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21,824 results, page 65
1. calculus
differentiate the following funcation: R(t) = (t + e^t)(4- sqrt (t))
2. calculus
fin derivative of the function (t+e^t)(4- sqrt t)
3. Calculus
help! ∫((ln x^2)/x)) dx u= ln x^2 du= 2 ln x dx (?) Not sure how to set this one up?
4. Calculus
Given f(x)=-10x2-42x-17 find f(x+h)
5. Calculus
For the function g(x) = xe power x, there is inflection point at?
6. Calculus
Need help with this complex number equation. 8-3i/-4-6i=
7. calculus
determine whether 9x-y+16 equals 13 49 y as a function of x
8. calculus
x=2cosa-cos2a,y=2sina-sin2a
9. Calculus
sif xyz=0 (-3x^2yz)^3 p=27x^7y^4z^6 then p=?
10. calculus
find cosx and sinx if tanx = 4
11. Calculus
I have an integration problem. INT(t+8)^2/t^4 dt
12. Calculus
find the limit of ((sqrt(x+4))-2)/x as x approaches 0.
13. calculus
Let f(x) be the function 6x^2-2x+8. Then the quotient (f(10+h)-f(10))/h can be simplified to ah+b for a and b eaquals what?
14. calculus
Lim Tan^2(3x)/2x^2 X-0 Find the Limit?
15. AP Calculus AB
how do you solve limit approaching 0 of (sin^2x) / (x)
16. calculus
Determine the following limit. lim (1/x-3) x->3
18. Calculus
Evaluate the indefinite integral: INT dx/(64+x^2)^2
19. Calculus
Determine the limit lim x->0 1-cos^2(3x)/x^2
20. calculus
How would you solve by completing the square? 3x^2 + X - 1/2= 0
21. calculus
Let P= (V^2)R / (R+r)^2 Calculate dP/dr, assuming that r is variable and R is constant.
22. calculus
john has\$1.70 . he has 3 more dimes than quarter.
23. Calculus
Find dy/dx by implicit differentiation: x^2-2xy+y^3 = c Thank you!
24. calculus
Find the limit as x approaches -4 of 4-lxl/4+x
25. calculus
What is the limit of this function as x approaches infinity? 2^x + x^3 / x^2 + 3^x
26. Calculus
Find all points of discontinuity f(x) = [x^2 + 5x - 2]/ [x^2 - 9]
27. Calculus
Find the points of discontinuity. f(x) = { 16/x^2 if x ≥ 2 { 3x - 2 if x < 2
28. calculus
find the derivative of (x^2+1) 3rd root of x^2+2
29. elements of calculus
find ds/dt if s= sqaureroot of t over (3t -1)
30. calculus
find the derivative with respect to x: sqrt(x+s) = (1/x) + (1/s)
31. Calculus
Given the following functions: F(u)= u^(5/2) and g(x)= x^6+1, find-f(g(x))=?, f'(u)=?, f'(g(x))=?, g'(x)=?,(f o g)'(x)=?
32. Calculus
find the critical numbers of the function: h(p)=p-1/p^2+4
33. Calculus
find d^2y/dx^2 (second derivative) for y=cos^2 4x
34. Calculus
Calculate the integral of: ((x^2)tan(x))/(1+((cos(x))^4))
35. calculus
How do you find the ariea between these curves? y=4x^2 y=7x^2 4x+y=3 x>=0?
37. calculus
tan−1(4x2y) = x + 5xy2
38. calculus
at what point is the function y=cos/(5x) continuous?
39. calculus
Find the inflection points of f''(x)= x(x+1)(x-2)^2. Explain
40. calculus
evaluate the limit 5x^(ln5+1)/(ln8x+1)
41. Calculus
Let x^3 + y^3 = 28. Find f''(x) at the point (3,1). Thank you very much for your help.
42. Calculus
Find f'(x) if f(x) = ln((8x+4)/(9x-7))^(1/2) Thank you very much for your time.
43. Calculus
Graph f(x) = 4^(x - 1) Please show work
45. calculus
How do I rond three decimal places?
46. Math - Calculus
find the derivative of: g(x) = log[(x^3+1)^3(x^3-1)^3)]
47. Calculus
evaluate the integral of dx/square root of 9-8x-x^2
48. calculus
find the rms value of the function i=15(1-e^-1/2t) from t=0 to t=4
49. calculus
find L- Hospital lim x_---- inf 1/e^(-x)=
50. calculus
find L- Hospital lim x-> inf f(x)=1/e^(-x)
51. calculus
find the average value of the function i=15[1-e^(-1/2t)] from t=0 to t=4 A. 7.5(1+e^-2) B. 7.5(2+e^-2) C. 7.5(2-e^-2) D. 7.5(3-e^-2)
52. calculus
find f(x -1) for the following function f(x) = sqrtx^2 - 3/ 6x + 6
53. calculus
w=4e^sqrt(s) Find the derivative of the function.
54. Calculus
FInd the area of the region between y=x^(1/2) and y=x^(1/5) for 0<x<1.
55. Calculus
Find the partial derivative for x. f(x,y)=1/SQRTx^2+y^2
56. calculus
What is the area under each of the given curves? y=e^x/2 ; x=0 x=1
57. calculus
find an antiderivative F(x) of f(x)=11x-sqrt(x)
58. calculus
Use logarithmic differentiation to find dy/dx for y=(1+x)^(1/x).
59. Calculus
How do you find the indefinite integral of (x^2-29x+5)/(x-4)^2(x^2+3)?
60. Calculus
Finding area under a curve From [1,3] 2x^2 -4x +1
61. Calculus
What are the critical numbers of the function: f(x)=2x^3-3x^2-36x+14 Thanks
62. calculus
Find the DERIVATIVE: h(x)= e^3x+2/e^3x-2 (quotient rule)
63. math/ calculus
given h(x)= cotx/x find h'(ð/2)
64. math/calculus
given h(x)= cotx/x find h'(ð /2)
65. math/calculus
find f"(x) if f(x)= cube root(x^2-2x+1)
66. calculus
dy\dx=x+1\square rod of x where y=5 x= 4 from chapter 5 (integration)
67. calculus
The derivative of f(x) = x^4/3 - x^5/5 attains its maximum value at x=
68. calculus
1] Reducing the Denominator: x + 1/m -m. 2] Factor the numerator.
69. integral calculus
ʃ (e^x – 4e^-x/e^x) dx need help...thnks
70. integral calculus
ʃ (e^x – 4e^-x/e^x) dx need help...thnks
71. math-calculus
|2x-3|-|2x+5| = 0 i please need helps with this.
72. math-calculus
solve for x. |3x-2| = |2x+5|+1 okay so for these one it be |3x-2| = |2x+5| because 1 times in with |2x+5| or it be (-2x-5)
73. math-calculus
|3-x|-1 = |4x+2| i try to do this and i not get correct answer which be -4/3 and 0.
74. math-calculus
|2x-3|-|2x+5| = 0 i please need helps with this.
75. math-calculus
solve for x. |3x-2| = |2x+5|+1 okay so for these one it be |3x-2| = |2x+5| because 1 times in with |2x+5| or it be (-2x-5)
76. Calculus 1
find an expression for a cubic function f if f(1)=6 and f(-1)=f(0)=f(2)=0
77. calculus
Solve the equation 14=65(19)x for x using logs.
78. Math(calculus)
Evaluate lim n( In(n+1)- In n ) as n--> infinity
79. calculus
let g(x)=integral x to (1/2) square root (t^3+1)dt .find g'(x)
80. calculus
integral(sin(2x)+5tan^2 x csc^2 x)dx=
81. calculus
integral(sin(2x)+5tan^2 x csc^2 x)dx=
82. calculus
f(x)=1/2sec2*3.14(t+1)-1/2 what does it look like graphed?
83. Calculus
∫[(x+3)/(x^2 +9)]=????? Can you give me a hint for solving it?
84. Calculus
f(x)=4-x^2-ln((1/2)x+1) find the tangent line approximation of f(-0.2)
85. Calculus
Find the domain of f(x) = (square root (x^2 - 16))/(x + 2)
86. Calculus (Antiderivatives)
What is the antiderivative of the following expression? 3x(x^2 + 7)^3
87. AP Calculus AB
What is the area of the region bounded by the graphs y=e^x, y=1, x=-1?
88. Calculus
Why is it important to determine continuity of the function?
90. Calculus
Determine the slope of the tangent to the curve y=2(x^2+x-1)^3 at (-1, -2)
91. Calculus AB
I have to find y' if ln(xy) = x + y. Where do I start? I'm not sure what to do with that ln(xy)
92. Calculus
If y= pie^2 then dy/dx = 2pie true or false ?
93. Calculus
Let f(x)=x^2-8x+16 and g(x) x^2-16. For what values of x is f(x)/g(x) increasing? Explain plz~
94. calculus
Determine the intervals on which f(x) is continuous f(x)=15sin(x^2+71)
95. Calculus
what is the curve given by the equation x^3=3xy^2+y^3=1 what is dy/dx?
96. college calculus
how do i solve sin(x)+2cos(2x)=0
97. calculus
find the domain of the function a) f(x,y) = e^(-xy) b) h,(u,v)= √(4-u^2 - v^2)
98. calculus
find the integral of x^2 + 1 over sqrt x from 1 to 2
99. calculus
Simplify with only positive exponents: (v^-1/2t)^-2(t^2/v^-2)^-3
100. calculus
Given f'(x)=3cosx-10sinx and f(0)= -6 Find f(x) | 2,848 | 7,373 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2018-26 | latest | en | 0.640297 |
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# Introduction to Graphing Calculators PowerPoint PPT Presentation
Introduction to Graphing Calculators. The “ON” button is at the lower left. You should see a little rectangle on the screen once you press “ON”. ON. To enter regular calculations, you enter the number and symbols such as To get the answer you press “ENTER” on the lower right. operations.
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Introduction to Graphing Calculators
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## Introduction to Graphing Calculators
• The “ON” button is at the lower left.
• You should see a little rectangle on the screen once you press “ON”.
ON
• To enter regular calculations, you enter the number and symbols such as
• To get the answer you press “ENTER” on the lower right
operations
Enter
numbers
• Press “CLEAR” to clear the screen
• Press “Y=” to get to the graphing utility
Y=
CLEAR
• Enter the equation
• Use the “X,T,θ,n” button to get the variable
X,T,θ,n
• Press the “GRAPH” key to see the graph
• Press the “WINDOW” key to change the window size to see better.
GRAPH
WINDOW
• What does “TRACE” do?
TRACE
• Let’s look at some more Graphs! | 421 | 1,704 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | longest | en | 0.866393 |
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Question Number 150841 by mathdanisur last updated on 15/Aug/21
$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }\:\mathrm{dx}\:=\:? \\$$
Answered by Olaf_Thorendsen last updated on 15/Aug/21
$${f}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{\mathrm{1}+{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\:{dx}\:\:\:\:\left(\mathrm{1}\right) \\$$ $$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{ax}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left(\mathrm{1}+{b}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}\:{dx} \\$$ $$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\right)\:{dx} \\$$ $$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left[\frac{\mathrm{arctan}\left({ax}\right)}{{a}}−\frac{\mathrm{arctan}\left({bx}\right)}{{b}}\right]_{\mathrm{0}} ^{\infty} \\$$ $$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\mathrm{2}{a}}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\left(\frac{\pi}{\mathrm{2}{a}}−\frac{\pi}{\mathrm{2}{b}}\right) \\$$ $$\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\:\frac{\pi}{{b}\left({a}+{b}\right)} \\$$ $$\Rightarrow\:{f}\left({a},{b}\right)\:=\:\frac{\pi}{{b}}\mathrm{ln}\left({a}+{b}\right)+\mathrm{C}\left({b}\right)\:\:\:\:\left(\mathrm{2}\right) \\$$ $$\\$$ $$\left(\mathrm{1}\right)\::\:{f}\left(\mathrm{0},{b}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{0}{x}^{\mathrm{2}} \right)}{\mathrm{1}+{b}^{\mathrm{2}} {x}^{\mathrm{2}} }\:{dx}\:=\:\mathrm{0} \\$$ $$\left(\mathrm{2}\right)\::\:{f}\left(\mathrm{0},{b}\right)\:=\:\frac{\pi}{{b}}\mathrm{ln}\left(\mathrm{0}+{b}\right)+\mathrm{C}\left({b}\right)\:=\:\:\frac{\pi}{{b}}\mathrm{ln}{b}+\mathrm{C}\left({b}\right) \\$$ $$\Rightarrow\:\mathrm{C}\left({b}\right)\:=\:−\frac{\pi}{{b}}\mathrm{ln}{b} \\$$ $$\\$$ $$\left(\mathrm{2}\right)\::\:{f}\left({a},{b}\right)\:=\:\frac{\pi}{{b}}\mathrm{ln}\left({a}+{b}\right)−\frac{\pi}{{b}}\mathrm{ln}{b} \\$$ $$\\$$ $${f}\left({a},{b}\right)\:=\:\frac{\pi}{{b}}\mathrm{ln}\left(\mathrm{1}+\frac{{a}}{{b}}\right) \\$$
Commented byamin96 last updated on 15/Aug/21
$${nice} \\$$
Commented bymathdanisur last updated on 15/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{cool} \\$$
Answered by mathmax by abdo last updated on 17/Aug/21
$$\Psi\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)}{\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:\Rightarrow\Psi^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2ax}^{\mathrm{2}} }{\left(\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dx} \\$$ $$=\mathrm{2a}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{ax}−\mathrm{i}\right)\left(\mathrm{ax}+\mathrm{i}\right)\left(\mathrm{bx}−\mathrm{i}\right)\left(\mathrm{bx}+\mathrm{i}\right)}\mathrm{dx} \\$$ $$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }\int_{−\infty} ^{+\infty} \:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{a}}\right)\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{a}}\right)\left(\mathrm{x}−\frac{\mathrm{i}}{\mathrm{b}}\right)\left(\mathrm{x}+\frac{\mathrm{i}}{\mathrm{b}}\right)}\mathrm{dx} \\$$ $$=\frac{\mathrm{1}}{\mathrm{ab}^{\mathrm{2}} }\left(\mathrm{2i}\pi\right)\left\{\mathrm{Res}\left(\Lambda,\frac{\mathrm{i}}{\mathrm{a}}\right)+\mathrm{Res}\left(\Lambda,\frac{\mathrm{i}}{\mathrm{b}}\right)\right\} \\$$ $$\mathrm{Res}\left(\Lambda,\frac{\mathrm{i}}{\mathrm{a}}\right)=\frac{−\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \left(\frac{\mathrm{2i}}{\mathrm{a}}\right)\left(−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\right)}=\frac{−\mathrm{1}}{\mathrm{2ia}\left(\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }\right)}=\frac{−\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} }{\mathrm{2ia}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)} \\$$ $$=\frac{−\mathrm{ab}^{\mathrm{2}} }{\mathrm{2i}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)} \\$$ $$\mathrm{Res}\left(\Lambda,\frac{\mathrm{i}}{\mathrm{b}}\right)\:=\frac{−\mathrm{ba}^{\mathrm{2}} }{\mathrm{2i}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)}\:\Rightarrow \\$$ $$\Psi^{'} \left(\mathrm{a}\right)=\frac{\mathrm{2i}\pi}{\mathrm{ab}^{\mathrm{2}} }\left\{−\frac{\mathrm{ab}^{\mathrm{2}} }{\mathrm{2i}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)}−\frac{\mathrm{ba}^{\mathrm{2}} }{\mathrm{2i}\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)}\right\} \\$$ $$=−\frac{\pi}{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }+\frac{\pi\mathrm{ba}^{\mathrm{2}} }{\mathrm{ab}^{\mathrm{2}} \left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)}=−\frac{\pi\mathrm{b}}{\mathrm{b}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)}\:+\frac{\pi\mathrm{a}}{\mathrm{b}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} \right)} \\$$ $$=\frac{\pi\left(\mathrm{a}−\mathrm{b}\right)}{\mathrm{b}\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)}=\frac{\pi}{\mathrm{b}\left(\mathrm{a}+\mathrm{b}\right)}\:\Rightarrow\Psi\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{b}}\mathrm{ln}\left(\mathrm{a}+\mathrm{b}\right)\:+\mathrm{C} \\$$ $$\Psi\left(\mathrm{0}\right)=\mathrm{0}=\frac{\pi\mathrm{lnb}}{\mathrm{b}}+\mathrm{c}\:\Rightarrow\mathrm{c}=−\frac{\pi\mathrm{lnb}}{\mathrm{b}}\:\Rightarrow \\$$ $$\Psi\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{b}}\mathrm{ln}\left(\mathrm{a}+\mathrm{b}\right)−\frac{\pi}{\mathrm{b}}\mathrm{ln}\left(\mathrm{b}\right)=\frac{\pi}{\mathrm{b}}\mathrm{ln}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right) \\$$
Commented bymathdanisur last updated on 18/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\$$ | 2,766 | 6,551 | {"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} | 3.828125 | 4 | CC-MAIN-2023-23 | latest | en | 0.173013 |
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IARCS home > OLYMPIAD
Solution to Problem 2: The Great Escape
```With this problem begins our first foray into what is called
Graph Theory. More about that later.
We say that a pair of buildings B_1 and B_2 are adjacent if the
director has determined that it is possible to jump from B_1 to
B_2 (and vice versa).
We know the building the hero starts his search from. Call this
building B and set S_0 = { B } and clearly every building in S_0
is reachable in 0 steps by the hero.
The set of buildings that can be reached in a single step can
computed as follows. It is the set of buildings that are adjacent
to B. Call this set S_1.
What about buildings that can be reached in 2 jumps (and not in
fewer than 2 jumps)? Every such building must be adjacent to
some building in S_1. However, not all buildings adjacent to
buildings in S_1 will be in S_2 (since they may already be in S1
or S0). For example, if we had 4 buildings B, C, D and E and the
following pairs are adjacent:
B C
B D
C D
D E
Then, S_0 = { B } and S_1 = { C,D }. Now, the buildings C, D, B,
and E have buildings in S_1 adjacent to them. However, since B, C
and D are already present in S_0 U S_1 (where U denotes the union
of sets), S_2 is the set { E }.
In general let S_i be the set of buildings that can be reached
from B by i jumps but not in i-1 or fewer jumps. Then,
S_0 = { B }
S_i = { C | C is adjacent to some building in S_{i-1} and
C is not in S_0, S_1, ... S_{i-1} }
So, we can compute S_0, S_1, ... in that order. When do we stop?
Since S_i contains only buildings that do not belong to S_0, S_1,
... S_{i-1}, we will soon run out of buildings and for some k,
S_k will be empty. As a matter of fact, k <= n if n is the
number of buildings. If S_k is empty then there are no buildings
that are reachable in k steps but not fewer. This in turn means
that there are no buildings that are reachable in k+1 steps and
not fewer and so on (Why?)
To illustrate this, consider the set of buildings B,C,D,E,F,G,H
with adjacency given by:
B C
B D
C D
D E
F G
H G
F H
Then S_0 = { B }
S_1 = { C,D }
S_2 = { E }
S_3 = { }
Thus, the buildings F,G and H are not reachable from B. C and D
are reachable in 1 step and E in two steps.
Now, how do we program this? Let the buildings be 1 ... n. We
keep an array D with D[i]=j indicating that "i belongs to S_j" or
in other words indicating that building i is at distance j from
the starting building.
We also assume that there are e adjacency relations and they are
stored in a two dimensional array E. That is, E is a two
dimensional array and The buildings E[i][1] and E[i][2], for i=1,
2, ... e are deemed to be close by the director.
So our above example with buildings B,C,D,E,F,G and H would now
consist of buildings 1,2,3,4,5,6,7 and the array E looks like:
1 2
1 3
2 3
3 4
5 6
7 6
5 7
where the first entry on row i is E[i][1] and the second entry is
E[i][2].
We now describe the algorithm to compute the shortest distance
from building b to all the other vertices.
for (i = 1 to n)
D[i] = MAXINT /* we use MAXINT > n
to denote "undefined" */
D[b]=0 /* We start at building b and it belongs to
S0 */
for (i=1 to n) { /* compute S_1,S_2,...,S_n */
for (j=1 to e) {
/* Check if each adjacency relation connects a building
in S_{i-1} to a building that is not in S_0,...S_{i-1}
Update S_i appropriately */
if ((D[E[j][1]] == (i-1) ) && (D[E[j][2]] == MAXINT))
D[E[j][2]]=i
/* You have found an as yet unvisited building that
is adjacent to an building in S{i-1}, so put that
in Si */
if ((D[E[j][2]] == (i-1) ) && (D[E[j][1]] == MAXINT))
D[E[j][1]]=i
/* You have found an as yet unvisited building that
is adjacent to an building in S{i-1}, so put that
in Si */
}
}
If the target is to reach building t, then D[t] gives us the
length of the shortest sequence of jumps that takes the hero to
building t. If D[t] is MAXINT then there is no route by which the
hero can reach the target.
Observations:
1) The above algorithm determines the shortest path from b to
every building and not just t!!
2) It also determines which buildings are reachable eventually
and which are not.
The structure of the above algorithm suggests that it takes
roughly n * e steps where n is the number if buildings and e is
the number of edges. We can do significantly better, in
pariticular this problem can be solved in roughly e steps. Before
we describe that solution, we want to introduce the terminology
of graph theory for future reference.
A graph, say G, consists of two sets, a set of vertices or nodes
(V) and a set of edges (E). The set of vertices could be given
finite set. Each edge is a set consisting of two vertices. We
may interpret the set of edges as describing which pairs of
vertices are "neighbours".
The buildings and close pairs define a graph. We take the set of
buildings to be the set of vertices and the set of edges consist
of all those pairs of vertices that the director has deemed fit.
The example with 7 vertices (1,2,...,7) is denoted by the graph
G = (V,E)
where
V = { 1,2,3,4,5,6,7 }
E = { {1,2},{1,3},{2,3},{3,4},{5,6},{5,7},{6,7} }
It is often convenient to draw graphs with boxes or circles to
represent the nodes and lines/curves connecting boxes/circles to
denote edges:
--- --- ---
| 1 |---------| 2 | | 7 |
--- /--- /---\
| / / \
| /-----/ / \
| / / \
---/ --- ---/ \ ---
| 3 |--------| 4 | | 5 |------------| 6 |
--- --- --- ---
Note that the placement of the boxes or the style of drawing the
edges are irrelevant. The "real" graph is given by the two sets V
and E and the picture is just to help us think.
A path in a graph is a sequence of vertices v1,v2,...,vk such
that {v1,v2}, {v2,v3} ... {v{k-1}vk} are edges. For example,
1,2,3,4 is a path in the above graph. The length of a path is the
number of edges in the path. For example, the length of the path
1,2,3,4 is 3.
The solution to the hero problem described above is actually a
solution to the following general problem on graphs: given a
vertex u, find the length of the shortest path from u to every
other vertex v in the graph. And the solution recast in terms of
arbitrary graphs reads as follows (and see that all I have to do
replace the word "building" by vertex and so on ..)
We assume that there are n vertices 1...n and e edges. We assume
further that the edges are stored in a two dimensional array E,
so that if E[i][1]=a and E[i][2]=b then {a,b} is an edge in the
graph.
for (i = 1 to n)
D[i] = MAXINT /* we use MAXINT > n
to denote "undefined" */
D[b]=0 /* We start at vertex b and it belongs to
S0 */
for (i=1 to n) { /* compute S_1,S_2,...,S_n */
for (j=1 to e) {
/* Check if any edge connects a vertex
in S_{i-1} to a vertex that is not in S_0,...S_{i-1}
Update Si appropriately */
if ((D[E[j][1]] == (i-1) ) && (D[E[j][2]] == MAXINT))
D[E[j][2]]=i
/* You have found an as yet unvisited vertex that
is adjacent to a verted in S_{i-1}, so put that
in S_i */
if ((D[E[j][2]] == (i-1) ) && (D[E[j][1]] == MAXINT))
D[E[j][1]]=i
/* You have found an as yet unvisited vertex that
is adjacent to a vertex in S_{i-1}, so put that
in S_i */
}
}
A vertex i is reachable from the vertex b (i.e. there is a path
starting at vertex b and ending at vertex i) if and only if D[i]
is not MAXINT. And if it is reachable then D[i] is the length of
the shortest path from b to i.
Note: The above algorithm is called a Breadth-First Search
algorithm (or BFS) as unravels the graph level by level, where
level i consisting of all the vertices at distance i from b.
This is in contrast to a different algorithm called Depth-First
Search.
Is the above algorithm the best we can do? The inner loop which
constructs S_i from S_{i-1} examines every edge. Is that
necessary? After all, elements of S_i are those unvisited
vertices (i.e. vertices that do no belong to S0, ... S{i-1}) that
are connected by an edge to some vertex in S_{i-1}. So it should
suffice to examine all the vertices that are connected by edges
to vertices in S_{i-1}. In other words, it should be enough to
consider only those edges that connect element of S_{i-1} to
other vertices.
If edges are stored in the array E as assumed in the earlier
solution, then it is not possible to examine the just this subset
without scanning through the entire array E.
The idea is to store the edges in what is called an "Adjacency
Matrix". Given a graph G with vertices 1,...,n, the edges of the
graph can be represented by a 2 dimensional array (or
equivalently a (n x n)-matrix) A defined as follows:
A[i][j] = 1 if vertex i and vertex j are connected by an
edge
A[i][j] = 0 otherwise.
Observation: If A[i][j] is 1 then A[j][i] will also be 1 (Since
our edges are subsets are undirected. i.e. an edge {i,j} and we
can use this to go from i to j as well as from j to i).
With this we rewrite the above algorithm as follows:
for (i=1 to n)
for (j=1 to n)
A[i][j]=0 /* Initialize the adjacency matrix*/
for (i=1 to e) { /* set the adjacency matrix based on E
A[E[i][1],E[i][2]]=1; if {a,b} is an edge set A[a][b]=1
A[E[i][2],E[i][1]]=1; and A[b][a]=1 */
}
for (i = 1 to n)
D[i] = MAXINT /* we use MAXINT > n
to denote "undefined" */
D[b]=0 /* We start at vertex b and it belongs to
S0 */
for (i=1 to n) { /* compute S1,S2,...,Sn */
for (j=1 to n) { /* scan vertices to find vertices
if (D[j] is i-1) in S{i-1} */
{ /* found one. look for its unvisited
neighbours*/
for (k=1 to n) {
if (A[j][k] == 1) and (D[k]==MAXINT)
/* Found one, so put it in Si */
D[k]=i
}
}
}
}
How much time does this take? The initialization takes time
roughly proportional to e + n*n. And now we have three levels of
nested loops and this is going to take roughly n*n*n steps and so
we get e + n*n + n*n*n which is roughly n*n*n (since e is not
more than n*n). Well, that is clearly not better than n*e which
can in the worst case be n*n*n!!!
What we needed to do was to directly identify the elements of
S_{i-1} (WITHOUT scanning the entire list of vertices) and then
for each such vertex find the list of neighbours (WITHOUT
scanning through the entire list of edges). The adjacency matrix
offers a solution to the second problem but not to the first.
Consider the following graph with 7 vertices:
G = (V,E)
where
V = { 1,2,3,4,5,6,7 }
E = { {1,2},{1,3},{2,3},{3,4},{3,6},{4,7},{6,5} }
Here, starting at vertex 1 we get
S0 = { 1 }
S1 = { 2,3 }
S2 = { 4,6 }
S3 = { 7,5 }
S4 = { }
The order in which we "visit" the vertices is <1 2 3 4 6 7 5> and
this sequence breaks nicely into S0,S1,... as {1}{2,3}{4,6}{7,5}.
This is not an accident and will always happen since we visit all
the elements of S_{i-1} before visiting the elements of S_i.
Moreover when we intend to explore the neighbours of a vertex v
in this sequence, we would have already explored all the
neighbours of all the vertices that appear before it in the
sequence.
Thus, if we keep track of the sequence in which vertices have
been visited and the last vertex whose neighbours have been
explored we can determine the next vertex whose neighbours need
to be explored.
Intially we have the sequence <1> and nothing is explored. So we
examine the neighbours of 1 and find that 2 and 3 are its
neighbours and we add them to the list to get <1.2 3> where I
have placed a . after 1 to indicate that 1 is the last vertex
whose neighbours have been explored. We examine the neighbours of
2 next and get <1 2.3 > (since the only neighbour of 2 is 3 and
it is already in the list.)
Considering 3 next we get <1 2 3.4 6> and then <1 2 3 4.6 7> and
then exploring neighbours of 6 we get <1 2 3 4 6.7> and finally
exploring the neighbours of 7 we get <1 2 3 4 6 7 5.> and now
there is nothing more to explore.
Observe that we need not carry along the vertices to the left of
the . as they play no further role. Thus, we can simply drop the
vertices to the left of the . in all these sequences. Thus, we
begin with <1>, exploring the neighbours of 1 we get <2 3> then
exploring the neighbours of 2 we get <3> and then exploring the
neighbours of 3 we get <4 6> and then <6 7> followed by <7> and
finally <>.
The way updated the sequences above has the following feature: we
add elements to the right and delete elements from the left.
This similar to what happens at the queue at a railway counter or
a cinema ticket counter. You enter at one end and leave from the
other. Thus, we need to keep the vertices in a queue, pick the
one at the head of the queue, explore its neighbours and and add
any unvisited ones to the tail of the queue and then delete the
head of the queue.
We can use an array Q and two integer variables H (for head) and
T (for tail) to do this as follows:
For instance the queue <2 7 6 3> would be represented by
i : 1 2 3 4
Q[i]: 2 7 6 3
H = 1 and T = 4
Now, if we add 4 to this queue, we will store it position T+1 (5)
and increment T to 5 to get:
i : 1 2 3 4 5
Q[i]: 2 7 6 3 4
H = 1 and T = 5.
Now if we delete an element from the head of this queue, the
resulting queue would look like:
i : 1 2 3 4 5
Q[i]: 2 7 6 3 4
H = 2 and T = 5.
Now if we add an element say 1, we would get:
i : 1 2 3 4 5 6
Q[i]: 2 7 6 3 4 1
H = 2 and T = 6.
At any point the queue consists of the elements Q[H],Q[H+1]
...Q[T]. Q[H] is at the head of the queue and Q[T] is at the
tail.
How is the empty queue represented? Simple, set T n
to denote "undefined" */
H = 1
T = 0 /* Initialize the queue to be empty */
D[b]=0 /* We start at vertex b and it belongs to
S_0 */
Q[T+1]=b
T=T+1 /* Insert b as the first element of the Queue*/
While ( T >= H ) { /* As long as the queue is not empty */
j = Q[H] /* Let us pick the first vertex in the queue
i.e. the one at the head of the queue */
for (k=1 to n) { /* See if it has unvisited neighbours */
if (A[j][k] == 1) and (D[k]==MAXINT) { /* Found one */
/* Since j belongs to S_{D[j]} k must belong
to S_{D[j]+1}. Update D[k] to indicate that*/
D[k]=D[j]+1
/* and insert it at the tail end of the queue*/
Q[T+1]=k
T=T+1
}
}
H = H+1 /* Since the neighbours of j have been considered
delete j from the queue
}
How long does this algorithm take? Notice that the inner "for"
loop is executed for a vertex j only when it is at the head of
the queue. Each vertex in the graph can be at the head at the
most once (If it ever enters the queue it will be at the head
exactly once and then it will be deleted from the queue). Thus
the inner for loop is executed at the most once for each vertex
in the graph and so the number of steps taken by this algorithm
is roughly n*n steps. This is a significant improvement on n*n*n
as well as n*e (since e can be as large as n(n-1)/2).
Can we do better? One place where we could optimize is the inner
for loop. To explore the neighbours of j we look at every vertex
k and check if it is a neighbour of j and then if it has been
visited. If we could somehow keep track of the neighbours of j
then we only need to examine whether each neighbour has been
visited. It turns out that this can be done and the resulting
algorithm takes roughly e steps. We postpone a description of
that idea to a different day. ```
Copyright (c) IARCS 2003-2020; Last Updated: 15 Mar 2005 | 4,694 | 15,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | latest | en | 0.970286 |
https://questions.examside.com/past-years/gate/question/let-zlefttrightxlefttrightast-ylefttright-where-denotes-conv-gate-ee-2017-set-1-marks-1-5a904d4e82b67.htm | 1,718,842,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00110.warc.gz | 428,435,942 | 30,374 | 1
GATE EE 2017 Set 1
+1
-0.3
Let $$z\left(t\right)=x\left(t\right)\ast y\left(t\right)$$, where "*" denotes convolution. Let C be a positive real-valued constant. Choose the correct expression for z(ct).
A
$$c.x\left(ct\right)\ast y\left(ct\right)$$
B
$$x\left(ct\right)\ast y\left(ct\right)$$
C
$$c.x\left(t\right)\ast y\left(ct\right)$$
D
$$c.x\left(ct\right)\ast y\left(t\right)$$
2
GATE EE 2016 Set 1
+1
-0.3
The value of $$\int_{-\infty}^{+\infty}e^{-t}\partial\left(2t-2\right)dt$$. where $$\partial\left(t\right)$$ is the Dirac delta function, is
A
$$\frac1{2e}$$
B
$$\frac2e$$
C
$$\frac1{e^2}$$
D
$$\frac1{2e^2}$$
3
GATE EE 2015 Set 1
Numerical
+1
-0
A moving average function is given by $$y\left(t\right)=\frac1T\int_{t-T}^tu\left(\tau\right)d\tau$$. If the input u is a sinusoidal signal of frequency $$\frac1{2T}Hz$$, then in steady state, the output y will lag u (in degree) by ________.
4
GATE EE 2013
+1
-0.3
For a periodic signal the $$v\left(t\right)=30\sin100t\;+\;10\cos300t\;+\;6\sin\left(500t\;+\;\frac{\mathrm\pi}4\right)$$ fundamental frequency in radians/s is
A
100
B
300
C
500
D
1500
EXAM MAP
Medical
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Number or ratio expressed as a portion of 100
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Basic Accuracy: 73.12% Submissions: 1866 Points: 1
Given two arrays of equal size N and an integer K. The task is to check if after permuting both arrays, we get sum of their corresponding element greater than or equal to k i.e Ai + Bi >= K for all i (from 0 to N-1). Return true if possible, else false.
Example 1:
Input :
a[] = {2, 1, 3},
b[] = { 7, 8, 9 },
k = 10.
Output :
True
Explanation:
Permutation a[] = { 1, 2, 3 }
and b[] = { 9, 8, 7 }
satisfied the condition a[i] + b[i] >= K.
Example 2:
Input :
a[] = {1, 2, 2, 1}, b[] = { 3, 3, 3, 4 }, k = 5.
Output :
False
Explanation:
Since any permutation won't give the answer.
You don't need to read input or print anything. Your task is to complete the function isPossible() which takes the array A[], B[], its size N and an integer K as inputs and returns the answer.
Expected Time Complexity: O(N. Log(N))
Expected Auxiliary Space: O(1)
Constraints:
1 ≤ N ≤ 105
1 ≤ K ≤ 1018
1 ≤ Ai, Bi ≤ 1017
We are replacing the old Disqus forum with the new Discussions section given below. | 341 | 1,067 | {"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} | 3.34375 | 3 | CC-MAIN-2021-39 | longest | en | 0.793468 |
http://nrich.maths.org/9546/index?nomenu=1 | 1,498,529,399,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320887.15/warc/CC-MAIN-20170627013832-20170627033832-00159.warc.gz | 281,980,444 | 3,044 | This problem is part of the Great Expectations: Probability through Problems collection. It is designed for classroom use; see the Teachers' Resources for a suggested classroom approach.
Every weekend, Team Beaver and Team Yeti play each other at 2-Goal Football - they play until two goals have been scored. Sometimes games last only a minute or two, sometimes they seem to go on for ever.
Over a 36-week season, which team is likely to win more games?
You can investigate this question through a practical experiment.
You can model this with a die with four yellow faces (Team Yeti) and two blue faces (Team Beaver).
If this is not available, use an ordinary die and let 1, 2, 3 and 4 correspond to a yellow face, and 5 and 6 to a blue face.
First investigate what happens for one game:
1. Throw the die. A yellow (1, 2, 3, 4) face means a goal to Team Yeti, a blue (5, 6) face means a goal to Team Beaver. Record this on the Results Worksheet - you might find it helpful to use a yellow colour for Team Yeti and a blue colour for Team Beaver.
2. Throw the die again, to see who scores the second goal, and record it on the Results Worksheet.
Who won? Or was it a draw? Compare your result with other people's - any surprises?
Now repeat the experiment 36 times in total, to get the results for a full season.
Are you surprised by your results? How do they compare with what you would expect?
Which team scores most often? Does this mean they will always win?
How often would you expect each team to score?
Did the Yetis win the proportion of games you would expect? How about the Beavers? Can you explain what happened?
You can explore the scenario in this problem using this interactive environment. | 400 | 1,717 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-26 | latest | en | 0.94764 |
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# Moments of Inertia - PowerPoint PPT Presentation
Moments of Inertia. Lesson 7.6. Review. Recall from previous lesson the first moment about y-axis The moment of inertia (or second moment) is the measure of the tendency of an object to resist change in motion. Moment of Inertia. For a system of n masses
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### Moments of Inertia
Lesson 7.6
Review
• Recall from previous lesson the first moment about y-axis
• The moment of inertia (or second moment) is the measure of the tendency of an object to resist change in motion
Moment of Inertia
• For a system of n masses
• If the masses were at the same distance r from the axis of rotation we have
• Radius of gyration about x-axis(where m is total mass of system)
Example
• Suppose we have 3g at (2,3), 4g at (-2,-4), and 3g at (-4,5)
• Find Iy
• Iy = 3*2 2 + 4*(-2)2 + 3*(-4)2 = 76
• Find Ix
• Ix = 3*32 + 4*(-4)2 + 3*52 = 166
• Find r0
• r0- = 4.9193
Density of region
Moment of Inertia for a Region
• Given a region bounded by curves of two functions and lines x = a, x = b
• The moment of inertia about the y-axis
f(x)
g(x)
x = b
x = a
f(x)
• Given the same region
• Radius of gyration, ry with respect to the y-axis is
g(x)
x = b
x = a
Try It Out
• Given the region bounded by y3 = x2 , y = 4 and the y-axis. Density = 4g/cm2
• Moment of Inertia about x-axis
Try It Out
• Given the region bounded by y3 = x2 , y = 4 and the y-axis. Density = 4g/cm2
For a Solid of Revolution
• Moment of inertia of a solid of revolution formed by generating a region around the y-axis
• The radius of gyration is
Example
• Consider region bounded by y = x2 , the y-axis, and y = 2 rotated
• What is the moment ofinertia about the x-axis
Interesting Application
• Sweet spot for a baseball bat
• What happens when you hit the ball …
• At point A?
• At point B or C?
Note the interesting description of this lab assignment from Colorado State
Assignment
• Lesson 7.6
• Page 299
• Exercises 1, 3, 5, 9, 11, 15, 17
• Second day7, 13, 19, 21, 23 | 788 | 2,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-05 | latest | en | 0.866193 |
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Full Version: Wormholes and infinite energy?
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I've been thinking, in OA, a lot of the WH need to be far from massive objects to be stable but some types of WH (the ones that archais use for in-processing node data transfer) can be located in very close proximity to these massive bodies. My question is, imagine a system with a massive body, one mouth of a relatively large WH (suficient to pass light beams or sub-atomic/atom sizeĀ particles) is anchored to the surface or near the massive body and the other mouth is a little far from the other but resonably close, then, imagine that you shoot a particle or a beam of light through the WH entering the massive body bound mouth, if the other mouth is positioned in a way that makes the particle/beam return and enter the massive body bound mouth again and repeat the process. AssumingĀ ideal conditions (no perturbances and perfect aiming), isn't possible to generate "infinite" energy? Since, in the case of a particle, because of the gravity of the massive body is stronger near one of the mouths of the WH, the particle would accelerate and exit the other mouth and repeat, and with light it would blue shift, would it be possible? Sorry for the massive quantity of words, I couldn't find a way to be clear with a smaller text.
Good question!
Although it is possible to extract energy from a wormhole using a range of different tricks, there is no such thing as a free lunch, and in most cases there is a kind of cosmic book-keeping that will take the extracted energy back out of the system somehow. So if you extract energy using a pair of wormholes the wormholes themselves will lose energy in some way, perhaps by shrinking or by losing orbital velocity. Someone with a better grasp of the maths involved would need to work out the exact results of the set-up you describe.
This wouldn't actually work.
In absolute terms because the way conservation laws work with WHs is that the mass-energy entering one mouth will cause that mouth to gain mass-energy (growing a bit in the process), while the 'exit' mouth will lose mass-energy (shrinking a bit in the process). As this continues, the exit mouth will get smaller and more unstable until eventually it can't be kept open and will implode, destroying the entire wormhole in the process (and most everything else in the vicinity as around 70% of the WHs mass-energy is radiated away).
Basically, this process is extracting mass-energy out of the exit wormhole mouth (and the wormhole insofar as the mouths are part of the same structure).
In OA terms, there would also be the problem that the type of wormholes you are using (comm-gates) have such intense internal tidal forces that only EM radiation (light) can pass thru them without being destroyed. Atoms or sub-atomic particles attempting to pass thru will be thoroughly shredded. Which wouldn't stop the mass-energy extraction effect, but would complicate things all around.
Hope this helps,
Todd
EDIT - So basically you could do this - but you couldn't extract 'infinite' energy from the process. The amount of energy you could extract would be bounded by the amount of mass-energy the exit wormhole mouth could lose before imploding. | 709 | 3,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-21 | longest | en | 0.9513 |
https://fr.mathworks.com/matlabcentral/cody/problems/44439-remove-the-air-bubbles-from-a-vector/solutions/1527544 | 1,590,490,040,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390755.1/warc/CC-MAIN-20200526081547-20200526111547-00247.warc.gz | 334,373,665 | 15,844 | Cody
# Problem 44439. Remove the air bubbles from a vector
Solution 1527544
Submitted on 13 May 2018 by avi kiko
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
filetext = fileread('bubbles.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
v = [1 3 0 5 0 -1]'; w_correct = [0 0 1 3 5 -1]'; assert(isequal(bubbles(v),w_correct))
w = 0 0 1 3 5 -1
3 Pass
v = [0 0 9 2 6]'; w_correct = v; assert(isequal(bubbles(v),w_correct))
w = 0 0 9 2 6
4 Pass
v = [1 3 5 -1]'; w_correct = v; assert(isequal(bubbles(v),w_correct))
w = 1 3 5 -1
5 Pass
v = [0 1 0 1 1 1 0]'; w_correct = [0 0 0 1 1 1 1]'; assert(isequal(bubbles(v),w_correct))
w = 0 0 0 1 1 1 1 | 324 | 816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.514429 |
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2021-02-11, 18:41 #1 bur Aug 2020 2·47 Posts Proth primes with k = 1281979 Because I was interested how sieving worked and also wanted to test Proth20 (proth testing for GPU), I started looking at 1281979 * 2^n +1. That k is large enough not to get into PG's various Proth subprojects. Also it's prime, which I find interesting since it means there can be a lot of small factors in 1281979 * 2^n + 1 so sieving will be quite efficient. Of course it also means the probability for smaller primes is low, so it evens out. And it's my birthday. ;) Sieving Range: 100 000 < n < 4 100 000 Factors: p < 290e12 I sieved with sr1sieve on a Pentium G870 which took about 6 months. I stopped when finding a factor took about 2 hours. 66272 candidates remained. I know now I should have sieved to much higher n, but I didn't know that when I began. Primality testing Software is LLR2, n < 1 200 000 was tested on the Pentium G870. Now I switched to a Ryzen 9 3900X. For n = 1 200 000 a test takes 460 s, for n = 2 300 000 it takes 1710 s, showing nicely how well the approximation "double the digits, quadruple the time" can be applied. At least as long as FFT and L3 sizes match. Results Code: For n < 1 000 000 k = digits 2 7 (not a Proth) 142 49 202 67 242 79 370 118 578 180 614 191 754 233 6430 1942 7438 2246 10474 3159 11542 3481 45022 13559 46802 14095 70382 21194 74938 22565 367310 110578 485014 146010 Next milestone is n = 1 560 000 at which point any prime would be large enough to enter Caldwell's Top 5000. Last fiddled with by bur on 2021-02-11 at 18:46
2021-02-12, 03:49 #2 LaurV Romulan Interpreter Jun 2011 Thailand 17·19·29 Posts Good job. Some random advice: that range is way too large. Split the range is 5-10 smaller ranges for sieving. You will be able to sieve higher and faster for each range, albeit the rate of eliminating the candidates per total range will be a bit slower, but you will compensate by having different sieve limits for each range, in such a way to equilibrate the LLR duration. Right now, your LLR may take (in a good computer) like 20 minutes per candidate for the start of the range, and 8 hours for the end of the range, so how do you calculate how much to sieve? Alternative, you may want to stop the LLR every few days/weeks and do some more sieving for a day or so, for the remaining candidates, to be sure that you always chose the task that eliminates the numbers faster. Write down how long it takes to eliminate candidates by sieving, and do a LLR test for a candidate which is down the list, about 5-10%. For example, the list is 3 thousand candidates, do a test for a candidate in position 200-300 in the list. Write down the time. Eliminate that candidate from the list once the LLR is done. Continue sieving on the list until you take the same time to find new factors. Then LLR the first 500-600 candidates. Repeat. There is nothing like "i should have sieve it higher", stop the LLR, take a text editor, get rid of the tested (LLR-ed) candidates at the beginning of the list, then sieve the remaining list as high as you want. In fact, is recommended to do this periodically, as your LLR time gets higher per candidate, in a certain point you will eliminate them faster by continuing the sieving process. You don't need to start the sieve from scratch, you can continue any time from where it left, to higher primes. Make a backup in case you screw up the editing, if you didn't do that before. The size of the sieving ranges needs to be not too large, and not too small. If too large, you will lose a lot of wall-clock time either by sieving too high (when you could possibly eliminate smaller candidates faster using LLR, therefore increasing the speed of the future sieving sessions) or sieving too low (when it would have been faster to eliminate larger candidates by more sieving). If too small, you will lose a lot of wall-clock time with the overhead of switching from sieving to LLR, sieve initialization, manual work, etc. Choosing the right range size is a matter of experience, system speed, number of cores, etc. That is why crus, for example, works in smaller ranges, 100k, 200k, etc. Last fiddled with by LaurV on 2021-02-12 at 03:56
2021-02-12, 06:54 #3
bur
Aug 2020
10111102 Posts
I read here and at PG that time per factor found increases only with sqrt(n), so I was explicitly told not to split the sieve. At the beginning I did just that because I thought it was the way to distribute sieving between several cores.
Quote:
There is nothing like "i should have sieve it higher"
By that I meant having candidates up to n = 16 400 000 in the sieve, not larger p. It would have taken only twice as long - at least to my understanding.
Does you method take that into account? Sorry if this is a stupid question, but I'm really not sure. And various people said large sieving files are good (put simply).
2021-02-12, 07:30 #4 VBCurtis "Curtis" Feb 2005 Riverside, CA 3·19·83 Posts Bur, I think you've got the right idea- make one sieve that goes up to an exponent big enough you're not sure you'll finish. If you're confident you'll get to 6M or 8M, I agree your sieve maybe could have included more n. On the bright side, you'll fully understand how big an exponent to sieve to if you make another sieve! If the sieve program you're using scales like the srsieve family (it likely does, since you refer to the same sqrt-n-range scaling), the plan is to sieve until the candidate-elimination rate is about double the time to test the smallest candidate. Then break off a chunk of candidates to LLR, and keep sieving the bigger ones. That gives well-sieved lists for LLR, and speeds the sieve up since the new sieve file has a smaller n-range (missing the smallest ones, I mean). I'm not sure why LaurV suggests splitting the range- that gives up a bunch of efficiency.
2021-02-12, 10:40 #5
kar_bon
Mar 2006
Germany
1011010000012 Posts
Quote:
Originally Posted by bur [CODE] For n < 1 000 000 k = (...)
1. You meant "n=" not "k="
2. You missed: 1281979*2^7894+1 is prime (2383 decimal digits), check your results please
3. Data inserted in the Wiki
Last fiddled with by kar_bon on 2021-02-12 at 10:41
2021-02-13, 07:33 #6 bur Aug 2020 10111102 Posts VBCurtis, yes, I'm using sr1sieve which was the fastest among the various srsieves for this task. Good advice to feed candidates from the sieve to LLR testing. I guess I would remove all those candidates that are faster to LLR than to sieve? And a general question, sometimes the LLR task will have to be interrupted (updates, etc). Is there a better way to continue afterwards than manually checking lresults for progress and removing those numbers from the LLR input file? Can it somehow check automatically which numbers are already tested in lresults? kar-bon, of course you're right. Fortunately, I only missed it when copying from my excel sheet and not a genuine miss. Still a bit embarassing... ;) Since I can't edit the original post anymore, here's the updated version of the table: Results Code: For n < 1 000 000 n = digits 2 7 (not a Proth) 142 49 202 67 242 79 370 118 578 180 614 191 754 233 6430 1942 7438 2246 7894 2383 10474 3159 11542 3481 45022 13559 46802 14095 70382 21194 74938 22565 367310 110578 485014 146010
2021-02-13, 07:57 #7
LaurV
Romulan Interpreter
Jun 2011
Thailand
17·19·29 Posts
Quote:
Originally Posted by VBCurtis I'm not sure why LaurV suggests splitting the range- that gives up a bunch of efficiency.
Nope. I mean, yes, if you think about the same sieving depth. But you will sieve different chunks to different depths, and that is where you GET speed. Say you work a single k, and you sieve very large large n to N, with some depth p to P (srXsieve notation). You sieve 100k primes per second and eliminate 5 candidates per hour, but this is misleading because you also eliminate large candidates, for which the LLR test would take hours, but also eliminate small candidates, which would be eliminated faster by LL test. That is why you have to test how the things are for your system. I will try to give a numerical example soon, let me make one first.
2021-02-13, 17:00 #8
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
3×19×83 Posts
Quote:
Originally Posted by bur VBCurtis, yes, I'm using sr1sieve which was the fastest among the various srsieves for this task. Good advice to feed candidates from the sieve to LLR testing. I guess I would remove all those candidates that are faster to LLR than to sieve? And a general question, sometimes the LLR task will have to be interrupted (updates, etc). Is there a better way to continue afterwards than manually checking lresults for progress and removing those numbers from the LLR input file? Can it somehow check automatically which numbers are already tested in lresults?
You don't pick up very much sieve speed by removing candidates- break off a chunk that tests in a reasonable time, something convenient for your manual labor. That is, you might sieve 100k to 4M, and then break off 100-150k for LLR while 150-4M keeps sieving. By the time you're up to 500k, you might break off 100k blocks instead- it doesn't matter a whole lot.
LLR maintains an .ini file that includes the line number it has tested. Don't edit the input file when restarting- it will always pick up where it left off.
LaurV seems to be comparing his idea to a plan that never breaks off small pieces when they're "ready" for LLR. I am confident that removing small candidates from the sieve when appropriate is much much faster than LaurV's splitting by N-range. He and I rarely disagree, and he will surely relish showing me I'm mistaken!
2021-02-18, 06:57 #9 bur Aug 2020 2·47 Posts I think the breaking-off-or-not depends on the time-scale. Splitting the range will speed up the sieving at the moment. However, in the long run overall sieving time would be shorter when sieving longer ranges. At least that's what I think. Anyway, one core freed up and I ran a sr1sieve for 2.7e6 < n < 4.1e6 and 314e12 < P < 325e12. Estimation was 24 factors in about 3 days or about 11000 s per candidate removed. To confirm, I ran it for 7 hours, about 10%, and found 4 factors, i.e. 6300 s per candidate removed. LLR2 takes 2500 s for n = 2.7e6, so I stopped the sieve and continued LLR. It seems for this n-range sieving doesn't make sense anymore. Of course I could extend the sieve, but for now I want to finish n < 4.1e6 and then see how and if I'll continue. Last fiddled with by bur on 2021-02-18 at 06:59
2021-03-09, 14:24 #10 bur Aug 2020 2×47 Posts All n < 2,400,000 tested No new primes to report. Is anyone interested in the lresults file with the residues? Code: For n < 2,400,000 n = digits 2 7 (not a Proth) 142 49 202 67 242 79 370 118 578 180 614 191 754 233 6,430 1,942 7,438 2,246 7,894 2,383 10,474 3,159 11,542 3,481 45,022 13,559 46,802 14,095 70,382 21,194 74,938 22,565 367,310 110,578 485,014 146,010 Due to the lack of new primes, here are some statistics: Largest n in progress = 3.20e6 # of digits = 963,000 (rank: ~ 1000) FFT size = 320k Avg. time = 3140 s Smallest n in progress = 2.45e6 # of digits = 737,500 (rank: ~ 1690) FFT size = 256k Avg. time = 1935 s In about 15 days Mega Prime territory will be reached with n = 3.33e6. In about 30 days all n < 3.2e6 will have been tested. Btw, since the office got a bit warm I have the CPU running with PPT = 128 W instead of 148 W. This decreased the package temperature from 74 °C to 68-70 °C while increasing computation times only by 1-2 %. That's not a bad tradeoff, I think.
2021-04-10, 06:52 #11 bur Aug 2020 10111102 Posts No new primes for n < 3,200,000 So I at least updated the stats: Largest n in progress = 3,575,000 # of digits = 1,075,000 (rank: ~ 405) FFT size = 386k Avg. time = 4230 s Smallest n in progress = 3,200,000 # of digits = 963,500 (rank: ~ 1025) FFT size = 336k Avg. time = 3565 s Around May, 1st all n < 3,600,000 will have been tested. Sometime in August or September the sieve will be exhausted at n = 4,100,000 which makes it about a year since I started ... and I hope at least one more prime will turn up. :)
Similar Threads Thread Thread Starter Forum Replies Last Post ET_ Proth Prime Search 9 2020-10-02 07:11 pepi37 Software 10 2020-09-06 17:26 lukerichards Number Theory Discussion Group 7 2018-01-20 16:47 kar_bon Riesel Prime Data Collecting (k*2^n-1) 6 2010-11-25 13:39 ixfd64 Lounge 1 2005-09-07 23:42
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https://www.jiskha.com/similar?question=1.3+%2F73.814+In+significant+digit+notation%2C+divid%2C+and+the+answer+to+two+significant+digits.+I+will+be+happy+to+critique+your+work&page=538 | 1,563,634,608,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526517.67/warc/CC-MAIN-20190720132039-20190720154039-00035.warc.gz | 722,855,540 | 25,342 | # 1.3 /73.814 In significant digit notation, divid, and the answer to two significant digits. I will be happy to critique your work
71,152 questions, page 538
1. ## dog behavior
please check my answer thanks so much :) When teaching the leave it command, when should you give the dog a reward? A. While its chewing the forbidden object so it will leave it and take the treat instead B. The instant it leaves the forbidden object C. As
asked by Maria on July 26, 2008
2. ## Calculus
yes! tnk u ok? It's actually (x->0.) Find the limit of cot(x)-csc(x) as x approached 0? Lim [cot(x) - csc (x)] ..x->0 = Lim [(cos x -1)/sin x] ..x->0 Use L'Hopital's rule and take the ratio of the derivatives: Lim (-sin x/cos x) = 0 x->0 thank you very
asked by Maria on May 6, 2007
a 1500 kg car is driven around a circle of radius 120 m and at a speed of 26 m/s. calculate the magnitude of the centripetal force of the car. What real force keeps the car moving in a circle? the answer is not one of the four fundamental forces, but way
asked by melanie on October 19, 2015
I ask a friend to watch my bike while I go shopping, and the bike gets stolen. I can sue my friend for damages but only if I can prove that my friend was: grossly negligent, or failed to exercise at least slight care, or failed to exercise reasonable care.
asked by bobby on June 26, 2012
5. ## math
A snowball was thrown from a cliff. The height above ground of the snowball is modeled by the function h(t) = -16t2 + 48t + 28, where h is height in feet, and t is time in seconds. How long was the snowball in the air? Did everything got (4t+7)(t+1) then
asked by Jalen on December 11, 2016
6. ## College Personal Finance
Annual interest of 6.0% paid if balance exceeds \$900, \$12 monthly fee if account falls below minimum balance, average monthly balance \$1,135, account falls below \$900 during 2 months. Round the answer to the nearest cent.
asked by Kendra on February 6, 2012
7. ## calc
Given the position function, s of t equals negative t cubed divided by 3 plus 13 times t squared divided by 2 minus 30 times t, between t = 0 and t = 9, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle
asked by Mel on January 8, 2016
8. ## Physics
What is the difference between scalar and vector quantities and what two methods are used to describe vectors? Answer: Scalar quantities measure only distance (speed) while vector quantities measure both distance and direction, ( which is specifically
asked by Lucina on January 6, 2015
9. ## Math
I have a problem. A theater runs 2 movies The first movie is 40 minutes long, the second is 60 minutes long. They start at 3:00. At what time will they start at the same time again? I know the answer is 5:00. I know it robably has something to do with 120
asked by Janice on September 20, 2009
10. ## Math
Determine whether each event is impossible, unlikely, as likely as not, likely or certain. Then tell whether the probability is 0, close to 0, 1/2, close to 1 or 1. 2. Randomly picking a green card from a standard deck of playing cards. My answers is that
asked by Patrick on March 27, 2018
11. ## math(help)
please help someone explain this please!! The length of a marathon is 138,435 ft.each stride,or step,by a particular runner is 3 ft long.does the number of strides the runner takes fit evenly into the length of the race?(EXPLAIN ANSWER PLEASE)
asked by matt on November 21, 2013
12. ## Physics
A 2 kg steel ball strikes a wall with a speed of 8.15 m/s at an angle of 54.4◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. If the ball is in contact with the wall for 0.243 s, what is the magnitude
asked by Seth on November 2, 2011
13. ## accounting
Compute inventory turnover for 2005 and 2004. The inventory balance at December 31, 2003, was \$294 million. Do the trend of net income from 2004 to 2005 and the change in the rate of inventory turnover tell the same story or a different story? Explain your
asked by mary on April 12, 2011
14. ## Physics
An electron is projected, with an initial speed of vi=1.10e+05 m/sec, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal
asked by bethel on February 28, 2013
15. ## Algebra
Jed decided he needs to start exercising more. On Monday, he runs 2 miles. on Tuesday, he runs 4 miles. On Wedsnesday he runs six miles. He continues with this pattern each day for five days, Monday through Friday. How many total miles did he run in the
asked by Anon on May 28, 2015
16. ## math
A cylinder container of radius 15 cm has some water in it. When a solid submerged into the water, the water level rises 1.2 cm. (a) Find, the volume of the water displace by the solid leaving your answer in terms of π (b) If the solid is a circular cone
asked by kudu on February 9, 2015
17. ## Chemistry
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 26.4 g of carbon were burned in the presence of 82.1 g of oxygen, 11.7 g of oxygen remained unreacted. What mass of carbon dioxide was produced? Express your answer to one
asked by Eric on September 14, 2016
18. ## math
What percent of 80 is 16? 20. How did they get that answer?? of means times. Is means equals. DecimalPercent x 80= 16 divide both sides by 80 decimal percent = 16/80= .2 percent=decimalpercent x 100 There is a formula for that -- but I never learned it. I
asked by Mack on June 26, 2007
19. ## Biology
The pulmonary artery, which pumps blood from the heart to the lungs, has an inner radius of 1.34 mm and is 5.0 cm long. If the pressure drop between the heart and lungs is 420.0 Pa, what is the volume flow rate of blood in the pulmonary artery? Give your
asked by sendy on November 22, 2018
20. ## Chemistry
Mothballs are composed of a chemical called naphthalene which has a formula of C10H8 a) what is the mass of a single naphthalene molecule in amu? b) what is the molar mass of naphthalene? c) what is the mass of a single naphthalene molecule in grams? I
asked by Kiana on March 22, 2016
21. ## Physics
An electron is projected, with an initial speed of vi=2.06e+05 m/sec, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal
asked by P on March 5, 2013
22. ## Statistics
Suppose 52 of the students in class have a calculator and 14 have a computer. It is known that 4 do not have either a computer or calc and there are 64 students in the class. Find the probability that a randomly chosen student has a computer AND a
asked by Kim on May 26, 2009
23. ## Physics
Halley comet which passes around the sun every 76 years has an elliptical orbit. when closest to the sun it is at a distance of 8.823 x 10^10 m and moves with a speed of 54.6 km/s. The greatest distance between hally's comet and the sun ( aphelion ) is
asked by Lonely on July 30, 2011
Janet Jones borrowed \$3,000 on 90-day 12 percent note. Janet paid \$250 toward the note on day 35. On day 80 she paid an additional \$400. Using the U.S. Rule her adjusted balance after her first payments is: Answer: \$2,785 Please do step by step. Thank you.
asked by Liz on January 29, 2012
25. ## Ap physics
fine sand moves through pipes just like fluid. a sandblasting machine moves fine sand through a pipe of inner diameter 2.8 cm at 75L/minutes. the sand comes out from two identical jets at speed 45 m/s. calculated the diameter of the holes through which the
asked by jay on December 6, 2015
i have to fill in the blank: adverbs of time tell ___ something happens. I wrote: when adverbs of frequency tell ___ something happens. i wrote : how often or is it also when? can you check this? i have an ARTICLES EXCERCISE 1) ARTICLES ARE USE WITH: a
asked by marko on December 7, 2011
27. ## English
Which sentence uses the active voice? A. Breakfast is always served at 8 a.m. sharp. B. Juice is included in all breakfast meals. C. Customers are served as quickly as possible. D. You can have anything that you find appealing. I don't really understand
asked by Klauus on May 8, 2019
28. ## Microeconomics
taxes may cause deadweight losses because________. a. they transfer purchasing power from buyers to the government. b. they lower the surplus in the market. c. they increase consumer surplus at the expense of producer surplus d. they transfer purchasing
asked by Juan on June 24, 2015
29. ## physics
Starting from rest, a(n)9 kg block slides 9.7 m down a frictionless ramp (inclined at 30◦ from the floor) to the bottom. The block then slides an additional 23.5 m along the floor before coming to stop. The acceleration of gravity is 9.8 m/s 2.Find the
asked by enrique on November 14, 2013
OMGosh Please help me (mom) so I can help my daughter with this word problem!!! Cam rode her bike 5 times as far as Dante did. Dante rode 187 meters farther than Michael did. Cam rode 15.25 kilometers. How many meters did Michael ride? Explain how you got
asked by Sara on April 10, 2014
31. ## Algebra
The instruction manual on the industrial oven suggests your oven temperature will increase by 45 degrees Fahrenheit per minute. When the oven is initially turned on, the temperature is 70 degrees Fahrenheit. What will the temperature of the oven be after 7
asked by Mary K on August 2, 2009
32. ## alg
ken griffy jr hit 23 home runs in the first 50 games of the season. If he hit homw runs at this pace all season, how many would he have at the end of the year? aSSUME A 168 games schedule. Round your answer to the nearest whole number of home runs.
asked by kid on July 12, 2010
33. ## geometry
Can anyone help me to solve this, please? A right prism has a polygonal base with n sides. Each side has a length of 5 in. The altitude of the prism is 20 in. If the base had n+1 sides, each still having a side of length 5 in, by how much, in square
asked by scott on August 23, 2011
34. ## math
Use the following to answer question 7: Point Bread Tractor A 20 0 B 18 1 C 14 2 D 8 3 E 0 4 Refer to the above schedule. Starting at point A, the opportunity cost of producing each successive unit of tractors is: a constant 2 units of bread 8, 6, 4, and 2
asked by Anonymous on April 18, 2014
35. ## Chemistry
Which of the following nitrogen-containing compounds would supply the most nitrogen per gram of fertilizer? a. iron azide, Fe(N3)2, MW = 140 g/mol b. sodium azide, NaN3, MW = 65 g/mol c. potassium nitrite, KNO2, MW = 85.1 g/mol d. potassium azide, KN3, MW
asked by Anonymous on November 8, 2015
36. ## Math for Educators
I am thinking of some #s less than 100. When I divide them by 2 there is a remainder of 1. When I divide them by 3 there is a remainder of 1. When I divide it by 4 there is a remainder of 1. However, when I divide by 5 there is no remainder. what are the
asked by Kelly on June 17, 2013
37. ## Math
Mia goes for a jog every morning. The line plot shows the number of hours she jogged for five days. 2/4=xx 3/4=xx 1=x Number of hours How many total hours did she jog for those five days? possible answer: 5/4; 10/4; 11/4; 14/4 hours.
asked by Student on April 20, 2019
38. ## accounting
What role might an accounting professional play in ethical corporate decision making? Should an accounting professional play a role even when he or she is not the final decision maker? Explain your answer.
asked by Gina on September 9, 2010
39. ## algebra
i have two answers to this problem so am not sure which way I need to do in or should say the order i should do it. lease help??????? problem is 7x-5+3x=6+x-10 what is the order to work this out there is no brackets to this problem 7x-x+3x=6+5-10
asked by Anonymous on April 29, 2007
40. ## physical science
A boy with a mass of 30 kg stands motionless at the end of a diving board 3 m above the floor. The boy has: Answer no energy. kinetic energy only. potential energy only. both kinetic and potential energy.
asked by Allison on September 10, 2011
41. ## math
A gear with 60 teeth is meshed to a gear with 40 teeth. If the larger gear revolves at 20 revolutions per minute, how many revolutions does the smaller gear make in a minute? not just an answer please (A)13 1/3 (B)3 (C)300 (D)120 (E)30
asked by Tomas on August 14, 2013
42. ## American government
Unlike ______, interest groups do not compete for ________office. A. Lobbyist; private B. Lobbyist; public C. Political parties; private D. Political parties; public Is the answer A?
asked by sallie on March 12, 2016
43. ## chemistry
rank these amino acid from most polar to least: His, Phe, Ser, Ile, Asp, Trp i don't think my answer is correct... Ser>Asp>His>Phe>Trp>Ile
asked by Anonymous on October 3, 2013
44. ## physics
A point charge of +4.6 micro-coulombs lies at x = 0 and a point charge of -7.3 micro-coulombs lies at x = +3.3 m. Where on the x axis, is the total electric potential equal to zero? Answer in meters.
asked by Mandaleigh on February 10, 2012
45. ## English
My uncle, along with my cousins, ________ to see my acrobatics recital; I hope they ________ the show. a) plan/enjoy b) plans/enjoy c) plans/enjoys d) plan/enjoys Answer A
asked by Abby on January 8, 2019
46. ## statistics
using the numbers only 0 through 10 only whole numbers with repeats allowed pick four numbers that would have the smallest standard deviation I said four zeros pick four numbers that would have the largest standard deviation I said 0 0 0 10 is there more
asked by statistics on July 28, 2009
47. ## English
Wait so is it: 5. Four years later, Rudolf achieved her greatest personal triumph, three gold medals in a single Olympics. *ANSWER* AP=three gold medals in a single Olympics WM=triumph
asked by Anne on January 9, 2010
48. ## Personal Dimension of Educatio
Select the sentence that has correct subject-verb agreement. 11. Neither my dad nor my siblings want to go to the family reunion this summer. 2. Neither my dad not my siblings wants to go to the family reunion this summer. My answer is 1
asked by Jackie on November 16, 2012
49. ## math?
1.How are the two statements given below related to each other? x: if you run for 10 minutes, then you will raise your heart rate. z: if you do not run for 10 minutes, then you will not raise your heart rate. A) Z is the contrapositive of X. B) Z is the
asked by tim on October 1, 2015
50. ## math
Use the data to answer the question: Favorite Sport to Watch: 25% soccer 33% baseball 42% football If 210 people said football was their favorite sport to watch, how many people were surveyed? Thank you!
asked by Moon and Magic on December 4, 2017
51. ## Math
Did I answer this problem right? A bag contains 13 marbles of which 10 are red. A marble is selected at random and replaced. A second marble is then selected at random. What is the probability that both marbles are red? 100/169. Thanks.
asked by B.B. on August 7, 2009
52. ## Math?
1.How are the two statements given below related to each other? x: if you run for 10 minutes, then you will raise your heart rate. z: if you do not run for 10 minutes, then you will not raise your heart rate. A) Z is the contrapositive of X. B) Z is the
asked by Tim on October 1, 2015
53. ## Math
A pole that is 3.2m tall casts a shadow that is 1.4m long. At the same time, a nearby tower casts a shadow that is 39.5m long. How tall is the tower? Round your answer to the nearest meter.
asked by Michelle on November 14, 2015
54. ## Physics
The velocity of the transverse waves produced by an earthquake is 5.08 km/s, while that of the longitudinal waves is 8.3312 km/s. A seismograph records the arrival of the transverse waves 55.9 s after that of the longitudinal waves. How far away was the
asked by Laura on May 8, 2011
55. ## Math
Alicia’s gross monthly pay is \$1,700. Alicia pays 10.5% of her monthly income in federal withholding. How much federal withholding does she pay? A \$10.50 B \$17.00 C \$161.90 D \$178.50 My Answer: C
asked by Suzy-- Su-Ji on January 20, 2016
56. ## Physics
At one end of the Venturi Tube the pressure and speed of flowing water is 21 kPa and 0.05m/s respectively. At another point the pressure is 5 kPa. Find the speed of the water at the second point. Answer in m/s
asked by Anonymous on May 30, 2016
57. ## math
Using the best-fit line below for prediction, answer the following questions: a) What would you predict the price of Product X in volume of 150 to be (approximately)? b) What would you predict the price of Product X in volume of 100 to be (approximately)?
asked by Jennifer on December 29, 2012
58. ## math
a triangular road sign has a height of 8 feet and a base of 16.5 feet. How much larger in ares is this sign than one with a height of 4 feet and a base of 8.25 feet? Please explain fully and show how you got the answer.
asked by audrey on May 17, 2011
59. ## Language Arts...Writing
Hello, I need some help with the following question. My answer is #1, is this corret? Thank You Find the sentence with the Active Voice? #1 The chairman told me that the legislature passed the bill. #2 I was told by the chairman that the bill was passed by
asked by Mandi on September 7, 2007
60. ## aig math
a factory had 1080 shirts and pants altogether.after donating 60 pants to charity the number of shirts it had was three times the number of pants left.how many pants did the factory have at first/ my answer is 315 am i right,if not please tell me why and
asked by lidia on August 19, 2012
61. ## chemisrty
Could anybody help me with this question "In practice the methanol production process is operated at temperatures of 250 to 300 degrees and at a pressure of 50 to 100 atmosphere (50 to 100 times normal atmospheric pressure) in the presence of a copper
asked by bex on March 12, 2007
62. ## physics (elena)
JUMPING OFF THE GROUND (10 points possible) A person of mass m jumps off the ground. Suppose the person pushes off the ground with a constant force of magnitude F for T seconds. What was the magnitude of the displacement of the center of mass of the person
asked by Anonymous on October 24, 2013
63. ## physics
A 120-V motor has mechanical power output of 3.00 hp. It is 94.0% efficient in converting power that it takes in by electrical transmission into mechanical power. (a) Find the current in the motor. I got the same answers,17.5 A, whether I used round
asked by blast on August 17, 2010
64. ## AP chem
An apparatus consists of a 4 L flask containing nitrogen gas at 32C and 885 kPa, joined by a valve to a 8 L flask containing argon gas at 32 C and 47.5 kPa. The valve is opened and the gases mix. What is the partial pressure of nitrogen after mixing?
asked by Sarah on March 9, 2012
65. ## phsics
A ball is rotating in a horizontal circle at the end of a string of length 3.8 m at an angular velocity of 8.5 rad/s. The string is gradually shortened to 3.3 m without any force being exerted in the direction of the ball’s motion. Find the new angular
asked by owen on November 19, 2014
66. ## Macroeconomics
How would i approach this question? I don't necessarily want an answer. A "how to" would be better. (Simple Spending Multiplier) For each of the following values for the MPC, determine the size of the simple spending multiplier and the total change in real
asked by Lauren on November 14, 2010
67. ## Adult Development and life Assessment
Ken is a successful landscrape designer who an visualize backyard plan and them create it. Ken shows a high degree of 1. bodily kinesthetic intelligence 2. spatial intelligences 3. interpersonal intelligences 4. intrapersonal intelligences My answer is
asked by Regina on November 25, 2012
68. ## Calculus
Graph of a piecewised function is f(x)=1, x>0 and f(x)=-1, x-3 I did the following: My f(x)=1, x>0 graph started from the point (0,1) and extended right infinitely. My (x)=-1, x-3 the limit is undefined. I know the correct answer is lim f(x)=-1 as x-->-3
asked by lania on September 9, 2011
69. ## math- commutative property
My daughter had this question for Hw and wasn't sure the correct answer. the expression was (4 7/8 +3 3/4)+5 1/8 which shows another way to write the expression using only the commutative property of addition A) 4 7/8+(3 3/4 +5 1/8) B)(5 1/8 +4 7/8) +3 3/4
asked by Lis on January 9, 2014
70. ## Chemisry
what mass of protons would be required to just neutralize the charge of 1.0g of electrons? how do I answer this? You know the mass of an electron. Calculate how many are in 1 gram. You will need the same number of protons. Knowing the mass of a proton
asked by Chris on February 8, 2007
71. ## physics- check
1. three point charges, q1, q2, q3, lie along the x-axis at x = 0,3,5, respectively. Calculate the magnitude and direction of the net electric force on q1 if q1 = +6 x 10^-6 C, q2 = +1.5 X 10^-6 C and q3 = -2.0 x 10^-6 C. my answre is 46.8 N to the right.
asked by todi on February 14, 2008
72. ## why doesn't anybody ever help me.
1. three point charges, q1, q2, q3, lie along the x-axis at x = 0,3,5, respectively. Calculate the magnitude and direction of the net electric force on q1 if q1 = +6 x 10^-6 C, q2 = +1.5 X 10^-6 C and q3 = -2.0 x 10^-6 C. my answre is 46.8 N to the right.
asked by todi on February 14, 2008
Manny loves to eat Cajun pizza. Each square inch of pizza is 8 calories. He orders a slice of pizza from a pizza with a diameter of 16 inches that is cut with a central angle of 20 degrees. How many calories does he eat? Answer A. about 89 calories B.
asked by Help me on February 15, 2012
74. ## Physics
A 10 kg block sits on a flat surface whose μs=0.60 and whose μk is 0.40 a. What horizontal force is required to get the block move? b.If we continue to apply the same force as in part a. what will the blocks acceleration be? In the answer: force=.6*mg to
asked by Lianne on August 26, 2012
75. ## chemistry
how to solve for a 28.4 L sample of methane gas is heated from 35.0° C to 76.0° C. The initial pressure of the gas is 1.00 atm at 35.0° C. Assuming constant volume, what is the final pressure of the gas? OK, I did the following: P₁ = 28.4 L P₂ = ?
asked by sandy on November 22, 2011
76. ## Chemistry
Enough of a monoprotic acid is dissolved in water to produce a 0.0195 M solution. The pH of the resulting solution is 6.24. Calculate the Ka for the acid. Also, the answer is not 1.7 * 10 ^ -11 The response is: You treated the initial concentration as 0.
asked by chem1 on April 22, 2012
77. ## Physics
A 4kg block, initially moving due east at 3 m/s, is acted upon by an impulse having magnitude 8Ns and direction due west. Is the final velocity of the block 2m/s west? This is how I came to that answer: v = impluse/mass so 8/4 = 2. Although this is the
asked by sarah on November 10, 2009
78. ## Psychology
The motives studied by Henry Murray and his colleagues are called higher order motives because: Answer a. They reflect motivations for a higher level of functioning. b. They are at a higher level of consciousness than other motives. c. They do not reflect
asked by Annmarie on February 22, 2014
79. ## calculus
A car leaves an intersection traveling east. Its position t sec later is given by x = t2 + t ft. At the same time, another car leaves the same intersection heading north, traveling y = t2 + 3t ft in t sec. Find the rate at which the distance between the
asked by sal on October 19, 2011
80. ## physics
Raul pushes a stalled car with a force of 204 N. If the required force decreases at a constant rate from 204 N to 44.0 N for a distance of 16.3 m in 16.0 s, calculate the average power required to move the car. 204-44=160 160*16.3/16 2608/16 answer: 163 W
asked by Jon on April 3, 2008
find the two ordered pairs of this equation. y=5/3x+8 then give the slope and y intercept. I know the slope is rise over run and is the y intercept 8? This is the answer I got but I am still really confused how do I get my two ordered pairs this makes no
asked by KiKi on March 28, 2010
A history class is comprised of 6 female and 9 male students. If the instructor of the class randomly chooses 5 students from the class for an oral exam, what is the probability that 2 female students and 3 male students will be selected? Round your answer
asked by Jonathan on August 30, 2016
83. ## College Chemistry
For the reaction that occurs in the voltaic cell Fe(s)|Fe2+(aq)||Cr3+(aq),Cr2+(aq)|Pt(s) use data from Appendix D in the textbook to answer the following questions: a)Determine the equation for the cell reaction b)Determine E0 cell c)Determine ΔG0
asked by Brandon on May 17, 2015
84. ## World History
Which of the following did reforms first try to end in the slave industry? A.Slavery in Caribbean agriculture. B.The culture of slavery in north america. C.The trade in slaves itself. D.Liberation and compensation of slaves. * this is a review for my test
asked by schoolfreak on November 2, 2011
85. ## physics
JUMPING OFF THE GROUND (10 points possible) A person of mass m jumps off the ground. Suppose the person pushes off the ground with a constant force of magnitude F for T seconds. What was the magnitude of the displacement of the center of mass of the person
asked by Anonymous on October 25, 2013
86. ## Chemistry
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.7L bulb, then filled it with the gas at 1.30 atm and 30.0 C and weighed it again. The difference in mass was 3.9 g. Identify the gas. I used Pv=nRT
asked by Kasey on July 1, 2012
87. ## math
The 3rd term of a geometric progression is nine times the 1st term.if the 2nd term is one twenty fourth the 5th term.find the 4th term.(there is a mistake in the 2nd to the last statement.help locate the mistake{final answer 4th term =48).i solved to this
asked by james on November 28, 2014
88. ## Psychology
The motives studied by Henry Murray and his colleagues are called higher order motives because: Answer a. They reflect motivations for a higher level of functioning. b. They are at a higher level of consciousness than other motives. c. They do not reflect
asked by Annmarie on February 22, 2014
89. ## math
Atmospheric pressure V decreases as altitude h increases. At a temperature of the pressure is 101.1kilopascals kPa at sea level, 86.4 kPa at h = 1 km, and 77.1 kPa at h = 2 km. Use a linear approximation to estimate the atmospheric pressure at an altitude
asked by Tina on March 30, 2007
90. ## first year chem
There are two naturally occurring isotopes of chlorine with ratios of 76% and 24%. One mole of PCl3 is made using a sample of isotopically pure chlorine gas (containing only one of the two isotopes). The product weighs 142 g. If chlorine gas containing the
asked by John on December 5, 2013
91. ## Physics
A point charge Q1=-2µC is located at x=0, and a point charge Q2=-8µC is placed at on the x -axis of a cartesian coordinate system.The goal of this problem is to determine the electric field,E(x)=E(x)x^ , at various points along the -axis. (a)What is E(x)
asked by Anonymous on February 25, 2013
92. ## Calculus
A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time t>0 its velocity v(t) is given by v(t)= t/(1+t^2). a). Determine the maximum velocity of the particle. Justify your answer. b). Determine the position of the
asked by Rasheda on January 29, 2014
93. ## math- Trigonometry
If cos degree equals to 0.8641 What is Sin degree? I have no idea how to find this. Please help me. I got help from two people, but I'm not getting the answer and how they got the numbers either. Someone says: cos^2+sin^2=1
asked by Simran on November 5, 2016
94. ## physics repost
Raul pushes a stalled car with a force of 204 N. If the required force decreases at a constant rate from 204 N to 44.0 N for a distance of 16.3 m in 16.0 s, calculate the average power required to move the car. 204-44=160 160*16.3/16 2608/16 answer: 163 W
asked by Jon on April 4, 2008
95. ## HMO
Please check my answer thank you You belong to an HMO By paying the HMO premium and being treated through the HMO you're helping to 1 provide more managed care options for enrollees 2. control plan costs for everyone 3. decrease managed care costs 4.
asked by Israel on January 19, 2008
96. ## Math
A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads? b) use your answer in a) to find the probability that there
asked by Steve on June 11, 2018
97. ## English
Indicate where each aspect of an index card should be placed when researching. Match the letter that corresponds to one of the items below to the correct location (1, 2, 3, 4, or 5) on an index card A.) Page number B.) Title of card (aspect of subject) C.)
asked by Cassie on April 21, 2014
98. ## algebra
simplify: sqrt (121/9) a) 11/3 b) sqrt (11/3) c) sqrt(11 / sqrt 3) d)3 sqrt (11) a is the answer
asked by anthony on June 24, 2010
99. ## College Discrete
An urn contains n > 0 white balls and m > 0 black balls. Consider the experiment where two balls are drawn without replacement. What is the probability that the balls are of the same color? What is the probability of this event, if the balls are drawn with
asked by Anonymous on October 16, 2012
100. ## English
true and false question are called Rejective Objective Dejective Subjective I think it's dejective Objective exams , subjective test contain More question Harder question Fewer question No good answer I think it's more question
asked by Linda on May 19, 2015 | 8,328 | 29,822 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-30 | latest | en | 0.94064 |
https://cboard.cprogramming.com/c-programming/46092-question.html | 1,513,409,400,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948585297.58/warc/CC-MAIN-20171216065121-20171216091121-00386.warc.gz | 514,683,181 | 11,863 | 1. Counting positive and negative numbers from a file
Hey guys,
I am stuck on a question in my C Programming Book and i was wondering if you guys can help anyways...
I need to write a program using a while/do-while/for loop command that counts positive and negative numbers in a list of integers from a file
eg.
5 (total number of integers in the list)
2
-3
4
7
-9
Ok I have...
Code:
``` int a, b, c, d, e, numbers;
int positives, negatives;
FILE *input;
input=fopen ("C:\\hw4-1in.dat.txt.", "r");
fscanf(input, "\r%d\n", &numbers);
printf( "%d", numbers);
fscanf(input, "%d\n", &a);
printf( "%d", a);
fscanf(input, "%d\n", &b);
printf( "%d", b);
fscanf(input, "%d\n", &c);
printf( "%d", c);
fscanf(input, "%d\n", &d);
printf( "%d", d);
fscanf(input, "%d\n", &e);
printf( "%d", e);
fclose(input);
if (a>0)
{
positives=1+positives;
}
else if (a<0)
{
negatives=1+negatives;
}
if (b>0)
{
positives=1+positives;
}
else if (b<0)
{
negatives=1+negatives;
}
if (c>0)
{
positives=1+positives;
}
else if (c<0)
{
negatives=1+negatives;
}
if (d>0)
{
positives=1+positives;
}
else if (d<0)
{
negatives=1+negatives;
}
if (e>0)
{
positives=1+positives;
}
else if (e<0)
{
negatives=1+negatives;
}
printf ("The total number of the positive integers is \n %d", positives);
printf ("The total number of the negative integers is \n %d", negatives);
}```
It keep saying windows encountered a problem when i run it...
Any help would be appreciated thanks!!
2. What part are you stuck on? Post what you're tried so far, so we can see where you're up to.
3. If that's all of your code then it looks like you're miising your 'main' function, which is what most of your code should probably be in, and is defined as an int. Take a look at some of the other examples in your programming book. | 560 | 1,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | latest | en | 0.760852 |
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# Discovering Some of Your "Yardsticks" Are Actually "Meter-sticks"
This lesson uses a simple discrepant event to demonstrate the underlying cause for early miscalculation of the size of the Milky Way galaxy. By standardizing the Cepheid period-luminosity relationship without recognizing there were two types of... (View More)
Audience: High school
Materials Cost: \$1 - \$5 per group of students
# Gamma Ray Burst Distribution on the Sky: The Plots Thicken
In this activity, students look at the distribution of aluminum foil balls arranged in a circle on the floor, and compare them to the distribution of gamma-ray bursts on the sky. This activity uses Gamma-ray Bursts as an engagement tool to teach... (View More)
Audience: High school
# Beam Me Up
In these activities, students investigate how gamma ray bursts emit energy in beams (as opposed to emitting light in all directions) and investigate the implications of this on the total number of gamma ray bursts seen in the universe. This activity... (View More)
Audience: High school
# Log Ruler
In this activity students construct Log Rulers, finely calibrated in base-10 exponents and numbers (logs and antilogs). They practice reading these scales as accurately as possible, listing all certain figures plus one uncertain figure. This is... (View More)
# Zooming In on Active Galaxies
This activity focuses on the question, What do active galaxies look like when viewed from different distances? Students work in small groups to learn about the small angle formula, construct a template, and use it to correctly measure the angular... (View More)
Audience: High school
Materials Cost: \$1 - \$5 per group of students
# Imágenes digitales
This is an activity about digital imagery and technology. Learners will create a picture using graph paper containing ones and zeroes and instruct others to reproduce it without showing it to them, just like digital technology. After, in groups,... (View More)
1 | 520 | 2,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-27 | longest | en | 0.858234 |
https://www.physicsforums.com/threads/germain-primes.225327/ | 1,575,609,285,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00226.warc.gz | 819,020,739 | 15,991 | # Germain primes
## Main Question or Discussion Point
Reference: www.mathpages.com/home367.htm[/URL]
On page 2 of reference the formula is given
(x+y)^p - x^p - y^p = pxy(x=y)Q(x,y) where Q(x,y) is a homogenous integer function of degree p-3.
If we insert a number of different value of p into the equation, it appears that
Q(x,y) = (x^2 = xy + y^2)^((p-3)/2)
Is there an easy way to prove this without getting lost in infinite series calculations, or is there a proof already in print?
Last edited by a moderator:
Related Linear and Abstract Algebra News on Phys.org
You seem in a much better position to investigate this than anybody else, particularly since your reference can not be found.
(x+y)^p - x^p - y^p = pxy(x=y)Q(x,y) where Q(x,y) is a homogenous integer function of degree p-3.
Are you sure you mean to write x=y? If so what is the point of Q(x,y)? Assuming you don't mean x=y, the Q(x,y) appears to be of degree p-2. Because we subtracted those of degree p, and pulled out a pxy from Q(x,y,) that leaves degree p-2.
Last edited:
Sorry for the errors. The correct web address is
www.mathpages.com/home/kmath367.htm[/URL]
The equal sign in the formula should have been "+" not "=". I neglected to hit the shift key.
I think p-3 is correct. When you substact x^p and y^p from the expansion, the results have a factor of pxy, so the x^p in now at the p-2 level. However, the remaining equation is divisible by (x+y), and this brings it to the p-3 level.
I'm an amateur looking for professional help. Is there an easy way to prove that
Q(x,y) = (x^2 + xy + y^2)^((n-3)/2) ?
Last edited by a moderator:
I think you are assuming too much from other people here. It does help to state the definition of Q(x,y)
$$Q(x,y) = \frac{(x+y)^P-(x^P+y^P)}{(xy)(x+y)p}$$
I thought it enough of a problem, letting p-1 = u, to show that p divides all terms: $$(x+y)^u -\frac{x^u+y^u}{x+y}$$
This can be found by induction on k: $$\frac{(p-1)!}{k!(p-1-k)!} \equiv (-1)^k Mod p$$
So you maybe looking at an induction problem on n.
Last edited:
Thank you. It's very much appreciated. | 610 | 2,090 | {"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} | 3.75 | 4 | CC-MAIN-2019-51 | latest | en | 0.891319 |
http://larrystake.blogspot.com/2011/07/valuations-and-fuzzy-math.html | 1,532,291,367,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00549.warc.gz | 207,354,475 | 30,492 | ## Thursday, July 21, 2011
### Valuations and fuzzy math
Continuing the commentary from the last post about valuing a property. My review of recent sales suggests that using the Property Appraiser's published "market value" is a pointless exercise unless it can be used to one's benefit against an uninformed opposing party in negotiations. With a margin of error of 45 to 50%, the PA's number should be considered for entertainment purposes only.
Dismissing the PA's number, to value a property we're left with either Miss Cleo or the comps. With apologies to fans of Miss Cleo I think recently sold comparable properties are the best starting point for determining another property's value. Identical condo units in the same building are perfect comps as are identical homes in the same neighborhood. The reality is that, in most cases, there aren't "identical" comps. In that case one has to look for the most similar and then make adjustments. The most commonly used starting metric for comparing similar properties is dollars per square foot. If a 1000 square foot unit closed last week for \$100,000 then another very similar unit with 1100 square feet should be worth close to \$110,000, all other things being equal. The big stinker there is the "all other things being equal" part. They never are, equal that is. We also need to consider that both units may have exactly identical garages, have access to the very same amenities and share an identical view. This makes a comparison based only only unit size (even with adjustments for condition) flawed from the get go. Place the properties in different buildings or neighborhoods and the comparisons become even less accurate. We can't accurately calculate the value of Starry Night based on the selling price in francs per square centimeter of Vincent's Bedroom in Arles . I think comparing Cape Winds to Sandcastles just as flawed although less extreme.
Consider that a direct ocean 2/2 with 1286 square feet and a garage at Sandcastles in Cocoa Beach will command no more rent per week than a Cape Winds 2/2 with 934 square feet and open parking in Cape Canaveral. Are they worth exactly the same because of identical potential income? I don't think so but I don't think a straight \$/sf comparison is remotely accurate either. Seems illogical to think that an otherwise similar 2/2 unit in one of the two buildings should sell for a 27% premium to the other.
My point in all this is to caution buyers and sellers about their or their agent's opinions of value. Putting a fair value on a property is not an exact science. If buying, do your best to get the lowest price possible and use all the tools available to substantiate your offers. Just be aware that in the end only you know what a property is worth to you. The commonly used valuation scales are all flawed. It's horseshoes. Closest one scores.
"Sheep only have 2 speeds - graze and stampede." _________________Colonel Dave Grossman
1. And the biggest fly in the ointment is---the unknown health of the economy/lack of discretionary income/local unemployment direction. This is what effects price in every type transaction--from a bag of peanuts to housing. All your points above are completely correct.I just felt compelled to add to the thought process.
2. Make my peanuts boiled, please. When the end comes I'll watch it from the beach or just beyond in the surf.
3. All I got is Lake Erie----The beach closes anytime it rains because of sewage back up--yeck.I'll take the ocean anytime.
4. Agreed, however I do think that the valuations offer some degree of value. There is such little turnover in some buidlings/areas that getting a good idea of market value is difficult, so in addition to some decent comps (and all real estate is unique), I see the valuations as offering some degree of assistance. 2011 preliminary valuations are out and they are generally down from the prior year. I realize that this is a stastical valuation and no way reflects each individual unit, but down is down. So if I'm interested in a unit and a unit in that same building sold last year, then I would think that given the trend in the valuation, the market value of last year's sold unit is somewhat less. I wouldn't commit to the exact % decrease in value as posted by the PA - but it is one indication. Of course the unit's view and interior status all matter, but given the same exact unit - if the trend is down in the PA - the trend should be down in market value in general.
5. Funny how we speak about prices going down in almost all areas of the country. Yea-it did tick up a tad for the last two months---but--year over year prices are down. But I have tried to follow the Cocoa Beach/Cape Canaveral markets for some time and have found the prices considerably higher than average.Its shocking to see some properties jump as much as 20-35+ percent.Example would be property on Tin Roof/Cape Canaveral.Those units were down to \$125K---now the asking price is almost \$200K.You can't tell me the market has turned on the current economic conditions of today. Gotta be overly medicated sellers and or agents or I have really missed the boat.Could be I did fall of that Turnip truck going down the road.
6. Looking at the asking prices is like browsing through the fiction section at the local library. What I find more telling are the actual closed sale prices along with the price changes that sellers are making.
When a property is overpriced by \$100k and they're making \$2k reductions in the asking price, you know you don't want to deal with them any time soon.
Looking forward to the July numbers and how much the sales pace has dropped off.
7. An extreme example of how these crazy over-pricing and reduction games can play out is the listing that I commented on a couple of years ago that was dropping the price every week by \$5000 or so. It was the house on the corner of Minutemen and Danube River. Listed in December 2007 for \$1,339,000. Price was dropped 35 times but it was always over-priced. Was eventually foreclosed and finally sold last month. Wells Fargo apparently caught the over-pricing bug associated with the house as their last asking price was \$373,900 when they finally accepted \$285,000.
When it's priced right it sells. We listed a Snug Harbor home 3 weeks ago and it closed 7 days after hitting the MLS. No price drops, just priced right to start with.
8. Library!!! In a few years kids will not know the whereabouts of a library---but that's fodder for another blog.Over the last few years I have placed offers on a few nice properties in Cocoa Beach/Cape Canaveral.Those offers were spurned and the property languished on the market for months. You probably guessed the outcome. The property finally sold for less or almost identical offers I put forth. I wonder about the true professionalism of the field of agents who in my eye view are leading buyers and sellers down the primrose path of falsehoods. I would have loved to relocate to the Cocoa Beach area for the winter months. It will be interesting to follow-- and I will-- prices for the next year or better.Could be the good fairy will leave the building and reality will regain the minds of sellers and many agents--not all agents--many are top tier--but way to many should probably be selling something other than real estate. I am compelled to say--Larry does a nice job and is probably as professional and honest as they come. His blog sheds light on the markets likes and dislikes. He tells it the way it is. Well its Friday---I've rambled enough---time to sit on the beach on Lake Erie.Whoops---its raining--the sewage odor will permeate-guess the front porch will have to do.
9. Some properties will never be properly priced yet plenty are. In Cocoa Beach and Cape Canaveral about 40 to 50 buyers a month find and purchase a condo for a number they are happy with. If you are reasonable with your expectations you can definitely find a deal you'll be satisfied with. The smell of the ocean was wonderful this morning, by the way.
10. Sure----rub it in-----next thing you will be telling me about is the beautiful sunrise over the Atlantic and sunsets over the Banana River.Can't blame ya though.Back to the porch.
11. After reading your blog, back issues and posts, I'm left keenly interested in snook fishing. It's gotta be the best past time in the world and you can eat well too. You are, living the good life :)
I'm new to shopping for a condo. Of course, your blog is helping tremendously but, I'm left hanging on at least one subject. I see corner units appear to be valued more than interior units. One owner told me the SE corner is more valuable. because the ocean breeze blows gently from the SE and it keeps the air circulating when the windows are open. I heard the lighting is better on a SE corner unit and saw the difference between NE and SE corner units at dusk. I also noticed wrap around balconies on corner units are not all created equal, but normally offer more outside space than interior units in the same buildings. I am perplexed over the value of a corner unit versus an interior unit in the same building. Is it merely buyer/seller preference or is there a no-kidding tax appraiser or lender appraiser value placed on a corner unit versus an interior unit? Now I got to go buy me a surf fishing rod and reel. Any recommendations?
12. I answered the corner unit question in a new post, October 8. For snook fishing from the beach I like 15 to 20 pound line on any good spinning reel. I don't like a long surf rod for snook fishing as you do a lot of casting and the long surf rods are too heavy for repeated casting. A stiff 7'to 8' rod is perfect. If you don't have live mullet use a diving artificial like a YoZuri minnow and work the trough right off the beach. I've caught most of my beach snook within a few feet of the beach. | 2,174 | 9,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-30 | latest | en | 0.942984 |
https://citizenmaths.com/length/point-to-cable-length-imperial | 1,623,974,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634576.73/warc/CC-MAIN-20210617222646-20210618012646-00349.warc.gz | 170,743,675 | 13,367 | # Point to Cable Length (Imperial) Conversions
From
Point
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
To
Cable Length (Imperial)
• Angstrom
• Cable Length (Imperial)
• Cable Length (International)
• Cable Length (U.S.)
• Centimeter
• Chain
• Cubit
• Decameter
• Decimeter
• Ell
• Fathom
• Finger
• Finger (cloth)
• Foot
• Furlong
• Gigameter
• Hand
• Hectometer
• Inch
• Kilofeet
• Kilometer
• League
• League (land)
• Light Day
• Light Hour
• Light Minute
• Light Second
• Light Year
• Line
• Marathon
• Megameter
• Meter
• Mickey
• Microinch
• Micrometer
• Micron
• Mil
• Mile
• Millimeter
• Myriameter
• Nail (cloth)
• Nanometer
• Nautical League
• Nautical Mile
• Pace
• Palm
• Parsec
• Pica
• Picometer
• Point
• Quarter
• Rod
• Rope
• Shaku
• Smoot
• Span
• Step
• Terameter
• Twip
• Yard
Formula 2 pt = 2 / 525312 cable-imp = 0.000003807261208576998 cable-imp
## How To Convert From Point to Cable Length (Imperial)
1 Point is equivalent to 1.9e-06 Cable Length (Imperial):
1 pt = 1.9e-06 cable-imp
For example, if the Point number is (8000000), then its equivalent Cable Length (Imperial) number would be (15.229).
Formula:
8,000,000 pt = 8000000 / 525312 cable-imp = 15.229 cable-imp
## Point to Cable Length (Imperial) conversion table
Point (pt) Cable Length (Imperial) (cable-imp)
10,000 pt 0.01904 cable-imp
20,000 pt 0.03807 cable-imp
30,000 pt 0.05711 cable-imp
40,000 pt 0.07615 cable-imp
50,000 pt 0.09518 cable-imp
60,000 pt 0.11422 cable-imp
70,000 pt 0.13325 cable-imp
80,000 pt 0.15229 cable-imp
90,000 pt 0.17133 cable-imp
100,000 pt 0.19036 cable-imp
110,000 pt 0.2094 cable-imp
120,000 pt 0.22844 cable-imp
130,000 pt 0.24747 cable-imp
140,000 pt 0.26651 cable-imp
150,000 pt 0.28554 cable-imp
160,000 pt 0.30458 cable-imp
170,000 pt 0.32362 cable-imp
180,000 pt 0.34265 cable-imp
190,000 pt 0.36169 cable-imp
200,000 pt 0.38073 cable-imp
210,000 pt 0.39976 cable-imp
220,000 pt 0.4188 cable-imp
230,000 pt 0.43784 cable-imp
240,000 pt 0.45687 cable-imp
250,000 pt 0.47591 cable-imp
260,000 pt 0.49494 cable-imp
270,000 pt 0.51398 cable-imp
280,000 pt 0.53302 cable-imp
290,000 pt 0.55205 cable-imp
300,000 pt 0.57109 cable-imp
310,000 pt 0.59013 cable-imp
320,000 pt 0.60916 cable-imp
330,000 pt 0.6282 cable-imp
340,000 pt 0.64723 cable-imp
350,000 pt 0.66627 cable-imp
360,000 pt 0.68531 cable-imp
370,000 pt 0.70434 cable-imp
380,000 pt 0.72338 cable-imp
390,000 pt 0.74242 cable-imp
400,000 pt 0.76145 cable-imp
410,000 pt 0.78049 cable-imp
420,000 pt 0.79952 cable-imp
430,000 pt 0.81856 cable-imp
440,000 pt 0.8376 cable-imp
450,000 pt 0.85663 cable-imp
460,000 pt 0.87567 cable-imp
470,000 pt 0.89471 cable-imp
480,000 pt 0.91374 cable-imp
490,000 pt 0.93278 cable-imp
500,000 pt 0.95182 cable-imp
510,000 pt 0.97085 cable-imp
520,000 pt 0.98989 cable-imp
530,000 pt 1.0089 cable-imp
540,000 pt 1.028 cable-imp
550,000 pt 1.047 cable-imp
560,000 pt 1.066 cable-imp
570,000 pt 1.0851 cable-imp
580,000 pt 1.1041 cable-imp
590,000 pt 1.1231 cable-imp
600,000 pt 1.1422 cable-imp
610,000 pt 1.1612 cable-imp
620,000 pt 1.1803 cable-imp
630,000 pt 1.1993 cable-imp
640,000 pt 1.2183 cable-imp
650,000 pt 1.2374 cable-imp
660,000 pt 1.2564 cable-imp
670,000 pt 1.2754 cable-imp
680,000 pt 1.2945 cable-imp
690,000 pt 1.3135 cable-imp
700,000 pt 1.3325 cable-imp
710,000 pt 1.3516 cable-imp
720,000 pt 1.3706 cable-imp
730,000 pt 1.3897 cable-imp
740,000 pt 1.4087 cable-imp
750,000 pt 1.4277 cable-imp
760,000 pt 1.4468 cable-imp
770,000 pt 1.4658 cable-imp
780,000 pt 1.4848 cable-imp
790,000 pt 1.5039 cable-imp
800,000 pt 1.5229 cable-imp
810,000 pt 1.5419 cable-imp
820,000 pt 1.561 cable-imp
830,000 pt 1.58 cable-imp
840,000 pt 1.599 cable-imp
850,000 pt 1.6181 cable-imp
860,000 pt 1.6371 cable-imp
870,000 pt 1.6562 cable-imp
880,000 pt 1.6752 cable-imp
890,000 pt 1.6942 cable-imp
900,000 pt 1.7133 cable-imp
910,000 pt 1.7323 cable-imp
920,000 pt 1.7513 cable-imp
930,000 pt 1.7704 cable-imp
940,000 pt 1.7894 cable-imp
950,000 pt 1.8084 cable-imp
960,000 pt 1.8275 cable-imp
970,000 pt 1.8465 cable-imp
980,000 pt 1.8656 cable-imp
990,000 pt 1.8846 cable-imp
1,000,000 pt 1.9036 cable-imp
2,000,000 pt 3.8073 cable-imp
3,000,000 pt 5.7109 cable-imp
4,000,000 pt 7.6145 cable-imp
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In town X, w watts of power can provide equal amounts of electricity f
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In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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14 Mar 2020, 18:29
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GMATBusters’ Quant Quiz Question -7
In town X, w watts of power can provide equal amounts of electricity for h houses. If g new houses were built in town X, which of the following would represent the difference between the new number of watts of power used in town X and the original number of watts of power used in town X if power is supplied equally to all of the homes?
A) wg/h
B) w^2g/h
C) w/h
D) w(h+g)/h
E) w(h-g)/h
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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14 Mar 2020, 18:30
w watts of power can provide equal amounts of electricity for h houses
power consumption per house = w/h
After g new houses were built in town X, total number of houses = g+h
Power consumption required = number of houses*power consumption per house = w/h *(g+h)
Difference = w/h *(g+h)- w = wg/h
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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14 Mar 2020, 19:38
Watts of power =w, no.of houses=h, so original no. of watts of power/home = w/h
g new homes are added, h+g, so new no. of watts of power/home = w/h+g
w/h+g - w/h
w*h- w (h+g)/ h (h+g)
wh-wh-wg/h (h+g)
-wg/h (h+g)
Not close to any option but i'd guess A as an answer.
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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14 Mar 2020, 20:15
Lets substitute values for each
Let w = 1000
Let h = 100
let g = 25
Original number of power used = 1000 watts / 100 house = 10 watts/house needed
Number of new houses = 100 + 25 = 125 houses
New number of watts used to power town = 125 houses * 10 watts/house = 125*10 = 1250 watts
Difference = 1250 - 1000 = 250 Watts
Out of our options , A gives us the answer = 250
w*g/h = (1000*25)/100 = 250
Ans: A
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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14 Mar 2020, 21:15
OA is supposedly C
If we take Watt power as 32 and H=4 per unit given to each house is 8
If we add 4 more house. So per unit supply will be 4 each (32/8)
So the difference between two = 32/4 = w/h
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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15 Mar 2020, 01:55
Given: In town X, w watts of power can provide equal amounts of electricity for h houses.
Asked: If g new houses were built in town X, which of the following would represent the difference between the new number of watts of power used in town X and the original number of watts of power used in town X if power is supplied equally to all of the homes?
w Power can be provided equally to h houses.
If g new houses were built in town X
w(h+g)/h Power can be provided equally to (h+g) houses
Difference = w(h+g)/h - w = wg/h
A) wg/h
B) w^2g/h
C) w/h
D) w(h+g)/h
E) w(h-g)/h
IMO A
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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15 Mar 2020, 02:32
w watts of power for h houses.
So, 1 house requires $$\frac{w}{h}$$ watts of power (equal power to all houses).
Now, g new houses built. So total houses are h+g.
Now if each house requires $$\frac{w}{h}$$ watts power, h+g houses will require (h+g)*$$\frac{w}{h}$$ watts power.
Difference in power
= $$\frac{w}{h}$$*(h+g) - w
= $$\frac{wg}{h}$$
Option: A
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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15 Mar 2020, 04:08
Each house has consumption=w/h
Including g houses ,total number of houses =g+h
New number of watt= (g+h)X w/h
Number of original watts used=w
Difference=(g+h)Xw/h-w=wg/h
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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15 Mar 2020, 06:29
In town X, w watts of power can provide equal amounts of electricity for h houses. If g new houses were built in town X, which of the following would represent the difference between the new number of watts of power used in town X and the original number of watts of power used in town X if power is supplied equally to all of the homes?
w/h= electricity per home
adding g new houses we need to provide g times the energy consumed by 1 home
wg/h
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink]
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15 Mar 2020, 07:11
1
Quote:
In town X, w watts of power can provide equal amounts of electricity for h houses. If g new houses were built in town X, which of the following would represent the difference between the new number of watts of power used in town X and the original number of watts of power used in town X if power is supplied equally to all of the homes?
A) wg/h
B) w^2g/h
C) w/h
D) w(h+g)/h
E) w(h-g)/h
I feel this was the trickiest of the lot.
Let's pick numbers .
w=1000
h=20
Each house uses = 1000/20 = 50
New houses g=30 (Assumption)
Each house still uses 50. New power in watts = 1000+1500=2500
Difference between 2500 and 1000 is 1500.
Keeping all this in mind,
A) wg/h = (1000*30)/20 = 500 Works
B) w^2g/h - Bigger value
C) w/h - 1000/20=50 - Nope
D) w(h+g)/h - NOpe
E) w(h-g)/h - Nope
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Re: In town X, w watts of power can provide equal amounts of electricity f [#permalink] 15 Mar 2020, 07:11 | 2,122 | 7,183 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-24 | latest | en | 0.909435 |
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Sunday December 18, 1994
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The caption says, "Dogbert teaches business math." Dogbert points to a diagram of an equation. A picture of Wally, Dilbert and Alice illustrates the equation, "Grunts equals zero." The caption says, "#1. Any job that can be done by two people . . ." The Boss stands behind two people. The caption continues, ". . . Can be done by one person for half the cost." The Boss yanks one of the workers out of his chair. The caption says, "#2. A bonus today is worth more than . . ." The Boss holds a large bag of money. The caption continues, ". . . The whole company tomorrow." An office building has a closed sign on it. The caption says, "#3. Your expense requirements for December can be calculated . . ." The Boss sits at his desk writing on a piece of paper. The caption continues, ". . . By taking what's left in the budget and multiplying by one." A delivery person asks the Boss, "Giraffe goes where?" Dogbert says, "Next week, a doctor with a flashlight shows us where sales projections come from."
Sunday July 09, 1995
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Dilbert approaches the Boss and says, "We desperately need another person on my project!" The Boss replies, "We're already over headcount. Get a contract employee." Dilbert says, "Okay, but they cost twice as much." Dilbert says, "Plus we need to buy a computer." The Boss says, "Rent one. We're over our capital budget." Dilbert replies, "Renting is expensive. We'll go over our expense budget." The Boss says, "I'll fire Wally. That will free up some cash." Dilbert throws his arms up in frustration and says, "Wally's on my project!" Dilbert shouts, "Forget it! I'll just work sixteen hours a day!!" The Boss thinks, "That worked out perfectly. I think I might be a genius or something." The Boss sits at his desk and thinks, "I wonder if I should have told him the project was cancelled last week."
Sunday September 10, 1995
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The Boss introduces a man to Dilbert and Wally. The Boss says, "We just hired Jack away from our competitor. He was their best manager." The Boss continues, "Jack will be in charge of project 'Goosefood.'" The Boss continues, "I'd like you two to brief Jack on the project." Jack, Dilbert and Wally sit at a conference table. Dilbert says, "Project 'Goosefood' has no budget and no management support." Wally says, "Your job is to build a global information network in two weeks." Dilbert says, "Failure is certain. Soon you will leave the industry in disgrace." Wally adds, ". . . Just like the other 'best managers' we hired from our competitors." Jack says, "Just our of curiosity, how did the project get its name?" Wally replies, "Let's just say that you're the goose food . . ."
Sunday October 15, 1995
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Dogbert sits across from the Boss's desk. The Boss asks, "Why do you want to be our new 'Long Range Planner,' Mister Dogbert?" Dogbert replies, "Because 'long range' is very far away . . ." Dogbert continues, ". . . Therefore it will be impossible to evaluate my performance." Dogbert continues, "If it's not too much to ask, I'd like to be on flex-time so you'll never know if I'm supposed to be at work." Dogbert continues, "I'll need an Internet connection at home so I can telecommute and not pollute. Because I give a hoot." Dogbert continues, "Also, I'd like to be in a group with lots of losers. That way I'll get the biggest riase when we're ranked." The Boss says, "You're hired. All of the other applicants demanded relocation expenses and I have no budget for that." Dogbert stands at a desk. Dilbert asks, "Why do I have to work while you just look for poodle graphics on the Internet?" Dogbert says, "Don't work too hard; I'd like a big raise."
Sunday November 19, 1995
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The Boss says to Dilbert and Wally, "We need to ship the V-1 product tomorrow. I promised our CEO he can announce it at the shareholder meeting." Wally says, "That's two months early!!" Dilbert says, "We haven't added any features yet!!" Dilbert says, "What would we ship? Our lab prototype is the only V-1 in existence!" Wally cries, "No . . . You wouldn't!" The Boss picks up the device and says, "I've scheduled press tours so you can do demos all next week." Wally asks, "On what?!" As the Boss walks away carrying the prototype, Dilbert says, "We'll need twenty thousand dollars to build another prototype!" The Boss says, "That reminds me; I froze the budget for the rest of the year." The Boss continues, "If there's anything you'd like me to do, don't hesitate to ask." Wally says, "Yeah, I'd like you to do something . . ." The Boss looks angry. Wally says, "Ooh . . . I think I should have hesitated to ask that."
Sunday January 07, 1996
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Dilbert sits at his desk thinking, "Why do I have a feeling of impending doom?" The Boss peeks into Dilbert's cubicle. The Boss says, "Good news!" Dilbert thinks, "Uh-oh." The Boss says, "You won't have to spend another lonely day in this tiny cubicle." Dilbert asks, "I'm getting an office?" The Boss replies, "Better! You're getting a roommate!" Dilbert shouts, "Why??? We've got plenty of empty cubicles! Our company owns the whole building!" The Boss says, "The finance department charges my budget for the square footage we use." Dilbert looks over the cubicle wall and says to the Boss, "It's a false savings! You're hurting the company!" The Boss walks away thinking, "All I hear is a faint buzzing." Dilbert sits at his desk thinking, "Oh, well. How bad could it be?" A man wearing a cowboy hat and carrying a can of beans and a radio says, "I hope you like baked beans and square-dancing as much as I do!"
Sunday July 28, 1996
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The caption says, "The Budget Trap." The Boss says, "I need a quick estimate for how much your next project will cost, Wally." Wally replies, "How should I know? You haven't even told me what my next project is." The Boss says, "That's okay. I only need a rough estimate for planning purposes." Wally says, "I see where this is going. You're going to turn my wild guess into a budget. Later I'll be blamed when it's wrong." The Boss replies, "No, no. I won't hold you to these numbers." Wally says, "Well . . . Okay, let's say two million dollars." The Boss says as he walks away, "Ooh . . . Can't afford that. I'll put you down for twenty thousand dollars." The caption says, "One year later . . ." The Boss sits at his desk and says to Wally, "You're way over budget. Can you show me the cause?" Wally replies, "It depends. Can mirrors reflect your image?"
Sunday June 29, 1997
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Dilbert stands across from the Boss's desk. He says, "I want a ten-percent raise." The Boss replies, "There's no budget for raises." Dilbert holds up a letter and says, "I have an offer from another company that will pay fifteen percent more." The Boss says, "I'll give you twenty percent if you stay." Dilbert says, "I thought you said there's no budget for raises." The Boss replies, "Well . . . It's supposed to be a secret but . . ." The Boss says, "Our policy is to give big raises to people who spend their time interviewing for other jobs." Dilbert tells Alice and Wally, "Good news! The secret company policy is to reward disloyalty!" They cheer and shout, "Yes! Yippee!" Wally asks, "What's the reward for leaving work early?" Dilbert replies, "He wouldn't show me the price sheet."
Sunday December 21, 1997
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Wally sticks his head into the boss' office. Wally says, "I'm back from training." Wally says, "I got a big binder." Wally holds out a big book. Wally says, "The training is already forgotten but the binder will last forever." Wally brings the binder to his chest. Wally says, "A living monument to temporary knowledge!" Wally says, "I'll put it in my cubicle with the others." Wally says, "Speaking of my cubicle, which direction is it?" The Boss points. Wally says, "Okay, thanks. That information should be in a binder." Dilbert says, "Did he approve funding for our project?" Wally says, "Not yet. Step one was to free up funds from the training budget."
Monday February 22, 1999
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The boss introduces a new man to Dilbert. The boss says, "Dilbert, meet our new sacrifial lamb." The boss says, "I filled our headcount vacancy so we have someone to dump after the next budget cut." The lamb says, "Should we shake hands?" The boss says, "I don't want to get attached." | 2,384 | 9,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-05 | longest | en | 0.957898 |
http://www.sharewareconnection.com/heat-transfer.htm | 1,386,615,124,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163995757/warc/CC-MAIN-20131204133315-00081-ip-10-33-133-15.ec2.internal.warc.gz | 518,610,151 | 9,036 | Home > By category > Education >Mathematics > Heat Transfer
Heat Transfer
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Numbat in a simulation
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(IEEE 802. | 1,818 | 8,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.74476 |
http://www.perlmonks.org/?node_id=509348 | 1,526,938,385,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864544.25/warc/CC-MAIN-20180521200606-20180521220606-00545.warc.gz | 439,619,958 | 6,051 | ### Re^2: CarTalk Puzzler
by Perl Mouse (Chaplain)
on Nov 17, 2005 at 09:51 UTC ( #509348=note: print w/replies, xml ) Need Help??
in reply to Re: CarTalk Puzzler
The trouble with enumerating the result - what if it had been 2,000,000,000,000,000,000,000 instead of 20,000?
2,000,000,000,000,000,000,000 == 2 * 1021 == 5 * 4 * 1020 == 5 * (2 * 1010)2.
So, all you need to do is list the squares of all numbers from 1 to sqrt(5) * 2 * 1010. It'll take a while to list, but you can do it with a one-liner.
Perl --((8:>*
Replies are listed 'Best First'.
Re^3: CarTalk Puzzler
by thor (Priest) on Nov 17, 2005 at 12:51 UTC
I think the OP was saying "what if the original solutions posted in this thread were used for <insert really large number here> bulbs?". The solutions posted near the top of the thread relied on walking the list numerous times, which is infeasible if the list is super large. The elegance in finding the mathematical property for the bulbs being on is that it does scale.
thor
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Be your own disciple, fan the sparks of will
For all of us waiting, your kingdom will come
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Notices? | 455 | 1,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.898567 |
https://www.infocomm.ky/bayes-theorem-the-formula-and-examples/ | 1,726,877,630,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701425385.95/warc/CC-MAIN-20240920222945-20240921012945-00047.warc.gz | 729,920,700 | 11,571 | # Bayes’ Theorem: The Formula and Examples.
. Bayes' Theorem. What makes something Bayesian? There are a few key things that make something Bayesian. First, Bayesian methods use probability to represent uncertainty. This is in contrast to other methods, which may use other measures, such as the median or mode. Second, Bayesians update their beliefs in light of new evidence. This is done by revising the probabilities assigned to different hypotheses in light of new data. Finally, Bayesian methods often involve making decisions based on expected utility, which takes into account both the likelihood of different outcomes and the value of those outcomes.
##### Why do we use Bayesian statistics?
Bayesian statistics are often used in financial analysis because they allow for the incorporation of prior information into the analysis. This is important in financial analysis because there is often a great deal of historical data available that can be used to inform the analysis. Bayesian statistics provide a framework for incorporating this prior information in a principled way.
There are a number of other reasons why Bayesian statistics may be used in financial analysis. For example, Bayesian methods can be used to deal with non-normal data, which is often encountered in financial data. Bayesian methods can also be used to deal with model uncertainty, which is also often encountered in financial analysis.
What is the difference between Bayesian and regular statistics? Bayesian statistics is a method of statistical inference that is based on Bayes' theorem. Bayes' theorem is a way to update the probability of an event occurring after observing new data. This new data can be data from another experiment or data from the same experiment that was not used in the original analysis.
Regular statistics is a method of statistical inference that is based on the traditional approach of using a sample to estimate the population parameters. This approach does not account for new data that may be observed after the initial analysis.
How is Bayes theorem used in machine learning? Bayes theorem is used in machine learning to calculate the probability of an event occurring, given that another event has occurred. For example, if we know that there is a 70% chance of event A occurring, and we also know that there is a 30% chance of event B occurring, then we can use Bayes theorem to calculate the probability of event A occurring given that event B has occurred.
Why is Bayesian better? There are a number of reasons why Bayesian methods may be preferable to other methods in financial analysis. First, Bayesian methods allow for the incorporation of prior information into the analysis, which can be particularly important in financial applications where data may be limited. Second, Bayesian methods can be used to construct models that are more flexible than those that can be constructed using other methods, which can be important in capturing the complex relationships that often exist in financial data. Finally, Bayesian methods can be used to generate predictions that are more accurate than those generated by other methods, which can be important in making investment decisions. | 590 | 3,213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-38 | latest | en | 0.953432 |
https://mailinglist.acer.edu.au/pipermail/rasch/2007-March/000346.html | 1,561,471,501,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999838.27/warc/CC-MAIN-20190625132645-20190625154645-00099.warc.gz | 495,459,025 | 2,910 | # [Rasch] Interpretation of Item Plot
Stephen Humphry stephen.humphry at uwa.edu.au
Mon Mar 26 15:41:32 EST 2007
```Hi Greg. You are quite correct interpreting this as signalling a problem.
Yes, it is possible it would work properly with only three categories,
although this is not guaranteed. Try recoding and look at the fit. If you
are able, look at conditional threshold probability curves (e.g. 2 given a 1
or 2). The case in which it is clearly justified to collapse two categories
is when there is 0 discrimination at a threshold.
The thresholds in your case are disordered. The parameter for threshold x
(t_x) in the model is the location at which it is equally likely a person
will response in adjacent ordered categories x-1 and x. For example, t_1 is
the location at which it is equally likely a person is equally likely to
respond in categories 0 and 1, t_2 is the location at which a person is
equally likely to respond in categories 1 and 2, etc.
In your example, taking the threshold estimates literally, the location at
which a person is equally likely to respond in categories 0 and 1 (t_1) is
*higher* than the location at which a person is equally likely to respond in
categories 1 and 2 (t_2). If the categories are ordered as you intended, t_1
(0/1 equally likely) shoud be lower than t_2 (1/2 equally likely) and then
there would be a region on the latent continuum in which x=1 is the most
likely response for any given person in the region (i.e. a region between
t_1 and t_2). There is no such region in which x=1 is the most likely in
your case, which reflects the disordering.
The disordering in your example is not pronounced, so it is possible that it
is due to instability of the threshold estimates if you don't have a large
data set. The category frequencies are not the key consideration. Rather,
the key consideration is how many people overall have estimates in the
region of the disordering.
Whether the thresholds are ordered as intended is an empirical question. The
structure of the Rasch model requires they are ordered. Ordered thresholds
are the basis for the definition of integer scoring of ordered categories,
given the formal structure of the model (Andrich, 1978, 2005). Disordered
thresholds imply the hypothesis of ordering has broken down somehow (if the
disordering is genuine).
Yes, your interpretation is correct. Yes it is possible that three
categories will work, though this is not guaranteed.
Regards,
Steve.
Andrich, D. (1978). A rating formulation for ordered response categories.
Psychometrika, 43, 357-74.
Andrich, D. (2005). The Rasch model explained. In Sivakumar Alagumalai,
David D Durtis, and Njora Hungi (Eds.) Applied Rasch Measurement: A book of
exemplars. Springer-Kluwer. Chapter 3, 308-328.
-----Original Message-----
From: rasch-bounces at acer.edu.au [mailto:rasch-bounces at acer.edu.au] On Behalf
Of Petroski, Greg
Sent: Saturday, 24 March 2007 1:10 PM
To: Eric Wong; rasch at acer.edu.au
Subject: [Rasch] Interpretation of Item Plot
To keep the size of this e-mail small I put an item plot on PhotoBucket.
http://i24.photobucket.com/albums/c1/Alpha_Mutt/ItemCategoryPlot.jpg
The item is ordinal with responses 0 to 4 with larger values indicating
greater ability. Note that two of the five catagories do not ever have the
highest probability of being endorsed. Am I correct in interpreting this as
suggesting a defective item -- one that might work as well with only three
categories?
Greg Petrsoki
U. of Missouri - Columbia
_______________________________________________
Rasch mailing list
Rasch at acer.edu.au
http://mailinglist.acer.edu.au/mailman/listinfo/rasch
``` | 929 | 3,670 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-26 | latest | en | 0.955005 |
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# R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 6 - Introduction to Euclid's Geometry
Our RS Aggarwal & V Agarwal Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 9 board exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 6 - Introduction to Euclid's Geometry.
All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RS Aggarwal & V Agarwal Textbook Solutions for CBSE Class 9 Mathematics will strengthen your fundamentals in this chapter and will help you in your attempts to score more marks in the final examination. CBSE Class 9 students can refer to our solutions any time — while doing their homework and while preparing for the exam.
Exercise/Page
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 6 - Introduction to Euclid's Geometry Page/Excercise MCQ
Solution 1
Correct option: (b)
Squares and circular altars were used for household rituals.
Whereas altars having shapes as combinations of rectangles, triangles and trapeziums were used for public worship.
Solution 2
Correct option: (b)
In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for public rituals.
Solution 3
Correct option: (c)
The Sriyantra consists of nine interwoven isosceles triangles.
Solution 4
Correct option: (b)
In Indus Valley Civilization (about 300 BC) the bricks used for construction work were having dimensions in the ratio is 4:2:1.
Solution 5
Correct option: (a)
The famous treatise 'The Elements' was divided into 13 chapters by Euclid.
Solution 6
Correct option: (b)
Euclid belongs to the country, Greece.
Solution 7
Correct option: (c)
Thales belongs to the country, Greece.
Solution 8
Correct option: (b)
Pythagoras was a student of Thales.
Solution 9
Correct option: (d)
A statement that requires a proof is called a theorem.
Solution 10
Correct option: (a)
'Lines are parallel if they do not intersect' is started in the form of a definition.
Solution 11
Correct option: (c)
Euclid stated that 'All right angles are equal to each other' in the form of a postulate.
This is Euclid's Postulate 4.
Note: The answer in the book is option (a). But if you have a look at the Euclid's postulate, the answer is a postulate.
Solution 12
Correct option: (d)
A pyramid is a solid figure, whose base is any polygon.
Solution 13
Correct option: (a)
The side faces of a pyramid are triangles.
Solution 14
Correct option: (c)
A solid has 3 dimensions.
Solution 15
Correct option: (b)
A surface has 2 dimensions.
Solution 16
Correct option: (a)
A point is an exact location. A fine dot represents a point. So, a point has 0 dimensions.
Solution 17
Correct option: (c)
Boundaries of solids are surfaces.
Solution 18
Correct option: (b)
Boundaries of surfaces are curves.
Solution 19
Correct option: (d)
The number of planes passing through three non-collinear points is 1.
Solution 20
Correct option: (d)
Axioms are assumed as universal truths in all branches of mathematics because they are taken for granted, without proof.
Solution 21
Correct option: (c)
Two lines are said to be parallel, if they have no point in common.
Options (a), (b) and (d) have a common point, hence they are not parallel.
In option (c), the floor and the ceiling of a room are parallel to each other is a true statement.
Solution 22
Correct option: (c)
In option (a), infinite number of line can be drawn to pass through a given point. So, it is not a true statement.
In option (b), only one line can be drawn to pass through two given points. So, it is not a true statement.
In option (c),
'If two circles are equal, then their radii are equal' is the true statement.
In option (d), A line has no end points. A line has an indefinite length. So, it is not a true statement.
Solution 23
Correct option: (c)
Option (a) is true, since we can pass an infinite number of lines through a given point.
Option (b) is true, since a unique line can be drawn to pass through two given points.
Consider option (c).
As shown in the above diagram, a ray has only one end-point. So, option (d) is true.
Hence, the only false statement is option (c).
Solution 24
Correct option: (c)
A point C is called the midpoint of a line segment , if C is an interior point of AB such that =.
Solution 25
Correct option: (d)
Observe the above figure. Clearly, C lies between A and B if AC + CB = AB.
That means, points A, B, C are collinear.
Solution 26
Correct option: (b)
Euclid's second axiom states that 'If equals are added to equals, the wholes are equal'.
Hence, when x + y = 15, then x + y + z = 15 + z.
Solution 27
Correct option: (a)
Euclid's first axiom states that 'Things which are equal to the same thing are equal to one another'.
That is,
A's age = B's age and C's age = B' age
A's age = C's age
## R S Aggarwal and V Aggarwal Solution for Class 9 Mathematics Chapter 6 - Introduction to Euclid's Geometry Page/Excercise 6
Solution 1
A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.
Example of Theorem: Pythagoras Theorem
Example of axiom: A unique line can be drawn through any two points.
Solution 2
(i) Line segment: The straight path between two points is called a line segment.
(ii) Ray: A line segment when extended indefinitely in one direction is called a ray.
(iii) Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.
(iv) Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.
(v) Half-line: A ray without its initial point is called a half-line.
(vi) Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.
(vii) Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.
(viii) Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.
Solution 3
(i) Six points: A,B,C,D,E,F
(ii) Five line segments:
(iii) Four rays:
(iv) Four lines:
(vi) Four collinear points: M,E,G,B
Solution 4
(i) and their corresponding point of intersection is R.
and their corresponding point of intersection is P.
(ii) and their point of intersection is R.
(iii) Three rays are:
.
(iv) Two line segments are:
.
Solution 5
(i) Three lines: Line AB, Line PQ and Line RS
(ii) One rectilinear figure: EFGC
(iii) Four concurrent points: Points A, E, F and B
Solution 6
(i) An infinite number of lines can be drawn to pass through a given point.
(ii) One and only one line can pass through two given points.
(iii) Two given lines can at the most intersect at one and only one point.
(iv)
Solution 7
(i) False
(ii) False
(iii) False
(iv) True
(v) False
(vi) True
(vii) True
(viii) True
(ix) True
(x) True
(xi) False
(xii) True
Solution 8
(ii) BL = BM
2BL = 2BM
AB = BC
## Browse Study Material
TopperLearning provides step-by-step solutions for each question in each chapter in the RS Aggarwal & V Aggarwal Textbook. Access the CBSE Class 9 Mathematics Chapter 6 - Introduction to Euclid's Geometry for free. The questions have been solved by our subject matter experts to help you understand how to answer them. Our RS Aggarwal Solutions for class 9 will help you to study and revise the whole chapter, and you can easily clear your fundamentals in Chapter 6 - Introduction to Euclid's Geometry now.
# Text Book Solutions
CBSE IX - Mathematics
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Let us get in touch with you | 2,082 | 8,226 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.922443 |
https://www.enotes.com/homework-help/prove-identity-cos-2-x-tan-x-2sinx-secx-cosx-sin-2-460372 | 1,660,614,948,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00108.warc.gz | 653,392,749 | 18,739 | # Prove the identity `cos^2 x . tan x = (2sinx)/(secx+cosx+sin^2 x. sec x)`
`cos^2 x . tanx = (2sinx)/(secx+cosx+sin^2 x . sec x)`
`cos^2 x . tanx`
`=> cos^2 x. (sinx/cosx) `
=cosx. sinx ----(1)
`(2.sinx)/(sec x+cosx+sin^2 x sec x)`
`(2.sinx)/((1/cos x) + (cosx) + (sin^2 x)/(cosx))`
`(2.sinx.cosx)/(1+cos^2 x+ sin^2 x)`
`(2. sinx . cosx)/(2)`
sinx.cosx--------(2)
As (1) = (2) So `cos^2 x . tanx = (2sinx)/(secx+cosx+sin^2 x . sec x)`
Approved by eNotes Editorial Team
The identity `cos^2 x* tan x = (2sinx)/(secx+cosx+sin^2 x* sec x) ` has to be proved.
`(2sinx)/(secx+cosx+sin^2 x* sec x) `
= `(2sinx)/(1/(cos x)+cosx+sin^2 x*1/(cos x)) `
= `(2sinx*cos x)/(1+cos^2x+sin^2 x) `
= `(2sinx*cos x)/(1+1)`
= `(2sinx*cos x)/2`
= `sinx*cos x`
= `(sin x/cos x)*cos x*cos x`
= `tan x*cos^2x`
This proves that ` cos^2 x* tan x = (2sinx)/(secx+cosx+sin^2 x* sec x)`
Approved by eNotes Editorial Team | 426 | 914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-33 | latest | en | 0.340538 |
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• Oregon Museum of Science and Industry (remove) | 676 | 2,877 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-26 | latest | en | 0.860759 |
https://physics.stackexchange.com/questions/61289/explanation-for-the-next-steps-of-chaplygin-dipole | 1,723,585,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641085898.84/warc/CC-MAIN-20240813204036-20240813234036-00558.warc.gz | 364,417,654 | 41,878 | # Explanation for the next steps of chaplygin dipole
this post is the Chaplygin dipole, it's an interesting issue.
Can someone explain me these steps in other words please? any Explanation of any step will help me, I hope that together I will understand all! please explain me the steps in other words because I didn't understand what is written
In 1903 Chaplygin published another remarkable paper (Chaplygin 1903) devoted to the motion associated with a compact vorticity distribution in a two-dimensional unbounded inviscid flow. In the introduction of that paper he gave a precise formulation of the problem: Consider an unbounded mass of incompressible fluid in which the motion is parallel to the OXY plane; let the motion outside some circular cylinder be irrotational, the velocity being equal to zero at infinity. The question is to find a distribution of vortex lines inside the cylinder that gives rise to a uniformly translating vortex column with a continuous velocity distribution and with a positive pressure all around.
As a first example of the solution Chaplygin considered in detail a case of rectilinear motion of a circular vortex of radius $a$ with a constant translation velocity $v_0$. By superimposing on the whole fluid fluid a uniform velocity $-v_0$ he obtained a stationary problem of a steady vortex cylinder placed in a potential flow with uniform velocity at infinity. By choosing the polar coordinate system $(r,\theta)$ , with the origin at the centre of the cylinder, the stream function $\psi_1$ for the potential flow around the cylinder is written as:
• Can you tell us more precisely what it is you want to understand and what help you need to get there? Commented Apr 16, 2013 at 15:25
• Indeed, for now it looks mostly like a math problem. I think the differential equation itself is solved 'easily'. Commented Apr 16, 2013 at 19:02
• @Bernhard, can you explain me please? thank you very much!! Commented Apr 17, 2013 at 20:58
• @ChristophB. I understood till (2.4) included. what's about the next steps please? thank you very much! :] Commented Apr 17, 2013 at 23:39
• How familiar are you with the equations of fluid dynamics and their resolution (involving partial differential equations)? Commented Apr 18, 2013 at 8:00
Equation (2.6) expresses that the two pieces of the total solution, the one inside the disk $\psi$ and the one outside $\psi_1$, have to match in some sense. The first part asks that the actual values of both solutions coincide on the circle, the second part asks that the derivatives be the same too. Nothing is said about higher order derivatives.
There'll probably be some latitude in choosing various constants once a general solution is found. Equation (2.6) should help fix these. Note that $\psi_1$ vanishes at $r=a$, so that $\psi$ must too. It so happens that the Bessel function $J_1$ vanishes when its argument equals $b$, so that's probably the reason why it appears under the form given in (2.7). We have that $\partial_r \psi_1 = v_0 (1+\frac{a^2}{r^2})\sin{(\theta)}$, so that at $r=a$ we have $\partial_r \psi_1\vert_{r=a} = v_0 2\sin{(\theta)}$. Matching $\partial_r \psi$ to that value explains why there's the constant $J'_1(b)^{-1}$. | 784 | 3,224 | {"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} | 3.390625 | 3 | CC-MAIN-2024-33 | latest | en | 0.926548 |
https://www.xiaohui.com/dev/mmx/mmx_p_28_2.htm | 1,547,814,263,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660070.15/warc/CC-MAIN-20190118110804-20190118132804-00537.warc.gz | 1,003,749,162 | 6,939 | # 28.2 Floating point instructions
Explanations:
Operands:
r = register, m = memory, m32 = 32 bit memory operand, etc.
Clock cycles:
The numbers are minimum values. Cache misses, misalignment, denormal operands, and exceptions may increase the clock counts considerably.
Pairability:
+ = pairable with FXCH, np = not pairable with FXCH.
i-ov:
Overlap with integer instructions. i-ov = 4 means that the last four clock cycles can overlap with subsequent integer instructions.
fp-ov:
Overlap with floating point instructions. fp-ov = 2 means that the last two clock cycles can overlap with subsequent floating point instructions. (WAIT is considered a floating point instruction here)
Instruction Operand Clock cycles Pairability i-ov fp-ov FLD r/m32/m64 1 + 0 0 FLD m80 3 np 0 0 FBLD m80 48-58 np 0 0 FST(P) r 1 np 0 0 FST(P) m32/m64 2 m) np 0 0 FST(P) m80 3 m) np 0 0 FBSTP m80 148-154 np 0 0 FILD m 3 np 2 2 FIST(P) m 6 np 0 0 FLDZ FLD1 2 np 0 0 FLDPI FLDL2E etc. 5 s) np 2 2 FNSTSW AX/m16 6 q) np 0 0 FLDCW m16 8 np 0 0 FNSTCW m16 2 np 0 0 FADD(P) r/m 3 + 2 2 FSUB(R)(P) r/m 3 + 2 2 FMUL(P) r/m 3 + 2 2 n) FDIV(R)(P) r/m 19/33/39 p) + 38 o) 2 FCHS FABS 1 + 0 0 FCOM(P)(P) FUCOM r/m 1 + 0 0 FIADD FISUB(R) m 6 np 2 2 FIMUL m 6 np 2 2 FIDIV(R) m 22/36/42 p) np 38 o) 2 FICOM m 4 np 0 0 FTST 1 np 0 0 FXAM 17-21 np 4 0 FPREM 16-64 np 2 2 FPREM1 20-70 np 2 2 FRNDINT 9-20 np 0 0 FSCALE 20-32 np 5 0 FXTRACT 12-66 np 0 0 FSQRT 70 np 69 o) 2 FSIN FCOS 65-100 r) np 2 2 FSINCOS 89-112 r) np 2 2 F2XM1 53-59 r) np 2 2 FYL2X 103 r) np 2 2 FYL2XP1 105 r) np 2 2 FPTAN 120-147 r) np 36 o) 0 FPATAN 112-134 r) np 2 2 FNOP 1 np 0 0 FXCH r 1 np 0 0 FINCSTP FDECSTP 2 np 0 0 FFREE r 2 np 0 0 FNCLEX 6-9 np 0 0 FNINIT 12-22 np 0 0 FNSAVE m 124-300 np 0 0 FRSTOR m 70-95 np 0 0 WAIT 1 np 0 0
Notes:
m) The value to store is needed one clock cycle in advance.
n) 1 if the overlapping instruction is also an FMUL.
o) Cannot overlap integer multiplication instructions.
p) FDIV takes 19, 33, or 39 clock cycles for 24, 53, and 64 bit precision respectively. FIDIV takes 3 clocks more. The precision is defined by bit 8-9 of the floating point control word.
q) The first 4 clock cycles can overlap with preceding integer instructions. See chapter 26.7.
r) clock counts are typical. Trivial cases may be faster, extreme cases may be slower.
s) may be up to 3 clocks more when output needed for FST, FCHS, or FABS.
文章评论
1: #include <stdio.h>
2:
3: int main()
4: {
004106B0 55 push ebp
004106B1 8B EC mov ebp,esp
004106B3 83 EC 4C sub esp,4Ch
004106B6 53 push ebx
004106B7 56 push esi
004106B8 57 push edi
004106B9 8D 7D B4 lea edi,[ebp-4Ch]
004106BC B9 13 00 00 00 mov ecx,13h
004106C1 B8 CC CC CC CC mov eax,0CCCCCCCCh
004106C6 F3 AB rep stos dword ptr [edi]
5: int x=12345;
004106C8 C7 45 FC 39 30 00 00 mov dword ptr [ebp-4],3039h
6:
7: float a=3458764513820540927.0;
004106CF C7 45 F8 00 00 40 5E mov dword ptr [ebp-8],5E400000h
8:
9: int c=123;
004106D6 C7 45 F4 7B 00 00 00 mov dword ptr [ebp-0Ch],7Bh
10: c=a;
004106DD D9 45 F8 fld dword ptr [ebp-8]
004106E0 E8 9F FF FF FF call __ftol (00410684)
004106E5 89 45 F4 mov dword ptr [ebp-0Ch],eax
11: return 0;
004106E8 33 C0 xor eax,eax
12: }
004106EA 5F pop edi
004106EB 5E pop esi
004106EC 5B pop ebx
004106ED 8B E5 mov esp,ebp
__ftol:
00410684 55 push ebp
00410685 8B EC mov ebp,esp
00410687 83 C4 F4 add esp,0F4h
0041068A 9B wait
0041068B D9 7D FE fnstcw word ptr [ebp-2] ;FNSTCW 将FPU控制字保存到xx，不检查非屏蔽浮点异常
0041068E 9B wait
0041068F 66 8B 45 FE mov ax,word ptr [ebp-2]
00410693 80 CC 0C or ah,0Ch ;修改FPU?
00410696 66 89 45 FC mov word ptr [ebp-4],ax
0041069A D9 6D FC fldcw word ptr [ebp-4]
0041069D DF 7D F4 fistp qword ptr [ebp-0Ch]
004106A0 D9 6D FE fldcw word ptr [ebp-2]
004106A3 8B 45 F4 mov eax,dword ptr [ebp-0Ch]
004106A6 8B 55 F8 mov edx,dword ptr [ebp-8]
004106A9 C9 leave
004106AA C3 ret
004106AB CC int 3
004106AC CC int 3
004106AD CC int 3
004106AE CC int 3
004106AF CC int 3
004106EF 5D pop ebp
004106F0 C3 ret
发表你的评论如果你想针对此文发表评论, 请填写下列表单: 姓名: * 必填 (Twitter 用户可输入以 @ 开头的用户名, Steemit 用户可输入 @@ 开头的用户名) E-mail: 可选 (不会被公开。如果我回复了你的评论，你将会收到邮件通知) 网站 / Blog: 可选 反垃圾广告: 为了防止广告机器人自动发贴, 请计算下列表达式的值: 8 x 2 + 1 = * 必填 评论内容: * 必填 你可以使用下列标签修饰文字: [b] 文字 [/b]: 加粗文字 [quote] 文字 [/quote]: 引用文字
Copyright © XiaoHui.com; 保留所有权利。 | 1,894 | 4,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-04 | longest | en | 0.646252 |
http://www.genericgamedev.com/general/snake-smooth-and-accurate-following-behaviour/ | 1,563,487,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525829.33/warc/CC-MAIN-20190718211312-20190718233312-00286.warc.gz | 219,997,933 | 14,306 | # Snake: Smooth and accurate following behaviour
Following another object is one of the most basic movement behaviours an item can exhibit – both in the real world, and in games.
There are many different ways in which objects can follow each other, and depending on the circumstances, different kinds of movement may be appropriate.
Today we will look at one particular kind: A number of objects following another in a trail at regular distances. The movement found in games like Snake.
However – unlike the original Snake and many of its spin-offs – we will neither constrain ourselves to a grid, or to fixed time steps.
Instead we want our solution to follow arbitrary paths with arbitrary accuracy.
## Constraints, requirements, basic implementation
The basic specifications of the system we are trying to develop are as follows:
• We have one head object, which is followed by a number of tail segments;
• Each segment follows the previous one at a fixed distance.
The second point implies – if this is not even its only meaning – that a segment will never be further away than the given distance from the one it is following.
This interpretation can be translated into code almost trivially:
``````void followParent()
{
var differenceToParent = this.parent.Position - this.position;
var distanceToParent = differenceToParent.Length;
if (distanceToParent > followDistance)
{
var tooFar = distanceToParent - followDistance;
var translation = differenceToParent.Normalized() * tooFar;
this.position += translation;
}
}
``````
Baring mathematical optimisations, this is pretty much as simple as it gets.
Here is a little animation of how this implementation looks in real-time:
Except – once the snake starts turning around, it’s tail seems to stop moving and collapse in on itself.
In some situations, that might be exactly what we want. But in this post it is not. We want our tail to follow the head smoothly, and without taking short cuts.
That is exactly what is happening here: We formulated – and more importantly implemented – our constraints too simply, which means that our tail segments are able to take a short cut. They move only the least distance they have to, in order to fulfil the distance requirement.
## Remembering where we were
Let us take a look by how far we were off from our desired behaviour. We can easily keep track of the positions of the head, and draw a path along this path, to visualise it.
We can clearly see how our trail remains reasonably accurate if the head does not change its direction of movement too much. However, as soon as we turn to any significant degree the trail folds in on itself, which results in large deviations from the actual path.
How can we improve this?
As I am sure is already clear, the answer is contained in the above animation: We can simply keep track of the exact path our head moves along and position the tail along that path.
Let us build this system step by step – solving any problems we may encounter as we encounter them.
## Leaving bread crumbs
At first, how exactly do we keep track of our previous positions? A simple list of vectors will do the trick:
``````List<Vector2> previousPositions = new List<Vector2>();
void update()
{
this.updateMovement();
}
``````
Actually, there really is no reason – and it probably is a bad idea in most cases – to keep a list of previous positions since the beginning of the game. We want to be able to add new positions to one end of the list, while easily removing them from the other.
At first it looks like we may want to use a queue for this. However, we also need to be able to enumerate the list, which is a feature a lot of queue implementations do not support. To keep things easy, we will go with a linked list.
``````LinkedList<Vector2> previousPositions = new LinkedList<Vector2>();
void update()
{
this.updateMovement();
}
``````
## Following the path
The next step is calculating the position of our segments from the recorded path.
Note that there are two possible approaches here: We could record a path for each segment, and have each following segment find the correct position on the path of the one it is following.
However, this will mean keeping track of a lot of redundant data – doing the same work again and again – and even more importantly it will be prone to the same accuracy errors as we see above – albeit to a smaller degree.
Instead, we will record only one path – that of the head – and have all parts follow this same path with an increasing distance.
Further note how the recorded path is really just a list of small line segments, each connecting two of the points. Our objective is to find the correct line segment for each part of our tail – and the correct position along the segment.
Let us assume that each part knows what distance it should keep from the head along the path. Knowing that we can loop over the recorded path adding up the lengths of segments, until we find the correct one. On that segment we can then interpolate the correct position:
``````void followHead(LinkedList<Vector2> path)
{
var distanceLeft = this.distanceToParent;
var segmentStart = path.First.Value;
foreach (var point in path.Skip(1))
{
var segmentEnd = point;
var segmentDiff = segmentEnd - segmentStart;
var segmentLength = segmentDiff.Length;
if (segmentLength >= distanceLeft)
{
var percentageAlongSegment = distanceLeft / segmentLength;
this.position = segmentStart +
segmentDiff * percentageAlongSegment;
break;
}
segmentStart = segmentEnd;
}
}
``````
While considerably more complex than what we had before, this method really is not all that bad.
If we replace our original movement with it, this is how it will look in game:
Perfect!
Yes. And no.
## Almost perfect
The solution above does indeed do what we want. It is all but perfect however.
First, we are still missing the functionality to cut off the unnecessary end of the path. Further, we are recording a new point every frame, which could lead to a very long list – especially if we are moving slowly compared to our frame rate. I am sure you can see what would happen if we dared to stand still.
However, these problems are relatively minor and fixed easily. We can get the distance of our last tail piece and remove any segments further away than that. In addition we can add a small minimum distance required before we add a new point to the list resolving this possible memory leak as well.
Secondly, there are a few special cases that we did not cover: What is the list is empty or has only a single item? What is the recorded path is shorter than the distance assigned to our tail segment?
Again, these are relatively trivial to resolve, if we decide on how we want our snake to behave in these cases.
## Performance considerations
The real problem I have with the code posted above is its runtime: We loop over the entire recorded path for every single part of the tail.
In other words, our runtime is O(nm), with n and m being the numbers of segments and tail parts respectively.
Note that the work we do inside the loop is also all but trivial: To calculate the length of each segment we need to use a square root. This is a significantly slower operation than simple algebra.
Arguably, the runtime is linear in both parameters and in fact it will not be a major problem for short tails, and paths made from reasonably few segments – something which we can influence with the minimum segment length mentioned above.
However, what if we do want snakes dozens of pieces long, with an accuracy of hundreds of segments?
For that matter, what if we simply care a lot about the runtime of our algorithm?
Fortunately for either case, we can optimise.
If we consider our tail pieces in the order they appear on the tail – starting at the head – each piece is further away from the head than the previous one. That means, that no piece is on a path segment closer to the head than any previous one either.
Using this fact, we can combine the loop over the line segments with that over the tail pieces. In pseudo code:
``````segment = first path segment
foreach tailPiece
{
segment = advance to segment that tailPiece is on
tailPiece.position = interpolate on segment
}
``````
The actual code is somewhat more complicated, especially if we include the edge cases and other behaviour mentioned above. Ignoring the treatment of a path too short however, we might get something like the following:
``````void followHead(List<TailPart> tail, LinkedList<Vector2> path)
{
// assuming at least one node in linked list
var segmentStart = path.First.Value;
var segmentEndNode = path.First.Next;
var lengthToSegmentEnd = 0f;
foreach (var part in tail)
{
var segmentEnd = Vector2.Zero;
var segmentDiff = Vector2.Zero;
var segmentLength = 0f;
var lengthToSegmentStart = lengthToSegmentEnd;
// advance to correct segment if needed
while (part.DistanceToParent > lengthToSegmentEnd)
{
if (segmentEndNode == null)
{
// path too short
// NullReferenceException inbound, if not handled
}
segmentEnd = segmentEndNode.Value;
segmentDiff = segmentEnd - segmentStart;
segmentLength = segmentDiff.Length;
lengthToEnd += segmentLength;
segmentEndNode = segmentEndNode.Next;
}
// interpolate position on segment
var distanceLeft = part.DistanceToParent - lengthToSegmentStart;
var percentageAlongSegment = distanceLeft / segmentLength;
part.position = segmentStart +
segmentDiff * percentageAlongSegment;
}
// cutting off unnecessary end of path
while (segmentEndNode != path.Last)
{
path.RemoveLast();
}
}
``````
Note that despite the nested loop our runtime has now changed to O(n+m) – significantly better.
The only other easy optimisation to make would be precalculating the lengths of each segment when recording them. The rest of the algorithm is about as close to optimal as we are going to get.
A variation on the above would be switching the order of the loops: The outer loop would iterate over the path, placing tail pieces using the inner loop.
While the two might perform different in a production case – depending on a variety of factors – they are equivalent in terms of their runtime.
## Conclusion
In this post we started with a naive implementation of our desired behaviour – quickly finding its flaws. By analyzing the same, we clarified our requirements, and developed a new solution which we iterated upon until reaching an optimal one.
I hope this approach has been interesting to you. Feel free to leave a comment and let me know what you think!
Also make sure to let me know if you use this code, or the presented algorithm in your own game/application. It is always great to know to have helped out.
Enjoy the pixels!
1. SaphireS says:
Very interesting, thanks :)
2. nanoexpertS says:
Where is the followDistance in the last piece of code? Is it not needed anymore?
• Paul Scharf says:
Hey!
You’re right, the `followDistance` variable is no longer used in the last pieces of code. Instead each part knows what distance it is supposed to keep from the head-piece, which is stored in `distanceToParent`.
Hope that makes it clearer! | 2,339 | 11,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2019-30 | latest | en | 0.945583 |
https://studylib.net/doc/9579755/ideal-gas | 1,638,165,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358688.35/warc/CC-MAIN-20211129044311-20211129074311-00032.warc.gz | 586,082,188 | 19,986 | Ideal gas
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```ERT 206/4
Thermodynamics
CHAPTER 2
The First Law and Other Basic Concepts
Miss. Rahimah Bt. Othman
Email: rahimah@unimap.edu.my
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
Define, discuss, apply and analyze internal energy, first law,
energy balance-closed system, thermodynamic state and state
function, equilibrium, the Phase Rule, reversible process,
constant-V and constant-P processes, enthalpy and heat
capacity.
3. Chapter 3: Volumetric properties of pure fluids
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
System
Closed system
- Known as controlled mass
system (mass constant).
- Allow energy flow (in/out)
and volume is not
necessarily constant as a
controlled.
- Often undergo processes
during the internal energy
of the system changes.
∆(Energy of the system) +
(∆Energy of surroundings) = 0
∆(Energy of the system) = Q+W
∆(Energy of the system)
=∆U+∆EK+ ∆EP
Open system
- Known as controlled
volume system.
- Allow mass and energy
cross through the
boundary of open system
(= controlled surface).
Total Energy
E = U+ KE+ PE
Microscopic Energy
Internal energy, U
- The internal energy of a system
comes from the kinetic energy of
the molecules motion and potential
energy from the atomic bonds.
- Eg: chemical energy, nuclear, sensible
energy, latent energy.
Makroscopic energy
- Kinetic energy, KE
- Potential energy, PE
Thermodynamic
state and State
functions
The phase rule
(F = 2 – π +N)
The
reversible
process
Heat Capacity
At constant Volume,
Cv and Constant
Pressure, CP
CLOSED
SYSTEM
Equilibrium
(Eg: heat, mechanical,
phase, chemical)
Entalphy
(H = U +PV)
Constant-V &
Constant-P
Processes
dmCV
(m) fs 0
dt
(2.25)
(m) fs m 3 m1 m 2
dmCV
( uA) fs 0
dt
H
(2.26)
u 2
g
z Q Ws
2 gc gc
(2.32a)
Energy Balance
at Steady-state
Flow Processes
Mass
Balance
OPEN
SYSTEM
Energy
Balance
Mass Balance
at Steady-state
Flow Processes
m
u2 A2 u1 A1 u A
V1
V1
V
d (mU ) cv
1 2
[(U u zg ) m] fs Q work rate
dt
2
d
(mU )CV
1
H u 2 zg m Q W
dt
2
fs
(2.28)
(2.27)
Example 2.9 (HEAT CAPACITY)
Air at 1 bar and 25oC (298.15 K) is compressed to 5 bar and 25oC (298.15 K)
by two different mechanically reversible processes:
(a) Cooling at constant pressure followed by heating at constant volume
(b) Heating at constant volume followed by cooling at constant pressure
Calculate Q and W and ∆U and ∆H of the air for each path.
Given ; Cv = 20.78 and CP = 29.10 Jmol-1K-1
Assume ;
- For air, PV/T is a constant
- At 25oC (298.15 K) and 1 bar the molar volume of air is 0.02479 m3mol-1.
Solution
In each case take the system as 1 mol of air contained in an imaginary
piston/cylinder arrangement. Because the process considered are
mechanically reversible, the piston is imagined to move in the cylinder
without friction. The final volume is;
V2 V1
P1
1
0.02479 0.004958 m3
P2
5
Example 2.9-cont’
(a) During the first step the air is cooled at the constant pressure of 1 bar until
the final volume of 0.004958 m3 is reached. The temperature of the air at
the end of this cooling step is;
Whence;
V2
0.004958
T ' T1 298.15
59.63 K
V1
0.02479
Q H C P T (29.10)(59.63 298.15) 6941 J
U H ( PV ) H PV
6941 (1 x 105 )(0.004958 0.02479) 4958 J
During the second step the volume is held constant at V2 while the air is
heated to its final state. By Eq. (2.19),
U Q CV T (20.78)(298.15 59.63) 4958 J
The complete process represents the sum of its steps. Hence,
Q 6941 4958 1983 J
And
U 4958 4958 0
Example 2.9-cont’
Because the first law applies to the entire process, H Q W , and
therefore,
whence;
0 1983 W
W 1983 J
Equation (2.15), H U PV , also applies to the entire process. But
T1 T2 , and therefore, P1V1 P2V2 . Hence, PV 0 and
H U 0
(b) Two different steps are used in this case to reach the same final state of
the air. In the first step the air is heated at a constant volume equal to its
initial value until the final pressure of 5 bar is reached. The air
temperature at the end of this step is;
P2
5
T ' T1 298.15 1490.75K
P1
1
For this step the volume is constant, and
Q U CV T (20.78)(1490.75 298.15) 24788 J
Example 2.9-cont’
In the second step the air is cooled at P = 5bar to its final state:
Q H C P T (29.10)( 298.15 1490.75) 34703 J
ΔU H PV H PV
34703 (5 x 105 )(0.004958 0.02479) 24788 J
For the two steps combined,
Q 24788 34703 9915 J
ΔU 24788 24788 0
W
and as before
U Q 0 9915 9915 J
H U 0
Conclusion:
The property changes ∆U and ∆H calculated for the given change in
state are the same for both paths. On the other hand the answers to
parts (a) and (b) show that Q and W depend on the path.
Example 2.11 (MASS ENERGY FOR OPEN SYSTEMS)
Liquid n-hexane flows at a rate of ṁ = 0.75 kg s-1 in a pipe with inside
diameter D = 5 cm. What are q, ṅ, and u? What would these quantities be for
the same ṁ if D = 2 cm? Assume for liquid n-hexane that ρ = 659 kg m-3.
Solution
We have q m
1
whence
M 1
n m
and
0.75 kg s 1
3 1
q
0.00114
m
s
3
659 kg m
0.75 kg s 10
n
1
g kg 1
1
8.703
mol
s
86.177 g mol 1
3
Given ṁ, these quantities are independent of D. The velocity, however,
depends on diameter through u = qA-1, where, for a circular cross-section,
A = (π/4)D2. For D = 5 cm,
2
A
whence
5 x 10
4
2
m 0.00196 m 2
0.00114 m 3s 1
1
u
0.582
m
s
0.00196 m 2
Example 2.11-cont’
Similarly, for D = 2 cm,
A 0.000314 m
2
and
0.00114
u
3.63 m s 1
0.000314
Example 2.14 (ENERGY BALANCE FOR OPEN SYSTEM)
An insulated, electrically heated tank for hot tank for hot water contains
190 kg of liquid water at 60 oC (333.15 K) when a power outage occurs. If
water is withdrawn from the tank at a steady state of ṁ = 0.2 kg s-1, how
long will it take for the temperature of the water in the tank to drop from 60 to
35 oC (333.15 K to 308.15 K)?
Assume:
(a) Cold water enters the tank at 10 oC (283.15 K)
(b) Heat losses from the tank is negligible.
For liquid water let CV = CP = C, independent of T and P.
Solution
Here, Q W 0 . Additional assumption;
(a) Perfect mixing of the contents of the tank.
(b) m(flow rate in) = m(flow rate out) = mcv = constant.
(c) Differences between inlet and outlet kinetic and potential energy can
be neglected.
Example 2.14-cont’
Equation (2.29) is therefore written as;
dU
dT
C
dt
dt
and
H H1 C (T T1 )
Where unsubscripted quantities refer to the contents of the tank and H1 is
the specific enthalpy of the water entering the tank. With CV = CP = C,
dU
m
m ( H H1 ) 0
dt
The energy balance then becomes, on arrangement,
dt
m dT
m T T1
Integration from t = 0 (where T = T0) to arbitrary time t yields
m T T1
t ln
m T 0T1
Example 2.14-cont’
Substitution of numerical values into this equation gives, for the conditions of
this problem,
t
190 308.15 283.15
ln
658.5 s
0.2 333.15 283.15
Thus, it takes about 11 minutes for the water temperature in the tank to drop
from 60 to 35 oC (333.15 K to 308.15 K).
Example 2.16
(ENERGY BALANCE FOR OPEN SYSTEM)
Air at 1 bar and 25oC (298.15 K) enters a compressor at low velocity,
discharges at 3 bar, and enters a nozzle in which it expands to a final velocity
of 600 ms-1 at the initial conditions of pressure and temperature. If the work of
compression is 240 kJ per kilogram of air, how much heat must be removed
during compression?
Solution
Because the air returns to its initial conditions of T and P, the overall
process produces no change in enthalpy of the air. Moreover, the potentialenergy change of the air is presumed negligible. Neglecting also the initial
kinetic energy of the air, we write Eq. (2.32a) as;
2
u
Q 2 Ws
2
Example 2.16-cont’
The kinetic-energy term is evaluated as follows:
2
1 2 1
m
m2
u2 600 180 000 2
2
2
s
s
m 2 kg
180 000 2 . 180 000 N m kg 1 180 kJ kg 1
s kg
Then
Q 180 240 60 kJ kg 1
Heat in the amount of 60 kJ must be removed per kilogram of air
compressed.
Example 2.17
(ENERGY BALANCE FOR STEADY-STATE FLOW PROCESSES OPEN SYSTEM)
Water 93.5oC (366.65 K) is pumped from a storage tank at the rate of 3.15 x
10-3 m3 s-1. The motor for the pump supplies work at the rate of 1.5 kW. The
water goes through a heat exchanger , giving up heat at the rate of 700 kW,
and is delivered to a second storage tank at an elevation 15 m above the first
tank. What is the temperature of the water delivered to the second tank?
Solution
This is a steady-state flow process for Eq. (2.3b) applies. The initial and final
velocities of water in the storage tank are negligible, and the term ∆u2 / 2gc
may be omitted. The remaining terms are expressed in units of kJ kg-1
through use of appropriate conversion factors. At 366.65 K the density of
water is 958 kg m-3 thus the mass flow rate is;
(3.15)(10-3)(958) = 3.018 kJ kg-1
from which we obtain
Q = - 700/3.018 = - 231.9 kJ kg-1
Example 2.17-cont’
Since 1 kW is equivalent to 1 kJ s-1, the shaft work is;
Ws = (1.5)(1) /3.018 = 0.497 kJ kg-1
If the local acceleration of gravity is taken as the standard value of 9.807 ms-2,
the potential-energy term becomes;
9.807 159.807 0.147 kJ kg 1
g
z
9.807 1000
gc
Equation (2.32b) now yields ∆H:
g
H Q Ws z 231.9 0.497 0.147 231.55 kJ kg 1
gc
The enthalpy of saturated water at 93.5 oC (366.65 K) is given in the steam
tables as 391.6 kJ kg-1. Thus
H H 2 H1 H 2 391.6 231.55
and
H 2 391.6 231.55 160.05 kJ kg 1
Example 2.17-cont’
The temperature of water having this enthalpy is found from the steam tables
to be;
t 38.2 o C
or
T 311.35 K
In this example Ws and (g/gc)∆z are small compared with Q, and for practical
purposes they could be neglected.
ERT 206/4
Thermodynamics
CHAPTER 3
Volumetric Properties of Pure Fluids
Miss. Rahimah Bt. Othman
Email: rahimah@unimap.edu.my
COURSE OUTCOME 1 CO1)
1. Chapter 1: Introduction to Thermodynamics
2. Chapter 2: The First Law and Other Basic Concepts
2. Chapter 3: Volumetric properties of pure fluids
DESCRIBE and EXPLAIN PVT behavior of pure substances,
Virial Equation of State, ideal gas, Virial EquationAPPLICATION, cubic equation of state, generalized correlations
for gases and liquids.
4. Chapter 4: Heat effects
5. Chapter 5: Second law of thermodynamics
6. Chapter 6: Thermodynamics properties of fluids
Objectives
• Introduce the concept of a pure substance.
• Discuss the physics of phase-change processes.
• Illustrate the P-v, T-v, and P-T property diagrams and
P-v-T surfaces of pure substances.
• Demonstrate the procedures for determining
thermodynamic properties of pure substances from
tables of property data.
• Describe the hypothetical substance “ideal gas” and the
ideal-gas equation of state.
• Apply the ideal-gas equation of state in the solution of
typical problems.
• Introduce the compressibility factor, which accounts
for the deviation of real gases from ideal-gas behavior.
• Present some of the best-known equations of state.
Pure Fluids/ Substance
A substance that has a fixed
chemical composition throughout is
called a Pure Substance.
Pure Substance:
- N2, O2, gaseous Air
-A mixture of liquid and gaseous
water is a pure substance, but a
mixture of liquid and gaseous air is
not.
PHASES OF A PURE SUBSTANCE
The molecules
in a SOLID are
kept at their
positions by
the large
spring like
inter-molecular
forces.
In a solid, the
attractive and
repulsive forces
between the
molecules tend to
maintain them at
relatively constant
distances from each
other.
The arrangement of atoms in different phases: (a) molecules are at relatively
fixed positions in a solid, (b) groups of molecules move about each other in the
LIQUID phase, and (c) molecules move about at random in the GAS phase.
Phase-Change Processes of Pure
Substance
Compressed liquid or a subcooled liquid:
A liquid that is not about to vaporize.
Saturated liquid: A liquid that is about
to vaporize.
Saturated vapor: A vapor that is about
to condense.
Saturated liquid-vapor mixture: the
liquid and vapor phases coexist in
equilibrium.
Superheated vapor: A vapor that is not
about to condense
T-v diagram for the heating process of water at
constant pressure.
2-1
Phase-Change Processes of Pure
Substance
Saturated temperature, Tsat: At a given
pressure, the temperature at which a pure
substance changes phase.
Saturated pressure, Psat: At a given
temperature, the pressure at which a pure
substance changes phase.
Latent heat: the amount of energy absorbed or
released during a phase-change process.
Latent heat of fusion: the amount of energy
absorbed during melting.
Latent heat of vaporization: the amount of
energy absorbed during vaporization.
PVT BEHAVIOR OF PURE SUBSTANCES/ FLUIDS
Critical point – highest
combination of pressure
and temperature where the
fluid exist in liq-vap
equilibrium
The 2-C line, also known as
vaporization curve is where
liquid-vapor is in equilibrium
The 1-2 line, also
known as sublimation
curve is where solidvapor is in
equilibrium
Triple point, three
phases exist in
equilibrium (F=0)
The 2-3 line, also known as
fusion curve is where solidliquid is in equilibrium
PV diagram
Boundaries in PT diagram becomes
region when illustrate with PV
diagram
Critical point becomes peak of the
curve
Triple point becomes horizontal line
Compressed liquid region
Saturated liquid line at boiling
temperature
Superheated vapor region
Saturated vapor line at condensation
temperature
T >Tc, the line do not cross the
boundary
Isotherms in sub-cooled/ compressed lliquid
region are steep, because liquid volumes
change little with large changes in pressure
SINGLE PHASE REGION
At single phase regions in PV diagram, there is a relation
connecting P,V and T.
This relation known as PVT equation of state; f(P,V,T)=0
If V is considered as a function of T and P, then V=V(T,P)
V
V
dV
dT
dP
T P
T T
1 V
Volume exp ansivity
V T P
1 V
V T T
isothermal compressib ility
The combination will yield;
dV
dT dP
V
3.4
Because the isotherms on the left side of PV diagram are very steep,
both β and κ are small.
Because of that, the liquid is known as incompressible fluid, where
both constants are equal to zero
However, this is just idealization, and in incompressible fluid, no
equation of state exist, since V is independent of T and P
If, we still want to calculate, for liquids, β is positive, and κ is
positive as well. Integration of 3.4 yield
ln
V2
(T2 T1 ) ( P2 P1 )
V1
Try examples 3.1
3.5
Virial Equation of State
The coefficients a(T), b(T), c(T), and so on, that are
functions of temperature alone are called virial
coefficients.
Ideal-Gas Temperature; Universal Gas Constant
Figure 3.4: Plot of PV vs. P for 4 gaseous at
triple-point temperature of water. The limiting
value of PV as P0 is the same for all of the
gaseous.
Limiting value (asterisk)
PV* = a = f(T)
This properties of gaseous is the basis for
establishing an absolute temperature scale.
The simplest procedure to define Kelvin scale:
1. (PV)* = a ≡ RT
2. (PV)*t = R x 273.16K(3.8)
3. PV * T / K
PV *
t
273.16 K
*
PV
T / K 273.16
PV *t
(3.9)
•In the limit, P0, molecules separated by finite distance
•Volumes becomes negligible compare with the total volume
of the gas, and intermolecular forces approach zero
• These condition define an IDEAL GAS state & Eq. 3.9
establishes
the ideal-gas temperature scale
• The proportionality constant R = universal gas constant
Through the use of conversion factors, R may be expresses in various units.
Commonly used values are given as above table.
2 Forms of the Virial
Equation
• Auxiliary thermodynamic property =
• = Compressibility factor
•
•
•
•
•
•
Z = 1 + B’P+C’P2 +D’P3 +…
Z= 1 + B/V +C/V2 + D/V3 +…
Virial expansion = Eq 3.11 & Eq 3.12
B’, C’, D’ …, B, C, D = virial coefficient
B’ = 2nd virial coefficient
C’ = 3rd virial coefficient…etc
PV
Z
RT
(3.10)
(3.11)
(3.12)
Eq 3.11 & Eq 3.12
B
B'
(3.13a )
RT
C B2
C'
( RT ) 2
(3.13b)
D 3BC 2 B 3
D'
( RT )3
(3.13c)
THE IDEAL GAS
• Because the Eq 3.12 arise on account of molecular
interactions, the virial coefficients B,C…etc = 0
were no such interaction to exist
Z 1 or
Z 1 or
PV RT
PV RT
• Phase Rule = Internal energy of a real gas is a
function of pressure as well as of temperature.
THE IDEAL GAS
• Equation of state: Any equation that
relates the pressure, temperature, and
specific volume of a substance.
• The simplest and best-known equation of
state for substances in the gas phase is the
ideal-gas equation of state. This equation
predicts the P-v-T behavior of a gas quite
accurately within some properly selected
region.
Ideal gas
equation of
state
R: gas constant
M: molar mass (kg/kmol)
Ru: universal gas constant
U = U(T) (Ideal gas)
Different substances have
(3.15) different gas constants.
Implied Property Relations for
an Ideal Gas (f(T) only)
• Heat Capacity for constant volume, Cv
U dU (T )
Cv
Cv(T )
dT
T v
(3.1)
• Eq 2.11 applied to an Ideal Gas
H U PV U (T ) RT H (T )
(3.17)
Implied Property Relations
for an Ideal Gas (f(T) only)
Heat Capacity for constant pressure, Cp
dH (T )
H
CP
CP
dT
T P
(3.18)
Useful relation between Cv & Cp
CP
dH dU
R CV R
dT dT
(3.19)
NOTE: This equation does not imply that Cp and Cv are themselves
constant for an ideal gas, but only that they vary with temperature in such a
way that their differences is equal to R
For any ∆ of state of an ideal gas, Eq 3.16 and Eq.3.18 lead to:
dU CV dT
(3.20a )
U CV dT
(3.20b)
dH C P dT
(3.21a )
H C P dT
(3.21b)
• T1 & T2
• a---b = Constant volume process
• a---c & a---d ≠ constant volume
The graph show that the ∆U= CV dT
Equation for Process
Calculation for Ideal Gas
Working equation of dQ and dW depend on which pair of these variables
is selected as independent
With P=RT/V,
dV
V
dP
dQ C P dT RT
P
dQ CV dT RT
dV
V
(3.22)
dW RT
(3.24)
dW RdT RT
(3.23)
dP
P
(3.25)
With V=RT/P and Cv given by Eq 3.19, dQ & dW written as Eq.3.24 & Eq.3.25
With T=PV/R, the work is simply dW=-PdV, and with Cv given by Eq.3.19,
CV
CP
dQ
VdP
PdV
R
R
(3.26)
Isothermal Process
• Q=-W
V2
P2
Q W RT ln
R ln
V1
P1
(const T ) (3.27)
Isobaric Process
Q H CP dT
(Const P) (3.28)
Isochoric Process
Q U CV dT
(Const V ) (3.29)
Adiabatic Process;
Constant Heat Capacities
dT
R dV
T
CV V
C P / CV
V1
dT
1
RdV
T2
T CV V T1 V2
similar to Eq.3.24 & Eq.3.26 lead to :
T2
T1
P2
P
1
R / CP
V1
P2
V
P1
2
C P / CV
Also,
TV 1 const. (3.30a)
PV const (3.30c)
TP (1 ) / const. (3.30b)
CP
CV
(3.31)
IMPORTANT: Eq 3.30 are restricted in application to ideal gases with
Constant Heat capacities undergoing mechanically reversible adiabatic
expansion or compression
For ideal gas, the WORK of any adiabatic closed-system process is given
by:
dW dU CV dT
W U CV T
(3.32)
Because RT1 = P1V1 and RT2 = P2V2,
W
RT2 RT1 P2V2 P1V1
1
1
(3.33)
Elimination of V2 from Eq 3.33 by Eq 3.30c, valid for
mechanically reversible process, lead to:
P1V1 P2
W
1 P1
1 /
RT P
1
2
1
1 P1
1 /
1
Polytropic Process
PV const
(3.35a )
TV 1 const
(3.35b)
TP (1 ) / const
RT1 P2
W
1 P1
(3.35c )
( 1) /
( ) RT1 P2
Q
( 1)( 1) P1
1
( 1) /
(3.36)
1
(3.37)
(3.34)
Polytropic Process
Isobaric process
Isothermal process
Adiabatic process
Isochoric process
APPLICATION OF
VIRIAL EQ.
All isotherms originate at Z=1 for P=0
PV
BP
Z
1
RT
RT
(3.38)
Applied to vapors at sub critical temperature up to their saturation
pressure
When the virial equation is truncated to 3 terms, the appropriate
form is:
PV
B C
Z
1 2
RT
V V
(3.40)
CUBIC EQ. OF STATE
1. The Van der Waals Equation of state
RT
a
P
2
V b V
(3.41)
2. A Generic Cubic Equation of State
RT
a(T )
P
V b V b (V b)
(3.42)
3. Determination of Eq-of State Parameters
P
0 (3.43)
V
T ;cr
2P
2 0 (3.44)
V T ;cr
4. Theorem of Corresponding State: Acentric Factor
1.0 log( Prsat )Tr 0.7
(3.48)
NOTE: All fluid having the same value of , when compared at the same Tr
And Pr have about the same value of Z, and all deviate from ideal gas behavior
To about he same degree
5. Vapor& Vapor-like Root of the Generic Cubic Eq of state
6. Liquid & Liquid-like Root of the Generic Cubic Eq of state
GENERALIZE
CORRELATION FOR GASES
GENERALIZE
CORRELATION FOR GASES
1. Pitzer Correlation for the
Compressibility Factor
2. Pitzer Correlation for the 2nd Virial
Coefficient
3. Correlations for the 3rd Virial
Coefficient
4. Condition of Approximate Vlidity of
the Ideal-Gas Equation
GENERALIZE CORRELATION
FOR LIQUID
Summary
• Pure substance
• Phases of a pure substance
• Phase-change processes of pure substances
– Compressed liquid, Saturated liquid, Saturated
vapor, Superheated vapor
– Saturation temperature and Saturation pressure
• Property diagrams for phase change processes
– The T-v diagram, The P-v diagram, The P-T
diagram, The P-v-T surface
• The ideal gas equation of state
– Is water vapor an ideal gas?
• Compressibility factor
• Other equations of state
Thank you
Prepared by,
MISS RAHIMAH OTHMAN
``` | 8,788 | 21,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-49 | latest | en | 0.767937 |
https://www.cip.ifi.lmu.de/~grinberg/algebra/gr-errata1.txt | 1,603,442,098,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-45/segments/1603107880878.30/warc/CC-MAIN-20201023073305-20201023103305-00559.warc.gz | 668,970,808 | 1,623 | A list of errata in the paper "A bijection between words and multisets of necklaces" by I. Gessel, A. Restivo, and C. Reutenauer ( http://people.brandeis.edu/~gessel/homepage/papers/necklace.pdf ) - page 1: "is hat the" -> "is that the". - page 2: "rows.Then" -> "rows. Then". - page 3: The 8-step proof of the bijectivity of $\Phi$ is somewhat broken. First of all, the "$\tau^i()$" in step 6 should be "$\tau^i(q)$". But more importantly, step 4 doesn't work as stated: we have $w_p^{\infty} \leq w_q^{\infty}$ rather than $w_p^{\infty} < w_q^{\infty}$ (there is no reason why a strict inequality should hold; we're in a multiset), and it's not a-priori clear that this uniquely determines the ordering of the $w_1, w_2, \ldots, w_n$ because the relation "$u^{\infty} \leq v^{\infty}$" between two words $u$ and $v$ is not a partial order (but merely a pre-order). Here is how I would fix this: The 8 steps need to be read in a different order. First read steps 1, 2, 3, 6 and 7 (the "$\tau^i()$" in step 6 should be "$\tau^i(q)$"). Then, read step 4 with the following caveat: you don't have $w_p^\infty < w_q^\infty$, but only $w_p^\infty \leq w_q^\infty$. As I said, the relation $u^\infty \leq v^\infty$ between two words $u$ and $v$ is not a partial order (just a total pre-order), so we cannot conclude from this that the $p$-th row of the tableau is $w_p$. However, $w_p$ and $w_q$ are primitive (as proven in step 7), and the relation $u^\infty \leq v^\infty$ between two *primitive* words $u$ and $v$ is a partial order (this isn't hard to check; it follows from the fact that two primitive words $u$ and $v$ satisfying $u^n = v^m$ for positive $n$ and $m$ must necessarily be equal). So we can conclude the claim of step 4 after all. Finally read steps 5 and 8. Note that this error persists in the published version of the paper ( https://doi.org/10.1016/j.ejc.2012.03.016 ). - page 4: "witht" -> "with". - page 4: "the define the inverse" -> "define the inverse". | 639 | 1,977 | {"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} | 3.609375 | 4 | CC-MAIN-2020-45 | latest | en | 0.895879 |
http://mathhelpforum.com/calculus/30169-help-improper-integrals.html | 1,481,340,979,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542938.92/warc/CC-MAIN-20161202170902-00325-ip-10-31-129-80.ec2.internal.warc.gz | 173,645,607 | 9,596 | # Thread: Help with improper integrals
1. ## Help with improper integrals
the integral from neg. infinity to 6 of re^(r/3)
thanks!
2. $\int_{-\infty}^6re^{\frac{r}{3}}dr=\lim_{a\to-\infty}\int_a^6rd^{\frac{r}{3}}=$
$\lim_{a\to-\infty}3re^{\frac{r}{3}}-9e^{\frac{r}{3}}|_{a}^{6}$
I used integration by parts | 127 | 312 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.510853 |
https://physics.stackexchange.com/questions/299972/is-the-force-due-to-buoyancy-a-conservative-force?noredirect=1 | 1,725,730,419,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00251.warc.gz | 436,526,697 | 42,062 | # Is the force due to buoyancy a conservative force?
In order to decide whether buoyancy is a conservative force, let us study do two thought experiments.
1. For objects with density less than the density of fluid, no matter how deep we place the object below the surface of the fluid, it comes up with some kinetic energy. This indicates that the force due to buoyancy is a conservative force with some potential energy associated with it.
2. However, for objects with density more than the density of fluid, no matter how deep we place the object below the surface of the fluid, it never comes up. This indicates that the force due to bouyancy is a non-conservative force.
Which of these two conclusions is correct?
• I was ready to offer a bounty , unfortunately my my reputation is just 1 less than 75( the privilege to offer bounty) :) Commented Dec 21, 2016 at 5:44
• You have just posted the query; wait for at least two days before thinking about bounty.
– user36790
Commented Dec 21, 2016 at 5:45
• Oh no, another "thoughtful" username... when will it end? :) Commented Dec 21, 2016 at 6:26
• @ConfusinglyCuriousTheThird Whose name mine or MAFIA36790 ? Commented Dec 21, 2016 at 6:28
• yours! There are a few (frequenting) users here with names close enough to cause confusion... Commented Dec 21, 2016 at 6:29
## 1 Answer
For an object that is fully submerged the buoyant force is constant (in both magnitude and direction) exactly as the usual "near the surface of the Earth" approximation to gravity. That part is clearly conservative.
To figure the behavior of a object while partly submerged we much rely on Archimedes' principle, which tells us that the buoyant force is proportional to the weight of the volume then displaced. The act of submerging the body may require a different distance depending on the shape and orientation of the body, so we can be suspicious.
Take for a concrete example a non-spherical, ellipsoidal object. By symmetry and construction it will require the same average force to insert it into the fluid at any angle, but that force will be applied over a different distance depending on orientation. The act of submerging the object itself requires different amounts of work depending on the orientation of the object.
But before we say "Ah! Ha!" we need to consider that the starting and ending conditions for putting it end the long way are separated by a larger total distance than those for putting it in the short way. To know if the force is conservative over those paths we have to make the starting and ending points the same. Allow the starting point of the COM to be a distance $a$ over the fluid surface and the ending point to be a distance $a$ under the surface (where $a$ is the semi-major axis of the ellipse in some cross-section). The work done by gravity plus buoyancy on a vertical path between those points is $$W_a = 2a\left(\rho_o - \frac{1}{2} \rho_f \right)V_o g\;,$$ where $V_o$ is the volume of the object and the densities $\rho_{o,f}$ are for the object and the fluid respectively. Compare to lowering the object oriented so that the minor axis $2b$ is vertical. The work done is \begin{align*} W_b &= \left[(a-b) \rho_o V_o g\right] + \left[2b\left(\rho_o - \frac{1}{2} \rho_f\right)V_o g \right]+ \left[(a-b) \left(\rho_o - \rho_f\right)g V_o\right] \\ &= \left[ \left((a-b) \rho_o\right) + \left(2b\left(\rho_o - \frac{1}{2} \rho_f\right) \right)+ \left((a-b) \left(\rho_o - \rho_f\right)\right)\right]V_o g \\ &= \left[ \left[ (a-b) + 2b + (a-b) \right] \rho_o - \left(b + (a-b)\right)\rho_f\right]V_o g \\ &= \left[ 2a\rho_o - a\rho_f\right]V_o g \\ &= 2a\left( \rho_o - \frac{1}{2}\rho_f\right)V_o g \;, \end{align*} exactly as before.
This shows that at least for sufficiently symmetric objects and considering the combination of gravity and buoyancy even the dunking process is conservative. Explicit extension to arbitrary shapes requires that the summation done here be performed as an integral with careful attention to the endpoints and the completely out of water and completely underwater parts of the movement, but one might also appeal to intuition about having to move the same amount of fluid upward to achieve submersion.
• Now that buoyant force is conservative for at least moderately symmetrical bodies , can we define a potential energy such that Work done by Buoyant force = - ( Change in potential energy) Commented Dec 21, 2016 at 6:26
• The question isn't "Can you define it?" but "Will it be sufficiently worthwhile to bother?", and because almost all fluid environments are heavily influenced by dissipative forces most people will probably answer "No." most of the time. The two important lessons here are the argument by analogy and the critical importance of paying attention to starting and ending points when asking about conservativeness. Commented Dec 21, 2016 at 6:32
• @dmckee nice answer. Ignoring the dissipative forces, and assuming a rigid object (e.g. for balloons this is not conservative), it seems like there should be stronger statements one could make. Something about smooth mapping of the buoyant force to the objects position and orientation or something. Commented Dec 21, 2016 at 6:41
• @ConfusinglyCuriousTheThird I feel that you are right, but I am too tired tonight for anything but the kind of straight ahead grind you see here. Commented Dec 21, 2016 at 7:03 | 1,349 | 5,397 | {"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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.935737 |
https://www.mapleprimes.com/questions/130141-Hyper-Geometric-Functions | 1,726,505,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00385.warc.gz | 817,240,057 | 23,801 | Question:Hyper geometric functions
Question:Hyper geometric functions
`Hi, I have and integral its results contains Beta function and hypergeometric function.I found a difficulty to program the hypergeometric function with Maple 13. is any one can solve this problemwhere `
∫₀^{(π/2)}((sin^{α}xcos^{β}x)/(√(1-k²sin²x)))dx=(1/2)B(((α+1)/2),((β+1)/2))F(((α+1)/2),(1/2);((α+β+2)/2);k²),α>-1,β>-1,∣k∣<1
The problem in written the following function F(((α+1)/2), (1/2); ((α+β+2)/2); k²)
` `
| 187 | 503 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.637456 |
https://brainmass.com/math/calculus-and-analysis/series-calculus-questions-22619 | 1,701,425,963,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00002.warc.gz | 172,877,440 | 7,237 | Purchase Solution
# Series of Calculus Questions
Not what you're looking for?
Please respond with a Microsoft Word document with the answers written in standard text. Thank you.
Series of Various Calculour Questions Attached.
You do not need to show your work for this one because I would simply like to compare your answers with mine so that I am sure that I did everything correct on mine. Please just write your answer after each number. I will know which problems I will have to study in detail with your help. Thank you very much.
You gave me this kind of format last time:
(12)
12. f'(x) = 1/(1-x)^2
Thanks!
1. Find all the critical numbers: .
2. Find all the extrema in the interval [0, ] if y = x + sin x.
3. Find the absolute maximum and absolute minimum of f on the interval (-1,2].
4. Consider . Find all values, c, in the interval [0,1] such that the slope of the tangent line to the graph of f at c is parallel to the secant line through the points (0,f(0)) and (1,f(1)).
5. Find all open intervals on which is decreasing.
6. Find the relative extrema of , including the designation of maximum or minimum.
7. A differentiable function f has only one critical number: x = -3. Identify the relative extrema of f at (-3,f(-3)) if f'(-4) = ½ and f'(-2) = -1.
8. Find all the intervals on which the graph of the function is concave upward: .
9. Find all points of inflection:
10. Find all horizontal asymptotes: .
11. Find
12. Calculate 3 iterations of Newton's Method to approximate the real zero of . Use as the initial guess and round to 4 decimal places after each iteration.
##### Solution Summary
The series of a calculus question is found.
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Understanding of how geometry applies to in real-world contexts
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In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
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Some questions on probability
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form. | 546 | 2,404 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.899115 |
https://www.reddit.com/user/serrated_pickle | 1,498,487,076,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320763.95/warc/CC-MAIN-20170626133830-20170626153830-00648.warc.gz | 910,336,939 | 22,302 | [–] 0 points1 point (0 children)
COMMENT!
[–] 21 points22 points (0 children)
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8760 hours
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One important factor that I would like to point out is herd immunity. Getting vaccinated does not only reduce the risk of getting sick yourself, but you also will help the reduce the chances of other getting sick as well. This is especially important for individuals that cannot receive the vaccine themselves.
[–] 0 points1 point (0 children)
515
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Crabman
[–] -1 points0 points (0 children)
Nice pic! Saw the same thing from my Shuksan summit.
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Upvote for Bell's!
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This would be a great pickup!
[–] 1 point2 points (0 children)
How can the net amount of entropy of the universe be massively decreased?
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AFRRIICAAA!
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Its called ANSYS, and yes I do know the math, and I have definitely done it before, but it has been a while since college so I don't have the formulas memorized any more. It is pretty easy for me to set this simple model up. It would take probably more time to look up the formulas and calculate the answer.
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I don't know about the math but my fancy analysis software says the total deformation is ~9mm. This makes a strain of .009m/213m = .004%.
http://imgur.com/uq1drKP
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I think you input the wrong Young's modulus. The problem says 200 GPA. Also I believe for the last question, you have to compare you final strain to the allowable strain of 3%.
[–] 3 points4 points (0 children)
I always imagined that the giant was strong enough to blast me into orbit.
[–] 0 points1 point (0 children)
I worked for Williams, they had some cool stuff. Imagine having a jet engine strapped to your back. Kind of scary if you think about it.
[–] 1 point2 points (0 children)
I run around that lake all the time! I was hoping to see myself in the video.
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Damn, I wish I got here sooner. One of my friends argued that he could use a fork and inflict as much damage to someone as if he had a knife. We called it the great fork stab debate.
[–] 0 points1 point (0 children)
Moved from Michigan.
I moved because my Fiance is from Washington and she wanted to move back to Seattle.
Seattle has an awesome combination of being able to do outdoor activities along with living in an interesting city. Get into hiking, climbing, backpacking, etc... you won't regret it.
You will find the drivers in seattle are very passive aggressive compared to the midwest and people will not get out of the left lane for anyone to pass. This is probably the worst part about Seattle.
[–] -1 points0 points (0 children)
These are the Dave's I know I know:
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Vaccines, and now my kids don't die.
[–] 0 points1 point (0 children)
Saw this thinking it was an article about military technology.
So my "chiropractor" gave me this today... by in WTF
[–] 2 points3 points (0 children)
As the SO of an epidemiologist, who currently helping with a measles outbreak because of the Anti-Vaccine movement, I say FUCK THIS PAMPHLET.
[–] 3 points4 points (0 children)
Literally, just said this out loud. | 922 | 3,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.931013 |
https://devsenv.com/example/-1106-leetcode-parsing-a-boolean-expression-solution-in-c,-c++,-java,-javascript,-python,-c-leetcode | 1,719,073,101,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00248.warc.gz | 167,249,426 | 52,222 | ## Algorithm
Problem Name: 1106. Parsing A Boolean Expression
A boolean expression is an expression that evaluates to either `true` or `false`. It can be in one of the following shapes:
• `'t'` that evaluates to `true`.
• `'f'` that evaluates to `false`.
• `'!(subExpr)'` that evaluates to the logical NOT of the inner expression `subExpr`.
• `'&(subExpr1, subExpr2, ..., subExprn)'` that evaluates to the logical AND of the inner expressions `subExpr1, subExpr2, ..., subExprn` where `n >= 1`.
• `'|(subExpr1, subExpr2, ..., subExprn)'` that evaluates to the logical OR of the inner expressions `subExpr1, subExpr2, ..., subExprn` where `n >= 1`.
Given a string `expression` that represents a boolean expression, return the evaluation of that expression.
It is guaranteed that the given expression is valid and follows the given rules.
Example 1:
```Input: expression = "&(|(f))"
Output: false
Explanation:
First, evaluate |(f) --> f. The expression is now "&(f)".
Then, evaluate &(f) --> f. The expression is now "f".
Finally, return false.
```
Example 2:
```Input: expression = "|(f,f,f,t)"
Output: true
Explanation: The evaluation of (false OR false OR false OR true) is true.
```
Example 3:
```Input: expression = "!(&(f,t))"
Output: true
Explanation:
First, evaluate &(f,t) --> (false AND true) --> false --> f. The expression is now "!(f)".
Then, evaluate !(f) --> NOT false --> true. We return true.
```
Constraints:
• `1 <= expression.length <= 2 * 104`
• expression[i] is one following characters: `'('`, `')'`, `'&'`, `'|'`, `'!'`, `'t'`, `'f'`, and `','`.
## Code Examples
### #1 Code Example with Java Programming
```Code - Java Programming```
``````
class Solution {
public boolean parseBoolExpr(String expression) {
char[] chars = expression.toCharArray();
Stack < Character> operations = new Stack<>();
Stack boolValues = new Stack<>();
for (int i = 0; i < chars.length; i++) {
char c = chars[i];
if (c == '!' || c == '&' || c == '|') {
operations.push(c);
}
else if (c == '(') {
boolValues.push('#');
}
else if (c == 't' || c == 'f') {
boolValues.push(chars[i]);
}
else if (c == ',') {
continue;
}
else {
List < Character> list = new ArrayList<>();
while (!boolValues.isEmpty()) {
char temp = boolValues.pop();
if (temp == '#') {
break;
}
}
boolValues.push(performOperation(list, operations.pop()));
}
}
return boolValues.peek() == 't' ? true : false;
}
private Character performOperation(List < Character> list, Character operation) {
if (operation == '|') {
return performOr(list);
}
else if (operation == '&') {
return performAnd(list);
}
else {
return list.get(0) == 't' ? 'f' : 't';
}
}
private Character performAnd(List < Character> list) {
boolean val = getBooleanValue(list.get(0));
for (int i = 1; i < list.size(); i++) {
val &= getBooleanValue(list.get(i));
}
return val ? 't' : 'f';
}
private Character performOr(List < Character> list) {
boolean val = getBooleanValue(list.get(0));
for (int i = 1; i < list.size(); i++) {
val |= getBooleanValue(list.get(i));
}
return val ? 't' : 'f';
}
private boolean getBooleanValue(Character character) {
return character == 't' ? true : false;
}
}
``````
Copy The Code &
Input
cmd
expression = "&(|(f))"
Output
cmd
false
### #2 Code Example with Javascript Programming
```Code - Javascript Programming```
``````
const parseBoolExpr = function(expression) {
const stack = []
for (let ch of expression) {
if (ch === '|' || ch === '&' || ch === '!') stack.push(ch)
if (ch === 't') stack.push(true)
if (ch === 'f') stack.push(false)
if (ch === ')') {
const tmp = []
while (stack.length) {
let t = stack.pop()
if (t === true || t === false) tmp.push(t)
else {
let res = tmp.pop()
if (t === '|') {
while (tmp.length) res = tmp.pop() || res
} else if (t === '&') {
while (tmp.length) res = tmp.pop() && res
} else if (t === '!') {
res = !res
}
stack.push(res)
break
}
}
}
}
return stack[0]
}
``````
Copy The Code &
Input
cmd
expression = "&(|(f))"
Output
cmd
false
### #3 Code Example with Python Programming
```Code - Python Programming```
``````
class Solution:
def parseBoolExpr(self, expression: str) -> bool:
stack = []
for c in expression:
if c == ')':
cache = []
while stack[-1] != '(':
cache.append(stack.pop())
stack.pop()
cur = stack.pop()
stack.append(all(cache) if cur == '&' else any(cache) if cur == '|' else not cache.pop())
elif c != ',':
stack.append(True if c == 't' else False if c == 'f' else c)
return stack.pop()
``````
Copy The Code &
Input
cmd
expression = "|(f,f,f,t)"
Output
cmd
true | 1,256 | 4,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.614862 |
http://ngss.nsta.org/DisplayStandard.aspx?view=pe&id=45 | 1,527,197,876,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866870.92/warc/CC-MAIN-20180524205512-20180524225512-00407.warc.gz | 208,341,544 | 12,398 | # Waves and Their Applications in Technologies for Information Transfer
### Students who demonstrate understanding can:
#### Performance Expectations
1. Use tools and materials to design and build a device that uses light or sound to solve the problem of communicating over a distance.
Clarification Statement and Assessment Boundary
A Peformance Expectation (PE) is what a student should be able to do to show mastery of a concept. Some PEs include a Clarification Statement and/or an Assessment Boundary. These can be found by clicking the PE for "More Info." By hovering over a PE, its corresponding pieces from the Science and Engineering Practices, Disciplinary Core Ideas, and Crosscutting Concepts will be highlighted.
### Connections to Engineering, Technology, and Applications of Science
By clicking on a specific Science and Engineering Practice, Disciplinary Core Idea, or Crosscutting Concept, you can find out more information on it. By hovering over one you can find its corresponding elements in the PEs.
## Planning Curriculum
### Common Core State Standards Connections
#### ELA/Literacy
• W.1.7 - Participate in shared research and writing projects (e.g., explore a number of “how-to” books on a given topic and use them to write a sequence of instructions). (1-PS4-4)
#### Mathematics
• 1.MD.A.1 - Order three objects by length; compare the lengths of two objects indirectly by using a third object. (1-PS4-4)
• 1.MD.A.2 - Express the length of an object as a whole number of length units, by layering multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. (1-PS4-4)
• MP.5 - Use appropriate tools strategically. (1-PS4-4)
## Resources & Lesson Plans
• More resources added each week!
A team of teacher curators is working to find, review, and vet online resources that support the standards. Check back often, as NSTA continues to add more targeted resources.
• In these lessons the students work as partners planning and designing a communication device that will signal across the gym or hallway from one partner to the other partner. The communication device must only use light and objects th ...
• This lesson is an activity where students make a model "landline" and investigate how sound travels.
• No community resources are available at this time. Do you have a great resource to share with the community? Click here.
Planning Curriculum gives connections to other areas of study for easier curriculum creation. | 575 | 2,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2018-22 | latest | en | 0.87783 |
https://plainmath.net/92860/diffraction-gratings-with-10-000-lines-p | 1,670,154,442,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710972.37/warc/CC-MAIN-20221204104311-20221204134311-00014.warc.gz | 484,731,284 | 12,138 | Diffraction gratings with 10,000 lines per centimeter are readily available.
Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away.Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm).
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Sanaa Hudson
Given information:
The number of lines on the grating,
The distance between the screen and slits,
The shortest wavelength,
The longest wavelength,
To find
the angles corresponding to first-order diffraction from these wavelengths.
The grating equation is given by
$d\mathrm{sin}\left(\theta \right)=n\lambda \phantom{\rule{0ex}{0ex}}\mathrm{sin}\left(\theta \right)=\frac{n\lambda }{d}\phantom{\rule{0ex}{0ex}}\theta ={\mathrm{sin}}^{-1}\frac{n\lambda }{d}$
where,
d=width of the slit
n=order of diffraction
$\lambda =$ wavelength of incident light
$\theta =$ angle of diffraction
The width of the slit is calculated as follows
Apply the above formula for the shortest wavelength
${\theta }_{\text{min}}={\mathrm{sin}}^{-1}\left(\frac{1×380×{10}^{-9}}{{10}^{-6}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(0.38\right)\phantom{\rule{0ex}{0ex}}={22.33}^{\circ }$
Now, apply the above formula for the longest wavelength
${\theta }_{\text{max}}={\mathrm{sin}}^{-1}\left(\frac{1×760×{10}^{-9}}{{10}^{-6}}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(0.76\right)\phantom{\rule{0ex}{0ex}}={49.46}^{\circ }$
The angle corresponding to the shortest wavelength is, ${\theta }_{\text{min}}={22.33}^{\circ }$
The angle corresponding to the longest wavelength is, ${\theta }_{\text{max}}={49.46}^{\circ }$ | 568 | 1,912 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-49 | latest | en | 0.774669 |
http://refrigeration.engineering/showthread.php/5964-How-to-figure-out-Strength-Test-Criteria-for-R404a-Refrigerant?s=81e7a537550f24db2703e10ecce9e07f&p=25251 | 1,601,192,086,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400265461.58/warc/CC-MAIN-20200927054550-20200927084550-00153.warc.gz | 106,602,876 | 13,181 | # Thread: How to figure out Strength Test Criteria for R404a Refrigerant?
1. ## How to figure out Strength Test Criteria for R404a Refrigerant?
Hi, I'am new to Refrigeration. I'am doing course which will allow me to work in the field as a Refrigeration Engineer. Good that website like that one exists. So I can ask someone about things I don't know.
My question is like in title of this topic.
How to count the Strength Test Pressure for split unit (Air/Con)?
2. Hi, Rob
Welcome to our Forum
3. Hi, Rob
First of all. You need to know in which geographic zone you live.
4. What you mean by zone?
5. Hi Rob here is how you work it out for R-404a for an air cooled condenser.
Low side 32 degrees = 13.91bar High side 55 degrees = 24.54bar
13.91x1.1=15.3bar) 24.54x1.1=27bar
13.91x1.5=20.7bar) 24.54x1.5=36.8bar
Range is between 15.3-20.7bar Range is between 27-36.8bar
hope that helps
6. Originally Posted by Rob
What you mean by zone?
By zone I mean the standard ambient temperature. Generally in the world we can find two different climate zones. The places where ambient temperature is taken as 43 Celsius (Asia, Africa, Australia, and some parts of middle and south America). Everywhere else the ambient is equal to 32 degree of Celsius.
32c Ambient 43c Ambient High pressure side Air cooled Condenser 55c 63c High pressure site Water Cooled Condenser 43c 43c Low pressure side 32c 43c
Now we need to find out the pressure of our refrigerant at 32C for low side pressure and 55C for high side pressure. To do that we need to use comparators.
Refrigerant R404a at 55 degree Celsius is under 24.54(Bar) of pressure and in low side is equal to 13.91(Bar).
We need to apply the formula to count a High side strength pressure.
The H.P Strength test = Ps(highside)x 1.1 and x 1.5
The H.P Strength test = 24.54 x 1.1 = 27 and 24.54 x 1.5 = 36.81
The H.P Strength test = 27(Bar)(minimum) and 36.81(Bar)(maximum)
Just before we apply the OFN in to the system for strength test we need to make sure that all devices on H.P side are able to take that high pressure. If one of the devices can't co-op with pressure of 36.81 (Bar) then we need to go for minimum Strength Test Criteria which in this scenario as we counted is equal to 27(Bar)
Formula to count Low Side Pressure, strength test.
The L.P. Strength test = Ps(Low side) x 1.1
The L.P. Strength test = 13.91 x 1.1
The L.P. Strength test = 15.3(Bar)
The Strength Test Criteria for low side is equal to 15.3(Bar)
*H.P.-High Pressure Side of refrigeration system.
*L.P.-Low Pressure Side of refrigeration system.
7. Thanks guys for all...
8. Thanks great topic
9. Ya nice topic cheers lads
10. Thanks great topic!
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Question 118 ptsGiven the system of linear equations:x'(t) = Xt) - z(t); x(o) = 1 y'(t)=x(t) + z(t); YO) = 0 2'(t)=x(t) + Xt) z(0) = -1Which or which...
Question
Question 118 ptsGiven the system of linear equations:x'(t) = Xt) - z(t); x(o) = 1 y'(t)=x(t) + z(t); YO) = 0 2'(t)=x(t) + Xt) z(0) = -1Which or which of the following functions is or are solution to it?x(t)=2-e-tx(t)=e -tYt)=et _ e-tYt) =cos(t) - sin(t) _ 1z(t) = ef(cos(t) - sin(t))z(t) =et _ 2None of the above answers is correct:
Question 11 8 pts Given the system of linear equations: x'(t) = Xt) - z(t); x(o) = 1 y'(t)=x(t) + z(t); YO) = 0 2'(t)=x(t) + Xt) z(0) = -1 Which or which of the following functions is or are solution to it? x(t)=2-e-t x(t)=e -t Yt)=et _ e-t Yt) =cos(t) - sin(t) _ 1 z(t) = ef(cos(t) - sin(t)) z(t) =et _ 2 None of the above answers is correct:
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http://www.physicsforums.com/showpost.php?p=3631764&postcount=21 | 1,410,917,948,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657120446.62/warc/CC-MAIN-20140914011200-00330-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 742,016,955 | 3,001 | View Single Post
Sci Advisor P: 906 this is how i count n-cycles in Sn (this works for any n, but i will use n = 4): since any 4-cycle in S4 involves 1, we may as well start with it: (1..... i have 3 choices for my next element (the image of 1): (1 2..... (1 3..... (1 4..... after i make my next choice, i'll have them all: (1 2 3 4) (1 2 4 3) (1 3 2 4) (1 2 4 2) (1 4 3 2) (1 4 2 3) that makes 6. in general, in Sn, we'll get (n-1)(n-2)....(2) = (n-1)! possible distinct n-cycles. | 197 | 483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2014-41 | latest | en | 0.918899 |
http://www.entropy-book.com/understanding-uncertainty/ | 1,632,188,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00265.warc.gz | 81,079,884 | 7,941 | # Understanding Uncertainty
Probability is a well-established mathematical branch of high importance. In mathematics probability is calculated with consistency with a set of axioms. Sometimes uncertainty is defined by the statisticians according to probability rules.
For example: Suppose Bob plans to dine with Alice in the evening: there is 1/10 chance the he will not be available. Since the total probability is 1 (Kolmogorov 2nd axiom), therefore there is 9/10 chance that they will dine together and 1/10 that they will not. If there is a chance of 1/2 that Bob will not be available, the total probability is still 1, but now it comprises of a probability of 1/2 for a joint dinner. In General, if the probability that Bob will not be available is p, it implies that the probability of the joint dinner is 1-p.
In this example some statisticians may say that the uncertainty of having a joint dinner in the first case is 10%, and 50% in the second. This is not correct.
Uncertainty is defined by its Shannon’s entropy and its expression for the joint dinner is,
-plnp-(1-p)ln(1-p).
Usually engineers use the logarithm in base 2 and the uncertainty is expressed in bits. If p=1/2 then the uncertainty is 1 bit (one or zero). If p=1/10 then the uncertainty is 0.46 bit, namely, it is little less than half a bit. The entropy is a physical quantity which is a function of a mathematical quantity p, but unlike mathematical quantities that exist in a formal mathematical space defined by its axioms, entropy is bounded by a physical law, the second law of thermodynamics. Namely, entropy tends to increase to its maximum.
The maximum value of S, in our example, is ln2 when p=1/2. Does it mean that nature prefers the chance of Bob not being available for dinner with Alice to be 1/2, where the entropy is at its maximum? The answer, surprisingly for a mathematician, is yes! If we will examine many events of this nature we will see a (bell-like) distribution that has a pick at the value p=1/2.
Similarly, the average of many polls in which one picks, randomly, 1 out of 3 choices, will be a distribution of 50%:29%:21% and not 33%:33%:33% as is expected from simple probability calculations. Laws of nature (the second law) can tell us something about the probabilities of probabilities. The function that describes the most probable distribution of the various events is called the distribution function.
The distribution functions in nature that are the result of the tendency of entropy to maximize are, among others:
• Bell like distributions for humans: mortality, heights, IQ etc.
• Long tail distributions for humans: Zipf law in networks and texts, Benford’s law in numbers, Pareto Law for wealth etc. | 630 | 2,729 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-39 | latest | en | 0.940321 |
https://numberworld.info/38495000000 | 1,606,623,518,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141196324.38/warc/CC-MAIN-20201129034021-20201129064021-00479.warc.gz | 423,359,111 | 4,533 | # Number 38495000000
### Properties of number 38495000000
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 5 * 5 * 5 * 7699
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):
8f67b15c0
Base 32:
13r7m5e0
sin(38495000000)
0.89799730426363
cos(38495000000)
-0.44000095628901
tan(38495000000)
-2.0408985285791
ln(38495000000)
24.373794199676
lg(38495000000)
10.585404323965
sqrt(38495000000)
196201.42711
Square(38495000000)
1.481865025E+21
### Number Look Up
Look Up
38495000000 which is pronounced (thirty-eight billion four hundred ninety-five million) is a amazing number. The cross sum of 38495000000 is 29. If you factorisate the number 38495000000 you will get these result 2 * 2 * 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 5 * 5 * 5 * 7699. The figure 38495000000 has 112 divisors ( 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, 100, 125, 160, 200, 250, 320, 400, 500, 625, 800, 1000, 1250, 1600, 2000, 2500, 3125, 4000, 5000, 6250, 7699, 8000, 10000, 12500, 15398, 15625, 20000, 25000, 30796, 31250, 38495, 40000, 50000, 61592, 62500, 76990, 78125, 100000, 123184, 125000, 153980, 156250, 192475, 200000, 246368, 250000, 307960, 312500, 384950, 492736, 500000, 615920, 625000, 769900, 962375, 1000000, 1231840, 1250000, 1539800, 1924750, 2463680, 2500000, 3079600, 3849500, 4811875, 5000000, 6159200, 7699000, 9623750, 12318400, 15398000, 19247500, 24059375, 30796000, 38495000, 48118750, 61592000, 76990000, 96237500, 120296875, 153980000, 192475000, 240593750, 307960000, 384950000, 481187500, 601484375, 769900000, 962375000, 1202968750, 1539800000, 1924750000, 2405937500, 3849500000, 4811875000, 7699000000, 9623750000, 19247500000, 38495000000 ) whith a sum of 95497802400. The number 38495000000 is not a prime number. The figure 38495000000 is not a fibonacci number. The figure 38495000000 is not a Bell Number. The number 38495000000 is not a Catalan Number. The convertion of 38495000000 to base 2 (Binary) is 100011110110011110110001010111000000. The convertion of 38495000000 to base 3 (Ternary) is 10200100210010121121202. The convertion of 38495000000 to base 4 (Quaternary) is 203312132301113000. The convertion of 38495000000 to base 5 (Quintal) is 1112314210000000. The convertion of 38495000000 to base 8 (Octal) is 436636612700. The convertion of 38495000000 to base 16 (Hexadecimal) is 8f67b15c0. The convertion of 38495000000 to base 32 is 13r7m5e0. The sine of 38495000000 is 0.89799730426363. The cosine of the number 38495000000 is -0.44000095628901. The tangent of the number 38495000000 is -2.0408985285791. The square root of 38495000000 is 196201.42711.
If you square 38495000000 you will get the following result 1.481865025E+21. The natural logarithm of 38495000000 is 24.373794199676 and the decimal logarithm is 10.585404323965. You should now know that 38495000000 is impressive number! | 1,250 | 2,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2020-50 | latest | en | 0.460775 |
https://webwork.maa.org/moodle/mod/forum/discuss.php?d=2950 | 1,695,879,244,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00809.warc.gz | 658,948,137 | 21,368 | ## WeBWorK Problems
### Intervals or Lists with Strings (and braces)
by Spyro Roubos -
Number of replies: 8
I want a correct answer to be a set of strings; however, it seems like I can only make my answer a "List" without braces.
Ex.: {Greece, Turkey, USA, Austria} <--being a correct answer.
I cannot make an interval with strings as it rejects sets with non-real entries.
I know that I can simply make braces around the answer blank, but I'd rather be consistent with how I have students answering these set theory questions (for other ones with numbers, their answer is an interval object where it's expected that they use braces).
Is there a way of doing this?...
Thank you,
Spyro
### Re: Intervals or Lists with Strings (and braces)
by Paul Pearson -
Hi Spyro,
I thought it might be possible to add 'constants' rather than 'strings' to the context, which worked except for the fact that the answer preview shows up using strings while what is entered shows up as decimals (try the code below to see what I mean). What concerns me about the code below is that it appears that the 'reduceConstants' flag is being ignored.
Best regards,
Paul Pearson
##############################
DOCUMENT();
"PGstandard.pl",
"MathObjects.pl",
"PGcourse.pl",
);
TEXT(beginproblem());
$showPartialCorrectAnswers = 1; Context("Interval"); Context()->constants->add('alpha'=>Real(pi/2),'beta'=>Real(2.236213)); Context()->flags->set( reduceConstants => 0, reduceConstantFunctions => 0, ); Context()->constants->set( alpha => {keepName => 1},beta => {keepName => 1});$answer = Compute("{alpha,beta}");
Context()->texStrings;
BEGIN_TEXT
Using set notation, what is the set of the first two letters of the
Greek alphabet?
\{ ans_rule(40) \}
END_TEXT
Context()->normalStrings;
ANS( $answer->cmp(requireParenMatch => 1) ); ENDDOCUMENT(); In reply to Spyro Roubos ### Re: Intervals or Lists with Strings (and braces) by Davide Cervone - Paul's suggestion works, but there are some improvements that could be made. First, the keepName and requireParenMatch values aren't needed, since these are the default, so they do nothing. Also, the reduceConstants settings are not needed, either. These only affect Formula objects, and (since everything is constant), your answer is a Set, not a Formula, so it is shown numerically (the Set doesn't know how it got its values, only what the values are). In order to show the student answer symbolically, you need to use formatStudentAnswers => "parsed" so that they are shown as formulas rather than constants. I'd also recommend adding a TeX rendering to the constants so that their names are displayed better. My version of the relevant part of Paul's code would be Context("Interval"); Context()->constants->add( alpha => pi/2, beta => 2.236213, ); Context()->constants->set( alpha => {TeX => '\text{alpha}'}, beta => {TeX => '\text{beta}'}, ); Context()->flags->set(formatStudentAnswer => "parsed");$answer = Compute("{alpha,beta}");
Context()->texStrings;
BEGIN_TEXT
Using set notation, what is the set of the first two letters of the Greek alphabet?
\{ans_rule(40)\}
END_TEXT
Context()->normalStrings;
ANS($answer->cmp); Note that there are still some subtle issues, here. First, since alpha and beta are treated as numbers, students could enter things like {sqrt(alpha^2),beta} and still get it right (not that anyone would enter this). Second, you won't get any error message for something like {alpha+1,beta}. Third, if you entered {alpha,x} you'd get an error about your second value not being a number (which is misleading, since you aren't asking the student to enter numbers). So it works for the most part, but can give some in appropriate error messages. It is also possible to do as you ask, and use a list of words. But handling the delimiters is a bit tricky (harder than it should be). Here is one approach: Context()->strings->add( alpha => {caseSensitive => 1}, beta => {caseSensitive => 1}, ); Context()->parens->set("{" => {type => "List", removable => 0});$answer = Compute("{alpha,beta}");
Context()->texStrings;
BEGIN_TEXT
Using set notation, what is the set of the first two letters of the Greek alphabet?
\{ans_rule(40)\}
END_TEXT
Context()->normalStrings;
ANS($answer->cmp( list_type=>"set", removeParens=>0, implicitList=>0, showParenHints=>1 )); In this case, we add alpha and beta as words (that must be in lower case -- you may want to change that if you don't care about using "Alpha" for "alpha"), and makes { produce lists, and not be removable. (Usually, if delimiters contain only one element, they are removed, as they are acting like parentheses in that case.) In the answer checker, we set removeParens => 0 to prevent the outermost list parentheses from being removed from the correct answer automatically (list answers usually don't have delimiters), implicitList => 0 (so that parentheses aren't added to an answer that is a single word), and showParenHints => a so there is a message about problems with the parentheses. This prevents the problems we had in Paul's case of using the words as numbers. One slight problem here is that if you enter {alpha,x} you are told that entries in a list must be the same type (we would like it to have called it a set rather than a list), and if you enter just {x} it is marked wrong with no message. Note that you also can't use unions or other set operations in this case, as you could in Paul's case. (Whereas Paul was using numbers to fake words, we are using lists to fake sets, so you will have to decide which set of problems you are willing to accept). Hope that helps. Davide In reply to Davide Cervone ### Re: Intervals or Lists with Strings (and braces) by Spyro Roubos - Thank you both! I feel like I understand it a bit better now. I wound up going with Davide's approach only because I had no way of hiding the numbers that the constants were defined to be. I have the problem exactly the way I want it with ONE tiny issue. I've attached a screen shot of the problem. What I'd like is to turn off the error message that notifies you that you got the 'nth' word incorrect (I know this is the flag 'showEqualErrors => 0')... But, I want a different error message to show in a certain case. If I set ShowEqualErrors to 0, no error messages are shown including the one in my custom answer checker. ...If I set it to 1, both error messages show. Is there a quick way of 'on-the-fly' changing an error message? Thank you both again, Spyro In reply to Spyro Roubos ### Re: Intervals or Lists with Strings (and braces) by Davide Cervone - Can you attach your code? It is probably in the way that you provide the error message. The "there is a problem with your nth word" comes from the trapping of errors in general, so you are probably issuing an error message, whereas you want to use one of the other message methods to set the nth warning message directly (so it won't be an untrapped error). But I can't tell you which one you need without looking at your code. In reply to Davide Cervone ### Re: Intervals or Lists with Strings (and braces) by Spyro Roubos - # Set Theory - Giving roster form from a chart. ######################################################################## DOCUMENT(); loadMacros( "PGstandard.pl", # Standard macros for PG language "MathObjects.pl", #"source.pl", # allows code to be displayed on certain sites. #"PGcourse.pl", # Customization file for the course ); # Print problem number and point value (weight) for the problem TEXT(beginproblem()); # Show which answers are correct and which ones are incorrect$showPartialCorrectAnswers = 0;
##############################################################
#
# Setup
#
#
Context("Numeric");
USA => {caseSensitive => 1},
Greece => {caseSensitive => 1},
China => {caseSensitive => 1},
Turkey => {caseSensitive => 1},
Egypt => {caseSensitive => 1},
Iceland => {caseSensitive => 1},
Brazil => {caseSensitive => 1},
);
Context()->parens->set("{" => {type => "List", removable => 0});
$Country[1] = "Brazil";$Country[2] = "China";
$Country[3] = "Egypt";$Country[4] = "Greece";
$Country[5] = "Iceland";$Country[6] = "Turkey";
$Country[7] = "USA"; # TRAP: Setting one countries price at exactly$1.25 which is not less than $1.25$I1 = random(1,7,1);
#Correct Countries
do {$C1 = random(1,7,1);} while ($C1 == $I1); do {$C2 = random(1,7,1);}
while (($C2 ==$C1) or ($C2 ==$I1));
do {$C3 = random(1,7,1);} while (($C3 == $C2) or ($C3 == $C1) or ($C3 == $I1)); #Question will ask which countries have prices less than$1.25
$answer = Compute("{$Country[$C1],$Country[$C2],$Country[$C3]}");$commonMistake = Compute("{$Country[$C1], $Country[$C2], $Country[$C3], $Country[$I1]}");
# Run through setting all prices initially to $0 # we do this so we can later differentiate between ones given correct values. for ($i = 1; $i <= 7;$i++) {
$price[$i] = 0;
}
#Give random but correct prices for Countries C1 - C3
# we want it to be realistic, so the lower bound is $0.70.$price[$I1] = 1.25;$price[$C1] = random(0.70, 1.24,0.01);$price[$C2] = random(0.70, 1.24,0.01);$price[$C3] = random(0.70, 1.24,0.01); # Run through the rest (these are all set to zero as default) and # give the rest prices outside of that range (1.25 or higher). for ($i = 1; $i <= 7;$i++) {
if ($price[$i] == 0) {
$price[$i] = random(1.26, 1.60,0.01);}
}
##############################################################
#
# Text
#
#
Context()->texStrings;
BEGIN_TEXT
The following average price for a can of Coca-Cola is shown below for a few countries:
$BR$BR
$BCENTER \{ begintable(2) \} \{ row( "Brazil", "$DOLLAR $price[1]") \} \{ row( "China", "$DOLLAR $price[2]") \} \{ row( "Egypt", "$DOLLAR $price[3]") \} \{ row( "Greece", "$DOLLAR $price[4]") \} \{ row( "Iceland", "$DOLLAR $price[5]") \} \{ row( "Turkey", "$DOLLAR $price[6]") \} \{ row( "USA", "$DOLLAR $price[7]") \} \{ endtable() \}$ECENTER
$BR$BR
In roster notation, give the set of countries that have average prices $BBOLD less than$EBOLD ${DOLLAR}1.25.$PAR
\{ans_rule(40)\}
END_TEXT
Context()->normalStrings;
##############################################################
#
#
#
sub mycheck {
my ($correct,$student, $ansHash) = @_; if ($student == $Country[$I1]) {
Value->Error("close...but remember $BBOLD less than${DOLLAR}1.25 $EBOLD"); } return$student == $correct; } ANS($answer->cmp(
list_type=>"set",
removeParens=>0,
implicitList=>0,
showParenHints=>1,
showLengthHints => 0,
checker=>~~&mycheck,
));
ENDDOCUMENT();
### Re: Intervals or Lists with Strings (and braces)
by Davide Cervone -
I think this is better handled using answer hints, as follows:
ANS($answer->cmp( list_type=>"set", removeParens=>0, implicitList=>0, showParenHints=>1, showLengthHints => 0, )->withPostFilter(AnswerHints( sub { my ($correct,$student) = @_; foreach my$name ($student->value) {return 1 if$name == $Country[$I1]}
return 0;
} => "Close ... but remember ${BBOLD}less than${DOLLAR}1.25 $EBOLD", sub { my ($correct,$student) = @_; my %used = (); foreach my$name ($student->value) { return 1 if$used{$name};$used{\$name} = 1;
}
return 0
} => "Countries should only be included once in your set"
))
);
While checking this, I realized that my approach didn't produce a warning if you used the same country more than once, so I've included a message for that as well.
Hope that does what you want.
Davide
### Re: Intervals or Lists with Strings (and braces)
by Spyro Roubos -
PERFECT!!
That's exactly what I want! Thank you so much! | 3,061 | 11,487 | {"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} | 2.984375 | 3 | CC-MAIN-2023-40 | longest | en | 0.849315 |
https://naijavarcity.com.ng/past-questions/physics/a-relative-density-bottle-has-a-mass-of-19-g-when-empty-when-it-is-completely-filled-with-water-its-mass-is-66-g-what-will-be-its-mass-if-completely-filled-with-alcohol-of-relative-density-0-8/17090/ | 1,708,885,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474641.34/warc/CC-MAIN-20240225171204-20240225201204-00109.warc.gz | 419,815,471 | 82,084 | ## A relative density bottle has a mass of 19 g when empty. When it is completely filled with water, its mass is 66 g. What will be its mass if completely filled with alcohol of relative density 0.8?
A relative density bottle has a mass of 19 g when empty. When it is completely filled with water, its mass is 66 g. What will be its mass if completely filled with alcohol of relative density 0.8?
A. 47 g
B. 52.8 g
C. 37.6 g
D. 56.6
Let mb=mass of empty bottle,
mw=mass of water only and
ma= mass of alcohol only
given; mb=19g
mb+ mw = 66g
mb+ ma = ?
R.d=0.8
R.d=mass of alcohol
⇒massofalcohol/massofequalvolumeofwater
⇒ mass of equal volume of water = mw=66-19=47g
⇒0.8 = ma/47
⇒ma=0.8×47 =37.6g
; mb + ma = 19+37.6=56.6g
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Notify of | 268 | 780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.890485 |
http://gizmodo.com/could-cinderellas-glass-slippers-physically-exist-witho-1671333073 | 1,503,124,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105304.35/warc/CC-MAIN-20170819051034-20170819071034-00296.warc.gz | 171,564,788 | 98,181 | It is delightful to have my masters degree in Mechanical Engineering put to use in resolving age old engineering problems.
One can never know the exact shape or size of the slippers that Cinderella wore, but one can hazard a guess that they must have looked something like:
Now, let's talk about failure. No, not about Cinderella's failure to keep her shoes on her feet, but about mechanical failure. Whenever we design something that needs to bear force, we test for various possible modes of failure and try to ensure that our object is strong against all of them.
Now, one possible way the slippers could break is by yielding to the compressive stress arising due to Cinderella's weight. But will that happen?
We can safely assume that Cindy didn't weigh more than than 50 kgs. I mean, her cousins were fat and ugly, so we have to leave them some room on the top, right? Let's assume this weight to be applied uniformly across the shoe. Note that, as Yehong Zhu pointed out in a comment, the toe region of a heeled shoe bears almost thrice as much force as the heel region, but it won't matter for our purposes.
Using a rough estimate of her foot size, her foot area comes out to be about
If 50 kgs of weight were to be applied uniformly across this area, the compressive stress developed in the material would be :
The Yield strength of ordinary glass for compressive stress is approximately 50 MPa, which is 3 orders of magnitude more than what Cinderella's weight can produce, so we can safely conclude that any regular glass can sustain it. Since the stress is so low, we don't even need to worry about the uneven loading on the shoe.
So, is she safe now? Can she safely dance at the ball without fear of tiny shards of glass cutting her skin and ruining her dress?
Not so soon, buddy!
There's another way her shoe could break, and this is due to the compressive stress due to the bending moment applied to her heel every time she walks.
Now, I don't want to be here all day, and I don't want to model her shoe in ANSYS, so I will make a few simplifying assumptions. Let her heel have a diameter 2 centimeters. and have a length of 6 centimeters from the tip to the point where it joins the rest of the shoe. The heel can now be modeled as a simple cantilever beam of circular cross section.
I'm in a bit of a hurry and I have to get back to reading The Casual Vacancy (2012 book), so I will defer to this for the actual calculation of the maximum bending stress. I will assume her stepping angle to be about 30°, which means that only half of her weight (500sin 30) would act in the normal direction to the heel (causing the bending). Plugging 250N as the bending force and the rest of the figures in place, we get the maximum bending stress in the heel to be 19 MPa. Note that this is dangerously close to our critical stress of . Even if we make a few more allowances by making the heel thicker or the stepping angle smaller, we cannot let our little princess veer so dangerously close to disaster.
In order for her to be safe enough, we would take a safety factor of at least 2, and also assume that the bending stress can go as high as 75 MPa. This means that her shoes need to be made of glass that has a yield strength of at least 150 MPa.
Safety glass (thermal toughened glass) seems to be a good bet. It has a yield strength of about 200 MPa and a higher Young's Modulus too, so I imagine Cinderella can use it safely without fear of it breaking just when she is shaking a leg with our awesome prince. Ideally, we would also want it mixed with something to make it less brittle, but I don't want to make it too different from glass or the answer becomes meaningless.
Steve Davis points out an interesting issue—what happens when she starts running out of the castle at midnight approaches?
When Cinderella runs, I expect the impact force to be 3-5 times that of the regular walking force (this is somewhat supported by the paper "Ground reaction forces at different speeds of human walking and running"). The shoes should be safe for these values.
We must also take into consideration the fact that Cinderella's dress would probably not let her take long strides. This would mean that her stepping angle would remain within safe limits, further ensuring that her shoes don't break.
Most importantly, she would be well-advised to develop a toe-first foot strike,which would totally solve the problem. This cannot be maintained for large distances, but would certainly take Cinderella out of the danger zone.
Bharat Jakati points out another caveat. What if the friction between her shoes and the ground/floor is so low that she slips? Well, we can assume that the flooring is either made of stone or is carpetted. The coefficient of friction for Glass on Stone is about 0.42, which is not very high, but is high enough for her to not slip. I couldn't get a value for the coefficient for glass on carpet, but I imagine it to be similar. | 1,094 | 4,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-34 | latest | en | 0.977023 |
https://cran.opencpu.org/web/packages/TrajDataMining/vignettes/TrajDataMining.html | 1,596,846,268,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737233.51/warc/CC-MAIN-20200807231820-20200808021820-00422.warc.gz | 264,947,504 | 15,774 | ## TrajDataMining
The TrajDataMining contains a set of methods for trajectory data preparation, such as filtering, compressing and clustering, and for trajectory pattern discovery. The meth- ods for data preparation are important to prepare trajectory data sets before the data mining phase. TrajDataMining contains the following methods:
• A speed filter that filters out trajectory observations whose speeds are above a user-defined maximum velocity ( ETIENNE , 2011).
• Two compression algorithms: (1) Douglas-Peucker which reduces trajectories by preserving spatial precisions ( DOUGLAS; PEUCKER , 1973), and (2) Open-window Meratnia-By which reduces trajectories by preserving spatiotempo- ral precisions ( MERATNIA; BY , 2004).
• Two algorithms to discover when objects stop and move, CB-SMoT ( PALMA et al. , 2008) and DB-SMoT ( ROCHA et al. , 2010), which can be used to semanti- cally enrich the trajectory data.
• A method, named Partner, that identifies objects that are moving together. We propose this method to recognize trajectories that stay together, based on trajectory distance time series analysis.
## Example data gives
``````library(TrajDataMining)
library(ggplot2)
library(sp)
library(magrittr)
suppressMessages(library(rgdal))
``````
## Plot of A1 data example
### A1 track its a trajectory of sea elephant
`````` df <- data.frame([email protected]@coords[,1],[email protected]@coords[,2])
ggplot(df,aes(x=df\$x,y=df\$y))+geom_path(aes(group = 1), arrow = arrow(),color='blue')+ggtitle("elephant sea Trajectory")+xlab("X")+ ylab("Y")+theme_classic()
``````
## Douglas Peucker Method
### Method that reduces a trajectory spatially
``````# get the maximum distance
max <- max([email protected]\$distance)
douglasp <- douglasPeucker(A1,max)
df <- data.frame([email protected]@coords[,1],[email protected]@coords[,2])
ggplot(df,aes(x=df\$x,y=df\$y))+geom_path(aes(group = 1), arrow = arrow(),color='blue')+ggtitle("Method Douglas Peucker")+xlab("X")+ ylab("Y")+theme_classic()
``````
## Speed Filter
### Method removes regions where speed was higher than the defined parameter
``````# get the maximum distance
sf <- speedFilter(A1,0.01)
``````
``````## [1] "2008-01-23 21:32:44"
## [1] "2008-01-24 01:31:26"
## [1] "2008-01-24 03:12:00"
## [1] "2008-01-24 11:24:42"
## [1] "2008-07-21 04:30:30"
## [1] "2008-07-21 11:22:30"
## [1] "2008-07-22 01:09:49"
## [1] "2008-07-25 13:19:46"
## [1] "2008-07-28 11:40:23"
## [1] "2008-07-29 01:23:43"
## [1] "2008-07-31 09:15:41"
## [1] "2008-11-19 12:38:34"
``````
`````` df <- data.frame([email protected]@coords[,1],[email protected]@coords[,2])
ggplot(df,aes(x=df\$x,y=df\$y))+geom_path(aes(group = 1), arrow = arrow(),color='blue')+ggtitle("Method Speed Filter")+xlab("X")+ ylab("Y")+theme_classic()
``````
## Partner
### Method to identify if two trajectories are partners
#### For to send this data for the database use the class DataSourceInfo.
`````` data <- FALSE
partner(A1,A2,110792,2277,0,data)
``````
``````## begintime endtime id1 id2
## 1 2008-02-04 21:26:54 2008-11-11 06:58:44 1 2
`````` | 981 | 3,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-34 | latest | en | 0.631875 |
https://themathlab.com/Standardized%20Test%20Prep/Practice%20Attic%20Middle%20School%20Level/g1.htm | 1,632,773,763,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00154.warc.gz | 583,844,877 | 7,902 | g1
is building a deck around his circular pool. The deck is going to be 4 feet wide and will go the entire way around the pool. The pool has a diameter of 21 feet. A diagram, not drawn to scale, is below.
Find the area of the deck.
*This is one of those SHOW ALL OF YOUR WORK AND EXPLAIN HOW YOU GOT YOUR ANSWER problems.
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Real Analysis Real Analysis Math Forum
February 26th, 2018, 01:43 PM #21
Senior Member
Joined: Oct 2009
Posts: 753
Thanks: 261
Quote:
Originally Posted by zylo I said Sn was not a natural number. I simply meant it to be an example of something constructed, or defined by, natural numbers. If it makes it easier, put spaces between. S$\displaystyle _{11}$ = 1 2 3 4 5 6 7 8 9 10 11 $\displaystyle lim_{n \rightarrow \infty}$ Sn =1 2 3,......,12578, .......... If none of {Sn} are unending, then there has to be a maximum value of n for Sn. There isn't.
Take Sn = n
S1 = 1, ending
S2 = 2, ending
S3 = 3, ending
S100 = 100, ending.
For which n is Sn non-ending?
Last edited by skipjack; February 27th, 2018 at 12:40 AM.
February 26th, 2018, 01:59 PM #22 Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999. P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............} What is the largest member of {Pn}? I am not asking what any particular member of {Pn} is, I am asking what the largest member is. Last edited by skipjack; February 27th, 2018 at 12:40 AM.
February 26th, 2018, 01:59 PM #23
Senior Member
Joined: Oct 2009
Posts: 753
Thanks: 261
Quote:
Originally Posted by Maschke Very interesting comment that I'd like to understand with your help. I dove into the hyperreals a couple of years ago. I read Terence Tao's articles on NSA including his awesome explanation of ultrafilters as voting systems. I know what a nonprincipal ultrafilter is. I read Lightstone's paper on his notation for hyperreals and I read a few other papers besides. There was a moment when I thought I had it all straight in my mind but apparently that clarity is gone now. I mention this so that you can calibrate the state of my knowledge. Can you please explain how $(S_n)$ has a different limit than $(n)$ in the hyperreals or hyperintegers? Thanks much.
Ah well, perhaps some interesting discussion can come from this rubbish thread anyway.
So, the real numbers are often defined to be equivalence classes of rational Cauchy sequences. The hyperreal construction is very similar, but instead of taking Cauchy sequences, we take all sequences (incidentally, if we start with Q and do the hyperreal construction on it, then the result will contain R, it is an alternative construction to R!!)
But the set of all sequences isn't a field, indeed, what would the inverse of (1,0,1,0,1,0,...) be? To solve this, we introduce the ultrafilter which decides for us which elements to identify and which not (this is problematic since ultrafilters cannot be constructed at all, and different choices of ultrafilters give different hyperreals!!)
A real sequence $(x_n)_n$ would then have as hyperreal limit the exact same sequence. I find this very elegant. And it gives some surprising other results:
If we interpret 0.999999... as the limit of 0.9999...9, then this limit will be LESS than one, less by an infinitesimal amount.
The limit of the sequence (1,2,3,4,...) would be the hyperreal (1,2,3,4,...) and this hyperreal is different from (1, 12, 123, 1234,...) since their agreement set (there is only one place where it agrees, the first place) is finite, hence cannot lie in the ultrafilter.
Now, I have to be fair and say that this is not really how limits are defined in non-standard analysis. In non-standard analysis, we "fix" the issue that 0.999...=/=1 by basically saying that the limit operator should forget about the infinitesimal part. But to me the infinitesimal part is the nicest part, so I like my interpretation.
Last edited by skipjack; February 27th, 2018 at 12:43 AM.
February 26th, 2018, 02:01 PM #24
Senior Member
Joined: Oct 2009
Posts: 753
Thanks: 261
Quote:
Originally Posted by zylo 0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999. P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............} What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
There is no largest member, and $P\notin \{P_n\}$. Am I right?
February 26th, 2018, 02:16 PM #25
Banned Camp
Joined: Mar 2015
From: New Jersey
Posts: 1,720
Thanks: 124
Quote:
Originally Posted by zylo 0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999. P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............} What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
So you have no response to my last post, above.
How did you manage to post your long post the instant afterI posted mine? Coincidence I guess.
"rubbish thread?" Can't do any better than that?
Last edited by zylo; February 26th, 2018 at 02:28 PM.
February 26th, 2018, 02:53 PM #26 Math Team Joined: May 2013 From: The Astral plane Posts: 2,093 Thanks: 852 Math Focus: Wibbly wobbly timey-wimey stuff. Alright, let's try a bit of a detour. Perhaps it has value, perhaps it doesn't. We know that the rationals are countable. The reals consist of the rational numbers and the irrational numbers. The rationals are countable, and thus the irrationals must have the same cardinality as the reals because the irrationals can't be a finite set. (I hope we agree on that anyway.) So let's construct the table using only the set of irrationals... we no longer need to consider the rationals and thankfully the issue of 0.9999999... = 1 is no longer an issue. For convenience, let's use the section (0, 1) of real numbers. Then our list to "count" the irrationals is of the form: 0.00222459875978... 0.39457935730945... 0.79874968734978... etc. We are back to Cantor's diagonal argument as the simplest way to work with the list, which I know that zylo will just looooove. This table seems to me to be a cleaner argument than trying to take "limits" of real numbers. -Dan Last edited by skipjack; February 27th, 2018 at 12:50 AM.
February 26th, 2018, 03:45 PM #27
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Originally Posted by zylo If none of {Sn} are unending, then there has to be a maximum value of n for Sn. Tkere isn't.
No there doesn't. $S_n$ terminates after $n$ elements. The infinite case $S$ doesn't terminate because it's infinite. If it terminated it would be finite.
February 26th, 2018, 03:55 PM #28
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Originally Posted by topsquark Then our list to "count" the irrationals is of the form: 0.00222459875978... 0.39457935730945... 0.79874968734978... etc. We are back to Cantor's diagonal argument as the simplest way to work with the list,
I don't think this guarantees that your diagonal constructs an irrational. Indeed, it seems easy to guarantee that it doesn't: make every element $0.00\ldots012345678910111213\ldots$ with $n$ zeros at the start of the $n$th element.
February 26th, 2018, 05:05 PM #29
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Originally Posted by zylo 0r, let Pn be the number of n-place decimals in [0,1) listed as .0000, .0001, ........, .9999 = 9999. P= $\displaystyle \lim_{n \rightarrow \infty}$Pn, a member of {Pn} ={9, 99, 999, 9999, ..............} What is the largest member of {Pn}. I am not asking what any particular member of {Pn} is, I am asking what the largest member is.
Let $t_n = 1 - \frac1n$. Clearly $t = \lim \limits_{n \to \infty} t_n = 2$. So what is the largest member of the sequence $\{t_n\}$?
Clearly, there isn't one. All $t_n \gt t_{n-1}$ and all $t_n \lt 2$. So $t \not \in \{t_n\}$.
February 26th, 2018, 07:14 PM #30
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Originally Posted by zylo So you have no response to my last post, above.
I did respond. Look again.
Quote:
"rubbish thread?" Can't do any better than that?
Why would I do better? I'm not here to insult you, or disprove you or humiliate you. I'm just here to discuss interesting mathematics. And this thread has no interesting mathematics. And it's made worse by very, very confusing notations. Again: not meant to be insulting or disprove you.
So why reply then? Hey, man, it's 5 am where I live and I'm replying to a math thread. Why do you think? Obviously: I can't sleep and I'm bored.
Last edited by skipjack; February 27th, 2018 at 12:47 AM.
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27 Feb 2009, 12:45
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Medical education in the United States has focused almost exclusively on curative medicine, while preventive care has been given scant attention. This is misguided. Medical schools should invest as much time in teaching their students how to prevent illness as in teaching them how to cure it.
Which of the following, if true, most strengthens the argument above?
(A) Many contagious diseases can be prevented with vaccines.
(B) In 1988, for every three cents the United States spent on prevention, it spent 97 cents on curative treatment.
(C) The number of students enrolled in medical school is the highest it has ever been.
(D) More people die each year from disease than from accidental causes.
(E) As the population grows, the number of doctors in certain specialties has not been keeping pace.
I disagree with the OA..but lets see if i am the only one reading it wrong!
[Reveal] Spoiler: OA
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27 Feb 2009, 13:37
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IMO: 2)
1) is tempting. It could be one of many reason why preventive care may be equally important.
However, this is justification of why preventive is equally important. Not necessarily strenthen the his argument: students need to invest as much time in prevenvtive care as in curarive care because the preventive care has been given little attention.
2) This provides direct evidence to author's initial claim that preventive care is getting little attention. This may not justify his argument, but i think it strenghen his argument.
3),4),5) no relevant.
What is OA?
Kudos anyone?
Last edited by cityjoy92 on 27 Feb 2009, 13:55, edited 2 times in total.
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27 Feb 2009, 13:48
IMO : 2
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27 Feb 2009, 14:45
FN wrote:
I disagree with the OA..but lets see if i am the only one reading it wrong!
Medical education in the United States has focused almost exclusively on curative medicine, while preventive care has been given scant attention. This is misguided. Medical schools should invest as much time in teaching their students how to prevent illness as in teaching them how to cure it.
Which of the following, if true, most strengthens the argument above?
1)Many contagious diseases can be prevented with vaccines.
curative is better. weakens
2)In 1988, for every three cents the United States spent on prevention, it spent 97 cents on curative treatment.
out of scope. Talks about how much US spents. Not how MS should work
3)The number of students enrolled in medical school is the highest it has ever been.
Good, But neither support nor weaken
4)More people die each year from disease than from accidental causes.
Again weaknes. If people are dying from disease then curative is better.
5)As the population grows, the number of doctors in certain specialties has not been keeping pace.
Doctors in certain specalities are not available, what happens? People do not get treatment for some thing that can be prevented and possibly die or suffer serious life long disability. So preventive care is losing ground.
Hence E
OA & OE?
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27 Feb 2009, 18:14
2 is the best
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28 Feb 2009, 02:23
It shld be B
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28 Feb 2009, 08:11
Hi mates,
uf! I don't see any clear answer...
I'd go with A
What's the conclusion? That medical schools should invest more in teaching to prevent diseases than to cure them. Well, A supports the argument by explaining that some diseases can be prevented
OA and Source?
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28 Feb 2009, 08:29
OA is A..
source is mgmat CR question bank..
i went with B
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10 Mar 2009, 20:50
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Wow!!
Lovely CR and I got it Right
B is a Trap
check the conclusion..only A strengthens it
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10 Mar 2009, 21:03
agree with A.
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10 Mar 2009, 21:15
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B is wrong because it talks about in 1988 we really dont know when this event took place.
Can some 1 clear my doubt.
Should i look for whole argument or just this Medical schools should invest as much time in teaching their students how to prevent illness as in teaching them how to cure it.
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10 Mar 2009, 22:30
Kaushik
we have to strengthen the ARGUMENT and
Argument is-->Medical schools should invest as much time in teaching their students how to prevent illness as in teaching them how to cure it
Now check A and B
A-->giving some alternatives
B-->Stating the fact again(We all know it from the first line whice states "Medical education in the United States has focused almost exclusively on curative medicine")
1)Many contagious diseases can be prevented with vaccines.
2)In 1988, for every three cents the United States spent on prevention, it spent 97 cents on curative treatment.
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Many contagious diseases can be prevented with vaccines
Is a vaccine preventive care or curative care?
I guess I completely got it wrong on the first go. It is preventive because one has not yet contracted the disease. It is curative only when some has contracted it and it is then cured.
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"The conclusion is that medical schools are misguided. The basis for this claim is that they pay
little attention to preventive medicine. The argument would be made stronger by a statement
concerning the benefits of preventive medicine.
(A) CORRECT. Vaccines are a type of preventive medicine that have known benefits, i.e.,
preventing contagious diseases.
(B) This statement supports the idea that the more time and money are spent on curative
medicine than on preventive medicine. However, it does not speak to the benefits of preventive
medicine."
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26 Jan 2010, 04:40
without a doubt its A.
this clearly indicates that the diseases that could have been prevented are still afflicting individuals, placing the blame fair and square on the overemphasis laid on the curative aspect of medicine.
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26 Jan 2010, 09:53
its Option A.
B is very tempting because it does sound similar to the conclusion but the catch is that it talks about a particular year 1988, the argument has no specific year the research was done.
whereas option A states that meny disease can be prevented. so if preventive cure is given more attention it would definately help prevent mant of the contagious disease.
so Option A best fits the conclusion.
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26 Jan 2010, 11:13
I completely agree with "Math & Verbal GMAT Forum Moderator".
At the first shot I thought vaccine is curative coz' it kinda cures what is contagious. Hmmm my bad
It cannot be 'B' because the argument says the EDUCATION BOARD gives less attention to preventive care. It cannot correlate to the money that US GOVERNMENT spend in health care. Also 1988 is a vague figure.
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26 Jan 2010, 11:16
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Medical education in the United States has focused almost exclusively on curative medicine, while preventive care has been given scant attention. This is misguided. Medical schools should invest as much time in teaching their students how to prevent illness as in teaching them how to cure it.
Which of the following, if true, most strengthens the argument above?
(A) Many contagious diseases can be prevented with vaccines.
Correct - Vaccination is preventitive medicine and hence this statement shows the knowledge of same can be more beneficial then curing people once they are infected.
(B) In 1988, for every three cents the United States spent on prevention, it spent 97 cents on curative treatment.
Irrelevant as the conclusion doesn't talk about any monetary factor
(C) The number of students enrolled in medical school is the highest it has ever been.
No significance for the conclusion
(D) More people die each year from disease than from accidental causes.
No help on preventive or curative medicine....
(E) As the population grows, the number of doctors in certain specialties has not been keeping pace.
Irrelevant with respect to conclusion
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26 Jan 2010, 11:24
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A ~ not the greatest answer but stays within the scope
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26 Jan 2010, 14:01
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Clearly its between A and B Even I selected B at first go but its actually out of scope...so it has to be A though not convincing.
Re: CR-Medical Education [#permalink] 26 Jan 2010, 14:01
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# Medical education in the United States has focused almost
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## B-spline approximation
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Computes the B-spline approximation from a set of coordinates. Supports periodicity and n-th order approximation.
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Computes the B-spline approximation from a set of coordinates (knots).
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
rng(1) % Set random seed for reproductility
XY=rand(5,2); % Random set of 5 points in 2D
BS2=BSpline(XY); % Default order=3 (quadratic)
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BSper=BSpline(XY,'periodic',true);
h=plot(XY(:,1),XY(:,2),'-o',BS2(:,1),BS2(:,2),BS3(:,1),BS3(:,2),'--',BSper(:,1),BSper(:,2),'-.');
set(h, 'LineWidth',2)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
### Cite As
Dorian Depriester (2021). B-spline approximation (https://www.mathworks.com/matlabcentral/fileexchange/71456-b-spline-approximation), MATLAB Central File Exchange. Retrieved .
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1. Jun 6, 2006
wScott
We're going to be starting them in a day or two, and I just wanted to know ahead of time what you guys might think we'll be learning with them, like formulae and that kind of stuff..
2. Jun 6, 2006
HallsofIvy
There's not a whole lot you have to know about "geometric sequences". A geometric sequence is one in which you go from one number to the next by always multiplying by the same thing.
Example: 3, 6, 12, 24, 48, 96,.... The point is that 6/3= 2, 12/6= 2, 48/24= 2, 96/48= 2. In other words, you start with the number 3 and proceed to just keep multiplying by 2. If we start counting terms with n= 0 (some people start with n= 1) then the "nth" term is 3(2)n or, in more general terms, arn where a is the first term and a is the "common multiplier"j. If you start counting with n= 1, then the nth term is arn-1:you have to subtract 1 to get back to 0.
Another nice property is this: suppose we add the terms (a geometric series rather than sequenc). For example if S= a+ ar+ ar2+ ar3, then S= a+ r(a+ ar+ ar2) where we've just lost one power. Okay, put it back in: S= a+ r(a+ ar+ ar2+ ar3- ar3. See how I added and subtracted the same thing? Now separate those: S= a+ r(a+ ar+ ar2+ ar3)+ ar4= a+ r(S)+ ar4 so
S- rS= S(1-r)= a+ ar4= a(1- r4). That is,
S(1-r)= a(1- r4) so S= a (1-r4)/(1- r).
More generally, the sum of the first n terms of a geometric sequence is
Sn= a(1-rn+1)/(1- r).
3. Jun 6, 2006
wScott
Well, it seems pretty simple to me, especially after the few arithmatic formulae we've learned over the past few days. Should be fun :) | 543 | 1,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-17 | longest | en | 0.921501 |
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# The dark side of Dubai
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The dark side of Dubai [#permalink]
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23 Apr 2009, 16:33
3
KUDOS
If anyone's choosing a career option in Dubai , please read this.
http://www.independent.co.uk/opinion/co ... 64368.html
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Re: The dark side of Dubai [#permalink]
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27 Apr 2009, 13:26
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Dubai is doomed for failure: over 1,000,000 mistreated migrant workers, building boom that is not supported by demand, city planning blunders, human development level on par with Eastern Europe, and limited freedom for not only the migrant workers & expats but also the few people there who are actually emirati. They put up a nice facade but when you actually veer off the beaten track you find that the "real dubai" is seen in the filth of residential Deira & the migrant worker shanties.
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24 Apr 2009, 04:22
pbanavara wrote:
If anyone's choosing a career option in Dubai , please read this.
http://www.independent.co.uk/opinion/co ... 64368.html
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24 Apr 2009, 07:13
great article, its totally one-sided and biased but interesting to read. appreciate it
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27 Apr 2009, 08:08
Heavily slanted but still an interesting read. Thanks for sharing!
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04 May 2009, 15:36
Did anybody hear the recent story about the pregnant lady in Dubai?
A pregnant woman (a foreigner) was driving in Dubai and was rear-ended. The jolt caused her to lose her unborn child.
So what happens? The courts charge HER with MANSLAUGHTER even though the accident wasnt her fault. Oh, and she was found guilty and fined.
Ohhhhhh Dubai. My disdain continues.
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Re: The dark side of Dubai [#permalink]
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12 Jan 2010, 17:54
Too bad...since so much money was spent there
Re: The dark side of Dubai [#permalink] 12 Jan 2010, 17:54
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https://lindseyvan.com/85ufoq75/ | 1,607,093,048,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00292.warc.gz | 379,419,668 | 8,791 | # Cut And Paste Math Worksheets First Grade
Slainie Héloïse October 21, 2020 Math Worksheet
The present generation seems to be blessed immensely with intellect and the benefits of mastering math are something worth considering. It is a well-known fact that math is not a subject that one learns by simply reading the problems and its solutions. In order to master the subject, earnest practice on multiple problems is the best way to go. However, not every person is bestowed with required materials like math worksheets to receive adequate amount of practice.
Great, fun and free math worksheets should be able to present a mathematical problem in different ways. Math is after all nothing more than a numeric expression of some of life’s simplest questions: How much money do I have left if I buy a soda? By the end of the week, how much of my daily allowance will I be able to save if I don’t?
I remember that with my Mom everything was somehow connected to math. She made me count the buttons in my shirt as she dressed me up, asked questions that demanded answers that are related to sums, like how many pair of shoes do you have? How many buttons are there on your Daddy’s shirt? Count all the furniture in the living room and several math games. All my toys were one way or the other math related. I had puzzles, and tons of things Mom had me do as games on daily basis at home to get me ready for kindergarten! In fact, she continued guiding me towards being math friendly throughout kindergarten and first grade during which time 1st grade math worksheets was my constant companion.
It is amazing the difference in effort you will get from worksheet to worksheet. Granted the amount of effort may vary immensely from year to year depending on the group of students you have. However for the most part, when a worksheet is needed to help drill down a procedure, standard, or lesson, its effectiveness can and will vary. Therefore it is our job as the teacher to make sure that when we need to utilize a worksheet, we provide the students with one that is as inspiring as can be.
A lot of the websites charge high dollar amount for these math worksheets. Considering your willingness to spend the high dollar, you run in to a new problem – the uniqueness of the material, a lot of websites offer sheets packed with identical problems leading the child to almost rote memorization of the problems and solution. This leads to deprivation of student and his/her ability to solve the problems logically. And some websites offer sheets in portable document format. And to open such sheets you are forced to download special software.
Practice makes perfect, Learning math requires repetition that is used to memorize concepts and solutions. Studying with math worksheets can provide them that opportunity; Math worksheets can enhance their math skills by providing them with constant practice. Working with this tool and answering questions on the worksheets increases their ability to focus on the areas they are weaker in. Math worksheets provide your kids’ the opportunity to analytical use problem solving skills developed through the practice tests that these math worksheets simulate.
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Nov 06, 2020 | 861 | 3,847 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-50 | latest | en | 0.974047 |
gentlemenrallye.de | 1,656,691,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103943339.53/warc/CC-MAIN-20220701155803-20220701185803-00126.warc.gz | 325,330,938 | 15,371 | How to find area of a triangle with 3 sidesFormulas for calculating the area of a triangle Calculation with given two sides and their angles. To calculate the area, the length of two sides and the angle can also be multiplied and divided by 2. $$\displaystyle A = \frac{ a · b · sin(γ)}{2}$$ $$\displaystyle A = \frac{ a · c · sin(β)}{2}$$I'm a newbie to Matlab and have been struggling on this question. Write a user-defined MATLAB function that determines the area of a triangle when the lengths of the sides are given. For the function name and arguments use [area] = triangle (a, b, c). Of triangle with the following sides: a. a = 10, b = 15, c = 7. b. a = 6, b = 8, c = 10.Enter the first side of the triangle: 5. Enter the second side of the triangle: 7. Enter the third side of the triangle: 9. Area of triangle = 17.41. Enter the first side of the triangle: 2. Enter the second side of the triangle: 7. Enter the third side of the triangle: 10. Not a valid triangle.To calculate Area of Triangle given sides, you need Side A (S a), Side B (S b) & Side C (S c). With our tool, you need to enter the respective value for Side A, Side B & Side C and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.Area = ½ × base (b) × height (h) Another formula that can be used to obtain the area of a triangle uses the sine function. It allows us to find the area of a triangle when we know the lengths of two sides and the size of angle between them. The formula is. Area of triangle = ½ ab sinC. Remember that the given angle must be between the two ...Copy. Sample Output: Input Side-1: 10 Input Side-2: 15 Input Side-3: 20 The area of the triangle is 72.61843774138907.Then, if we find the length of one of its sides, we can find all three sides, including OD. Proof. Here's how you find the area of a circle inscribed in an equilateral triangle: (1) OE = OD = r //radii of a circle are all equal to each other (2) BE=BD // Two Tangent theorem (3) BEOD is a kite //(1), (2) , defintion of a kiteToday, we will learn how to calculate surface area of a triangle. We will use the formula for surface area: area = 1 /2 x base x height. Therefore, we must have base and height of the given triangle. We will prompt the user to input base and height of given triangle. Then we will use the above mentioned formula to calculate area of the given ...From the figure we can write, Area of trapezium = Area of parallelogram AECD + Area of area of triangle CEF. Area of trapezium = height + (sum of parallel sides) Area of trapezium = 3√21 × (25 + 13) Area of trapezium = 3√21 × 19 = 57√21. ∴ Area of trapezium = 57√21 cm². Example 2: A field is in the shape of a trapezium whose ...1. Algorithm: find area of triangle using hero's formula in java (example) Read the sides of triangle from console. Check given sides of triangle can form a valid triangle. Any two sides of a triangle should be greater than third side. If yes, We will find out the area of triangle. Find the semi perimeter of triangle. semiPerimeter = (a+ b ...The image below shows an isosceles triangle. The two base angles are equal to each other. b=2A/h. The formula of the area of the isosceles triangle is equal to half the product ofWe can now plug in our problem's values for a, b and c, get the area of triangle ΔADO, and then multiply by 4 to get the area of the parallelogram. Solution (1) a= 11 //given (2) b=10 //AC=20, given, the diagonals of a parallelogram bisect each other so AO=10 (3) c=12 //BD=24, given, the diagonals of a parallelogram bisect each other so DO=12Create one 'Scanner' object. Take the input of the user as double using the Scanner class and save it in a variable called 'side'. Calculate the area using formula ' (√ 3 / 4) * side * side ' . For √ 3, use Math.sqrt (3). The result we got on step 3 returns a double value. It will be something like 15.4456789 .Calculate height in centimeters: 6 feet, 3 inches = 75 inches x 2.54 cm/inch = 190.5 cm. Multiply height by weight and divide by 3600. (190.5 cm x 95.45 kg) ÷ 3600 = 5. Take the square root of 5 = 2.24 m2". Note: I give credit to the person who did the calculations.So, now let us name the sides as a = 10, b = 24 and c = 26. The first step has to be to solve R.H.S. of equation 1 which is given above. a (square) + b (square) = 100 + 576 = 676. Then let us take L.H.S, and this gives us: c (square) = 262 = 676. We can thus prove that the left-hand side is equal to the right-hand side.A=1/2 (5 cm) (2.5 cm + 6 cm) A= (2.5 cm) (8.5 cm) A= 21.25 square centimeters. Thus, the area of our trapezoid is 21.25 square centimeters. Now that we know how to find the area of a triangle and the area of a trapezoid, let's do an activity utilizing the new concepts that we have just learned. Activity: marvel legend figureirobot walmartfallout 4 modpacksonic 3 air modsiphone accesoriesnew line in latex
Using Area To Find the Height of a Triangle. Now that you know the area of the triangle pictured above, you can plug it into triangle formula A=1/2bh to find the height of the triangle. In this case, the base would equal half the distance of five (2.5), since this is the shortest side of the triangle.How to find the area of a parallelogram? Step 1: Determine all the sides of irregular shape, Make sure all the sides are in same unit. Step 2: Draw the area on a piece of paper using the measurements you obtained. Remember your drawing is to scale. Step 3: Divide the drawing into different shapes. The easy ones are Square and rectangle, circles and triangle could be a bit tricky.The formula for finding the area of an obtuse triangle is the same as for other triangles, area = 1/2 x (base x height). How do you find the area of an irregular triangle? The area of a scalene triangle with base b and height h is given by 1/2 bh. If you know the lengths of all three sides, you can calculate area using Heron's Formula without ...It is given as: A + B + C = 180. Calculate the length of its base and height. Step 1: Calculate the midpoint of one of the sides of the triangle. formula to find area = (1/2) b h. = (1/2) x Base x Height. Everything in trigonometry seems to revolve around the 90-degree triangle and its ratios. The area of an equilateral triangle is the amount of two-dimensional space inside it. If we know the length of any side of equilateral triangle, then we can use following formulae to calculate it's area. Area of Equilateral Triangle = (√ 3 /4)S2. Where, S is the length of each side of triangle.C++ Program to Calculate Area of Triangle. Calculate Area of Triangle with given three sides in C++ : Heron's Formula for the area of a triangle (Hero's Formula) A method for calculating the area of a triangle when you know the lengths of all three sides. Let a,b,c be the lengths of the sides of a triangle. Heron's formula states that the ...A method for calculating the area of a triangle when you know the lengths of all three sides. Let a,b,c be the lengths of the sides of a triangle. The area is given by: Area = √ p ( p − a ) ( p − b ) ( p − c ) where p is half the perimeter, or a + b + c 2 Try this Drag the orange dots to reshape the triangle. In this Python program, we will learn how to find the area of an equilateral triangle. What is the Equilateral Triangle? In geometry, an equilateral triangle is a triangle in which all three sides are equal. In the familiar Euclidean geometry, an equilateral triangle is also equiangular; that is, all three internal angles are also congruent to each other and are each 60°.Find the area of a triangle with side lengths 2, and . 0. 414. 1. Find the area of a triangle with side lengths \ (\frac43,\) 2, and \ (\frac83.\)Area of a Triangle Formula. The area of a triangle, knowing its three sides, is expressed by Heron's formula:. In this formula, a, b, and c represent the lengths of the three sides. s represents the "semiperimeter" or half the perimeter:JS task: Write a JavaScript function that calculates a triangle's area by its 3 sides. The input comes as 3 (three) number arguments, representing one sides of a triangle. The output should be printed to the console. Example: Input: 2 3.5 4 Output: 3.4994419197923547A = 1 2 b h. Area of a triangle is equal to half of the product of its base and height. The height of a triangle is the perpendicular distance from a vertex to the base of the triangle. Any of the 3 sides of a triangle can be used as a base. It all depends on where the height is drawn. If you are given the sides of an isosceles or equilateral ...A sector of a circle is essentially a proportion of the circle that is enclosed by two radii and an arc. Given a radius and an angle, the area of a sector can be calculated by multiplying the area of the entire circle by a ratio of the known angle to 360° or 2π radians, as shown in the following equation: area =. θ. 360.4.Using when three sides are given Basically, In order to calculate the area, you need to find out the Height of the triangle. If you don't know the height or you may have no idea how to find out the height of the triangle, then you can use the below program to calculate the area of a triangle.Example 2: Find the area of a triangle whose sides and the angle between them are given as following: a = 5cm and b = 7cm C = 45 o. Solution: Area of a triangle = ½ · a · b · sinC. Area = ½ × 5 ×7 × 0.707 (since sin 45 ° = 0.707) Area = ½ × 24.745 = 12.3725 m 2 . Example 3: Find the area (in m 2) of an isosceles triangle, whose sides ...Calculate the length of a leg if given other sides and angles ( a b ) : Calculate the length of a hypotenuse if given legs and angles at the hypotenuse ( c ) : Calculate the length of sides of a right triangle using Pythagorean theorem ( c a b ) :just to check whether term sqrt(a^2+b^2-2*a*b*cos(theta)), in formula for "perimeter", may give correct answer (altrough i have already known that it can''t); for example a=b=2^32, theta =1/2^48, just 26 significant figgures, out of 50, due to cosine being too close to 1, better method: c=sqrt((a-b)^2+4*a*b*(sin(theta/2))^2), but nobody uses it, …Heron's formula is easiest as it "requires no arbitrary choice of side as base or vertex as origin, contrary to other formulas for the area of a triangle:" A = √s(s − a)(s − b)(s − c) where s = p / 2 is half of the perimeter p = a + b + c (called the semiperimeter of the triangle).We know that the formula that is used to find the area of a triangle with 3 sides is, Area =√[s(s-a)(s-b)(s-c)], where 'a', 'b', 'c' are the three sides and 's' is the semi perimeter of the triangle. In this case, a = 8; b = 11, c = 13, and the semi-perimeter is, s = 8 + 11 + 13 = 32/2 = 16 Question 2: Perimeter of the equilateral triangle is 63 cm find the side of the triangle. Solution: Perimeter of an equilateral triangle = 3×side. 3×side = 64. side = 63/3. side = 21 cm. Question 3: Find the measure of the third side of a right-angled triangle if the two sides are 6 cm and 8 cm. Solution: Perpendicular = 6 cm. Base = 8 cm michigan state football coaching staffdust ruffle queenfuel pump driver module f150vitamin ahoppewine searcher comthe glencoe clubapple watch won't erase all content and settingsbutterbean the boxeravery design and print | 3,061 | 11,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-27 | latest | en | 0.843865 |
https://us.metamath.org/nfeuni/lenc.html | 1,709,216,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00299.warc.gz | 581,028,749 | 9,551 | New Foundations Explorer < Previous Next > Nearby theorems Mirrors > Home > NFE Home > Th. List > lenc GIF version
Theorem lenc 6223
Description: Less than or equal condition for the cardinality of a number. (Contributed by SF, 18-Mar-2015.)
Hypothesis
Ref Expression
lenc.1 A V
Assertion
Ref Expression
lenc (M NC → (Mc Nc Ax M x A))
Distinct variable groups: x,M x,A
Proof of Theorem lenc
Dummy variables f g p q y are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 elncs 6119 . 2 (M NCy M = Nc y)
2 ncex 6117 . . . . . . 7 Nc y V
3 ncex 6117 . . . . . . 7 Nc A V
42, 3brlec 6113 . . . . . 6 ( Nc yc Nc Ap Nc yq Nc Ap q)
5 elnc 6125 . . . . . . . . . . 11 (p Nc ypy)
6 bren 6030 . . . . . . . . . . 11 (pyf f:p1-1-ontoy)
75, 6bitri 240 . . . . . . . . . 10 (p Nc yf f:p1-1-ontoy)
8 elnc 6125 . . . . . . . . . . 11 (q Nc AqA)
9 bren 6030 . . . . . . . . . . 11 (qAg g:q1-1-ontoA)
108, 9bitri 240 . . . . . . . . . 10 (q Nc Ag g:q1-1-ontoA)
117, 10anbi12i 678 . . . . . . . . 9 ((p Nc y q Nc A) ↔ (f f:p1-1-ontoy g g:q1-1-ontoA))
12 eeanv 1913 . . . . . . . . 9 (fg(f:p1-1-ontoy g:q1-1-ontoA) ↔ (f f:p1-1-ontoy g g:q1-1-ontoA))
1311, 12bitr4i 243 . . . . . . . 8 ((p Nc y q Nc A) ↔ fg(f:p1-1-ontoy g:q1-1-ontoA))
14 f1of1 5286 . . . . . . . . . . . . . . . 16 (g:q1-1-ontoAg:q1-1A)
15143ad2ant2 977 . . . . . . . . . . . . . . 15 ((f:p1-1-ontoy g:q1-1-ontoA p q) → g:q1-1A)
16 simp3 957 . . . . . . . . . . . . . . 15 ((f:p1-1-ontoy g:q1-1-ontoA p q) → p q)
17 f1ores 5300 . . . . . . . . . . . . . . 15 ((g:q1-1A p q) → (g p):p1-1-onto→(gp))
1815, 16, 17syl2anc 642 . . . . . . . . . . . . . 14 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (g p):p1-1-onto→(gp))
19 f1ocnv 5299 . . . . . . . . . . . . . . 15 (f:p1-1-ontoyf:y1-1-ontop)
20193ad2ant1 976 . . . . . . . . . . . . . 14 ((f:p1-1-ontoy g:q1-1-ontoA p q) → f:y1-1-ontop)
21 f1oco 5308 . . . . . . . . . . . . . 14 (((g p):p1-1-onto→(gp) f:y1-1-ontop) → ((g p) f):y1-1-onto→(gp))
2218, 20, 21syl2anc 642 . . . . . . . . . . . . 13 ((f:p1-1-ontoy g:q1-1-ontoA p q) → ((g p) f):y1-1-onto→(gp))
23 f1ocnv 5299 . . . . . . . . . . . . 13 (((g p) f):y1-1-onto→(gp) → ((g p) f):(gp)–1-1-ontoy)
24 vex 2862 . . . . . . . . . . . . . . . . 17 g V
25 vex 2862 . . . . . . . . . . . . . . . . 17 p V
2624, 25resex 5117 . . . . . . . . . . . . . . . 16 (g p) V
27 vex 2862 . . . . . . . . . . . . . . . . 17 f V
2827cnvex 5102 . . . . . . . . . . . . . . . 16 f V
2926, 28coex 4750 . . . . . . . . . . . . . . 15 ((g p) f) V
3029cnvex 5102 . . . . . . . . . . . . . 14 ((g p) f) V
3130f1oen 6033 . . . . . . . . . . . . 13 (((g p) f):(gp)–1-1-ontoy → (gp) ≈ y)
3222, 23, 313syl 18 . . . . . . . . . . . 12 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (gp) ≈ y)
33 elnc 6125 . . . . . . . . . . . 12 ((gp) Nc y ↔ (gp) ≈ y)
3432, 33sylibr 203 . . . . . . . . . . 11 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (gp) Nc y)
35 imass2 5024 . . . . . . . . . . . . 13 (p q → (gp) (gq))
36353ad2ant3 978 . . . . . . . . . . . 12 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (gp) (gq))
37 f1ofo 5293 . . . . . . . . . . . . . 14 (g:q1-1-ontoAg:qontoA)
38 foima 5274 . . . . . . . . . . . . . 14 (g:qontoA → (gq) = A)
3937, 38syl 15 . . . . . . . . . . . . 13 (g:q1-1-ontoA → (gq) = A)
40393ad2ant2 977 . . . . . . . . . . . 12 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (gq) = A)
4136, 40sseqtrd 3307 . . . . . . . . . . 11 ((f:p1-1-ontoy g:q1-1-ontoA p q) → (gp) A)
42 sseq1 3292 . . . . . . . . . . . 12 (x = (gp) → (x A ↔ (gp) A))
4342rspcev 2955 . . . . . . . . . . 11 (((gp) Nc y (gp) A) → x Nc yx A)
4434, 41, 43syl2anc 642 . . . . . . . . . 10 ((f:p1-1-ontoy g:q1-1-ontoA p q) → x Nc yx A)
45443expia 1153 . . . . . . . . 9 ((f:p1-1-ontoy g:q1-1-ontoA) → (p qx Nc yx A))
4645exlimivv 1635 . . . . . . . 8 (fg(f:p1-1-ontoy g:q1-1-ontoA) → (p qx Nc yx A))
4713, 46sylbi 187 . . . . . . 7 ((p Nc y q Nc A) → (p qx Nc yx A))
4847rexlimivv 2743 . . . . . 6 (p Nc yq Nc Ap qx Nc yx A)
494, 48sylbi 187 . . . . 5 ( Nc yc Nc Ax Nc yx A)
50 vex 2862 . . . . . . . 8 x V
51 lenc.1 . . . . . . . 8 A V
5250, 51nclec 6195 . . . . . . 7 (x ANc xc Nc A)
5350eqnc 6127 . . . . . . . . 9 ( Nc x = Nc yxy)
54 elnc 6125 . . . . . . . . 9 (x Nc yxy)
5553, 54bitr4i 243 . . . . . . . 8 ( Nc x = Nc yx Nc y)
56 breq1 4642 . . . . . . . 8 ( Nc x = Nc y → ( Nc xc Nc ANc yc Nc A))
5755, 56sylbir 204 . . . . . . 7 (x Nc y → ( Nc xc Nc ANc yc Nc A))
5852, 57syl5ib 210 . . . . . 6 (x Nc y → (x ANc yc Nc A))
5958rexlimiv 2732 . . . . 5 (x Nc yx ANc yc Nc A)
6049, 59impbii 180 . . . 4 ( Nc yc Nc Ax Nc yx A)
61 breq1 4642 . . . . 5 (M = Nc y → (Mc Nc ANc yc Nc A))
62 rexeq 2808 . . . . 5 (M = Nc y → (x M x Ax Nc yx A))
6361, 62bibi12d 312 . . . 4 (M = Nc y → ((Mc Nc Ax M x A) ↔ ( Nc yc Nc Ax Nc yx A)))
6460, 63mpbiri 224 . . 3 (M = Nc y → (Mc Nc Ax M x A))
6564exlimiv 1634 . 2 (y M = Nc y → (Mc Nc Ax M x A))
661, 65sylbi 187 1 (M NC → (Mc Nc Ax M x A))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 176 ∧ wa 358 ∧ w3a 934 ∃wex 1541 = wceq 1642 ∈ wcel 1710 ∃wrex 2615 Vcvv 2859 ⊆ wss 3257 class class class wbr 4639 ∘ ccom 4721 “ cima 4722 ◡ccnv 4771 ↾ cres 4774 –1-1→wf1 4778 –onto→wfo 4779 –1-1-onto→wf1o 4780 ≈ cen 6028 NC cncs 6088 ≤c clec 6089 Nc cnc 6091 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1546 ax-5 1557 ax-17 1616 ax-9 1654 ax-8 1675 ax-13 1712 ax-14 1714 ax-6 1729 ax-7 1734 ax-11 1746 ax-12 1925 ax-ext 2334 ax-nin 4078 ax-xp 4079 ax-cnv 4080 ax-1c 4081 ax-sset 4082 ax-si 4083 ax-ins2 4084 ax-ins3 4085 ax-typlower 4086 ax-sn 4087 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-nan 1288 df-tru 1319 df-ex 1542 df-nf 1545 df-sb 1649 df-eu 2208 df-mo 2209 df-clab 2340 df-cleq 2346 df-clel 2349 df-nfc 2478 df-ne 2518 df-ral 2619 df-rex 2620 df-reu 2621 df-rmo 2622 df-rab 2623 df-v 2861 df-sbc 3047 df-nin 3211 df-compl 3212 df-in 3213 df-un 3214 df-dif 3215 df-symdif 3216 df-ss 3259 df-pss 3261 df-nul 3551 df-if 3663 df-pw 3724 df-sn 3741 df-pr 3742 df-uni 3892 df-int 3927 df-opk 4058 df-1c 4136 df-pw1 4137 df-uni1 4138 df-xpk 4185 df-cnvk 4186 df-ins2k 4187 df-ins3k 4188 df-imak 4189 df-cok 4190 df-p6 4191 df-sik 4192 df-ssetk 4193 df-imagek 4194 df-idk 4195 df-iota 4339 df-0c 4377 df-addc 4378 df-nnc 4379 df-fin 4380 df-lefin 4440 df-ltfin 4441 df-ncfin 4442 df-tfin 4443 df-evenfin 4444 df-oddfin 4445 df-sfin 4446 df-spfin 4447 df-phi 4565 df-op 4566 df-proj1 4567 df-proj2 4568 df-opab 4623 df-br 4640 df-1st 4723 df-swap 4724 df-sset 4725 df-co 4726 df-ima 4727 df-si 4728 df-id 4767 df-xp 4784 df-cnv 4785 df-rn 4786 df-dm 4787 df-res 4788 df-fun 4789 df-fn 4790 df-f 4791 df-f1 4792 df-fo 4793 df-f1o 4794 df-2nd 4797 df-txp 5736 df-ins2 5750 df-ins3 5752 df-image 5754 df-ins4 5756 df-si3 5758 df-funs 5760 df-fns 5762 df-trans 5899 df-sym 5908 df-er 5909 df-ec 5947 df-qs 5951 df-en 6029 df-ncs 6098 df-lec 6099 df-nc 6101 This theorem is referenced by: ce0lenc1 6239 nchoicelem13 6301
Copyright terms: Public domain W3C validator | 3,954 | 7,170 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-10 | latest | en | 0.089647 |
https://www.physicsforums.com/threads/lightnings-lethality-why-doesnt-it-kill.631639/ | 1,695,297,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00531.warc.gz | 1,043,985,798 | 19,643 | Lightning's Lethality: Why Doesn't it Kill?
• mugenshiyo
mugenshiyo
Does anybody know why lightning doesn't kill any living thing it touches if it is supposed to be four or five times hotter than the sun and has a huge amount of voltage? The heat alone should be able to kill living things instantly, right?
Does anybody know why lightning doesn't kill any living thing it touches if it is supposed to be four or five times hotter than the sun and has a huge amount of voltage? The heat alone should be able to kill living things instantly, right?
Wait, what? There have been reports of people killed by lightning! What do you mean that it doesn't kill any living thing it touches? Where do you get such an idea?
Zz.
I think (but am not sure) that he is asking how come it doesn't *always* kill whatever it strikes.
If so, it just depends on how much of the current goes through the critical parts of the body (brain and heart). If for some reason most of the current bypasses the person's core (or animal's core), then they may survive the strike...
I think (but am not sure) that he is asking how come it doesn't *always* kill whatever it strikes.
If so, it just depends on how much of the current goes through the critical parts of the body (brain and heart). If for some reason most of the current bypasses the person's core (or animal's core), then they may survive the strike...
Yes, that is what I meant. Sorry for not phrasing the question right.
I get what your saying. I was wondering the exact process though. I understand that all the current doesn't travel through the body, even with a direct hit by lightning, but if the lightning is so hot, that should be independent of the current, right? The heat should have exploded the water inside your body the same way it explodes clothes outside the body and inside trees, destroying them.
Unless when they say that the temperature of a lightning bolt is five times the sun, they really mean that it has the potential to reach five times the heat of the sun if the current is allowed to move freely through whatever it touches.
Just trying to sound it out, but when I think about it a lightning strike is not that much more lethal than a gunshot unless the survivors simply aren't ever hit by a direct lightning bolt. That would explain some things.
Just like if you take a blow torch to your hand, it doesn't kill you, it demolishes your hand. With lightning, the heat only spikes where the current travels, so of it doesn't hit vitals, it's less likely to kill you.
Lightning can take different paths depending on where it finds ground, or the lowest potential path.
Damage to the human body:
Lightning affects the many electrochemical systems in the body. People struck by lightning can suffer from nerve damage, memory loss, personality change, and emotional problems. There is a national support group for lightning and electric shock survivors.
An example is some single nerve cells, such as those extending from the brain to the foot, can be as long as 6 feet or more. These types of cells are most prone to lightning damage due to the instantaneous potential difference across the length of the cell as lightning begins to enter the body.
The intense heat of the lightning stroke can turn sweat instantly to steam and the tremendous pressure of the steam has been known to blow people's boots, shoes, and clothing off them. In places where metal is in contact with or close proximity to the body, such as jewelry or belt buckles, burn marks are found. Likewise, burn marks are found in places where the body had been sweaty, such as the feet, underarms, and chest.
http://www.nssl.noaa.gov/primer/lightning/ltg_damage.html
http://www.nssl.noaa.gov/primer/lightning/ltg_damage.html
I read through that link and it instructs to take safe shelter, but it doesn't really describe safe shelter.
If I'm up on a mountain, what should I do to be safe? It says stay away from trees, but I know you also don't want to isolate yourself.
I know from a physical perspective, lighting will strike the path of least resistance. Air has a very high resistivity compared to most solids/liquids, so being higher up is therefore dangerous. I'd guess that human resistivity is lower than that of wood, or rock, so you'd need to be significantly lower than the surrounding wood/rock, so that your elevation difference will outweigh the lower resistivity of your body.
I read through that link and it instructs to take safe shelter, but it doesn't really describe safe shelter.
If I'm up on a mountain, what should I do to be safe? It says stay away from trees, but I know you also don't want to isolate yourself.
I know from a physical perspective, lighting will strike the path of least resistance. Air has a very high resistivity compared to most solids/liquids, so being higher up is therefore dangerous. I'd guess that human resistivity is lower than that of wood, or rock, so you'd need to be significantly lower than the surrounding wood/rock, so that your elevation difference will outweigh the lower resistivity of your body.
It provides links to many sources, you need to read about the links.
For example that page links to this one for safety.
Level-3: When lightning threatens, go to a safer location. Do not hesitate. The lightning casualty lore is replete with tales of persons just about to make it to safety when they were struck. Even a few extra minutes lead time can be life saving.
What is a safer location? The safest place commonly available during a lightning storm is a large, fully enclosed, substantially constructed building, e.g. your typical house, school, library, or other public building. Substantial construction also implies the building has wiring and plumbing, which can conduct lightning current safely to ground. However, any metal conductor exposed to the outside must not be touched precisely because it could become a lightning conduit. Once inside, stay away from corded telephones, electrical appliances, lighting fixtures, ham radio microphones, electric sockets and plumbing. Don’t watch
lightning from open windows or doorways. Inner rooms are generally preferable from a safety viewpoint.
If you can’t reach a substantial building, an enclosed vehicle with a solid metal roof and metal sides is a reasonable second choice. As with a building, avoid contact with conducting paths going outside. Close the windows, lean away from the door, put your hands in your lap and don’t touch the steering wheel, ignition, gear shifter or radio. Convertibles, cars with fiberglass or plastic shells, and open-framed vehicles are not suitable lightning shelters.
continued...
http://www.ametsoc.org/POLICY/Lightning_Safety_Article.pdf
I've often wondered this too why lighting doesn't kill you with the heat if nothing else. My guess is that perhaps this is the temperature directly on the lighting arc itself. Air has very high resistance, people have less resistance so not as much heat would be generated passing current through you.
Also air is very low density so it may burn you where the lightning hits you but it would dissipate quickly like tin foil coming out of the oven.
I think I've heard people indeed do have burn where the lighting goes in and out which would make sense.
Also I've read they're often burnt where they had moisture on their bodies from the moisture vaporizing. While it burns them it also probably protects them from more serious injury by throwing off the heat.
It reminds me of trying to strike an arc with a stick welder and sticking the rod the arc strike instantly melts the metal and sticks the rod and the base metal together. If you leave it there for a couple seconds it'll get hot from the current flow but not that hot because the real heat comes from the current passing through the air gap between the two metals.
I've often wondered this too why lighting doesn't kill you with the heat if nothing else. My guess is that perhaps this is the temperature directly on the lighting arc itself. Air has very high resistance, people have less resistance so not as much heat would be generated passing current through you.
Also air is very low density so it may burn you where the lightning hits you but it would dissipate quickly like tin foil coming out of the oven.
I think I've heard people indeed do have burn where the lighting goes in and out which would make sense.
Also I've read they're often burnt where they had moisture on their bodies from the moisture vaporizing. While it burns them it also probably protects them from more serious injury by throwing off the heat.
It reminds me of trying to strike an arc with a stick welder and sticking the rod the arc strike instantly melts the metal and sticks the rod and the base metal together. If you leave it there for a couple seconds it'll get hot from the current flow but not that hot because the real heat comes from the current passing through the air gap between the two metals.
You could read the answers in post #6.
There are 2 possible ways lightning can damage you. 1 is the immediately arrhythmia in the heart from the lightning, which why ER's will always check your ECG first. 2 is the burn itself, which is not actually the surface burn. The electrical current travels in the body and slowly releases heat and burns on the inside, which takes some time and also is the most severe kind.
Lightning is not intrinsically hot. It generates heat as a result of its passage through resistive media that dissipate the energy. In the human body the highest resistance and heat occurs on the skin. Inside the body the resistance is relatively low so less heat is dissipated internally. The lightning also spreads out inside the body which reduces the resistance further and dissipates the heat over a larger volume. Of course if the strike is powerful enough it can still cook you. | 2,068 | 9,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.95868 |
https://math.answers.com/Q/What_is_33_percent_as_a_simplified_fraction | 1,642,888,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303884.44/warc/CC-MAIN-20220122194730-20220122224730-00034.warc.gz | 439,408,627 | 70,914 | 0
# What is 33 percent as a simplified fraction?
Wiki User
2012-09-14 11:27:17
Best Answer
= 33%/100%
= 33/100 in fraction
Wiki User
2012-09-14 11:27:17
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Still have questions? | 137 | 379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.882661 |
https://learn.careers360.com/school/question-find-the-period-of-the-function-fxsin-pi-x-n-cos-pi-x-n-1-38043/ | 1,631,959,543,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00663.warc.gz | 436,261,387 | 110,854 | # Find the period of the function f(x)=sin (pi x / n!) - cos (pi x / n+1!
Answers (1)
Solution: We have ,
$f(x)=\sin \frac{\pi x}{n!}-\cos \frac{\pi x}{(n+1)!}$
$\\ \\ \sin \frac{\pi x}{n!}\rightarrow T_{1}=(\frac{2\pi}{\frac{\pi}{n!}}) =2n!\\ \\ \cos \frac{\pi x}{(n+1)!}\rightarrow T_{2}=(\frac{2\pi}{\frac{\pi}{(n+1)!}})=2(n+1)!\\ \\ \Rightarrow \hspace{1cm}T=2(n+1)!$
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Questions | 476 | 1,594 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.605875 |
https://engineering.stackexchange.com/questions/13332/how-to-couple-two-pistons-to-a-rotating-shaft-so-that-the-piston-heads-move-like | 1,701,315,192,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00581.warc.gz | 269,105,076 | 41,999 | # How to couple two pistons to a rotating shaft so that the piston heads move like this?
Assume two vertical pistons with their flat head facing upward.
Is it possible to couple them to one rotating shaft in a way that the position vs. time diagram gets like this? (This is one complete cycle)
Each color being representing a piston.
The equations are:
Red: $$2 \left|\cos(\pi t)\right| - \frac12$$
Blue: $$2 \left|\sin(\pi t)\right| - \frac12$$ (The first with a phase difference of $\pi \over 2$)
I think because of the Absolute values, this can not be a simple crankshaft.
Can anyone help?
EDIT:
Actually I want to get as close as possible to this diagram of motion, which is the ideal theoretical case. Maybe this can help you for a better approximation.
• That instantaneous velocity direction change is a killer. would require infinite acceleration. Jan 19, 2017 at 16:59
• @joojaa Then we could soften the edge so that it becomes seamless. Would $\sin^2(\pi t)$ be fine enough? Is sin squared motion possible to be achieved using mechanical structures?
– AHB
Jan 19, 2017 at 17:55
• How exact do you need to be with any of this? A little insight as to why they have to have that position profile could help.
– JMac
Jan 19, 2017 at 18:09
• @JMac I added what you want.
– AHB
Jan 19, 2017 at 18:32
• You can do a lot with 4-bar linkages. Note that the vertical movement of the feet (ignoring the horizontal movement) is very close to the profile you're looking for. Jan 20, 2017 at 13:29 | 401 | 1,503 | {"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 | 0.925816 |
https://astronomy.stackexchange.com/questions/13590/is-it-ever-possible-that-the-graviation-from-the-mass-of-kinetic-rotational-ener | 1,716,193,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00183.warc.gz | 90,552,340 | 37,877 | # Is it ever possible that the graviation from the mass of kinetic rotational energy will overcome centrifugal force?
This is, I think, somewhat intuitively obvious, but this comment got me thinking about this:
what is gravitational force?
"Rotation speed can create centrifugal force opposing gravity and making things lighter," Rotation also stores energy, and energy is mass, per $E=mc^2$. A body spinning sufficiently fast will exert higher gravity, e.g. a slowly-spinning neutron star will have a weaker gravitational pull than equivalent neutron star that spins very fast."
Intuitively, my answer is "no way", added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation. Even with a Neutron Star I would think it's impossible, but I'm not 100% sure, so I thought I'd ask.
If we ignore relativity from motion but not $E=mc^2$, as that's the crux of the question
Centrifugal force = $m v^2/r$
Kinetic energy of rotation = $1/2 I w^2 = 1/5 m v^2$
Mass equivalent of kinetic energy = $1/5 m v^2/c^2$
gravitational force $g=\frac{Gm}{r^2}$
so if we apply the gravitational force of kinetic energy
$g = G(1/5 m * v^2/c^2) / r^2$, or, simplified, $G*m*v^2 / r^2 c^2$
and we compare the two equations Centrifugal $m v^2/r$ additional gravitational $G*m*v^2 / r^2 c^2$
we can remove $m*v^2$ from both on the top and 1 $r$ from the bottom
additional gravitation ratio to centrifugal force = $G/r * c^2$
$G$ and $c$ are numerical. $G$ is very small, $c$ is very big and the ratio grows smaller as the radius grows larger.
gravitational constant: $6.67408 × 10^{-11} m^3 kg^{-1} s^{-2} c = 2.998 x 10^8 m s-2 the ratio, unless my math is broken, centrifugal force to additional gravity from added mass by kinetic energy of motion =$1/r * 1.35 * 10^{27}$, so you'd need a hugely small r, almost a plank length or a singularity where the added gravity from kinetic energy of rotation would overcome the "flicking off the surface" or centrifugal force. When I work out the units, I get meters per kilogram, which I don't think is right. The units should cancel out with a ratio of two forces in opposite directions, so I think I made an error, but I don't see where I made it. My question is two fold. 1) is my math broken? and if so, where? and 2) is the added mass from kinetic rotational energy ever relevant, say in a very rapidly rotating Neutron star? Could it ever assist the Neutron star in collapse or add gravity?, I can see how it could add to flattening, as rotation flattens objects naturally and perhaps, add a speck of gravity on the polls where centrifugal force is zero, but logically, I think, rotational energy would end up spinning any non black hole object apart, long before the added energy of rotation had enough mass to make a measurable difference on gravity. Is my sense right or is there a situation where added mass from energy of rotation could overcome the centrifugal force? • This site supports TeX style maths, You could make your question prettier by TeX'ing the maths. Feb 7, 2016 at 11:43 • Yes, your maths is broken. Questions like these cannot be addressed without GR. But yes, rapid rotation does increase the radius of a neutron star and increase the maximum mass of a neutron star that can be supported by a given eqn of state. Centrifugal force always wins for lower mass neutron stars. Feb 7, 2016 at 21:33 • "added mass by rotational velocity I would think, could never be extensive enough to counteract the "flicking off" or centrifugal force of very fast rotation. Even with a Neutron Star I would think it's impossible" - You are right with the neutron star. It would increase gravity and as result reduce impact of centrifugal force but not enough to prevent escape with sufficient angular velocity. Only if you increase the density even more (and get a black hole) this ceases to be the case. (and how black hole, which is a singularity=point can rotate... that's a whole can of worms.) – SF. Feb 7, 2016 at 21:34 • BTW, there's (probably) one body between a neutron star and black hole, density-wise: a quark star. But we haven't discovered any yet (conditions for one to form are exceedingly rare) so we don't exactly know if it would satisfy conditions of your question. – SF. Feb 8, 2016 at 23:50 • I think you are misunderstanding the statement you quote about the centrifugal force making things lighter, it wasn't meant to be about kinetic energy being equivalent to mass and therefore exerting gravity, it's just a much simpler statement about fictitious forces in rotating coordinate systems in Newtonian physics. In a rotating coordinate system centered on the center of the Earth, any object experiences a centrifugal force which is in the opposite direction as gravity, so it can partly cancel the gravitational force from their perspective. Apr 14, 2016 at 17:58 ## 1 Answer In some sense, the answer is trivially 'yes': Due to spacetime curvature, there exist no circular orbits below$\frac{3}{2}r_s$, with$r_s$the Schwarzschild radius. So whenever you get an object sufficiently dense, that its radius is below its$\frac{3}{2}r_s$, the mass equivalent to any nonzero rotational energy overcomes the centrifugal force of an orbit at$\frac{3}{2}r_s$. Below$\frac{3}{2}r_s\$ even zero rotational energy overcomes the centrifugal force, since the latter is then pointing inward.
This doesn't require a Black Hole. | 1,352 | 5,479 | {"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} | 3.75 | 4 | CC-MAIN-2024-22 | latest | en | 0.881426 |
http://www.jiskha.com/display.cgi?id=1359474310 | 1,498,171,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319912.4/warc/CC-MAIN-20170622220117-20170623000117-00623.warc.gz | 578,664,718 | 3,808 | # physics
posted by on .
A golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0°. If the hang time of the golf ball is 3.44 seconds, what is the range of the golf ball?
• physics - ,
t= 2vₒ•sinα/g
L=vₒ²•sin2α/g,
• physics - ,
94 | 92 | 263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-26 | latest | en | 0.768014 |
https://myhomeworkgeeks.com/note-please-select-consumer-foods-for-a-qnt-561-course/ | 1,618,277,890,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038071212.27/warc/CC-MAIN-20210413000853-20210413030853-00082.warc.gz | 522,776,261 | 22,769 | # Note: Please select Consumer Foods for a QNT 561 course
Note: Please select Consumer Foods for a QNT 561 course
This signature assignment is designed to align with specific program student learning outcome(s) in your program. Program Student Learning Outcomes are broad statements that describe what students should know and be able to do upon completion of their degree. The signature assignments might be graded with an automated rubric that allows the University to collect data that can be aggregated across a location or college/school and used for program improvements.
Purpose of Assignment
The purpose of this assignment is for students to synthesize the concepts learned throughout the course. This assignment will provide students an opportunity to build critical thinking skills, develop businesses and organizations, and solve problems requiring data by compiling all pertinent information into one report.
Assignment Steps
Resources: Microsoft Excel®, Signature Assignment Databases, Signature Assignment Options, Part 3: Inferential Statistics
Scenario: Upon successful completion of the MBA program, say you work in the analytics department for a consulting company. Your assignment is to analyze one of the following databases:
Manufacturing
Hospital
Consumer Food
Financial
Select one of the databases based on the information in the Signature Assignment Options.
Provide a 1,600-word detailed, statistical report including the following:
Explain the context of the case
Provide a research foundation for the topic
Present graphs
Explain outliers
Prepare calculations
Conduct hypotheses tests
Discuss inferences you have made from the results
This assignment is broken down into four parts:
Part 1 – Preliminary Analysis
Part 2 – Examination of Descriptive Statistics
Part 3 – Examination of Inferential Statistics
Part 4 – Conclusion/Recommendations
Part 1 – Preliminary Analysis (3-4 paragraphs)
Generally, as a statistics consultant, you will be given a problem and data. At times, you may have to gather additional data. For this assignment, assume all the data is already gathered for you.
State the objective:
What are the questions you are trying to address?
Describe the population in the study clearly and in sufficient detail:
What is the sample?
Discuss the types of data and variables:
Are the data quantitative or qualitative?
What are levels of measurement for the data?
Part 2 – Descriptive Statistics (3-4 paragraphs)
Examine the given data.
Present the descriptive statistics (mean, median, mode, range, standard deviation, variance, CV, and five-number summary).
Identify any outliers in the data.
Present any graphs or charts you think are appropriate for the data.
Note: Ideally, we want to assess the conditions of normality too. However, for the purpose of this exercise, assume data is drawn from normal populations.
Part 3 – Inferential Statistics (2-3 paragraphs)
Use the Part 3: Inferential Statistics document.
Create (formulate) hypotheses
Run formal hypothesis tests
Make decisions. Your decisions should be stated in non-technical terms.
Hint: A final conclusion saying “reject the null hypothesis” by itself without explanation is basically worthless to those who hired you. Similarly, stating the conclusion is false or rejected is not sufficient.
Part 4 – Conclusion and Recommendations (1-2 paragraphs)
Include the following:
What are your conclusions?
What do you infer from the statistical analysis?
State the interpretations in non-technical terms. What information might lead to a different conclusion?
Are there any variables missing?
What additional information would be valuable to help draw a more certain conclusion?
Format your assignment consistent with APA format.
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# MGT201 Complete Solution Assignment#01 Spring 2013
MGT201 Complete Solution Assignment#01 Spring 2013
Spring%202013_MGT201%20solution.docx
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### Replies to This Discussion
* MGT 201 1ST Assignment solution*
Presented by :
Payback period:
Years Cash flow Cumulative cash flow 0 (500,000) (500,000) 1 (800,000) (1300,000) 2 150,000 (1150,000) 3 200,000 (950,000) 4 250,000 (700,000) 5 300,000 (400,000) 6 300,000 (100,000) 7 300,000 200,000 8 300,000 500,000 9 300,000 800,000 10 300,000 1100,000
Payback period: = 6 + (100,000/300,000)
= 6 +0.333
Payback period = 6.33 years
Net present value:
Year Cash flow Present value discount factor (14%) Present value 0 (500,000) 1.00 ( 500,000) 1 (800,000) 0.877 (701600) 2 150,000 0.769 115,350 3 200,000 0.675 135,000 4 250,000 0.592 148,000 5-10 300,000 2.304 691,200 Net present value (112,050)
Because the net present value is negative the project is unacceptable.
Year Cash flow Discount factor (12%) Present value Discount factor (11%) Present value 0 (500,000) 1.00 (500,000) 1.00 (500,000) 1 (800,000) 0.893 (714,400) 0.901 ( 720,800) 2 150,000 0.797 119,550 0.812 121,800 3 200,000 0.712 142,400 0.731 146,200 4 250,000 0.636 159,000 0.659 164,750 5-10 300,000 2.613 783,900 2.787 836,100 ( 9,550) 48,050
=0.01 [ X [0.11 48050, IRR O, 0.12 (9550)]48050]57600
= X/0.01=48050/57600
X = (0.01)*(48050)/57600
=0.00834
X = 0.00834
= X + 0.11
=O.OO834+0.11
IRR =0.11834 OR 11.834%
Because the internal rate of return is less than the required rate of return the project would not be acceptable.
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47 Slides--Turing Machines
# 47 Slides--Turing Machines - CS103 HO#47 Slides-Turing...
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CS103 HO#47 Slides--Turing Machines May 14, 2010 1 T R R u v x y z Context-Free Grammars Long strings force repeated variables in their derivations. ... ... y z x DFA’s Long strings force repeated states. The idea behind the pumping lemmas Strings that are long enough satisfy the pumping lemmas. If we can find one that doesn’t, the language must not be regular or context-free. Theorem 2.34. Pumping Lemma for Context-Free Languages If A is a context-free language, then there is a number p (the pumping length) such that if s is any string in A of length at least p, then s may be divided into five pieces s = uvxyz satisfying the conditions 1. for each i 0, uv i xy i z A, 2. |vy| > 0, and 3. |vxy| p. T R R u v x y z T * u R z * u v R y z * u v x y z T R R u v R y z v x y uv 2 xy 2 z T R u z x uv 0 xy 0 z Example 2.37. B = { a n b n c n | n 0 } is not context-free. Assume B is CF. Let s = a p b p c p , where p is the pumping length. a p b p c p We don’t know exactly where vxy falls, but |vxy| p and |vy| 0. At least one of v, y is not empty, suppose it is v. v might contain just one symbol, in which case pumping s unbalances the number of a's, b's, and c's, or v might contain 2 symbols, in which case pumping s gets symbols out of order. So s cannot be pumped, and B is not context-free. Suppose s = uvxyz Typo in Sipser: Page 125, third and fourth lines of the third paragraph.
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Ask a homework question - tutors are online | 583 | 2,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.811041 |
http://stackoverflow.com/questions/18996156/cell-mean-imputation?answertab=active | 1,394,718,546,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678676855/warc/CC-MAIN-20140313024436-00017-ip-10-183-142-35.ec2.internal.warc.gz | 133,694,513 | 15,272 | Cell Mean Imputation
I wish I was better at R, but I need some help with something pretty basic.
I am having some problems writing a function that will do cell mean imputation. The data I am currently working with has 3 columns and the way I currently have the function written, the mean for the observed values in the 3rd column is imputed into all of the NAs in all 3 columns. How can I fix this? Thank you!
``````cellmean.imp <- function(a){
for (i in 1:dim(a)[2]){
new=replace(a, is.na(a), mean(a[, i], na.rm=TRUE))
}
return(new)
}
``````
Sorry, I forgot to add: I am trying to impute the mean of the observed values for the 1st column into the NAs in the first column, then the mean of the observed values for hte 2nd column into the NAs in the second column, and so on.
-
Is `a` a matrix or a data.frame? Also, you make it sound like the 3rd column plays a special role, while it does not in your code: can you please clarify? – flodel Sep 25 '13 at 4:34
@flodel - I think they want the mean for each column imputed into the `NA`s in each corresponding column. Something like: `apply(x,2,function(x) {x[is.na(x)] <- mean(x,na.rm=TRUE); x} )` – thelatemail Sep 25 '13 at 5:53
If I understand you correctly, you want to put the mean of third column into every NA in the matrix
``````x <- matrix(rnorm(30),10,3)
``````
Introduce a few NAs
``````x[3,1] <- NA
x[4,1] <- NA
x[5,2] <- NA
x[6,3] <- NA
``````
Replace them with the third column mean
``````x[is.na(x)] <- mean(x[,3],na.rm=TRUE)
``````
-
``````x[,3][is.na(x[,3]] <- mean(x[,3], na.rm=TRUE) | 489 | 1,567 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-10 | latest | en | 0.895764 |
http://mathhelpforum.com/calculus/67975-fourier-series.html | 1,529,805,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00382.warc.gz | 195,916,209 | 11,046 | 1. ## Fourier series
Hey guys.
I wrote the problem in word, it's much easier for me
2. Originally Posted by asi123
The problem is the other one, $\displaystyle (x^3)$, how can I get $\displaystyle \sum_{n=1}^\infty\frac1{n^3}$ from that?
You'll be famous if you succeed. $\displaystyle \sum_{n=1}^\infty\frac1{n^3}$ is Apéry's constant, for which there is no known closed form. Even less is known about $\displaystyle \sum_{n=1}^\infty\frac1{n^5}$.
3. Originally Posted by Opalg
You'll be famous if you succeed. $\displaystyle \sum_{n=1}^\infty\frac1{n^3}$ is Apéry's constant, for which there is no known closed form. Even less is known about $\displaystyle \sum_{n=1}^\infty\frac1{n^5}$.
Well, the question is find the sum of the series $\displaystyle \sum_{n=1}^\infty\frac1{n^5}$, are you sure that can not be done?
4. Originally Posted by asi123
Well, the question is find the sum of the series $\displaystyle \sum_{n=1}^\infty\frac1{n^5}$, are you sure that can not be done?
He is sure. And so am I.
Some research for you to do:
Riemann zeta function.
$\displaystyle \zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$.
Very little is known about $\displaystyle \zeta(2n + 1)$ apart from what has already been mentioned for the case n = 1.
5. Originally Posted by mr fantastic
He is sure. And so am I.
Some research for you to do:
Riemann zeta function.
$\displaystyle \zeta(2n)=\frac{(-1)^{k+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$.
Very little is known about $\displaystyle \zeta(2n + 1)$ apart from what has already been mentioned for the case n = 1.
Just in case the original poster want to see $\displaystyle \zeta(2n)$ derivation, it is here.
6. Originally Posted by asi123
Hey guys.
I wrote the problem in word, it's much easier for me | 564 | 1,754 | {"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} | 3.671875 | 4 | CC-MAIN-2018-26 | latest | en | 0.930919 |
http://theanalyticpoem.net/taxonomy/ | 1,603,214,411,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874026.22/warc/CC-MAIN-20201020162922-20201020192922-00498.warc.gz | 97,584,448 | 14,168 | # taxonomy
Like architecture, geometry constructs dimensional relationships and associations; in the process, it can also build a network of impressions. For example, anything can be mapped through a zero-dimensional point, as if forced through a tiny hole in space not unlike a black hole singularity, to reemerge on the “other side” in a different zone (the coordinate plane), unchanged but turned. Metaphorically, the subject now has a new viewpoint. The transformation, called point reflection, actually rotates the pre-image around the origin by an amount of exactly 180 degrees. We can consider that network inception arises with this timeless point in zero-dimensional space, which is actually just a location. The property of a singular point is an identity because of its isolated nature, an unconnected vertex that is equivalent to a self-aware state – “I am” and “there exists” characterize and define it. Mathematically, this property is known as reflexive, A = A. Thus, the main attribute of this network property is that of perception; an analogy from Nature is a seed, the potential for growth as a nascent vitality. With unconnected points as individual phenomena, the intellect is within them, but there is no context, no relationship among them, as Abbott described in Flatland:
“That point is a Being like ourselves, but confined to the non-dimensional Gulf. He is himself his own World, his own Universe; of any other than himself he can form no conception; he knows not Length, nor Breadth, nor Height, for he has had no experience of them; he has no cognizance even of the number Two; nor has he a thought of Plurality; for he is himself his One and All, being really Nothing.”
Using this taxonomy as a general structure, the next dimension results when two vertices are connected in space, forming a geometric segment, analogous to a stem rising from the soil. A geometric transformation using a line is known as a reflection, which maps an object across the line of symmetry, as if flipping it over a fence, i.e., if A = B, then B = A. Symmetrical objects are prevalent in Nature: most animals have bilateral symmetry where a line of reflection can be drawn through their bodies, producing two matching halves. A line represents not only a bridge, serving as the connecting edge between two or more vertices, but also polarity since the vertices may be conceptually opposed; yet even then, a stronger bond is established between the formerly isolated points as the structure increases. The property of a line is therefore duality, and its attribute is relation. Thus, selected affinities in art, science, and Nature which have dichotomy characteristics will be represented: is/is not, light/dark, good/evil, odd/even, smooth/lumpy, plus/minus, etc.
If the progression of dimensions is like that of subatomic particles, having rapidly formed during an inflationary period, then the advent of one-dimensional structures necessitates the introduction of a temporal unit. Time now exists along the infinitely narrow band – it just goes back and forth between the two nodes. However, there is still no surface to the object; this happens later in geometric evolution when another vertex is added noncollinearly, creating the simplest polygon, a triangle. It is easy to see this possibility only from a viewpoint standing outside the given dimension: we have achieved a transition to another dimensional level that likewise requires new sources and perceptions.
Two-dimensional structures are flat areas with no thickness, whose main network property in this poem is complexity. As the system has increasingly grown by adding vertices and connecting edges, the form is now essentially an infinite sheet of surface, extending in all planar directions. Once a third vertex was added and connected outside the line’s boundary, a triangle was formed (it has the minimum number of vertices and edges to make a planar figure) – a more intricate structure from Nature is a leaf. There was also simultaneously created a closed pathway or circuit that is directional, and thus also temporal, where the first and last vertices are the same, like connecting dots. It is similar to both the transitive property, if A = B, and B = C, then A = C, and the logical syllogism which is also a directed process, forming a conclusion given two or more premises. Examples of two-dimensional affinities are: direction, orientation, transit, periodicity, resonance, pattern, cycle, triangulation, as well as polygon associations such as stability, equilibrium, mosaic, tessellation, etc. Thus, within the planar figure, there is a continuous function of relay among the nodal elements.
As a further development, more than one transformational event could occur in planar space and higher dimensions. A composition is a procedure in which two or more transformations take place, one following the other, wherein the image of the first transformation becomes the pre-image of the next one, allowing the transfer of identities. It has been proven that every isometry (a transformation that preserves distance between points) can be expressed as the composition of at most three line reflections. For example, a translation is the composition of two reflections across parallel lines and a rotation is the composition of two reflections across intersecting lines.
While any object’s identity is aligned with the zero-dimensional analog, volume is the three-dimensional space that it occupies, i.e., the projection of Nature as a set of impressions. Area (two-dimensions) is its surface, and perimeter (one-dimension) is the surrounding linear enclosure. Two planes form a line when they intersect, and multiple intersecting planes connect to form a volume in space. Intersecting edges within planes in turn create new vertices and the groups of formerly separate relations and circuits become incorporated into a composite mass. This sum total is a wholly formed entity with no spatial limitations, as a tree might grow. The property of a volume is therefore found in the concept of unity, and the attribute of this network property is that of summation. Examples of three-dimensional affinities include cohesion, integration, recognition, and equanimity. Unlike other transformations, a dilation is not an isometry as the pre-image can enlarge or diminish the image, according to a scale factor. A constant of k > 1 will increase the distance between all vertices in the structure, and if 0 < k < 1, then the image will collapse. If k = -1 the image will vanish through the origin, turning 180 degrees as it does so, and reemerge unchanged in size and in another location:
Since the world that we inhabit is three-dimensional, then the foundation of our universe requires the integration of all previous dimensions, including time, without which recollection, progress, movement, and growth would not be possible. Motion of objects can only be measured relative to each other and stationary sources, such as a fixed distance, or the background of stars (which itself is in motion). If a tree represents volume, then a grove becomes a symbol of concurrence: space and time operate together as matter and memory to define all things, places, and events; this lends a network property of ability. The four-dimensional structure is called the worldline, a representation of the spacetime continuum. It allows moment and eternity, presence and absence – in short, our ability to exist in a world that is best suited to our means, knowing of no other.
Because the speed of light is constant, Einstein predicted that in a moving frame of reference objects approaching near-light speeds would undergo transformations in the form of time dilation, length contraction, and a relativistic increase in mass. Since all forms of electromagnetic radiation already travel at this light speed, the Lorentz Transformations predict the impossibility of spaceships attaining light speeds, but they do not account for objects that always travel faster than light, such as the alleged tachyon particle, whose lower speed limit is that of light, or the neutrino, also predicted to travel faster than light. Undoubtedly, such possible encounters would result in a rethinking of the laws of cause and effect: could simultaneous events be separated into dimensions greater than four? Since present theory postulates that the Universe began as a singularity, then this point would also have no time-value, existing indefinitely as the location of potential space. It would then seem reasonable to suggest that coincident to the post-inception inflationary period geometric vertices as dimensions, including time, were sequentially created in conjunction with subatomic particles – it would also grant the possibility that any additional whole or fractional dimensions beyond spacetime were simultaneous to this event as well, but presently remain unknown.
Since the zero-dimension is comprised of one vertex, the first dimension requires a minimum of two vertices, and the second dimension needs only three vertices, a pattern can be seen to develop where each dimensional state is equal to the required minimum number of vertices minus one. There is also a specific symmetry aligned with the numbers in a section of Pascal’s triangle, as detailed by mathematician George Pólya:
“The analogy between segment, triangle, tetrahedron has many aspects. A segment is contained in a straight line, a triangle in a plane, a tetrahedron in space. Straight line-segments are the simplest one-dimensional bounded figures, triangles the simplest polygons, tetrahedrons the simplest polyhedrons. The segment has 2 zero-dimensional bounding elements and its interior is one-dimensional. The triangle has 3 zero-dimensional and 3 one-dimensional bounding elements and its interior is two-dimensional. The tetrahedron has 4 zero-dimensional, 6 one-dimensional, and 4 two-dimensional bounding elements, and its interior is three-dimensional.”
2 1
3 3 1
4 6 4 1
By extension, we can enlarge this pattern to include a time vertex, originating with the third dimension but in a separate row, along with further dimensions comprising multiple temporal and spatial vertices, all aligned with the binomial expansion of Blaise Pascal. To maintain the symmetrical pattern, such additional units of space and time become fractional amounts; and as hypothetical transparent vertices they could geometrically describe multi-spatial and polytemporal features existing undetected among the known dimensions. Only through the viewpoint of a meta-observer, standing outside the familiar worldline, can sources and events in these interstices or quasi-dimensions be directly witnessed:
1
2 1
3 3 1
4 6 4 1
5 10(3/5) 10(2/5) 5(1/5) 1
6 15(4/6) 20(3/6) 15(2/6) 6(1/6) 1
If these other dimensions exist, then they contain authentic impressions, which we only partially apprehend, referenced by each archetype and source-entity – it is the domain of irreducible phenomena. And since it is defined as beyond our experiences (a priori), we can call the potential field a transcendental state. Occasionally this world is revealed through projections and glimpses of non-ordinary reality: dreams, apparitions and phantasms, delusions, mirages, hallucinations, premonitions, precognition and déjà vu feelings, ecstatic visions and exultations, revelations, epiphanies, miracles, and states of grace. Because these experiences are more generally “felt” than understood, such as sensing a “presence,” traditionally it has been easy to ascribe them to the realm of another dimension. These experiences are ubiquitous and resonate worldwide in religion, art, and philosophy, and also include their more ancient forms of spiritualism, animism, shamanism, and magic; for that reason, if no other, they must be also be an aspect Nature by their predominance among humanity. Returning to the pre-science era of natural philosophy, linked with cultural studies of ancient myths as Joseph Campbell, Frazier, and others have done, there are often found unusual explanations, representations, and the means for expressing these eccentric or supernatural phenomena as actual events. Those experiences should be compared to contemporary models by referring their similarities to a common historical baseline in the idealization, as Husserl suggested, which is valid for geometry and all other spiritual structures. In terms of writing the analytic poem, the baseline is the intersection of common associations in Nature, and the idealization refers to the dimension of source-entities – affinities should also be drawn from it to complete the network.
To freely speculate, this transcendental state is a density of impressions, not unlike Jung’s idea of an inherited collective unconscious, but here in this context, an alive and active source. This realm, then, is the reservoir of universal intellect (νους) and the medium for transmigration (μετεμψύχωσις) – the relic of past lives, the vessel of present thought, and the fabric of future possibilities:
“The sluggish cream wound curdling spirals through her tea. Better remind her of the word: metempsychosis. An example would be better. An example.
The Bath of the Nymph over the bed. Given away with the Easter number of Photo Bits: Splendid masterpiece in art colours. Tea before you put milk in. Not unlike her with her hair down: slimmer. Three and six I gave for the frame. She said it would look nice over the bed. Naked nymphs: Greece: and for instance all the people that lived then.
He turned the pages back.
–Metempsychosis, he said, is what the ancient Greeks called it. They used to believe you could be changed into an animal or a tree, for instance. What they called nymphs, for example.”
Through the framework of geometry, achieving that next step in this taxonomy requires only an additional vertex to gain another dimension, now called v+. The vertex unit for this multi-dimensional network is a membrane – a hypergeometry that is an extension of all prior networks and affinities, consisting of myriad cognitive nodes folded and packed into an edgeless density. The v+ network is made by grafting variables of cause, potential, and alternative reality. It is a compression of multiple worlds no longer containing any unique identities, the membrane wholly emerging as a new singularity beyond all previously known experiences. Although it is formally detached from our own world, all features of dimensions zero through four are present with it, just as a flat panel is part of a box, an edge is part of the box’s panels, and its corners define the edge’s length. A popular depiction of an object found in this dimension is the hypercube, described by astronomer Carl Sagan and others as a cube-within-a-cube, where all edges are of equal length and all angles between the edges are 90 degrees. The hypercube model (or tesseract, from L’Engle’s, A Wrinkle in Time) is a three-dimensional representation, i.e., merely the projected shadow of the authentic one that is beyond our capability to perceive:
Leonard Euler not only formalized graph theory, but he is also known for his comprehensive assessment of two-dimensional polygons through a formula that today still bears his name, Euler’s Formula: V – E + F = 1. Simply stated, the number of vertices minus the number of edges, plus the number of faces for any polygon or map, will always equal one (all zero dimensional units, minus one dimensional units, plus two dimensional units, is one).
And, for the next dimensional level, polyhedra, the formula is: V – E + F = 2. Can we assume that the fourth spatial dimension will likewise have the same pattern for the hypercube, V – E + F = 3 ? More generally, our modified version of Euler’s Formula aligns all units for each dimensional state, 0 through v+ :
V – E + F = S + T – Mx + 1, where S and T represent space and time values and Mx is a manifold unit.
If it is possible, then, to depict such higher dimensional levels architecturally, even knowing that they are false projections and poorly represented in our world, we can also assume that their poetic metaphors are not only valid, but perhaps more substantial and powerful as they are inherent to ourselves, arising from mind and spirit. With these tools, a comprehensive map could be made from diverse sources in geometry, art, philosophy, and science enabling connections among all dimensional levels – including transcendental states. The analytic poem aligns those selected affinities, initially subjective to each observer, but ultimately aspiring to convey universal impressions, as a comprehensive statement of Nature.
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sources:_____________________________________________________________________
I am: In the Bible, book of Exodus (3:14): “I am that I am” (אֶהְיֶה אֲשֶׁר אֶהְיֶה) was the response given to Moses when he asked God His name. From the Hebrew letters, without vowels, is the word YHWH (יהוה), pronounced as “Yahweh.”
Edwin Abbott Abbott, “Flatland: A Romance of Many Dimensions,” p. 93-94. Princeton University Press, Princeton, 1991.
worldline image: 320px-World_line2.svg.pngen.wikipedia.org
George Polya, “How to Solve It. A New Aspect of Mathematical Method.” Princeton University Press, 1945, p.44
James Joyce, “Ulysses.” Vintage Books, New York, 1990, page 64
density: “More is unknown than is known. We know how much dark energy there is because it affects the Universe’s expansion…It turns out that roughly 70% of the Universe is dark energy. Dark matter makes up about 25%. The rest – everything on Earth, everything ever observed with all of our instruments, all normal matter – adds up to less than 5% of the Universe.”
NASA, Astrophysics press release: http://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy/
hypercube image: 190px‑Hypercube.svg.png en.wikipedia.org
see Carl Sagan, “Cosmos – The Edge of Forever –The 4th Dimension,” 1980. | 3,891 | 18,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.957785 |
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PA.CC.7.NS. The Number System
Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers.
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Description
Computerized Tomography, used every day in hospitals to produce internal images of patients, is based on the inversion of the Radon transform. In two spatial dimensions, this transform maps a function to the collection of its integrals along all lines through the plane. More general 'X-ray' transforms can be defined by integrating along different families of curves, with applications to, e.g., seismology. These transforms are examples of integral-geometric operators and their inversion is what constitutes an inverse problem. The task of inverting these transforms can be approached by addressing analytical questions such as injectivity and stability, and by attempting to find inversion formulas.
In
this mini-course, we will first discuss the analytical framework behind the study of general inverse problems. We will then focus on studying three prototypes of integral-geometric inverse problems whose study can be made rather explicit via Fourier-based techniques: the Funk transform on the sphere, the Radon transform on the plane, and the X-ray transform on the unit disk. Time permitting, numerical examples and/or a discussion on possible generalizations will be given.
Learning outcomes
Students familiarize themselves with the model behind X-ray tomography. They also become familiar with the prototypical questions related to the analysis of inverse problems (injectivity, stability, inversion) and the methods to address these questions in the context of integral-geometric problems, gaining exposure to a mix of functional-analytic, Fourier-analytic and geometric concepts along the way.
Description of prerequisites
Basic functional analysis.
Study materials
Bibliography: Lecture notes will be provided.
Other available resources:
- F. Natterer, The mathematics of Computerized Tomography, Society for Industrial and Applied Mathematics, 2001.
- G. Paternain, M. Salo and G. Uhlmann, Geometric Inverse Problems with emphasis on two dimensions, 2021.
- J. Ilmavirta and F. Monard, Integral geometry on manifolds with boundary and applications, The Radon
Transform: the first 100 years and beyond. Radon series on computations and applied mathematics 22 (2019). Editors: R. Ramlau, O. Scherzer.
- J. Ilmavirta, Analysis and X-ray tomography, lecture notes (2017), arXiv:1711.06557.
Completion methods
Method 1
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Pass - fail
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Language:
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Study methods:
Lectures and exercises
Study materials:
Bibliography: Lecture notes will be provided.
Other available resources:
- F. Natterer, The mathematics of Computerized Tomography, Society for Industrial and Applied Mathematics, 2001.
- G. Paternain, M. Salo and G. Uhlmann, Geometric Inverse Problems with emphasis on two dimensions, 2021.
- J. Ilmavirta and F. Monard, Integral geometry on manifolds with boundary and applications, The Radon
Transform: the first 100 years and beyond. Radon series on computations and applied mathematics 22 (2019). Editors: R. Ramlau, O. Scherzer.
- J. Ilmavirta, Analysis and X-ray tomography, lecture notes (2017), arXiv:1711.06557. | 806 | 3,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.869381 |
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# Waves and Sound - bowlesphysics.com
Waves and Sound AP physics B. What is a wave A WAVE is a vibration or disturbance in space. A MEDIUM is the substance that all Sound Waves travel through and need to have in order to move. Two types of Waves The first type of wave is called Longitudinal. Longitudinal Wave - A fixed point will move parallel with the wave motion 2 areas Compression- an area of high molecular density and pressure Rarefaction - an area of low molecular density and pressure Two types of Waves The second type of wave is called Transverse. Transverse Wave - A fixed point will move perpendicular with the wave motion. Wave parts(recall demo for simple harmonic motion )- crest, trough, wavelength, amplitude, frequency, period Wave Speed You can find the speed of a wave by multiplying the wave's wavelength in meters by the frequency (cycles per second). Since a cycle is not a standard unit this gives you meters/second. Example A harmonic wave is traveling along a rope.
Waves and Sound AP Physics B. What is a wave A WAVE is a vibration or disturbance in space. A MEDIUM is the substance that all SOUND WAVES travel through and need to have in order to move. Two types of Waves The first type of wave is called Longitudinal. Longitudinal Wave - A fixed point will move parallel with the wave motion 2 areas …
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1 Waves and Sound AP physics B. What is a wave A WAVE is a vibration or disturbance in space. A MEDIUM is the substance that all Sound Waves travel through and need to have in order to move. Two types of Waves The first type of wave is called Longitudinal. Longitudinal Wave - A fixed point will move parallel with the wave motion 2 areas Compression- an area of high molecular density and pressure Rarefaction - an area of low molecular density and pressure Two types of Waves The second type of wave is called Transverse. Transverse Wave - A fixed point will move perpendicular with the wave motion. Wave parts(recall demo for simple harmonic motion )- crest, trough, wavelength, amplitude, frequency, period Wave Speed You can find the speed of a wave by multiplying the wave's wavelength in meters by the frequency (cycles per second). Since a cycle is not a standard unit this gives you meters/second. Example A harmonic wave is traveling along a rope.
2 It is observed that the oscillator that generates the wave completes vibrations in s. Also, a given maximum travels 425 cm along a rope in s . What is the wavelength? cycles 40. f = = = Hz sec 30. x v= = = m/s t 10. vwave vwave = f = = m f Standing Waves A standing wave is produced when a wave that is traveling is reflected back upon itself. There are two main parts to a standing wave: Antinodes Areas of MAXIMUM AMPLITUDE. Nodes Areas of ZERO. AMPLITUDE. Sound Waves Sound Waves are a common type of standing wave as they are caused by RESONANCE. Resonance when a FORCED vibration matches an object's natural frequency thus producing vibration, Sound , or even damage. One example of this involves shattering a wine glass by hitting a musical note that is on the same frequency as the natural frequency of the glass. (Natural frequency depends on the size, shape, and composition of the object in question.).
3 Because the frequencies resonate, or are in sync with one another, maximum energy transfer is possible. Sound Waves The production of Sound involves setting up a wave in air. To set up a CONTINUOUS Sound you will need to set a standing wave pattern. Three LARGE CLASSES of instruments Stringed - standing wave is set up in a tightly stretched string Percussion - standing wave is produced by the vibration of solid objects Wind - standing wave is set up in a column of air that is either OPEN or CLOSED. Factors that influence the speed of Sound are density of solids or liquid, and TEMPERATURE. Closed Pipes Have an antinode at one end and a node at the other. Each Sound you hear will occur when an antinode appears at the top of the pipe. What is the SMALLEST length of pipe you can have to hear a Sound ? You get your first Sound or encounter your first antinode when the length of the actual pipe is equal to a quarter of a wavelength.
4 This FIRST Sound is called the FUNDAMENTAL FREQUENCY or the FIRST HARMONIC. Closed Pipes - Harmonics Harmonics are MULTIPLES of the fundamental frequency. In a closed pipe, you have a NODE at the 2nd harmonic position, therefore NO. Sound is produced Closed Pipes - Harmonics In a closed pipe you have an ANTINODE at the 3rd harmonic position, therefore Sound is produced. CONCLUSION: Sounds in CLOSED pipes are produced ONLY at ODD HARMONICS! Open Pipes OPEN PIPES- have an antinode on BOTH ends of the tube. What is the SMALLEST length of pipe you can have to hear a Sound ? You will get your FIRST Sound when the length of the pipe equals one-half of a wavelength. Open Pipes - Harmonics Since harmonics are MULTIPLES of the fundamental, the second harmonic of an open pipe will be ONE WAVELENGTH. The picture above is the SECOND harmonic or the FIRST OVERTONE. Open pipes - Harmonics Another half of a wavelength would ALSO produce an antinode on BOTH ends.
5 In fact, no matter how many halves you add you will always have an antinode on the ends The picture above is the THIRD harmonic or the SECOND. OVERTONE. CONCLUSION: Sounds in OPEN pipes are produced at ALL. HARMONICS! Example The speed of Sound Waves in air is found to be 340. m/s. Determine the fundamental frequency (1st harmonic) of an open-end air column which has a length of cm. v = 2lf 340 = 2( ) f f = HZ. Example The windpipe of a typical whooping crane is about long. What is the lowest resonant frequency of this pipe assuming it is a pipe closed at one end? Assume a temperature of 37 C. [( )(37)] + 331 = m/s v = 4lf v = 4( ) f f = Hz | 1,338 | 5,951 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.890068 |
http://mathhelpforum.com/differential-geometry/148549-proof-recurring-number-must-rational-print.html | 1,526,845,733,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863684.0/warc/CC-MAIN-20180520190018-20180520205925-00010.warc.gz | 173,414,420 | 2,890 | # Proof that recurring number must be rational?
• Jun 10th 2010, 04:00 AM
Nappy
Proof that recurring number must be rational?
Let m,n be natural numbers with n >= 1 and 0 < m <= n. Im asked to give a clear and concise argument that shows that the decimal expansion of m/n found by long division is recurring.
Can anyone help me out with this one, came up on a recent past paper.
Cheers
Nappy
• Jun 10th 2010, 04:20 AM
CaptainBlack
Quote:
Originally Posted by Nappy
Let m,n be natural numbers with n >= 1 and 0 < m <= n. Im asked to give a clear and concise argument that shows that the decimal expansion of m/n found by long division is recurring.
Can anyone help me out with this one, came up on a recent past paper.
Cheers
Nappy
If you ignore the decimal point there are only n possible remainders at each stage of the log-division process. Once a remainder is repeated all the subsequent multipliers on to of the long division will repeat as will the remainders. hence eventually the decimal expansion will become recuring. | 251 | 1,035 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.940302 |
https://web.math.pmf.unizg.hr/glasnik/vol_44/no1_01.html | 1,542,465,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117165412-00047.warc.gz | 785,667,755 | 3,375 | ### Bo He
Sichuan Provincial Key Laboratory of Computer Software, Sichuan Normal University, Chengdu, 610068, P. R. China
e-mail: hebo-one@hotmail.com
e-mail: bhe@live.cn
Abstract. In this remark, we use some properties of simple continued fractions of quadratic irrational numbers to prove that the equation
(x3 - 1)/(x - 1) = (yn - 1)/(y - 1), x, y, n N, x > 1, y > 1, n > 3, n odd,
has only the solutions (x,y,n) = (5,2,5) and (90,2,13).
2000 Mathematics Subject Classification. 11A55, 11D09, 11D61.
Key words and phrases. Goormaghtigh equation, continued fraction, congruence.
Full text (PDF) (free access)
DOI: 10.3336/gm.44.1.01
References:
1. A. Baker, Bounds for the solutions of the hyperelliptic equation, Proc. Cambridge Philos. Soc. 65 (1969), 439-444.
MathSciNet CrossRef
2. Y. Bugeaud, T. N. Shorey, On the number of solutions of the generalized Ramanujan-Nagell equation, J. Reine Angew. Math. 539 (2001), 55-74.
MathSciNet CrossRef
3. Y. Bugeaud, T. N. Shorey, On the diophantine equation (xm-1)/(x-1)=(yn-1)/(y-1), Pacific J. Math. 207 (2002), 61-75.
MathSciNet
4. J. Coates, An effective p-adic analogue of a theorem of Thue II: The greatest prime factor of a binary form, Acta Arith. 16 (1969/1970), 399-412.
MathSciNet
5. H. Davenport, D. J. Lewis, and A. Schinzel, Equations of the form f(x) = g(y), Quart. J. Math. Oxford Ser. (2) 12 (1961), 304-312.
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6. A. Dujella, Continued fractions and RSA with small secret exponent, Tatra Mt. Math. Publ. 29 (2004), 101-112.
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7. A. Dujella and B. Ibrahimpasic, On Worley's theorem in Diophantine approximations, Ann. Math. Inform. 35 (2008), 61-73.
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8. A. Dujella and B. Jadrijevic, A family of quartic Thue inequalities, Acta Arith. 111 (2004), 61-76.
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9. P. Fatou, Sur l'approximation des incommensurables et les series trigonometriques, C. R. Acad. Sci. (Paris) 139 (1904), 1019-1021.
10. R. Goormaghtigh, L'Intermédiaire des Mathématiciens, 24 (1917), 88.
11. I. Niven, H. S. Zuckerman and H. L. Montgomery, An Introduction to the Theory of Numbers, John Wiley & Sons, Inc., New York, 1991.
MathSciNet
12. M. Le, On the Diophantine equation (x3-1)/(x-1)=(yn-1)/(y-1), Trans. Amer. Math. Soc. 351 (1999), 1063-1074.
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13. M. Le, Exceptional solutions of the exponential Diophantine equation (x3-1)/(x-1)=(yn-1)/(y-1), J. Reine Angew. Math. 543 (2002), 187-192.
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14. M. Le, On Goormaghtigh's equation (x3-1)/(x-1)=(yn-1)/(y-1), Acta Math. Sinica (Chin. Ser.) 45 (2002), 505-508.
MathSciNet
15. T. Nagell, The diophantine equation x2 + 7 = 2n, Ark. Mat. 4 (1961), 185-187.
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16. Yu. V. Nesterenko and T. N. Shorey, On an equation of Goormaghtigh, Acta Arith. 83 (1998), 381-389.
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17. R. Ratat, L'Intermédiaire des Mathématiciens, 23 (1916), 150.
18. A. Schinzel, On two theorems of Gelfond and some of their applications, Acta Arith. 13 (1967/1968), 177-236.
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19. A. Schinzel, An improvement of Runge's theorem on diophantine equations, Comment. Pontificia Acad. Sci. 2 (1969), 1-9.
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20. T. N. Shorey, On the equation axm - byn = k, Nederl. Akad. Wetensch. Indag. Math. 48 (1986), 353-358.
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21. T. N. Shorey, Integers with identical digits, Acta Arith. 53 (1989), 187-205.
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22. C. L. Siegel, Approximation algebraischer Zahlen, Math. Z. 10 (1921), 173-213.
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23. C. L. Siegel, The integer solutions of the equation y2 = axn + bxn-1 + ... + k, (Under the pseudonym X), J. London Math. Soc. 1 (1926), 66-68.
24. A. Thue, Über Annäherungswerte algebraischer Zahlen, J. Reine Angew. Math. 135 (1909), 284-305.
25. R. T. Worley, Estimating |α - p/q|, J. Austral. Math. Soc. Ser. A 31 (1981), 202-206.
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26. P. Z. Yuan, On the diophantine equation (x3-1)/(x-1)=(yn-1)/(y-1), J. Number Theory 112 (2005), 20-25.
MathSciNet CrossRef | 1,459 | 3,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-47 | latest | en | 0.562411 |
https://discourse.julialang.org/t/calling-a-function-to-change-a-struct/98846 | 1,685,610,439,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00342.warc.gz | 244,755,837 | 6,832 | # Calling a function to change a struct?
I am trying to use a function in order to modify a struct that is passed to it as a variable. Below is a minimal bit of code I have put together to illustrate my confusion.
``````struct a
x::Int
y::Int
end
function switchVariables!(b::a)
newB = a(b.y, b.x)
b = newB
println(b)
end
b = a(1, 2)
println(b)
switchVariables!(b)
println(b)
``````
The two `println` calls outside of the `switchVariables!` function return `a(1, 2)`, whereas the one inside the function returns `a(2, 1)`. So it looks like the proper operation is being done to `b` inside the function, but not outside of it. I’ve poked around with variants like `b = deepcopy(newB)` but have not had any luck. Would someone be able to explain to me what is going on here? I recognize that for this minimal example, it would probably be better to just return the desired modified struct rather than explicitly trying to modify `b` - while the same might be true for the actual code that I’m working on, I would at least like to understand why the above code doesn’t work the way I would naively expect it to. Thanks!
Type
``````?Ref
``````
at the Julia prompt.
1. The struct created is immutable, you cannot modify it.
2. Doing `b = new struct`, inside the function, is just assigning the label `b` to a new struct.
3. That label is local to the scope of the function.
You need to use `b = switch(b)` to assign the label `b` of the outer scope to the new struct created in the function. But that is not mutating anything. Usually that’s fine and the way to go, but you can also use a mutable struct and get actually the behavior you were expecting.
Ps. Think of an immutable struct as a number. You are doing:
``````function switch(b)
newb = 2
b = newb
return b
end
b = 1
switch(b)
``````
You will see that `b` does not change outside `switch`, because the label `b` of the outer scope was bound to the value `1` and that was not changed in that scope. You need to do here `b = swich(b)`. The same goes for immutable objects, like your struct. In any case, you are not modifying the immutable number 1, which is an immutable value, corresponding to your struct.
7 Likes
To get the behaviour you ask for, you should do this:
``````mutable struct a
x::Int
y::Int
end
function switchVariables!(b::a)
b.x, b.y = b.y, b.x
end
``````
Unless you declare the struct as `mutable` it will be immutable, i.e. the content can’t be modified. This has a number of advantages for performance. But if you want modifiable struct you must explicitly use the `mutable` keyword. The way you wrote your function would have worked if arguments were “called by name”, Julia does not support that.
4 Likes
``````global b = newB
``````
If you can get `@set` to turn iteration-based destructuring assignment `b.x, b.y = b.y, b.x` to instantiation `a(b.y, b.x)` then please provide the code because AFAIK Accessors.jl can’t do that yet. It’s not possible to do the changes separately because both the previous x and y fields must be retrieved before replacing either.
``````julia> @set b.x = b.y
a(2, 2)
julia> @set b.x, b.y = b.y, b.x
(2, 1)
``````
EDIT: Answered my own question, but tbf, it is a longstanding missing feature in Accessors.jl and the predecessor Setfield.jl. `setproperties` is only mentioned in the docs under an explicitly experimental `mapproperties`, so it could easily be gone in minor revisions. I don’t know the internal workings of why `@set` can’t just do what `setproperties` can.
``````julia> b
a(1, 2)
julia> setproperties(b, (x = b.y, y = b.x))
a(2, 1)
``````
Note also that Accessors, or Setfield, do not mutate immutable data, really, they are just convenience macros to redefine new immutable structs. The issue the OP has would be the same with it:
``````
julia> using Accessors
julia> struct A
i
j
end
julia> function set_i(a)
a = @set a.i = 0
return a
end
set_i (generic function with 1 method)
julia> b = A(1,1)
A(1, 1)
julia> set_i(b)
A(0, 1)
julia> b # doesn´t change
A(1, 1)
``````
1 Like
Yeah you’re absolutely right! | 1,150 | 4,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2023-23 | latest | en | 0.945396 |
https://www.edupil.com/question/numbers-between-4000-and-5000-12-18-21-and-32/ | 1,623,686,216,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487612537.23/warc/CC-MAIN-20210614135913-20210614165913-00074.warc.gz | 679,303,322 | 15,575 | # The numbers between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is :
1. 4032
2. 4302
3. 4203
4. 4023
Anurag Mishra Professor Asked on 29th December 2015 in
Explanation:-
Factors of 12 = 2 x 2 x 3
Factors of 18 = 2 x 3 x 3
Factors of 21 = 3 x 7
Factors of 32 = 2 x 2 x 2 x 2 x 2
Then, L C M of 12, 18, 21 & 32 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 7
= 2016
The multiple of 2016 = 2016 x 2
= 4032
Then, the number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is 4032.
Hence, the correct answer is option (1) 4032.
Monis Rasool Professor Answered on 29th December 2015. | 269 | 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} | 4 | 4 | CC-MAIN-2021-25 | latest | en | 0.919481 |
https://papertowriters.com/how-does-this-relate-to-your-answer-in-part-a-2-answers-below/ | 1,675,585,810,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500250.51/warc/CC-MAIN-20230205063441-20230205093441-00445.warc.gz | 449,392,245 | 15,687 | Suppose that you are a monopolist who produces gizmos, Z, with the total cost function C (Z) =F + 50Z, where F represents the firm’s fixed cost. Your marginal cost is MC = 50. Suppose also that there is only one consumer in the market for gizmos, and she has the demand function P = 60 – Z.
a) If you use a constant per-unit price for gizmos, what price maximizes your profits? What is the smallest value of F such that you could earn positive profits at this price?
b) Suppose instead that you charge a per-unit price equal to marginal cost, that is, P = MC = 50. How many units would the customer purchase at this price? Illustrate your answer in a graph (featuring the individual demand curve and marginal cost).
c) Now consider charging the customer a “subscription fee” of S in addition to a usage fee. If you set the usage fee as in part
(b), what is the largest fixed fee you could charge the consumer, while ensuring that she is willing to participate in this market?
d) For what values of F will you be able to earn positive profits if you follow the pricing strategy you outlined in part (c)? How does this relate to your answer in part (a)?
e) Suppose now that there are N consumers in the market for gizmos, each with the individual demand function P = 60 – Z. Expressing your answer in terms of N, how large can the fixed costs F be for you to still earn positive profits if you use the above nonlinear pricing strategy. | 331 | 1,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-06 | latest | en | 0.945085 |
https://osiris.utwente.nl/student/OnderwijsCatalogusSelect.do?selectie=cursus&cursus=202001178&collegejaar=2021&taal=en | 1,685,790,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00514.warc.gz | 495,877,707 | 5,711 | Close Help Print
Course module: 202001178
202001178Linear Algebra
Course info Schedule
Course module202001178
Credits (ECTS)3
Course typeCourse
Language of instructionEnglish
Lecturer(s)
Lecturer dr. A. Antoniadis Examiner dr. A. Antoniadis Contactperson for the course dr. A. Antoniadis Examiner dr. J.B. Timmer Lecturer ir. L. Weedage
Starting block
1A
Application procedureYou apply via OSIRIS Student
Registration using OSIRISYes
Aims
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } The student can: work with systems of linear equations, vectors, matrices, subspaces of the n-dimensional real space, and explain the connections between these concepts, work with determinants, eigenvalues, eigenvectors, linear transformations and connect them with the previous concepts.
Content
body { font-size: 9pt; font-family: Arial } table { font-size: 9pt; font-family: Arial } In Linear Algebra we mainly focus on systems of linear equations (so-called linear systems). Many real-life situations can be modelled as a linear system. Examples are networks (traffic networks, data networks, electrical networks, etc.), economic models, chemical reactions, cryptography (coding of messages), scheduling, computer graphics, and GPS. Linear Algebra starts with an introduction of linear systems which will be described using a (coefficient) matrix. You learn how to solve linear systems systematically, using a row reduction technique on the coefficient matrix. Thereafter, we focus on operations for vectors and matrices, such as addition, multiplication, inverse and transpose. These operations are fundamental in Linear Algebra. Next, we deal with sets of vectors with very nice properties: subspaces. It turns out that the properties of subspaces tell us a lot about the structure of solution sets of linear systems. Here the concepts of linear combination, linear independence, basis and dimension play an important role. We also introduce the concept of determinant of a square matrix. We explore its properties and show some interesting interpretations. Further, we deal with eigenvectors and eigenvalues of a square matrix. These concepts play a crucial role in discrete dynamic systems, which arise in many scientific fields. Finally, we examine linear transformations and their properties. Some well-known applications in geometry will be treated as well. Throughout, much emphasis is laid on the relations among the various concepts.
Assumed previous knowledge
Some experience with vectors, lines, planes, and systems of linear equations.
Participating study
Master Electrical Engineering
Participating study
Master Embedded Systems
Participating study
Master Mechanical Engineering
Participating study
Master Systems and Control
Participating study
Master Sustainable Energy Technology
Required materials
Course material
Lecture notes (dictaat)
Recommended materials
-
Instructional modes
Colstructie Guided self study
Tests
Written exam
Close Help Print | 618 | 3,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.83053 |
https://www.jiskha.com/display.cgi?id=1366811479 | 1,516,310,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887621.26/warc/CC-MAIN-20180118210638-20180118230638-00471.warc.gz | 940,458,373 | 3,906 | # physics
posted by .
A standard man climbs 10 m up a vertical rope. How much energy in calories is dissipated as heat in a single climb if 20% of the total energy required is used to do the work?
• physics -
pls answer someone
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17. A 60.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. Approximately how much total work is done against gravity by the student during the climb?
8. ### Kinesology
A person performs a work task at a mechanical efficiency of 20 percent and at total energy output of 15kcal/min. How much energy(kcal/min) is lost as heat?
9. ### Science
A person performs a work task at a mechanical efficiency of 20 percent and at total energy output of 15kcal/min. How much energy(kcal/min) is lost as heat?
10. ### Physics- Help!!
1. What is the heat required in kilocalories to convert 2 kg of ice at 0°C completely into steam at 100°C?
More Similar Questions | 550 | 2,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-05 | latest | en | 0.921989 |
www.renttoowncenter.com | 1,618,895,445,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039375537.73/warc/CC-MAIN-20210420025739-20210420055739-00286.warc.gz | 1,036,692,305 | 22,057 | # What Size Air Conditioner Do I Need?
If you’re wondering what size of air conditioner you need, this post is for you. Here’s how to determine what size air conditioner unit will best fit your needs. The right air conditioner helps you cool a room efficiently. An undersized unit won’t cool adequately while one too large will not remove enough humidity, leaving the air feeling damp. The first thing to consider is “How big a space are you trying to cool?” Measuring your space helps determine what size AC unit will best fit your needs. To calculate the square footage of your space, multiply the length of your room by the width of your room. Measure from the inside of your room. Use the simple equation of Length In Feet x Width In Feet = Square Footage. Compare the square footage of your room to the list below. This is how much you can expect a typical window air conditioner unit will cool in total square footage.
– 150 to 350 sq ft: look for a 5,000- to 8,000-BTU unit
– 350 to 550 sq ft: look for an 8,000- to 12,000-BTU unit
– 550 to 1,050 sq ft: look for a 12,000- to 18,500-BTU unit
– 1,050 to 1,600 sq ft: look for an 18,500- to 25,000-BTU unit
Does the room you are trying to cool receive a lot of direct sunlight? If so you may need to step up one size to cool it efficiently.
It is also important to determine the power source. Most homes are equipped to run 110 volt units but some have wiring that will support 220 volt units. Larger units often require 220 volts to operate and many homes are not wired to support 220 volt units. These units will cool a larger area but take special plugs and wiring. Let us help you make sure the unit you pick matches the voltage output in your home.
Lastly, make sure the unit will fit into the window you want to put it in. A lot of times the customer will select a large unit only to find out that the window they want it installed in is too big or small to support the AC unit. At Triad Leasing, we hand select quality A/C units to offer you the best in cooling power, energy savings and value. Our selection includes cool features like effortless temperature sensing remote controls which allows you to see, set and maintain room temperature from across the room. Multi-speed fans offer you three different speeds for cooling flexibility. If you’re an allergy sufferer, ask us about the potential benefits of clean air ionizers designed to remove pollen and impurities from the air. Many models offer low power start up which can save you money by conserving energy. We have the window air conditioner you need to quickly cool a room on hot days and with quiet operation to keep you cool without keeping you awake. Many of our air conditioning window units feature extra long cords which make extension cords unnecessary.
If you’re shopping for a window air conditioner unit, come in to Triad Leasing, let us help you determine the right size air conditioner to meet your needs.
Shop Window Air Conditioners | 668 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2021-17 | longest | en | 0.920924 |
https://www.sparrho.com/item/methods-for-practical-worst-test-definition-and-debug-during-block-based-statistical-static-timing-analysis/d2c1c7/ | 1,638,108,203,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00524.warc.gz | 1,095,375,997 | 31,583 | # Methods for practical worst test definition and debug during block based statistical static timing analysis
Imported: 13 Feb '17 | Published: 18 Jan '11
Nathan C. Buck, Eric A. Foreman, James C. Gregerson, Jeffrey G. Hemmett
USPTO - Utility Patents
## Abstract
Methods for analyzing timing of an integrated circuit using block-based static statistical timing analysis and for practical worst test definition and debug. The method includes building a timing graph, determining a slack for each of the nodes in the timing graph, and identifying a statistically worst slack for at least one of the nodes. The method further includes replacing this statistically worst slack with a proxy worst slack.
## Description
### FIELD OF THE INVENTION
The invention relates generally to integrated circuit design and, in particular, to statistical static timing analysis for analyzing the timing characteristics of an integrated circuit, before physically fabricating the integrated circuit, by propagating arrival and required arrival times of an integrated circuit in a probabilistic or statistical fashion.
### BACKGROUND OF THE INVENTION
Static timing analysis (STA) is a common method of computing the expected timing of a digital circuit to identify problem areas of an integrated circuit during the design phase and in advance of actual fabrication. The timing of the integrated circuit is simulated to determine if it meets the timing constraints and, therefore, is likely to operate properly if fabricated in accordance with the tested design.
Deterministic STA (DSTA) propagates timing quantities, such as arrival times, required arrival times, and slews, along with any other timing related quantities (guard times, adjusts, asserts, etc.), as single valued deterministic data. DSTA only covers a single corner of a space of process variations with each individual timing run. A corner is a set of input values for parameters that may include temperature of the circuit, input voltage, and various manufacturing parameters of an integrated circuit. In order to evaluate the impact that a given parameter will have on timing, multiple DSTA timing runs must be executed with parameters that affect timing set at several maximum and minimum corners, such as high and low temperature, high and low voltages, and various processing conditions. For example, DSTA timing runs may compare a corner characterized by a combination of high input voltage, a high operating temperature, and the worst manufacturing parameters with a corner characterized by a combination of a low input voltage, a low operating temperature, and the best manufacturing parameters. As a check of the performance of the integrated circuit design, many or all of the corners may be run and the integrated circuit design adjusted until all of the corners pass the timing tests. These results reflect the extreme performance bounds of the integrated circuit and may require numerous timing runs to fully explore the space of process variations. Even then, the results may be overly pessimistic and misleading for optimization tools.
Statistical static timing analysis (SSTA) propagates timing quantities as statistical distributions instead of as single valued deterministic data. In contrast to the DSTA approach that only predicts a single corner of the space of process variations with each timing run, a single timing run using SSTA predicts the performance of the integrated circuit over the entire space of process variations. Consequently, in order to close timing, a single SSTA timing run may replace multiple DSTA timing runs. For example, assuming N parameters (i.e., sources of variation) and two corners per parameter, then 2N corners would have to be analyzed by individual DSTA timing runs to match the effectiveness of a single SSTA run. Hence, SSTA is far more computationally efficient than DSTA. A test run that passes in a single process corner under a DSTA run may actually fail in one or more other performance-limiting corners in the process space, which SSTA would reveal. SSTA methods can, therefore, reduce the exaggerated pessimism inherent in DSTA.
SSTA operates on a timing graph comprised of nodes, which represent points at which signal transitions can occur, and edges that connect incident nodes. Timing values are computed for the timing graph at each node based upon arrival times (ATs), which define the time (or the time distribution) at which a given signal arrives at a timing point, and required arrival times (RATs), which defines the time (or the time distribution) at which the signal is required to get to the timing point, in order to meet the timing requirements. These ATs and RATs are used to compute timing metrics in the form of slacks at nodes (RAT minus AT for late mode and AT minus RAT for early mode). A negative value for either a late mode slack or an early mode slack indicates a timing constraint violation.
Typically, slack data is only collected for timing qualification at a subset of all of the timing graph nodes, sometimes referred to as test points, or endpoints. These endpoints are frequently defined as those nodes that are either primary outputs of the circuit, or those that control data propagation (e.g. the input to storage elements such as latches). There frequently are multiple slacks incident at these nodes, due to both the backward propagated RAT data from upstream nodes, and also due to any timing tests that may be present on those nodes (e.g. setup or hold tests). These slacks may be called edge slacks, as they arise from the timing and test segments that form the graph edges incident on the node of interest. The final slack value for that node is defined as the minimum of all edge slacks, as this is the limiting slack that can cause timing failure.
In deterministic timing analysis, a ‘traditional’ deterministic minimum operation is used, where the smallest of all the incident edge slacks is used to define the nodal slack value. For this reason, an individual edge slack can never be worse than the node slack, and the node slacks always capture the worst timing constraint violations.
In block-based SSTA, statistical minimum and maximum operations used to calculate a node slack for the propagated statistical distributions. As the input data to these operations are functions (statistical distributions), the output of a statistical minimum or maximum operation is also a function. These propagated distributions will typically overlap such that each one will produce a minimum (or maximum) some percentage of the time over numerous samples. Therefore the output function is defined as a linear combination of all of the inputs, with each input weighted by the probability that it will produce either the minimum or the maximum result. Therefore, failing results in a timing report cannot be easily correlated with the actual failing test that produced the fail, which increases the difficulty to trace failing tests as required to resolve timing issues.
Moreover, in block-based SSTA reporting of the slack distribution, an edge slack for a propagated distribution can be worse than the slacks of the nodes connected by the edge because of the statistical maximum and minimum operations performed on the propagated distributions. Problems arise because the timing correctness of a circuit design is often determined by ensuring that all edge slacks are non-negative or, more generally, are greater than some user-specified threshold. Because slacks of the nodes connected by an edge may be higher than the slack of the edge, this leads to a situation in which a timing constraint violation is reported on an edge. However, the slacks of the nodes connected by the edge, which the designers and optimization tools generally use, forming the sink slack do not indicate a timing constraint violation. Therefore, designers and optimization tools may be confused or mislead because of the slacks of the edges are not considered.
Accordingly, there is a need for an improved method for practical worst test definition and debug during block-based SSTA that overcomes these and other deficiencies of conventional block-based SSTA.
### SUMMARY OF THE INVENTION
In an embodiment of the invention, a method is provided for analyzing timing of an integrated circuit using static statistical timing analysis. The method includes building a timing graph having a plurality of nodes and a plurality of edges connecting the nodes, forward propagating a first set of probabilistic distributions in the timing graph, and statistically calculating an arrival time of the first set of probabilistic distributions at each of the nodes. The method further includes backward propagating a second set of probabilistic distributions in the timing graph and statistically calculating a required arrival time of the second set of probabilistic distributions at each of the nodes. A statistically worst slack is identified for at least one of the nodes from the arrival times and the required arrival times either through a slack calculation based upon the arrival times and the required arrival times or through required arrival time calculations. The method further includes replacing the statistically worst slack for at least one of the edges with a proxy worst slack.
### DETAILED DESCRIPTION
With reference to FIG. 1, a process flow 10 for a statistical static timing analysis (SSTA) tool is shown. The SSTA tool utilizes a block-based approach that propagates one probability distribution from each node to the next node and, so on until it reaches a sink node using a statistical maximum operation for early mode arrival times (setup timing constraint) or a statistical minimum operation for late mode arrival times (hold timing constraint).
In block 12, inputs are supplied to the SSTA tool. Specifically, the SSTA tool reads and flattens a netlist representing the structure of the circuit to be analyzed, a set of timing assertions, a set of canonical delay models governing the sources of variation in the space of process variations, sensitivities relating the delay attributable to each individual parameter, and statistical information describing the sources of variation. The timing assertions, each of which can be either deterministic or probabilistic, may include arrival times at the primary inputs, required arrival times at the primary outputs, information about the phases of the clock, and details of external loads that are driven by the primary outputs. The canonical delay model, which may be a parameterized first-order expansion, permits the SSTA tool to determine the delay of a gate or wire as a function not only of traditional delay-model variables (like input slew or rise/fall time, and output load) but also as a function of the sources of variation.
In block 14, the SSTA tool constructs a timing graph from the netlist. The timing graph contains nodes at which signal transitions occur and edges connecting various incident nodes. A typical static timing analysis tool analyzes a synchronous design description for timing violations by breaking down the design into individual timing paths in the timing graph. The timing graph consists of nodes and edges representing delays in the circuit incurred when a logical transition (either from low to high or from high to low) is transmitted through a circuit component, such as a gate or wire. Each path has a source node (i.e., a place in a design description where data is launched by a reference signal edge) and a sink node (i.e., a place in the design description where data is captured by a reference signal edge). The static timing analysis tool calculates a signal propagation delay corresponding to each individual timing path in the timing graph.
With reference to FIG. 2, a representative timing graph is shown that includes a source node 26, a sink node 28, a plurality of nodes 31-37 connected between the source node 26 and sink node 28, and edges 40-51. Nodes 31-34 represent primary inputs of the circuit represented by the timing graph and nodes 36, 37 represent primary outputs.
In block 16, the SSTA tool computes early mode and later mode ATs at each node by forward propagating signals in the form of probabilistic distributions from the source node to the sink node. ATs are typically stored on the nodes of the timing graph and delays of individual gates and wires are typically stored on the edges of the timing graph. Signals are propagated through edges by adding edge delays to signal ATs. When multiple ATs are propagated along different edges to each of the nodes, the AT at the node is computed as the statistical maximum (for late mode ATs) or the statistical minimum (for early mode ATs) of the incoming edge delays. A late mode AT is the earliest time at which the corresponding signal is guaranteed to be stable at its correct logical value after the signal has traversed any of the possible paths of the integrated circuit. An early mode AT is the earliest time at which the corresponding signal can change from its stable logical value during the previous clock cycle. The output of a node cannot change earlier than the early mode AT.
At each node, the mean and variance of the statistical maximum AT is calculated using either analytical formulas or numerical methods, understood by a person having ordinary skill in the art of SSTA. Each output parameter sensitivity is computed by linearly combining all input data (upstream AT+respective edge delays), with each input being weighted by it's respective tightness probability. Generally, a tightness probability is the probability that the random variable is the greatest of a set. Arrival tightness probabilities are determined that reflect the probabilities that the arrival time at each of the nodes may be dominated by one or the other of the incoming edges. The first and second moments of the resulting distribution are matched by selecting an appropriate mean and random sensitivity.
In block 18 and after the conclusion of forward propagation, probabilistic distributions representing RATs are backward propagated by the SSTA tool through the timing graph. When multiple RATs are propagated along different edges to a node, the RAT at the node is computed as the statistical maximum (for early mode RATs) or the statistical minimum (for late mode RATs) of the RATs of the incoming edges. A late mode RAT indicates the latest time within the clock period of a given clock phase that a signal transition may occur at the node without violating a timing constraint. An early mode RAT indicates the earliest time within the clock period of a given clock phase that a signal transition may occur at the node without violating a timing constraint. Delays are subtracted from RATs in backward propagation as timing progresses from the sink node back to the source node. Required arrival tightness probability of an edge is defined as the probability that the RAT at a node is determined by that edge.
In addition to the propagated RATs described above, the RAT calculation also considers any timing constraints that are incident on the current node. These include things such as setup and hold constraints at latch inputs, as well as pulse width tests, loop cut tests, clock gating tests, etc. Setup timing constraints enforce a maximum delay on a timing path relative to the reference signal path. Hold timing constraints enforce a minimum delay on a timing path relative to the reference signal path. Each constraint also contributes a RAT indicating the timing results of that constraint on the current incident node. These constraints may be handled in a manner analogous to propagated RATs, as they can also be represented as additional edges in a timing graph (either in embodiment, or theoretically), although these are virtual edges between whose ends a test result is calculated, rather than a true physical path for timing propagation. So, the data considered when calculating a minimum slack (or max/min RAT) considers all incident constraint data in addition to all incident propagated data.
In block 20, the SSTA tool uses a difference between the ATs and RATs to compute timing metrics in the form of slacks at each of the nodes (RAT minus AT for late mode and AT minus RAT for early mode), which are defined thus so that a negative slack value will always indicate a violation of a setup or hold timing constraint. A positive slack indicates an amount of time by which a violation of a timing constraint is avoided by a path delay.
Conversely, block 20 may be skipped, with no explicit slack calculation being performed during timing propagation. It is understood by a person having ordinary skill in the art that calculation of the worst RAT will always yield the worst slack. That is, in early mode, calculation of the max RAT will always produce the same solution as explicit calculation of the min slack, and in late mode, calculation of the min RAT will also produce the correct min slack solution. Therefore, the slack calculations may be postponed until they are required for reporting or optimization purposes, if this approach is preferred. The RAT and slack are interchangeable, and either may be propagated with equivalent results, so the embodiments of the invention applies equally well to either explicit min slack calculations, or to max/min RAT calculations.
In block 22, a proxy worst slack is substituted for the statistical worst slack of at least one of the nodes in the timing graph, as opposed to using the worst slack derived from the statistical minimum of all slacks of the edges incoming into the node, and the proxy worst slack is reported in the timing report, as described below. The proxy worst slack, which is flexible in definition, permits the designer to chase a specific issue in the integrated circuit is currently limiting design closure in the timing graph.
In one embodiment of the invention, the proxy worst slack for a set of edges incoming into a node, may consist of the slack of the edge exhibiting the largest tightness (i.e., the most probable edge to cause a fail). In another embodiment of the invention, the proxy worst slack for a set of edges incoming into a node may consist of the slack of the specific edge that produces a worst projected slack. This proxy worst slack may differ, often significantly, from the most probable edge to cause a fail. In yet another embodiment of the invention, the proxy worst slack for a set of edges incoming into a node may consist of one or more slacks of one or more edges with unusually large parameter sensitivities. The parameter sensitivity must exceed a user-defined limit, which is defined by the user on a per-parameter basis, to be considered unusually large. In other embodiments of the invention, the proxy worst slack for a set of edges incoming into a node may consist of specific test types including, but not limited to, setup or hold timing constraints. The proxy worst slack may also be defined as some linear combination of the inputs, using some weighting factors other than the tightness probabilities typically used (e.g., user defined, or based on criticality of the incident paths).
The proxy worst slack may be defined during reporting, fixup, or optimization, using various combinations of the different embodiments, as required to focus on the problem at hand rather than potentially dismiss the node as acceptable (i.e., the case when the statistical minimum slack at the node is passing in the timing graph). That is, the proxy worst slack can be any of the individual criteria in the different alternative embodiments, or may comprise some scaled combination of the individual criteria.
The proxy worst slack may be cached during backward propagation in the initial timing run, which may improve computational efficiency. When the statistical minimum of all slacks for the edges is performed (or max/min for early/late mode RATs is performed), one or more individual slacks of the edges are identified as the statistically worst slack for the corresponding node based upon the user selected criteria for the proxy worst slack and some minimal information identifying these statistically worst slacks for the edges is stored into a side data structure. The data is updated with any incremental change in the RAT at each node. The data may then be used in a flexible manner during reporting, optimization, and fixup, optionally replacing the true edge slack in a user defined manner. Alternatively, the proxy worst slacks may be determined on the fly in real time during reporting, optimization, and fixup, via dynamic re-analysis of all incident slacks on the edges. When the proxy worst slack is desired on edges, timing runs on all edges incident on a node are looped over and those with the desired traits are flagged for use as the proxy worst slack for the edges.
In an alternative embodiment and in addition to being reported in the timing report, the proxy worst slack associated with a node may be used during backward propagation as a substitute for the actual slack at a node. The backward propagation of the proxy worst slack may aid in tracing specific timing problems to their sources along a path. The backward propagation of the proxy worst slack may also provide automated optimization and fixup tools with the appropriate data required to help close timing (in a manner totally contained inside the SSTA tool, with no changes required in the other tools consuming this data). This is embodied by replacing the actual RAT data with RAT data corresponding to the proxy worst slack, and then invalidating the RAT to trigger a backward propagation. In an incremental SSTA tool, the re-propagation is performed with a minimum overhead, as the re-calculation only affects those cones of logic affected by the RAT change, and only occurs when a timing data query is made within those cones. Similarly, if slack data itself is propagated, the proxy worst slack itself may be propagated.
In block 24, a timing report is produced by the SSTA tool. Information generated in the timing report may be used to constrain timing paths during circuit synthesis to reduce timing constraint violations. The timing report typically includes arrival times, required arrival times, slacks, and slews at each node of the integrated circuit expressed as probability distributions. Specifically, the information in the timing report may include, but is not limited to, mean value and variance for each timing quantity, a parameterized representation of the distribution of each timing quantity, a graphical representation of the distribution of each timing quantity, and a correlation report between these various timing quantities. Various automatic audits, such as checking for excessive parameter sensitivities, may be built into the timing report. Excessive parameter sensitivities typically must be reduced in order to improve the robustness of the circuit. The timing report also includes the proxy worst slacks for the nodes that include incoming edges, which are determined as outlined hereinabove.
Applicants hereby incorporate by reference herein the entire disclosure of U.S. Publication No. 2005/0065765, published Mar. 24, 2005 from Ser. No. 10/666,353 filed Sep. 19, 2003 and entitled “System and Method for Statistical Timing Analysis of Digital Circuits”, for additional description of the SSTA process.
While the invention has been illustrated by a description of various embodiments and while these embodiments have been described in considerable detail, it is not the intention of the applicants to restrict or in any way limit the scope of the appended claims to such detail. Additional advantages and modifications will readily appear to those skilled in the art. Thus, the invention in its broader aspects is therefore not limited to the specific details, representative apparatus and method, and illustrative example shown and described. Accordingly, departures may be made from such details without departing from the spirit or scope of applicants' general inventive concept.
## Claims
1. A method for analyzing timing of an integrated circuit using a static statistical timing analysis (SSTA) tool, the method comprising:
computing a timing graph from a netlist using the SSTA tool, the timing graph having a plurality of nodes and a plurality of edges connecting the nodes;
in response to forward propagating a first set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool an arrival time of the first set of probabilistic distributions at each of the nodes;
in response to backward propagating a second set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool a required arrival time of the second set of probabilistic distributions at each of the nodes;
identifying with the SSTA tool a statistically worst slack for a node in the timing graph from the arrival times and the required arrival times either through a slack calculation based upon the arrival times and the required arrival times or through required arrival time calculations;
determining one or more of the slacks of the edges arriving at the node in the timing graph with a parameter sensitivity that exceeds a user-defined limit;
equating a proxy worst slack to the slack of the edge that produces a worst projected slack: and
generating a timing report with the SSTA tool that replaces the statistically worst slack with the proxy worst slack for at least one of the edges arriving at the node in the timing graph at which the statistically worst slack is identified during backward propagation.
computing a timing graph from a netlist using the SSTA tool, the timing graph having a plurality of nodes and a plurality of edges connecting the nodes;
in response to forward propagating a first set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool an arrival time of the first set of probabilistic distributions at each of the nodes;
in response to backward propagating a second set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool a required arrival time of the second set of probabilistic distributions at each of the nodes;
identifying with the SSTA tool a statistically worst slack for a node in the timing graph from the arrival times and the required arrival times either through a slack calculation based upon the arrival times and the required arrival times or through required arrival time calculations;
determining one or more of the slacks of the edges arriving at the node in the timing graph with a parameter sensitivity that exceeds a user-defined limit;
equating a proxy worst slack to the slack of the edge that produces a worst projected slack: and
generating a timing report with the SSTA tool that replaces the statistically worst slack with the proxy worst slack for at least one of the edges arriving at the node in the timing graph at which the statistically worst slack is identified during backward propagation.
2. A method for analyzing timing of an integrated circuit using a static statistical timing analysis (SSTA) tool, the method comprising:
computing a timing graph from a netlist using the SSTA tool, the timing graph having a plurality of nodes and a plurality of edges connecting the nodes;
in response to forward propagating a first set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool an arrival time of the first set of probabilistic distributions at each of the nodes;
in response to backward propagating a second set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool a required arrival time of the second set of probabilistic distributions at each of the nodes;
identifying with the SSTA tool a statistically worst slack for a node in the timing graph from the arrival times and the required arrival times either through a slack calculation based upon the arrival times and the required arrival times or through required arrival time calculations;
calculating a tightness probability for each of the edges;
comparing the tightness probabilities to identify a largest tightness probability;
equating a proxy worst slack to the slack for the edge exhibiting the largest tightness probability; and
generating a timing report with the SSTA tool that replaces the statistically worst slack with the proxy worst slack for at least one of the edges arriving at the node in the timing graph at which the statistically worst slack is identified during backward propagation.
computing a timing graph from a netlist using the SSTA tool, the timing graph having a plurality of nodes and a plurality of edges connecting the nodes;
in response to forward propagating a first set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool an arrival time of the first set of probabilistic distributions at each of the nodes;
in response to backward propagating a second set of probabilistic distributions in the timing graph, statistically calculating with the SSTA tool a required arrival time of the second set of probabilistic distributions at each of the nodes;
identifying with the SSTA tool a statistically worst slack for a node in the timing graph from the arrival times and the required arrival times either through a slack calculation based upon the arrival times and the required arrival times or through required arrival time calculations;
calculating a tightness probability for each of the edges;
comparing the tightness probabilities to identify a largest tightness probability;
equating a proxy worst slack to the slack for the edge exhibiting the largest tightness probability; and
generating a timing report with the SSTA tool that replaces the statistically worst slack with the proxy worst slack for at least one of the edges arriving at the node in the timing graph at which the statistically worst slack is identified during backward propagation. | 5,696 | 30,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-49 | latest | en | 0.883882 |
https://koraexpert.com/qa/quick-answer-how-many-types-of-data-types-are-there.html | 1,620,876,636,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00117.warc.gz | 377,141,710 | 7,921 | # Quick Answer: How Many Types Of Data Types Are There?
## What are basic data types?
Basic Data TypesInteger.
An integer number, from -2147483648 to 2147483647.Double or Real.
A floating-point value, for instance, 3.14.
String.
Any textual data (a single character or an arbitrary string).
Boolean.
A value that is either True , or False .
Date/Time.
A value that stores a date, time or both date and time.
Object.
Variant..
## What data type is age?
Mondal[1] suggests that age can be viewed as a discrete variable because it is commonly expressed as an integer in units of years with no decimal to indicate days and presumably, hours, minutes, and seconds.
## What is data type and explain their types?
A data type is a type of data. … Some common data types include integers, floating point numbers, characters, strings, and arrays. They may also be more specific types, such as dates, timestamps, boolean values, and varchar (variable character) formats.
## What are the 4 types of data?
In this paper he claimed that all measurement in science was conducted using 4 different types of scales that he called Nominal, Ordinal, Interval and Ratio. This paper essentially unified Qualitative data (Nominal data and Ordinal data) and Quantitative data (Interval data and Ratio data).
## What is the data type of 1?
1 is an integer, 1.0 is a floating-point number. Complex numbers are written in the form, x + yj , where x is the real part and y is the imaginary part. Here are some examples. Notice that the float variable b got truncated.
## What is double data type?
double: The double data type is a double-precision 64-bit IEEE 754 floating point. … This data type represents one bit of information, but its “size” isn’t something that’s precisely defined. char: The char data type is a single 16-bit Unicode character.
## What is primary data type?
Primary data type , also known as ‘primitive data type’ or ‘fundamental data type’, are the built-in data types that are provided by the programming language. It defines the most basic data like int, char, float,etc.
## What is data types in Java?
Data type specifies the size and type of values that can be stored in an identifier. … Data types in Java are classified into two types: Primitive—which include Integer, Character, Boolean, and Floating Point. Non-primitive—which include Classes, Interfaces, and Arrays.
## What type of data is money?
The money data type is an abstract data type. Money values are stored significant to two decimal places. These values are rounded to their amounts in dollars and cents or other currency units on input and output, and arithmetic operations on the money data type retain two-decimal-place precision.
## What are the two basic types of data?
Understanding Qualitative, Quantitative, Attribute, Discrete, and Continuous Data TypesAt the highest level, two kinds of data exist: quantitative and qualitative.There are two types of quantitative data, which is also referred to as numeric data: continuous and discrete.More items…•
## What type of data is yes or no?
In research activities a YES/NO scale is nominal. It has no order and there is no distance between YES and NO. There are also highly sophisticated modelling techniques available for nominal data. An ordinal scale is next up the list in terms of power of measurement.
## What are the types of data types?
Common data types include:Integer.Floating-point number.Character.String.Boolean.
## What are the 5 main data types?
Most modern computer languages recognize five basic categories of data types: Integral, Floating Point, Character, Character String, and composite types, with various specific subtypes defined within each broad category.
## What is main data type of function?
Main data type of function are – integer, floating point, character and strings. Integer – Integers holds values which are numbers. The numbers are whole numbers and negative numbers. They can not be fraction.
## What is string data type?
A string is generally considered as a data type and is often implemented as an array data structure of bytes (or words) that stores a sequence of elements, typically characters, using some character encoding. String may also denote more general arrays or other sequence (or list) data types and structures. | 910 | 4,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.900561 |
http://math.tutorpace.com/geometry/circle-graph-online-tutoring | 1,513,176,569,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948527279.33/warc/CC-MAIN-20171213143307-20171213163307-00799.warc.gz | 179,464,459 | 8,624 | # Circle Graph
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
A circle is set of all the points that are in the same plane and equidistant from a central point. The radius of a circle is a line segment that joins center of the circle and any point on the circle. The circle graph is a graph drawn in the shape of a circle and is divided into sectors. Each of these sectors represents a part of a data set. These sectors may be mostly represented as percentage. This circle graph is also called pie chart. This circle graph is a very useful tool and may be better explained by the following examples.
Example 1: A collection of 100 people are sitting in a room. These are 3 colored chairs; 40 sitting on red, 30 sitting on blue, and 30 sitting on green. Calculate the percentage of people on these 3 types of chairs.
Solution: Total no of people are 100
Percentage on red chair = 40/100 x 100= 40%
Percentage on blue chair = 30/100 x 100= 30%
Percentage on green chair = 30/100 x 100= 30%
Example 2: In a traffic jam, we have 30 Zen, 20 Santro, 25 Nano and 15 WagonR. Find the percentage of the types of cars.
Solution: Total cars = 30+20+25+15 = 90
Percentage of Zen = 30/90 x 100 = 33.33%
Percentage of Santro = 20/90 x 100 = 22.22%
Percentage of Nano = 25/90 x 100 = 27.77%
Percentage of WagonR = 15/90 x 100 = 16.66% | 393 | 1,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-51 | latest | en | 0.900575 |
http://www.ask.com/web?q=For+a+Given+Wave+If+the+Frequency+Doubles+the+Wavelength&o=2603&l=dir&qsrc=3139&gc=1 | 1,475,346,599,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738663142.93/warc/CC-MAIN-20160924173743-00200-ip-10-143-35-109.ec2.internal.warc.gz | 329,935,711 | 16,541 | Web Results
## Ch 11, Waves - eiu.edu
If the frequency of a wave is doubled, what happens to its wavelength? ... What will happen to the wavelength of these waves if you move your finger up and ...
## Waves Review - with Answers - The Physics Classroom
www.physicsclassroom.com/reviews/waves/wavesans.cfm
As a wave travels into a medium in which its speed increases, its wavelength would ____ .... In this problem, it is given that v=5.00 m/s and f = 2.00 Hz. Substitution and .... If the frequency of a wave is doubled and if the speed remains constant, ...
## Free Flashcards about Physics Unit 1 test - StudyStack
www.studystack.com/flashcard-568258
If the wavelength is 2mm and frequency is doubled, the wavelength becomes ... Attenuation is teh reduction in ______ and ______ as a wave travels ... Can absorption be greater than attenuation in a given medium at a given frequency? No.
## If wavelength is double then what is change in frequency and wave ...
www.quora.com/If-wavelength-is-double-then-what-is-change-in-frequency-and-wave-speed
The frequency of the wave is set by whatever is driving the oscillation in the medium. Examples ... Using the above relation, if the wavelength of the light is doubled, assuming the frequency to be constant, then the speed of the light would also ...
## New Page 1 [www.pstcc.edu]
www.pstcc.edu/departments/natural_behavioral_sciences/Web Physics/Chapter016.htm
When a string, fixed at both ends, is given a vertical hit by a stick, a dent appears ... If you hold end A of the above string and try to give it a continuous ... Calculate the wavelength of a sound wave which frequency is 1324 Hz at STP conditions.
## Cause and effect relationship between wave speed frequency ...
www.slideshare.net/jan_parker/cause-and-effect-relationship-between-wave-speed-frequency-wavelength
Feb 6, 2012 ... The wavelength of The frequency the vibrating Cause and .... V=f x λ• If wavelength is held constant and frequency is doubled, what ...
## Chapter 11 Concepts - Physics
physics.weber.edu/johnston/genphsx/solns/concepts11.htm
The wavelength of the fundamental standing wave on a cello string depends on which ... However, given that wavelength, you can produce different frequencies of ... acceleration of a point on the string: Like above, if the frequency increases, ...
## Ch 20 slides - Cal Poly
www.calpoly.edu/~jfernsle/Classes/PHYS132/Knight slides/Ch20 slides.pptx
If you double the wavelength l of a wave on a string, what happens to the wave speed v and the wave frequency f? ... B. v is doubled and f is unchanged. .... If the rope is given SHM with frequency 2 Hz, how many cycles (wavelengths) are ...
## Standing Waves on a String
hep.physics.indiana.edu/~rickv/Standing_Waves_on_String.html
Oct 11, 1999 ... The lowest standing wave frequency is called the fundamental or first harmonic. ... Alternatively, we say that the wavelength of the fundamental is twice the ... If we either increase the wave speed along the string or decrease ...
## Waves in a Slinky | Department of Physics and Astronomy ...
If the first coil of the slinky is given a single back-and-forth vibration, then we call the .... For a given velocity, if you increase the frequency, the wavelength will be ...
### Waves Review with Answers #2 - The Physics Classroom
www.physicsclassroom.com
If the frequency of the source is 2.00 Hz and the wave speed is 5.00m/s then the ... In this problem, it is given that v=5.00 m/s and f = 2.00 Hz. Substitution and .... of a wave is doubled and if the speed remains constant, its wavelength is ____.
### How does wavelength change as frequency increases? | Socratic
socratic.org
Jul 8, 2015 ... The wavelength decreases as the frequency increases. ... The frequency f of a wave is the number of complete waves that pass a point in a given time. Above we ... You can see that the wavelength is halved when the frequency is doubled. ... Since v is constant, if f increases, λ must decrease, and vice versa.
### If the speed of a wave stays constant, but the wavelength is ... - Quora
www.quora.com
On top of all the great answers already given here: This is exactly what happens when a wave ... Clearly, if v increases and the frequency stays the same, the wavelength must increase by an equivalent factor. 600 Views · View Upvotes. | 1,008 | 4,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-40 | latest | en | 0.899513 |
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# In many corporations, employees are being replaced by
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31 Aug 2011, 03:30
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In many corporations, employees are being replaced by automated equipment in order to save money. However, many workers who lose their jobs to automation will need government assistance to survive, and the same corporations that are laying people off will eventually pay for that assistance through increased taxes and unemployment insurance payments.
Which of the following, if true, most strengthens the author's argument?
A. Many workers who have already lost their jobs to automation have been unable to find new jobs.
B. Many corporations that have failed to automate have seen their profits decline.
C. Taxes and unemployment insurance are paid also by corporations that are not automating.
D. Most of the new jobs created by automation pay less than the jobs eliminated by automation did.
E. The initial investment in machinery for automation is often greater than the short-term savings in labor costs.
[Reveal] Spoiler: OA
Last edited by JarvisR on 18 Jul 2015, 23:59, edited 1 time in total.
OA updated
If you have any questions
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31 Aug 2011, 06:18
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DeeptiM wrote:
In many corporations, employees are being replaced by automated equipment in order to save money. However, many workers who lose their jobs to automation will need government assistance to survive, and the same corporations that are laying people off will eventually pay for that assistance through increased taxes and unemployment insurance payments.
Author's Argument: Automation may not result in reduced expenses in the corporations; on the contrary, laying-off workers may adversely affect them because they will have to pay more in taxes and for unemployment insurance claims.
Which of the following, if true, most strengthens the author's argument?
A. Many workers who have already lost their jobs to automation have been unable to find new jobs.
As author warned, the corporations are indeed bearing the monetary responsibilities for cutting jobs as the people they laid-off are jobless for extended period. Correct.
B. Many corporations that have failed to automate have seen their profits decline.
Weakens. Opposing author's calculated skepticism with automation, corporations that employed it actually saved their own skin.
C. Taxes and unemployment insurance are paid also by corporations that are not automating.
Corporations Counter-argument: So what Mr. Author, what are we to lose. Automation or no automation, we're anyway to bear the burden of both the things you just mentioned. We'll go ahead with our plan anyway; thanks for the heads-up though.
D. Most of the new jobs created by automation pay less than the jobs eliminated by automation did.
Automation were employed for some reasons and this is one among those. Automation did bring down the expenses that the corporations had been paying off as salaries. Weakens.
E. The initial investment in machinery for automation is often greater than the short-term savings in labor costs.
Corporations may be seeking long-term benefits. This is out of scope.
Ans: "A"
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Re: In many corporations, employees are being replaced by [#permalink]
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17 Jan 2012, 17:17
fluke wrote:
A. Many workers who have already lost their jobs to automation have been unable to find new jobs.
As author warned, the corporations are indeed bearing the monetary responsibilities for cutting jobs as the people they laid-off are jobless for extended period. Correct.
Ans: "A"
good reasoning.
First statement says companies are SAVING money. Second statement contradicts with that by introducing the word 'However' with the fact that companies have to pay money for the unemployment they created. With this, A is the only one that fits the bill.
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Re: In many corporations, employees are being replaced by [#permalink]
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25 Jan 2012, 09:02
Very stright A
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27 Jan 2012, 16:20
+1 A
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Hello from the GMAT Club VerbalBot!
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02 Sep 2015, 11:06
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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In many corporations, employees are being replaced by [#permalink]
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13 Apr 2016, 13:06
Chembeti wrote:
fluke wrote:
A. Many workers who have already lost their jobs to automation have been unable to find new jobs.
As author warned, the corporations are indeed bearing the monetary responsibilities for cutting jobs as the people they laid-off are jobless for extended period. Correct.
Ans: "A"
good reasoning.
First statement says companies are SAVING money. Second statement contradicts with that by introducing the word 'However' with the fact that companies have to pay money for the unemployment they created. With this, A is the only one that fits the bill.
I am not sure i follow the reasoning. Please elaborate a bit
To me option A seemed like this: If people laid off are unable to find new jobs, it is most likely because companies are not willing to hire people as they found a better and more profitable alternative in automation, and are more willing to do so.
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17 Dec 2016, 14:35
A. However, D has somewhat similar logic, but is wrong because it does not say how much less the new jobs pay.
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17 Jan 2017, 00:42
This question I feel should be in the form of Find the Assumption. Any thoughts?
Posted from my mobile device
Re: In many corporations, employees are being replaced by [#permalink] 17 Jan 2017, 00:42
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Display posts from previous: Sort by | 2,300 | 9,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2017-04 | latest | en | 0.952198 |
https://numberworld.info/149212 | 1,660,955,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573849.97/warc/CC-MAIN-20220819222115-20220820012115-00649.warc.gz | 387,128,218 | 3,900 | # Number 149212
### Properties of number 149212
Cross Sum:
Factorization:
2 * 2 * 7 * 73 * 73
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
246dc
Base 32:
4hms
sin(149212)
-0.88415157477207
cos(149212)
0.4672001635574
tan(149212)
-1.8924470574665
ln(149212)
11.913123392472
lg(149212)
5.1738037515831
sqrt(149212)
386.27969141543
Square(149212)
### Number Look Up
Look Up
149212 which is pronounced (one hundred forty-nine thousand two hundred twelve) is a special number. The cross sum of 149212 is 19. If you factorisate 149212 you will get these result 2 * 2 * 7 * 73 * 73. 149212 has 18 divisors ( 1, 2, 4, 7, 14, 28, 73, 146, 292, 511, 1022, 2044, 5329, 10658, 21316, 37303, 74606, 149212 ) whith a sum of 302568. 149212 is not a prime number. 149212 is not a fibonacci number. The number 149212 is not a Bell Number. The number 149212 is not a Catalan Number. The convertion of 149212 to base 2 (Binary) is 100100011011011100. The convertion of 149212 to base 3 (Ternary) is 21120200101. The convertion of 149212 to base 4 (Quaternary) is 210123130. The convertion of 149212 to base 5 (Quintal) is 14233322. The convertion of 149212 to base 8 (Octal) is 443334. The convertion of 149212 to base 16 (Hexadecimal) is 246dc. The convertion of 149212 to base 32 is 4hms. The sine of the number 149212 is -0.88415157477207. The cosine of 149212 is 0.4672001635574. The tangent of the number 149212 is -1.8924470574665. The root of 149212 is 386.27969141543.
If you square 149212 you will get the following result 22264220944. The natural logarithm of 149212 is 11.913123392472 and the decimal logarithm is 5.1738037515831. that 149212 is unique figure! | 661 | 1,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.689564 |
https://www.gmattutor.com/practice/?qbid=38 | 1,632,094,074,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00117.warc.gz | 825,802,258 | 5,835 | # Free GMAT Test Questions
Question 1 of 1
ID: GMAT-PS-7
Section: Quantitative Reasoning - Problem Solving
Topics: Geometry; Triangles
Difficulty level: Medium
(Practice Mode: Single selected Question » Back to Overview)
If in the figure below, triangles ABC and FGE are isosceles, what is the measure of $\angle$FGE? (Figure: Not to scale)
A$15^\circ$
B$35^\circ$
C$45^\circ$
D$75^\circ$
E$110^\circ$ | 124 | 405 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-39 | longest | en | 0.7659 |
https://www.gospel10.com/how-to-calculate-bmi-a-comprehensive-guide-to-understanding-weight-status/ | 1,721,371,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00468.warc.gz | 681,432,396 | 56,130 | # How to Calculate BMI: A Comprehensive Guide to Understanding Weight Status
Calculating Body Mass Index (BMI) using a calculator is a simple yet effective way to assess an individual’s weight status.
BMI is a measure of body fat based on height and weight. It is used to categorize individuals into different weight groups, including underweight, normal weight, overweight, and obese. Understanding BMI is crucial for maintaining a healthy weight and reducing the risk of chronic diseases.
The calculation of BMI dates back to the 19th century when Adolphe Quetelet, a Belgian statistician, developed the Quetelet Index, which later became known as BMI. This index gained widespread recognition in the 20th century as a tool for assessing weight status and screening for potential health risks.
## How to Calculate BMI in Calculator
Understanding the essential aspects of calculating Body Mass Index (BMI) using a calculator is crucial for accurately assessing weight status and potential health risks.
• Formula: BMI = weight (kg) / height (m)
• Units: Kilograms and meters
• Categories: Underweight, normal weight, overweight, obese
• Health Risks: Obesity-related diseases
• Limitations: Does not account for muscle mass or body composition
• Calculator: Online or mobile app
• Accuracy: Ensure correct input of height and weight
• Interpretation: Consider age, gender, and other health factors
These aspects provide a comprehensive understanding of BMI calculation, enabling individuals to accurately assess their weight status and make informed decisions about their health.
### Formula
At the heart of calculating BMI using a calculator lies the formula BMI = weight (kg) / height (m). This formula serves as the cornerstone for assessing weight status and understanding potential health risks.
• Units of Measurement
The formula relies on the metric system, utilizing kilograms for weight and meters for height. This ensures consistency and accuracy in calculations across different regions and populations.
• Weight and Height
The formula requires accurate input of weight and height. Weight should be measured in kilograms, while height should be measured in meters. Precise measurements are essential for obtaining a reliable BMI.
• Calculation Process
To calculate BMI, simply divide the weight in kilograms by the square of the height in meters. This calculation yields a numerical value that falls within specific BMI categories.
• Interpretation
Once BMI is calculated, it is crucial to interpret it correctly. BMI categories range from underweight to obese, providing insights into weight status and potential health implications.
Understanding these facets of the BMI formula empowers individuals to accurately calculate their BMI using a calculator. This knowledge supports informed decision-making regarding weight management and overall health.
### Units
In the context of calculating BMI using a calculator, the units of measurement play a critical role in ensuring accurate and reliable results. The formula BMI = weight (kg) / height (m) specifies the use of kilograms for weight and meters for height. This adherence to the metric system is essential for several reasons.
Firstly, the metric system is the international standard for scientific measurements, ensuring consistency and comparability across different countries and regions. By utilizing kilograms and meters, BMI calculations can be easily understood and interpreted globally. Moreover, the metric system is based on decimal units, making calculations and conversions straightforward and minimizing the risk of errors.
Secondly, the use of kilograms and meters aligns with the scientific definitions of weight and height. Weight, measured in kilograms, represents the force exerted on an object due to gravity. Height, measured in meters, represents the vertical distance from the base to the top of an object. These units provide precise and meaningful measures that are directly applicable to the BMI formula.
In practice, using kilograms and meters for BMI calculations is essential for obtaining accurate results. For instance, if weight is measured in pounds and height in inches, the calculated BMI will be incorrect and potentially misleading. Therefore, it is imperative to ensure that the correct units are used throughout the calculation process.
### Categories
Categorizing individuals into different weight groups is a crucial step in understanding BMI and its implications for health. The four primary categoriesunderweight, normal weight, overweight, and obeseprovide a framework for assessing weight status and potential health risks.
• Underweight
Individuals with a BMI below 18.5 are considered underweight. This category may indicate nutrient deficiencies, eating disorders, or underlying medical conditions.
• Normal Weight
A BMI between 18.5 and 24.9 is considered normal weight. This range is generally associated with a lower risk of chronic diseases and optimal health.
• Overweight
Individuals with a BMI between 25 and 29.9 are considered overweight. This category may indicate an increased risk of developing weight-related health problems.
• Obese
A BMI of 30 or higher is considered obese. Obesity is a major risk factor for various chronic diseases, including heart disease, stroke, type 2 diabetes, and some types of cancer.
Understanding these categories is essential for interpreting BMI results and making informed decisions about weight management. Individuals who fall into the underweight or obese categories may require further evaluation and lifestyle modifications to improve their health outcomes.
### Health Risks
Understanding BMI and its implications for health is crucial, as obesity is a major risk factor for various chronic diseases. Obesity-related diseases pose a significant threat to global health, affecting countless individuals and straining healthcare systems.
• Cardiovascular Disease
Obesity increases the risk of heart disease, stroke, and high blood pressure. Excess weight puts a strain on the heart and blood vessels, leading to plaque buildup and potential blockages.
• Type 2 Diabetes
Obesity is a significant risk factor for type 2 diabetes, a chronic condition affecting the body’s ability to regulate blood sugar levels.
• Cancer
Obesity has been linked to an increased risk of certain types of cancer, including breast, colon, and endometrial cancer.
• Musculoskeletal Disorders
Excess weight can put stress on joints and bones, leading to osteoarthritis, back pain, and other musculoskeletal issues.
The link between BMI and obesity-related diseases highlights the importance of maintaining a healthy weight. Calculating BMI using a calculator is a simple yet effective way to assess weight status and identify potential health risks. By understanding the risks associated with obesity, individuals can make informed decisions about their lifestyle and seek appropriate medical advice when needed.
### Limitations
The formula used to calculate BMI does not take into account muscle mass or body composition. This is a limitation because muscle weighs more than fat, so individuals with a high muscle mass may have a higher BMI than individuals with a lower muscle mass but the same amount of body fat. Additionally, BMI does not distinguish between different types of body fat, such as subcutaneous fat (which is stored under the skin) and visceral fat (which is stored around the organs). Visceral fat is more harmful to health than subcutaneous fat, but BMI does not account for this difference.
Despite these limitations, BMI is still a useful tool for assessing weight status and obesity risk. It is simple and inexpensive to calculate, and it can be used to track changes in weight over time. BMI is also a good indicator of overall health risks, such as heart disease, stroke, type 2 diabetes, and some types of cancer.
When interpreting BMI results, it is important to keep in mind the limitations of the formula. Individuals with a high muscle mass or a high proportion of visceral fat may have a higher BMI than individuals with a lower muscle mass or a lower proportion of visceral fat but the same amount of body fat. Additionally, BMI does not provide information about overall health or fitness. For these reasons, it is important to use BMI in conjunction with other measures of health, such as waist circumference, body fat percentage, and blood pressure.
### Calculator
Calculating Body Mass Index (BMI) using a calculator is a convenient and widely adopted method for assessing weight status and potential health risks. Online and mobile app calculators offer a user-friendly and accessible means to perform this calculation.
• Simplicity and Accessibility
Online and mobile app calculators eliminate the need for manual calculations or complex formulas. With a few simple inputs, users can obtain their BMI results instantly.
• Portability and Convenience
Mobile app calculators are readily available on smartphones, allowing for BMI calculations on the go, making it convenient for individuals to track their weight status regularly.
Some calculators offer additional features such as BMI history tracking, progress charts, and personalized recommendations, providing users with a comprehensive weight management tool.
• Integration with Health Apps
Many BMI calculators integrate with popular health apps, enabling users to seamlessly track their BMI alongside other health metrics, facilitating a holistic approach to health monitoring.
By leveraging the convenience and accessibility of online and mobile app calculators, individuals can effortlessly calculate their BMI, empowering them to make informed decisions about their weight management and overall well-being.
### Accuracy
In the context of calculating Body Mass Index (BMI) using a calculator, accuracy reigns supreme. Precise input of height and weight forms the cornerstone of a reliable BMI calculation, directly influencing the accuracy of the results. Errors in these inputs can lead to misleading BMI values, potentially undermining informed decision-making about weight management and health.
Consider the following real-life example: An individual who enters their height as 5 feet 10 inches instead of 5 feet 8 inches will result in an overestimation of their BMI. This discrepancy can potentially categorize them into a higher weight group, leading to unnecessary concern or inaccurate assessment of their weight status. Conversely, underestimating height or overestimating weight can lead to an underestimation of BMI, potentially masking potential health risks.
To ensure accurate BMI calculations, it is imperative to measure height and weight with precision. Utilize a stadiometer or tape measure for height measurements, ensuring proper posture and a flat surface. For weight measurements, employ a calibrated scale, standing upright with minimal clothing. By adhering to these simple yet crucial steps, individuals can confidently obtain accurate BMI results, empowering them to make informed choices about their health and well-being.
### Interpretation
When interpreting BMI results, it is crucial to consider age, gender, and other health factors to gain a comprehensive understanding of weight status and potential health risks. Age and gender influence BMI ranges, as muscle mass and body composition vary with these factors. For instance, older adults tend to have lower BMI values due to age-related muscle loss, while athletes may have higher BMI values due to increased muscle mass.
Other health factors, such as pregnancy, certain medical conditions, and medications, can also affect BMI interpretation. For example, pregnancy leads to an increase in BMI due to the weight of the fetus and amniotic fluid. Certain medical conditions, such as fluid retention and thyroid disorders, can also impact weight and BMI values.
By considering age, gender, and other health factors in BMI interpretation, individuals can obtain a more accurate assessment of their weight status and potential health risks. This understanding empowers them to make informed decisions about weight management, lifestyle modifications, and seeking appropriate medical advice when necessary.
In summary, interpreting BMI results requires consideration of age, gender, and other health factors to provide a comprehensive evaluation of weight status and health risks. This understanding is essential for making informed decisions about weight management and overall well-being.
### Frequently Asked Questions (FAQs) on BMI Calculator
This section addresses commonly asked questions and clarifies important aspects related to calculating BMI using a calculator. These FAQs aim to provide comprehensive information and dispel any misconceptions.
Question 1: What is the formula for calculating BMI?
Answer: BMI is calculated using the formula: BMI = weight (kg) / height (m)2, where weight is in kilograms and height is in meters.
Question 2: What units should I use for weight and height?
Answer: For accurate results, it is crucial to use kilograms for weight and meters for height, as the formula is designed for these metric units.
Question 3: How do I interpret my BMI results?
Answer: BMI categories provide insights into weight status. Underweight: BMI < 18.5; Normal weight: 18.5 – 24.9; Overweight: 25 – 29.9; Obese: BMI 30.
Question 4: What are the limitations of BMI?
Answer: BMI does not consider muscle mass or body composition, so individuals with higher muscle mass may have higher BMI values despite being within a healthy weight range.
Question 5: Can I use a BMI calculator on my phone?
Answer: Yes, there are many reputable BMI calculator apps available for smartphones, offering convenience and portability.
Question 6: Should I consult a healthcare professional about my BMI?
Answer: While BMI calculators provide a general assessment, it is always advisable to consult a healthcare professional for personalized advice and to discuss any concerns related to weight or overall health.
These FAQs offer essential insights into calculating and interpreting BMI using a calculator. Understanding these aspects empowers individuals to make informed decisions about their weight management and health.
The following section delves into the health implications associated with different BMI categories, providing further guidance on weight management and maintaining a healthy lifestyle.
### Tips for Maintaining a Healthy Weight
Maintaining a healthy weight is crucial for overall well-being. These actionable tips provide guidance on adopting healthy habits and achieving weight management goals.
Tip 1: Set Realistic Goals
Start with small, achievable goals to avoid discouragement. Focus on gradual, sustainable weight loss rather than drastic measures.
Tip 2: Follow a Balanced Diet
Incorporate nutrient-rich foods from all food groups. Fruits, vegetables, lean protein, and whole grains provide essential vitamins, minerals, and fiber.
Tip 3: Engage in Regular Physical Activity
Aim for at least 150 minutes of moderate-intensity exercise or 75 minutes of vigorous-intensity exercise per week.
Tip 4: Prioritize Sleep
Sleep deprivation can lead to hormonal imbalances and increased appetite. Aim for 7-9 hours of quality sleep each night.
Tip 5: Manage Stress
Chronic stress can trigger unhealthy eating habits. Engage in stress-reducing activities such as exercise, yoga, or meditation.
Tip 6: Seek Professional Help When Needed
If struggling with weight management, don’t hesitate to consult a healthcare professional or registered dietitian for personalized guidance.
Key Takeaways
These tips emphasize the importance of setting realistic goals, adopting a balanced diet, engaging in regular physical activity, prioritizing sleep, and managing stress. By implementing these habits, individuals can improve their overall health and well-being.
Transition to Conclusion
Adopting these tips can pave the way for sustainable weight management and a healthier lifestyle. The following section explores additional strategies for maintaining a healthy weight and preventing weight-related health risks.
### Conclusion
This comprehensive guide has delved into the intricacies of calculating BMI using a calculator, emphasizing the importance of precise measurements and accurate interpretation. BMI serves as a valuable tool for assessing weight status and potential health risks, with clear implications for maintaining a healthy lifestyle.
Key insights include the significance of considering age, gender, and other health factors when interpreting BMI results. While BMI provides a general assessment, it is crucial to consult healthcare professionals for personalized advice and to address any underlying health concerns. Additionally, adopting healthy habits, such as a balanced diet, regular exercise, and stress management, plays a vital role in weight management and overall well-being. | 3,128 | 17,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-30 | latest | en | 0.894451 |
https://vuo.org/comment/6503 | 1,585,808,704,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506673.7/warc/CC-MAIN-20200402045741-20200402075741-00027.warc.gz | 736,239,791 | 15,213 | Hello - I was wondering if someone could suggest a way in Vuo (or point to a sample comp or tutorial or documentation) for me to be able to take a numeric value that is changing (going up and down and up again etc.) over time and use that numeric value such that it can trigger something (perhaps toggling between 2 states, perhaps incrementing another numeric value) when a specified threshold is crossed but only when the threshold is crossed in a specific direction (for example only when the value crosses the threshold as it is increasing) Although it may not be necessary I just made this crude graphic to help illustrate what I mean...
## graph.jpg
The green curved line is the changing input value (left to right is time passing), black line is the threshold, green and reddish shading represent 2 states. See how it's only toggling between the 2 states crossing the threshold on the way up ? What is a good way to get that behavior in Vuo ? Thanks for any info.
### Not sure I understand your
Not sure I understand your question correctly but if I do, would the joined comp be something ?
### Thanks for the response. I
Thanks for the response. I was making it more complicated than it really is. Your comp offers a simple and effective way to do it. Thanks again, I appreciate it.
### If you want to make it more
If you want to make it more complicated though, you could use an enqueue node set to two items, get them, and compare them as the boolean operation. Then you would have a directional threshold ;)
### use an enqueue node set to
use an enqueue node set to two items, get them, and compare them
yeah, I had been kind of imagining something like that... maybe something like that would allow me to additionally determine how fast the number was rising (or falling) when it crossed the threshold and it might be fun to be able to make use of that info ?
### Shouldn't be too hard, I
Shouldn't be too hard, I would imagine something like this would work; if (item1 - item2) < [item1 AND threshold], do something. | 443 | 2,042 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-16 | latest | en | 0.958267 |
http://www.businessinsider.com/how-much-should-you-invest-in-marketing-2014-3 | 1,527,288,470,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867220.74/warc/CC-MAIN-20180525215613-20180525235613-00305.warc.gz | 356,521,026 | 31,066 | If you're wondering what your marketing budget should be, you're not alone. This is the million-dollar question: How can you be sure you're spending the right amount of money on the right types of marketing? Many say it's an art, not a science. Others argue that there is a clear equation that can help you to calculate exactly how much of your marketing dollars to place where. Simply figure out the equation, enter your variables, and voila.
In truth, it's a little bit of both. With a little thought, a little math, a little data, and a little creativity, it is possible to have a good idea of how much to invest in marketing for the highest possible return. In terms of "return," I have a strong finance view. Simply put, the role of marketing is to create leads and business opportunities. You should always return to this metric.
The key to ROI marketing is to not only determine your marketing budget, but to consistently building your company revenue. Your ROI always needs to link back to actual sales.
## Return on Investment Marketing
ROI marketing is a measurement tool. It measures how much profit you make on a given marketing investment. To figure out the return on your investment, you need to identify a few figures to plug into your ROI formulas (as long as you are consistent, you can define your terms however you choose):
• Marketing investment: Media cost or production cost
## The Components of ROI Marketing
There are six key components of ROI marketing:
1. Understanding lifetime customer value. Once you know this, you can begin to figure out how much you should expect to spend on new customer acquisition. To calculate the lifetime value of a customer, you need to identify the following variables: average annual revenue per customer, average gross profit margin (before any marketing expense), cost of capital, and average number of years per customer.
2. Estimating target acquisition cost per customer cost. Look at your company data. Take the total cost of your marketing budget and divide by the number of customers you won with this investment. This is your historic acquisition price.
3. Determining your marketing budget. Divide your target revenue by the average customer revenue. Then multiply this number by the target acquisition price. Once you have your ROI goal and overall annual revenue goal, calculate your targeted marketing spend.
4. Predicting which tactic will help you to realize your customer acquisition goals. Use the by-product of your calculations to make some informed decisions as to which marketing strategy will be most successful in helping you to achieve your goals.
5. Setting your marketing ROI goal. Once you have established your ROI threshold, stick to it. If a marketing initiative isn't hitting the threshold, cut it. Put your marketing dollars where you know they will have a greater impact.
## ROI Marketing: More Than a Measurement
How you choose to track your marketing spend and calculate your ROI marketing can differ from company to company. It's important that you make the effort to add some rigor to your marketing activities. Even if your calculations aren't exact, they can still show you clear trends of which marketing activities are getting real results and which aren't. And again, results means actual sales.
Ultimately, ROI marketing is more than a measure, it's also a philosophy. But you can't implement ROI marketing without making a larger organizational change. This is no small task.
Some early-stage startups with limited funding might view marketing as a low priority; an unnecessary cost. In fact, marketing is not just an outlay of capital. It's an investment back into your company — not a drain on it. ROI marketing helps you to justify your investments, supporting the old adage that you need money to make money.
A version of this post originally appeared on the author's blog.
David Ehrenberg is the founder and CEO of Early Growth Financial Services, an outsourced financial services firm that provides early-stage companies with day-to-day transactional accounting, CFO service, tax, and valuation services and support. He's a financial expert and startup mentor whose passion is helping businesses focus on what they do best. Follow David @EarlyGrowthFS. | 835 | 4,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.933022 |
https://milesquarefabricstudio.com/dh3d6b09/eD6058tK/ | 1,623,537,994,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586465.3/warc/CC-MAIN-20210612222407-20210613012407-00630.warc.gz | 347,764,442 | 17,041 | Math Grade 5: Math Worksheet Math Coloring Worksheets 2nd Grade Free Color By Number 3rd Common Core Pdf Addition Forcond Subtraction For Second Coloring Worksheets Grade Free Color By Number Common Core. Fun Kids Worksheets Area Model Division 5th Grade Worksheets kids worksheet jokes | Milesquarefabricstudio
# Math Worksheet Math Coloring Worksheets 2nd Grade Free Color By Number 3rd Common Core Pdf Addition Forcond Subtraction For Second Coloring Worksheets Grade Free Color By Number Common Core
Published at Wednesday, 16 September 2020. Math Grade 5. By .
Two of the best options are Omega Math and ALEKS Math. Both of these programs are well-developed online math programs. Omega Math covers Pre-Algebra, Algebra I and II, as well as Geometry, and ALEKS is a full program for grades Kindergarten through High School, including Trigonometry, Statistics, and Accounting. There are some differences in presentation style, but both programs cover the material thoroughly, and all that a student needs to do is log in, have their pencil and paper nearby, and begin their study. Omega Math tends to be better equipped for students who catch on to math skills fairly easily and are motivated to streamline their work. Students log in to their course, view a PowerPoint lesson, and work through homework problems on their own. Feedback is given and students can also complete worksheets for extra practice. Chapter tests are provided, scored immediately, and parents can track the progress throughout the course by viewing simple charts and grade books making it very parent-friendly.
On the first day of each month the canteen cooks hamburgers. The Cook needs 1/10 of a kg of mince, 1/8 of a lettuce and 1/5 of a lettuce for each burger. How much of each item does the Cook need to have on hand if every student (except 5) order a hamburger? It takes Principal Jones 20 minutes to walk from his house to the school. If he walks to the school in the morning and home in the afternoon. How long will he walk in total in 1 school week? How long will he walk in total in a 9 week term? The school day starts at 9:00 a.m. First break is from 11:00 a.m to 11:20 a.m. The second break is from 1:00 p.m to 1:40 p.m. The school day finishes at 3:00 p.m. How much time do the students spend in class in one day? How much time do the students spend in class in one week? How much time do the students spend on break in one day? How much time do the students spend on break in one week?
According to the research, solid early mathematics skills are the strongest predictor of future academic success; greater than early reading skills, attention skills, and socioeconomic factors! So what early math skills are most important? The research focuses on "school-entry math skills" such as understanding small numbers (up to 30), quantities, and simple shapes. In addition, being able to count, compare, sort, and describe objects (up to 30) are considered core kindergarten math skills. A host of online tools are available to help young children improve their math and reading skills and technology has made great strides in the past several years in areas related to children has educational software. However, many of the technical advances may be overwhelming for young learners. Websites with 3D graphics and online virtual worlds ("edutainment") may be a useful tool for older children looking to build skills while having fun on the computer. However, younger children can be easily distracted by the overuse of technology in many of these programs.
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### Math Worksheet Math Coloring Worksheets 2nd Grade Free Color By Number 3rd Common Core Pdf Addition Forcond Subtraction For Second Coloring Worksheets Grade Free Color By Number Common Core
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FL Cuts HPN To 4 Flights In Winter
Enilria From Canada, joined Feb 2008, 9849 posts, RR: 15Posted Fri Aug 1 2008 06:46:28 UTC (7 years 6 months 1 week 5 days 6 hours ago) and read 2695 times:
FL's new HPN schedule has just come out... 13/week to ATL 10/week to MCO 5/week to PBI 28/week = 4 /day average That is in effect through the Winter. Previously they were flying... 25/week to ATL 7/week to FLL 14/week to MCO 14/week to PBI 60/week = ~9/day This comes at about the same time B6 goes from 6 flights to 11 at HPN. So FL is backing down in HPN? I wonder if it will turn out like SWF. Two flights to ATL is probably too few to be sustainable.
Jaysan From India, joined Apr 2008, 104 posts, RR: 0 Reply 1, posted Fri Aug 1 2008 07:54:12 UTC (7 years 6 months 1 week 5 days 5 hours ago) and read 2626 times:
Quoting Enilria (Thread starter):FL's new HPN schedule has just come out... 13/week to ATL 10/week to MCO 5/week to PBI
ATL 2 daily for 14 weekly
MCO 2 flights a day x23 for 10 weekly
PBI 1 daily x23 for 5 weekly
Total per week - 29/weekly
Enilria From Canada, joined Feb 2008, 9849 posts, RR: 15 Reply 2, posted Fri Aug 1 2008 08:23:17 UTC (7 years 6 months 1 week 5 days 4 hours ago) and read 2595 times:
Quoting Jaysan (Reply 1):ATL 2 daily for 14 weekly MCO 2 flights a day x23 for 10 weekly PBI 1 daily x23 for 5 weekly Total per week - 29/weekly
There is probably an X6 on one of the ATL's for the period I'm looking at which represents the difference.
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http://www.jiskha.com/display.cgi?id=1241682844 | 1,498,419,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00030.warc.gz | 543,684,129 | 3,982 | # Science (Please Respond SOON!)
posted by .
I think I'm using the wrong formulas or SOMETHING... Here's the question:
"Two equal charges are separated by 3.7 ´ 10^-10 m.
The force between the charges has a magnitude of 2.37 ´ 10^-3 N. What is the magnitude of q on the charges?"
Here's my work:
F = kc(q1q2/r^2)
2.37 x 10^-3 = 8.99 x 10^9 • q1q2 / (3.7 x 10^-10)^2
F • r^2 = kc • q1q2
F • r^2 / kc = q1q2
2.37 x 10^-3 • (3.7 x 10^-10)^2 / 8.99 x 10^9 = q1q2
q1q2 = 3.61 x 10^-32
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
E = kc • q/r^2
2.37 x 10^-3 = 8.99 x 10^9 • q/(3.7 x 10^-10)^2
2.37 x 10^-3 / 8.99 x 10^9 = q/(3.7 x 10^-10)^2
2.37 x 10^-3 / 8.99 x 10^9 • (3.7 x 10^-10)^2 = q
q = 3.61 x 10^-32
PLEASE, PLEASE, PLEASE tell me what I'm doing wrong!!!
Thank you very much. :)
• Science (Please Respond SOON!) -
I answered below. You calculated q^2
Take the square root to get q.
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http://stackoverflow.com/questions/11788565/most-pythonic-and-efficient-way-of-nesting-a-list-in-pairs?answertab=oldest | 1,394,739,927,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678678233/warc/CC-MAIN-20140313024438-00091-ip-10-183-142-35.ec2.internal.warc.gz | 131,949,056 | 17,200 | # Most pythonic (and efficient) way of nesting a list in pairs
my list is:
``````mylist=[1,2,3,4,5,6]
``````
I would like to convert mylist into a list of pairs:
``````[[1,2],[3,4],[5,6]]
``````
Is there a pythonic way of doing so? List comprehension? Itertools?
-
`[mylist[2*n:2*n+2] for n in xrange(len(mylist)/2)]`
This solution combines the use of list comprehensions and slicing to extract pairs in sequence from the original list, and build a list of the slices.
Alternatively, `[mylist[n:n+2] for n in xrange(0, len(mylist), 2)]` which is the same except `xrange` counts by twos instead of the slices. Thanks to Steven Rumbalski for the suggestion.
And now for something completely different: here is a solution (ab)using `zip` and an ephemeral function instead of intermediate assignment:
``````>>> (lambda i: zip(i, i))(iter(mylist))
[(1, 2), (3, 4), (5, 6)]
``````
-
More complicated than it needs to be -- just step `xrange` by 2 and most of the math goes away. – Steven Rumbalski Aug 3 '12 at 3:03
Feedback incorporated. – wberry Aug 3 '12 at 4:20
Yeppers, list comprehension is my usual way of doing it:
``````>>> groupsize = 2
>>> [mylist[x:x+groupsize] for x in range(0,len(mylist),groupsize)]
[[1,2],[3,4],[5,6]]
>>> groupsize = 3
>>> [mylist[x:x+groupsize] for x in range(0,len(mylist),groupsize)]
[[1,2,3],[4,5,6]]
``````
I use `range` for portability, if you are using python 2 (you probably are) change the `range` to `xrange` to save memory.
-
An alternate way:
``````zip( mylist[:-1:2], mylist[1::2] )
``````
Which produces a list of tuples:
``````>>> zip(mylist[:-1:2],mylist[1::2])
[(1, 2), (3, 4), (5, 6)]
``````
If you really want a list of lists:
``````map(list, zip(mylist[:-1:2],mylist[1::2]))
``````
-
Check out the "grouper" recipe from the itertools documentation:
``````def grouper(n, iterable, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
``````
-
-1. This will give 5 pairs on a 6 item sequence, which is incorrect. If you want an itertools recipe, it should be `grouper`. – Steven Rumbalski Aug 3 '12 at 3:09
You're absolutely correct. Fixed as suggested. – Evan Grim Aug 3 '12 at 3:14
My preferred technique:
``````>>> mylist = [1, 2, 3, 4, 5, 6]
>>> mylist = iter(mylist)
>>> zip(mylist, mylist)
[(1, 2), (3, 4), (5, 6)]
``````
I usually use generators instead of lists anyway, so line 2 usually isn't required.
-
For added fun, change the zip to `zip(*[mylist]*groupsize)` to allow for generic group size. Of course, the zip way is always going to have a problem with handling lists that have a length that is not divisible by the desired group size. – Josiah Aug 3 '12 at 3:10
This seems like the "most pythonic" way to do this. – Brian M. Hunt Aug 3 '12 at 3:14
@BrianM.Hunt I don't know that I would agree with that. The zen of python says "Simple is better than complex", and if you don't understand exactly how zip and iterators work, that could hardly be called simple. It's certainly the most clever way, though. – Josiah Aug 3 '12 at 3:28
@Josiah I kind of agree with you, although If one can zip two different generators together, one knows they can zip a generator with itself sequentially. I have found this technique to be fairly safe. especially if you were to go for (x, y) in zip(points, points), for example it can be obvious – robert king Aug 3 '12 at 6:33 | 1,080 | 3,490 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2014-10 | latest | en | 0.733516 |
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APยฎ๏ธ Calculus BC (2017 edition)
Course: APยฎ๏ธ Calculus BC (2017 edition)ย >ย Unit 1
Lesson 9: Trigonometric limits & squeeze theorem
Limit of sin(x)/x as x approaches 0
In this video, we prove that the limit of sin(θ)/θ as θ approaches 0 is equal to 1. We use a geometric construction involving a unit circle, triangles, and trigonometric functions. By comparing the areas of these triangles and applying the squeeze theorem, we demonstrate that the limit is indeed 1. This proof helps clarify a fundamental concept in calculus.
Want to join the conversation?
• how did one come up with such a complex proof ?
i mean how did one guess that we must first take a part of circle cut it into triangles and make an inequality
also the whole proof was making no sense until we reached the last step ......
sorry if my question is stupid or if i am missing some basic skills to come up with a proof like this!
• whoever did this really clever theorem didn't made it by accident or just because he wanted to know whats the limit of sinx/x without any previous knowledge. It is enough to see the graph of the function to see that sinx/x could be 1. NOW, that's the first step. then, the mathematician must figure a formal and irrefutable theorem for the limit to be commonly accepted. It must be in fact hard to find some relationship like this but these people are just committed with their work but they are normal just like you.
• It seems clear that the area of the wedge is greater than the red triangle, but how do we know they can be equal? Same with the blue triangle being greater than or equal to the wedge.
• As ฮธ approaches 0, all the three areas approach each other, try to think about it visually. Maybe that is the reason we put an equality sign there.
• why does Sal only consider the first and fourth quadrant? he doesn't do an restrict the values of theta.
• Sal was trying to prove that the limit of sin x/x as x approaches zero. To prove this, we'd need to consider values of x approaching 0 from both the positive and the negative side. So, for the sake of simplicity, he cares about the values of x approaching 0 in the interval (-pi/2, pi/2), which approach 0 from both the negative (-pi/2, 0) and the positive (0, pi/2) side. The interval (-pi/2, pi/2) spans only the first and fourth quadrants. This is why Sal discards the rest.
• To Sal Khan,
I believe that more important than learning a concept itself is learning how it came to be, learning how it was discovered, learning how to THINK like that. You can just teach the proof just because you have learnt the proof. But please, if possible, teach how to think like a mathematician. Teach how we can, by ourselves, prove something like this.
I mean, that proof made no sense in the beginning. Half the time, I didn't have a single clue to what he was doing or more importantly, 'why' he was doing it.
Nobody's a genius mathematician by birth (well except for some maybe). You can't just think of proofs like that out of thin air! Please tell me, I want to know, of some way to actually think of doing bizarre things like that to prove what we want. I think that remembering the proof after someone else has taught me is kinda useless.
• Math major here.
Your objection is well-placed, and you're correct that proof-writing is a skill that is mostly improved by practice. Before you continue, there are two unfortunate truths to keep in mind:
1. The United States math curriculum places almost zero emphasis on proof writing or proof comprehension (and the things that pass for proofs in geometry are a joke). Sal's primary target audience is U.S. grade-schoolers and high-schoolers, and so the videos on Khan Academy cater to these admittedly bad standards. Before Common Core became widespread, it was common to see KA videos presenting the exact same topics in the exact same order as your own teacher.
2. The culture in the mathematics community dictates that once you've written a proof, you 'polish' it to make it as short and concise as possible. The onus is very much put on the reader of the proof to slog through it word-by-word, and not on the writer to be clear.
All that said, the only thing that Sal really pulled out of a hat was the idea to compare the different areas in the figure. To draw the figure in the first place was fairly natural, because how else can we interpret the sine function if not by the unit circle? And once he had made the decision to compare areas, the proof was fairly straightforward algebra.
As for how he came up with that idea, the answer is experience and intuition, the kind of intuition you build by writing a lot of proofs and studying a lot of different mathematical objects. If you really want to learn this type of thinking, a standard place to start is the book 'How to Solve It' by George Pรณlya.
• Why were the inequalities switched?
• Because he took the reciprocal of each function/number. The same way the reciprocal of three is 1/3, the reciprocals of each function (other than one) were less than the original value. Thus, the inequalities needed to be switched. Think of how 1 is less than three, which is less than four, but if you take the reciprocals you switch the order of inequalities. (1 is greater than 1/3, which is greater than 1/4). Hope this helps!
• Why were the absolute value signs necessary for inclusion in the beginning and middle parts of the process but discarded at the end? At he mentions that they are discarded because we're only concerned with the 1st and 4th quadrants but why is that? Please explain.
• Sal uses absolute values because we're dealing with areas and areas must be positive. In the instance that ฮธ is negative, we'd have negative area since sine and tangent of ฮธ would be negative. So, he took the absolute values of sine and tangent so that the areas would be positive regardless of ฮธ.
Now, why did he discard the absolute values?
1. In the case of |sin(ฮธ)|/|ฮธ|
Well,
sin(ฮธ)/ฮธ = sin(-ฮธ)/-ฮธ = -sin(ฮธ)/-ฮธ
So, |sin(ฮธ)|/|ฮธ| = sin(ฮธ)/ฮธ
He can therefore discard the absolute values.
2. In the case of cos(ฮธ)
Since -ฯ/2 < ฮธ < ฯ/2,
cos(ฮธ) is always positive and so for this constraint, |cos(ฮธ)| = cos(ฮธ).
So the absolute values can be discarded
• How about lim of sin(x)/x as x approaches infinity?
• The numerator is bounded between 1 and -1, while the denominator goes to infinity, so the limit is 0.
• At why is the limit of 1 is 1 when theta approaches zero?
• The lim(1) when ฮโ0 means: on the graph y=1, what does the y-coordinate approach when the x-coordinate (or in this case ฮ) approach 0. But on the graph y=1, the y-coordinate is always 1 no matter what the x-coordinate is. In other words, lim(k) as ฮโn = k, where k,n are any real numbers.
• At why does Sal multiply theta/2 pi by pi | 1,681 | 7,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-50 | latest | en | 0.959163 |
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# Proportions to Find Base b
## Cross - multiply to find a missing part of a proportion, base, b.
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## Proportions to Find Base b
by CK-12 //at grade
Explains how to use proportions and cross products to solve for the base as well as the percent and the amount.
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https://www.canal-u.tv/chaines/ifourier/bettina-eick-computational-group-theory-cohomology-of-groups-and-topological-2 | 1,701,647,448,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100518.73/warc/CC-MAIN-20231203225036-20231204015036-00267.warc.gz | 785,990,828 | 12,559 | Cours/Séminaire
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Bettina EICK - Computational group theory, cohomology of groups and topological methods 3
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Aurel PAGE - Cohomology of arithmetic groups and number theory: geometric, asymptotic and computati…
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In this lecture series, the first part will be dedicated to cohomology of arithmetic groups of lower ranks (e.g., Bianchi groups), their associated geometric models (mainly from hyperbolic geometry) | 938 | 3,626 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.736753 |
https://eranraviv.com/statistical-shrinkage-4-covariance-estimation/ | 1,709,188,375,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474784.33/warc/CC-MAIN-20240229035411-20240229065411-00764.warc.gz | 224,360,129 | 19,253 | Statistical Shrinkage (4) – Covariance estimation
A common issue encountered in modern statistics involves the inversion of a matrix. For example, when your data is sick with multicollinearity your estimates for the regression coefficient can bounce all over the place.
In finance we use the covariance matrix as an input for portfolio construction. Analogous to the fact that variance must be positive, covariance matrix must be positive definite to be meaningful. The focus of this post is on understanding the underlying issues with an unstable covariance matrix, identifying a practical solution for such an instability, and connecting that solution to the all-important concept of statistical shrinkage. I present a strong link between the following three concepts: regularization of the covariance matrix, ridge regression, and measurement error bias, with some easy-to-follow math.
Contents
What seems to be the problem officer?
A covariance matrix is positive semi-definite if and only if for all possible vectors (check this post for the why), and it also means that the determinant of the matrix . And if you remember that , it directly follows that all the eigenvalues must be positive. If any of them is negative or if any of them is too close to zero, the inversion operation will become problematic, akin to how dividing any number by almost-zero results in disproportionately large value.
So what do we do if not all eigenvalues are positive? We make them positive! One way to do that is by way of a process known as diagonal loading. We add a positive value (usually small) to the diagonal elements of the matrix .
Why adding a constant to the diagonal of results in the same constant added to the vector of eigenvalues? Good question. Here is a short proof:
Let be a matrix, and a positive scalar. Consider , where is the identity matrix. Let be an eigenvector of with corresponding eigenvalue . Before we start a quick reminder that by definition: if then are the eigenvectors of and are the eigenvalues of . Now, consider the action of on :
So what is shown is that has the same eigenvectors, and its eigenvalues are increased by exactly , which concludes the proof.
Negative or small eigenvalues are shifted upwards which helps inversion. You may wonder (as I’m sure you do 🙂 ): must we push all eigenvalues? I mean, it’s only the small or negative which are the culprits of instability.
The answer is no, we don’t have to increase all the eigenvalues. In fact, if you wish to remain as close as possible to the original matrix, you can choose to increase only the problematic eigenvalues. This is exactly what nearest-positive-definite type of algorithms do. You apply the minimum eigenvalues shift needed to reach positive definiteness.
Ridge regression minimize the Residual Sum of Squared error, but with a penalty on the size of the coefficients:
This is the resulting RR (for ridge regression) coefficients:
Why? because when is a scaled version of the original data matrix (as it is with ridge regression) then is the covariance of the scaled data (up to the division by the number of observations).
Ok, so ridge regression has something to do with diagonal loading (and so increasing the eigenvalues of the covariance matrix), why does this mean that we shrink the vector of regression coefficients? Another good question, let’s doctor some econometrics to help us understand this.
In the measurement error bias I have already shown that the fact that you add noise to the explanatory variable shrink the coefficient so I don’t repeat it here. The main equation from that post is:
For our purpose here you can simply think of it as increasing the variance of the explanatory variable by a constant (say ). That exactly means that you increase the diagonal entry (again, in the covariance matrix) which corresponds to that explanatory variable.
What is interesting to see, I think, is that while in the econometric literature measurement error is considered a serious problem, in the statistical learning literature we introduce measurement error intentionally, to our advantage. Of course, I steer clear from the inference versus prediction never-ending debate.
Summary
Armed with this understanding, you can use it for estimating the entries of the covariance matrix individually. This approach is not widely used is primarily due to the numerical instabilities which are very likely to follow. But if you can handle numerical instability effectively, there are significant benefits to estimating individual elements of the covariance matrix as opposed to estimating the entire covariance structure in one go.
For those of you old enough to remember, paraphrasing John Hannibal Smith from the A-team: “I love it when optimization, econometrics and statistical learning come together”. | 937 | 4,834 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-10 | longest | en | 0.905515 |
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