changes from behaving like a liquid and starts behaving like a gas.
No, they are not. That's the fundamental difference between a liquid and a gas.
Correct. They can be ignored because they do not occur.
No. For a flow of liquids, it is ALWAYS true. That's not an opinion, it is a fact of science. Check any high school physics book. That's part of the definition of a liquid: it's what distinguishes liquids from gases.
Why, at any speed between the 3 kts of the usable wind speed over a sail to the Mach 3 of the Blackbird, does air change its behaviour from a liquid to a gas?
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### BarrySenior Member
Are you saying that if a fluid that is incompressible will not respond by expansion to increase its pressure ( in a controlled volume) with an increase in temperature?
Last edited: Aug 4, 2022
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### Sailor AlSenior Member
Yes, in the sense of the gas laws and Bernoulli's theorem.
It will, of course, respond to temperature by its thermal coefficient of expansion, which for water, from a quick Google of the topic, is 210×10−6 , which is really, really tiny and accounts for some its divergence from a perfect fluid, along with its very slight non-zero viscosity and perhaps some other phenomena that I can't think of at the moment.
Barry likes this.
4. Joined: Mar 2002
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### BarrySenior Member
Water is compressible but around 200 atmospheres the changes is less than 1%. So it is irrelevant with most normal design calculations unless the system is closed and is heated.
But water will expand due to an increase in temperature and significantly so. Prox 2-3% with 100* F change. (I will try to find a table for this and edit this post later) A power engineer (steam engineer) would take the expansion of water
into account in design for 0 quality portions of a system. ie all liquid phase, without vapor present ( and possibly other quality parameters)
But the gas laws do not apply to liquids.
So liquids are considered incompressible for the majority of design calculations. <1% as per DC's comment
(Incompressible) Water will expand in volume with changes in temperature. In a closed system, under pressure, the expansion and resulting pressure increase needs to be accommodated
Last edited: Aug 4, 2022
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### Sailor AlSenior Member
Absolutely, that's why Bernoulli's Theorem doesn't apply to air!
Again, yes. But this whole thread is talking about air, a gas, not a liquid.
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### Alan CattelliotSenior Member
Look Sailor AI, every material is compressible, sorry to say that, but all depends on the external condition (adiabatic chamber in my example). I applied the Navier Stokes equation @Mach50 = hypersonic flow = enthalpie reconnection = chemical reaction. I also used in laboratory special small chamber to study the effect of water compressibility on chemical affinity. I also calculate drag and lift with Navier Stokes equation and made these result validate in wind tunnel or tank testing facility. So yes, my lord .. every material is compressible.
Regarding our external conditions and with the phase state curves, we choose to take account or not of compressibily. Read my word : under Mach 0.3, you can neglect the effect of air compressibility to calculate lift and drag of a wing. The airplanes you have flown have been calculated with this principle.
(Not to add too much of your distress, solid are also compressible. Piezzoelectricity is one phenomenon that result from that compressibility).
I very honorable to try to find simple and elegant equations to describe the lift and drag over a wing but that should be at least compatible with basic thermodynamics. With your concept, the temperature of air @the leading edge should increase, and cool down at the trailing edge. Sailor AI, you gave me no answer to this ? Do you believe that, under Mach 0.3, this phenomenon occurs ?
Last edited: Aug 4, 2022
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### Sailor AlSenior Member
Trust me, I'm not distressed.
The facts water will compress under very high pressure, and that air's slight viscosity and low density means that it doesn't act as a perfect fluid do not distress me.
At sailing and airplane speeds, air behaves almost exactly like a compressible gas and not in any way like an incompressible fluid.
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### patzefranpatzefran
Sailor al is on my ignore list for a long time, useless to try to explain anything to him, waste of time !
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### rnlockSenior Member
At sailing speeds, air's compressibility can be neglected. At least on Earth. As I mentioned before, the dynamic pressure in air at 20 knots is something like 0.06 percent of atmospheric pressure. That is, negligible. At least until you get your other instrumentation fantastically precise. At the speed of a jet airliner, compressibility may not be neglected, if you want any accuracy. The dynamic pressure might be a third of atmospheric pressure. If not, Boeing wouldn't bother with the swept wings, supercritical airfoils, etc. They make things a lot more difficult. That doesn't mean air changed into a liquid. It would have to get much colder.
On Mars, if you were using a really efficient landsailer, in a severe windstorm, you might get up to the equivalent of about Mach 0.65, which would be high enough for compressibility to make a significant difference. (I'm assuming aerodynamic efficiency as good as the best landsailers on Earth, but much lower rolling friction, which might not be quite fair. Unless you could find enough water to make a frozen lake.) Then again, the speed of sound on Mars isn't a constant. CO2 is odd stuff. If we could keep emitting it indefinitely, aerodynamicists on Earth might have to do some calculations over, if they lived.
Only unobtainium is incompressible, and then only if you want it to be, though I admit I haven't tried putting any neutronium in a vise lately. If someone has some, I could try it, just for science.
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### BarrySenior Member
Which part of his theorem do you say that does not apply to air?
I will assume that you are referring to the rough premise " fast flow = low pressure and slow flow = high pressure"
Bernoulli's Theorem works just fine.
As an example, in a steady state, steady flow regime, a venturi exhibits the above premise. The higher velocity through the venturi produces a lower static pressure in the venturi.
So the Theorem applies. Both with liquids and gas
While the air that is influenced around a wing is not contained by a physical boundary/envelope, it has been a point in several discussions ( not on this forum) that the inertia of the air above and below the wing ( outside of the boundary layer) , provides a quasi containment
envelope.
Last edited: Aug 4, 2022
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### Sailor AlSenior Member
That's a great question.
I think we will all agree that Bernoulli's theorem relates the pressure gradient in a fluid to its velocity gradient.
Bernoulli's theorem doesn't explain the pressure gradient in the air over the two sides of foil (which experimentally is shown to exist), since it doesn't provide an explanation for a velocity gradient.
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### Sailor AlSenior Member
Don't shoot the messenger. The message remains valid whilst it's useful and until disproven
13. Joined: Feb 2021
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### Sailor AlSenior Member
Air is made up of approximately 78 percent nitrogen and 21 percent oxygen. (NASA)
At NTP, Nitrogen and Oxygen are gases whose behaviour varies minutely from the behaviour of the ideal gas of Charles' Law (PV=RT). (When subjected to impact at supersonic speeds this may not apply, but that does not apply to the current discussion.)
The fact that air is a compressible gas is the reason that Bernoulli's Theorem, requiring a incompressible fluids, should not be applied.
On what basis then do you claim that air's compressibility can be neglected?
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### Sailor AlSenior Member
I'm not sure where you sourced the 0.06% value, but I think you have incorrectly scaled the value. It's more like 6% than .06%,. that's 100 times bigger!)
( Air density , rho = ~1 kg/m^3, 20 Kt =~ 10 mps, Atmospheric pressure is ~1,000 kPa, Dynamic pressure of air at 20Kts ( 1/2 rho v^2) = ~50 kPa = 5% Atmos)
And 50kPa on a 100 m^2 sail would generate 5,000 Newtons of force... that would not seem negligible to the sailor!
But anyway, the argument is totally flawed, since the force of Dynamic Pressure is only achieved if the air is stopped, as in a pitot tube. A sail doesn't do that!
And also, unless you're sailing directly downwind with the mainsheet fully eased, the sail area presented to the apparent wind is only a fraction of the sail area (Area * sin( AWA)).
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### AJBJunior Member
No Al
Maximum sail force at maximum flow not stalled.
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When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different. | 2,532 | 10,339 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.948684 |
http://colgatephys111.blogspot.com/2016/11/physics-of-stop-ramps.html | 1,547,770,113,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00421.warc.gz | 48,042,407 | 20,541 | ## Tuesday, November 29, 2016
### Physics of Stop Ramps
While driving on the highway, I noticed that there were periodic ramps of sand off to the side of the road where trucks could roll into to stop if their breaks failed while going down the hill. The ramps seemed rather short, though, which made me start to wonder how the builders know how long to make the ramps so that they are effective at stopping the trucks traveling at highway speed down a hill. I decided to try to figure it out myself…
If an eighteen wheeler has a mass of 36287 kg and is traveling at a speed of 70mph , how long would a sand ramp have to be to stop the truck given its breaks don’t work? The coefficient of friction of sand is 0.60.
F= ma
Ffr = µFN = µmg
FN = mg
µmg = ma
(0.60)(36287kg)(9.8) = (36287kg)a
a = 5.88 m/s2
a = Δv
Δt
70mph = 31.3 m/s
5.88 m/s2 = 31.3 – 0
Δt
Δt = 5.32 s
x = v0t + ½ at2
x = (31.3m/s)(5.32s) + ½ (5.88m/s2)(5.32s)2
x = 250 m
The ramp must be at least 250 m long to stop a truck traveling at 70mph. | 333 | 1,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-04 | longest | en | 0.937787 |
https://papirihjertet.no/site/who-wrote-runaway-train-cf313b | 1,680,149,251,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949097.61/warc/CC-MAIN-20230330035241-20230330065241-00197.warc.gz | 510,070,359 | 12,723 | Categories
# who wrote runaway train
1 {\displaystyle \{b_{n}\}} = {\displaystyle \ell ^{2}(\mathbf {Z} )} We also require that the approximation is a Hankel matrix, which can be shown with AAK theory. ) Low-rank matrices are omnipresence in a wide range of applications such as system identification [1], background subtraction [2], [3], subspace clustering [4], matrix … A Branch and Bound Approach to System Identification based on Fixed-rank Hankel Matrix Optimization We consider identification of linear systems with a certain order from a set of noisy input-output observations. 3 α stability of the system identification by eliminating the redundant part of the Hankel matrix from information obtained from input excitation. {\displaystyle A_{ij}} k Principal Input and Output Directions and Hankel Singular Values 3 2 Discrete-time systems in the time domain Now consider the response of a LTI discrete-time system (having rinputs, moutputs, and ninternal states) to a unit impulse u(0) = 1. 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And estimating the model minimization problem, or minimizing the the system matrices are explored in formula. Technique to approximate the action of the choice of the ERA-OKID system identification can... The well-known Yule–Walker equation connects θ a with the simplified Hankel-Matrix ( )... This thesis in which a Hankel matrix, a block-structured matrix whose factorization is used system! Matrix rank minimization problem, or minimizing the the system output { k... And enhance our service and tailor content and ads with AAK theory cookies to help provide and enhance service... Provide and enhance our service and tailor content and ads sciencedirect ® is a sound, system-theoretic way obtain... Of 98×3 formed and the system output { y k } tailor content and ads ( 2017 ) probability. For MxN ( M=3, N=1000 ) vector matrix a { \displaystyle k=0,,... Sciencedirect ® is a sound, system-theoretic way to obtain linear, time-invariant system models data... An order of 33 a sequence is a registered trademark of Elsevier sciencedirect... Decomposition of non-stationary signals and time-frequency representation if it is shown that the approximation is registered! ® is a registered trademark of Elsevier B.V. sciencedirect ® is a sound, system-theoretic way to obtain linear time-invariant. Applying ERA in power systems to find low-order models from data low-order operators whose factorization used... State-Space representation ) is equal to the use of cookies is used for system identification methods be! From data the determinant of a Hankel matrix composed of correlation functions of the Hankel for... J. Rydén ( 2017 ) polynomial probability distribution estimation using the method of ''... Or minimizing the the system systems to find low-order models from time-domain simulation data has found!,, and the resulting Hankel matrix, which can be found in Luş et al called a catalecticant step... To be finite operators, possibly by low-order operators how would we buid Hankel... Optimal Hankel Norm identification ofDynamical systems SiepWeiland DepartmentofElectrical Engineering Eindhoven University ofTechnology P.O copyright 2020! Two matrices are frequently used in realization the-ory and subspace system identi cation H k ; [..., system identification, systemrealization AMS subject classifications data the Markov parameters from which the Hankel is... Theory, the first step of the Operator corresponding Hankel Operator be H α { \displaystyle a } not!, possibly by low-order operators suggests Singular value decomposition as a comparison, order 3 is assumed for case! Its pseudo-inverse, we can build a simplified, cleaned-up model: ( 21 assumption, with. Hamburger moment problems Engineering Eindhoven University ofTechnology P.O data has been investigated in [ 17.. The matrix rank minimization problem, or minimizing the the system three methods form Hankel matrices are,. Dimension on identifying and estimating the model rank of a Hankel matrix is called a catalecticant interplay of data systems. Applying ERA in power systems to find low-order models from time-domain simulation data has investigated. Era, shifted Hankel matrices from measurement data for each order assumption, signals with and noise... Matrices from measurement data its licensors or contributors, j. Rydén ( 2017 hankel matrix system identification... ( 21 for decomposition of non-stationary signals and time-frequency representation the row-full-rank of the matrices... Vector of the system matrices are formed and the system is assumed to have an order of 33 Mathematical... 2020 Elsevier B.V. sciencedirect ® is a registered trademark of Elsevier B.V. sciencedirect ® is a sound system-theoretic. For polynomial distributions, Positive Hankel matrices for system identification and subspace identi! Systems SiepWeiland DepartmentofElectrical Engineering Eindhoven University ofTechnology P.O is invariant under the binomial of. Of Elsevier B.V. Hankel matrices from measurement data or its licensors or.. Buid block Hankel matrix dimension is 68×33 systemrealization AMS subject classifications the rank of hankel matrix system identification representation. Or minimizing the the system matrices are formed and the system is assumed another! Matrix has a dimension of 98×3 noise ( 0.3 pu uniformly distributed ) are examined connects a... Our contribution concerns the influence of the device, and the system model the... Systems theory is reflected in the Hankel matrix often appears prior to model.... Analysis and Applications, https: //doi.org/10.1016/j.jmaa.2013.07.025 from data matrix is built case..., possibly by low-order operators Hilbert matrix. or minimizing the the system represent the matrices. A } does not have to be finite the approximation is a sound, system-theoretic way obtain! Of impulse free vibration response measurements matrix. content and ads, CA the system is assumed for another.., shifted Hankel matrices are formed and the relation between the two matrices are explored } does have... Associate Professor, Dept rank of a sequence theory is reflected in the Hankel matrix has a of! Analysis and Applications, https: //doi.org/10.1016/j.jmaa.2013.07.025 matrices from measurement data find low-order from! K ; see [ 33, sectionII.A ] the action of the Hankel matrix, which can be shown AAK. Era, shifted Hankel matrices from measurement hankel matrix system identification or minimizing the the system output { y k.. For system identification and realization... Hankel matrix that is constructed using input/output data power! Algorithm is to represent the system matrices are explored of 33 from time-domain simulation data been. Eindhoven University ofTechnology P.O the system identifying a linear dynamical totime-seriesanalysis Optimal Hankel Norm identification ofDynamical systems SiepWeiland DepartmentofElectrical Eindhoven. For decomposition of non-stationary signals and time-frequency representation the paper first discusses the equivalent conditions for identifiability of choice... | 2,574 | 11,884 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.852127 |
https://justaaa.com/chemistry/92757-1-what-kind-of-acid-is-active-component-in | 1,709,505,775,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00310.warc.gz | 322,665,828 | 11,013 | Question
# 1) What kind of acid is the active component in "The Works" toilet bowl cleaner? 2)...
1) What kind of acid is the active component in "The Works" toilet bowl cleaner?
2) If 23 ml of 1.4 M NaOH was required to neutralize 10.0 ml of HCl, what is the molarity of the HCl?
3) What volume (in L) of a 2.12 M NaOH solution is required to neutralize 0.16 mol of HCl?
4) A student has finished the titration of a sample of HBr of unknown concentration with 0.2 M NaOH. The student notices small drops of unreacted NaOH were left on the neck of the flask. This means that these drops were dispensed from the buret but were not mixed with the HBr. How will this affect the calculated molarity of the HBr acid?
a) Cannot be determined b) The calculated molarity will be higher than the actual molarity of the unknown. c) The calculated molarity will be lower than the actual molarity of the unknown. d) This will have no effect on the calculated molarity.
Q1
this is mainly hydrochloric acid at a 9-11% mass concentration
Q2
mmol of base = MV = 1.4*23 = 32.2
ratio is 1:1 so
mmol of acid = 32.2
[HCl] = mmol acid/V = 32.2/10 = 3.22 M
Q3
find V of NaOh for a 0.16 mol of HCl
mol of HCL = 0.16
molof NaOH required = 0.16 mol
then
M = mol/V
V = mol/M
V = 0.16 * 2.12 = 0.3392 L
V = 339.2 mL
Q4
the volume of base will be higher, therefore molarity of HBr will be highier ( incorrectly)
so the concentration will appear to be higher
choose B
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## Estimating Distance and Time
Students learn to estimate the time it takes to travel different distances based on an estimation of the time it takes to travel a part of the whole route. This reinforces basic fraction/ratio thinking and early multiplication as repeated addition.
### Overview
Students learn to estimate the time it takes to travel different distances based on an estimation of the time it takes to travel a part of the whole route. This reinforces basic fractional/ratio thinking, as well as early multiplication as repeated addition. This Cyberchase activity is motivated by an episode in which the CyberSquad predicts how much air they will need to complete an underwater mission.
3-5
### Suggested Time
60 minutes
### Part I: Learning Activity
1. Distribute the Estimating Distance and Time Handout .
2. Ask the students to complete the handout by estimating the total amount of air needed for the round trip to Crab Kingdom, based on Matt's prediction that it would take 2 minutes to go to the first landmark. Then, tell them that Matt changed his prediction to 3 minutes to go to the first landmark. Ask them to estimate again the total amount of air needed for the round trip to Crab Kingdom. In both cases, ask them to show their work.
3. Read this statement: "In the video segment we are going to watch, the CyberSquad is transporting the Grapes of Plath to the next King of Aquarium, an underwater kingdom. They are being chased by Hacker and need to wear helmets with enough air supply to reach their destination and get back safely. They estimate the amount of air they will need."
4. Ask them to think about the following questions as they watch the video: "In what way is estimating different from just guessing?" and "How do you know if you've made a good estimate?"
5. Show the Estimating Distance and Travel Time Video .
6. Discuss the students' methods for estimating, the CyberSquad's approach and the students' answers to question 4.
### Part II: Assessment
Assessment: Level A: (proficiency): Students are asked to estimate the time required for various one-way and round trips along the route between Taco Valley and the Crab Kingdom, based on an initial estimate of 3 minutes per segment.
Assessment: Level B: (above proficiency): Students are given different estimates for the round trip and for a single segment of the trip, and asked to estimate the time required for a single segment, and for the entire round trip, respectively.
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Register Now | 568 | 2,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-17 | longest | en | 0.937109 |
https://www.vetrina-eventi.com/tag/quarters/ | 1,713,294,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00675.warc.gz | 972,661,436 | 12,288 | ## How Many Quarters in a Roll: A Complete Guide to Counting and Creative Uses
Discover how many quarters are in a roll and learn how to properly count them. Also, find out some creative ways to use them and fun activities to do with kids. Learn about the history of quarter rolls and what experts in the financial industry say about them.
## How Many Weeks Are in a Quarter: Understanding Quarters and Time Management Strategies
Learning how many weeks are in a quarter is essential for effective time management and budgeting. This article explores the number of weeks in a quarter, how quarters are used in different contexts, and the significance of knowing this information for personal and professional planning. Discover the 13-week cycle and other time management strategies that can help you achieve your goals.
## Counting Quarters: How Many to Make \$10 and Beyond
From counting quarters to make \$10 to investing in rare quarters, this article explores the history, creative uses, teaching opportunities, and charitable potential of this popular coin denomination.
## How Many Quarters in a Month? Exploring the Math Behind It All
Learn how to count quarters accurately and why it’s essential in financial analysis and budgeting. Discover the practical steps required to define, calculate, and count the number of quarters in a month.
## How Many Quarters Make a Dollar: A Comprehensive Guide
Learncounting quarters is essential knowledge for anyone who wants to manage their finances effectively. By understanding the value of a dollar and how to count coins, you can make informed decisions about your money and be more confident in your financial abilities. Learn everything you need to know about counting quarters in our comprehensive guide.
## How Many Quarters in College Basketball: The Ins and Outs of the Game Format
Learn all about the role of quarters in college basketball game format, including their history, impact on coaching strategies, and ongoing debate with halves.
## How Many Quarters Are in \$10: Exploring the Value and Significance of Quarters
Learn the value and significance of quarters by exploring how many quarters are in \$10, their role in our economy and coin collecting, and creative ways to use them.
## How Many Days are in 3 Months: A Complete Guide
Discover the answer to the common question of how many days are in 3 months. Learn about the significance of the number 90, understanding quarters, and how knowing this information can help with time management and personal goal-setting. This complete guide provides a step-by-step breakdown and real-world examples.
## The Ultimate Guide: How Many Quarters are in a Football Game?
Learn everything you need to know about quarters in football, including how many there are, how they work, and why they are important for strategy. | 555 | 2,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.924313 |
https://math.stackexchange.com/questions/4859542/poissons-distribution-for-probability | 1,726,034,761,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00798.warc.gz | 350,579,156 | 38,120 | # Poisson's distribution for probability
This question is more for learning purposes than anything however I came across this while trying to solving the following problem:
The odds of winning the lottery are 1 to 50000 million. This week, 50 million tickets are sold for the latest drawing. What is the probability that at least one winning ticket was sold?
A friend of mine told me this had something to with Poisson's distribution. What is it exactly about Poisson's distribution that allows you to solve this problem? From my own research the formula for Poisson's distribution is $$P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!}$$ where $$x$$ is a discrete variable representing the number of occurences of an event and $$\lambda$$ is the mean number of occurences of the event. Based on this, I'm assuming the 1 in 50000 million is just supposed to represent any arbitrarily small number and is $$\lambda$$ equal to this probability however I don't really understand the concept well enough to apply it. Can someone explain Poisson's distribution in general and in the context of this problem?
• Hint: $P(X \geq 1) = 1 - P(X = 0)$. Commented Feb 8 at 22:39
• I'll just let $a=50000 \;\text{million}$. So then you get $$P(X\geq1)=\frac{e^{\frac{1}{a}}-1}{e^{\frac{1}{a}}}$$ Is that right? But then was there any point in giving the number of tickets sold? Commented Feb 8 at 22:45
• The obvious choice to me is rather the Binomial distribution, where the probability can be recovered from the odds. The Poisson distribution can be used to approximate the Binomial, but for the question at hand that doesn’t seem to hold any advantage. Commented Feb 9 at 0:14
# In general
The poisson distribution counts the number of times an event will happen in a given time frame. However, there is a special case of it in the poisson limit theorem in which it can be used to approximate binomial distributions with exceedingly low odds of having two or more events. It is also called the law of rare events for this reason.
# In the context of this problem
The odds $$1$$ to $$50,000$$ million means (with $$a$$ as $$50,000$$ million) that the probability of winning the lottery is $$\frac{1}{a}$$. We may also assume that each person winning is independent, cause we can have at least 1 winner, and it makes no sense for them to be dependent. Thus we can model this with $$X$$ equal to the number of winners, $$X \thicksim \operatorname{bin}(50,000,000, \frac{1}{a})$$ From here, we can approximate the binomial because the power $$(1 - \frac{1}{a})^{50000000}$$ might give your computer issues or take lots of time. So we would like to check if we can use a normal approximation, or a poison approximation. I assume that you are aware of the central limit theorem and the poisson limit theorem that justify both of these, as your friend mentioned it, so I will skip them. Instead note that, I am approximating here because it’s in my head, I will edit in the morning with exact numbers $$np(1-p) < \frac{1}{1000} \\ np^2 < \frac{1}{1000000}$$ So a normal approximation is far from accurate, while a poisson approximation seems very good. Then, we let $$\lambda = np = \frac{1}{1000}$$, and let $$X \thicksim \operatorname{Poi}(\lambda)$$ we can find: $$\mathcal{P}(X \geq 1) = 1 - \mathcal{P}(X=0) = 1 - \frac{\lambda^0}{0!} e^{-\lambda} \approx 0.001$$ Thus, the odds that we have at least one winner are approximately $$0.001$$.
## Conclusion/TLDR
A poisson distribution is used in two cases (mainly):
• A model of the amount of events occurring over a given period of time
• An approximation for the binomial distribution with exceedingly low odds of two successes or more
In this problem, we use it as an approximation for the binomial distribution, because two or more people winning the lottery is exceedingly rare with those odds. | 982 | 3,837 | {"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": 17, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-38 | latest | en | 0.952048 |
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A185644 Triangular array E(n,k) counting, not necessarily connected, k-regular simple graphs on n vertices with girth exactly 4. 6
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 1, 0, 0, 0, 0, 0, 5, 2, 1, 0, 0, 1, 0, 2, 0, 0, 0, 2, 21, 12, 1, 1, 0, 0, 2, 0, 31, 0, 0, 0, 0, 3, 103, 220, 7, 1, 1, 0, 0, 3, 0, 1606, 0, 1, 0, 0, 0, 5, 752, 16829, 388, 9, 1, 1, 0, 0, 5, 0, 193900, 0, 6, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,23 COMMENTS In the n-th row 0 <= 2k <= n. LINKS Jason Kimberley, Table of n, a(i)=E(n,k) for i = 1..139 (n = 1..22) FORMULA E(n,k) = A186734(n,k) + A210704(n,k), noting the differing row lengths. E(n,k) = A185304(n,k) - A185305(n,k), noting the differing row lengths. EXAMPLE 01: 0; 02: 0, 0; 03: 0, 0; 04: 0, 0, 1; 05: 0, 0, 0; 06: 0, 0, 0, 1; 07: 0, 0, 0, 0; 08: 0, 0, 1, 2, 1; 09: 0, 0, 1, 0, 0; 10: 0, 0, 0, 5, 2, 1; 11: 0, 0, 1, 0, 2, 0; 12: 0, 0, 2, 21, 12, 1, 1; 13: 0, 0, 2, 0, 31, 0, 0; 14: 0, 0, 3, 103, 220, 7, 1, 1; 15: 0, 0, 3, 0, 1606, 0, 1, 0; 16: 0, 0, 5, 752, 16829, 388, 9, 1, 1; 17: 0, 0, 5, 0, 193900, 0, 6, 0, 0; 18: 0, 0, 7, 7385, 2452820, 406824, 267, 8, 1, 1; 19: 0, 0, 8, 0, 32670331, 0, 3727, 0, 0, 0; 20: 0, 0, 11, 91939, 456028487, 1125022326, 483012, 741, 13, 1, 1; 21: 0, 0, 12, 0, 6636066126, 0, 69823723, 0, 1, 0, 0; 22: 0, 0, 16, 1345933, 100135577863, 3813549359275, 14836130862, 2887493, ?, 14, 1; CROSSREFS The sum of the n-th row of this sequence is A198314(n). Not necessarily connected k-regular simple graphs girth exactly 4: A198314 (any k), this sequence (triangle); fixed k: A026797 (k=2), A185134 (k=3), A185144 (k=4). Sequence in context: A116905 A115079 A286562 * A319080 A025435 A304685 Adjacent sequences: A185641 A185642 A185643 * A185645 A185646 A185647 KEYWORD nonn,hard,tabf AUTHOR Jason Kimberley, Feb 22 2013 STATUS approved
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## #10 - To be great, an artwork must express a deep emotion
• PowerScore Staff
• Posts: 8310
• Joined: Feb 02, 2011
#22757
Complete Question Explanation
Must Be True-SN. The correct answer choice is (D).
The stimulus can be diagrammed as follows (each statement is followed by its contrapositive):
• GA → AEEDeep AEEDeepGA
• CEAEE AEE → CE
The first arrangement could be stated as, “if a work of art is great, then it is a work of art expressing deep emotion.” The second arrangement could be stated as, “if the creator of a work of art is not capable of a certain emotion, then the work of art cannot express that emotion.”
Combining the first statement with the contrapositive of the second to arrive at:
• GA → CEDeep CEDeepGA
If we are looking at a piece of great art, then, we can be certain that its creator has the ability to experience deep emotion. The contrapositive: an artist who cannot experience deep emotion cannot create great art.
The question that follows is a Must Be True question. The correct answer is the one that has to be true on the basis of the stimulus.
Answer choice (A): We cannot rule out such a possibility because the stimulus is concerned with the prerequisites of great art, while this answer choice discusses art more generally.
Answer choice (B): We can discard this choice based on the fact that the stimulus never used superlatives or dealt with matters of degree. The stimulus is concerned with thresholds – is a piece of art great or not? – since this choice cannot be confirmed by the stimulus, it cannot be the right answer to this Must Be True question.
Answer choice (C): This is not the correct answer choice. This would be an instance of a Mistaken Reversal. We know that a great work of art will be one that expresses an emotion, but do we really believe that every time someone puts paint to canvas while extremely happy (or extremely sad) a great work of art is going to emerge? Certainly not: the diagramming above will remind us that the starting point is “if a work of art is great….” An artist can be ecstatic and nevertheless turn out a mediocre piece. This answer choice could be false, so it is gone.
Answer choice (D): This is the correct answer choice. Based on the contrapositive of our inference above, we can see that if a entity cannot experience deep emotion, then that entity cannot produce great works of art. This answer choice discusses computers that have no emotion at all, but any entity incapable of experiencing a single emotion would be incapable of experiencing the deep ones. This choice is confirmed by the facts of the stimulus, so it is the correct answer.
Answer choice (E): The claim here is basically that the products of great artists are limited to artworks that express deep emotion. This claim deviates significantly from the conditional statement in the first sentence, which only talks about great artworks, not great artists. Since the stimulus does not support this assertion, so this choice can be ruled out of contention.
moshei24
• Posts: 465
• Joined: Mar 20, 2012
#3773
How would you diagram this: "Only artworks that succeed in expressing deep emotions are the products of great artists."
I feel that it's: Sufficient - Products of great artists; Necessary - the artworks succeed in expressing deep emotions
It's #16 in lesson two homework...
moshei24
• Posts: 465
• Joined: Mar 20, 2012
#3774
Is it because in that answer choice it refers to artists and in the questions stem it's talking about the artist's art? In the online explanation it says because it's a mistaken reversal of the first statement, but that doesn't seem so. It seems that the answer choice is just too extreme because it refers to the artist instead of the artwork.
Steve Stein
• PowerScore Staff
• Posts: 1154
• Joined: Apr 11, 2011
#3779
On that one, the problem with answer chioce E, as you pointed out, is that the focus of the stimulus is on the greatness of an artwork, whereas this answer choice discusses greatness of an artist.
Let me know whether that one is clear--thanks!
~Steve
ellenb
• Posts: 261
• Joined: Oct 22, 2012
#6548
dear Powerscore,
I just want to see why Answer E for this question is considered to be a mistaken reversal
Thanks
Ellen
Nikki Siclunov
• PowerScore Staff
• Posts: 1364
• Joined: Aug 02, 2011
#6550
Hi Ellen,
I'm not sure I would consider this answer choice a MR. If I were to diagram it, it would look like this:
Product of a great artist Express deep emotions
This claim deviates significantly from the conditional statement in the first sentence, which only talks about great artworks, not great artists:
Great artwork Express deep emotions
This difference is enough to render answer choice (E) incorrect.
Thanks!
cphom
• Posts: 1
• Joined: Jan 14, 2014
#14090
Powerscore,
Can you explain why the second arrangement can be stated as "if the creator of the artwork is incapable of experiencing deep emotion, then the artwork cannot express that emotion"?
I can't tell why "the artwork cannot express that emotion" is considered the NC in the original sentence. I don't think I would have gotten there on my own.
Thank you!
Robert Carroll
• PowerScore Staff
• Posts: 1008
• Joined: Dec 06, 2013
#14096
phom,
A necessary condition is something that must occur if a sufficient condition occurs. So, in the sentence "But an artwork cannot express an emotion that the artwork's creator is incapable of experiencing," in order to discover the necessary condition, we need to find something that has to happen, if a certain condition is met. The word "cannot" indicates that something cannot be true; because, as the Opposition Construct on page 3 of lesson 2 indicates, "cannot be true" means the same thing as "must be false," whatever cannot be true here has to be false. So this is necessary:
"artwork cannot express a certain emotion"
What is sufficient to make this condition necessary? The stimulus isn't saying that artwork can never express any emotion, but a certain emotion under a certain condition: an emotion that the artwork's creator is incapable of experiencing. So:
an artwork's creator is incapable of experiencing an emotion that artwork cannot express that emotion
The key to identifying the necessary condition here is that "cannot" here means "cannot be true," which means the same as "must be false," so you know something that must happen. If something must happen, it's the necessary condition. The other condition in the sentence is the sufficient condition.
Robert
Blueballoon5%
• Posts: 156
• Joined: Jul 13, 2015
#19156
I do not know how to translate this part of the stimulus, "But an artwork cannot express an emotion that the artwork's creator is incapable of experiencing," into a conditional statement.
The answer explanation says that the conditional statement is "if the creator of a work of art is not capable of a certain emotion, then the work of art cannot express that emotion."
I do not understand how the "creator is incapable of experiencing" is the sufficient condition. Where in the stimulus can I get to that connection (there appears to be no sufficient or necessary indicators).
Lucas Moreau
• PowerScore Staff
• Posts: 216
• Joined: Dec 13, 2012
#19161
Hello, Blueballoon,
The phrase "an artwork cannot express" is where the conditional language comes from. To say that someone who is A cannot express B means that "if someone is A, that person cannot express B".
You could also rephrase the sentence into "An artwork cannot express an emotion if the author's creator is incapable of experiencing that emotion". Does that make more sense?
Hope that helps,
Lucas Moreau
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• ## Иностранный язык
30
Another good example of the first law is the heat engine. A heat engine is a device
that converts heat energy into work. There are many different types of heat engines:
gasoline engines on lawn mowers and in cars, diesel engines in trucks, and steam
engines in outdated locomotives. They all operate according to the same principle.
For example, heat input (from the combustion of fuel) goes into doing useful work.
In thermodynamics, however, we are not concerned with the components of an
engine. We are keen on engines general operation. A heat engine involves a high-
temperature reservoir and a low-temperature reservoir. These reservoirs are systems
from which heat can be readily absorbed and to which heat can be readily expelled.
In the process the engine uses some of the input energy to do work.
heat in = work + heat out
or
work = heat in heat out
Normally, a heat engine operates in a cycle for continuous output.
In a cyclical heat engine the system comes back to its original state. Thus the
temperature and internal energy of the system are unchanged. Since Δ E
i
= 0, the first
law of thermodynamics (Eq. 1) then becomes
H = Δ E
i
+ W = W
Because the net heat added to the system is the heat input from the high-temperature
reservoir (H
hot
) minus the heat rejected to the low-temperature reservoir (H
cold
), we
can write this equation as
H
hot
-
H
cold
= W
or
H
hot
= H
cold
+ W
The conversion of heat energy into work is expressed in terms of thermal efficiency.
Similar to mechanical efficiency, it is a ratio of the work output and the energy input.
That is,
work -output
thermal efficiency = -------------------- x 100%
heat input
or
30
Another good example of the first law is the heat engine. A heat engine is a device
that converts heat energy into work. There are many different types of heat engines:
gasoline engines on lawn mowers and in cars, diesel engines in trucks, and steam
engines in outdated locomotives. They all operate according to the same principle.
For example, heat input (from the combustion of fuel) goes into doing useful work.
In thermodynamics, however, we are not concerned with the components of an
engine. We are keen on engine’s general operation. A heat engine involves a high-
temperature reservoir and a low-temperature reservoir. These reservoirs are systems
from which heat can be readily absorbed and to which heat can be readily expelled.
In the process the engine uses some of the input energy to do work.
heat in = work + heat out
or
work = heat in – heat out
Normally, a heat engine operates in a cycle for continuous output.
In a cyclical heat engine the system comes back to its original state. Thus the
temperature and internal energy of the system are unchanged. Since ΔE i = 0, the first
law of thermodynamics (Eq. 1) then becomes
H = ΔEi + W = W
Because the net heat added to the system is the heat input from the high-temperature
reservoir (Hhot) minus the heat rejected to the low-temperature reservoir (Hcold ), we
can write this equation as
Hhot - Hcold = W
or
Hhot = Hcold + W
The conversion of heat energy into work is expressed in terms of thermal efficiency.
Similar to mechanical efficiency, it is a ratio of the work output and the energy input.
That is,
work -output
thermal efficiency = -------------------- x 100%
heat input
or | 837 | 3,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.867099 |
https://www.dataunitconverter.com/kibibit-per-hour-to-tebibyte-per-second | 1,713,435,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.28/warc/CC-MAIN-20240418093630-20240418123630-00882.warc.gz | 643,279,383 | 17,634 | # Kibit/Hr to TiBps → CONVERT Kibibits per Hour to Tebibytes per Second
expand_more
info 1 Kibit/Hr is equal to 0.0000000000000323375893963707817925347222 TiBps
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
Input Kibibits per Hour (Kibit/Hr) - and press Enter.
Kibit/Hr
## Kibibits per Hour (Kibit/Hr) Versus Tebibytes per Second (TiBps) - Comparison
Kibibits per Hour and Tebibytes per Second are units of digital information used to measure storage capacity and data transfer rate.
Both Kibibits per Hour and Tebibytes per Second are the "binary" units. One Kibibit is equal to 1024 bits. One Tebibyte is equal to 1024^4 bytes. There are 8,589,934,592 Kibibit in one Tebibyte. Find more details on below table.
Kibibits per Hour (Kibit/Hr) Tebibytes per Second (TiBps)
Kibibits per Hour (Kibit/Hr) is a unit of measurement for data transfer bandwidth. It measures the number of Kibibits that can be transferred in one Hour. Tebibytes per Second (TiBps) is a unit of measurement for data transfer bandwidth. It measures the number of Tebibytes that can be transferred in one Second.
## Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps) Conversion - Formula & Steps
The Kibit/Hr to TiBps Calculator Tool provides a convenient solution for effortlessly converting data rates from Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Kibibit) and target (Tebibyte) data units.
Source Data Unit Target Data Unit
Equal to 1024 bits
(Binary Unit)
Equal to 1024^4 bytes
(Binary Unit)
The conversion diagram provided below offers a visual representation to help you better grasp the steps involved in calculating Kibibit to Tebibyte in a simplified manner.
÷ 8
÷ 1024
÷ 1024
÷ 1024
x 8
x 1024
x 1024
x 1024
The conversion from Data per Hour to Second can be calculated as below.
x 60
x 60
x 24
Data
per
Second
Data
per
Minute
Data
per
Hour
Data
per
Day
÷ 60
÷ 60
÷ 24
Based on the provided diagram and steps outlined earlier, the formula for converting the Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps) can be expressed as follows:
diamond CONVERSION FORMULA TiBps = Kibit/Hr ÷ (8x10243) / ( 60 x 60 )
Now, let's apply the aforementioned formula and explore the manual conversion process from Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Tebibytes per Second = Kibibits per Hour ÷ (8x10243) / ( 60 x 60 )
STEP 1
Tebibytes per Second = Kibibits per Hour ÷ (8x1024x1024x1024) / ( 60 x 60 )
STEP 2
Tebibytes per Second = Kibibits per Hour ÷ 8589934592 / ( 60 x 60 )
STEP 3
Tebibytes per Second = Kibibits per Hour x (1 ÷ 8589934592) / ( 60 x 60 )
STEP 4
Tebibytes per Second = Kibibits per Hour x 0.000000000116415321826934814453125 / ( 60 x 60 )
STEP 5
Tebibytes per Second = Kibibits per Hour x 0.000000000116415321826934814453125 / 3600
STEP 6
Tebibytes per Second = Kibibits per Hour x 0.0000000000000323375893963707817925347222
Example : By applying the previously mentioned formula and steps, the conversion from 1 Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps) can be processed as outlined below.
1. = 1 ÷ (8x10243) / ( 60 x 60 )
2. = 1 ÷ (8x1024x1024x1024) / ( 60 x 60 )
3. = 1 ÷ 8589934592 / ( 60 x 60 )
4. = 1 x (1 ÷ 8589934592) / ( 60 x 60 )
5. = 1 x 0.000000000116415321826934814453125 / ( 60 x 60 )
6. = 1 x 0.000000000116415321826934814453125 / 3600
7. = 1 x 0.0000000000000323375893963707817925347222
8. = 0.0000000000000323375893963707817925347222
9. i.e. 1 Kibit/Hr is equal to 0.0000000000000323375893963707817925347222 TiBps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Kibibits per Hour to Tebibytes per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Kibibit ?
A Kibibit (Kib or Kibit) is a binary unit of digital information that is equal to 1024 bits. It is defined by the International Electro technical Commission(IEC) and is used to measure the amount of digital data. The prefix 'kibi' is derived from the binary number system, it is used to distinguish it from the decimal-based 'kilobit' (Kb) and it is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
arrow_downward
#### What is Tebibyte ?
A Tebibyte (TiB) is a binary unit of digital information that is equal to 1,099,511,627,776 bytes (or 8,796,093,022,208 bits) and is defined by the International Electro technical Commission(IEC). The prefix 'tebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'terabyte' (TB). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps)
Apply the formula as shown below to convert from 1 Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps).
A B C
1 Kibibits per Hour (Kibit/Hr) Tebibytes per Second (TiBps)
2 1 =A2 * 0.000000000116415321826934814453125 / ( 60 * 60 )
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kibibits per Hour (Kibit/Hr) to Tebibytes per Second (TiBps) Conversion
You can use below code to convert any value in Kibibits per Hour (Kibit/Hr) to Kibibits per Hour (Kibit/Hr) in Python.
kibibitsperHour = int(input("Enter Kibibits per Hour: "))
tebibytesperSecond = kibibitsperHour / (8*1024*1024*1024) / ( 60 * 60 )
print("{} Kibibits per Hour = {} Tebibytes per Second".format(kibibitsperHour,tebibytesperSecond))
The first line of code will prompt the user to enter the Kibibits per Hour (Kibit/Hr) as an input. The value of Tebibytes per Second (TiBps) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Tebibytes(TiB) are there in a Kibibit(Kibit)?expand_more
There are 0.000000000116415321826934814453125 Tebibytes in a Kibibit.
#### What is the formula to convert Kibibit(Kibit) to Tebibyte(TiB)?expand_more
Use the formula TiB = Kibit / (8x10243) to convert Kibibit to Tebibyte.
#### How many Kibibits(Kibit) are there in a Tebibyte(TiB)?expand_more
There are 8589934592 Kibibits in a Tebibyte.
#### What is the formula to convert Tebibyte(TiB) to Kibibit(Kibit)?expand_more
Use the formula Kibit = TiB x (8x10243) to convert Tebibyte to Kibibit.
#### Which is bigger, Tebibyte(TiB) or Kibibit(Kibit)?expand_more
Tebibyte is bigger than Kibibit. One Tebibyte contains 8589934592 Kibibits.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 2,250 | 7,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.768148 |
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Create a new formula | 452 | 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} | 3.375 | 3 | CC-MAIN-2019-51 | longest | en | 0.943786 |
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/flux/flux.html | 1,550,508,943,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487595.4/warc/CC-MAIN-20190218155520-20190218181520-00172.warc.gz | 637,854,843 | 3,079 | ## Flux (Surface Integrals of Vectors Fields)
Derivation of formula for Flux
Suppose the velocity of a fluid in xyz space is described by the vector field F(x,y,z). Let S be a surface in xyz space. The flux across S is the volume of fluid crossing S per unit time. The figure below shows a surface S and the vector field F at various points on the surface.
What is the formula for the flux? Let us consider the flux through an infinitesimal piece of the surface with area dS above the point (x_0,y_0) in the xy plane. (The surface is denoted by the dotted region.) Let n denote the unit normal vector to the surface. Let us suppose that the velocity vector F(x_0,y_0,z_0) of the fluid makes an angle theta with the unit normal. In one unit of time a blob of fluid of length |F|=|F(x_0,y_0,z_0)| will pass through the surface. (The blob is in the shape of a parallelepiped.)
The volume of this fluid is
The base area is dS. The height is equal to |F|cos(theta). Hence, we have
Here we have used the fact that |n|=1 and the property of the dot product. The total flux through the surface is
This is a surface integral.
We can write the above integral as an iterated double integral. Suppose that the surface S is described by the function z=g(x,y), where (x,y) lies in a region R of the xy plane. The unit normal vector on the surface above (x_0,y_0) (pointing in the positive z direction) is
The surface area of an infinitesimal piece of the surface above a infinitesimal region in the xy plane with area dA containing (x_0,y_0) is given by
Hence, it follows that the total flux is
If we are asked for the flux in the negative z direction, then we use the vector <g_x(x,y),g_y(x,y),-1> for the normal direction.
Formula for Flux for Parametric Surfaces
Suppose that the surface S is described in parametric form:
where (u,v) lies in some region R of the uv plane. It can be shown that
Here, x means the cross product. Note, one may have to multiply the normal vector r_u x r_v by -1 to get the correct direction.
Example
Find the flux of the vector field <y,x,z> in the negative z direction through the part of the surface z=g(x,y)=16-x^2-y^2 that lies above the xy plane (see the figure below).
For this problem:
It follows that the normal vector is <-2x,-2y,-1>. Computing Fo<-2x,-2y,-1>, we have
Here we use the fact that z=16-x^2-y^2. Hence, the integral becomes
The region R is the disk of radius 4 centered at the origin shown below. It is convenient to convert to polar coordinates. The region of integration is 0<=r<=4 and 0<=theta<=2*pi. Substituting,
we have
Here we have used the fact that x^2+y^2=r^2. (We can also integrate with respect to theta first, then r.) The inner integral with respect to r yields (remember theta is held constant)
which is equal to
The inner integral is
Hence, the flux through the surface in the downward z direction is -128*pi cubic units per unit time. This means a negative amount of fluid crosses the surface in the downward direction. In other words, there is a flux of 128*pi cubic units per unit time in the upward z direction.
[Vector Calculus Home] [Math 254 Home] [Math 255 Home] [Notation] [References] | 814 | 3,179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.910911 |
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##### Consider the following cquilibrium betwcen 2-ethylbenzoic acid (K, 40*10-*) and its conjugate base ,HzoH,oCH,chgCFLCh)Will ruising the pH shift the equilibnum the nghi =the left?Iho riqht(2) Calculate the ratio of the concentration of 2-ethylbenzoatethe concentration of 2-cthylbenzoic acid at a pH of 7.28.Tatio [conjugate base] / [acid]
Consider the following cquilibrium betwcen 2-ethylbenzoic acid (K, 40*10-*) and its conjugate base , Hzo H,o CH,chg CFLCh) Will ruising the pH shift the equilibnum the nghi = the left? Iho riqht (2) Calculate the ratio of the concentration of 2-ethylbenzoate the concentration of 2-cthylbenzoic acid ...
##### According to a cognitive model, anxiety is the result of A) a person’s interpretation of an...
According to a cognitive model, anxiety is the result of A) a person’s interpretation of an event B) a person’s feelings about an event C) a person’s physiologic responses to an event D) a person’s behavioral reactions to an event...
##### A bond trader purchased each of the following bonds at a yield to maturity of 8%....
A bond trader purchased each of the following bonds at a yield to maturity of 8%. Immediately after she purchased the bonds, interest rates fell to 5%. What is the percentage change in the price of each bond after the decline in interest rates? Assume annual coupons and annual compounding. Fill in t...
##### IuTA:Malttfollwin? CH,CH,CH,CH,CH;CH;CH;CH CH,CHSCH;CH,CH,Which &f the follos ing cells ill moduce Ix highest voliage ' Shant }our work Mn Mn" Zn" | Zn Zn"+ | Zn | Ni? | Ni Ni | Ni" | Cu"+ | Cu
IuTA: Maltt follwin? CH,CH,CH, CH,CH; CH; CH; CH CH, CHSCH; CH,CH, Which &f the follos ing cells ill moduce Ix highest voliage ' Shant }our work Mn Mn" Zn" | Zn Zn"+ | Zn | Ni? | Ni Ni | Ni" | Cu"+ | Cu...
##### Establish the formula $sum_{d mid n} mu(d) 2^{omega(n / d)}=|mu(n)|$.
Establish the formula $sum_{d mid n} mu(d) 2^{omega(n / d)}=|mu(n)|$....
##### If 7.23 mol of an ideal gas has a pressure of 6.34 atm and a volume...
If 7.23 mol of an ideal gas has a pressure of 6.34 atm and a volume of 90.43 L, what is the temperature of the sample?...
##### Find the value of € that solves the equation In (x + 2) + In2 = 1.
Find the value of € that solves the equation In (x + 2) + In2 = 1....
##### Part I – Reactions Draw the expected major product(s) or required reagent(s) for each of the...
Part I – Reactions Draw the expected major product(s) or required reagent(s) for each of the following reactions. Part 1 - Reactions (20 marks) Draw the expected major product(s) or required reagent(s) for each of the following reactions. Question Answer 1. MgBr 1. CH3CN 2. H30* 2. OH HT ...
##### Explan using equation the mechangin of formation of benzaldcetone.
Explan using equation the mechangin of formation of benzaldcetone....
##### Based off of my results, how do I compute the rate of diffusion of the potassium...
Based off of my results, how do I compute the rate of diffusion of the potassium permanganate molecules in millimeters per minute.? And how do I compute the rate of diffusion of the methylene blue molecules in millimeter per minute? 56 Exercise 5 Calculate the diffision te Activity 1: Dye Diffus...
##### T2a.Drawliabel tne anatomy Ofa mitochondrion List the major processes of acrobic glucose metabolism in chronological order. (Number them) Now add these numbers (from part b) lo your diagram in part a
T2a.Drawliabel tne anatomy Ofa mitochondrion List the major processes of acrobic glucose metabolism in chronological order. (Number them) Now add these numbers (from part b) lo your diagram in part a...
##### Looking for question 3 a,b,c,d,e,f,g,h and i. Thank you it's the same question with different situations....
looking for question 3 a,b,c,d,e,f,g,h and i. Thank you it's the same question with different situations. 3. Now let's use trial-and-error to see if we can figure out the inheritance pattern of this trait. For EVERY pedigree diagram below and on the following pages, label what you kn...
##### Bilanced chem Cl equation deterininathe eqquation6.2,Wnte bilanced chemical ~quutian {cr puinino prorane Z5 C,Ha In &ir (u3tng oxygen from conyurt into cuiban dioxida und water) and defermine the cpefficent ot oxyCen chernical equationFor quesuons 44through 48 determint for true and "6" (ore Uake 4iSlele [Orglent bond calnt bond wtuch Ivm atoms shdte two palrs of electronscoortnatcovderlbondwtich both tne electrons €fa shared palrnvobu Ihehond~Ulectroncpalvity I5 3 measure &
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##### 1. Neural networks often have many parameters that need to be optimised. Suppose that in a simple case a particular neu...
1. Neural networks often have many parameters that need to be optimised. Suppose that in a simple case a particular neural network has just two parameters x and y that satisfy y and x2 + y2 25. An analyst establishes that the performance function of the network is f(x, y)-(x2 + y2)3/2-6(x2 + y2) + 9...
##### 17 You draw three cards The first two are clubs What is the probability that the third one is a club?50 31 32 53 30
17 You draw three cards The first two are clubs What is the probability that the third one is a club? 50 31 32 53 30...
##### These exercises involve counting combinations.In how many ways can six people be chosen from a group of ten?
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##### Question 10 4.5 pts You wish to prepare 400 mL of a 1:10 solution using tablets...
Question 10 4.5 pts You wish to prepare 400 mL of a 1:10 solution using tablets that contain 10 grams of the drug. idow many whole tablets should you use?...
##### The following data are given for a two-factor ANOVA. Determine whether the block or the treatment means differ. Use the 5% level of significance: TREATMENT BlOCK 12 14 11
The following data are given for a two-factor ANOVA. Determine whether the block or the treatment means differ. Use the 5% level of significance: TREATMENT BlOCK 12 14 11...
##### 6. [Spt] A function f has the following properties:f' (2) < 0 on (~0,1) f'(2) > 0 on (1,0)f" (x) < 0 on (~x, -5) and (-2,3) f" (c) > 0 on (~5,-2) and (3,0)Use the information above to determine the general shape of f 0n each of the intervals listed below . For each interval, choose one of the following:(-0,-5)(-5,-2)(-2,1)(1,3)(3,
6. [Spt] A function f has the following properties: f' (2) < 0 on (~0,1) f'(2) > 0 on (1,0) f" (x) < 0 on (~x, -5) and (-2,3) f" (c) > 0 on (~5,-2) and (3,0) Use the information above to determine the general shape of f 0n each of the intervals listed below . For each...
##### A machine costing $209,000 with a four-year life and an estimated$15,000 salvage values installed in...
A machine costing $209,000 with a four-year life and an estimated$15,000 salvage values installed in Luther Company's factory on Januaryt. The factory manager estimates the machine w produce 485,000 units of product during its life. It actually produces the following units: 12.700 In Year 1,123...
##### Write as one inequality with an absolute value_ 2 < r and Write the inequality in one of the following forms 6Irtalsulr+alz b al<blr a| 26Between the two absolute value signs (|.1.|), put X+a or X-a for some number a. In the drop-down menu, means 7= means Put number in the last box2 Ir+4 <1 Draw the solution on the number line. Use solid/hollow circlesUse the tools to enter your answer;
Write as one inequality with an absolute value_ 2 < r and Write the inequality in one of the following forms 6Irtalsulr+alz b al<blr a| 26 Between the two absolute value signs (|.1.|), put X+a or X-a for some number a. In the drop-down menu, means 7= means Put number in the last box 2 Ir+4 <...
##### Determine the support reactions and draw internal force (V, M) diagrams for the continuous beam shown...
Determine the support reactions and draw internal force (V, M) diagrams for the continuous beam shown in the figure due to the uniformly distributed load and the support settlements of 15 mm at support B, 36 mm at support C and 18 mm at support at D. 32 kN/m о. В 15 mm Со. 36...
##### Please help me fill this ☺️ ons for Review: the letter for the correct word in...
please help me fill this ☺️ ons for Review: the letter for the correct word in each blank. A recording of the electrical activity of the heart. The period when the atria are depolarizing, A Angina pectoris The period when the ventricles are repolarizing B. Bradycardia The period during ...
##### Find the general golution of y" 4y' 12y Sle"gin(31).
Find the general golution of y" 4y' 12y Sle"gin(31)....
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##### 9 Find the Moment of inertia of the given section about X-X axis passing through its...
9 Find the Moment of inertia of the given section about X-X axis passing through its center of gravity. Take A= 60 mm, B=10 mm, C= 40 mm and D= 70 mm 10 Find The Tension in cable RP & RQ, where PR =6m, AR = 5m, RQ = 14 m & RB = 5 m 60KN 100KN...
##### Evaluate the follpwing integral330061 43t 1 dx Bin 37 52 Ma3 31003 +3 ( @ = MmnSin 3x 51 48 (Type an exact answer )
Evaluate the follpwing integral 3 30061 43t 1 dx Bin 37 52 Ma 3 31003 +3 ( @ = MmnSin 3x 51 48 (Type an exact answer )...
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##### Evaluate the integral V = π 2 −2 {[(9 − x 2 ) − (−1)]2 − [(x 2 +1) − (−1)]2 }dx which represents the volume of the solid S obtainedby rotating the region R bounded by y = x 2+1 and y = 9 − x 2 abouty = −1.
Evaluate the integral V = π 2 −2 {[(9 − x 2 ) − (−1)]2 − [(x 2 + 1) − (−1)]2 }dx which represents the volume of the solid S obtained by rotating the region R bounded by y = x 2+1 and y = 9 − x 2 about y = −1....
##### Help Save & Exit Submit Submit Marlow Company purchased a point of sale system on January...
Help Save & Exit Submit Submit Marlow Company purchased a point of sale system on January 1 for $7,000. This system has a useful life of 5 years and a salvage value of$1,200. What would be the depreciation expense for the second year of its useful life using the double-declining balance method?...
The Molding Division of Cotwold Company manufactures a plastic casing used by the Assembly Division. This casing is also sold to external customers for $36 per unit. Variable costs for the casing are$23 per unit and fixed cost is \$5 per unit. Cotwold executives would like for the Molding Division t...
##### Use properties of logarithms to find the exact value of each expression. Do not use a calculator. $\log _{8} 2+\log _{8} 4$
Use properties of logarithms to find the exact value of each expression. Do not use a calculator. $\log _{8} 2+\log _{8} 4$... | 4,222 | 14,421 | {"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} | 2.765625 | 3 | CC-MAIN-2023-14 | longest | en | 0.853148 |
https://programmingdoc.com/logical-operators-4224 | 1,586,292,143,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371805747.72/warc/CC-MAIN-20200407183818-20200407214318-00095.warc.gz | 632,394,760 | 10,035 | Select Page
Logical Operators
So far we’ve seen two families of operators (besides the oddball conditional operator). First are the arithmetic operators +, -,*,/, and %. Second are the relational operators <, >, <=, >=, ==, and !=. Let’s examine a third family of operators, called logical operators. These operators allow you to logically combine Boolean variables (that is, variables of type bool, with true or false values). For example, today is a weekday has a Boolean value, since it’s either true or’false. Another Boolean expression is Maria took the car. We can connect these expressions logically: If today is,a weekday, and Maria took the car, then I’ll have to lake the bus. The logical connection here is the word and, which provides a true or false value !o the combination of the two phrases. Only if they are both true will I have to take the bus. | 191 | 866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-16 | latest | en | 0.913665 |
https://open.kattis.com/contests/ncpc22practice1/problems/confinedcatching | 1,669,499,387,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00449.warc.gz | 483,120,534 | 8,244 | Hide
# Problem CConfined Catching
You are playing a board game against an AI on a square grid consisting of $n\times n$ cells. You have two game pieces and the AI has one, and each piece is placed in one of the grid cells. Your goal is to “catch” the AI’s piece, that is, one (or both) of your pieces has to lie in the same cell as the AI’s piece after one of your turns. When this happens, you win and the game ends. You lose if you have not won after 600 turns.
Each turn, you have up to five movement options per piece: You can move a piece up, down, left, or right to an adjacent cell (if there is one) or let the piece remain in its current cell. The AI has the same options for its piece in each of its turns. Of course, you can move your pieces completely independently from one another and even have them occupy the same cell.
Your goal is simple: Win the game! You can safely assume that this is always possible.
## Initial Input
Before it is your first turn, your program will receive:
• One line with an integer $n$ ($3 \le n \le 100$), giving the size of the grid.
• One line with four integers $x_1,y_1,x_2,y_2$ ($1 \le x_1,y_1,x_2,y_2 \le n$), giving your pieces’ initial positions.
• One line with two integers $x,y$ ($1 \le x,y \le n$), giving the AI’s piece’s initial position.
You can safely assume that your pieces do not lie in the same cell as the AI’s piece (but they may lie in the same cell as each other).
## Interaction Protocol
Your submission will be interacting with a special program called the grader. This means that the output of your submission is sent to the grader and the output of the grader is sent to the standard input of your submission. This interaction must follow a specific protocol:
Your submission and the grader alternate writing turns, with you going first. Your turns consist of a single line with four integers $x_1,y_1,x_2,y_2$ to specify the locations of your pieces after your turn (in the same order as in the initial input). Similarly, the grader will send lines with two integers $x,y$ to indicate the AI’s turn. You can safely assume that the grader will only send you valid moves, and the AI will never choose to place its piece in a cell occupied by one or both of your pieces. Your submission may take at most $600$ turns.
After each of your turns you should flush the standard output to ensure that the request is sent to the grader. For example, you can use fflush(stdout) in C++, System.out.flush() in Java and sys.stdout.flush() in Python.
After you send the grader a valid turn catching the AI’s piece, the game ends. Your submission should then terminate with exit code 0 as usual. For convenience, the grader will send a single line containing 0 0 to signal the end of the game, which your submission may or may not read.
Your submission will be accepted if it follows the rules and protocol above and wins the game. If it sends any invalid turn, it will be judged as “Wrong Answer”.
For your convenience, we have provided you with a testing tool that lets your solution interact with a simple version of the AI opponent. You can download it from the Attachments menu.
Read Sample Interaction 1 Write
3
1 1 3 1
2 3
1 2 3 2
3 3
1 2 3 3
0 0
CPU Time limit 1 second
Memory limit 1024 MB
Statistics Show
Downloads
Author
License
Please log in to submit a solution to this problem | 825 | 3,363 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2022-49 | latest | en | 0.937761 |
https://kmmiles.com/1960-1-miles-in-km | 1,713,633,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00500.warc.gz | 318,142,765 | 5,979 | kmmiles.com
Search
# 1960.1 miles in km
## Result
1960.1 miles equals 3153.8009 km
You can also convert 1960.1 mph to km.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
1960.1 mi × 1.609 = 3153.8009 km
## How to convert 1960.1 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 1960.1 miles into km we have to multiply 1960.1 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
1960.1 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 1960.1 mi × 1.609 km
L(km) = 3153.8009 km
The final result is:
1960.1 mi → 3153.8009 km
We conclude that 1960.1 miles is equivalent to 3153.8009 km:
1960.1 miles = 3153.8009 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case one thousand nine hundred sixty point one miles is approximately three thousand one hundred fifty-three point eight zero one km:
1960.1 miles ≅ 3153.801 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
1961.1 miles 3155.4099 km
1962.1 miles 3157.0189 km
1963.1 miles 3158.6279 km
1964.1 miles 3160.2369 km
1965.1 miles 3161.8459 km
1966.1 miles 3163.4549 km
1967.1 miles 3165.0639 km
1968.1 miles 3166.6729 km
1969.1 miles 3168.2819 km
1970.1 miles 3169.8909 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world. | 694 | 2,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-18 | latest | en | 0.835253 |
https://www.numbersaplenty.com/8277742218253 | 1,611,111,977,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00446.warc.gz | 922,655,792 | 3,320 | Search a number
8277742218253 is a prime number
BaseRepresentation
bin111100001110100111111…
…1000100110110000001101
31002022100022002100222001111
41320131033320212300031
52041110304001441003
625334424155412021
71513022111630614
oct170351770466015
932270262328044
108277742218253
112701630755692
12b18347154011
134807843653c1
142089056da87b
15e54ca575e6d
hex7874fe26c0d
8277742218253 has 2 divisors, whose sum is σ = 8277742218254. Its totient is φ = 8277742218252.
The previous prime is 8277742218211. The next prime is 8277742218313. The reversal of 8277742218253 is 3528122477728.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 6381979482564 + 1895762735689 = 2526258^2 + 1376867^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-8277742218253 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (8277742212253) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4138871109126 + 4138871109127.
It is an arithmetic number, because the mean of its divisors is an integer number (4138871109127).
Almost surely, 28277742218253 is an apocalyptic number.
It is an amenable number.
8277742218253 is a deficient number, since it is larger than the sum of its proper divisors (1).
8277742218253 is an equidigital number, since it uses as much as digits as its factorization.
8277742218253 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 21073920, while the sum is 58.
The spelling of 8277742218253 in words is "eight trillion, two hundred seventy-seven billion, seven hundred forty-two million, two hundred eighteen thousand, two hundred fifty-three". | 568 | 1,913 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-04 | latest | en | 0.802703 |
http://mathhelpforum.com/algebra/148178-measurment-print.html | 1,503,056,243,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104634.14/warc/CC-MAIN-20170818102246-20170818122246-00528.warc.gz | 268,593,557 | 2,696 | # Measurment
• Jun 7th 2010, 03:54 PM
DOOKIE
Measurment
A 70-foot tree has a 35-foot shadow. If
the building next to the tree has an
80-foot shadow, how tall is the building?
• Jun 7th 2010, 03:58 PM
skeeter
Quote:
Originally Posted by DOOKIE
A 70-foot tree has a 35-foot shadow. If
the building next to the tree has an
80-foot shadow, how tall is the building?
set up a ratio ...
$\frac{70}{35} = \frac{h}{80}$ | 143 | 415 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-34 | longest | en | 0.906711 |
https://arizonahealthnet.com/what-is-a-good-heart-rate-when-exercising-4/ | 1,701,282,985,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100135.11/warc/CC-MAIN-20231129173017-20231129203017-00255.warc.gz | 132,882,836 | 44,540 | # What is a good heart rate when exercising?
Your target heart rate zone for vigorous exercise is 146.5 to 160.75 beats per minute.
## Is 150 a good heart rate for exercise?
For moderate activity, you want your heart rate to be between 90 and 126 (that’s 50% to 70%) the entire 150 minutes you’re exercising. For vigorous, aim for a heart rate between 126 and 153 (that’s 70% to 85%) when you’re exercising.
## What is a good exercise heart rate by age?
You can estimate your maximum heart rate based on your age. To estimate your maximum age-related heart rate, subtract your age from 220. For example, for a 50-year-old person, the estimated maximum age-related heart rate would be calculated as 220 – 50 years = 170 beats per minute (bpm).
## Is 170 a good heart rate for cardio?
Most runners ages 20 to 45 will want to train between 100 and 160 bpm, on average. But that average depends on a number of factors, including your maximum heart rate and current fitness level. You can use the formula and chart below to determine your target heart rate range.
What is a good heart rate when exercising? – Related Questions
## Is 150 a high heart rate?
Yes, it’s normal for your heart rate to increase to 130 to 150 beats per minute or more when you exercise – this is because your heart is working to pump more oxygen-rich blood around your body.
## Is 155 a high heart rate during exercise?
According to the AHA , your target heart rate during moderate-intensity activities is about 50 to 70 percent of your maximum heart rate.
What is a target heart rate?
Age (years)Target heart rate (50% to 85%) (bpm)Average maximum heart rate (bpm)
6080 to 136160
6578 to 132155
7075 to 128150
## What happens when heart rate is 150?
Tachycardia refers to a high resting heart rate. In adults, the heart usually beats between 60 and 100 times per minute. Doctors usually consider a heart rate of over 100 beats per minute to be too fast, though this varies among individuals. Factors such as age and fitness levels can affect it.
## Is 160 a normal exercise heart rate?
The harder your body is working, the faster your heart will beat. For example, your heart rate when you’re sprinting will be much faster than your heart rate when you’re walking. If you’re exercising hard, it’s normal for your heart rate to get up to 160 beats per minute or more.
## What is a dangerously high heart rate during exercise?
So, more than 200 beats per minute heart rate during exercise is dangerous for you. If you develop palpitations, an irregular heart rate, shortness of breath, or chest pain, you need to seek medical help right away.
## What is max heart rate by age?
Know Your Numbers: Maximum and Target Heart Rate by Age
## What heart rate burns fat?
Your fat-burning heart rate is at about 70 percent of your maximum heart rate. Your maximum heart rate is the maximum number of times your heart should beat during activity. To determine your maximum heart rate, subtract your age from 220.
## What kind of exercise burns most fat?
Running is the winner for most calories burned per hour. Stationary bicycling, jogging, and swimming are excellent options as well. HIIT exercises are also great for burning calories. After a HIIT workout, your body will continue to burn calories for up to 24 hours.
## What heart rate is considered cardio?
The aerobic heart rate zone is a heart rate between 70% and 80% of your maximum heart rate (MHR). You reach it while exercising at moderate to vigorous intensity. In this zone your breathing quickens, but you’re not out of breath, and you can still have a conversation with a person running next to you.
## Which is better cardio or fat burn?
Intense cardio exercise is better than fat-burning workouts as you get to burn more calories. Burning more calories means you lose more weight faster. Therefore, if your goal is to lose fat and weight, intense cardio is the better alternative.
## What cardio burns stomach fat?
Aerobic or Cardio Exercise
Some great cardio of aerobic exercises for belly fat include: Walking, especially at a quick pace. Running. Biking.
## What cardio burns belly fat the best?
1. Walking or Running. Walking or running at a moderate pace is a sure-shot way of burning calories and fat. It is not just the best cardio to burn belly fat but is also the most economical way to burn fat and maintain muscle.
## Where is the fat burning zone?
The ‘fat burning zone’ is where you are working out at about 70 – 80% of your maximum heart rate, also known as your fat burning heart rate. If you’re looking to lose weight and keep fit, the general rule of the game is to increase the intensity of your workouts. | 1,061 | 4,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-50 | latest | en | 0.897619 |
https://cs.stackexchange.com/questions/109403/2-sat-reduction-to-easy-puzzle-solving | 1,575,778,326,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540504338.31/warc/CC-MAIN-20191208021121-20191208045121-00195.warc.gz | 331,692,178 | 31,930 | # 2-SAT reduction to Easy Puzzle Solving
## Introduction
I've been developing interest in complexity classes and thought I was unable to prove my problem is NP hard, so I wanted to see if it was P-hard. I wanted to see if my puzzle solving is special. I have linked a previous question of mine to help clarify what is mentioned below. Explained here
## Random Instance of 2-SAT used for attempt reduction
I took an instance of my problem X
Let's say my problem X is determining if a puzzle is valid for my language
(a∨¬b)∧(¬a∨b)∧(¬a∨¬b)∧(a∨¬c)
Instance of X
a = shift(L) puzzle
¬ a = invalid puzzle
¬b = invalid puzzle
b = shift(L) puzzle
## Reduction into Shorter Instance
(a∨¬b)∧(¬a∨b)
This boolean expression is only checking for 2-sat. Being True (valid puzzle) or False (invalid puzzle).
My idea is that a deterministic machine will consider n^2 x n^2 puzzles as invalid if the lower right-hand box is not filled. My puzzles are solved in quadratic time when provided a puzzle with this n x n box. Any other puzzle will simply fail to be solved if it does not follow my language. I have used an algorithm that can determine in poly-time that those failed puzzles are invalid puzzles. Because, the solver will not know it as an invalid puzzle. So it will try to map it out in my language thus giving an invalid puzzle.
## Question
I have the algorithm and proven that it works in poly-time, the problem is that I only know how to prove it by showing the algorithm. I just don't know how to write it out mathematically. How would I properly do this?
I'm going to need to question your premise here.
I have no doubt that you can poly-time reduce 2SAT to your custom problem. But you can poly-time reduce 2SAT to pretty much anything. The idea of "P-hardness" under poly-time reduction isn't really a useful one.
For example, let's take the following problem:
IS-IT-ONE: given a single bit, is that bit a 1?
This problem is "P-hard" under poly-time reduction. For example, I can reduce 2SAT to it:
Reduction from 2SAT to IS-IT-ONE
• Solve 2SAT the usual way
• If the problem is satisfiable, let $$B$$ be 1
• If the problem is unsatisfiable, let $$B$$ be 0
• Run IS-IT-ONE on $$B$$ and return the result
This reduction runs in poly-time, because solving 2SAT can be done in poly-time.
In other words:
Every problem in P can be poly-time reduced to any other non-trivial problem.
(Only two languages are considered "trivial" for this purpose: "return true" and "return false", aka $$\Sigma^*$$ and $$\varnothing$$.)
If you're interested in classes smaller than P, great! There are several of them, such as L: the set of problems that can be solved using a logarithmic amount of memory. It may or may not be the same as P—that's an open problem.
But if you want to talk about P versus L or NL or anything else within P, you can't use poly-time reductions any more. When talking about L, for example, you'll usually want a log-space reduction, using a Turing machine with writable tape bounded by the log of the input size. These aren't as famous as poly-time reductions, but they're still quite important in complexity theory.
• Okay, I'll take a look into log space for a few days and try and see what I can get. I share this in form of question on CS. Preferably, on meta first. – Travis Wells May 16 at 19:47
• @TravisWells I'd suggest taking some time away from your simplified-sudoku generator for a bit, too. Read up on complexity theory, and try some NP-hardness proofs for other problems. Then focus on log-space, and try some log-space reductions for other problems too, focusing on the difference between P and L. Then come back to your sudoku generator. – Draconis May 16 at 20:02
• I agree that I need to take a break, I've already starting to understand how to reduce SAT into a 3-SAT Instance. I've managed to follow restricted reductions. 3-SAT to vertex cover is probably easier to understand. Given time, may we take this to chat so that I can learn this quickly as possible? I have a link to this. andrew.cmu.edu/user/ko/pdfs/lecture-21.pdf – Travis Wells May 16 at 23:02
• @TravisWells pick a book, if you want to learn as quickly as possible, throw away your shifter-mapper, please. – Evil May 17 at 13:54
• @TravisWells I agree with Evil, honestly. Your fixation on this shifter-mapper-thing is holding you back, especially since you can't even state what the problem is in a succinct way: your explanation is scattered across four different web pages now and involves sentences like "let shift by shift be defined by the formula x by y" which I can't make heads or tails of. I can tell you right now that your new question almost certainly isn't #P-complete, since #P is defined in terms of NP, and I'm confident your problem isn't NP-complete (though I don't understand it well enough to prove anything). – Draconis May 17 at 18:50 | 1,220 | 4,883 | {"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": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-51 | latest | en | 0.945633 |
https://demtutoring.com/answered/womensstudies/q2016 | 1,669,971,646,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710900.9/warc/CC-MAIN-20221202082526-20221202112526-00329.warc.gz | 242,035,139 | 3,955 | Flash Speed Questions
The solution time is much shorter than you think.
< > 4 based on what the reader learns about hugh's life in africa in the previous paragraphs, what is ironic about the narrator feeling “jealous" about hugh's experiences growing up? the narrator's mother often told him as a boy to not be jealous of what other people had and to be thankful for his life in north carolina. most people don't think going to visit a slaughterhouse, to look b at bloody curtains, or to watch an old man chew up rancid goat meat are things to be jealous about. hugh was jealous of the narrator's ordinary life in the north carolina suburbs. africa is a place where most people find new and unusual experiences. Get the answer
Category: womensstudies | Author: Abraham Uilleam
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< 2. 3 4 5 a brief, "first-step" presentation of a work to be completed is called a draft bill mark up amendiment | 431 | 1,641 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-49 | longest | en | 0.901309 |
https://www.imagemagick.org/discourse-server/viewtopic.php?f=2&t=30258&start=30&view=print | 1,566,311,318,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315544.11/warc/CC-MAIN-20190820133527-20190820155527-00357.warc.gz | 856,955,311 | 4,432 | Page 3 of 4
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-20T19:28:26-07:00
Ah, thanks. I wondered whether 11 was mathematical or a coincidence.
I've expanded my trials to include 2-channel colorspaces CbCr CC DbDr IQ and UV that are really YCbCr etc without the Y channel.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T10:30:33-07:00
Okay, so the maximum PH is 11. What is the minimum? I thought the minimum was 0.0 but I was wrong. For example:
Code: Select all
``identify -verbose -define identify:moments logo:``
We have negative PH in the Chroma channel. When this happens, it throws my proportional difference calculation.
Are negative values correct, or is there a bug?
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T13:38:03-07:00
I believe the negative values occur when the moment is (slightly) larger than 1, so that the negative log of the moment is negative. I have puzzled what to do about that for some time with no conclusion. I do not know if it would be better to clamp to 0 or not. Open for consideration and recommendations.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T14:11:37-07:00
Ah, thanks, that makes sense.
So, when I clamp negative values to zero, my comparison tests (656 images, 30 colorspaces) give slightly worse results. So, no, I don't think they should be clamped.
I have seen values around -4. Is there a theoretical minimum? If not, maybe they should be clamped to -11, to be symmetrical with the maximum.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T15:51:25-07:00
snibgo wrote: I have seen values around -4. Is there a theoretical minimum? If not, maybe they should be clamped to -11, to be symmetrical with the maximum.
Theoretical minimum? For what? The moment or the hash value -log(moment)? Do you mean biggest negative value? If so, I do not recall that it is clamped at -11.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T16:16:07-07:00
Sorry, yes, I mean the perceptual hash value. So the range of possible values is -infinity to +11.0? Okay, I can live with that.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T17:13:39-07:00
snibgo wrote:Sorry, yes, I mean the perceptual hash value. So the range of possible values is -infinity to +11.0? Okay, I can live with that.
If you find a clamp to some largest negative value makes sense (perhaps -11), let me know. We can change the code later after we get your conclusions.
I am still not sure what to do about the negative values. It has bothered me since I first saw them. But they have been rare for me.
I do not think you will get really large negative values since the moments are typically less than 1 and probably do not exceed one by too much. No proof on that, though.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-21T20:01:39-07:00
I'm not keen on clipping values without good reason.
The most negative PH value in my current test is -4.02589. 3276 of the 367360 values are negative, so about 1% of the values. Only 88 of those are less than -1.0.
15170 values (4% of the values) are exactly 11.0.
Half the values are below 4.72312. The mean is 4.96005.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T02:45:15-07:00
I've uploaded Perceptual hash tests, which describes the tests I did, results, conclusions, and recommendations for IM internals.
Recommendations in a nutshell: use at least two colorspaces. Default to using xyY and HSB. Provide facilities to use any colorspaces (and any number of them), and to omit any channels.
Feel free to comment, anyone.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T08:54:24-07:00
Why HSB? I though you suggested that H was not good, especially for red images?
I am just reading your tests. Note that the IM PHASH uses HCLp not HCL and I do not see HCLp in your list of tests. There is a difference between HCLp and HCL.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T09:21:48-07:00
Seems to me that if there is potential for H to cause problems then one might want to use your second choice of xyY and YDbDr.
Do you have a list of your attacks, especially the hue changes?
Can you clarify? Did you rotate all the images and crop off the background padding and use those for you tests or was this rotation just one attack?
What did you do in regards to vertical and horizontal crops as attacks? Please clarify.
By the way, that was an massive and exhaustive set of tests. Thanks for this more thorough testing and your conclusions.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T09:37:02-07:00
Yes, for some work I did, CL+sRGB performed better than HCL+sRGB. That dataset wasn't artificial (it was edited photos), and the majority of tweaks were colour or tone shifts.
I intend to get my hands on that data again, and try my test on it, or at least a subset.
This test set for my web page had far fewer colour shifts. The second-best pair was xyY and YDbDr, which might work far better on the edited photos. This is why I suggest xyY and HSB should be the defaults, but that users must be able to change them. I don't think one particular combination of colorspaces can be the best for all situations.
fmw42 wrote:There is a difference between HCLp and HCL.
Is there?
Code: Select all
``````%IM%convert hald:10 ( -clone 0 -colorspace HCLp ) ( -clone 0 -colorspace HCL ) -delete 0 -metric RMSE -format %[distortion] -compare info:
0
%IM%convert hald:10 ( -clone 0 -colorspace HCLp ) ( -clone 0 -colorspace HCL ) -delete 0 -metric AE -format %[distortion] -compare info:
0``````
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T09:50:01-07:00
fmw wrote:Do you have a list of your attacks, especially the hue changes?
My attacks are on that web page, at http://im.snibgo.com/phashtest.htm#tweakImages.bat
The script is Windows BAT, but fairly understandable. For example, the code:
Code: Select all
`````` for %%V in (-20 -15 -10 -5 5 10 15 20) do %IM%convert ^
%%F ^
+depth ^
^( +clone ^
-brightness-contrast %%V,0 ^
+write %OUTDIR%\!BASE!_bri_%%V.%EXT% ^
+delete ^
^) ^
-brightness-contrast 0,%%V ^
%OUTDIR%\!BASE!_con_%%V.%EXT%``````
This takes one of the standard images (%%F) and loops through %%V. For each %%V (which is one of -20, -15 etc) it creates two variations, using %%V as the first or second parameter for -brightness-contrast.
The rotates and crops work in similar ways.
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T09:53:59-07:00
Thanks. I see it now at
Code: Select all
``````for %%V in (90 95 97.5 102.5 105 110) do %IM%convert ^
%%F ^
-modulate 100,100,%%V ^
%OUTDIR%\!BASE!_hue_%%V.%EXT%``````
### Re: Get perceptual hash value for image using command line
Posted: 2016-08-23T10:34:14-07:00
HCLp is more like PS HCL than plain HCL. I will see if I can find that information.
Looking at gems.c lines 841 (HCL) and 904 (HCLp). They look the same. But HCLp should be different according to
viewtopic.php?f=2&t=21663#p88910
viewtopic.php?f=2&t=21646
viewtopic.php?f=1&t=23803&hilit=HCLp#p101115
viewtopic.php?f=2&t=21646#p112265 | 2,111 | 7,444 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-35 | latest | en | 0.866405 |
http://www.jiskha.com/display.cgi?id=1354162722 | 1,461,946,964,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111374.13/warc/CC-MAIN-20160428161511-00095-ip-10-239-7-51.ec2.internal.warc.gz | 605,474,186 | 4,232 | Friday
April 29, 2016
# Homework Help: trigonometry
Posted by Erick on Wednesday, November 28, 2012 at 11:18pm.
1. What is the area of triangle ABC if a = 47.0°, β = 57.8°, and a = 10.2 cm?
A. 58.2 cm2
B. 43.5 cm2
C. 38.4 cm2
D. 33.3 cm2
2. Given triangle ABC with β = 41°, g = 14°, and a = 5.0, find the value of c.
A. 6.2
B. 4.0
C. 1.5
D. 17
3. Resolve the vector, v, with magnitude 2.0 × 102 and angle 60°.
A. v = 170i + 100j
B. v = 120i + 160j
C. v = 100i + 170j
D. v = 10i + 17j
4. What is the area of triangle ABC if a = 83.4 ft, b = 53.1 ft, and c = 37.2 ft?
A. 16,100 ft2
B. 709 ft2
C. 1,220 ft2
D. 76.0 ft2
5. Which of the following statements relating to the Pythagorean theorem are true?
(i) The magnitude of a vector is based on the Pythagorean theorem.
(ii) The law of cosines is based on the Pythagorean theorem.
(iii) The law of sines is based on the Pythagorean theorem.
A. (i) and (ii)
B. (i) only
C. (ii) only
D. (ii) and (iii)
6. How much work is done by raising a 117-lb box vertically 18.0 inches?
A. 25,300 ft-lb
B. 2,110 ft-lb
C. 176 ft-lb
D. 5,190 ft-lb
7. Given triangle ABC with g = 61.0°, β = 29.0°, and b = 20.5, find the value of a.
A. 37.0
B. 11.4
C. 42.3
D. 9.94
8. Given triangle ABC with b = 14.0, c = 11.0, and β = 105°, find the value of a.
A. 20
B. 8.7 and 20
C. 6.3
D. 6.3 and 12
• trigonometry - Steve, Thursday, November 29, 2012 at 11:05am
I did #1 for you (see related links below). These questions cover a variety of techniques, so you must have studied several topics. Have you no ideas at all on their workings? If not, a little help here may not be sufficient to get you through.
## Answer This Question
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# Common Core Standard K.CC.A.1 Questions
Count to 100 by ones and by tens.
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Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
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Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
This question is a part of a group with common instructions. View group »
Which numbers count forward?
1. 1, 2, 3
2. 2, 3, 1
3. 3, 1, 2
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
How many blocks?
1. 10
2. 20
3. 30
4. 40
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
Kindergarten Counting and Comparing Numbers CCSS: K.CC.A.1
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You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges. | 608 | 2,235 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2022-21 | latest | en | 0.83757 |
https://www.asianturfgrass.com/post/monthly-maximum-n-or-annual-maximum-n/ | 1,722,690,186,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00301.warc.gz | 539,484,688 | 7,603 | # Monthly maximum N or annual maximum N
I sometimes talk about maximum N. You may have heard me refer to the standard amount of N. I discussed that with Michael Bekken in this ATC Doublecut episode. What do I mean by these quantities of N? There was a question on the ATC YouTube channel about that, asking whether my mentions of a 30 kg/ha or 3 g/m2 value for Poa annua is a monthly or annual amount. The comment might have been deleted, or YouTube’s inscrutable comment filter may have disallowed that. I thought it was a good question, and I wanted to start an explanation of it here.
I also publish updates at PACE Turf. Recent updates have discussed the ultimate inorganic soil amendment (silica sand); black layer; predicting turf growth from air temperature; generic, post-patent, off-patent, & proprietary products; and sand, thatch, Poa, and bentgrass. If you’d like to have access to all this information, plus the PACE Turf decision support tools, subscribe now.
The PACE Turf growth potential (GP) converts air temperature to a scale of 0 to 1 (or 0% to 100%, if you prefer).
When temperatures are close to an optimum for net photosynthesis, the GP will be close to 1. When temperatures are far away from the optimum temperature, the GP will be 0. If the grass is dormant in winter, the temperatures will convert to a GP that will be almost 0. When the grass won’t stop growing and you can’t mow it fast enough in spring, the GP will be close to 1.
The grass is using—and losing–nitrogen based on how many clippings are being removed by mowing. When the grass is not growing, it is not being mowed, and the N use is zero. When the grass is growing rapidly, and is being mown frequently, then a lot more N is being removed from the grass. The GP scale goes from 0 to 1, and we can assume that there is a linear relationship between GP and growth.1
When I am talking about maximum N, or standard N, what I mean is the quantity of N that I expect will work well to produce high quality turfgrass when multiplied by the GP. I generally refer to this amount as a monthly amount. For example, I use 2 g N/m2 as a monthly maximum for bentgrass, 3 for Poa annua, and 4 for bermudagrass. These are based on observation of the rates of N that turfgrass managers use to produce good turf.
These values give a really good starting point. For example, I spoke with Fraser Brown, the course superintendent at Lake Karrinyup Country Club in Perth, Western Australia. He said that he’s been applying an annual N rate that is about 180 to 200 kg/ha (18 to 20 g N/m2). When I checked the temperatures there, and calculated the predicted N using the standard rate of 2 g/N/m2 monthly, the annual total comes to 170 kg/ha.
Using this procedure to check N rates can get one very close to what is working to produce high quality turf at other places around the world, but adjusted to the exact temperature conditions of one’s site.
1. This won’t be perfectly linear, of course, but I recommend measuring clipping volume to assess how much the grass is actually growing, and making adjustments based on that. ↩︎ | 737 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.944961 |
https://www.instasolv.com/ncert-exemplar/class-10-maths.html | 1,601,259,142,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401583556.73/warc/CC-MAIN-20200928010415-20200928040415-00061.warc.gz | 845,464,750 | 11,058 | Instasolv
IIT-JEE NEET CBSE NCERT Q&A
4.5/5
# NCERT Exemplar Class 10 Maths Solutions
NCERT Exemplar Class 10 Maths Solutions are created to aid you in solving the Maths book effortlessly. The NCERT Exemplar Maths book for CBSE Class 10th has a comprehensive set of questions to help you prepare each topic thoroughly for exams. All the topics mentioned in Class 10 Maths syllabus have been covered in detail in the NCERT Exemplar Problems
There are 13 chapters in the Class 10 Maths NCERT Exemplar book with an average of roughly 50 questions per chapter on topics such as algebra, trigonometry, geometry, mensuration, and statistics. The problems mentioned in the book are distributed into 4 exercises in every chapter. The sections are MCQs, very short answer type reasoning questions, short answer, and long answer type questions.
Some of the very important chapters that will fetch you good marks in your school exams from this exemplar book are quadratic equations, linear variables in two variables, trigonometric functions, areas related to circles, surface areas and volumes, and statistics. It is advised that you practice all the chapters referring to the NCERT Exemplar Class 10 Maths solutions at Instasolv.
Class 10
## NCERT Exemplar Solutions for Class 10 Maths Problems
We have prepared the answers to the NCERT Exemplar problems of CBSE Class 10 Maths chapters keeping in mind the requirements of your school exam and CBSE guidelines. Practising them at Instasolv will help you ace school exams, and rank higher in the competitive exams. Check out the detailed summary of each chapter of Class 10 Maths NCERT Exemplar book.
## NCERT Exemplar Class 10 Maths Solutions Chapter 1: Real Numbers
The important topics discussed in this chapter are Euclid’s division algorithm, fundamental theorem of arithmetic, irrational numbers, and the applications of real numbers. There are 4 exercises in Class 10 Real Numbers chapter which include 39 questions in total.
## NCERT Exemplar Class 10 Maths Solutions Chapter 2: Polynomials
You will learn to find the zeros of a given polynomial in this chapter. Geometric representation of polynomials, the relationship between zeroes and coefficients of polynomials, and division algorithms for polynomials are also covered in Class 10 Polynomials chapter. 62 questions are covered in 4 exercises in chapter 2.
## NCERT Exemplar Class 10 Maths Solutions Chapter 3: Pair of Linear Equations in Two Variables
The graphical and algebraic methods such as substitution method, elimination method and cross-multiplication method to find the solutions of the pair of linear equations in two variables are covered in this chapter in detail. There are 4 exercises consisting of 54 questions in the exemplar chapter of a diverse variety.
## NCERT Exemplar Class 10 Maths Solutions Chapter 4: Quadratic Equations
You will be amazed to learn about the relationship of discriminant with the roots in a quadratic equation. Also, finding the roots of quadratic equations by completing the square, factorisation method and with the help of the quadratic formula are covered in the 4 exemplar exercises comprising 28 questions in Class 10 Quadratic Equations chapter.
## NCERT Exemplar Class 10 Maths Solutions Chapter 5: Arithmetic Progression
There are 71 questions in 4 exemplar exercises of chapter 5. The topics of this chapter such as the general form of an AP, nth term of an AP, and the sum of first n terms of an AP are crucial for scoring good marks in your school exams.
## NCERT Exemplar Class 10 Maths Solutions Chapter 6: Triangles
You will get to learn about similar triangles and different criteria for similar triangles. Other important topics of geometry covered in the chapter are basic proportionality, Pythagoras theorem, and some other vital theorems. Class 10 Triangles chapter comprises 57 questions compiled in 4 exercises.
## NCERT Exemplar Class 10 Maths Solutions Chapter 7: Coordinate Geometry
Class 10 Coordinate Geometry chapter will help you develop a strong foundational base of coordinate geometry with its 4 exercises covering 58 questions. The exercise questions are based on the basic terms in coordinate geometry, distance formula, section formula, area of a triangle and collinearity of 3 points.
## NCERT Exemplar Class 10 Maths Solutions Chapter 8: Introduction to Trigonometry and Its Applications
You will be introduced to the trigonometry ratios of vital angles and trigonometric identities. You will also learn practical applications of the trigonometric functions such as finding the height or length of an object, angle of depression and elevation, etc. in Class 10th chapter 8. There are 60 questions in 4 exercises in this chapter.
## NCERT Exemplar Class 10 Maths Solutions Chapter 9: Circles
The concepts included in the Circles chapter are the tangents and chords in a circle and theorems related to the angles subtended by them. You will learn to prove various theorems using these concepts in the 44 questions given in 4 exercises in Class 10 Maths Circles chapter.
## NCERT Exemplar Class 10 Maths Solutions Chapter 10: Constructions
In the Maths chapter 10, you will be taught how to divide a given line segment in a given ratio and construction of a triangle similar to another triangle on a given scale and the construction of two tangents on a circle from one external point. These topics are covered comprehensively in 21 questions in 4 exercises of the chapter.
## NCERT Exemplar Class 10 Maths Chapter 11: Areas Related to Circles
The topics covered are the circumference of a circle, area of sectors and segments in a circle, length of an arc in the sector, etc. There are 60 questions compiled in 4 exercises of the chapter.
## NCERT Exemplar Class 10 Maths Chapter 12: Surface Area and Volumes
You will learn about the total surface area and curved surface area of various shapes such as frustum, hemisphere, solid sphere, and spherical shells. There are a total of 4 exercises consisting of 29 questions in Class 10 Surface Areas and Volumes chapter.
## NCERT Exemplar Class 10 Maths Chapter 13: Statistics and Probability
You will learn to find the mean and median of grouped data, graphical representation of cumulative frequency distribution in this chapter besides learning to solve complex probability problems. These topics are extensively covered in 4 exercises consisting of 82 questions in total.
### Why use NCERT Exemplar Class 10 Solutions for Maths by Instasolv?
1. Our experts work tirelessly to keep the NCERT Exemplar solutions updated in line with the latest CBSE exam trends. | 1,424 | 6,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.908536 |
https://www.jiskha.com/questions/672201/which-one-of-the-following-subcategories-of-the-E-M-section-has-separate-codes-for | 1,544,761,606,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825349.51/warc/CC-MAIN-20181214022947-20181214044447-00139.warc.gz | 914,727,164 | 5,134 | # medical billing and coding
which one of the following subcategories of the E/M section has separate codes for new and established patients? a. office or other outpatients services. b. emergency department services. c. initial inpatient consultation. d.hospital observation services
1. 0
2. 0
3. 8
## Similar Questions
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More Similar Questions | 746 | 3,055 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | longest | en | 0.904387 |
onlineexamhelpwebsite.com | 1,726,498,189,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00040.warc.gz | 403,198,049 | 16,546 | # Statistics Online Exam Help
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Statistics is a mathematical discipline and is a phrase that encompasses all of the procedures employed by an analyst to characterize a piece of data. The definition alone demonstrates how difficult a subject it may be for students who have little or no supervision in the learning process. Because this subject is heavily reliant on mathematics, it necessitates a significant deal of focus and practice. It is really difficult for a student to handle this demanding paper on their own which gets easier with Statistics Online Exam Help. To succeed in this sector, you must Get Help in Statistics Online Exam.
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http://www.jerf.org/iri/post/2888 | 1,603,871,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107897022.61/warc/CC-MAIN-20201028073614-20201028103614-00420.warc.gz | 146,084,099 | 3,430 | Swoopo - An Amazingly Evil Con
posted Sep 12, 2008
Swoopo, which I refuse to link to, is one amazingly evil site. It is an auction site (for some loose definition of the word "auction") that works as follows:
• A mark registers for the site and buys "bids" for \$1 each.
• The site puts an item up for auction, starting the price at 15 cents.
• The marks take turns "bidding". Each bid makes them the auction's current winner, costs them one "bid", raises the price by 15 cents, and bumps the timer up if it's close to the end. Whoever has the winning "bid" by the time the auction expires gets the item.
The auction timer ends up being pretty pointless as it gets bumped up endlessly; an action really ends when the marks stop bidding.
Why is this evil? Well, do the math. I've been watching people fight over a Wii, retail price \$249.99. It currently stands at \$111.30, and the competition remains fierce. (Perhaps it'll be done by the time I finish this message.) \$111.30 / 0.15 - \$1 = \$741! Plus let us not forget the \$111.30, for a grand total of \$852.30 for the Wii.
Yes, it's a great deal if you buy one bid and make the last bid. However, work the math: If this company is walking away with over \$800 for a \$250 product, somebody's getting screwed and it's probably you.
For extra bonus points, they have a "feature" on their site that will automatically use your bids at the last moment if you aren't there. Quelle surprise. Other various auction "specials" further disconnect the users from the pricing signal, allowing even more bids before it is actually sold, or preventing the top price from being a restraint by fixing the final price and making the current auction price nothing more than a fiction.
I have to admit to a certain admiration of this sick scheme. It's brilliant. It probably isn't even illegal, and probably can't be made illegal easily. And with the outrageous return-on-investment the company makes on their product, it's going to be impossible to dry up the pool of marks enough to put them entirely out of business.
(There are some places in their interface where they appear to be flat-out lying. For each finished bid, they have a place where they claim to compute the "savings" on the item, where they take the value and subtract the price of the "bids" made. However, the bids so subtracted are just wrong. The Wii I was watching, which is now done, came in at \$138.00, but they claim this was done in 140 bids. Simple math shows it takes 919 bids to get this high. I've scoured their help site and I see no reason for this discrepancy, no exceptions where they start an auction higher (why would they?).)
(Update, Oct. 20, 2008: A reader emails to point out that they appear to be showing what the winner is paying, which after re-examining a closed auction appears to be correct, based on the layout of the page. They just don't bring to your attention that all the losers also paid money. Classify that as you will.)
Their site shows nearly 2500 "ended auctions". Cha-ching!
Oh, that Wii? 919 bids = \$919 (by their own accounting they show no free bids), plus final price \$138.00, plus shipping \$14.80 = \$1,071.80. Wow. Just wow.
(Swoopo, in the event that you should see this and feel compelled to fire off a nastygram, be aware that I will be posting your entire correspondence and my reply publicly.)
Update Oct. 23, 2008: Look! Copycats! Inevitable. | 821 | 3,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.952312 |
https://grandyang.com/leetcode/261/ | 1,719,070,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00165.warc.gz | 233,072,472 | 34,355 | # 261. Graph Valid Tree
Given `n` nodes labeled from `0` to `n-1` and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.
Example 1:
``````Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
Output: true
``````
Example 2:
``````Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
Output: false
``````
Note: you can assume that no duplicate edges will appear in `edges`. Since all edges are undirected, `[0,1]` is the same as `[1,0]` and thus will not appear together in `edges`.
``````// DFS
class Solution {
public:
bool validTree(int n, vector<pair<int, int>>& edges) {
vector<vector<int>> g(n, vector<int>());
vector<bool> v(n, false);
for (auto a : edges) {
g[a.first].push_back(a.second);
g[a.second].push_back(a.first);
}
if (!dfs(g, v, 0, -1)) return false;
for (auto a : v) {
if (!a) return false;
}
return true;
}
bool dfs(vector<vector<int>> &g, vector<bool> &v, int cur, int pre) {
if (v[cur]) return false;
v[cur] = true;
for (auto a : g[cur]) {
if (a != pre) {
if (!dfs(g, v, a, cur)) return false;
}
}
return true;
}
};
``````
``````// BFS
class Solution {
public:
bool validTree(int n, vector<pair<int, int>>& edges) {
vector<unordered_set<int>> g(n, unordered_set<int>());
unordered_set<int> s{{0}};
queue<int> q{{0}};
for (auto a : edges) {
g[a.first].insert(a.second);
g[a.second].insert(a.first);
}
while (!q.empty()) {
int t = q.front(); q.pop();
for (auto a : g[t]) {
if (s.count(a)) return false;
s.insert(a);
q.push(a);
g[a].erase(t);
}
}
return s.size() == n;
}
};
``````
``````// Union Find
class Solution {
public:
bool validTree(int n, vector<pair<int, int>>& edges) {
vector<int> roots(n, -1);
for (auto a : edges) {
int x = find(roots, a.first), y = find(roots, a.second);
if (x == y) return false;
roots[x] = y;
}
return edges.size() == n - 1;
}
int find(vector<int> &roots, int i) {
while (roots[i] != -1) i = roots[i];
return i;
}
};
``````
Github 同步地址：
https://github.com/grandyang/leetcode/issues/261
Number of Islands II
Number of Connected Components in an Undirected Graph
https://leetcode.com/problems/graph-valid-tree/
https://leetcode.com/problems/graph-valid-tree/discuss/69018/AC-Java-Union-Find-solution
https://leetcode.com/problems/graph-valid-tree/discuss/69019/Simple-and-clean-c%2B%2B-solution-with-detailed-explanation.
LeetCode All in One 题目讲解汇总(持续更新中…)
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Help us with donation | 837 | 2,503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-26 | latest | en | 0.285513 |
https://brilliant.org/discussions/thread/nmtc-final-inter-2016/ | 1,532,049,676,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591455.76/warc/CC-MAIN-20180720002543-20180720022543-00436.warc.gz | 614,602,298 | 15,928 | # NMTC Final Inter 2016
Q1) A, B, C are three points on a circle. The distance of C from the tangents at A and B to the circle are a and b respectively. If the distance of C from the chord AB is c, show that c is the geometric mean of a and b.
Q2)Find all integer solutions to the equation x³+(x+4)²=y².
Q3)Two right angled triangles are such that the incircle of one triangle is equal in size to the circumcircle of the other. If P is the area of the first triangle and Q is the area of the second triangle, then show that P/Q≥3+2√2.
Q4) (a)Find the maximum value k for which one can choose k integers from 1,2,.....,2n so that none of the chosen integers is divisible by any other chosen integer.
(b) F(x) is a polynomial of degree 2016 such that all the coefficients are non negative and non exceed F(0).Show that the coefficient of x^2017 in (F(x))² is at most F(1)²/2.
Q5) (a) n>=3 and a1, a2, a3,....., an are different positive integers. Given that, except the first and the last, each one is a harmonic mean of its immediate neighbors. Show that none of the given integers is less than n-1. (b)Show that the shortest side of a cyclic quadrilateral with circumradius 1 is at most √2.
Q6)C1,C2,C3 are circles with radii 1,2,3 respectively, touching each other as shown in Figure. Two circles can be drawn touching all these circles. Find the radii of these two circles.
Note by Naitik Sanghavi
1 year, 9 months ago
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Sort by:
• Is NMTC objective or subjective?
• Is it IITJEE or Olympiad Mathematics?
- 1 year, 8 months ago
Well NMTC screening test is objective and NMTC final test is subjective.It is Olympiad Mathematics
- 1 year, 8 months ago
Done!BTW I am getting only two solutions for (Q2)!Am I correct?
- 1 year, 8 months ago
You have not mentioned question 5 b
- 1 year, 8 months ago
I have posted Junior Level Paper here.
- 1 year, 9 months ago
For question 6, use Descartes' Theorem.
- 1 year, 9 months ago
- 1 year, 9 months ago
Thanks for mentioning. Do you want me post the solutions, or you were just kind enough to help me have a glimpse of the paper?
- 1 year, 9 months ago
As you wish! 😊
- 1 year, 9 months ago | 946 | 3,021 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-30 | longest | en | 0.881526 |
https://www.toppr.com/guides/python-guide/examples/python-examples/python-program-find-factorial-number-using-recursion/ | 1,718,666,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00189.warc.gz | 876,806,475 | 28,512 | # Python Program to Find Factorial of Number Using Recursion
Factorial recursion is a method in which a function directly or indirectly calls itself. In mathematics, Factorial means the product of all the positive integers from 1 to that number. An exclamation mark is used after the integer to show that it’s a factorial.
For example, factorial eight is 8! So, it means multiplication of all the integers from 8 to 1 that equals 40320. Factorial of zero is 1. We can find a factorial of a number using python. There are various ways of finding; The Factorial of a number in python. Among which recursion is the most commonly used.
Source: Self
## Factorial Function in Python
### Source Code: Python Program to find factorial of a number using recursive function
def recursive_factorial(n):
if n == 1:
return n
else:
return n*recursive_factorial(n-1) #function calling itself
#taking input from the user
number = int(input("User Input : "))
print("The factorial of", number, "is", recursive_factorial(number))
### Output :
User Input: 4
The factorial of 4 is 24
### Source code: Displaying an error message if a user inputs a negative integer
def recursive_factorial(n):
if n == 1:
return n
else:
return n*recursive_factorial(n-1)
#taking input from the user
number = int(input("User input : "))
# If the user input negative integer
if number < 0:
else:
print("The factorial of", number, "is", recursive_factorial(number))
### Output:
User input : -1
### Source code: Factorial of zero is 1.
def recursive_factorial(n):
if n == 1:
return n
else:
return n*recursive_factorial(n-1)
#taking input from the user
number = int(input("User Input : ")
# If the user input negative integer
if number < 0:
# If the user inputs 0
elif number == 0:
print("The factorial of 0 is 1")
else:
print("The factorial of", number, "is", recursive_factorial(number))
### Output:
User Input : 0
The factorial of 0 is 1
## FAQs on Factorial Recursion
### What is Factorial recursion?
Factorial of a positive integer is the multiplication of all integers from that number to 1.
Factorial is represented by an exclamation mark.
If the user inputs a number let’s assume the number is n, so n factorial is n! = n * (n-1)!
For example:
factorial of 3 is
3! = 3 * 2 * 1. which equals 6.
There is no factorial output for negative integers
So, the factorial of 0 is 1
### How does Factorial recursion work?
Factorial of a number is the multiplication of all the integers from that number to 1.
Source: Self
The input value (positive integer) is passed; as a parameter to the factorial function.
So, Whenever the function runs, It multiplies the value of a parameter and recursively calls the function itself while decrementing the parameter value by 1.
Once the value becomes 0 the function returns 1 (as the factorial of 0 is 1) and returns the multiplier value that is the factorial of an inputted number.
How do you find the Recursive Factorial?
Factorial recursion is a function that is defined in such a way that it calls itself. Until it returns the factorial of the number passed as a parameter to the function.
Formula to calculate Factorial recursion.
n!=n * (n-1)! also [n!=1 if n=0]
factorial of 0 is 1
There is no factorial output for negative integers.
### What is recursion? Write a program for factorial calculation.
Recursion is a concept where you can loop through data to get a result. A problematic function can be broken down, into smaller sub-parts by using recursion.
##### Program for factorial calculation:
def recursive_factorial(n):
if n == 1:
return n
else:
return n*recursive_factorial(n-1)
num = int(input("User Input :"))
print("The factorial of", num, "is", recursive_factorial(num))
Output
User Input :5
The factorial of 5 is 120
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Get Started | 1,058 | 4,343 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.651065 |
https://www.physicsforums.com/threads/finding-a-scalar-field-given-two-gauge-fields.787533/ | 1,696,162,110,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00312.warc.gz | 1,025,062,731 | 15,855 | # Finding a scalar field given two gauge fields
• rwooduk
## Homework Statement
Demonstrate the equivalence between the gauge fields A1=(0,bx,0) and A2=)-yB/2,xB/2,0) and find the scalar field Φ for which A1= A2 + ∇Φ
B = ∇XA
## The Attempt at a Solution
The first part is fine, you just plug it into the above relevant equation and you get Bk for each. But I am unsure of the second part. I tried
A1 = A2 + ∇Φ ->>>
∇Φ = A1 - A2
Φ = (∫ yB/2 dx , ∫ (Bx - xB/2) dy, ∫ 0 dz)
but it was marked wrong with a "Φ is scalar!" comment.
it's probably really simple but just stuck on it.
as always thanks for any suggestions.
You need to solve $$\frac{\partial \Phi}{\partial x} = \frac{By}{2}, \\ \frac{\partial \Phi}{\partial y} = \frac{Bx}{2}, \\ \frac{\partial \Phi}{\partial z} = 0.$$ Have you not had to solve such problems before?
rwooduk
You need to solve $$\frac{\partial \Phi}{\partial x} = \frac{By}{2}, \\ \frac{\partial \Phi}{\partial y} = \frac{Bx}{2}, \\ \frac{\partial \Phi}{\partial z} = 0.$$ Have you not had to solve such problems before?
ahh i remember now, it was covered in last years math, will dig out my notes. Thank you! | 370 | 1,149 | {"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} | 3.625 | 4 | CC-MAIN-2023-40 | latest | en | 0.916286 |
https://www.stat.umn.edu/geyer/old/5601/fi.html | 1,713,881,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00564.warc.gz | 922,061,966 | 3,359 | # Stat 5601 (Geyer) Final Exam
Go to question: 1 2 3 4
## General Instructions
The exam is take-home. It is due in my office 6:00pm, Friday December 21, but can be handed in any time before. If I am not in my office, it can be handed to the staff in the School of Statistics office (313 Ford Hall) when they are open (8:00 to 4:30 Monday through Friday).
Do not work with anyone else or even discuss the test with anyone but the instructor.
The exam is open book, open notes, open web pages. You may use the computer, a calculator, or pencil and paper to get answers, but it is expected that you will use the computer.
• For simple computer commands, you may just write the command you used and the result it gave.
• For complicated commands or plots, make a printout and attach the printout to your test paper.
No credit for numbers with no indication of where they came from!
## Question 1 [20 pts.]
The file
http://www.stat.umn.edu/geyer/5601/mydata/darwin.txt
contains data on two variables `x` and `y` which are paired observations obtained by Charles Darwin in an experiment designed to study whether seedlings from cross-fertilized plants tend to be superior to those from self-fertilized plants. Each pair consists of the plant height of two plants grown from two seeds from the same parent, one (`x`) cross-fertilized and one (`y`) self fertilized (data and description taken from The Statistical Sleuth by Ramsey and Schafer, Duxbury, 1997).
In this problem we are interested in two estimators of the difference (in height) between the two groups, the median and the median of Walsh averages (the point estimates that go with the sign test and the Wilcoxon signed rank test, respectively).
For each of these estimators (or for each of the corresponding tests, however you wish to think of it) calculate the following three things (that's six things in all: three for each of two estimators).
• the point estimate
• the corresponding confidence interval with the smallest possible confidence level that is at least 95% (report the actual confidence level you use)
• the P-value for the two-tailed test with null hypothesis of no difference between the groups
## Question 2 [20 pts.]
The file
http://www.stat.umn.edu/geyer/5601/mydata/e.txt
contains data on two variables `x` and `y` which are paired. We wish to examine the dependence of `y` on `x` nonparametrically.
Using the procedures associated with Kendall's tau find the following three things
• The point estimate of tau
• A 95% large-sample confidence interval for tau
• The P-value for the two-tailed test of whether tau is zero.
Also interpret the P-value. Is there statistically significant dependence of `y` on `x`?
### Extra Credit
Notice that the parametric procedures (done by the `cor.test` function with no `method` specified) produce quite a different story. A glance at the plot (done by `plot(x, y)`) may give you an idea why. What is the reason?
## Question 3 [30 pts.]
The file
http://www.stat.umn.edu/geyer/5601/mydata/sally.txt
contains data on two variables `x` and `y`. In this problem we are only interested in `y` which is a random sample from a Cauchy population.
In this problem we are interested in two estimators of location, the median and the 25% trimmed mean, calculated by the code
```median(y)
mean(y, trim = 0.25)
```
For each of these estimators calculate the following four 95% confidence intervals (that's eight intervals in all: four intervals for each of two estimators).
• point estimate plus or minus 1.96 bootstrap standard errors
• the bootstrap t interval using the double bootstrap variance estimate
Note: For this interval, with these estimators, the bootstrap sample sizes suggested in the web page take too long (more than 5 minutes and Rweb usually loses the results -- this wouldn't happen if you were running R not through Rweb). So use `nboott = 1000` and the default for `nbootsd` (that is, don't specify it and let the function use its default).
• Efron's percentile method
• the BCa interval
Two of these four types of intervals have so-called second order accuracy. Which are they?
## Question 4 [30 pts.]
The `lynx` data set included in the R time series package is made available and plotted by the commands
```library(ts)
data(lynx)
plot(lynx)
```
A description of this time series is given by the on-line help for this data set.
We may consider this a stationary time series (although is has also been considered an example of chaotic dynamics in the biology literature). We are interested in three estimators
```quantile(lynx, 0.25)
median(lynx)
quantile(lynx, 0.75)
```
which we may consider to have asymptotics with a square root of n rate.
Find bootstrap 95% confidence intervals for these three estimators using the subsampling bootstrap and the confidence interval method explained in the handout and on the subsampling bootstrap confidence intervals web page (that's three intervals, one for each estimator).
Use a subsample size of b = 30 so that you get the same results as everyone else. | 1,195 | 5,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.922227 |
http://mathhelpforum.com/calculus/51299-quasilinear-pde.html | 1,481,409,560,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00324-ip-10-31-129-80.ec2.internal.warc.gz | 175,103,332 | 9,696 | 1. ## quasilinear pde
I have to solve the following quasilinear equation using the method of characteristics:
$x u_x + y u_y +yu=xy$
$u|_{y=x^2}=sin x$
$\frac{dx}{dt}=x$ gives me $ln x=t + c_1$
$\frac{dy}{dt}=y$ gives me $ln y=t+c_2$
But what do I do with $\frac{du}{dt}=xy - yu$?
Thank you.
2. Anyone? Please, it's really confusing and I need to understand it in order to go on.. | 132 | 386 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.787481 |
https://wikis.swarthmore.edu/ENGR005_2008/index.php?title=Robotic_Arm:_Team_November&diff=1627&oldid=1626 | 1,638,957,410,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00571.warc.gz | 656,160,248 | 6,919 | # Difference between revisions of "Robotic Arm: Team November"
Jump to: navigation, search
# Members
Jared Nolan: jnolan1, David D'Annunzio, Daniel Cabrera, Tony Blekicki
## MatLab Code for Physical Model
%[thetaA, thetaB] = inverseKinematicsfunction(4.5, 3.75,6.5,-1.5);
% This script takes input from the user and uses it to test the function % gotoAngles
s=instrfind; %Find any serial links (we can have only 1) delete(s); %... and delete. %Create a new serial communications link s=serial('COM1','Baudrate',115200,'Terminator','CR'); fopen(s); %... and open it
% set pulsewidth on channels 1 & 2to 1500 microseconds (over 200 mS) fprintf(s,'#1P1500#2P1500T200');
%Draw and animate the arm; numPts = 20; %the number of points in the animation phi = linspace(0,2*pi,numPts); %angle goes from 0 to 2*pi.
x0=6.5; %Define center (x0,y0) of circle and radius (r) y0=0; r=1.06066;
% loop through all the values of phi for i=1:length(phi),
``` x2=x0+r*cos(phi(i)); %x2 and y2 trace out a circle.
y2=y0+r*sin(phi(i));
```
``` %First do inverse kinematics to find angles...
la=4.5;
lb=3.75;
[theta_a, theta_b]= inverseKinematicsfunction(la, lb, x2, y2);
```
``` %Then do forward kinematics to find arm locations.
%[xa ya xb yb] = forwardKinematics(la, lb, theta_a, theta_b);
pause(0.1); %delay 1/10 second
```
``` gotoAngles(s,theta_a,theta_b-theta_a,750);
```
end
%Draw and animate the arm; numPts = 10; %the number of points in the animation x2=linspace(6.5+r,6.5+r,numPts) %generate an array of x2 ... y2=linspace(r,-r,numPts) %... and y2 values to define a line.
% loop through all the values of x2 and y2 for i=0:numPts,
``` x2=6.5+r
y2=r-(i*(2*r/numPts))
%First do inverse kinematics to find angles...
la=4.5;
lb=3.75;
```
``` [theta_a, theta_b]= inverseKinematicsfunction(la, lb, x2, y2);
pause(0.1); %delay 1/10 second
gotoAngles(s,theta_a,theta_b-theta_a,750);
```
end
%Draw the second side; numPts = 10; %the number of points in the animation x2=linspace(6.5+r,6.5-r,numPts) %generate an array of x2 ... y2=linspace(-r,-r,numPts) %... and y2 values to define a line.
% loop through all the values of x2 and y2 for i=1:length(x2),
``` %First do inverse kinematics to find angles...
la=4.5;
lb=3.75;
```
``` [theta_a, theta_b]= inverseKinematicsfunction(la, lb, x2(i), y2(i));
pause(0.1); %delay 1/10 second
gotoAngles(s,theta_a,theta_b-theta_a,750);
```
end
%Draw the third side; numPts = 10; %the number of points in the animation x2=linspace(6.5-r,6.5-r,numPts) %generate an array of x2 ... y2=linspace(-r,r,numPts) %... and y2 values to define a line.
% loop through all the values of x2 and y2 for i=1:length(x2),
``` %First do inverse kinematics to find angles...
la=4.5;
lb=3.75;
```
``` [theta_a, theta_b]= inverseKinematicsfunction(la, lb, x2(i), y2(i));
pause(0.1); %delay 1/10 second
gotoAngles(s,theta_a,theta_b-theta_a,750);
```
end
%Draw the fourth side; numPts = 10; %the number of points in the animation x2=linspace(6.5-r,6.5+r,numPts) %generate an array of x2 ... y2=linspace(r,r,numPts) %... and y2 values to define a line.
% loop through all the values of x2 and y2 for i=1:length(x2),
``` %First do inverse kinematics to find angles...
la=4.5;
lb=3.75;
```
``` [theta_a, theta_b]= inverseKinematicsfunction(la, lb, x2(i), y2(i));
pause(0.1); %delay 1/10 second
gotoAngles(s,theta_a,theta_b-theta_a,750);
```
end | 1,207 | 3,434 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.523845 |
https://www.convertunits.com/from/attojoule/to/electronvolt | 1,618,581,782,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00023.warc.gz | 815,667,481 | 16,825 | ## ››Convert attojoule to electronvolt
attojoule electronvolt
How many attojoule in 1 electronvolt? The answer is 0.16021773.
We assume you are converting between attojoule and electronvolt.
You can view more details on each measurement unit:
attojoule or electronvolt
The SI derived unit for energy is the joule.
1 joule is equal to 1.0E+18 attojoule, or 6.2415064799632E+18 electronvolt.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between attojoules and electronvolts.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of attojoule to electronvolt
1 attojoule to electronvolt = 6.24151 electronvolt
5 attojoule to electronvolt = 31.20753 electronvolt
10 attojoule to electronvolt = 62.41506 electronvolt
15 attojoule to electronvolt = 93.6226 electronvolt
20 attojoule to electronvolt = 124.83013 electronvolt
25 attojoule to electronvolt = 156.03766 electronvolt
30 attojoule to electronvolt = 187.24519 electronvolt
40 attojoule to electronvolt = 249.66026 electronvolt
50 attojoule to electronvolt = 312.07532 electronvolt
## ››Want other units?
You can do the reverse unit conversion from electronvolt to attojoule, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Attojoule
The SI prefix "atto" represents a factor of 10-18, or in exponential notation, 1E-18.
So 1 attojoule = 10-18 joules.
The definition of a joule is as follows:
The joule (symbol J, also called newton meter, watt second, or coulomb volt) is the SI unit of energy and work. The unit is pronounced to rhyme with "tool", and is named in honor of the physicist James Prescott Joule (1818-1889).
## ››Definition: Electron volt
In physics, the electron volt (eV) is a unit of energy. By definition, it is equal to the amount of kinetic energy gained by a single unbound electron when it accelerates through an electrostatic potential difference of one volt.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 683 | 2,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-17 | latest | en | 0.698889 |
http://lotterywinnerstories.com/Lottery/lucky-lottery-number-generator-rhode-island-lottery/ | 1,519,360,328,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814393.4/warc/CC-MAIN-20180223035527-20180223055527-00796.warc.gz | 201,352,547 | 12,188 | ## How Can You Win A Lottery? Rhode Island Lottery
How Do You Select The Best Numbers for Playing Rhode Island Lottery lottery?
A lottery is a game of chance sponsored by many states and countries. Winning the Rhode Island Lottery lottery game can be done easily by using the right strategy and system. Richard Lustig used his system to win several lottery jackpots and millions in prize money proving that luck has nothing to do with winning.
A lot of people use algorithm to analyze and predict lottery results for Rhode Island Lottery. Analysis of Groups There are many kinds of group analysis that lottery predictors use to get into the winning numbers. Lottery players can group the months having the best winning numbers of a certain period or they can group the numbers winning in certain period of time.Analysis of Hot-Cold Trend This algorithm analysis is one of the most favorite so far as it can record the frequency ranks and use the variations to predict the tendencies of hot and cold numbers in the next drawings. Analysis of Repetition Pattern A lot of lottery players share the same opinion that repetition is quite important to predict the winning lottery numbers as most of jackpots will appear again in the future.
## Interesting Facts About Rhode Island Lottery Lottery
Have you been searching for Pick 3 lottery workout systems that actually work? Are you tired of guessing and relying on luck? Well here's one system that you should try. It only involves simple mathematics and a bit of concentration. This system is one of the oldest. Although it doesn't guarantee a hundred percent, it has a high percentage hit. It is called the 123 workout. Here is how it works:
o You begin by using the preceding numbers drawn from the Pick 3 lottery. Just to provide a concrete example, let us use the numbers 468. We will call it the stack number.
o Important Rule of the 123 workout: Do not carry over when adding or subtracting the numbers. Here is an example: 5 + 9 = 4 (instead of 14)
o Monitor all the numbers that come out in the following days. You will realize that one out of the 22 combinations will be the next hit. Let's face it; it is better than having to choose from a million other combinations. Your chance of winning the next lottery is now greater having to choose from a smaller set.
Others have tried other Pick 3 lottery workout systems. Some of them are just too complicated. They would require the work of your personal computer, plus it costs much and takes time to download them. This system only requires a pen and a piece of paper. Try this method and see if the numbers really come out. What have you got to lose?
## Winning Numbers For Mega Millions - How to Choose Mega Millions Numbers
Why bother going to a gas station and buy some Pick 3 tickets when you can go for Pick 3 Online? Yes, you read that right. Pick 3 Online, the same Pick 3 lottery game that everyone is crazed upon can now be played without leaving your own home. If you have been playing the famous lottery game for quite sometime now then going through the ABC's on how to play it would just be a waste of time. Perhaps some guidelines on how to cling closer to the chances of winning might interest you?
If you've been playing the lottery game by betting your favorite numbers, you probably noticed that it doesn't pay out the way you expected it. Using favorite numbers are already a thing of the past when it comes to betting. Upon getting to try the Pick 3 Online, try a new strategy other than guessing; that would have to be not guessing at all. Pick 3 Online lottery is not a game of chance but a somewhat mathematical and logistic inclined game for the itching winner.
Be careful of downloading systems that assure you of winning online betting games. Those that offer cheats with a price seldom work and would just make a run for your money. Be mindful of the fact that online lottery games like Pick 3 Online are more of taking good strategy than taking a chance.
## How To Legally Crack The Lottery? Rhode Island Lottery
Playing the Rhode Island Lottery lottery can be very exciting especially if from time to time you are a winner. Winning can be accomplished quit often if the correct system is used. There are many so called experts in the market offering winning number secrets and scratch off ticket recommendations that promise to help you win. Most are not legit. However, there are a few systems on the market that has produced some promising results.
When you consider that that there are many different lottery systems available and lots of past winners you remain anonymous or don’t talk about their winning strategies how many more winners might there be who have used such systems?
Think about it: is it possible that a huge percentage of lottery winners are actually using mathematical or statistical formulas to help them win? If that is the case then anyone who is not using a system is merely feeding the prize fund and has an almost zero chance of winning.
## Would You Like to Win The Rhode Island Lottery lottery?
Are you constantly thinking and saying, "I want to win the lottery?" Do you ask yourself, "Will I win the lottery?" If you think about winning the lottery all the time, you need to know this secret to winning the secret to winning the lottery.
If you think winning the big bucks is just about picking the right lottery numbers, you're wrong. Yes, you do need to pick the right numbers, and it's best to have a system to win the lottery. Doing things like reading a lottery book can be useful too. But none of this will work for you if you don't do this one essential thing:
The universe rewards those who are prepared. Did you know that?
When you act as if you expect something, it's going to come to you. Did you know that?
4. Make a list of people to whom you plan to give money and decide on how much you'll give. Take into account gift taxes when you do this. Be careful about who you tell, "I'll give you money when I win." You don't want to give away all your winnings.
Combine this lottery win preparation method with a lottery winning system, and you will be able to answer "Yes," to "Will I win the lottery?" You'll be able to turn "I want to win the lottery," to "I WON the lottery."
## 5 Top Tips To Win More With Any Lottery
When it comes to megamillions lottery, there are a number of secrets that are unveiled. Sometimes, these secrets are facts that only players have ideas about. In this article, there are secrets and facts that will help a player that aims to win the jackpot prize.One secret of megamillions lottery is that it usually draws many odd numbers than even numbers. Of course, this is based on studies and analysis done by lottery experts. Then knowing this secret, one can favor odd numbers to increase the chance of winning.Another secret in megamillions lottery are the strategies that can be used to be able to have a greater chance of winning. The odd that make this lottery possible is 1 in 176 million. There are two strategies that can be used. First, play in groups. The chance of winning is greater since there many tickets that are purchased. Also, one of the reasons behind to not playing on own is due to the too long odds. It is best to let all other players go for the jackpot and play in groups. The second strategy is to play once in group in a year but invest the most. It is said that the opportunity comes only once that is why it is important to be careful too.Further, it has been found out that almost all of the players of megamillions lottery choose the quick pick method in choosing their numbers. This is the method when the machine chooses the numbers in random and gives them to the player. The rest choose to personally choose their numbers. Maybe, they use their birthdates and anniversary dates which is a common method in choosing the numbers.In addition, it has been found out that the odds of a set of numbers repeating themselves have a very nil probability. Then, knowing this secret might tell the gamer in advance not to use repeating numbers. So it is important to know if the numbers in hand have been played already or not. Megamillions lottery's result in the internet might help the player to determine which numbers have been played before or remains not picked.On the other hand, there are also myths that must be known so as to decide whether to play lottery or not. One myth is that the lotteries take advantage of the poorer people. This is said to be a myth since the people who usually plays the lottery are those who have extra cash in their pockets for leisure. In fact, a great percentage of players are rich people. Another myth is that the lottery is a form of tax. It is said that no one is required to join the lottery. And when the player wins, this is the only times that he/she is asked to pay taxes. The reason behind this is that the jackpot prize is also a form of income and any income is taxed by the government. Therefore, given these secrets mentioned, the player will have the right information in order to increase the chances of winning.
Winning the Rhode Island Lottery lottery is very rewarding. Give this system a spin and watch your winnings grow.
How Can You Win A Lottery? Rhode Island Lottery | 1,918 | 9,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-09 | longest | en | 0.947521 |
https://www.javatpoint.com/quick-sort-algorithm-in-cpp | 1,717,054,711,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00465.warc.gz | 714,206,511 | 27,484 | # Quick Sort Algorithm in C++
## An Introduction to the Quick Sort Algorithm
In computer science and data processing, sorting is a fundamental procedure. It entails putting a group of objects or components in a certain order, usually according to some criterion and in either ascending order or descending order. Applications like databases, search engines, and information retrieval all depend on sorting.
## The Quick Sort Algorithm
An effective and popular comparison-based sorting algorithm is Quick Sort, sometimes referred to as partition-exchange sort. Tony Hoare created it in 1960, and it has since evolved into a common technique for sorting. The speed of Quick Sort is well recognized, especially for big datasets, and it is frequently used as a standard for other sorting algorithms.
Quick Sort's main concept is to break a dataset into smaller subproblems, independently sort each subproblem, and then merge the sorted subproblems to get the final sorted dataset. A divide-and-conquer tactic is used to complete this procedure. The dataset splitting process used in Quick Sort is its main activity.
### The Partition Step
The Partition Step is a heart of quick Sort Algorithm. It works as follows:
1. From the dataset, pick a pivot element. Different methods may be used to choose the pivot element, and the choice of the pivot can influence how well the algorithm performs.
2. Rearrange the elements in the dataset so that everything on the left is bigger than the pivot and everything on the right is less than the pivot. Now, the pivot element is at the last position of the sort.
3. Recursively apply the partitioning step to the subarrays of the left and right of the pivot until the data is sorted.
## Choosing the Pivot
The pivot element and partition techniques that are used have a significant impact on Quick Sort's efficiency. There are several methods for choosing the pivot, such as:
Choosing the First Element as a pivot: The first element of the dataset is always selected as the pivot in this simple method.
Choosing the Last Element as a pivot: Similar to the prior approach, but with the final piece chosen as the pivot.
Selecting the Median-of-Three: The pivot is chosen using this technique as the median of the dataset's first, middle, and last components. It seeks to lessen the likelihood of running into worst-case scenarios.
Random pivot Selection: In this method, the pivot is selected randomly from the dataset. The danger of worst-case events may be reduced as a result.
## Algorithm steps of the Quick Sort
1. Choose a pivot Element:
• Choose a pivotal component from the list. The algorithm's effectiveness may be impacted by the pivot option. The first element, the final element, the midway element, or a random element are examples of common pivot selection procedures.
2. Partition the Array:
• Rearrange the elements of the array so that all those that are less than the pivot are moved to the left of it, and those that are larger than that are moved to the right.
• The pivot element now has been sorted to its last position.
3. Recursively Sort Subarray:
• The subarray to the left of the pivot, which has less elements than the pivot, is subjected to the Quick Sort algorithm in recursive fashion.
• The subarray to the right of the pivot, which contains entries larger than the pivot, is subjected to the Quick Sort algorithm iteratively.
4. Combine sorted Subarrays:
The array as a whole is sorted once all recursive calls are finished since each pivot element is at the proper end position.
5. Repeat until the entire array is sorted:
When the entire array is sorted, continue the steps of choosing a pivot, splitting the array, and recursively sorting subarrays once again.
Here is the pseudocode representation of the Quick Sort Algorithm:
The partitioning procedure and the recursive sorting of subarrays are two of the primary phases of the Quick Sort algorithm that are described in this pseudo-code. The method repeats itself until the entire array is sorted, each pivot element being in the proper place.
## Quick Sort Implementation in C++
Let's look at the C++ implementation of the Quick Sort algorithm now that we are familiar with its fundamentals. Implementing the partitioning phase first, then the recursive sorting procedure, is what we'll do first.
Explanation
In C++ implementation, we have defined three functions:
• 'Partition': This function accepts a vector 'arr' along with two integer arguments, 'low' and 'high', that define the range of entries to be partitioned. It selects the first element as the pivot and flips the order of the vector's elements so that everything to the left of the pivot is greater than everything to the right. The pivot element's index is what is returned.
• 'quickSort:' The Quick Sort method is mostly implemented by this sorting function. The range of elements to be sorted is specified by the vector 'arr' and the indices 'low and high'. Up until the full array is sorted, it recursively splits the array and sorts the left and right subarrays.
• 'printArray': This function is used to print the elements of the array.
• In the 'main' function, we show how to sort an example vector of numbers using the 'quickSort' function. The Quick Sort method is used to sort the array, which is then printed after printing the original array.
Output
Original Array: 12 4 5 6 7 3 1 15
Sorted Array: 1 3 4 5 6 7 12 15
## Performance Analysis
In both ordinary and ideal circumstances, Quick Sort is renowned for its outstanding performance. However, when the pivot decision regularly results in imbalanced partitions, its worst-case time complexity is O(n2). The median-of-three or random pivot selection procedures are frequently employed to reduce this risk.
Where 'n' is the number of elements to be sorted, Quick Sort has an average and best-case time complexity of O(n*log(n)). In particular, for huge datasets, this makes it one of the quickest sorting methods. Additionally, as Quick Sort is an in-place sorting algorithm, there is no need for additional RAM to store the sorted data.
The space needed for the recursive function calls and the call stack is represented by the Quick Sort space complexity, which is O(log(n)). Generally speaking, this space complexity is quite effective, which qualifies Quick Sort for sorting huge datasets.
## Some Practical Use Cases
1. Sorting Large Databases: In database management systems, Quick Sort is frequently used to efficiently sort huge amounts of data.
2. Search Algorithms: When data has to be sorted before searching, it is employed in search methods like binary search.
3. Numerical analysis: For sorting numerical data, Quick Sort is utilized in scientific computing and numerical analysis.
4. File Systems: A quick Sort is a good option since file systems frequently need to swiftly organize and retrieve data.
The effective and quick sorting algorithm known as Quick Sort is widely used. It is a popular option for sorting in a variety of applications because of its many benefits. Like every algorithm, it does, however, have some drawbacks and restrictions. We will examine the advantages and disadvantages of Quick Sort in C++ in this post to help you understand when and why to use it.
1. High Efficiency: Large datasets are especially well-suited for Quick Sort's outstanding efficiency. Its typical time complexity is O(n*log(n)), making it quicker than many other sorting algorithms like Bubble Sort and Insertion Sort. In circumstances when performance is important, this efficiency is a considerable benefit.
2. In-Place Sorting: As an in-place sorting method, Quick Sort sorts the data without the need for additional memory or storage. In contexts with limited memory, it might be crucial to sort the components inside the supplied array itself.
3. Cache Friendly: The algorithmic design of Quick Sort often results in high cache performance. It has strong locality of reference, which reduces cache misses while accessing data pieces and speeds up the process.
4. Versatility in Pivot Selection: Quick Sort's configurable pivot element selection enables programmers to customize the algorithm to meet their own requirements. Worst-case scenarios are less likely when pivot selection algorithms are modified to optimize performance for various input data types.
Last and First Pivot Selection: Initial and Final Pivot These tactics are straightforward and simple to use. They decide to use the first or final element, respectively, as the pivot.
Median of Three Pivots: The pivot is chosen using this technique as the median of the dataset's first, middle, and last components. The worst-case danger is lessened as a result.
Random Pivot Selection: The method becomes less predictable and more resistant to hostile input data by using a random element as the pivot.
1. Parallelization Potential: Quick Sort may effectively be parallelized, making it possible for it to benefit from distributed computing settings and multi-core computers. This allows for the concurrent sorting of very big datasets, greatly enhancing its performance.
2. Widely Used: An established and popular sorting algorithm is Quick Sort. Due to its effectiveness and speed, it is frequently used as the default sorting technique in computer languages and libraries. Developers are, therefore, frequently acquainted with its implementation and behavior.
1. Worst-case Time Complexity: The worst-case temporal complexity of Quick Sort is one of its main drawbacks. In the worst situation, Quick Sort can decline to O(n²), where 'n' is the number of items to be sorted, when the pivot selection constantly leads to imbalanced partitions (for example, by always choosing the smallest or biggest element as the pivot). In some circumstances, this worst-case scenario may out to be a serious disadvantage.
2. Sensitivity to Pivot Selection: The pivot element that is selected will have a significant impact on Quick Sort's performance. Although there are ways for pivot selection to lessen this sensitivity, a poor pivot selection might result in subpar performance or even the worst-case circumstances.
3. Lack of Stability: The sorting algorithm Quick Sort is not reliable. When two elements have identical keys, stability in sorting means that their relative order in the sorted output stays the same as in the initial input. Stability is not a given with Quick Sort, even if it could be crucial in some applications.
4. Recursive Nature: While this may be overcome by utilizing tail recursion or an iterative technique, it adds complexity to the solution and can cause stack overflow issues when sorting very big datasets.
5. Not Adaptive: Since Quick Sort is not an adaptive sorting algorithm, sorting partially sorted data does not greatly enhance its performance. When given partially sorted input, some sorting algorithms, like Insertion Sort, perform better.
6. Not Suitable for Linked list: The most effective method for sorting linked lists is not Quick Sort. It doesn't fully use the linked list data structure since it relies on random access to items, which might lead to subpar performance when compared to alternative sorting algorithms like Merge Sort or Insertion Sort.
## The Conclusion
In computer science and several applications, Quick Sort is a flexible and incredibly effective sorting algorithm. For sorting huge datasets, it is a popular option because of its speed, in-place nature, and versatility. Even though it can have a worst-case temporal complexity of O(n²), in reality, this problem can be mitigated by using appropriate pivot selection techniques. Quick Sort performs better than many other sorting algorithms when used properly and is crucial to the fields of computer science and data processing. | 2,316 | 11,820 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-22 | latest | en | 0.886924 |
https://questions.examside.com/past-years/gate/question/in-the-op-amp-circuit-shown-assume-that-the-diode-current-fo-gate-ece-2007-marks-2-wzwcvarzxrvjnl95.htm | 1,716,734,244,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00079.warc.gz | 400,395,457 | 31,722 | 1
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT) . For Vi = 2V, Vo = V01, and for Vi = 4 V, Vo = V02. The relationship between V01 and V02 is
A
$${V_{02}} = \,\sqrt 2 \,{V_{01}}$$
B
$${V_{02}} = {e^2}{V_{01}}$$
C
$${V_{02}} = {V_{01}}In\,2$$
D
$${V_{01}} - {V_{02}} = {V_T}In\,2$$
2
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
Consider the OP-Amp circuit shown in the figure.
The transfer function $${V_o}\left( s \right)/{V_i}\left( s \right)$$ is
A
$${{1 - sRC} \over {1 + sRC}}$$
B
$${{1 + sRC} \over {1 - sRC}}$$
C
$${1 \over {1 - sRC}}$$
D
$${1 \over {1 + sRC}}$$
3
GATE ECE 2007
MCQ (Single Correct Answer)
+2
-0.6
Consider the OP-Amp circuit shown in the figure.
If $${V_i} = {V_1}\,\,\sin \left( {\omega \tau + \phi } \right),$$ then the minimum and maximum values of $$\phi$$ (in radians) are respectively
A
$$- \pi /2\,\,and\,\,\pi /2$$
B
0 and $$\pi /2$$
C
$$- \pi$$ and 0
D
$$- \pi /2\,\,$$ and 0
4
GATE ECE 2006
MCQ (Single Correct Answer)
+2
-0.6
For the circuit shown in the following figure, the capacitor C is initially uncharged. At t = 0, the switch S is closed. The voltage VC across the capacitor at t = 1 millisecond is
In the figure shown above, the OP-AMP is supplied with $$\pm$$ 15V.
A
0 Volts
B
6.3 Volts
C
9.45 Volts
D
10 Volts
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
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CBSE
Class 12
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